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Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge 5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge 5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\ge 0$, hence $\frac{x^2+y^2+z^2}{3}\ge \frac{5}{3}$...???
I'm not sure about this, can someone help?
| Viewed geometrically, $x+y+z=5$ is a plane and $x^2+y^2+z^2=a^2$ is a sphere with radius $a$ at the origin; the smallest (positive) $a$ for which the sphere touches the plane corresponds to the shortest distance between the plane and the origin, and the point of tangency represents the smallest possible value of $x^2+y^2+z^2$. Intuitively this is achieved at $x=y=z=\frac53$, whence the result follows (indeed $x^2+y^2+z^2\ge\frac{25}3$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sum_{n=1}^{\infty} a_n$ is a series and {$s_n$} is the sequence of partial sums. If $s_n = \frac{n^2+1}{4n^2-3}$, find $\sum_{n=1}^{\infty} a_n$ Only idea I have is to use $a_n = s_n-s_{n-1}$, which would mean $a_n = \frac{n^2 + 1}{4n^2 - 3}-\frac{(n-1)^2 + 1}{4(n-1)^2 -3}$. But I'm not sure if I'm thinking in the right direction. Or would $a_n =\lim s_n$? Is that only for when I'm finding the sum?
| Thanks to a commenter @MartinR, I was able to find $a_n$. Originally I was confused as to what the question was asking to find $a_n$ as a rule or a single value/sum, and Martin's comment helped me realize that we're looking for a single value.
$$\sum_{n=1}^{\infty} a_n = \lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \frac{n^2 + 1}{4n^2 - 3}.$$
We can multiply and divide the quotient by $n^2$ to simplify it $$\lim_{n \rightarrow \infty} \frac{n^2\cdot\frac{n^2 + 1}{n^2}}{n^2\cdot\frac{4n^2 - 3}{n^2}} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n^2}}{4 - \frac{3}{n^2}}.$$
I know that the quotient of a limit is equal to the quotient of the limits. So
$$\lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n^2}}{4 - \frac{3}{n^2}} = \frac{\lim_{n \rightarrow \infty} \left(1+\frac{1}{n^2}\right)}{\lim_{n \rightarrow \infty}\left(4 - \frac{3}{n^2}\right)}.$$
I also know that the limit of a sum/difference is the sum of the limits, we can rewrite what we currently have as: $$\frac{\lim_{n \rightarrow \infty} 1+ \lim_{n \rightarrow \infty}\frac{1}{n^2}}{\lim_{n \rightarrow \infty}4 - \lim_{n \rightarrow \infty}\frac{3}{n^2}}.$$
$\lim_{n \rightarrow \infty} 1 = 1$ and $\lim_{n \rightarrow \infty}\frac{1}{n^2} = 0$, $\lim_{n \rightarrow \infty} 4 = 4$ and $\lim_{n \rightarrow \infty}\frac{3}{n^2} = 0$. So $$\frac{\lim_{n \rightarrow \infty} 1+ \lim_{n \rightarrow \infty}\frac{1}{n^2}}{\lim_{n \rightarrow \infty}4 - \lim_{n \rightarrow \infty}\frac{3}{n^2}} = \frac{1 + 0}{4 - 0} = \frac{1}{4}.$$
In conclusion, $\sum_{n=1}^{\infty} a_n = \frac{1}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.
The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.
This is what I tried below.
I am not sure how to use the minimum value information for the $F(x)$ function.
I tried to find $c$ in terms of $a$ to help simplify the problem, but I don't think it helps.
I think I might have to set up some sort of system of equations with the two pieces of information.
| Following your work with $F(2) = 4 \implies c = 8a^2$. Thus: $F(x) = \dfrac{x^4}{4} -2a^2x^2+8a^2= \left(\dfrac{x^2}{2} - 2a^2\right)^2+8a^2 - 4a^4\ge 8a^2-4a^4$. This minimum value equals to $0$. So $8a^2 - 4a^4 = 0\implies 4a^2(2-a^2)=0\implies a = 0,\pm \sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}.$ I want to evaluate the following integral ($\theta_0>0$)
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta
\end{equation*}
So I thought about $d(\cos\theta)=-\sin\theta d\theta$, then
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta = - \int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}
\end{equation*}
My question is about that step. Is that legit? I mean, that integral is doubly improper at both limits. How can I ensure that it is integrable?
| Spectree’s step is correct and can be carried on as follows:
$$\begin{aligned}\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta &= - \int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}\\ &\stackrel{y=\cos \theta}{=} -\int_{a}^{-1} \frac{d y}{\sqrt{a-(1-a) y-y^{2}}}, \text{ where }a=\cos \theta_{0}.
\\&=\int_{-1}^{a} \frac{d y}{\sqrt{\left(\frac{1+a}{2}\right)^{2}-\left(y+\frac{1-a}{2}\right)^{2}}}
\\&=\left[\sin ^{-1}\left(\frac{y+\frac{1-a}{2}}{\frac{1+a}{2}}\right)\right]_{-1}^{a}\\&=\pi,
\end{aligned}$$
which is certainly independent of $\theta_{0}.$
$$\text{********}\tag*{} $$
Alternate method:
$$
\begin{aligned}
I &=-\int_{a}^{-1} \frac{d y}{\sqrt{a-y} \sqrt{1+y}} \\
&=-2 \int_{a}^{-1} \frac{1}{\sqrt{a-y}} d(\sqrt{1+y}) \\
&=-2 \int_{a}^{-1} \frac{1}{\sqrt{a+1-(\sqrt{1+y})^{2}}} d(\sqrt{1+y}) \\
&=2\left[\sin ^{-1}\left(\frac{\sqrt{1+y}}{\sqrt{a+1}}\right)\right]_{-1}^{a}\\ &=\pi
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A point $(X, Y )$ is chosen randomly from the unit circle. Let $R = \sqrt{(X^2 +Y^2)}$. Find the joint density function of the random vector $(X, R)$ A point $(X, Y )$ is chosen randomly from the unit circle.
Let $R = \sqrt{(X^2 +Y^2)}$.
Find the joint density function of the random vector $(X, R)$
My try:
$R=\sqrt{X^{2}+Y^{2}} \implies 0\leq R\leq\sqrt{2} $
$F_{R}(r)=\begin{cases}
\frac{r}{\sqrt{2}} & \text{if }0\le r\le\sqrt{2},\\
0 & \text{if }r<0,\\
1 & \text{if }r>\sqrt{2}.
\end{cases}$
$F'_{R}=f_{R} \implies f_{R}(r)=\begin{cases}
\frac{1}{\sqrt{2}} & \text{if }0\le r\le\sqrt{2},\\
0 & \text{otherwise.}
\end{cases}$
First of all, is this correct ?
I suppose that i have to use this theorem: $f_{X|R}(x|r)=\frac{f(x,r)}{f_R(r)}$ , how can i find $f_{X|R}(x|r)?$
Thanks!
| $R = \sqrt{X^2 + Y^2}$ where $X$ and $Y$ are uniformly distributed on a unit disk i.e. $x^2 + y^2 \leq 1$. So you should have $0 \leq r \leq 1$ and not $0 \leq r \leq \sqrt2$.
I will use the Jacobian method.
Use the fact that $ \displaystyle f_{XY}(x,y) = \frac{1}{\pi}~$ over unit circle.
As $r = \sqrt{x^2 + y^2}$, $y = \pm \sqrt{r^2 - x^2}$. First note that we must have $r \geq |x|$.
Now also note that we can multiply the density function by $2$ given the symmetry above or below $y = 0$. The jacobian itself is $ \displaystyle \frac{r}{\sqrt{r^2 - x^2}}$.
$\displaystyle f(r, x) = 2 ~|J| ~f(x, y)$
i.e $ ~\displaystyle f(r, x) = \frac{2r}{\pi \sqrt{r^2-x^2}}, |x| \lt r \lt 1, -1 \lt x \lt 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When is the integer between twin primes a perfect square? Explain why there exists an integer $n$ such that $4n^2 − 1 = p$ Suppose that $$ and $ + 2$ are both prime numbers.
a) Is the integer between $$ and $ + 2$ odd or even? Explain your answer.
All prime numbers are odd except for $\pm 2.$
So the integer between the primes $p$ and $p + 2$ i.e. $p + 1,$ is always even for $p \neq \pm 2;$
and odd for $p = \pm 2.$
b) Assume additionally that the integer between the primes $$
and $ + 2$ is a perfect square.
Also $(2n)^2=4n^2, (2n+1)^2=4n^2+4n+1$
Explain why there exists an integer $n$ such that $4^2 − 1 = .$
c) By considering $(2 − 1)(2 + 1),$ find the only possible value of $.$
Part (a) seems straight forward, although I imagine there's a more formal explanation required?
Any tips on parts (b) and (c), any obvious deductions/observations i need to consider?
| Given$\quad A^2+B^2=C^2\quad $ the most common means of generating Pythagorean triples is Euclid's formula
$A=m^2-k^2,\quad B=2mk, \quad C=m^2+k^2.\quad$ We can restrict this formula to generating only primitive triples and odd square multiples of primitives if we let
$\quad m=(2n-1+k),\quad $ i.e.
$A=(2n-1+k)^2-k^2,\quad
B=2(2n-1+k)k, \quad
C=(2n-1+k)^2+k^2.$
Which expands to
\begin{align*}
A=&(2n-1)^2+&2(2n-1)k&\\
B=& &2(2n-1)k& +2k^2\\
C=&(2n-1)^2+&2(2n-1)k&+2k^2
\end{align*}
and, if we let $\space k=1,\space$ the formula collapses to
$$A=4n^2-1\quad B=4n\quad C=4n^2+1$$
Which generates all triples where $\space C-A=2\space$, et seq
$$(3,4,5)\quad (15,8,17)\quad (35,12,37)
\quad (63,16,65)\quad (99,20,101)\quad \cdots
$$
note that, for all of these, $C-A=2$
and
$\dfrac{C+A}{2}$ is a perfect square
If we factor A we get
$$
A=(2n-1)^2+2(2n-1)k\quad = (2 n - 1) (2 k + 2 n - 1)
$$
and we can see that A is always composite unless $n=1$, leaving the only solution where $P$ and $P+2$ can both be prime is $(3,4,5)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\underbrace{\int\frac{\sqrt{1-x}}{x} d x}_{K}\right]
$$
To deal with $J$, we use rationalization instead of substitution. $$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=2 \int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=2 \int \frac{d(\sqrt{1+x})}{x}+2 \sqrt{1+x} \\
&=2 \int \frac{d(\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+2 \sqrt{1+x} \\
&=\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1} \right| +2 \sqrt{1+x}+C_{1}
\end{aligned}
$$
$\text {Replacing } x \text { by } -x \text { yields }$
$$
\begin{array}{l} \\
\displaystyle K=\int \frac{\sqrt{1-x}}{-x}(-d x)=\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+2 \sqrt{1-x}+C_{2}
\end{array}
$$
Now we can conclude that $$
I=\sqrt{1+x}+\sqrt{1-x}+\frac{1}{2}\left(\ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|\right)+C
$$
My question is whether there are any simpler methods such as integration by parts , trigonometric substitution, etc…
Please help if you have. Thank you for your attention.
| Seek $f$ so$$\frac{\sqrt{1+x}+\sqrt{1-x}}{2x}=f(1+x)-f(1-x),$$e.g.$$f(y):=\frac{\sqrt{y}}{2(y-1)}.$$You want to evaluate$$\int(f(1+x)-f(1-x))dx=F(1+x)+F(1-x)+C$$with $F^\prime(y)=F(y)$.
Just about any concise solution technique will exploit the above facts. Your approach is equivalent to next taking $y=z^2$, so$$F(y)=\int\frac12\left(2+\frac{1}{z-1}-\frac{1}{z+1}\right)dz=\sqrt{y}+\frac12\ln\left|\frac{\sqrt{y}-1}{\sqrt{y}+1}\right|+C.$$I doubt there's anything much simpler than this, but what's preferable is up to taste. In terms of trigonometric substitutions, you may like $y=\cos^2u$ so$$fdy=\frac12(\csc u-\sin u)du$$or $y=\sec^2u$ so$$fdy=\frac12(\cot u+\tan u)du,$$depending on the range of $y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many squares with vertices on $x^2y^2 =1$ are possible? We can form a square with vertices on the curve $x^2 y^2 = 1$ and which does not intersect this curve.
Can we get any other square with vertices on $x^2 y^2 = 1$ and which does not intersect this curve? Can anyone please tell me elaborately?
| Yes. Rotating and scaling so as to keep the vertices on the curve, we get a square with vertices at $\pm(a,1/a)$ and $\pm(-1/a,a)$ with $a>0$. We can rotate clockwise until there is an intersection when the slope of the side from $(a,1/a)$ to $(-1/a,a)$ is less than the slope at $(a,1/a)$. This condition on $a$ may be expressed as
$$\frac{1/a-a}{a+1/a}\geq -\frac{1}{a^2},$$
or equivalently,
$$a^4 -2a^2-1\leq 0.$$
There is a similar condition for counterclockwise rotation, but rather than repeat the calculation, we can note that counterclockwise is the same as clockwise rotation followed by exchanging $x$ and $y$ coordinates (in this case $a\mapsto 1/a$). Hence, we also have the condition
$$a^{-4} -2a^{-2}-1\leq 0,$$
or equivalently,
$$a^{4} +2a^{2}-1\geq 0.$$
Solving for these two equations, we get $\sqrt{\sqrt{2}-1}\leq a\leq \sqrt{1+\sqrt{2}}$.
EDIT: We can show that these are the only such squares. Firstly suppose a square could be inscribed in the first quadrant. Let $(x_i,y_i)$ be the coordinates of the vertices with their $x$ coordinate in increasing order. Then the chords from $(x_1,y_1)$ to $(x_2,y_2)$ and from $(x_3,y_3)$ to $(x_4,y_4)$ have the same slope. By the intermediate value theorem, there are two points $c_1,c_2$ with $0<x_1<c_1<x_2<x_3<c_2<x_4$ so that the slope of the curve at those points are the same. Hence,
$$-\frac{1}{c_1^2}=-\frac{1}{c_2^2}$$
or $c_1=c_2$, which is a contradiction. Thus, at least one vertex must lie in another quadrant. The curve in the first quadrant is convex, so if at least two vertices lie in the first quadrant, then all of the vertices must lie in the convex hull of the curve in the first quadrant, as otherwise there would be an intersection with the curve. As such, at most one vertex can lie in the first quadrant, and by symmetry with the other quadrants, we conclude that each vertex must lie in a different quadrant. Now suppose we have a vertex $(a,a^{-1})$ in quadrant I, and a vertex $(b,-b^{-1})$ in quadrant II. Then there is a vertex $(b+a^{-1}+b^{-1},b-b^{-1}-a)$ in quadrant III. Since it lies on the curve in quadrant III, we have
$$(b+a^{-1}+b^{-1})(b-b^{-1}-a)=1$$
However,
$$(b+a^{-1}+b^{-1})(b-b^{-1}-a)=b^2-b^{-2}-ab+a^{-1}b-a^{-1}b^{-1}-ab^{-1}-1$$
Hence,
$$b^2-b^{-2}-ab+a^{-1}b-a^{-1}b^{-1}-ab^{-1}-2=0$$
Multiplying by $-ab^3$, we get
$$-ab^5+ab+a^2b^4-b^4+b^2+a^2b^2+2ab^3=(ab+1)(ab^3+ab-b^4+b^2)=0$$
Letting $c=ab$, we get
$$(c+1)(c(b^2+1)-b^4+b^2)=0$$
Then, either
$$c=-1\quad\text{or}\quad c=\frac{b^4-b^2}{b^2+1}$$
Now, suppose $c$ is the latter. Then if $b\leq-1$, $$a=\frac{b^3-b}{b^2+1}\leq 0,$$ which is a contradiction since $a>0$. However, if $b>-1$, then $$b+a^{-1}+b^{-1}=\frac{b^4+b^2}{b^3-b}> 0$$
which is a contradiction since $b+a^{-1}+b^{-1}<0$. Thus, we conclude that $$c=ab=-1$$
which characterizes squares with a center that coincides with the origin.
| {
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$\binom{n}{r}$ where $r$ can only be chosen in groups of $k$ I just started my learning in combinatorics, and I know that number of ways to choose $r$ items from $n$ items without order is given by $\binom{n}{r}= \frac{n!}{r!(n-r)!}$. However, I couldn't figure out the number of ways to choose $r$ items from $n$ items, where the $r$ chosen items are restricted such that they can be chosen only in groups less than or equal to $k$ $(k<=r)$ each time.
Perhaps my above description is unclear, so let me try to present it in a more concrete way: on a cartesian plane, in each move you can either go $(+1,0)$ or $(0,+1)$ (move in x or y direction for 1 positive unit), how many valid shortest paths (only walking $n$ steps in total) are there from $(0,0)$ to $(r,n-r)$, if you can only walk in the $x$ direction for at most $k$ consecutive times before changing direction?
|
The number of valid paths of length $n$ consisting of $r$ horizontal $(1,0)$-steps and $n-r$ vertical $(0,1)$-steps with at most $k$ consecutive horizontal steps is
\begin{align*}
\color{blue}{\sum_{q=0}^{\left\lfloor\frac{r}{k+1}\right\rfloor}\binom{n-(k+1)q}{r-(k+1)q}\binom{n-r+1}{q}(-1)^q}\tag{1}
\end{align*}
where the symbol $\left\lfloor\cdot\right\rfloor$ denotes the floor function.
Example: Let's look at an example $n=6, r=4, k=2$.
We have a total of $\binom{n}{r}=\binom{6}{4}=15$ paths of length $6$ with four horizontal steps. The restriction $k=2$ gives according to (1)
\begin{align*}
\sum_{q=0}^{1}\binom{6-3q}{4-3q}\binom{3}{q}(-1)^q&=\binom{6}{4}\binom{3}{0}-\binom{3}{1}\binom{3}{1}\\
&=15-9\\
&\,\,\color{blue}{=6}
\end{align*}
valid paths. The six valid paths are marked blue in the list below where $a$ encodes a horizontal $(1,0)$-step and $b$ a vertical $(0,1)$-step.
\begin{align*}
\begin{array}{ccc}
\mathrm{aaaabb}\qquad&\qquad \mathrm{abaaab}\qquad&\qquad \mathrm{baaaab}\\
\mathrm{aaabab}\qquad&\qquad \color{blue}{\mathrm{abaaba}}\qquad&\qquad \mathrm{baaaba}\\
\mathrm{aaabba}\qquad&\qquad \color{blue}{\mathrm{ababaa}}\qquad&\qquad \color{blue}{\mathrm{baabaa}}\\
\color{blue}{\mathrm{aabaab}}\qquad&\qquad \mathrm{abbaaa}\qquad&\qquad \mathrm{babaaa}\\
\color{blue}{\mathrm{aababa}}\qquad&&\qquad \mathrm{bbaaaa}\\
\color{blue}{\mathrm{aabbaa}}\qquad
\end{array}
\end{align*}
The following proof might be somewhat beyond OPs current level. Its just to give an impression how (1) can be shown.
Proof: We use generating functions in order to derive (1) and start with generating functions of binary Smirnov words. These are words with no equal consecutive characters. In terms of paths these are paths with no equal consecutive steps. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.) Encoding $(1,0)$-steps with $x$ and $(0,1)$-steps with $y$ the generating function of Smirnov words is
\begin{align*}
\left(1-\frac{x}{1+x}-\frac{y}{1+y}\right)^{-1}\tag{2}
\end{align*}
The coefficient of $x^ry^{n-r}$ in the generating function above gives all words of length $r+(n-r)=n$ with no runs of $(1,0)$-steps or $(0, 1)$-steps with length $>1$. This is a convenient starter but we need some adaptations. At first we note that the runs of $(0,1)$-steps are not restricted. We therefore substitute occurrences of $(0,1)$-steps with one or more $(0,1)$-steps.
\begin{align*}
y\quad\to\quad y+y^2+y^3+\cdots=\frac{y}{1-y}
\end{align*}
Each of the wanted paths has at most $k$ horizontal $(1,0)$-steps. We respect this by substituting
\begin{align*}
x\quad\to\quad x+x^ 2+\cdots+x^k=\frac{x\left(1-x^k\right)}{1-x}
\end{align*}
We obtain using the generating function (2) and the substitutions above the wanted generating function
\begin{align*}
\color{blue}{A_k(x,y)}&=\left(1-\frac{\frac{x\left(1-x^k\right)}{1-x}}{1+\frac{x\left(1-x^k\right)}{1-x}}-\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}\right)^{-1}\\
&=\left(1-\frac{x\left(1-x^k\right)}{1-x^{k+1}}-y\right)^{-1}\\
&\,\,\color{blue}{=\frac{1}{1-y\,\frac{1-x^{k+1}}{1-x}}\,\frac{1-x^{k+1}}{1-x}}\tag{3}
\end{align*}
Since we are looking for paths with length $n$ with $r$ $(1,0)$-steps, and $n-r$ $(0,1)$-steps, we are looking for the coefficient of $x^{r}y^{n-r}$ in $A_k(x,y)$.
We obtain denoting with $[x^r]$ the coefficient of $x^r$ in a series the number of valid paths as
\begin{align*}
\color{blue}{[x^{r}y^{n-r}]A_k(x,y)}\tag{4}
\end{align*}
We start calculating (4) by expanding (3) in powers of $y$. We obtain
\begin{align*}
\color{blue}{[x^{r}}&\color{blue}{y^{n-r}]A_k(x,y)}=[x^{r}y^{n-r}]\frac{1}{1-y\,\frac{1-x^{k+1}}{1-x}}\,\frac{1-x^{k+1}}{1-x}\\
&=[x^{r}y^{n-r}]\sum_{q=0}^\infty\left(\frac{1-x^{k+1}}{1-x}\right)^{q+1}y^q\tag{5.1}\\
&=[x^r]\left(\frac{1-x^{k+1}}{1-x}\right)^{n-r+1}\tag{5.2}\\
&=[x^r]\sum_{q=0}^{\infty}\binom{-(n-r+1)}{q}(-x)^q\left(1-x^{k+1}\right)^{n-r+1}\tag{5.3}\\
&=[x^r]\sum_{q=0}^{\infty}\binom{n-r+q}{q}x^q\left(1-x^{k+1}\right)^{n-r+1}\tag{5.4}\\
&=\sum_{q=0}^r\binom{n-r+q}{q}[x^{r-q}]\left(1-x^{k+1}\right)^{n-r+1}\tag{5.5}\\
&=\sum_{q=0}^r\binom{n-q}{r-q}[x^{q}]\left(1-x^{k+1}\right)^{n-r+1}\tag{5.6}\\
&=\sum_{q=0}^{\left\lfloor\frac{r}{k+1}\right\rfloor}\binom{n-(k+1)q}{r-(k+1)q}[x^{(k+1)q}]\left(1-x^{k+1}\right)^{n-r+1}\tag{5.7}\\
&\,\,\color{blue}{=\sum_{q=0}^{\left\lfloor\frac{r}{k+1}\right\rfloor}\binom{n-(k+1)q}{r-(k+1)q}\binom{n-r+1}{q}(-1)^q}\\
\end{align*}
and the claim (1) follows.
Comment:
*
*In (5.1) we apply the geometric series expansion.
*In (5.2) we select the coefficient of $y^{n-r}$.
*In (5.3) we apply the binomial series expansion of $(1-x)^{-(n-r+1)}$.
*In (5.4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (5.5) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper limit of the series to $r$ since other indices do not contribute.
*In (5.6) we note that in the binomial expansion of $\left(1-x^{k+1}\right)^{n-r+1}$ the powers of $x$ are multiples of $k+1$. We respect this by substituting $q\to(k+1)q$ in (5.7) and select finally the coefficient of $x^{(k+1)q}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do next? I can't come up with anything else...
| As a general principle, sometimes things look simpler if we perform a suitable substitution: let $u = \sqrt[3]{t}$, so that $t = u^3$ and $$A(t) = A(u^3) = \frac{1 - u^3}{1 - u} + \frac{1 + u^3}{1 + u}.$$ Now it becomes obvious that we either need to factor the numerators, or put everything over a common denominator. In the first approach, the difference of cubes factorization
$$1 \pm u^3 = (1 \pm u)(1 \mp u + u^2)$$
yields
$$A(u^3) = (1 + u + u^2) + (1 - u + u^2) = 2(1 + u^2),$$
hence $$A(t) = 2(1 + t^{2/3}).$$
In the second case,
$$\begin{align}
A(u^3) &= \frac{(1-u^3)(1+u) + (1+u^3)(1-u)}{(1-u)(1+u)} \\
&= \frac{(1 + u - u^3 - u^4) + (1 - u + u^3 - u^4)}{1-u^2} \\
&= \frac{2(1-u^4)}{1-u^2} \\
&= \frac{2(1+u^2)(1-u^2)}{1-u^2} \\
&= 2(1+u^2),
\end{align}$$ which is the same as the first approach. When dealing with rational powers, it is often easier to visualize and manipulate the expression if we use a carefully chosen substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Surface delimited by an elliptical cylinder and z+y = 9. We're covering Stokes' Theorem in class right now, and I can't understand anything. I'm struggling to solve even the most trivial examples. Here's one I can't solve: Calculate the surface integral of $\mathbf{F}(x,y,z) = (z,x,y)$ where the surface $S$ is the surface delimited by $\frac{x^2}{4}+\frac{y^2}{9} = 1$ and $z+y=9$. What do I do? I feel like I should solve for $x$ and $y$ and then find a parametric curve in $z$. But then what's the surface associated with it? Should I solve $\frac{x^2}{4}+\frac{y^2}{9} - 1 = z + y - 9$ instead? Would that be a surface? I have no idea what is going on. A hint towards a general approach to solve these types of problems would help also.
| I am going to interpret the question as computing the line integral of $\mathbf{F}(x, y, z)=(z, x, y)$ over the closed curve given by the intersection of the elliptical cylinder $\frac{x^2}{4}+\frac{y^2}{9} = 1$ and the plane $y+z=9$. The reason I interpret it like this is twofold: the elliptical cylinder and the plane do not "delimit" any finite surface, and the alternative interpretation of computing the surface integral over some ellipse requires computation of an "anti-curl", which is less straightforward. In any case, the alternative interpretation can be solved in the same way by noting that $(xz, xy, yz)$ is an anti-curl of $\mathbf{F}$.
The intersection of the elliptical cylinder and the plane is a closed curve $C$ that can be parameterized as
$$(x, y, z)=(2\cos\theta, 3\sin\theta, 9-3\sin\theta),\qquad \theta\in[-\pi,\pi]$$
Calculation using Stokes' theorem
Stokes' theorem says
$$\oint_C\mathbf{F}(x, y, z)\cdot d\boldsymbol{\ell}=\iint_S \left(\nabla\times\mathbf{F}(x, y, z)\right)\cdot d\mathbf{S}$$
where $S$ is any surface whose boundary is $C$.
The curl of $\mathbf{F}$ is
$$\nabla\times\mathbf{F}(x, y, z)=(1, 1, 1)$$
We choose $S$ to be the surface in the plane $y+z=9$ that is delimited by the curve $C$. A parameterization of $S$ is
$$(x, y, 9-y),\qquad x\in[-2, 2], y\in\left[-\frac{1}{2}\sqrt{36-9x^2},\frac{1}{2}\sqrt{36-9x^2}\right]$$
and the differential of surface is
$$d\mathbf{S}=(0, 1, 1)\:dy\:dx$$
Note that the norm of $d\mathbf{S}$ must be the area of the surface element, which is $\sqrt{2}$ in this case.
We compute the surface integral:
\begin{align*}
\iint_S \left(\nabla\times\mathbf{F}(x, y, z)\right)\cdot d\mathbf{S} & = \int_{x=-2}^{2} \int_{y=-\frac{3}{2}\sqrt{4-x^2}}^{\frac{3}{2}\sqrt{4-x^2}} (1, 1, 1) \cdot (0, 1, 1) \:dy\: dx \\
& = \int_{x=-2}^{2} \int_{y=-\frac{3}{2}\sqrt{4-x^2}}^{\frac{3}{2}\sqrt{4-x^2}} 2 \:dy\: dx \\
& = \int_{x=-2}^{2} 6\sqrt{4-x^2} \:dy\: dx \\
& = 12\pi.
\end{align*}
Direct computation of the line integral
The line integral can also be computed directly. The differential of length is
$$d\boldsymbol{\ell}=(-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta$$
so the line integral is
\begin{align*}
\oint_C\mathbf{F}(x, y, z)\cdot d\boldsymbol{\ell} & = \int_{-\pi}^{\pi} \mathbf{F}(2\cos\theta, 3\sin\theta, 9-3\sin\theta)\cdot (-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta \\
& = \int_{-\pi}^{\pi} (9-3\sin\theta, 2\cos\theta, 3\sin\theta)\cdot (-2\sin\theta, 3\cos\theta, -3\cos\theta)\:d\theta \\
& = \int_{-\pi}^{\pi} \left[-18\sin\theta + 6\sin^2\theta + 6 \cos^2\theta - 9\sin\theta\cos\theta \right]\:d\theta \\
& = 12\pi.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if a+b+c+d=5m, what are the sets of values of a,b,c,d for which 5 divides none of them? Let's assume $a+b+c+d$ is an integer multiple of $5$, that is
$$a+b+c+d=5m-------(1)$$
For all integer $m,$
where $5$ does not divide any of $a,b,c$ and $d$. What are all the sets of values of $a,b,c,d$?
The sets I could find are:
$$5w+1, 5x+1, 5y+1, 5z+7.$$
$$5w+1, 5x+1, 5y+4, 5z+4.$$
$$5w+1, 5x+2, 5y+3, 5z+4.$$
$$5w+2, 5x+2, 5y+2, 5z+4.$$
$$5w+2, 5x+2, 5y+3, 5z+3.$$
$$5w+3, 5x+3, 5y+3, 5z+1.$$
$$5w+4, 5x+4, 5y+4, 5z+3.$$
For all integers $w,x,y,z.$
Are the sets I have listed above sufficient for (1)? If not, please give a counter example.
| You missed $5w+3,5x+4,5y+4,5z+4$. Also your first set can be rendered with the last term as $5z+2$.
We may render each term as $\in\{1,2,3,4\}\bmod 5$ and require four such numbers to add up to a multiple of $5$. The sum must then be $\in\{5,10,15\}$.
Thus we now have a restricted partitioning problem. Clearly only $1+1+1+2=5$ and $3+4+4+4=15$ meet the restrictions with their respecive sums. The remaining sums must equal $10$ with the minimum term $\in\{1,2\}$ (i e. less than the average which would be $5/2$) and the maximum term $\in\{3,4\}$ (greater than the average). By pairing each possible minimum with each possible maximum we do indeed generate just the last five partitions indicated in the question ($1+1+4+4,1+2+3+4,...$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof verification: Given $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, show that $\lim _{x\to 0}\frac {f(x)}{x}=0$. Given that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$.
Proof: Let $\epsilon\gt 0$ be fixed.
\begin{align*}
\frac {f(x)}x&=\sum_{k=0}^n\color{blue}{\frac{f(x/2^k)-f(x/2^{k+1})}{x/2^{k+1}}}(1/2^{k+1})+\frac{\color{green}{f(x/2^{n+1})}}{x}\\
&=\sum_{k=0}^n\color{blue}{g_k(x)}(1/2^{k+1})+\frac{\color{green}{h_n(x)}}{x}\\
\text{There exists $N$ such that }\\ \left|\frac {f(x)}x\right|&\le \sum_{k=0}^N|g_k(x)|\frac 1{2^{k+1}}+\epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\frac 1{|x|}\epsilon|x|\\
\text{It follows that } \\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\epsilon\\
\text{Taking $N\to \infty$,it follows that}\\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\tag 1
\end{align*}
Since $\epsilon\gt 0$ is arbitrary, it follows by $(1)$ that $\liminf \left|\frac {f(x)}x\right|= \limsup\left|\frac {f(x)}x\right|=0=\lim_{x\to 0} \frac{f(x)}x. \;\;\; \blacksquare$
Is my proof correct? Thanks.
| Let $\epsilon > 0$ be arbitrary, and let $x_\epsilon$ be small enough so that
$
\Bigg| \dfrac{f \big( x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg|
\le \epsilon
$ for all $|x| \le x_\epsilon$.
Then,
$$
\Bigg| \dfrac{f \big(x \big)}{x} \Bigg| =
$$
$$
\dfrac{1}{2} \Bigg|
\dfrac{f \big(x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} +
\dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}}
\Bigg| \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg|
\Bigg] \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {4} \big)}{\frac {x} {4}} \Bigg|
\Bigg]
\Bigg] \le \ldots \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\dfrac{1}{2} \Bigg[
\epsilon + \ldots +
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{\frac {x} {2^{n}}} \Bigg|
\Bigg]
\Bigg]
\Bigg] =
$$
$$
\sum_{k=1}^n \dfrac{\epsilon}{2^k} +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \le
$$
$$
\sum_{k=1}^\infty \dfrac{\epsilon}{2^k} +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| =
$$
$$
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \xrightarrow{n \rightarrow \infty}
$$
$$
\epsilon + 0 =
$$
$$
\epsilon
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula? Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
| If you use complex analysis, then there is a pole at $x = i$
If we take the contour of the semi-circle in the upper half-plane, then when $n\ge 1$ the integral along the semi-circle vanishes. That means that we can focus on the residual at $i.$
$\int_0^\infty \frac {1}{(x^2 + 1)^n} \ dx = \frac {\pi i}{(n-1)!} \frac {d^{n-1}}{dx^{n-1}} \frac {1}{(x+i)^n}$ evaluated at $i$
$\frac {\pi (2n-2)!}{((n-1)!)^2(2)^{2n-1}}$
Since it seem like you are interested:
$\int_0^\infty \frac {1}{x^n + 1}\ dx = \frac {\pi \csc \frac {\pi}{n}}{n}$
And, I am thinking that $\int_0^\infty \frac {1}{(x^m + 1)^n}\ dx$ will give something along the lines of $\frac {\pi \csc \frac {\pi}{m}}{m^n} {2n-2 \choose n-1}.$ But, I haven't worked it out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
$S_n = \frac{12}{(4-3)(4^2-3^2)} + \frac{12^2}{(4^2-3^2)(4^3-3^3)} + ... + \frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$
So I know that in order to find the series we need to change the form because we can cancel the terms $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} = \frac{A}{4^n-3^n} + \frac{B}{4^{n+1}-3^{n+1}}$$
But my question is how do we get $A = \frac{4^n+3^n}{2}$ and $B = \frac{4^{n+1}+3^{n+1}}{2}$
When I find the common denominator and add them together I got $12^n = A(4^{n+1}-3^{n+1}) + B(4^n-3^n)$. I tried simplifying but it can't seem to work.
Can anyone show me how to find A and B? Thank you in advance.
| $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} =\frac {4^n\cdot3^n}{(4^n-3^n)(4\cdot4^n-3\cdot3^n)}$$
Divide top and bottom by $3^{2n}$ and write $x=\frac{4^n}{3^n}$ and we get
$$\frac{x}{(x-1)(4x-3)}$$
In partial fractions, this is $$\frac{1}{x-1}-\frac{3}{4x-3}=\frac{1}{x-1}-\frac{1}{\frac43x-1}$$
So the expression is $$\frac{1}{(\frac43)^n-1}-\frac{1}{(\frac43)^{n+1}-1}$$
or, equivalently, $$\frac{3^n}{4^n-3^n}-\frac{3^{n+1}}{4^{n+1}-3^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)} $ I used all the trig and log tricks but still can't compute this limit
$$\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}.$$
I tried the following:
\begin{align*}\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}
&= \lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(x)}
\\ &= \lim_{x\to0}\frac{\ln\left(1+4\sin^2(x)\cos^2(x)\right)}{\sin^2(x)} .\end{align*}
I tried to substitute $\sin^2(x)$ but still wandering.
| Just using the standard limit $\lim_{t\to 0}\frac{\ln (1+t)}{t} = 1$ you get
\begin{eqnarray*}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}
& = & \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)}\cdot \frac{\sin^2(2x)}{\sin^2 x} \\
& = & 4\cos^2 x \cdot \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)} \\
& \stackrel{x\to 0}{\longrightarrow} & 4\cdot 1 \cdot 1 = 4
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? How to calculate integral $\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? I got this integral by using Abel-Plana formula on series $\sum_{n=0}^\infty \frac{1}{(n+1)^2}$. This integral can be splitted into two integrals with bounds from 0 to 1 and from 1 to infinity and the both integrals converge, so does the sum. I checked with WolframAlpha and the value of the integral is $\frac{-9 + \pi^2}{24}$, but I don't know how to compute it. Also, I tried to write $\frac{2xdx}{(1+x^2)^2}=d\frac{1}{x^2+1}$ and then tried to use partial integration, but didn't succeded.
Any help is welcome. Thanks in advance.
| Here is a brutal-force computation using contour integral.
Let $\operatorname{Log}(\cdot)$ be the logarithm with the branch cut $[0, \infty)$ so that $\operatorname{Arg}(z) \in (0, 2\pi)$. Also, for $s > 0$ we define
$$ I(s) = \int_{0^+ i}^{\infty + 0^+ i} f_s(z) \, \mathrm{d}z, \quad \text{where} \quad f_s(z) = \frac{e^{s\operatorname{Log}(z)}}{(e^{2\pi z}-1)(z^2+1)^2}. $$
Then by noting that the contour integral of $f_s(z)$ along the square with the corners $(\pm1 \pm i)(N+\frac{1}{2})$ vanishes as $N \to \infty$, we get
\begin{align*}
(1 - e^{2\pi i s}) I(s)
&= \int_{0^+ i}^{\infty + 0^+ i} f_s(z) \, \mathrm{d}z + \int_{\infty + 0^- i}^{0^- i} f_s(z) \, \mathrm{d}z
= 2\pi i \sum_{n \neq 0} \mathop{\mathrm{Res}}_{z=ni} f_s(z).
\end{align*}
Some tedious computation yields:
$$ 2\pi i \mathop{\mathrm{Res}}_{z=ni} f_s(z) = \begin{cases}
e^{i\pi s/2} \left( \frac{is^2}{8} - \frac{3is}{8} + \frac{3i}{16} - \frac{i\pi^2}{12} + \frac{\pi(s-1)}{4} \right), & n = 1, \\
e^{3i\pi s/2} \left( \frac{is^2}{8} - \frac{3is}{8} + \frac{3i}{16} - \frac{i\pi^2}{12} - \frac{\pi(s-1)}{4} \right), & n = -1, \\
i e^{i\pi s/2} \frac{n^s}{(n^2-1)^2}, & n \geq 2, \\
i e^{3i\pi s/2} \frac{|n|^s}{(n^2-1)^2}, & n \leq -2. \\
\end{cases} $$
Assuming in addition that $s \notin \mathbb{Z}$, we get
\begin{align*}
I(s)
&=
+ \frac{e^{i\pi s/2} + e^{3i\pi s/2}}{1 - e^{2\pi i s}} \left( \frac{is^2}{8} - \frac{3is}{8} + \frac{3i}{16} - \frac{i\pi^2}{12} \right) \\
&\quad + \frac{e^{i\pi s/2} - e^{3i\pi s/2}}{1 - e^{2\pi i s}} \frac{\pi(s-1)}{4} \\
&\quad + \frac{i(e^{i\pi s/2} + e^{3i\pi s/2})}{1 - e^{2\pi i s}} \sum_{n=2}^{\infty} \frac{n^s}{(n^2-1)^2}
\end{align*}
Letting $s \to 1$,
\begin{align*}
I(1)
&= \left( \frac{1}{32}+\frac{\pi ^2}{24} \right) - \frac{1}{4} - \frac{1}{2} \sum_{n=2}^{\infty} \frac{n}{(n^2-1)^2}
\end{align*}
Then the desired answer follows from
$$ \sum_{n=2}^{\infty} \frac{n}{(n^2-1)^2}
= \frac{1}{4} \sum_{n=2}^{\infty} \left( \frac{1}{(n-1)^2} - \frac{1}{(n+1)^2} \right)
= \frac{5}{16}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find power series solution for $y''-xy=0$ close to $x=0$. Find power series solution for $y''-xy=0$ close to $x=0$.
$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-x\sum_{n=0}^\infty a_nx^n=0$
Then,
$a_{n+2}(n+2)(n+1)-a_{n-1}=0 \implies a_{n+2}(n^2+3n+2)-a_{n-1}=0 \implies a_{n+3}=a_n\cdot \frac{1}{(n+3)(n+2)}$
I got two problems:
$(1)$ The answer is $a_{n+3}=(n+3)(n+2)a_n$ , where am I wrong ?
$(2)$ How can I find the general power series solution ?
| If you look at the pattern (not very simple, I agree), there is a closed form for the coefficients
$$a_n=\frac{3^{-2 n/3}\,\, \Gamma \left(\frac{2}{3}\right)}{ \Gamma \left(\frac{n+2}{3}\right) \Gamma \left(\frac{n+3}{3}\right)}\left(a_0+a_1 \cos \left(\frac{2 \pi n}{3}\right)+a_2\sin \left(\frac{2 \pi n}{3}\right) \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of the square harmonic series I stumbled across the following series reviewing some HW from a few years ago
$\sum_{i=1}^{n}\left(\sum_{j=i}^{n}\frac{1}{j}\right)^2$
i.e.
$(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n})^2+(\frac{1}{2}+\ldots+\frac{1}{n})^2+\ldots+(\frac{1}{n})^2$
This series equals $2n-\sum_{i=1}^{n}\frac{1}{i}$, which I have confirmed with some code. I am curious if anyone can give a hand in trying to show this relation. So far, writing $\sum_{i=1}^{n}\frac{1}{i}$ as $S_n$, I have rewritten the sum as
$S_n^2+(S_n-S_1)^2+(S_n-S_2)^2+\ldots +(S_n-S_{n-1})^2$
But have been stuck at dead ends using this approach. Any thoughts or hints would be greatly appreciated.
| Proof by induction. For $n = 1$ it is easy to show that this holds. Assume the induction hypothesis for some $n$. Then we want to show that
$$
\sum_{p = 1}^{n + 1} \biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
\;=\;
2 (n + 1) - \sum_{p = 1}^{n + 1} \frac{1}{p}
\text{.}
$$
We have
$$
\biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
\;=\;
\biggl( \sum_{q = p}^{n} \frac{1}{q} + \frac{1}{n + 1}\biggr)^2
\;=\;
\biggl( \sum_{q = p}^{n} \frac{1}{q} \biggr)^2 + \frac{2}{n + 1} \sum_{q = p}^{n} \frac{1}{q} + \frac{1}{(n + 1)^2}
$$
and so
$$
\sum_{p = 1}^{n + 1} \biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
\;=\;
\sum_{p = 1}^{n} \biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
+
\frac{1}{(n + 1)^2}
\;=\;
\sum_{p = 1}^{n} \biggl( \sum_{q = p}^{n} \frac{1}{q} \biggr)^2
+
\frac{2}{n + 1} \sum_{p = 1}^n \sum_{q = p}^n \frac{1}{q}
+
\frac{1}{n + 1}
.
$$
Here the last double sum equals
$$
\sum_{p = 1}^n \sum_{q = p}^n \frac{1}{q}
\;=\;
\sum_{q = 1}^n \sum_{p = 1}^q \frac{1}{q}
\;=\;
\sum_{q = 1}^n 1
\;=\;
n.
$$
Now using our assumption, we find
$$
\sum_{p = 1}^{n + 1} \biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
\;=\;
\sum_{p = 1}^{n} \biggl( \sum_{q = p}^{n} \frac{1}{q} \biggr)^2
+
2
-
\frac{1}{n + 1}
\;=\;
2 n + 2
-
\sum_{p = 1}^{n} \frac{1}{p}
-
\frac{1}{n + 1}
$$
which equals what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding extreme values of $ f :\mathbb R^2\rightarrow \mathbb R$, when the determinant $\Delta = AC - B^2 = 0$. We generally rely on the result:
[Let $ f$ be a real valued function from $\mathbb R^2$ with continuous partial derivatives at a stationary point $\vec a$ in $\mathbb R^2$. Let
$A = D_{11}f(\vec a)$,
$ B = D_{12}f(\vec a) = D_{21}f(\vec a) $
$C = D_{22}f(\vec a)$,
and let $\Delta = AC - B^2$ . Then,
(a) If $\Delta>0 $ and $A>0$, $f$ has a relative minimum at $\vec a$
(b) If $\Delta>0 $ and $A<0$, $f$ has a relative maximum at $\vec a$
(c) If $\Delta<0 $ , $f$ has a saddle point at $\vec a$ ]
But when $\Delta = 0, f$ can have a local minimum of a local maximum or a saddle point at $\vec a$.
What are the other ways to proceed when we have $\Delta = 0$ or $A = 0$?
| Consider the problem:
Find and classify the extreme values (if any) of the function: $$f(x,y) = y^2+x^2y+x^4$$
Here we have,
$D_1f = \partial f/\partial x = 2xy +4x^3$
$D_2f = \partial f/\partial y = 2y +x^2$
$D_{11}f = \partial^2 f/\partial x^2 =2y +12x^2$
$D_{12}f = \partial^2 f/\partial x \partial y =2x$
$D_{22}f = \partial^2 f/\partial y^2 =2$
For a stationary point (x,y) we want $D_1f(x,y) =0$, and $ D_2f(x,y)=0$. Then,
$D_1f(x,y) =0 \Rightarrow2xy +4x^3 = 0 \Rightarrow x(2y+4x^2)=0$ (equation 1)
$D_2f(x,y) =0 \Rightarrow2y +x^2 = 0 \Rightarrow 2y=-x^2$ (equation 2)
Substituting the value of $2y$ in equation 1, we get, $3x^3=0 \Rightarrow x=0 \Rightarrow y=0$ and so the origin $(0,0)$ is the only stationary point for $\vec f$.
But for this point,
$A = D_{11}f((0,0)) = 2(0)+12(0)^2 = 0$
$B = D_{12}f((0,0)) = 2(0) = 0$
$C = D_{22}f((0,0)) = 2$
i.e. both $A = 0$ and $\Delta = AC - B^2 = 0$. Thus, no direct conclusion can be drawn.
However (using completing the square method),
$f(x,y) - f(0,0) = [y^2+x^2y+x^4] -[(0)^2+(0)^2(0)+(0)^4]$
$= (y)^2+x^2(y)+x^4 $
$= (y)^2+2(x^2/2)(y)+x^4 $
$= (y)^2+2(x^2/2)(y)+ (x^2/2)^2 - (x^2/2)^2 + x^4 $
$= [(y)^2+2(x^2/2)(y)+ (x^2/2)^2] + [x^4 - (x^4/4)] $
$= [y + (x^2/2)]^2 +(3/4)x^4 $
$\geq 0$ , for any $(x,y) \in \mathbb R^2$.
Therefore $f(x,y) \geq f(0,0)$ for all $(x,y) \in \mathbb R^2$ where $(0,0)$ is the only stationary point of f. Thus it is the global minima of f and the minimum value of f is $0 (=f(0,0))$.
The inference here is that in such cases one must always try to analyze not the function $f(\vec x)$ in particular, but the difference $f(\vec x) - f(\vec a)$ and see if something meaningful can be inferred from there.
Thank You!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
n-th root in limits I'm having trouble with this kind of series: $$a_n = \sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})$$
I tried to make something like perfect square to simplify those $7$th root junk which is obviously impossible, and as $n$ tends to infinity L'Hopital's rule also can't solve the problem in this form.
The multiple choice:
*
*$$\lim_{n\to\infty}a_n=\frac{9}{7}$$
*$$\lim_{n\to\infty}a_n=+\infty$$
*$$a_n\sim\frac{9}{7}n^{-5/14},\quad n\to+\infty$$
| One more, multistorey, without series, way
$$\sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})=$$
$$=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-\sqrt[7]{1-\frac{4}{n}}\right)
=\\
=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-1-\sqrt[7]{1-\frac{4}{n}}+1 \right) =\\
=\sqrt{n}\sqrt[7]{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot \frac{5}{n}-\frac{\sqrt[7]{1-\frac{4}{n}}-1}{\frac{4}{n}}\cdot \frac{4}{n}\right)
=\\
=\frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot 5+\frac{\sqrt[7]{1-\frac{4}{n}}-1}{-\frac{4}{n}}\cdot 4\right) \sim \frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{5}{7}+\frac{4}{7}\right) = \\
=\frac{9}{7n^{\frac{5}{14}}}$$
where it was used $\frac{(1+x)^\alpha-1}{x}\to \alpha, x\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.
Proof: $$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})
$$$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$
As $\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$
The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.
I.e. $$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$
$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$
and the RHS will reduce down to $\sqrt{3}$. Hence LHS=RHS.
Some things that I've noticed about this method of proof:
*
*It could be used to (incorrectly) prove that $$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$
So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?
*
*Instead of proving (*), wouldn't this method of proof actually prove that? $$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$
for some $k\in \mathbb{Z}$ which we must find. In this case being when $k=0$.
| Using complex numbers:
$$
\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})
= \arg((5+\sqrt{3}i)(2+\sqrt{3}i))
= \arg(7+7\sqrt{3}i)
= \arctan(\sqrt{3})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
I found the similar question in Quora. There was slightly a mistake.
$$(\tan\theta+\sin\theta)^2-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\tan\theta\sin\theta-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{\tan^2\theta-\sin^2\theta}-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
I just said that "we have to take $4\sqrt{m(\tan\theta-\sin\theta)}=4\sqrt{mn}$ to satisfy $\tan\theta-\sin\theta=n$ but I don't like it, I just want direct derivation."
And in quora they just took $4\sqrt{m}$ common and wrote $$(\tan\theta-\sin\theta)^2+4\sqrt{m}(\sqrt{\tan\theta-\sin\theta-n})-n^2=0$$
which is totally wrong
cause $\sqrt{\tan\theta-\sin\theta-n}\neq\sqrt{\tan\theta-\sin\theta}-\sqrt{n}$
| Assume that $\theta\in (0,\pi/2)$ and hence $\tan\theta,\,\sin\theta>0$.
If $w=\sin\theta$, then $\tan\theta=\frac{w}{\sqrt{1-w^2}}$ and
$$
m=w+\frac{w}{\sqrt{1-w^2}}=\frac{w+w\sqrt{1-w^2}}{\sqrt{1-w^2}}
$$
Now
$$
F(n)=m^2-n^2-4\sqrt{mn}, \quad n\ge 0,
$$
is strictly decreasing and continuous $F(0)=m^2>0$ and $F(\infty)=-\infty$. Hence, it vanishes for a unique $n$. It suffices to show that
$$
F(\tan\theta-\sin\theta)=F\left(\frac{w-w\sqrt{1-w^2}}{\sqrt{1-w^2}}\right)=0
$$
which is straight-forward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why is $13^n\cdot 14 + 13\cdot 2^{2n}$ divisible by $9$ for any nonnegative integer $n$? Prove that statement
$13^n\cdot14+13\cdot2^{2n}$ is divisible by $9$ for any natural $n$
So, the first step is to represent $13^n$ as sum of
$(9+4)^n$ and what the next one?
Give me a hint, please.
I've no idea where to go for now.
Thanks!
| One can always do it by induction.
$13^{n+1} \cdot 14 + 13\cdot 2^{2(n+1)}=$
$13^n\cdot 14\times 13 + 13\cdot 2^{2n}\times 4=$
$13^n\cdot 14\times(9+4) + 13 \cdot2^{2n}\times 4=$
$13^n\cdot 14\times 9 + 4\times (13^n\cdot 14+ 13 \cdot2^{2n})$
which is a multiple of $9$ if $13^n\cdot 14+ 13 \cdot2^{2n}$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$ I am trying to evaluate this integral:
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$$
I coverted the region by using polar coordinate system for xy-plane and it became:
$$\displaystyle\int_{0}^{1}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz+\displaystyle\int_{0}^{1}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta}}r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}drd\theta dz$$
After IBP
$$\int r\sqrt{r^2+z^2}\arctan{\sqrt{r^2+z^2}}dr=\frac{1}{3}(r^2+z^2)^{\frac{3}{2}}\arctan{\sqrt{r^2+z^2}}-\frac{1}{6}(r^2+z^2)+\frac{1}{6}\ln(r^2+z^2+1)+C $$ and replace upper and lower limits, it is scary-looking.
My experience in solving this tyle of integral is still weak, I think there is a way to get rid of one variable from the beginning integral, but i can't figure it out. Hope everyone can help me, thanks.
| A simpler way using symmetry.
Firstly denote $r=\sqrt{x^2+y^2+z^2}$ and split the integrand
$$
I=\int_0^1\int_0^1\int_0^1r\arctan r~dV=\frac\pi2\underbrace{\int_0^1\int_0^1\int_0^1r~dV}_{I_1}-\underbrace{\int_0^1\int_0^1\int_0^1r\operatorname{arccot} r ~dV}_{I_2}
$$
The former is the average distance from a point in a unit cube to the origin. Use its symmetry to split the cube into 6 parts, in each of which the integral is identical. Substitute $p=xz,q=yz$ in one of them and perform the easy integral over $z$
$$
I_1=6\int_0^1\int_0^z\int_0^y\sqrt{x^2+y^2+z^2}dxdydz\\
=6\int_0^1\int_0^1\int_0^1z^3\sqrt{p^2+q^2+1}dpdqdz=\frac32\int_0^1\int_0^1\sqrt{p^2+q^2+1}dpdq
$$
Now switch to polar coordinate $(p,q)\to (\rho,\theta)$
$$
\begin{align}
&I_1=\frac32\int_0^1\int_0^1\sqrt{p^2+q^2+1}dpdq
\\ =&\frac32\int_0^{\pi/4}\int_0^{\sec\theta}\sqrt{\rho^2+1}~\rho d\rho d\theta
\\ =&\frac12\int_0^{\pi/4}d\theta\cdot\int_0^{\sec\theta}\frac32\sqrt{\rho^2+1}~d(\rho^2+1)
\\ =&\frac12\int_0^{\pi/4}\frac{\sqrt{1+\cos^2\theta}^3}{\cos^3\theta}-1~d\theta
\\ =&\frac12\left(\arcsin\left(\frac{\sin \theta }{\sqrt{2}}\right)+\frac{\sin\theta~\sqrt{1+\cos^2\theta}}{2\cos^2\theta }\right.
\\&\left.\left.+\log \left(\frac{1+\sin\theta\sqrt{1+\cos^2\theta}}{\cos^2\theta}\right)\right)\right|_0^{\pi/4}-\frac\pi8
\\ =&\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\frac12\log \left(2+\sqrt3\right)
\end{align}
$$
The last integral has an elementary antiderivative and can be evaluated quickly substituting $s=\sin\theta$.
By the identity $\displaystyle\operatorname{arccot}\lambda=\int_0^1\frac{\lambda ~dw}{\lambda^2+w^2}$ and the symmetry of the latter
$$
\begin{align}
&I_2=\int_0^1\int_0^1\int_0^1r\cdot\left(\int_0^1\frac{r~dw}{w^2+r^2}\right)dV
\\ =&\int_0^1\int_0^1\int_0^1\int_0^1\frac{x^2+y^2+z^2}{x^2+y^2+z^2+w^2}dxdydzdw
\\ =&3\int_0^1\int_0^1\int_0^1\int_0^1\frac{x^2}{x^2+y^2+z^2+w^2}dxdydzdw
\\ =&\frac34\int_0^1\int_0^1\int_0^1\int_0^1\frac{x^2+y^2+z^2+w^2}{x^2+y^2+z^2+w^2}dxdydzdw
\\ =&\frac34
\end{align}
$$
Combining the results above and the desired integral follows
$$
I=\frac{\pi \sqrt{3}}{8}-\frac{\pi ^2}{48}+\frac{\pi}{4} \log \left(2+\sqrt{3}\right)-\frac34
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$
Evaluate $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$$
Wolfram alpha gives $\dfrac{-7}{20}$.
Here is my work
For $x \to 0 $
$\tan x \sim x$
$\sin(x+\tan x) \sim \sin2x$
$\sin2x \sim 2x$ (I wasn't sure about this but I evaluated the limit and got $1$.)
$\sin(x+\tan x) \sim 2x$
$\sin(x+\tan x)-2x\cos x \sim 2x(1-\cos x)$
$(1-\cos x) \sim x$
So I got that
$\dfrac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2} \sim \dfrac{3x}{x(\sin x^2)^2}$
which means I got $\infty$.
Did make a mistake somewhere?
And if so can this problem be solved by this approach or I need to try something else (like l'hopitals or something)?
| You have messed up with asymptotic relations. You must be really really careful when you use them.
In this limit, we know from the denominator that we have to search for a $5$ degree expansion. Namely:
$$x\cdot(\sin(x^2))^2\,\,\sim\,\, x^5$$
Now, we set things ready for Tyalor-MacLaurin series:
$$\sin(t)=t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5)$$
$$\tan(t)=t+\frac{1}{3}t^3+\frac{2}{15}t^5+o(t^5)$$
$$\cos(t)=1-\frac{1}{2}t^2+\frac{1}{24}t^4+o(t^4)$$
Now, we can work on t numerator:
$$\sin(x+\tan(x))-2x\cos(x)=\sin\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)-2x\left(1-\frac{1}{2}x^2+\frac{1}{24}x^4+o(x^4)\right)=2x+\frac{1}{3}x^3+\frac{1}{120}x^5-\frac{1}{6}\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)^3+\frac{1}{120}\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)^5-2x+x^3-\frac{1}{12}x^5+o(x^5)=-\frac{7}{20}x^5+o(x^5)$$
Note that, here, I haven't done all the calculations in order to exapand the $3$ and $5$ powers of quadrinomials, but I have kept only the most significant (grade $1$, $3$ and $5$).
So:
$$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}=\lim_{x\to 0}\frac{-\frac{7}{20}x^5+o(x^5)}{x^5}=-\frac{7}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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In how many ways we can place 9 different balls in 3 different boxes such that in every box at least 2 balls are placed?
In how many ways we can place $9$ different balls in $3$ different boxes such that in every box at least $2$ balls are placed?
Approach 1:
Let $x_1$, $x_2$, $x_3$ denote the number of balls in boxes $1$,$2$, and $3$ respectively. Now,solving $x_1+x_2+x_3 =9$ for $2≤x_1≤x_2≤x_3≤5$ ($5= 9-$(min of $x_1$) $-$ (min of $x_2$)), we get $(x_1,x_2,x_3) ≡ (2,2,5),(2,3,4),(3,3,3)$
Now, we have to divide $9$ objects in these groups and then distribute into boxes. So the required answer is $$\frac{9!}{2!2!5!}×\frac{1}{2!}×3! +\frac{9!}{2!3!4!}×3! + \frac{9!}{3!3!3!}×\frac{1}{3!}×3!$$ which simplifies to be $\boxed{11508}$. This is the correct answer.
Approach 2 (PIE):
At least $2$ $=$ all $-$ at most $1$. Now, considering $1$ box contains no ($0$) elements, the number of cases corresponding to that is ${3 \choose 1}×2^9$ and the number of cases when $2$ boxes contain no elements will be ${3 \choose 2}×1^9$. Now, number of cases when $1$ box contains only $1$ element is ${3 \choose 1}×{9 \choose 1}×2^8$ (I.e., number of ways to choose 1 box × number of ways to choose 1 ball × remaining distribution) and the number of cases when 2 boxes contain only 1 element is ${3 \choose 2}×{9 \choose 2}×1^7$ . So, using PIE, the required answer should be $$3^9 -({3 \choose 1}2^9 - {3 \choose 2}1^9) - ({3 \choose 1}{9 \choose 1}2^8 - {3 \choose 2}{9 \choose 2}1^7)$$ that is $11346$. This answer is wrong but the "closeness" of the answer to the correct one indicates that I have deducted some cases. Please point out my mistake in approach 2.
| There are two small mistakes otherwise your work using the second approach is also correct. It should be -
$ \displaystyle \small 3^9 - \underbrace{\left[{3 \choose 1}2^9 - {3 \choose 2}1^9 \right]}_{\Large {(a)}} - \underbrace{\left[{3 \choose 1}{9 \choose 1} (2^8 - \color { blue } {2}) - \color { blue } {2} \cdot {3 \choose 2}{9 \choose 2}1^7 \right]}_{\Large (b)}$
Explanation on what I added in $ \color {blue} { \text {blue}}$ in $(b)$:
$1)$ Please note $2^8$ also includes two arrangements where one of the two boxes is empty but those are already counted in $(a)$. So we need to subtract those two arrangements.
$2)$ Please note when we choose two balls for two boxes, there are $2$ ways to place them in two boxes such that each box has one ball.
| {
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Domain of solution of differential equation $y’=\frac{1}{(y+7)(t-3)}$ If we are given the initial-value problem $$\frac{dy}{dt}=\frac{1}{(y+7)(t-3)}, y(0)=0$$
I want to solve said initial value problem & state the domain of the solution. I also want to observe what happens when $t$ approaches the limits of the solution’s domain.
Via separation of variables, one obtains: $$\implies(y+7) \space dy = \frac{dt}{t-3}\implies\frac{1}{2}y^2+7y=\ln|t-3|+c_1$$ $$\iff y^2+14y+49=2\ln|t-3|+(2c_1+49)$$
Now call $C=2c_1+49$, then $$(y+7)^2=2\ln|t-3|+C\implies y(t) = \pm\sqrt{2\ln|t-3|+C}-7$$
Substituting the initial condition:
$$0=\pm\sqrt{2\ln|0-3|+C}-7\iff 7 = \pm\sqrt{2\ln(3)+C}$$
This shows we must choose the positive square root in order for our solution $y(t)$ to pass through the initial condition & solve the IVP. Then $$C=49-2\ln(3)$$ so that
$$y(t)=\sqrt{2\ln\left|\frac{t-3}{3}\right|+49}-7$$
We know from the original differential equation that $y(t)\neq-7$ & $t\ne3$. Thus: $$-7\neq \sqrt{2\ln\left|\frac{t-3}{3}\right|+49}-7 \iff\ln\left|\frac{t-3}{3}\right|\neq-\frac{49}{2}\iff t\neq\pm 3\exp\left(-\frac{49}{2}\right)+3$$
Does this tell us that the domain of the solution has to be $(-\infty,-3\exp\left(-\frac{49}{2}\right)+3)$ in order for $t=0$ to be on the domain of this solution (so the solution passes through the initial condition)?
Would we just say $y(t)\to0$ as $t\to-3\exp\left(-\frac{49}{2}\right)+3$?
Then finally $y(t)=\sqrt{2\ln\left(1-\frac{t}{3}\right)+49}-7$, for $t\in(-\infty, -3\exp\left(-\frac{49}{2}\right)+3)$.
| The equation in question is $$y'(t)=\frac1{[y(t)+7](t-3)}.$$ Thus, there is a singularity at $t=3,$ which suggests that the solutions will have a domain that is a subset of $(-\infty,3)\cup(3,\infty).$ To solve the equation, we consider $$[y(t)+7]y'(t)=\frac1{t-3}.$$ Notice that $$\left(\frac{y^2}2+7y\right)'(t)=[y(t)+7]y'(t).$$ Therefore, we have that $$\frac{y(t)^2}2+7y(t)=\begin{cases}\ln(3-t)+A&t\lt3\\\ln(t-3)+B&t\gt3\end{cases}.$$ Given the initial condition $y(0)=0,$ one has that $$\frac{y(0)^2}2+7y(0)=0=\ln(3)+A,$$ implying $A=-\ln(3).$ Therefore, $$\frac{y(t)^2}2+7y(t)=\begin{cases}\ln(3-t)-\ln(3)&t\lt3\\\ln(t-3)+B&t\gt3\end{cases},$$ which is equivalent to $$y(t)^2+14y(t)=\begin{cases}2\ln(3-t)-2\ln(3)&t\lt3\\2\ln(t-3)+2B&t\gt3\end{cases},$$ which is equivalent to $$[y(t)+7]^2=\begin{cases}2\ln(3-t)-2\ln(3)+49&t\lt3\\2\ln(t-3)+2B+49&t\gt3\end{cases}.$$ Here, it gets complicated. It is required that $$2\ln(3-t)-2\ln(3)+49\geq0$$ and $$2\ln(t-3)+2B+49\geq0.$$ This implies, respectively, that $$\ln(3-t)\geq\ln(3)-\frac{49}2$$ and $$\ln(t-3)\geq{B}-\frac{49}2.$$ Therefore, $$3-t\geq3\exp\left(-\frac{49}2\right)$$ and $$t-3\geq\exp\left(B-\frac{49}2\right)=C\exp\left(-\frac{49}2\right)$$ with $C\gt0.$ Therefore, $$t\leq3-3\exp\left(-\frac{49}2\right)$$ and $$t\geq3+C\exp\left(-\frac{49}2\right).$$ However, at the endpoints, the function is not differentiable, so the domain of the function is $\left(-\infty,3-3\exp\left(-\frac{49}2\right)\right)\cup\left(3+C\exp\left(-\frac{49}2\right),\infty\right).$
With this in mind, the solutions to the equation are given by the two families $$y(t)=-7-\sqrt{f(t)}$$ and $$y(t)=-7+\sqrt{f(t)},$$ where $f$ is the family of functions $$f(t)=\begin{cases}2\ln(3-t)-2\ln(3)+49&t\lt3\\2\ln(t-3)+2B+49&t\gt3\end{cases}.$$
| {
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show this equation $3x^4-y^2=3$ has no integer solution show this diophantine equation
$$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer solution,then $3|y$,let $y=3y'$,then we have
$$x^4-1=3y'^2\tag{1}$$
and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if $x^4\equiv 0\pmod 5$or $5|x$,then $(1)$ it is clear no solution
(2)if $(x,5)=1$,then $x^4-1\equiv 0\pmod 5$,there also exsit $y'$ such $5|3y'^2$,so How to solve my problem? Thanks
| I tried the following.
$3x^4-y^2=3$
$3x^4-3=y^2$
$3(x^2-1)(x^2+1)=y^2 \tag{1}$
If $y^2$ is divisible by $3$ then one of the factors of $(x^2-1)(x^2+1)$ should be divided by $3$ to make the right hand side a perfect square. So $x$ can't be $0 \pmod{3}$. Now if we divide 3 from the both hand side we can see the following.
$$ (x^2-1)(x^2+1)=3^{\alpha}\cdot B^{\gamma} \tag{2} $$
Note that $\alpha$ is odd and $B$ is not necessarily a prime factor.
We've already argued $x\not\equiv 0\pmod{3}$, so $x\equiv +1\pmod{3}$ or $x\equiv -1\pmod{3}$.
If $x\equiv +1\pmod{3}$ then $x^2-1=3^{2\lambda}\cdot D^\eta$ and $x^2+1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
If $x\equiv -1 \pmod{3}$, then $x^2+1=3^{2\lambda}\cdot D^\eta$ and
$x^2-1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
Hope this helps.
| {
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$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.
| Fix a positive integer $n$ and define
$$
f_n(x):=e^x-1-x-x^2/2-\cdots-x^n/n!.
$$
Then $f^\prime_n(x)=e^x-1-x-\cdots-\frac{x^{n-1}}{(n-1)!}$, which is positive (by induction) on $(0,\infty)$.
Hence $f_n$ in strictly increasing on $(0,\infty)$. In particular, this gives the known inequality
$$
\forall x>0, \quad
e^x>1+x+\cdots+\frac{x^n}{n}.
$$
Suppose that $z$ satisfies $e^z=z^2+z$. Then
$$
z^2+z=e^z=f_n(z)+1+z+\cdots+\frac{z^{n}}{n!}>f(0)+1+z+\cdots+\frac{z^n}{n!}=1+z+\cdots+\frac{z^n}{n!}
$$
for all $n$. Now pick $n=1$ and you know that $z>1$. Set $y:=z-1 \in (0,\infty)$, so that
$$
y^2+3y+2=z^2+z=e^z=e\cdot e^y.
$$
Using the above inequality with $n=2$ we obtain
$$
e\cdot e^y> e\left(1+y+\frac{y^2}{2}\right)>y^2+3y+2,
$$
where the last inequality is equivalent to
$$
y^2+2y\left(1-\frac{1}{e-2}\right)+2>0,
$$
or, equivalently,
$$
\left(y+1-\frac{1}{e-2}\right)^2+\left(2-\left(1-\frac{1}{e-2}\right)^2\right)>0
$$
However, the first square is nonnegative and the second one is positive.
| {
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Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$
The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve for $A+B$ (what we want).
How can I do that?
| Note that
$$\sqrt[3]{40+11\sqrt{13}}=\frac{5+\sqrt{13}}{2}$$
$$\sqrt[3]{40-11\sqrt{13}}=\frac{5-\sqrt{13}}{2}$$
| {
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Differential Equation by substitution How can I solve:
$$
yy'+x=\sqrt{x^2+y^2}
$$
I tried:
$$
y\frac{dy}{dx}+x=\sqrt{x^2+y^2}$$
let $v = x^2+y^2$,
$dv=2xdx+2ydy$
then I don't know what's next
I tried isolating the dy
$$dy=\dfrac {(dv-2xdx)}{2\sqrt {(v-x^2)}}$$
and substituting back to the equation
$$\sqrt {(v-x^2)}\dfrac {(dv-2xdx)}{2\sqrt {(v-x^2)}}+x=\sqrt v$$
which is I think wrong
| Notice that the equation only requires that $y$ be differentiable for real numbers $x$ such that $y(x)\neq0.$ Therefore, $x^2+y(x)^2\gt0.$ Let $D=\{x\in\mathbb{R}:y(x)\neq0\}.$ Let $w:D\to\mathbb{R}$ with $w(x)=x^2+y(x)^2.$ Hence $y(x)y'(x)+x=\sqrt{x^2+y(x)^2}$ is equivalent to $w'(x)=2\sqrt{w(x)},$ where $w(x)\gt0.$ Thus $$\frac{w'(x)}{2w(x)}=1=(\sqrt{w})'(x),$$ which is equivalent to $$\sqrt{w(x)}=x+C,$$ which requires $x\gt-C.$ Thus $D_C=\{x\in\mathbb{R}:x\gt-C\}.$ Therefore, $w(x)=(x+C)^2,$ which means that $$x^2+y(x)^2=(x+C)^2=x^2+2Cx+C^2,$$ equivalent to $$y(x)^2=2Cx+C^2.$$ Now, $y(x)^2\gt0,$ so $2Cx+C^2=C(2x+C)\gt0,$ which means $C\gt0$ and $x\gt-\frac{C}2,$ or $C\lt0$ and $x\lt-\frac{C}2.$ The latter case is impossible, since $-C\gt0,$ and $x\gt-C,$ but in that case, $\frac{-C}2\lt-C.$ Therefore, it necessarily is the case that $C\gt0.$ Therefore, $$y(x)=\sqrt{2C}\sqrt{x+\frac{C}2},$$ implying $$y(x)y'(x)=\sqrt{2C}\sqrt{x+\frac{C}2}\frac1{2\sqrt{x+\frac{C}2}}=\sqrt{\frac{C}2}=\frac{\sqrt{C}}{\sqrt{2}},$$ and so $$y(x)y'(x)+x=x+\frac{\sqrt{C}}{\sqrt{2}}.$$ Meanwhile, $$\sqrt{x^2+y(x)^2}=x+C.$$ Therefore, $$C=\frac{\sqrt{C}}{\sqrt{2}},$$ equivalent to $$\sqrt{C}^2-\frac1{\sqrt{2}}\sqrt{C}=\sqrt{C}\left(\sqrt{C}-\frac1{\sqrt{2}}\right)=0,$$ meaning $$\sqrt{C}=\sqrt{\frac12},$$ hence $C=\frac12.$ It turns out that $C=0$ also works, but makes $y(x)=0$ and $y'(x)=0.$ As such, the two solutions are $y(x)=0$ and $y(x)=x+\frac14.$
| {
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Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2$ for $0 \le a, b, c, d \le 1$
Let $0\le a,b,c,d\le 1$, prove that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2.$$
I’ve solved it, but I want a solution without derivatives. My solution is posted below.
The part of the expression concerning $a$ is $\frac{a}{1+b}+ \frac{d}{1+a}$, the second derivative of which is $0\le\frac{2d}{1+a^3}$. This means that to reach its maximum value, there must be $a=0$ or $a=1$. Similarly, $b,c,d$ must be either $1$ or $0$. Plug in the values to find it does not exceed 2. Done.
| Note that, for all $x\in [0, 1]$,
$$1 - x/2 - \frac{1}{1 + x}
= \frac{x(1 - x)}{2(1 + x)}\ge 0.$$
We have
\begin{align*}
&\frac{a}{1 + b} + \frac{b}{1 + c}
+ \frac{c}{1 + d} + \frac{d}{1 + a}\\
\le\, & a (1 - b/2) + b(1 - c/2) + c(1 - d/2) + d(1 - a/2)\\
=\, & a + b + c + d - \frac12(ab + bc + cd + da)\\
=\,& a + b + c + d - \frac12(a + c)(b + d)\\
=\,& \frac12 (a + c) (2 - b - d) + b + d\\
\le\, &\frac12\cdot 2\cdot (2 - b - d) + b + d\\
=\,& 2
\end{align*}
where we have used $a + c \le 2$ and $b + d \le 2$.
We are done.
| {
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Proving $2\cos(k\theta) = \sum_{r=0}^{\lfloor k/2\rfloor} c(k,r) (2\cos \theta)^{k-2r}$ for $c(k,r)$ as defined
Let $k$ be a positive integer. Define for $n\ge 1, 0\leq r\leq \lfloor n/2\rfloor$, the integers $c(n,r)$ so that $c(1,0) = 1, c(2,0) = 1, c(2,1) = -2$ and for $n\ge 3$,
$$\begin{cases}
c(n,0)\phantom{/2} = 1,\\
c(n,r)\phantom{/2} = c(n-1, r) - c(n-2, r-1), &1\leq r\leq (n-1)/2,\\
c(n,n/2) = (-1)^{n/2} 2, &\text{ if $n$ is even}\end{cases}$$
Prove that $2\cos(k\theta) = \sum_{r=0}^{\lfloor k/2\rfloor} c(k,r) (2\cos \theta)^{k-2r}$ for any real number $\theta$.
I was thinking of doing a proof by induction, but it seems very complicated to do.
If $k=1, 2$, then
$$2\cos \theta = 2\cos\theta \quad\text{and}\quad 2\cos (2\theta) = (2\cos\theta)^2 + (-2)(2\cos \theta)^{2-2}$$
so the claim holds. So assume the claim holds for all $1\leq n < m$ and $m\ge 3$.
Suppose $m$ is odd. We want to show that
$$2\cos m\theta = \sum_{r=0}^{(m-1)/2} c(m,r) (2\cos \theta)^{m-2r}$$ We know by induction that
$$2\cos(m-1)\theta = \sum_{r=0}^{(m-1)/2} c(m-1, r)(2\cos \theta)^{m-1-2r}$$ and
$$2\cos(m-2)\theta = \sum_{r=0}^{(m-3)/2} c(m-2, r)(2\cos\theta)^{m-2-2r}$$
But then if I use the cosine addition formula on
$$2\cos m\theta = 2\cos ((m-1)\theta + (m-1)\theta - (m-2)\theta)$$ then I get an expression involving sines for which I don't have an explicit formula. Also, if I were to simplify the resulting expression, it would be very complicated.
Edit: Here's my attempt for even m, based on the first answer for odd m. There's likely a shorter approach though (see Mindlack's comment).
If m is even, then $2\cos ((m-1)\theta) = \sum_{r=0}^{(m-2)/2} c(m-1, r) (2\cos \theta)^{m-1-2r}$ and $2\cos ((m-2)\theta) = \sum_{r=0}^{(m-2)/2} c(m-2, r) (2\cos\theta)^{m-2-2r}$. We have
$\begin{align}
2\cos m\theta &= 2\cos \theta \cdot 2\cos ((m-1)\theta) - 2\cos((m-2)\theta)\\
&= \sum_{r=0}^{(m-2)/2} c(m-1, r) (2\cos \theta)^{m-2r} - \sum_{r=0}^{(m-2)/2}c(m-2, r)(2\cos \theta)^{(m-2) - 2r}\\
&= c(m-1, 0)(2\cos \theta)^m +\sum_{r=0}^{(m-4)/2} c(m-1, r + 1)(2\cos \theta)^{m-2 - 2r} -\sum_{r=0}^{(m-2)/2}c(m-2, r)(2\cos \theta)^{m-2-2r}\\
&= (2\cos \theta)^m - c(m-2, \frac{m-2}2) + \sum_{r=0}^{(m-4)/2}c(m, r+1)(2\cos \theta)^{m-2-2r}\\
&= \sum_{r=1}^{(m-2)/2}c(m, r)(2\cos \theta)^{m-2r} + c(m, m/2) +(2\cos \theta)^m\\
&= \sum_{r=0}^{m/2} c(m,r)(2\cos \theta)^{m-2r}\end{align}$
| Remarks: The proof for even $m$ is similar. So I omit it.
The proof for odd $m$:
By the inductive hypothesis, we have
$$2\cos (m - 1)\theta = \sum_{r=0}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 1 - 2r}$$
and
$$2\cos (m - 2)\theta = \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}.$$
Using the identity $\cos m \theta + \cos (m - 2)\theta = 2\cos \theta \cos (m-1)\theta$, we have
\begin{align*}
2\cos m x &= 2\cos\theta \cdot 2 \cos (m - 1)\theta - 2\cos (m - 2)\theta\\
&= \sum_{r=0}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 2r} - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= c(m - 1, 0)(2\cos \theta)^m + \sum_{r=1}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 2r} \\
&\qquad - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 1)/2 - 1} c(m - 1, 1 + r)(2\cos \theta)^{m - 2 - 2r} \\
&\qquad - \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 3)/2} [c(m - 1, 1 + r) - c(m - 2, r)](2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=0}^{(m - 3)/2} c(m, 1 + r)(2\cos \theta)^{m - 2 - 2r}\\
&= (2\cos \theta)^m + \sum_{r=1}^{(m - 3)/2 + 1} c(m, r)(2\cos \theta)^{m - 2r}\\
&= (2\cos \theta)^m - c(m, 0)(2\cos \theta)^m + \sum_{r=0}^{(m - 1)/2} c(m, r)(2\cos \theta)^{m - 2r}\\
&= \sum_{r=0}^{(m - 1)/2} c(m, r)(2\cos \theta)^{m - 2r}
\end{align*}
where we have used $c(m - 1, 0) = c(m, 0) = 1$.
The desired result follows.
| {
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Generalization of the law of tangents for a cyclic quadrilateral The law of tangents is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides.
Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle, and $\alpha$, $\beta$ and $\gamma$ be the angles opposite those three respective sides. The law of tangents states that
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{a-b}{a+b}\tag{1}$$
The law of tangent can be used in any case where two sides and the included angle, or two angles and a side, are known.
Although Viète gave us the modern version of the law of tangents, it was Fincke who stated the law of tangents for the first time and also demonstrated its application by solving a triangle when two sides and the included angle are given (see Wu - The Story of Mollweide and Some Trigonometric Identities)
A proof of the law of tangent is provided by Wikipedia (see here).
Generalization. Let $a$, $b$, $c$ and $d$ be the sides of a cyclic convex quadrilateral. Let $\angle{DAB}=\alpha$ and $\angle{ABC}=\beta$, then the following identity holds
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\tag{2}$$
Proof. Using the sum-to-product formulas we can rewrite the left-hand side of $(2)$ as follows
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{\sin\frac12(\alpha-\beta)\cos\frac12(\alpha+\beta)}{\cos\frac12(\alpha-\beta)\sin\frac12(\alpha+\beta)}=\frac{\sin{\alpha}-\sin{\beta}}{\sin{\alpha}+\sin{\beta}}.$$
The area of a cyclic quadrilateral can be expressed as $\Delta=\frac12(ad+bc)\sin{\alpha}$ (see $(12)$ at Killing three birds with one stone) and similarly for the other angles. Then substituting, simplifying and factorizing we have
$$\begin{align*}\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}&=\frac{\frac{2\Delta}{ad+bc}-\frac{2\Delta}{ab+cd}}{\frac{2\Delta}{ad+bc}+\frac{2\Delta}{ab+cd}}=\frac{ab-ad+cd-bc}{ab+ad+cd+bc}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\end{align*}.$$
$\square$
The formula $(2)$ reduces to the law of tangent for a triangle when $c=0$.
A related result can be found at A generalization of Mollweide's formula (rather Newton's).
Crossposted at MO.
Question: Is this generalization known?
| The Usual Law of Tangents
Applying the formulae for the Sum of Sines and the Sum of Cosines, we get
$$
\begin{align}
\frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}
&=\frac{2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}{2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}\\[6pt]
&=\tan\left(\tfrac{A+B}2\right)\tag1
\end{align}
$$
Substituting $B\mapsto-B$ in $(1)$ and then dividing by $(1)$ gives
$$
\frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag2
$$
If we let $a$ and $b$ be the sides opposite angles $A$ and $B$ respectively, the Law of Sines leads to
$$
\frac{a-b}{a+b}=\frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}\tag3
$$
Equating $(2)$ and $(3)$, we get the Law of Tangents
$$
\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag4
$$
where $a$ and $b$ are the sides of a triangle opposite angles $A$ and $B$ respectively.
Inscribed Generalization
The power of the point $S$ is equal to both
$\overline{SA}\,\overline{SD}=\overline{SB}\,\overline{SC}$; therefore,
$$
\frac{\overline{SA}}{\overline{SB}}=\frac{\overline{SC}}{\overline{SD}}\tag5
$$
and since $\angle ASB=\angle CSD$, SAS says that
$$
\triangle ASB\simeq\triangle CSD\tag6
$$
Therefore,
$$
\frac{\overline{SA}}{\overline{SC}}=\frac{\overline{SB}}{\overline{SD}}=\frac ac\tag7
$$
Thus, we have
$$
\begin{align}
b&=\overline{SB}-\overline{SC}=\overline{SB}-\overline{SA}\frac ca\tag{8a}\\[6pt]
d&=\overline{SA}-\overline{SD}=\overline{SA}-\overline{SB}\frac ca\tag{8b}
\end{align}
$$
From $(8)$, we can solve
$$
\begin{align}
\overline{SA}&=a\frac{ad+bc}{a^2-c^2}\tag{9a}\\[3pt]
\overline{SB}&=a\frac{ab+cd}{a^2-c^2}\tag{9b}
\end{align}
$$
Now we are ready to apply the usual Law of Tangents from $(4)$:
$$
\begin{align}
\frac{\tan\left(\frac{\alpha-\beta}2\right)}{\tan\left(\frac{\alpha+\beta}2\right)}
&=\frac{\overline{SB}-\overline{SA}}{\overline{SB}+\overline{SA}}\tag{10a}\\
&=\frac{(ab+cd)-(ad+bc)}{(ab+cd)+(ad+bc)}\tag{10b}\\[6pt]
&=\frac{(a-c)(b-d)}{(a+c)(b+d)}\tag{10c}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work because I lose equality case.
| There exists a very beautiful graphic solution.
We have $ |z| \le 1$ the unit disk bounded by the green circle, with centre $C_1$.
Let's assume: $|z-1| = r$, drawn as the purple circle $(C_2,r)$.
Note that we must have $0 \le r \le 2 $.
Now our inequality is $|z^2-1||z-1|^2 \le 3\sqrt{3} $ which is equivalent to: $|z+1||z-1|^3 \le 3\sqrt{3}$.
Applying the above conditions we are left to prove $|z+1| \le \dfrac{3\sqrt{3}}{r^3} $.
Geometrically, $|z+1|$ is the distance between $z$ and the point $C_3=-1$. We must now show this distance will always be less than or equal to $\dfrac{3\sqrt{3}}{r^3} .$
We can now see that the allowed $z$ under our conditions lie on the arc $AB$. The points $A$ and $B$ both have maximum distance to $C_3$.
An important thing to note is $\angle C_2AC_3 = \frac{\pi}{2} \implies {AC_2}^2 + {AC_3}^2 = {C_3C_2}^2$.
As, $AC_2 = r$ and ${C_3C_2} = 2,$ we get: $ AC_3 =\vert z+1\vert = \sqrt{4-r^2}$.
Hence we should prove that:
$\sqrt{4-r^2} \le \dfrac{3\sqrt{3}}{r^3}$ which is equivalent to $\sqrt{4-r^2} \le 3\sqrt{3}$, for $r > 0 $.
(The case $r=0$ is trivial.)
Using the derivative of the function $\displaystyle f(r) = r^3 \sqrt{4-r^2}$, we can see $f(r)$ attains a global maximum at
$$\displaystyle r =\sqrt{3} \implies f(r) \le 3\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.
| The line $y = m x + c$ has the parametric equation
$ (x, y) = (0, c) + t (\cos \theta, \sin \theta) $
where $\theta$ is the angle between the line and the positive $x$-axis.
and $m = \tan \theta $
Intersecting this line with the parabola $y = x^2$ yields
$ c + t \sin \theta = t^2 \cos^2 \theta $
which has two solutions
$t = \dfrac{1}{2 \cos^2 \theta } ( \sin \theta \pm \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} ) $
The absolute values of $t$ are (assuming $c \ge 0 $ )
$ | t_1 | = \dfrac{1}{2 \cos^2 \theta } ( \sin \theta + \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} ) $
$ | t_2 | = \dfrac{1}{2 \cos^2 \theta } ( \sqrt{ \sin^2 \theta + 4 c \cos^2 \theta} - \sin \theta ) $
The difference between these two absolute values is equal to $1$, hence,
$ \dfrac{ |\sin \theta | }{ \cos^2 \theta } = 1 $
So that
$ \cos^2 \theta = 1 - \sin^2 \theta = | \sin \theta | $
which becomes
$ 1 - | \sin \theta |^2 = | \sin \theta | $
Therefore,
$ | \sin \theta | = \frac{1}{2} ( -1 + \sqrt{5} ) $
Therefore,
$ \theta = \pm \sin^{-1} \left( \dfrac{ -1 + \sqrt{5}}{2} \right) $
Now,
$\cos^2 \theta = 1 - \sin^2 \theta = | \sin \theta | = \dfrac{-1 + \sqrt{5}}{2}$
Hence,
$\begin{equation} \begin{split}
m^2 &= \tan^2 \theta = \sec^2 \theta - 1 \\
&= \dfrac{1}{\cos^2\theta} - 1 \\
&= \dfrac{2}{-1 + \sqrt{5}} - 1 = \dfrac{\sqrt{5} + 1 }{2} - 1 \\
&= \boxed{\dfrac{\sqrt{5} - 1}{2}} \end{split}\end{equation}$
But, what if $ c \lt 0 $ ?
Clearly, (by graphing the situation), both $t_1$ and $t_2$ will be positive, or both negative. Their positive difference is
$ \Delta t = 1 = \dfrac{\sqrt{ \sin^2 \theta + 4 c \cos^2 \theta }} { \cos^2 \theta } $
Hence, we must have
$ \sin^2 \theta + 4 c \cos^2 \theta = \cos^4 \theta $
or
$ \cos^4 \theta + (1 - 4 c) \cos^2 \theta - 1 = 0 $
Solving for $\cos^2 \theta$ from this equation
$ \cos^2 \theta = \dfrac{1}{2} ( 4 c - 1 + \sqrt{ (4 c - 1)^2 + 4 } ) $
Hence,
$ \begin{equation} \begin{split}
m^2 &= \tan^2 \theta = \dfrac{1}{\cos^2 \theta} - 1 \\
&= \dfrac{2}{ 4 c - 1 + \sqrt{ (4 c - 1)^2 + 4 }} - 1 \\
\end{split}\end{equation}$
i.e.
$m^2 = \dfrac{ 3 - 4 c - \sqrt{ (4 c - 1)^2 + 4 }}{ 4 c - 1 + \sqrt{(4c - 1)^2 + 4 }} $
simplifying further, by eliminating the surd in the denominator,
$ m^2 = \boxed{ \dfrac{ -1 - 4 c + \sqrt{ (4 c - 1)^2 + 4 } }{2} }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equation of the plane that this $x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}$ curve lies
Prove that all points of the given curve lie in one plane, and find
the equation of that plane:
$$x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}.$$
If the given curve lies in one plane, then
$$a\left(\frac{1+t}{1-t}\right)+b\left(\frac{1}{1-t^2}\right)+c\left(\frac{1}{1+t}\right)+d=0.$$
Solving this I get $2a=c,a=d,a=\frac{-b}{4}.$
How from this find equation of plane? Or maybe I did something wrong?
When putting values back into equation of plane I get
$$ax-4ay+2az+a=0.$$ Now, problem is I can't cancel $a$ here because first of all, I need to prove that such plane exists.
| We have the following parameteric curve
$x(t) = \dfrac{1 + t}{1 - t} , y(t) = \dfrac{1}{1 - t^2} , z(t) = \dfrac{1}{1 + t} $
As the OP pointed out, we can simply assume that the curve lies on the plane
$$ a x + b y + c z + d = 0 $$
And then try and find $a,b,c,d$. Using this method, and multiplying through by $(1 - t^2) $ results in
$ a (1 + t)^2 + b + c (1 - t) + d (1 - t^2) = 0 $
Collecting terms containing $t^2 , t $ and the constant terms we get
$ t^2 ( a - d ) + t ( 2 a - c ) + ( a + b + c + d) = 0 $
Taking $d = 1 $ , then
$ a = 1, c = 2 , b = -4 $
Thus this curve does indeed lie on the plane
$ x - 4 y + 2 z + 1 = 0 $
which has been pointed out by @RKK
A more involved method that is not necessarily better, is to differentiate $(x(t), y(t), z(t))$ to obtain
$T(t) = (x'(t), y'(t), z'(t) ) = \left( \dfrac{ 2 }{(1 - t)^2 }, \dfrac{2 t}{(1 - t^2)^2} , -\dfrac{1}{(1 + t)^2 } \right) $
And then differentiate $T(t)$ to obtain
$T'(t) = ( \dfrac{ 4 }{ (1 - t)^3 } , \dfrac{ 2 + 6 t^2 }{ (1 - t^2)^3 } , \dfrac{ 2 }{(1 + t)^3 }) $
Now, we find the cross product $T(t) \times T'(t) $
$ T(t) \times T'(t) = \begin{vmatrix} \mathbf{i} && \mathbf{j} && \mathbf{k} \\
\dfrac{ 2 }{(1 - t)^2 } && \dfrac{2 t}{(1 - t^2)^2} && -\dfrac{1}{(1 + t)^2 } \\
\dfrac{ 4 }{ (1 - t)^3 } && \dfrac{ 2 + 6 t^2 }{ (1 - t^2)^3 } && \dfrac{ 2 }{(1 + t)^3 } \end{vmatrix} $
$ = \mathbf{i} \left( \dfrac{ 4 t }{ (1 - t^2)^2 (1 + t)^3 } + \dfrac{ 2 + 6 t^2 }{ (1 + t)^2 (1 - t^2)^3 } \right) \\
+ \mathbf{j} \left( - \dfrac{4}{ (1 + t)^2 (1 - t)^3 } - \dfrac{4}{(1 - t)^2 (1 + t)^3 } \right)
+ \mathbf{k} \left( \dfrac{ 2 (2 + 6 t^2)}{(1 - t)^2 (1 - t^2)^3 } - \dfrac{ 8t }{ (1 - t)^3 (1 - t^2)^2} \right) $
And this simplifies to,
$ =\dfrac{2}{(1 - t^2)^3} \left( \mathbf{i} - 4 \mathbf{j} + 2 \mathbf{k} \right) $
Hence the vector $(1, -4, 2) $ is always normal to the curve, which means the curve lies in the plane $ x - 4 y + 2 z + d = 0 $
To find $d$ , substitute $t = 0$, you get the point $(1, 1, 1) $ on the plane, hence
$ d = - (1 - 4 + 2) = 1 $
Finally the equation of the plane is $ x - 4 y + 2 z + 1 = 0 $
But the method of the OP and @RKK is much simpler than the second method that involved the tangent vector and its derivative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412630",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Common region between an icosahedron and a dodecahedron This is admittedly one of the hard problems I've come across. It involves the common region (intersection) between two dual platonic solids: icosahedron, and dodecahedron.
The question is as follows:
A dodecahedron and an icosahedron intersect as shown below. Their edge lengths are such that the intersection (i.e. taking the points that are inside the dodecahedron and inside the icosahedron) results in the polyhedron shown below (second image), where all faces are regular pentagons and hexagons of side length $1$. Find the volume of this polyhedron which is almost like a soccer ball.
| As @David K observed, by considering the edge of the icosahedron, it is segmented into 3 equal parts, each being of length $1$. Hence the edge length of the icosahedron is $3$.
The second concern is the edge length of the dodecahedron. Consider a face of the dodecahedron. Suppose the edge length is $a$, and the small pentagon contained within it is of side length $1$. Therefore the distance $w$ between a vertex of the small pentagon and the edge of the dodecahedron is
$w = (\dfrac{a}{2}) \cot(36^\circ) -(\dfrac{1}{2}) \csc(36^\circ) \hspace{25pt}(1)$
and we have
$ 2 w \sin\left( \frac{1}{2} \cos^{-1}\left(\dfrac{1}{\sqrt{5}} \right) \right) = 1\hspace{25pt}(2)$
Using $(1), (2)$:
$ c_1 a + c_2 = 1 $
where
$ c_1 = 1.17082039 $
$ c_2 = - 1.44721359$
Therefore the edge length of the dodecahedron is
$ a = \dfrac{ 1 - c_2 }{c_1} = 2.09016994$
Finally we note from the figure, that for the intersection solid, there are $12$ unit pentagons and $20 $ unit hexagons. Their areas are
$A_{Pent} = 5 (\frac{1}{2}) ( (\frac{1}{2} \csc(36^\circ) )^2 \sin(72^\circ) = 1.72047740 $
$A_{Hex} = 6 ( \frac{1}{2}) (1)^2 \sin(60^\circ) ) = 2.59807621$
The distance (pyramid height) between the center of the solid and the hexagons is $3$ time the inradius of the unit icosahedron, and this comes to
$ h_{Hex} = 3 \left( \dfrac{\sqrt{3}}{12}(3+\sqrt{5}) ) \right) =2.26728394$
and the distance (pyramid height) between the center of the solid and the pentagons is
$a$ times the inradius of the unit dodecahedron,
$h_{Pent} =(2.09016994) \left( \frac{1}{2} \sqrt{ \dfrac{5}{2} + \dfrac{11}{10} \sqrt{5}} \right) = 2.32743843 $
Putting it all together, the volume is
$ \text{V} = \left(\dfrac{1}{3}\right) ( 12 A_{Pent} h_{Pent} + 20 A_{Hex} h_{Hex} ) = \boxed{55.2877} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the right hand side into $ab(a-b)$ and equated with the expansion earlier, giving the following inequality: $a^2+ab+b^2 \geq ab$. However by AM-GM, it follows that $a^2+b^2 \geq 2ab$ which my statement above does not follow. Help would be much appreciated.
| Alternative approach:
$0 < (a - b)^3 = (a^3 - b^3) - (3)(a^2b - ab^2) \implies$
$(a^3 - b^3) > 3(a^2b - ab^2).$
Further, since $a > b,$
you know that $a^2b > ab^2 \implies (a^2b - ab^2) > 0.$
Therefore $3(a^2b - ab^2) > (a^2b - ab^2)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generate an explicit formula for: $a_n=a_{n-1}+n^2-5n+7, a_3=2$ So I'm currently solving this math homework problem and I've set up a recursive formula:
$a_n=a_{n-1}+n^2-5n+7; a_3=2$
Right now, I'm having trouble as to where to start. I was listing out the cases but they didn't get me very far: $a_3=2,a_4=5,a_5=12,a_6=25, \dots$
Thanks!
| We may observe that:
$$
a_n-a_{n-1}=n^2-5n+7.
$$
Then we could have:
$$
\begin{align}
&a_n-a_3\\
=&\sum_{k=4}^n{a_k-a_{k-1}}\\
=&\sum_{k=4}^n{k^2-5k+7}\\
=&\left[ \left( \sum_{k=1}^n{k^2} \right) -1^2-2^2-3^2 \right] -\frac{5\left( 4+n \right) \left( n-3 \right)}{2}+7\left( n-3 \right)\\
=&\left[ \frac{n\left( n+1 \right) \left( 2n+1 \right)}{6}-14 \right] -\left( \frac{5}{2}n^2+\frac{5}{2}n-30 \right) +\left( 7n-21 \right)\\
=&\frac{1}{3}n^3-2n^2+\frac{14}{3}n-5,
\end{align}
$$
(note that the above summation has $(n-3)$ items).
So the explicit formula could be obtained by $a_n=(a_n-a_3)+a_3$:
$$
a_n=\frac{1}{3}n^3-2n^2+\frac{14}{3}n-3.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to expand the numerator (Harmonic numbers) in terms of integral, to get:
$\displaystyle \tag*{} \sum \limits_{n=1}^{\infty} \frac{(-1)^n}{n2^n\binom{2n}{n}} \int _{0}^{1} \frac{x^n - x^{2n}}{1-x} \ \mathrm dx$
And I found that with the help of series expansion of $\sin^{-1}(x)$ and subsituting $x = i \sqrt{x} /8 $ where $i^2=1$
$\displaystyle \tag*{} -2(\sinh^{-1} (\sqrt{x}/8))^2 = \sum \limits_{n=1}^{\infty} \frac{(-1)^nx^n}{n^22^n \binom{2n}{n}} $
But this has $n^2$ in the denominator, which makes it complicated. EDIT: we can eliminate $n^2$ by differentiating and multiplying by $x$ as mentioned in the comments. But now, how can we solve our sum since $H_{2n}-H_n$ is numerator?
And I have the general formula for generating sum:
$ \displaystyle \tag*{} \sum \limits _{n=1}^{\infty} \frac{x^n}{n^y \binom{2n}{n}}$
And this doesn't have $n$ in the denominator and also it has closed-form $\forall \ y \geq 2$
Maybe if there is a way of expressing the denominator in the form integral, the sum can be changed in evaluating the double integral. I think there are other easy ways (such as using Hypergeometric functions)? Any help would be appreciated.
EDIT 2:
From the help of comments and a quora user,
$\DeclareMathOperator{\arcsinh}{arcsinh}$
By @Bertrand87, we have:
$\displaystyle \tag{1} H_{k} - H_{2k} + \ln (2) = \int _{0}^{1} \frac{x^{2k}}{1+x} \ \mathrm dx$
To make use of this, we express our sum as follows:
$\displaystyle \tag*{} S = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}} + \sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}}$
We know
$\displaystyle \tag*{} 2\arcsin^2(x) = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k}}{k^2 \binom{2k}{k}}$
We differentiate both sides w.r.t $x$ both sides,
$\displaystyle \tag*{} \frac{2 \arcsin(x)}{\sqrt{1-x^2}} = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k-1}}{k \binom{2k}{k}}$
Now, we multiply both sides by $(2x)$ and define $x:= ix/ \sqrt{8}$ to get:
$\displaystyle \tag{2} \frac{-2x \arcsinh (x/ \sqrt {8})}{\sqrt{8}\sqrt{1+x^2/8}} = \sum \limits_{k=1}^{\infty} \frac{(-1)^k x^{2k}}{k2^k \binom{2k}{k}}$
We now multiply both sides by $-1/(1+x)$ and integrate from $0$ to $1$ and arrive at:
$\displaystyle \tag*{} \frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}}$
Similarly, from $(2)$ if we let $x=1$ and multiply both sides by $\ln 2$, it yields:
$\displaystyle \tag*{} \frac{-2 \arcsinh (1/ \sqrt{8}) \ln 2}{ \sqrt{8} \sqrt{1 + 1/8}} =\sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}} \approx -0.1601$
Now, our only problem is to evaluate the integral:
$\displaystyle \tag*{} \boxed{\frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx} $
Can anyone help me with this integral?
| Simplify the integral
\begin{align}
&\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx\\
=& \>\sqrt2 \left(\sinh^{-1}\frac x{\sqrt8}\right)^2\bigg|_0^1
- \int_0^1\frac{\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,\overset{x=\sqrt2(y-\frac1y)}{dx}\\
=&\>\frac1{2\sqrt2}\ln^22-2\sqrt2 \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\tag1
\end{align}
With $\int_0^1 \frac{\ln t}{1+t}dt=-\frac{\pi^2}{12}$ and $\int_0^1 \frac{\ln t}{1-t}dt=-\frac{\pi^2}{6}$
\begin{align}\int_0^{1/2} \frac{\ln t}{1-t}dt\overset{ibp}=&
-\ln^22+\int_0^{1/2} \frac{\ln (1-t)}{t}\overset{t\to1-t}{dt}\\
=&\>\frac12\left(-\ln^22+\int_0^{1} \frac{\ln t}{1-t} dt\right)
=-\frac12\ln^22 -\frac{\pi^2}{12}
\end{align}
\begin{align}
& \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\\
=& \>\frac13\bigg(\int_1^{\sqrt2}\frac{\ln y}{y-\frac1{\sqrt2}}\overset{y=\frac1{\sqrt2t}}{dy}- \int_1^{\sqrt2} \frac{\ln y}{y+{\sqrt2}} \overset{y={\sqrt2t}}{dy}\bigg)\\
=&\>\frac13\bigg( -\frac38\ln^22-\int_0^1 \frac{\ln t}{1+t}dt+\int_0^{1/2}\frac{\ln t}{1-t}dt-\int_0^{1/\sqrt2} \frac{2t\ln t}{1-t^2}\overset{t^2\to t}{dt} \bigg)\\
=&\>-\frac18\ln^22+\frac{\pi^2}{36}
+\frac16 \int_0^{1/2}\frac{\ln t}{1-t}dt
= \frac{\pi^2}{72}-\frac5{24}\ln^22
\end{align}
Plug into (1) to obtain
$$\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx=\frac{2\sqrt2}{3}\left(\ln^22-\frac{\pi^2}{24}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Maximize $z$ over $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$ Suppose that $x$, $y$, and $z$ are real numbers such that $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$, but this doesn't say anything. I also tried some different ways to apply Cauchy, but they all didn't do much.
Thanks in advance!!
| Here is a solution avoiding calculus.
Use cylindrical coordinates.
$x = r \cos\theta, y = r \sin\theta, z = z, r \geq 0$
From $x^2 + y^2 + z^2 = 6$, we get
$$z^2 = 6 - r^2 \tag1$$
From $x + y + z = 3$, we get
$ \displaystyle r = \frac {3-z}{\cos\theta + \sin\theta} \tag2$
From $(1), z \leq \sqrt6~$ and so we know $(3 - z)$ is positive.
As we know, $(1)$ is a sphere centered at the origin. The lower the radius of the circle parallel to xy-plane on the sphere, the higher the value of $|z|$.
From $(2)$, we see that at any given $z$, of all the circles parallel to xy-plane that intersect the plane $x + y + z = 3$, the one with minimal $r$ is one when $~(\cos \theta + \sin \theta)~$ is maximum. We know that maximum value of $~(\cos\theta + \sin\theta)~$ is $~\sqrt2~$. That leads to $~ \displaystyle r = \frac{3-z}{\sqrt2}~$. Now plugging into $(1)$ and solving should give us the maximum and minimum value of $z$.
$ \displaystyle z^2 = 6 - \frac{(3-z)^2}{2}$
Simplifying, $z^2 - 2 z - 1 = 0$
Solving, $z = 1 + \sqrt 2~$ is the maximum and $z = 1 - \sqrt 2~$ is the minimum.
To find maximum of $(\cos\theta + \sin\theta)$, we maximize $(\cos\theta + \sin\theta)^2$.
$(\cos\theta + \sin\theta)^2 = 1 + \sin2\theta \leq 2 \implies \cos\theta + \sin\theta \leq \sqrt2$. The maximum occurs at $\theta = \pi/4$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$, so $n^2-4 = k(2n+1)$, for some $k \in \Bbb Z$. But, I did not be able to find $n$ from here.
Any ideas? Thanks in advanced.
| First way: Using the Extended Euclidean Algorithm in $\Bbb Q[x]$.
Notice that
\begin{equation*}
15 = 4(3n^2+4n+5) - (6n+5)(2n+1).
\end{equation*}
Hence, if $2n+1$ divides $3n^2+4n+5$, then it also divides $15$.
Thus, $2n+1 \in \{\pm 1, \pm 3, \pm 5, \pm 15\}$, i.e.,
\begin{equation*}
n \in \{-8,-3,-2,-1,0,1,2,7\}.
\end{equation*}
Second way: Just using the Elementary number theory.
From your approach, we have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
Now, let $k=2n+1$, then $2n \equiv -1 \pmod{k}. \ldots (1)$
We want $(2n+1) \mid (n^2-4)$, i.e., $n^2-4 \equiv 0 \pmod{k}. \ldots (2)$
Multiplying both side in $(2)$ by $4$, we have
\begin{align*}
4n^2 - 16 \equiv 0 \pmod{k} &\Leftrightarrow (2n)^2 - 16 \equiv 0 \pmod{k} \\
&\Leftrightarrow (-1)^2 - 16 \equiv 0 \pmod{k} \qquad \qquad (\text{by $(1)$}) \\
&\Leftrightarrow -15 \equiv 0 \pmod{k}.
\end{align*}
Hence, $2n+1 \mid 15$. Thus, $2n+1 \in \{\pm 1, \pm 3, \pm 5, \pm 15\}$, i.e.,
\begin{equation*}
n \in \{-8,-3,-2,-1,0,1,2,7\}.
\end{equation*}
Now, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum possible value of $x+y-2xy$ over reals. Find the minimum possible value of $x+y-2xy$, if $x+y\ge 3xy,~ 2(x+y)\ge 1+3xy~$ and $~0\le x\le y\le 1.$
My guess is $x=y=\frac{2}{3},$ gives the minimum value of $\frac{4}{9}$.
I thought this manipulation might be useful,
$$x+y-2xy=\frac{3(x+y)-6xy}{3}\ge \frac{1}{3}$$
But, $x+y-2xy=\frac{1}{3}$ would imply $x+y=3xy$ and $2(x+y)=1+3xy\implies x+y=1$ and $xy=\frac{1}{3}.$ This system does not have real solutions and hence $\frac{1}{3}$ is not an attainable value.
Another idea was to substitute $0\le x+y=s\le 2~$ and $~0\le xy=p\le 1$, transforming the problem into finding $\min(s-2p)$ given the conditions,
$$s\ge 3p, ~ 2s\ge 1+3p, ~ s^2\ge 4p.$$
| I think that your second idea works.
Let $k:=s-2p$, and $f : p=\frac 14s^2, g : p=\frac 23s-\frac 13$ and $h : p=\frac s3$.
Then, $f$ and $g$ intersect at $A(2/3,1/9)$ and $C(2,1)$, and $f$ and $h$ intersect at $O(0,0)$ and $B(4/3,4/9)$.
So, one can see that
*
*for $s\lt\frac 23$, $p\leqslant \frac 23s-\frac 13$
*for $\frac 23\leqslant s\leqslant \frac 43$, $p\leqslant \frac 14s^2$
*for $\frac 43\lt s$, $p\leqslant \frac s3$
Since the slope of the line $k=s-2p$, i.e. $p=\frac 12s-\frac k2$ is $\frac 12$, the minimum possible value of $k$, i.e. the largest $p$-intercept is attained when the line passes through $A$ and $B$.
So, the minimum possible value of $k$ is $k=s-2p=\frac 23-2\times \frac 19=\color{red}{\frac 49}$ which is attained when $(s,p)=(2/3,1/9),(4/3,4/9)$, i.e. when $(x,y)=(1/3,1/3),(2/3,2/3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove that the sum of the elements in two subsets is the same
Eight consecutive positive integers are partitioned into two subsets such that the sum of the squares in each subset is the same. Prove that the sum of the elements in each subset is also the same, assuming that the smallest element is at least 56.
The method below is mostly a brute force approach.
Is there a better method than brute force?
For a set $S, $ let $f(S)$ denote the sum of the squares of its elements and let $g(S)$ denote the sum of the elements of S.
Let the integers be $x + 1,\cdots, x + 8$. The sum of the squares in each subset must be $\frac{1}2 \sum_{i=1}^8 (x+i)^2 = \frac{1}2 (8x^2 + 72x + 204) = 4x^2 + 36x+102.$
First we need to show that the partitions must in fact have the same number of elements. If one partition has more than $4$ elements, then its sum of squares will be at least $(x+1)^2 + (x+2)^2 + (x+3)^2 + (x+4)^2 + (x+5)^2 = 5x^2 + 30x + 55,$ which is too large (for $x\ge 11,$ since we would have $x^2 - 6x - 47 > 0$). So no subset can have more than 4 elements, implying than both subsets must have exactly 4 elements.
One of the subsets, say $S_1$, must contain $x+8$, and so the sum of squares of the remaining three elements in this case must be $3x^2 + 20 x +38.$ Let $S_2$ be the other subset. Suppose this subset also contains $x+7$. Then the remaining two elements have a sum of squares equal to $2x^2 + 6x -11.$ The minimum possible sum of squares of two elements is $(x+1)^2 + (x+2)^2 = 2x^2 + 6x + 5$, so we cannot have $x+7$ in $S_1$. If $x+6$ is in $S_1$, then $f(S_1\backslash \{x+8, x+6\}) = 2x^2 +8x+2$. The only two possible remaining elements are $x+1$ and $x+2$, as $(x+1)^2 + (x+3)^2 > 2x^2 + 8x+2$, and is the second smallest sum of squares of elements. But then $f(S_1) < 2X^2 + 8x+2$, a contradiction.
So suppose $x+5\in S_1$. $f(S_1\backslash \{x+8,x+5\} ) = 2x^2 + 10x + 13,$ and in this case one can verify that $S_1 = \{x+8, x+5, x+2, x+3\}$, since all other choices of two remaining elements give either a smaller or larger sum of squares than $2x^2 + 10x+13$. In this case, $g(S_1) = 4x+18$, so $g(S_1) = g(S_2)$.
We know $x+7$ and $x+6$ are in $S_2 $. So we need to find the remaining two elements of $S_2$. If $x+5$ is in $S_2$, then the square of the remaining element in $S_2$ must be $ 4x^2 + 36x + 102 - ((x+7)^2 + (x+6)^2 + (x+5)^2) = x^2 - 8,$ which is clearly impossible.
So the only possibility is $S_1 = \{x+8, x+5, x+2, x+3\}$.
| I have a simple proof if the smallest number, name it $a$, is greater than $56$. We have $$(a+a_1)^2+...+(a+a_4)^2=(a+a_5)^2+...+(a+a_8)^2,$$ where $a_1,...a_8\in \{0,...,7\}$. Hence $$2a(S_1-S_2)=a_5^2+...a_8^2-a_4^2-...-a_1^2,$$ where $S_1=a_1+...+a_4, S_2=a_5+...+a_8$. So $$|S_1-S_2|< (7^2+6^2+5^2+4^2-3^2-2^2-1)/112=1,$$ thus $S_1=S_2$, the two sums are equal, and the two sums of the $a+a_i$ are equal as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $x \in \mathbb R$ and $a \neq 0$. Throughout this derivation, I will use the definition: $\sqrt{x^2}=|x|$.
Here is my derivation, and I have placed a $\color{red}{\dagger}$ next to the part that I would like some clarificaiton about:
$ax^2+bx+c =0 \iff x^2+\frac{bx}{a}+\frac{c}{a}=0 \iff (x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})=0 \iff (x+\frac{b}{2a})^2+(\frac{4ac-b^2}{4a^2})=0 $
Bringing the right summand over to the right side of the equation:
$(x+\frac{b}{2a})^2=(\frac{b^2-4ac}{4a^2}) \iff \left| x+\frac{b}{2a}\right|=\sqrt{b^2-4ac}\cdot\sqrt{\frac{1}{4a^2}} \iff \left| x+\frac{b}{2a}\right|=\left|\frac{1}{2a} \right| \cdot \sqrt{b^2-4ac} \quad \quad \color{red}{\dagger}$
My confusion stems from how the final expression above is equivalent to the syntax "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" .
For $\color{red}{\dagger}$, we have 4 total cases:
*
*$ a \lt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \lt 0$ and $x+\frac{b}{a} \geq 0$
*$ a \gt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \gt 0$ and $x+\frac{b}{a} \geq 0$
Case 1: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
Case 2: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 3: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 4: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
From the four scenarios, is the way we get to "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" simply by noting that for a fixed $a$, we have $x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ or } x=\frac{-b-\sqrt{b^2-4ac}}{2a}$? ...where the '$\text{ or }$' here is denoting the logical or.
At this point, we define "$x=\pm \alpha$" as meaning $x = \alpha \text { or } x=-\alpha$...therefore meaning that $x=\beta\pm \alpha$ is equivalent to $x=\beta+\alpha \text{ or } x=\beta - \alpha$.
Is that the proper understanding?
| Your analysis is spot on, up until the point where you concluded that
$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \tag1 $$
I didn't bother reading after that, because there is no point. All that is necessary from the point of (1) above is to consider that any equation of the form
$$[f(x)]^2 = C$$
may be dissected as follows:
*
*If $C < 0$ then there can't be any real roots. So, without loss of generality (assuming that you are only concerned with real roots), assume that $C \geq 0.$
*Then, you must have that $f(x) = \pm \sqrt{C}$.
Bang, that makes it: game over right there.
So, no further analysis is needed.
This begs the question: what happens if $C < 0$.
Then, if you let $D = \sqrt{-C}$, then the equation becomes
$$[f(x)] = \pm (iD) ~: ~i = \sqrt{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425762",
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"source": "stackexchange",
"question_score": "2",
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How to prove $ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$ I'm trying to prove that: $$ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$$ What have I done so far is, I'm getting an x from left hand side of equation:
$$
x \in (3\Bbb N + 1) \cap (5\Bbb N + 3)
$$
Then I'm simplifying it, like this:
$$
x \in (3\Bbb N + 1) \land x \in (5\Bbb N + 3)
$$
$3\Bbb N + 1$ is basicly $1 \pmod{3}$, and $(5\Bbb N + 3)$ is basicly $3 \pmod{5}$, thus:
$$
x \equiv 1 \pmod{3} \; and \; x\equiv 3 \pmod{5}
$$
Use CRT (i.e. https://en.wikipedia.org/wiki/Chinese_remainder_theorem) to conclude result for left hand side:
$$
a_1 = 1, \; m_1 = 5, \; y_1 \equiv 2 \pmod{3}
$$
After that, Use CRT to conclude result for right hand side:
$$
a_2 = 3, \; m_2 = 3, \; y_2 = 2 \pmod{5}
$$
Then plug in:
$$
\begin{split}
x &\equiv a_1m_1y_1 + a_2m_2y_2 \pmod{M} \\
& \equiv 1\cdot 5\cdot 2 + 3\cdot 3 \cdot 2 \pmod{3\cdot 5} \\
& \equiv 28 \pmod{15}
\equiv 13 \pmod{15}
\end{split}
$$
then find for the right side;
$$
x \in (15\Bbb N + 13) \implies x = 13 \pmod{15}
$$
Did I prove it right? Is there an easier way to do this?
| Let $n=3x+1$ and assume $x=5y+z$, then $n=15y+3z+1$. If $n\equiv 3 \pmod{5}$, then $z\equiv -1 \pmod5$. Thus we can take $z=-1$ and obtain the result you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426113",
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"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in the numerator and no square in the denominator. I am unsuccesful even in using $\arcsin^2x$ expansion.
However, we know a well known result:
$\displaystyle \tag*{} \sum \limits_{k=1}^{\infty} \frac{\binom{2n}{n}}{k^24^k} = \frac{\pi^2}{6} - 2\ln^2(2)$
So this somehows tells (?) maybe we can decompose our sum into two parts with one being Basel sum. Any help would be appreciated, thanks.
| Using @RobPratt's elegant solution
$$S_1(x)=\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 }x^k=2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;4 x\right)$$ Simplifying
$$S_1(x)=2 x \left(\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4
x}\right)}{x}-\frac{\left(\log \left(1+\sqrt{1-4 x}\right)-\log (2)\right)^2}{2
x}\right)$$
$$S_1(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-i \sqrt{15}\right)\right)+\left(\tan
^{-1}\left(\sqrt{15}\right)-i \log (2)\right)^2$$
$$\color{blue}{S_1(4)=\frac {\pi^2}6-2 \log ^2(2)}$$
$$S_2(x)=\sum_{k=1}^\infty (-1)^k \frac{\binom{2k}{k}}{k^2 }x^k=-2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;-4 x\right)$$ Simplifying
$$S_2(x)=-2 x \left(\frac{\left(\log \left(1+\sqrt{1+4 x}\right)-\log (2)\right)^2}{2
x}-\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1+4 x}\right)}{x}\right)$$
$$S_2(4)=2 \text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{17}\right)\right)-\log ^2(2)-\log
\left(1+\sqrt{17}\right) \text{csch}^{-1}(4)$$
$$\color{blue}{S_2(4)=2 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\sinh ^{-1}(1)-\log (2)\right)^2}$$
Combining all of the above
$$\color{red}{2(S_1(4)+S_2(4))=4 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\frac{\pi ^2}{3}-6 \log
^2(2)+2 \sinh ^{-1}(1) \left(2 \log (2)-\sinh ^{-1}(1)\right)}$$
| {
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"url": "https://math.stackexchange.com/questions/4427827",
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"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$ using L'Hôpital's rule I need to evaluate the following limit (using L'Hôpital's rule).
$$\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$$
I expressed $\tanh(x)$ as $\frac{\sinh(x)}{\cosh(x)}$:
$$\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\frac{\sinh(x)}{\cosh(x)}}-\frac{1}{x}\bigg)=\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{\cosh(x)}{\sinh(x)}-\frac{1}{x}\bigg)$$
$$=\lim_{x \rightarrow 0}\frac{3}{x} \bigg(\frac{x\cosh(x)-\sinh(x)}{x\sinh(x)}\bigg)$$
But I do not know how to continue.
EDIT:
Continuing,
$$\lim_{x \rightarrow 0}\frac{3}{x} \bigg(\frac{x\cosh(x)-\sinh(x)}{x\sinh(x)}\bigg)=\lim_{x \rightarrow 0}3 \bigg(\frac{x\cosh(x)-\sinh(x)}{x^2\sinh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)+x\sinh(x)-\cosh(x)}{2x\sinh(x)+x^2\cosh(x)}\bigg)=3\lim_{x \rightarrow 0}\bigg(\frac{\sinh(x)}{2\sinh(x)+x\cosh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{2\cosh(x)+\cosh(x)+x\sinh(x)}\bigg)=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{2\cosh(x)+\cosh(x)+x\sinh(x)}\bigg)$$
$$=3\lim_{x \rightarrow 0}\bigg(\frac{\cosh(x)}{3\cosh(x)+x\sinh(x)}\bigg)=3\times \frac{1}{3}=1$$
Your help would be appreciated. THANKS!
| Solution :
$$\lim _{x\to \:0}\left(\frac{3}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)$$
Taking out constant :
$$\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)$$
We get :
$$=3\cdot \lim _{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)$$
Simplify :
$$\frac{1}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)=\frac{x-\tanh \left(x\right)}{x^2\tanh \left(x\right)}$$
We get :
$$=3\cdot \lim _{x\to \:0}\left(\frac{x-\tanh \left(x\right)}{x^2\tanh \left(x\right)}\right)$$
Apply L'Hopital's Rule first time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{1-sech ^2\left(x\right)}{2x\tanh \left(x\right)+sech ^2\left(x\right)x^2}\right)$$
Apply L'Hopital's Rule second time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{2sech ^2\left(x\right)\tanh \left(x\right)}{-2x^2sech ^2\left(x\right)\tanh \left(x\right)+4xsech ^2\left(x\right)+2\tanh \left(x\right)}\right)$$
Apply L'Hopital's Rule third time :
$$=3\cdot \lim _{x\to \:0}\left(\frac{2\left(-2sech ^2\left(x\right)\tanh ^2\left(x\right)+sech ^4\left(x\right)\right)}{4x^2sech ^2\left(x\right)\tanh ^2\left(x\right)-2x^2sech ^4\left(x\right)-12xsech ^2\left(x\right)\tanh \left(x\right)+6sech ^2\left(x\right)}\right)$$
Plug the value of $x=0$ :
$$=3\cdot \frac{2\left(-2sech ^2\left(0\right)\tanh ^2\left(0\right)+sech ^4\left(0\right)\right)}{4\cdot \:0^2sech ^2\left(0\right)\tanh ^2\left(0\right)-2\cdot \:0^2sech ^4\left(0\right)-12\cdot \:0\cdot sech ^2\left(0\right)\tanh \left(0\right)+6sech ^2\left(0\right)}$$
Simplify the answer and we will get :
$$=1$$
So, we calculated that :
$$\lim _{x\to \:0}\left(\frac{3}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving system of three equations in three unknowns for electric circuit analysis problem My electrical engineering circuit analysis textbook presents the following problem:
Find the currents and voltages in the circuit shown in Fig. 2.28.
Answer: $v_1 = 6 \ \text{V}$, $v_2 = 4 \ \text{V}$, $v_3 = 10 \ \text{V}$, $i_1 = 3 \ \text{A}$, $i_2 = 500 \ \text{mA}$, $i_3 = 2.5 \ \text{A}$
For those that are interested, I will begin by giving the circuit analysis explanation of what is going on. We have two loops here. For the first loop, we have $-10 \ \text{V} + 2i_1 + 8i_2 = 0$. For the second loop, we have $4i_3 - 6 \ \text{V} - 8i_2 = 0$, where I have $8i_2$ by Ohm's law instead of $-8i_2$, because the current $i_2$ is actually going from + to - for the $8 \ \Omega$ resistor (and despite this being counter to the clockwise direction of the loop, my understanding is that this results in a positive current for the purpose of the Ohm's law calculation). Finally, if we designate the top middle node, then we have that $i_1 - i_2 - i_3 = 0$.
This gives us the following system of three equations in three variables:
$$-10 \ \text{V} + v_1 + v_2 = 0 = -10 \ \text{V} + 2i_1 + 8i_2 \ \Rightarrow i_2 = \dfrac{10 \ \text{V} - 2i_1}{8} = \dfrac{5 \ \text{V} - i_1}{4}$$
$$-6 \ \text{V} - v_2 + v_3 = 0 = -6 \ \text{V} - (8i_2) + 4i_3 = -6 \ \text{V} - 8i_2 + 4i_3$$
$$i_1 - i_2 - i_3 = 0$$
I tried solving this as follows:
$$-6 \ \text{V} - 8\left( \dfrac{5 \ \text{V} - i_1}{4} \right) + 4i_3 = 0 \ \Rightarrow -6 \ \text{V} - 10 \ \text{V} + 2i_1 + 4i_3 = 0 \\ \Rightarrow 2i_1 = 16 \ \text{V} - 4i_3 \ \Rightarrow i_1 = 8 \text{V} - 2i_3$$
$$(8 \ \text{V} - 2i_3) - i_2 - i_3 = 0 \ \Rightarrow 8 \ \text{V} - 3i_3 - i_2 = 0 \ \Rightarrow i_2 = 3i_3 - 8 \ \text{V}$$
I don't understand what I'm doing wrong and what the correct way to proceed from here is. Since we've already utilised every equation in the system, if we continue with further substitutions, are we not double-counting?
| It is easier to write the three equations and use Gaussian Elimination, the equations being
$$ 2i_1 + 8i_2 + 0 i_3 = 10 \ \text{V}\\ 0 i_1 - 8i_2 +4i_3 = 6 \ \text{V} \\ i_1 - i_2 - i_3 = 0 \text{V}\\ $$
Method 1: We have the augmented matrix
$$ \left(
\begin{array}{ccc|c}
2 & 8 & 0 & 10 \\
0 & -8 & 4 & 6 \\
1 & -1 & -1 & 0 \\
\end{array}
\right)$$
Using RREF, we reduce to
$$\left(
\begin{array}{ccc|c}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & \dfrac{1}{2} \\
0 & 0 & 1 & \dfrac{5}{2} \\
\end{array}
\right)$$
From this result, we have
$$i_3 = \dfrac{5}{2}, i_2 = \dfrac{1}{2}, i_1 = 3$$
Method 2: We can also use elimination from the augmented system, lets add equation one and two
$$2 i_1 + 4 i_3 = 16$$
Next, subtract $8$ times equation three from equation two
$$-8 i_1 + 12 i_3 = 6$$
Now we have two equations in two unknowns and we can add four times equation one to equation two
$$28 i_3 = 70 \implies i_3 = \dfrac{5}{2}$$
It is easy to move forward from here.
Method 3: Your substitution approach is also perfectly fine, we have
$$i_2 = \dfrac{5-i_1}{4}$$
Substituting that into the second equation
$$-6 - 8\left(\dfrac{5-i_1}{4} \right) + 4 i_3 = 0$$
This reduces to
$$2i_1 + 4 i_3 = 16$$
Next, substitute the same into equation three
$$i_1 - \left(\dfrac{5-i_1}{4} \right) -i_3 = 0$$
this reduces to
$$5 i_1 - 4 i_3 = 5$$
Now you have two equation in two unknowns, adding them
$$7i_1 = 21 \implies i_1 = 3$$
| {
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"url": "https://math.stackexchange.com/questions/4429285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$x^4+y^4+z^4=\frac{m}{n}$, find $m+n$. $x^4+y^4+z^4$=$m\over n$
x, y, z are all real numbers,
satisfying $xy+yz+zx=1$ and $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$
m, n are positive integers and their greatest common divisor is 1. Calculate m+n.
My thinkings so far are as follows:
*
*to operate such that we can arrange 5,12,13 into some kind of pythagorean triplets.
*to do whole square of x+1/x or x^2 + 1/x^2 such that
the third term is a constant and can be moved to other side.
| Notice if $(x,y,z)$ satisfies the two conditions, so does $(-x,-y,-z)$. Furthermore, it is clear $x,y,z$ have the same sign. Since $(x,y,z)$ and $(-x,-y,-z)$ gives the same $\frac{m}{n}$, we only need to consider the case $x,y,z > 0$.
Take three numbers $\alpha,\beta,\gamma \in (0,\frac{\pi}{2})$ such that
$$x = \tan\alpha, y = \tan\beta, z = \tan\gamma$$
Recall the three angle summation formula for tangent, we have
$$\begin{align}\tan(\alpha+\beta+\gamma)
&= \frac{\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta - \tan\beta\tan\gamma - \tan\gamma\tan\alpha}\\
&= \frac{x+y+z - xyz}{1 -xy - yz -zx}\\&= \infty\end{align}$$
This implies $\alpha + \beta + \gamma = (N + \frac12)\pi$ for some integer $N$.
Since $\alpha + \beta + \gamma \in (0,\frac32\pi)$, this forces $N = 0$ and
hence $\alpha + \beta + \gamma = \frac{\pi}{2}$.
Notice
$$\frac{2}{x+\frac1x} = \sin(2\alpha),\quad
\frac{2}{y+\frac1y} = \sin(2\beta)\quad\text{ and }\quad
\frac{2}{z+\frac1z} = \sin(2\gamma)$$
the condition
$$5\left(x+\frac1x\right) = 12\left(y + \frac1y\right) = 13\left(z + \frac1z\right)$$
is equivalent to
$$\sin(2\alpha) : \sin(2\beta) : \sin(2\gamma) = 5 : 12 : 13 \tag{*1}$$
Since $5^2 + 12^2 = 13^2$, we find
$$\sin^2(2\alpha) + \sin^2(2\beta) = \sin^2(2\gamma)\tag{*2}$$
With a little bit of algebra, one can verify when $\alpha + \beta + \gamma = \frac{\pi}{2}$, one has the "cosine rule":
$$\cos(2\gamma) = \frac{\sin^2(2\alpha) + \sin^2(2\beta) - \sin^2(2\gamma)}{2\sin(2\alpha)\sin(2\beta)}$$
$(*2)$ tell us $\cos(2\gamma) = 0 \implies \gamma = \frac{\pi}{4} \implies z = \tan\gamma = 1$. Plug this into $(*1)$, we get
$$
\begin{cases}
\sin(2\alpha) = \frac{5}{13}\\
\sin(2\beta) = \frac{12}{13}
\end{cases}
\quad\implies\quad
\begin{cases}
\cos(2\alpha) = \frac{12}{13}\\
\cos(2\beta) = \frac{5}{13}
\end{cases}
$$
This leads to
$$\begin{align}
x &= \tan \alpha = \sqrt{\frac{1-\cos(2\alpha)}{1 + \cos(2\alpha)}}
= \sqrt{\frac{1-\frac{12}{13}}{1+\frac{12}{13}}} = \frac15\\
y &= \tan \beta = \sqrt{\frac{1-\cos(2\beta)}{1 + \cos(2\beta)}}
= \sqrt{\frac{1-\frac{5}{13}}{1+\frac{5}{13}}} = \frac23\\
\end{align}
$$
As a result
$$\frac{m}{n} = x^4 + y^4 + z^4 = \frac1{5^4} + \frac{2^4}{3^4} + 1^4 = \frac{60706}{50625}$$
This leads to $m + n = 60706 + 50625 = 111331$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n \to \infty}a_n$ where $a_1=1$ and $a_{n+1}=a_n+\frac{1}{2^na_n}$. There are some attempts as follows.
Obviously, $\{a_n\}$ is increasing. Thus $a_n\ge a_1=1$. Therefore
$$a_{n+1}-a_n=\frac{1}{2^na_n}\le \frac{1}{2^n},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\le a_1+\sum_{k=1}^{n-1}\frac{1}{2^k}=2-\frac{1}{2^{n-1}}.$$
Hence, $a_n\le 2$. Similarily,
$$a_{n+1}-a_n=\frac{1}{2^na_n}\ge \frac{1}{2^{n+1}},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\ge a_1+\sum_{k=1}^{n-1}\frac{1}{2^{k+1}}=\frac{3}{2}-\frac{1}{2^{n}}.$$
Put the both aspects together, we have
$$\frac{3}{2}-\frac{1}{2^n}\le a_n\le 2-\frac{1}{2^{n-1}}.$$
But this does not satisfy the applying conditions of the squeeze theorem. Any other solutions?
| Here is a short proof that the limit is between 1.810 and 1.828. This can be improved to get better precision, I believe.
Squaring both sides of the recurrence, we get
$$
a_{n+1}^2 = a_n^2 + \frac{2}{2^n} + \frac{1}{4^n a_n^2} \tag{1}
$$
and so, summing,
$$
a_n^2 = a_1^2 + \sum_{k=1}^{n-1} (a_{k+1}^2 - a_k^2) = 1+\sum_{k=1}^{n-1}\frac{2}{2^k} + \sum_{k=1}^{n-1}\frac{1}{4^k a_k^2} \tag{2}
$$
and so the limit $\ell$ satisfies
$$
\ell^2 = 1 + \sum_{k=1}^\infty \frac{2}{2^k} + \sum_{k=1}^\infty \frac{1}{4^ka_k^2} = 3 + \sum_{k=1}^\infty \frac{1}{4^ka_k^2} \tag{3}
$$
Since we have established that $1 \leq a_n \leq \ell$ for every $n\geq 1$, we get (also using in the LHS that $a_1=1$ to get better precision):
$$
3 + \frac{1}{4} + \frac{1}{\ell^2}\sum_{k=2}^\infty \frac{1}{4^k} \leq \ell^2 \leq 3 + \sum_{k=1}^\infty \frac{1}{4^k} \tag{4}
$$
that is, since $\sum_{k=1}^\infty \frac{1}{4^k}=\frac{1}{3}$,
$$
\frac{13}{12}+\frac{1}{12\ell^2} \leq \ell^2 \leq \frac{10}{3} \tag{5}
$$
which gives $\frac{39+\sqrt{1569}}{24}\leq \ell^2 \leq \frac{10}{3}$, i.e.,
$$
1.810\approx\boxed{\sqrt{\frac{39+\sqrt{1569}}{24}}\leq \ell \leq \sqrt{\frac{10}{3}}} \approx 1.828
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Is there a solution for $(\frac{3}{x+y+z})^n+(\frac{3}{x+y+z})^{5-n}<2$? Is there a solution for
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n}<2$$
where $n\in\mathbb Z$ and $x,y,z>0, xyz=1$.
This is the part of my attempts for my homework, that I stucked.
This is obvious $x+y+z≥3\implies \frac {3}{x+y+z}≤1$. But there is no any restriction for integer $n$. Therefore, I stucked. I tried Wolfram Alpha, even WA couldn't solve.
| Hint
By AM-GM inequality:
$$x+y+y\ge3(xyz)^{1/3}\Rightarrow \frac{3}{x+y+z}\le1.$$
The equality holds only when $x=y=z=1$.
It means that, for $0\le n \le 5$, we have
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n} \le 2.$$
You want to know if there is a solution for
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n} <2.$$
Think about the situation where you reach the sum equal to $2$.
Think on the interval $0\le n \le 5$.
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{\sin x} dx = 7~\zeta(3) ,\text { equivalently }\ \int_{0}^{1}\frac{x - x^{2} }{ \sin(\pi x)}dx = 7\frac{\zeta (3)}{\pi^{3}},$$ has been solved here.)
| Integrate by parts twice
\begin{align}
\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx
=& \>2 \int_0^{ \frac\pi2}\frac{x^2(\pi-x)^2}{\sin^2 x} dx \\
=&\>4 \int_0^{\frac\pi2}(\pi^2x-3\pi x^2+2x^3)\cot x\> dx\\ =&\> 4\int_0^{\frac\pi2} (6\pi x-6x^2 -\pi^2)\ln(2\sin x)dx
\end{align}
Then, utilize the known results $\int_0^{\frac\pi2} \ln(2\sin x)dx=0$, $\int_0^\frac{\pi}{2}x\ln(2\sin x)dx= \frac7{16}\zeta(3)$ and $\int_0^{\frac\pi2} x^2\ln(2\sin x)\,dx= \frac{3\pi}{16}\zeta(3)$ to arrive at
\begin{align}
\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx
=6\pi\zeta(3),
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Integrate $(1-\sqrt{x})/(1+\sqrt{x})$
Integrate $(1-\sqrt{x})/(1+\sqrt{x})$.
So, this was on my test, and I got
$$4(1+\sqrt{x})-4\ln(1+\sqrt{x})-x+c$$
as a result using symbolab. But my teacher gave the result shown below (his solution is shown here, and transcribed below).
$$\int \frac{1-\sqrt{x}}{1+\sqrt{x}} \, dx \;=\; 6\left(1-\sqrt{x}\right)-\left(1-\sqrt{x}\right)^2-2\ln\big|1-\sqrt{x}\big|+c.$$
I'm trying to understand what is wrong in symbolab or in his result, but I really can't find out.
Teacher's solution:
Let $u = 1+\sqrt{x}$. Then $1-\sqrt{x} = 2-u$, and $dx = 2(u-1)du$. Then
\begin{align*}
\int \frac{1-\sqrt{x}}{1+\sqrt{x}} \, dx
&= \int \frac{(2-u)2(u-1)}{u} \, du
= 2 \int \frac{3u-u^2-2}{u} \, du
\\&= \int 6 \, du - 2 \int u \, du - 2 \int \frac{1}{u} \, du
\\&= 6u - u^2 - 2\ln|u| + C
\\&= 6(1-\sqrt{x}) - (1-\sqrt{x})^2 - 2\ln|1-\sqrt{x}| + C
\end{align*}
| Your teacher made 2 separate mistakes. First when splitting the integrals:
$$2\int\frac{3u-u^2-2}{u}du=\int6\,du-2\int u \,du-4\int \frac1udu$$
They wrote a 2 instead of 4 in the final term. Then when resubstituting the expression for $u$ they used $1-\sqrt{x}$ instead of $1+\sqrt{x}$:
$$6(1+\sqrt{x})-(1+\sqrt{x})^{2}-4\ln\left(1+\sqrt{x}\right)+c$$
Indeed, this is equivalent to $4(1+\sqrt{x})-4\ln(1+\sqrt{x})-x+K$ as:
$$6(1+\sqrt{x})-(1+\sqrt{x})^{2}-4\ln\left(1+\sqrt{x}\right)+c=
6+6\sqrt{x}-1-2\sqrt{x}-x-4\ln\left(1+\sqrt{x}\right)+c$$
$$=5+4\sqrt{x}-x-4\ln\left(1+\sqrt{x}\right)+c=4(1+\sqrt{x}-\ln\left(1+\sqrt{x}\right))-x+1+c$$
And finally, let $K=1+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt{(5y+2)(y-2)}+1$$
then $f(x,y)=x+\frac{1}{2y}=\frac{2xy+1}{2y}=\frac{\sqrt{(5y+2)(y-2)}+2}{2y}=g(y)$,$$\therefore g'(y) = \frac{2 + 2 y - \sqrt{-4 - 8 y + 5 y^2}}{y^2 \sqrt{-4 - 8 y + 5 y^2}}$$
and $2+2y = \sqrt{5y^2-8y-4} \Rightarrow y = 8+6\sqrt{2}$, with the help of graphing calculator ,i know this is the maximum point. $$\therefore f_{\text{max}}(x,y)=g(8+6\sqrt{2})=\frac{3\sqrt{2}}{2}-1$$
Is there any other way to work out it?
| Here is another way. Change variables to replace $x$ with $z$ under the relation $z=2xy\iff x=\frac{z}{2y}$. So now you are maximizing $\frac{z+1}{2y}$ subject to $(z-1)^2=(5y+2)(y-2)$.
Lagrange multipliers can handle this. You have a system of three equations in three variables $y,z,k$:
$$\left\{\begin{align}
(z-1)^2&=(5y+2)(y-2)\\
-\frac{z+1}{2y^2}&=k(-10y+8)\\
\frac{1}{2y}&=2k(z-1)
\end{align}\right.$$
Dividing sides of the latter two equations:
$$-\frac{z+1}{y}=\frac{-5y+4}{z-1}$$
So now we have a system of two equations (the first equation above and this new equation, each rearranged algebraically):
$$\left\{\begin{align}
z^2-2z+1&=5y^2-8y-4\\
-z^2+1&=-5y^2+4y
\end{align}\right.$$
Add these: $$-2z+2=-4y-4$$
So $z-1=2y+2$, which we can substitute back into the original constraint:
$$(2y+2)^2=(5y+2)(y-2)\implies0=y^2-16y-8$$ from which we solve $y=8\pm6\sqrt{2}$. Then $z=2y+3=19\pm12\sqrt{2}$. And finally
$$\begin{align}
\frac{z+1}{2y}&=\frac{20\pm12\sqrt{2}}{16\pm12\sqrt{2}}\\
&=\frac{5\pm3\sqrt{2}}{4\pm3\sqrt{2}}\cdot\frac{4\mp3\sqrt{2}}{4\mp3\sqrt{2}}\\
&=\frac{2\mp3\sqrt{2}}{-2}=\pm\frac32\sqrt{2}-1
\end{align}$$
$\frac32\sqrt{2}-1$ is a maximum, and $-\frac32\sqrt{2}-1$ is a minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
I have checked on google for $ \sum_{n=1}^\infty \frac{1}{n^2}$ and i do not suppose to know that is equal to $\frac{\pi^2}{6}$.
and I checked how The sum can be given explicitly of $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
and the result was show that needs the Fourier series and we haven't learned that yet.
So, I have no idea how to prove that equal to $\frac{3}{4}S$ without know the exactly the sum
Thanks.
| Hint:
\begin{align}
S & = \sum_{n=1}^\infty \frac{1}{n^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum_{n=1}^\infty \frac{1}{(2n)^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{1}{4}S
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the value of expression $Q = \frac{x + 1}{y}$ when $xy > 1$ and expression $P = x + 2y + \frac{5x + 5y}{xy - 1}$ reaches its maximum value.
Consider two positives $x$ and $y$ where $xy > 1$. The maximum value of the expression $P = x + 2y + \dfrac{5x + 5y}{xy - 1}$ is achieved when $x = x_0$ and $y = y_0$. Calculate the value of expression $Q = \dfrac{x_0 + 1}{y_0}$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Isn't being demotivated one of the worst feelings in the world?
Let $x = \cot\alpha \left(n\pi < \alpha < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$ and $y = \cot\beta \left(n\pi < \beta < n\pi + \dfrac{\pi}{2}, n \in \mathbb Z\right)$. It could be implied that $P = \cot\alpha + 2\cot\beta + \dfrac{5}{\cot(\alpha + \beta)}$.
Hmmm~ never-mind, this leads to nowhere.
Perhaps creating new variables, specifically $a = x + 1 \ (a > 2)$ and $b = \dfrac{1}{y} \ (b > 0)$ might actually be a clue. Then, we have that $a - 1 > 2b$ and $$\begin{aligned} P = a + 5b - 1 + \dfrac{2}{b} + \dfrac{5(b^2 + 1)}{a - b - 1} &= 6b + \dfrac{2}{b} + (a - b - 1) + \dfrac{5(b^2 + 1)}{a - b - 1}\\ &\ge 2\left[3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}\right] \end{aligned}$$
Function $f(b) = 3b + \sqrt{5(b^2 + 1)} + \dfrac{1}{b}$ contains its first derivative as $f'(b) = 3 + \dfrac{b\sqrt{5}}{\sqrt{b^2 + 1}} - \dfrac{1}{b^2}$ and the equation $f'(b) = 0$ has roots $b = \dfrac{1}{2}$ and $b = -\sqrt{\dfrac{\sqrt 5 - 1}{2}}$
We can then draw the table of variations for function $f(b)$.
As $f(b) \ge 6, \forall b > 0$, it can be inferred that the minimum value of $P$ is $2 \times 6 = 12$, achieved when $$\left\{ \begin{aligned} a > 0, b > 0&, a - 1 > 2b\\ (a - b - 1)^2 &= 5(b^2 + 1)\\ b &= \dfrac{1}{2} \end{aligned} \right. \iff \left\{ \begin{aligned} a &= 4\\ b &= \dfrac{1}{2} \end{aligned} \right. \implies Q = ab = 4 \times \dfrac{1}{2} = 2$$
Wait... minimum? That's not what the question asks for, although that's one of the given options.
Anyhow, I'm still figuring out how to tackle this problem. Should I give in to the asymmetry or otherwise? As always, thanks for reading, (and even more so if you could help), have a wonderful tomorrow~
By the way, the options are $\sqrt 2, \sqrt 3, 2$ and $1$, so I could have been fooled there.
| Just note that since the points $xy=1$ are excluded, the maximizers/minimizers will be a stationary points in the open set defined by the condition $x y>1$. Once you determine the stationary points (some are complex numbers, rule them out), you'll see that the only stationary point with $x,y>0$ and $xy>1$ is $(x,y)=(3,2)$. Computing the Hessian matrix, it is in fact true that it is a minimizer.
It was clear right from the start that there could not be a maximizer... when we approach the line $xy = 1$ function values tend to $+\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{1\le iLet $n \ge 2$ be a an integer and $x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$
Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}$$
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\sum_{1\le i<j\le 1}\frac{x_ix_j}{(1-x_i)(1-x_j)}=\frac{1}{2}\left(\sum_{k=1}^n\frac{x_k}{1-x_k}\right)^2-\frac{1}{2}\sum_{k=1}^n\frac{x_k^2}{(1-x_k)^2}$$
This formula was derived in @GCab's Answer.
First let $a_k=x_k/(1-x_k)$
so we want to prove
$$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le \frac{n(n-1)}{(2n-1)^2}$$
But since $$\frac{x_k}{1-x_k}<2x_k\implies \sum_{k=1}^na_k<1 \quad (1)$$
Hence $$\left(\sum_{k=1}^na_k\right)^2\le \sum_{k=1}^na_k$$
Meaning $$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le\sum_{k=1}^na_k(1-a_k)$$
Now consider the following function $$f(x)=\frac{x}{1-x}\left(1-\frac{x}{1-x}\right)$$
$f$ is concave on $(0,1)$ and by the tangent line trick we have $$f(x)\le f'(a)(x-a)+f(a)$$
set $a=1/2n$ to get $$a_k(1-a_k)\le\frac{4n^2\left(2n-3\right)}{\left(2n-1\right)^3}\left(x_k-\frac{1}{2n}\right)+ \frac{2(n-1)}{(2n-1)^2}$$
Now we sum to finish $$\sum_{k=1}^na_k(1-a_k)\le \frac{2n(n-1)}{(2n-1)^2}$$
Maybe by tweaking $(1)$ a little bit we can get rid of this factor of $2$
| You should be able to look for the maximum, by the method of Lagrange. We set
$$F(x_1,...,x_n) \equiv \sum_{i\neq j} \frac{x_i x_j}{(1-x_i)(1-x_j)} \stackrel{\text{wts.}}{\leq} \frac{n(n-1)}{(2n-1)^2} \, ,$$
and define the Lagrange-function as $$L(x_1,...,x_n) = F(x_1,...,x_n) - \lambda \left(x_1+...+x_n-1/2\right) \,.$$
At an extremum we have $$\nabla L = 0 \qquad , \qquad \nabla=\begin{pmatrix} \partial_{x_1} \\ \vdots \\ \partial_{x_n} \\ \partial_\lambda\end{pmatrix} \, ,$$
in coordinates
\begin{align}\frac{2}{(1-x_k)^2} \sum_{\substack{i=1 \\ i\neq k}}^n \frac{x_i}{1-x_i} &= \lambda \quad\text{for}\quad k=1,...,n \tag{1} \\
\sum_{i=1}^n x_i &= 1/2 \tag{2} \, .\end{align}
Multiplying Equation (1) by $\frac{(1-x_k)^2}{2}$ and adding $\frac{x_k}{1-x_k}$ to both sides gives
$$c=\sum_{i=1}^n \frac{x_i}{1-x_i} = \frac{\lambda}{2} (1-x_k)^2 + \frac{x_k}{1-x_k} \tag{3}$$
where the sum on the LHS is now independent on $k$ and we can consider it as a constant $c$. We now sum (3) from $k=1$ to $n$
\begin{align}
nc&=\frac{\lambda}{2} \left( n - 1 + \sum_{k=1}^n x_k^2\right) + c \\
\Rightarrow \quad c&= \frac{\lambda}{2} + \frac{\lambda}{2(n-1)} \sum_{k=1}^n x_k^2 > \frac{\lambda}{2} \tag{4}
\end{align}
which we will need to use in a second. Rearranging Equation (3) gives a monic polynomial in $x_k$
$$x_k^3 - 3x_k^2 + \left( 3 - \frac{2+2c}{\lambda} \right)x_k + \frac{2c}{\lambda} - 1 = 0 \, . \tag{5}$$
A cubic can have either only 1 real solution, in which case there is nothing to show, since then $x_1=...=x_n$ follows, or it can have 3 real solutions. In this case there appears to be a great variety of combinations of the three roots for the various $x_k$. However, since the product of the three roots is equal to $1-\frac{2c}{\lambda}<0$ by Equation (4) and (5), either all three roots are negative (discard), or only one root is negative and we are left with two choices for the various $x_k$. But then, since the sum of all three roots is equal to $3$ by Equation (5) and one root is negative, the sum of the two positive roots must be strictly greater than $3$. This implies, that at least one of them is strictly greater than $1/2$, which stands in contradiction to Equation (2). Thus, we are left with only one valid real solution satisfying Equation (2) and $$x_1=...=x_n=x$$
follows. It is then easy to see that $x=1/2n$.
It remains to show, that this extremal point is actually a maximum. The hessian of $L$ is given in coordinates by
$$\frac{1}{2} \, {\rm Hess}_{k,l} (L) = \frac{2\delta_{k,l}}{(1-x_k)^3} \sum_{i\neq k} \frac{x_i}{1-x_i} + \frac{1-\delta_{k,l}}{(1-x_k)^2(1-x_l)^2} \qquad \text{for} \qquad k,l=1,...,n \\
\frac{1}{2} \, {\rm Hess}_{n+1,k} (L) = \frac{1}{2} \, {\rm Hess}_{k,n+1} (L) = -1 \qquad \text{for} \qquad k=1,...,n \\
\frac{1}{2} \, {\rm Hess}_{n+1,n+1} (L) = 0 \, .$$
With $x_k=x=1/2n$ the first line becomes
$$\frac{1}{2} \, {\rm Hess}_{k,l} (L) = \frac{1-\frac{\delta_{k,l}}{n}}{\left(1-\frac{1}{2n}\right)^4} \qquad \text{for} \qquad k,l=1,...,n$$
and in matrix-form this looks like
$$\frac{1}{2} \, {\rm Hess} (L) = \frac{1}{\left(1-\frac{1}{2n}\right)^4} \begin{pmatrix}
1-1/n & 1 & \dots & 1 & -1 \\
1 & 1-1/n & \dots & 1 & -1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 1 & \dots & 1-1/n & -1 \\
-1 & -1 & \dots & -1 & 0 \end{pmatrix} \, .$$
Then, defining the variational vector $\epsilon=(\epsilon_1,...,\epsilon_n,\epsilon_\lambda)^t$, we find for the second variation
$$\frac{1}{2} \sum_{k,l=1}^{n+1} \epsilon_k \, {\rm Hess}_{k,l} (L) \, \epsilon_l = - \frac{\sum_{k=1}^n \epsilon_k^2/n}{\left(1-\frac{1}{2n}\right)^4}<0$$ $\forall \epsilon \in {\mathbb R}^{n+1}$ satisfying
$$\sum_{k=1}^n \epsilon_k = 0 \, ,$$
which is a consequence of Equation (2).
| {
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In a school of $1025$ students, $400$ of them ($\frac{1}{5}$ of the boys and $\frac{4}{7}$ of the girls) cannot swim. How many boys are there? Not sure how to set up an equation for this problem.
A school has $1025$ students. A total of $400$ students cannot swim. This consists of $\frac{1}{5}$ of the boys and $\frac{4}{7}$ of the girls. If $x$ boys can swim, write an equation for $x$ and solve it. How many boys are there in the school?
The answer is the following equation:
$$625-x=\frac{3}{4}\left(400-\frac{x}{4}\right)$$
I understand that $625-x$ represents the number of girls who can swim but I cannot wrap my head around how the $\frac{3}{4}(400-\frac{x}{4})$ came about. Initially I thought that $400-\frac{x}{4}$ must mean the number of girls who cannot swim but if that was the case, the equation should be $400-\frac{x}{5}$ since $\frac{1}{5}$ of the boys cannot not swim.
If someone can help me to make sense of the equation in the answer, that would be much appreciated!
| What the equation is saying is that the number of girls who cannot swim is $3/4$ of the number of girls who can swim. This is true since the fraction of girls who can swim is
$$\frac{3}{7} = \frac{3}{4} \cdot \frac{4}{7}$$
As you observed, $625 - x$ is the number of girls who can swim. Since $x$ is the number of boys who can swim, $x/4$ is the fraction of boys who cannot swim since
$$\frac{1}{5} = \frac{1}{4} \cdot \frac{4}{5}$$
Hence, the term
$$400 - \frac{x}{4}$$
is the number of girls who cannot swim since there are a total of $400$ students who cannot swim and $x/4$ of those are boys. Since the number of girls who can swim is $3/4$ of the number of girls who cannot swim, we obtain
$$625 - x = \frac{3}{4}\left(400 - \frac{x}{4}\right)$$
Note: Another way to solve the problem is to let $b$ be the total number of male students and $g$ be the total number of female students. We are told that there are $1025$ students in the school, so
$$b + g = 1025$$
We are also told that $400$ students cannot swim, including $1/5$ of the boys and $4/7$ of the girls cannot swim, so
$$\frac{1}{5}b + \frac{4}{7}g = 400$$
You can solve that system of equations to determine the number of boys in the school, then multiply the result by $4/5$ to find the number of boys in the school who can swim. You should find that the two methods yield the same result.
| {
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How to find the explicit solution of $y'(x)= |y(x)|(1-y(x)) \frac{x^3}{1+x^4}$ Given,
$$y'(x)= |y(x)|(1-y(x)) \frac{x^3}{1+x^4}$$ with initial conditions $y(0) = 2$
I know that the y(x) can be found using Bernoulli's method such that
$$\frac{y'}{y^2} -\frac{x^3}{y(1+x^4)} = -\frac{x^3}{1+x^4} $$
Assuming $\frac{1}{y} = v$
Thus,
$$I.F ={\mathrm e}^{\int_0^x\frac{x^3}{1+x^4}} dx$$
or,$$ I.F. = (1+x^4)^\frac{1}{4} $$
Thus,
$$ \frac{(1+x)^{\frac{1}{4}}}{y} =\int_0^x {\frac{x^3(1+x^4)^{\frac{1}{4}}}{1+x^4}} dx $$
and this is where I got stuck. How shall I proceed from here?
Somehow the answer is $$y(x) = \frac {(2^4)\sqrt{1+x^4}}{-1 + 2(1+x^4)^{\frac{1}4}}$$
I also observe that this can be solved using the separable variable method but I still have trouble getting to the answer. I will be grateful if anyone can help me out with this.
| The derivative of $(1+x^4)^{1/4}$ is $x^3(1+x^4)^{-3/4}$, which is exactly the integrand you are in doubt of.
The equation is separable, so a better substitution would use and preserve that pattern by setting $v=y^{-1}-1$. Then
$$
v'=-y^{-2}y'=-sv\frac{x^3}{1+x^4}, ~~~ s=sign(y(0)).
$$
This is obviously again separable
$$
\ln(v)=\frac{-s}4\ln(1+x^4)+c
\\
v=C(1+x^4)^{-s/4}
$$
etc.
| {
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Prove that for positive reals $x,y,z$, $x^6+y^6+z^6 + 6x^2y^2z^2 \geq 3xyz(x^3+y^3+z^3)$. I am not sure if the inequality is true. My first attempt was to try AM-GM inequality in clever ways. I also tried Schur's inequality which gives
$$
x^6+y^6+z^6 + 6x^2y^2z^2 \geq (x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2)
$$
but it is not true that
$$
(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2) \geq 3xyz(x^3+y^3+z^3)
$$
Edit: I found a variant and the solutions there work for this problem too: $a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$
| The problem is equivalent to prove that
$a^2+b^2+c^2+6 \ge 3(a+b+c)$, $abc=1$
But the last inequality is true because
$$LHS - RHS = \frac{(a+b+c-3)^2+a^2+b^2+c^2+2abc+1 -2(ab+bc+ca)}{2} \ge 0$$
Note that the inequality $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$ is well-known.
| {
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What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be minimum, it gave me an absurd relation like $-6=6$.
How can we solve this? Why is the differentiation method not working here? Please help !!!
$f(x)= \frac{x^2+3x-6}{x^2+3x+6}$
$f'(x)= \frac{(x^2+3x+6)(2x+3)-(x^2+3x-6)(2x+3)}{(x^2+3x+6)^2}$
Then equated this to $0$ to find the value of $x$.
Thanks in advance !!!
| Another didactic way, we prove that $-11/5 \le y=f(x)<1, x \in \Re$.
Let $y=\frac{x^2+3x-6}{x^2+3x+6}$ and check that $x^2+3x+6 >0$
Then we can write $(y-1)x^2+3(y-1)x+6(y+1)=0, x \in \Re$
$\implies B^2 \ge 4AC \implies 9(y-1)^2\ge 24(y+1)(y-1).$
Case 1: $y<1$ we get $9y-9 \le 24y+24 \implies y \ge -11/5$
and hence $-11/5 \le y <1$.
Case 2: when $y>1$ gives $y\le -11/5$ a Null.
| {
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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
x^2 && \hbox{} \\[-3pt]
x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
\underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
2 +i.x^2
\end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
| The equation $x^3 - x^2 + 2 = 0$ has three solutions. Therefore we can write it as $(x-a)(x-b)(x-c)=0$. Expanding the second expression and equating terms we get:
$a+b+c = 1$; $ab+bc+ac=0$; $abc=-2$
Now substitute $a = 1+i$. We get $b+c = -i$ and $bc=-1+i$. Which is easily solved.
| {
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Why does my solution of the closed form formula for the Fibonacci sequence differ from the actual solution by a sign? I have $a_{n}=a_{n-1}+a_{n-2}$ with $a_0=0$ and $a_1=1$.
I've found the generating function $G(x)=\frac{x}{1-x-x^2}$. Solving $1-x-x^2=0$ gives the solutions $\alpha_1=\frac{-1+\sqrt{5}}{2}$ and $\alpha_2=\frac{-1-\sqrt{5}}{2}$. Then, using partial fractions, I found: $$\frac{x}{1-x-x^2}=\frac{\frac{1}{\sqrt5}\alpha_1}{x-\alpha_1}-\frac{\frac{1}{\sqrt{5}}\alpha_2}{x-\alpha_2},$$
so that
\begin{align}
G(x)&=\frac{1}{\sqrt{5}}( \frac{\alpha_1}{x-\alpha_1} - \frac{\alpha_2}{x-\alpha_2})\\
& =\frac{1}{\sqrt{5}} (\frac{1}{\frac{1}{\alpha_1}x-1}- \frac{1}{\frac{1}{\alpha_2}x-1})\\
&= \frac{1}{\sqrt{5}} (\frac{1}{1-\frac{1}{\alpha_2}x} - \frac{1}{1-\frac{1}{\alpha_1}x})\\
& = \frac{1}{\sqrt{5}} (\sum((\frac{1}{\alpha_2})^n-(\frac{1}{\alpha_1})^n)x^n,
\end{align}
So that the solution is $a_n=\frac{1}{\sqrt{5}}((\frac{1}{\alpha_2})^n-(\frac{1}{\alpha_1})^n)$.
$\frac{1}{\alpha_1}=\frac{1+\sqrt{5}}{2}$ and $\frac{1}{\alpha_2}=\frac{1-\sqrt{5}}{2}$. Substituting these values get us:
$$a_n=\frac{1}{\sqrt{5}}((\frac{1-\sqrt{5}}{2})^n-(\frac{1+\sqrt{5}}{2})^n),$$
which actually differs from the correct answer only by a negative sign.
| Hint: It seems we have
\begin{align*}
\frac{x}{1-x-x^2}=\color{blue}{-}\frac{\frac{1}{\sqrt5}\alpha_1}{x-\alpha_1}\color{blue}{+}\frac{\frac{1}{\sqrt{5}}\alpha_2}{x-\alpha_2}
\end{align*}
| {
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Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $ I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.
Let $ x, y \in \mathbb{R} $. Therefore:
$ \lfloor{x}\rfloor \le x $
$ \lfloor{y}\rfloor \le y $
$ \Rightarrow $
$ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le x + y $
| $ \lfloor{x}\rfloor $ is an integer $ \leq x;\ \lfloor{y}\rfloor $ is an integer $ \leq y.$
Therefore, $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y.$
Since $\lfloor{x+y}\rfloor$ is the greatest integer $ \leq x+y,$ and $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y,$ it follows that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \leq \lfloor{x+y}\rfloor.$
| {
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
~
EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
| Since $x$ and $y$ are positive integers, we can use AM-GM inequality. Also, notice that $3y = 5(20 -x)$, therefore $y$ is a multiple of $5$ or $xy$ is a multiple of $5$.
So, $$\frac{5x + 3y}{2} \ge \sqrt{15xy} $$
or, $$50 \ge \sqrt{15xy}$$
By squaring both sides, we ge:$$2500 \ge 15xy$$
or $$166.67 \ge xy$$
Since $xy$ is an integer less than $166.67$ and a multiple of $5$, so you can check that the greatest value of $xy$ is $165$.
| {
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How to prove this summation of floor function I am supposed to show that
$$
\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots = n
$$
For all postive integers $n \geq 1 $. (Note that this summation is not infinite, since $0 < \frac{n+2^k}{2^{k+1}} < 1$ for large $k$.
I can't seem to find a good way to tackle this exercise.
| The statement is trivially true for $n=1$ and $n=2$. Suppose it is true for any positive integer less than $n$, and let's prove it is true for $n$. We have two cases:
$n=2m$ is even. Then:
$$
\begin{array}{cll}&\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots\\=&
\left\lfloor \frac{2m+1}{2} \right\rfloor + \left\lfloor \frac{2m+2}{2^2} \right\rfloor + \left\lfloor \frac{2m+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{2m+2^k}{2^{k+1}} \right\rfloor + \cdots
\\=&
m + \left(\left\lfloor \frac{m+1}{2} \right\rfloor + \left\lfloor \frac{m+2}{2^2} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{k-1}}{2^k} \right\rfloor + \cdots\right)\\=& m + m &\text{ (inductive hypothesis)}\\ =&2m\\=&n\end{array}
$$
$n=2m+1$ is odd. This case is only slightly more complicated:
$$
\begin{array}{cll}&\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots\\=&
\left\lfloor \frac{2m+2}{2} \right\rfloor + \left\lfloor \frac{2m+2+1}{2^2} \right\rfloor + \left\lfloor \frac{2m+2^2+1}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{2m+2^k+1}{2^{k+1}} \right\rfloor + \cdots
\\ =&
(m + 1) + \left(\left\lfloor \frac{m+1+1/2}{2} \right\rfloor + \left\lfloor \frac{m+2+1/2}{2^2} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{k-1}+1/2}{2^k} \right\rfloor + \cdots\right)\\
\\ =&
(m + 1) + \left(\left\lfloor \frac{m+1}{2} \right\rfloor + \left\lfloor \frac{m+2}{2^2} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{k-1}}{2^k} \right\rfloor + \cdots\right)&\text{ the additional }1/2\text{ does not change anything}\\=& m+1 + m&\text{ (inductive hypothesis)}\\ =&2m+1\\=&n\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series solution for $x^2y''-x(x+6)y'+10y=0$ I have to solve this differential equation:
$$x^2y''-x(x+6)y'+10y=0$$
by using this method and I am stuck at this step.
Please help me to solve it.
Here is my attempt:
https://i.stack.imgur.com/XglDm.jpg
| The given differential equation is a secondorder linear ordinary differential equation. We are required to find a series solution to this problem.
Step 1: Assume that an infinite series solution exists such that
$$
y = \sum_{n = 0}^{\infty}c_{n}x^{n}
$$
Step 2: Calculate $y''$, $y'$:
$$
y' = \sum_{n=1}^{\infty}nc_{n}x^{n-1}\\
y'' = \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}
$$
Step 3: Substitute $y''$, $y'$, and $y$ in the given differential equation:
$$
\begin{align}
&x^2y''-x(x+6)y'+10y&=0\\
\implies &x^2(\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}) - x(x+6)\sum_{n=1}^{\infty}nc_{n}x^{n-1} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - (x+6)\sum_{n=1}^{\infty}nc_{n}x^{n} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - x\sum_{n=1}^{\infty}nc_{n}x^{n} -6\sum_{n=1}^{\infty}nc_{n}x^{n} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - \sum_{n=1}^{\infty}nc_{n}x^{n+1} -\sum_{n=1}^{\infty}6nc_{n}x^{n} + \sum_{n = 0}^{\infty}10c_{n}x^{n} &= 0
\end{align}
$$
All terms, except one, are in terms of $x^n$:
$$
\sum_{n=0}^{\infty}nc_{n}x^{n+1} = \sum_{n=1}^{\infty}(n-1)c_{n-1}x^{n}
$$
It would be nice if we could take the same limits for all the sums. We are in luck because we can. Upon changing all lower limits to $n=1$:
$$
\begin{align}
\implies &\sum_{n=1}^{\infty}n(n-1)c_{n}x^{n} - \sum_{n=1}^{\infty}nc_{n}x^{n+1} -\sum_{n=1}^{\infty}6nc_{n}x^{n} + \sum_{n=1}^{\infty}10c_{n}x^{n} +10c_{0} &= 0\\
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n}x^{n} - (n-1)c_{n-1}x^{n} -6nc_{n}x^{n} + 10c_{n}x^{n}\right] +10c_{0}&= 0\\
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n}\right]x^{n} +10c_{0}&= 0
\end{align}
$$
We are required to make an assumption that $c_{0}=0$:
$$
\begin{align}
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n}\right]x^{n} &= 0\\
\implies & n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n} &= 0\\
\implies & (n(n-1) -6n + 10)c_{n} = (n-1)c_{n-1}&\\
\implies & c_{n} = \dfrac{n-1}{n(n-1) -6n + 10}c_{n-1}&n>1\\
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $S=\{\overline{0},\overline{5},\overline{10}\}\subset\mathbb{Z}_{15}$ isomorphic to$\mathbb{Z}_{3}$ as rings? My book asked me to prove that it is in fact. I generated the tables for addition
\begin{array}{c|ccc}
+ & 0 & 5 & 10 \\
\hline 0 & 0 & 5 & 10 \\
5 & 5 & 10 & 0 \\
10 & 10 & 0 & 5
\end{array} \begin{array}{c|ccc}
+ & 0 & 1 & 2 \\
\hline 0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1
\end{array}
and for the product
\begin{array}{c|ccc}
\cdot & 0 & 5 & 10 \\
\hline 0 & 0 & 0 & 0 \\
5 & 0 & 10 & 5 \\
10 & 0 & 5 & 10
\end{array}
\begin{array}{l|lll}
. & 0 & 1 & 2 \\
\hline 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 \\
2 & 0 & 2 & 1
\end{array}
If I choose 0 $\mapsto 0$, $10 \mapsto 1$, because 10 and 1 are units of those rings, respectively and $5 \mapsto 2$, there is no problem to prove that $f(xy)=f(x)f(y)$, but
$$f(5+5)=f(10)=1 \neq 2=1+1=f(5)+f(5)$$
Did I do anything wrong?
| You are almost right.
$f(x+y)=f(x)+f(y)$ also holds as
$$f(5+5)=f(10)=1 =2+2=f(5)+f(5)$$
The remaining argument to prove the isomorphism is to note that $f$ has inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Find all positive integer solutions for $3^x-2^y=1$.
Find all positive integer solutions for $3^x-2^y=1$.
Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-2 \implies 3(3^{x-1}-1)=2(2^{y-1}-1)$ now from this clearly $2$ is not a multiple of $3$ so it must be that $$2^{y-1}-1 \equiv 0 \pmod 3 \implies 2^{y-1} \equiv 1 \pmod 3$$ and the order of $2$ modulo $3$ is $2$ so $y-1$ is a multiple of two say $y-2=2t$.
Plugging this back to the lhs of the equation we get $$3(3^{x-1}-1)=2(2^{2t}-1)=2(2^2-1)(2^{2(t-1)}+ \dots +1) = 6(2^{2(t-1)}+ \dots +1)$$ now $3$ is not a multiple of $6$ so we must have that $$3^{x-1}-1\equiv 0 \pmod 6 \implies 3^{x-1} \equiv1 \pmod 6$$ but noticing that the only way this can happen is if $x-1=0$ i.e. $x=1$, but this contradicts our assumption that $x \ge2$ thus there are no solutions for $x \ge 2$.
Is this a correct argument? I think there are multiple ways to do the problem, but this seemed most natural to me.
| You used two times the following false statement:
$ax \equiv 0 \pmod m$ and $a$ is not multiple of $m$ so $x \equiv 0 \pmod m$
The correct statement is
$ax \equiv 0 \pmod m$ and $a$ is coprime to $m$ so $x \equiv 0 \pmod m$
The first time you used it was with $a=2$ and $m=3$ so you got a correct conclusion, but the second time was with $a=3$ and $m=6$ so that conclusion is wrong.
From $$3(3^{x-1}-1) = 6(2^{2(t-1)}+ \dots +1)$$ it follows $$3^{x-1}-1= 2(2^{2(t-1)}+ \dots +1)$$ and you may conclude that $3^{x-1}-1 \equiv 0 \pmod {\color{red}2}$ , not $\pmod 6$ as you got.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the provided question of AM-GM inequality in the description, I'm not able to figure out which manipulation was used here, in the solution?
Let $x, y, z \in \Bbb R_{> 0}$ such that $xyz=1$. Prove that $$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}$$
Now, in the first step they suggested that we can use AM-GM in the following form
$$\frac{x^3}{(1+y)(1+z)} + \frac{(1+y)}{8} + \frac{(1+z)}{8} \ge \frac{3x}{4}$$
I am wondering what manipulation they did in the first step to write the following expression as stated above? A little help is appreciated.
| The expression wasn't rewritten, those other two terms were more or less poofed into existence. Implicitly we're saying
$$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)}$$ $$ = \frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} + \frac{(1+y)}{8} + \frac{(1+z)}{8} - \frac{(1+y)}{8} - \frac{(1+z)}{8}$$
Why does that help? Because cyclicly adding and subtracting those two terms and then applying AM-GM on all three resulting expressions gives
$$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} \geq \frac{3(x+y+z)}{4} - \frac{(3+x+y+z)}{4}$$
$$ = \frac{x+y+z}{2}-\frac{3}{4}$$
Can you finish the last step?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ integer. Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$
Clearly if $n=0$ then we will have $20$ but I couldn't decide that how can I find the other integers. Any hint?
If I say that $x=\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ then we have $200+2\sqrt{10^4-n}=x^2$
| If
$$
x=\sqrt{100-\sqrt n}+\sqrt{100+\sqrt n}\tag{1}
$$
then squaring both sides gives us
$$
x^2=200+2\sqrt{100^2-n}
$$
Rearranging and squaring again gives
$$
n=100x^2-\frac14x^4\tag{2}
$$
For this to be an integer greater than zero, $x$ must be non-zero and even.
From (1) we have that $\sqrt n\leq 100$. This gives us $x>\sqrt {200}\approx14.2$, so since $x$ is an integer, $x\geq 16$. Also from (1), $n>0$ implies that $x<20$, so again by the integer condition, $x\leq 18$.
There are only two possibilities, $x=16$ or $x=18$. (2) is decreasing for $x>10$, so we have $x=18$ as the solution, and $n=6156$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. The question asks to find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. I have done the following work below. Could you please provide feedback on whether or not I have done the question correctly? Thanks!
Since the planes are not scalar multiples of each other, they are not parallel or coincident, so solve the system of equations:
$$
x+2y+5z=3\\
2x+3y=1\\
$$
$$
-2x-4y-5z=-6\\
2x+3y=1\\
$$
$$
-y-5z=-5
$$
$$
\textrm{Let }z=s\\
-y-5s=-5\\
-y=-5+5s\\
y=5-5s, s=\frac{y-5}{5}
$$
$$
2x+3(5-5s)=1\\
2x+15-15s=1\\
2x=-14+15s\\
x=-7+\frac{15}{7}s, s=\frac{x+7}{\frac{15}{7}}=\frac{7x+49}{15}
$$
$$
\therefore\textrm{The symmetric equations for the intersection are:}\\
\frac{7x+49}{15}=\frac{y-5}{5}=z
$$
Edit: I realized the above solution is wrong. I will edit my solution later to my corrected one when I have time.
| Another way to do it is to find 2 points which should lie on this line;
Let $z=0$
$$x+2y=3\\
2x+3y=1\\x=-7,y=5,z=0 $$
Let $z=1$ $$x+2y=-2\\
2x+3y=1\\x=13,y=-5,z=1$$
So line through these 2 points is:
$$\frac{x+7}{20}=\frac{y-5}{-10}=\frac{z}{1}$$
where $<13,-5,1>-<-7,5,0>=<20,-10,1>$ represents the direction ratios of the line
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum value of the multiple variable function Show that the maximum value of$$xy(z-h)\left (\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right )$$is$$\left (\frac{2}{5}h\right )^5\frac{ab}{c^4}.$$
| Take the partial derivatives with respect to $x$, $y$, and $z$, then set them to $0$. For $x$,
$$y(z-h)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)+xy(z-h)\frac{2x}{a^2}=0\\y(z-h)\left(\frac{3x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)=0$$
Similarly for $y$:
$$x(z-h)\left(\frac{x^2}{a^2}+\frac{3y^2}{b^2}-\frac{z^2}{c^2}\right)=0$$
For $z$:
$$xy\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)-xy(z-h)\frac{2z}{c^2}=0\\xy\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{3z^2}{c^2}+\frac{2zh}{c^2}\right)=0$$
You know that $x\ne0$, $y\ne 0$, and $(z-h)\ne 0$ (otherwise the expression is zero). Then you have a simpler system:
$$\frac{3x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\\
\frac{x^2}{a^2}+\frac{3y^2}{b^2}-\frac{z^2}{c^2}=0\\\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{3z^2}{c^2}+\frac{2zh}{c^2}=0$$
Subtracting the first two equations above will give $$\frac{x^2}{a^2}=\frac{y^2}{b^2}$$
Adding the first two equations yield:
$$4\frac{x^2}{a^2}+4\frac{y^2}{z^2}=2\frac{z^2}{c^2}$$
Therefore $$\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac14\frac{z^2}{c^2}$$
Now plug this expression into the third simplified equation. You will get an equation in $z$. It is quadratic, but one solution $z=0$ is not good (since that yields $x=y=0$). So it's a linear equation that will give you $z$ as a function of $h$ and $c$. Then find $x$ and $y$, and plug them into the original expression.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$ 20! = (2^{2} \cdot 5)(19)(2 \cdot 3^{2})(17)(2^{4})(3 \cdot 5)(2 \cdot 7)(13)(2^{2} \cdot 3)(11)(2 \cdot 5)(3^{2})(2^{3})(7)(2 \cdot 3) (5) (2^{2}) (3) (2) $$
$$ = 2^{18} (5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $$
so it is left to compare $2^{42}$ and $(5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $.
We write the prime factors nicely as:
$$ 3^{8}5^{4}7^{2}(11) (13)(17)(19) $$
Notice
$(3)(11) > 2^{5}$,
$(13)(5)>2^{6}$,
$(19)(7) >2^{7}$,
$17 > 2^{4}$, so we now focus on
$$3^{7}5^{3}7 = 2187(5^{3})7 > 2048(5^{3})7 = 2^{11}875 >2^{11}512 = 2^{20} $$
So we have that the prime factors is larger than $2^{42}$.
| $$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$
$$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$
So it suffices to show:
$$3^7\cdot 5^4\cdot 7^2>2^{25}.$$
Now, $5\cdot 7>2^5,$ and $3^2>2^3,$ so it suffices to showing:
$$75=3\cdot 5^2>2^{6}=64$$
This shows: $$\frac{20!}{2^{60}}=\left(1+\frac{103}{4096}\right)
\left(1+\frac1{32}\right)\left(1+\frac3{32}\right)^2\left(1+\frac18\right)^3\left(1+\frac{11}{64}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the roots of $x^5+x^2-9x+3$ I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$
I can take it from here employing the quadratic and cubic formula, but how can I get the same answer w/o using any software?
| It can be factorize by adding and subtracting $3x^3$
$$x^5+x^2-9x+3=x^5-3x^3+x^2+3x^3-9x+3=x^2(x^3-3x+1)+3(x^3-3x+1)=(x^2 + 3) (x^3 - 3 x + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.0951875 ... $,
and its possible closed form is $ - \frac18\ln2\ln3$. But I don't know how to do it.
Thanks in advance.
| After integration by parts we have
\begin{gather*}
I=\int_{0}^{1/2}\frac{x}{3+4x^2}\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\,\mathrm{d}x= \underbrace{\left[\dfrac{1}{8}\ln\left(\frac{3+4x^2}{4}\right)\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\right]_{0}^{1/2}}_{=0}-\\[2ex]
\dfrac{1}{8}\int_{0}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\left(\dfrac{1}{\ln(1/2+x)(1/2+x)}+\dfrac{1}{\ln(1/2-x)(1/2-x)}\right)\,\mathrm{d}x= \\[2ex]
-\dfrac{1}{8}\int_{-1/2}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\dfrac{1}{\ln(1/2+x)(1/2+x)}\,\mathrm{d}x.
\end{gather*}
Now substitute $1/2+x=y$. We get
\begin{equation*}
I= -\dfrac{1}{8}\int_{0}^{1}\dfrac{\ln(y^2-y+1)}{y\ln(y)}\,\mathrm{d}y.
\end{equation*}
Finally substitute $y=e^{-t}$. Then
\begin{equation*}
I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-2t}-e^{-t}+1)}{t}\,\mathrm{d}t=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-3t}+1)-\ln(e^{-t}+1)}{t}\,\mathrm{d}t.
\end{equation*}
But this is a Frullani integral. See
https://en.wikipedia.org/wiki/Frullani_integral
Put $f(t)=\ln(e^{-t}+1)$. Then
\begin{equation*}
I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{f(3t)-f(t)}{t}\,\mathrm{d}t=\dfrac{1}{8}(f(\infty)-f(0))\ln\left(\dfrac{3}{1}\right) =-\dfrac{1}{8}\ln(2)\ln(3).
\end{equation*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I can then just apply the definition of the area of a surface
$$A(S) = \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy$$
I consider the parametrization $\Phi (x,y) = (x, y, \sqrt{x^2 + y^2 - 1}) \ $. Then $$\begin{cases} \Phi_x = (1,0,\displaystyle \frac{-x}{\sqrt{x^2 + y^2 - 1}}) \\ \Phi_y = (0,1,\displaystyle \frac{-y}{\sqrt{x^2 + y^2 - 1}})\end{cases} \Longrightarrow \Phi_x \times \Phi_y = (\frac{x}{\sqrt{x^2 + y^2 - 1}},\frac{y}{\sqrt{x^2 + y^2 - 1}},1)$$
Then
$$|| \Phi_x \times \Phi_y||= \displaystyle \sqrt{\frac{x^2}{x^2 + y^2 - 1} + \frac{y^2}{x^2 + y^2 - 1} + 1} = \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1} $$
As we work in a symettric surface, we'll consider $z \in [0, \sqrt 3]$ and simply multiply the result by two. Then, the parametrization goes from $D$ to $\mathbb R^3$, $\Phi : D \subset \mathbb R^2 \rightarrow \mathbb R^3$, being $D$ the following domain
$$D = \lbrace (x,y) \in \mathbb R^2 : 1 \leq x^2 + y^2 \leq 4 \rbrace$$
Thus we get
$$A(S) = 2 \cdot \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy = 2 \cdot \iint_D \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1}\ dx \ dy $$
Using polar coordinates, $\begin{cases} x = r \cdot \cos \theta \\ y = r \cdot \sin \theta \end{cases} : r \in [1,2] \ \& \ \theta \in [0, 2\pi]$ we get the following integral
$$A(S) = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \displaystyle \sqrt{\frac{r^2 \cos^2 \theta + r^2 \sin^2 \theta}{r^2 \cos^2 \theta + r^2 \sin^2 \theta - 1} + 1} \ dr \ d\theta = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr \ d\theta$$
$$ = 4 \pi \cdot \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr $$
The problem is that I reach the integral above that I don´t know how to tackle. I think I may have done something wrong along the process since this is a question extracted from an university exam where no computers nor calculators were avilable. Any help?
| Continue with
\begin{align}
A= &\ 4\pi \int_{1}^2 r \sqrt{\frac{r^2}{r^2 - 1} + 1} \ \overset{t=r^2-1}{dr }\\
=& \ 2\pi \int_0^3 \sqrt{\frac{2t+1}t}\ dt\overset{ibp}=
2\pi\bigg(\sqrt{t(2t+1)}\bigg|_0^3+\frac12\int_0^3 \frac1{\sqrt{t(2t+1}}dt\bigg)\\
=& \ 2\pi\bigg( \sqrt{21}+\frac1{\sqrt2}\sinh^{-1} \sqrt{2t}\bigg|_0^3\bigg)= 2\pi \bigg( \sqrt{21}+\frac1{\sqrt2}\sinh^{-1} \sqrt{6}\bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the average of rolling two dice and only taking the value of the lower dice roll? My question is related to this question, with the exception that I'm looking for the average when taking only the lower of two dice rolls.
My question is:
What is the average of rolling two dice and only taking the value of the lower dice roll?
This formula is used for the "take higher roll" case:
$$
E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47
$$
...and includes the term "$2(x−1)+1$". Could someone please explain how that term would have to be changed so that it applies for "take lower roll"? From the original example, I'm somewhat confused where "$x-1$" comes from, although it is mentioned in the comments.
PS: I would have written this question as a comment to the original question, but I lack the points to be allowed to write comments.
| Let the random variable $X$ be defined as the maximum value when two fair six-sided dice are rolled. We wish to find $E[X]$.
Since there are $k$ possible values which are at most $k$, there are $k^2$ ordered pairs of values in which both dice display a number which is at most $k$. To ensure that the maximum value is $k$, we must subtract the $(k - 1)^2$ cases in which both values are at most $k - 1$. Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the maximum value is $k$ is
$$\Pr(\max = k) = \frac{k^2 - (k - 1)^2}{36} = \frac{k^2 - (k^2 - 2k + 1)}{36} = \frac{2k - 1}{36}$$
Since the maximum value can assume any integer value from $1$ to $6$, the expected value for the maximum value is
\begin{align*}
E[X] & = \sum_{k = 1}^{6} k\Pr(X = k)\\
& = \sum_{k = 1}^{6} \frac{k(2k - 1)}{36}\\
& = \sum_{k = 1}^{6} \frac{2k^2 - k}{36}\\
& = \frac{1}{18} \sum_{k = 1}^{6} k^2 - \frac{1}{36}\sum_{k = 1}^{6} k\\
\end{align*}
Use the formulas
\begin{align*}
\sum_{k = 1}^{n} k^2 & = \frac{n(n + 1)(2n + 1)}{6}\\
\sum_{k = 1}^{n} k & = \frac{n(n + 1)}{2}
\end{align*}
with $n = 6$ to finish the calculation.
Let the random variable $Y$ be defined as the minimum value when two fair six-sided dice are rolled. We wish to find $E[Y]$.
On a six-sided die, there are $7 - k$ values which are at least $k$. Hence, there are $(7 - k)^2$ ordered pairs of values in which both dice display a number which is at least $k$. To ensure that the minimum value is $k$, we must subtract the $[7 - (k + 1)]^2 = (6 - k)^2$ cases in which both dice display a number which is at least $k + 1$. Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the minimum value is $k$ is
$$\Pr(\min = k) = \frac{(7 - k)^2 - (6 - k)^2}{36} = \frac{49 - 14k + k^2 - (36 - 12k + k^2)}{36} = \frac{13 - 2k}{36}$$
Since the minimum value can assume any integer value from $1$ to $6$, the expected value for the minimum value is
$$E[Y] = \sum_{k = 1}^{6} k\Pr(Y = k) = \sum_{k = 1}^{6} \frac{k(13 - 2k)}{36}$$
I will leave it to you to finish the calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
| Putting another answer here, a simpler one. We start by multiplying both sides of the equation by $x\sqrt{x^2-1}$. This gives $$(x^4-3x^2+x+1)\sqrt{x^2-1}+(x^5-4x^3+x^2+3x-1)=0$$
Subtracting this from the original equation, we get directly:
$$\sqrt{x^2-1}=x^3-2x+1$$
We can see that $x=1$ is a solution and other solution must be for $|x|>1$.
We now look at $f(x)=x^3-2x+1-\sqrt{x^2-1}$. Its derivative is $f'(x)=3x^2-2-\frac x{\sqrt{x^2-1}}$.
For $x<-1$, $f'(x)$ always positive, so there is only one solution, which is $x=-\sqrt2$.
For $x>1$, we check that $f''(x)$ is always positive. So we can have only 2 solutions. We already have $x=1$ so the only one left is $x=\sqrt2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 5
} |
parametrization of the result of $\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N$ as a function of $N$ As stated in the title, what's the result as a function of N for
$$\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N 1$$
| Summing from $a$ to $b$ where $b\geqslant a-1$ has $b-a+1$ summands
$$\sum_a^b 1 = \underbrace{1+1+\cdots 1}_{\textstyle b-a+1 \text{ times}} \tag0$$
where the empty sum equals $0$ by definition. Hence
$$\begin{align}
\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N 1
&= \sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1} (N-j)\\
&= \sum_{i=1}^{N-2}\left(N\sum_{j=i+1}^{N-1}1 - \sum_{j=i+1}^{N-1}j\right)\\
&\stackrel{(1)}= \sum_{i=1}^{N-2}\left(N(N-1-i) - \frac N2(N-1) + \frac{i}2(i+1)) \right)\\
&= \frac12 \sum_{i=1}^{N-2}\left(N^2-N +i(1-2N) +i^2 \right)\\
&=: \frac12(A+B+C)
\end{align}$$
The last summand can be divided into tree parts $A$, $B$ and $C$:
*
*$A$ not depending on the summation index $i$,
*$B$ depending (linearly) on $i$ and
*$C$ depending (linearly) on $i^2$, see here
The first one is computaed using $(0)$ again:
$$A= (N-2)(N^2-N) = N(N-1)(N-2)$$
The second one uses
$$\sum_{k=1}^n k= \frac n2(n+1) \tag 1$$
and the third one uses
$$\sum_{k=1}^n k^2= \frac n6(n+1)(2n+1) \tag 2$$
$$\begin{align}
B &= (1-2N)\sum_{i=1}^{N-2}i \stackrel{(1)}= \frac12(1-2N)(N-1)(N-2) \\
C &\stackrel{(2)}= \sum_{i=1}^{N-2}i^2 = \frac16(N-1)(N-2)(2N-3) \\
B+C &= -\frac23 N(N-1)(N-2)\\
\sum = \frac12(A+B+C) &= \frac16 N(N-1)(N-2)\\
\end{align}$$
which are the Tetrahedral Numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
What does passing the base 3 test mean here? I have got this one:
Let $n$ be a positive odd integer passing the base 3 test, i.e. $3^n\equiv 3\pmod n$. Prove that $\frac{3^n-1}{2}$ also passes the base 3 test.
My attempt:
$3^n\equiv 3\pmod n$
$\Rightarrow 3^n-1\equiv 2\pmod n$
$\Rightarrow \frac{3^n-1}{2}\equiv 1 \pmod n$
Thus, $3^{\frac{3^n-1}{2}}\equiv3^1\pmod n$
Is that what it means to pass the base 3 test for $\frac{3^n-1}{2}$?
| For simpler algebra, let
$$r = \frac{3^n-1}{2} \tag{1}\label{eq1A}$$
Note the problem is actually asking to prove that, given $n$ passes the test, then $r$ also passes the "base $3$ test", i.e., $r$ is a positive, odd integer and $3^r \equiv 3 \pmod{r}$.
Since $n$ is a positive, odd integer, then $r \ge 1$ and $3^n \equiv (-1)^n \equiv 3 \pmod{4}$, so $3^n-1 \equiv 2 \pmod{4} \; \; \to \; \; \frac{3^n-1}{2} \equiv 1 \pmod{2}$, i.e., $r$ is a positive, odd integer too. From \eqref{eq1A}, we also have
$$3^n \equiv 1 \pmod{r} \tag{2}\label{eq2A}$$
Next, let
$$m = \operatorname{ord}_{r}(3) \; \; \to \; \; 3^m \equiv 1 \pmod{r} \tag{3}\label{eq3A}$$
where $\operatorname{ord}_{r}(3)$ is the multiplicative order of $3$ modulo $r$. Since $n$ passes the "base $3$ test", and \eqref{eq2A} shows that $m \mid n$, we have
$$3^n \equiv 3 \pmod{n} \; \; \to \; \; 3^n \equiv 3 \pmod{m} \tag{4}\label{eq4A}$$
As you've already basically done, $3^n - 1 \equiv 2 \pmod{m} \; \to \; r = \frac{3^n-1}{2} \equiv 1 \pmod{m}$. This means there's an integer $k$ where $r = km + 1$. Thus, using this and \eqref{eq3A}, we get that
$$3^r \equiv 3^{km + 1} \equiv \left(3^m\right)^k(3) \equiv 3 \pmod{r} \tag{5}\label{eq5A}$$
This proves that $r = \frac{3^n-1}{2}$ also passes the "base $3$ test".
| {
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How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
| Personally I'm somewhat inclined to use the fifth row of Pascal's triangle: $$
1\\
1\quad 1 \\
1\quad 2\quad 1\\
1\quad 3\quad 3\quad 1\\
1 \quad 4\quad 6\quad 4 \quad 1\\
1\quad 5\quad 10\quad 10\quad 5 \quad 1$$
Thus $$(1-x)^5=1-5x^1+10x^2-10x^3+5x^4-x^5$$.
So, $$x^4-5x^5+10x^6-10x^7+5x^8-x^9$$ is the integrand.
We get $$ 1/5-5/6+10/7-10/8+5/9-1/10=-19/30+10/56+41/90=-16/90+10/56=-8/45+5/28=1/1260$$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
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A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.
| Let $$f(x)=\sqrt{2+\sqrt{2-x}}$$ and
$$g(x)=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
See the fig below where $y=f(x)$ blue and $y=g(x)$ green and $y=x$ (red).
The solution of $x=f(x)$ is given by the intersection of red and blue lines at $x=1.6180...=\frac{1+\sqrt{5}}{2}$ (the same golden number)
The root of $x=g(x)$ is also given intersection of red and green at the golden number.
But the roots of the equation $f(x)=g(x)$ is given by two points of intersection of blue and green lines, one at the golden number and the other at $x=-0.8959...$
Also it turns out that the common root of $x=f(x), x=g(x), f(x)=g(x)$ is at $\frac{1+\sqrt{5}}{2}$ and not at $\frac{\sqrt{5}-1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to make this proof rigorous by introducing partition of numbers? Question: Let $n$ be a positive integer and $ H_n=\{A=(a_{ij})_{n×n}\in M_n(K) : a_{ij}=a_{rs} \text{ whenever }
i +j=r+s\}$. Then what is $\dim H_n$?
Proof:
For $n=2$
$H_2=\begin{pmatrix} a_{11}& a_{12}\\a_{12}&a_{22}\end{pmatrix}$
Matrix of sum of indices $J_n$:
$J_2=\begin{pmatrix}\color{red}{2} & \color{red}{3} \\{3}&\color{red}{4}\end{pmatrix}$
$\dim(H_2) =3=2\cdot 2-1$
For $n=3$
$H_3=\left(\begin{array}{cc|c} a_{11}& a_{12}&a_{13}\\ \hline a_{12}&a_{13}&a_{23}\\a_{13}&a_{23}&a_{33}\end{array}\right)$
$J_3=\begin{pmatrix}\color{red}{2} & \color{red}{3} &\color{red}{4}\\ 3 &4&\color{red}{5}\\4&{5}&\color{red}{6}\end{pmatrix}$
$\dim(H_3) =5=2\cdot 3-1$
$J_n=\left(\begin{array}{ccc|c} \color{red}{2}& \color{red}{3}&\color{red}{\ldots}&\color{red}{n+1}\\ \hline 3&4&\ldots&\color{red}{n+2}\\4&5&\ldots&\color{red}{n+3}\\\vdots&\vdots&\ldots&\color{red}{\vdots}\\n+1&n+2&\ldots&\color{red}{n+n}\end{array}\right)$
$\dim(H_n) =2n-1$
$\dim(H_n) = |\{2,3,\ldots,2n\}|=2n-1$
How to make this proof rigorous by introducing partition of numbers?
From the work I have done so far, it is clear that I can find the basis and dimension. But I think the matrix of indices can simplify the problem easily. I need some nice tricks involving elementary number theory specially integer partitions on the matrix of indices.
| Sums of indices can go from $2$ to $2n$. Define, for $2\le k\le n$ the matrix $M_k$ that has $1$ where $i+j=k$ and $0$ elsewhere.
It should be easy to prove that $\{M_k:2\le k\le 2n\}$ is a basis of your subspace. That's all, you don't need to count the number of nonzero entries in $M_k$.
Let's look at a typical matrix in the set when $n=3$. The possible sums are $2,3,4,5,6$ and we get
$$
\begin{bmatrix}
a & b & c \\
b & c & d \\
c & d & e
\end{bmatrix}=
a\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}+
b\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}+
c\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}+
d\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}+
e\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
$$
This should give the idea of how to find the basis in the general case.
| {
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How to prove that $\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$? $$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$$
Equality seems to be when x = 1
I have managed to show that the derivative is 0 at x = 1, and that this is a minimum (by the second derivative test), but I am stuck on how to show that this is the only minimum.
If $f(x)$ is defined as the LHS, then we have
$$\begin{array}{l}f\left(x\right)=\sqrt{\frac{x^2-1+2}{x+1}}+\frac{2}{\sqrt{x}+1}=\sqrt{x-1+\frac{2}{x+1}}+\frac{2}{\sqrt{x}+1}\\
f^{\prime}\left(x\right)=\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}-\frac{2}{\left(\sqrt{x}+1\right)^2}\cdot\frac{1}{2\sqrt{x}}=0\\
\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\\
\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\left(x+1\right)^2-2\right)=2\left(x+1\right)^2\sqrt{\frac{x^2+1}{x+1}}\\
x\left(x+1+2\sqrt{x}\right)^2\left(\left(x+1\right)^2-2\right)^2=2\left(x+1\right)^4\left(\frac{x^2+1}{x+1}\right)\end{array}$$
At this point the computations become unrealistic without enlisting the help of wolfram alpha to expand everything. I did this, and was left with a massive polynomial which had the root x = 1.
Here is a graph
| Apply Cauchy-Schwarz inequality twice followed by an application of AM-GM inequality:
$\sqrt{\dfrac{x^2+1}{x+1}}+\dfrac{2}{\sqrt{x}+1}\ge \dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{x}+1}\ge\dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{2(x+1)}}\ge 2\sqrt{\dfrac{\sqrt{x+1}}{\sqrt{2}}\cdot \dfrac{2}{\sqrt{2(x+1)}}}=2$. Thus the minimum value is $2$ and this is achieved when $x = 1$.
| {
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Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$
Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$
My try: $a^3b^3c^3(a+b)(b+c)(c+a)\leq8a^2b^2c^2$ so $(a^2bc+ab^2c)(ab^2c+abc^2)(abc^2+a^2bc)\leq 8ab.bc.ca$. Then choose $ab=x,bc=y,ca=z$. Problem become $x,y,z>0$, $x+y+z=3$. Prove that $(x+y)(y+z)(z+x)\leq8xyz$. It's so wrong.
So I think maybe this inequality is wrong, I try $a=\dfrac{1}{2},b=1,c=\dfrac{5}{3}$ and it still right. Can you help me for any ideas? Thank you!
| Your idea is briliant indeed. Put $x = ab, y = bc, z = ac$, then $x+y+z = 3$, and we prove: $(x+y)(y+z)(z+x)\le 8$. By AM-GM inequality: $(x+y)(y+z)(z+x) \le \left(\dfrac{x+y+y+z+z+x}{3}\right)^3= 2^3 = 8$. Equality occurs when $x+y=y+z=z+x \implies x = y = z = 1\implies a = b =c = 1$.
| {
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Regular heptagon area formula Be a regular heptagon. Can anyone demonstrate this formula?
I know:
Let apothem = ap
side = l
$ S=\frac{p.ap}{2}\implies S = \frac{7l.ap}{2}\\
\triangle OBC: OC^2 = a^2+l^2\\
\triangle PAB: b^2+l^2 = PB^2\\
OC^2-a^2=PB^2-b^2$
but i can't go on
|
Hints: You can use this formula to find area of regular polygon:
$$A=\frac {n\cdot l^2}{4 tan \frac {\pi}n}$$
where n is number of sides, $l$. is length of side. For $l=1$ and $n=7$ we get $A=3.63..$
Now you can find a as :
$\angle KBC=\angle OFH$
because their rays are perpendicular on each other. So we have:
$\triangle KBC\sim \triangle HOF$
$\frac{KB=a}{FH}\frac {BC=1}{OF=r}$
$r=\frac{IF=1/2}{\sin \frac{2\pi}{14}}$
You finally find $ r\approx 1.2$ and $a=0.78$. For b we have:
$b=FL+a$
You have to show $\angle GFM=\angle HFE\approx 25.7^o$ and in triangle FGM you find:
$FL\approx 0.97\Rightarrow b=0.97+0.87\approx 1.76$
$A=\sqrt 7 a\cdot b\approx 3.63$
That is the formula is correct.
I you want to find formula, then we have:
Sum of angles of heptagon s is:
$s=(2\times 7-4)\frac{\pi}2=5\pi$
$\angle ABC=\frac{5\pi}7$
$\angle KBC=\frac{5\pi}7-\frac{\pi}2=\frac{3\pi}{14} $
$a=1\cdot \cos KBC=\cos\frac{3\pi}{14}\space\space\space\space (1)$
$b_1=FH= FE\cos EFH$
$\angle EFH=\frac {5\pi}7-\frac {\pi}2-\frac{\angle EFH}2$
$\Rightarrow \angle EFH=\frac {\pi}7$
$\Rightarrow b_1=FH=\frac{\pi}7$
$b=a+b_1=\cos \frac{\pi}7+\cos\frac {3\pi}{14}\space\space\space\space (2)$
$A=\frac{7\cdot \cos\frac{\pi}7}{4\cdot \sin \frac {\pi}7}\space\space\space\space (3)$
now you have to find $A=\sqrt 7 ab$ from relation (1) , (2) and (3).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
Product of $i^4+4$ is a perfect square The following is the Problem $50$ (Folklore) from this pdf:
Determine all positive integers $k$ such that $\prod_{i=1}^k (i^4+4)$ is a perfect square.
I realised that $i^4+4$ is just the Sophie-Germain identity, so the problem resolves to proving $2(k^2+2k+2)(k^2+1)$ is square, and I realised $(k^2+2k+2)(k^2+1)=(k^2+k+1)^2+1,$ so we want to prove $(k^2+k+1)^2+1=2n^2$ (because if $2x$ is a perfect square, $x$ is $n^2$). This makes me think of the Pell equation, but it seems to be too complicated.
Any help would be much appreciated, thanks in advance.
| I suppose this question requires to find all integers $k$ satisfying the statement than proving that statement is true ($k=1$ is false since it evaluates to 5)
So we can start from $2(k^2+1)(k^2+2k+2)$ which you found. For perfect squares we need two factors of the number to be equal:
$2(k^2+1)=k^2+2k+2_{---(1)}\\OR\\2(k^2+2k+2)=k^2+1_{---(2)}\\OR\\(k^2+1)(k^2+2k+2)=2_{---(3)}$
Solving $(1)$, we get $k=2 \: or \: 0$
Solving $(2)$, we get $k=-3 \: or \: -1$
Solving $(3)$, we get $k=0 \: or \: -1$
We can reject the solutions 0, -1 and -3 since $k\geqslant1$, therefore the only answer is 2.
Edit: Still figuring out how to find the integers k for non-equal factors like $2\times8=4^2$
For others who don't know how to find $2(k^2+1)(k^2+2k+2)$
$i^4 + 4\\= i^4 + 4i^2 + 4 - 4i^2\\=(i^2+2)^2-(2i)^2\\=(i^2+2i+2)(i^2-2i+2)\\=[(i+1)^2+1][(i-1)^2+1]$
So the question resolves to finding integers k making $\prod_{i=1}^{k}[(i+1)^2+1][(i-1)^2+1]$ a perfect square
Observe that for the graphs $L_1:y=(x+1)^2+1$ and $L_2:y=(x-1)^2+1$, $L_1$ is just 2 units on the left of $L_2$ which means substituting some n+2 into $L_2$ gives the same result as substituting n into $L_1$
Then we can get back to the given expression:
$\prod_{i=1}^{k}(i^4+4)\\=\prod_{i=1}^{k}[(i+1)^2+1][(i-1)^2+1]$
For $k\geqslant3$, there will be $k-2$ squares in the middle of the expansion. So the expressions unknown to be perfect squares are the two terms of both ends:
i.e. $[(1-1)^2+1][(2-1)^2+1] \\ [(k-1+1)^2+1][(k+1)^2+1]\\=2(k^2+1)(k^2+2k+2)$
| {
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"question_score": "5",
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Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
| A proof based on the definition. Let $\epsilon>0$ such that $\epsilon<1$. We are seeking a $\delta$ such that if $|x|<\delta$ we get:
$\dfrac{x^{2}}{x^{2}+2x+1}<\epsilon^{2}$ which is equivalent to,
$\dfrac{\epsilon^{2}-\epsilon}{1-\epsilon^{2}}<x<\dfrac{\epsilon^{2}+\epsilon}{1-\epsilon^{2}}$. Now take $\delta=\dfrac{1}{2}\dfrac{\epsilon-\epsilon^{2}}{1-\epsilon^{2}}$.
If $|x|<\delta$ we get $-\delta<x<\delta$.
Clearly $\delta<\dfrac{\epsilon^{2}+\epsilon}{1-\epsilon^{2}}$. Also
$-\delta>\dfrac{\epsilon^{2}-\epsilon}{1-\epsilon^{2}}$. Thus $|x|<\delta$ implies $\left|\dfrac{x}{x+1} \right|<\epsilon$.
If $\epsilon_{1}\geq\,1$,then clearly for the same $\delta$ we get :
$\left|\dfrac{x}{x+1} \right|<\epsilon$$<1\leq\epsilon_{1}$ and the result is proved!!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to use Mittag Leffler expansion (if possible) to solve the following integrals? In a physics related context I've been trying to solve the following two integrals:
(i)
$$
\text{int}_1 = \int_{-\infty}^\infty \csc\left( \frac{\pi+2 i z}{2 \sqrt{2} } \right) \text{sech}^2(z) \,dz
$$
(ii)
$$
\text{int}_2 = \int_{-\infty}^\infty \frac{\csc\left( \frac{\pi+2 i z}{2 \sqrt{2} } \right) \text{sech}^2(z)}{\pi + 2 i z} \,dz
$$
My strategy to solve them was to expand $\csc(a+bz)$ and/or $\text{sech}^2(z)$ using Mittag-Lefflers-theorem and then to swap the order of integration and summation. Unfortunately, I don't see any way to simplify anything in those expressions and the product of the two series is quite nasty as well (does it even converge?).
Maybe one can use the fact that only the real parts are even and therefore non-vanishing for the bounds of integration.
Do you have any other strategy to solve those integrals or an idea how to use the approach that I tried?
| To solve $int_2$, I tried the following using Mittag-Lefflers' theorem and the corresponding expansions of $\csc(x)$ and $\text{sech}^2(x)$ given by
$$
\text{csc}(x) = \underbrace{\frac{1}{x}}_{A_1} + \underbrace{2 x \sum_{k=1}^\infty \frac{(-1)^k}{x^2 - (\pi k)^2}}_{A_2}, \forall \frac{x}{\pi} \notin \mathbb{Z} \\
\text{with } \ x = \frac{\pi + 2 i z}{2 \sqrt{2}}
$$
and
$$
\text{sech}^2(z)= \underbrace{\frac{8 (\pi^2 - 4 z^2}{(\pi^2 + 4 z^2)^2}}_{B_1} - \overbrace{\sum_{k=1}^{\infty} \frac{1}{\left( z - (k+\frac{1}{2}) i \pi \right)^2}}^{B_{21}} - \overbrace{\sum_{k=1}^{\infty} \frac{1}{\left( z + (k+\frac{1}{2}) i \pi \right)^2}}^{B_{22}}, \forall \frac{i z}{\pi} \pm \frac{1}{2} \notin \mathbb{Z} \ .
$$
Applying those to the problem, the integrand of $int_2$ now consists of 5 contributions:
$$
int_2 = I_2 = \int_{-\infty}^\infty \frac{A_1 B_1 - A_1 B_2 + B_1 A_2 - A_2 B_{21} - A_2 B_{22}}{\pi+2 iz} \,dz \ .
$$
Using Cauchy's product formula $\left( \sum_{i=1}^{\infty} a_i \right)\left( \sum_{j=1}^{\infty} b_j \right) = \sum_{k \geq 1} \sum_{l=1}^k a_l b_{k+1-l}$, I get for the addends $A_2 B_{2i}$ (summation is now from $k: 1 \to \infty$ (first sum) and $l: 1\to k$ (second sum))
$$
A_2 B_{21} = \frac{(-1)^l}{\left(-\pi ^2 l^2+\frac{1}{8} (\pi +2 i z)^2\right) \left(z-i \pi \left(k-l+\frac{3}{2}\right)\right)^2}
$$
and
$$
A_2 B_{22} = \frac{(-1)^l}{\left(-\pi ^2 l^2+\frac{1}{8} (\pi +2 i z)^2\right) \left(z+i \pi \left(k-l+\frac{3}{2}\right)\right)^2} \ .
$$
Swapping the order of summation and integration and performing the integrals, I get the results
$$
\text{int}(A_1 B_1) = \frac{2 \sqrt{2}}{\pi ^2} \ ,\\
\text{int}(A_1 B_2) = -\frac{2 \sqrt{2}}{\pi ^2 (k+1)^3} \ ,\\
\text{int}(B_1 A_2) = -\frac{(-1)^k \left(4 k^2+2 \sqrt{2} k+1\right)}{2 k^3 \left(\pi \sqrt{2} k+\pi \right)^2} \ ,\\
\text{int}(A_2 B_{21}) = \frac{(-1)^l}{2 \pi ^3 l^2 \left(2 \left(\sqrt{2}-1\right) (k+1) l+(k+1)^2+\left(3-2 \sqrt{2}\right) l^2\right)} \text{and}\\
\text{int}(A_2 B_{22}) = \frac{(-1)^l \left(\frac{2}{(k-l+2)^2}-\frac{1}{\left(k+\left(\sqrt{2}-1\right) l+2\right)^2}\right)}{2 \pi ^3 l^2} \ .
$$
Performing the sums, I get (using Lerch $\Phi$-function and Riemann $\zeta$ function)
$$
\sum_{k=1}^\infty \text{int}(A_1 B_2) = -\frac{2 \sqrt{2} (\zeta (3)-1)}{\pi ^2} \ \text{and} \\
\sum_{k=1}^\infty \text{int}(B_1 A_2) = \frac{-8 \Phi \left(-1,1,1+\frac{1}{\sqrt{2}}\right)+3 \zeta (3)-\sqrt{2} \zeta \left(2,\frac{1}{4} \left(\sqrt{2}+2\right)\right)+\sqrt{2} \zeta \left(2,\frac{1}{4} \left(\sqrt{2}+4\right)\right)+\log (256)}{8 \pi ^2} \ .
$$
For the last two terms in the integral
$$
\sum_{k=1}^{\infty} \sum_{l=1}^k \text{int}(A_2 B_{21}) \quad \text{and} \quad \sum_{k=1}^{\infty} \sum_{l=1}^k \text{int}(A_2 B_{22})
$$
I don't find those nice solutions... Plotting two double series' $\sum_k^{k_{max}}\sum_{l}^k$ for $1 \leq k \leq k_{max} = 200$ however, one could get the intuition, that the expressions converge to something (see figure below) - Does anyone have an idea of how to find the corresponding expressions and do you think that this approach is correct?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$, with $n$=odd? In previous post, I got the answer that $\gcd \left(\frac{p^2+1}{2}, \frac{p^5-1}{2} \right)=1$, where $p$ is prime number.
I am looking for more general case, that is for $p$ prime,
When is $\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1$ ?
where $t=r+s$ such that $\gcd(r,s)=1$.
I am excluding the cases $r=1=s$.
In the previous post, it was $r=2,~t=5$. So $t=5=2+3=r+s$ with $s=3$ so that $\gcd(r,s)=1$.
In this current question:
Since $\gcd(r,s)=1$, we also have $\gcd(r,t)=1$.
I have the following intuition:
Case-I:
Assume $r$=even and $s$=odd so that $\gcd(r,s)=1$ as well as $\gcd(r,t)=1$. We can also assume $r$=odd and $s$=even.
I think the same strategies of previous post can be applied to show that $$\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1.$$
Case II:
The problem arises when both $r$ and $s$ are odd numbers so that $t=r+s$ is even.
If I take $r=3, ~s=5$, then $t=8$.
For prime $p=3$, $\frac{p^r+1}{2}=\frac{3^3+1}{2}=14$ and $\frac{p^t-1}{2}=\frac{3^8-1}{2}=3280$ so that the gcd is $2$ at least.
For other primes also we can find gcd is not $1$.
So I think it is possible only for Case I, where among $r$ and $s$, one is odd and another is even so that $t$ is odd.
In other word, $t$ can not be even number.
But I need to be ensured with a general method.
So the question reduces to
How to prove $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$ ?
provided $\gcd(m,n)=1$ and $n$ is odd number and $p$ is prime number.
Thanks
| Suppose that
$$\gcd\left(\frac{p^m + 1}{2}, \frac{p^n - 1}{2}\right) \neq 1.$$
There are two cases:
*
*Both $p^m + 1$ and $p^n - 1$ are multiples of $4$.
*Both $p^m + 1$ and $p^n - 1$ are multiples of the same odd prime $q$.
In case 1, $p^m + 1 \equiv 0 \pmod 4$ implies that $p^m \equiv 3 \pmod 4$, so that $p \equiv 3 \pmod 4$ and $m$ is odd. But then $p^n \equiv 3 \pmod 4$ as well, since $n$ is odd. This contradicts the fact that $p^n - 1$ is a multiple of $4$.
In case 2, let $r$ be the order of $p$ modulo $q$. Since $p^m \equiv -1 \pmod q$, $r$ must be even (as $r \mid 2m$, but $r \not \mid m$). But this makes it impossible for $p^n \equiv 1 \pmod q$, since $n$ is odd and thus cannot be a multiple of $r$.
Since neither case is possible, it must be that $\gcd\left(\frac{p^m + 1}{2}, \frac{p^n - 1}{2}\right) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $|ax^2+bx+c|\le100$ for all $|x|\le 1$, What is the maxima for $|a|+|b|+|c|$ Here is a similar problem posted before. I try to use this method to solve this problem.
$$f(-1)=a-b+c, f(0)=c, f(1)=a+b+c$$
So we have $|c|=|f(0)|\le 100$
$$|2a|=|f(-1)-2f(0)+f(1)|\le|f(-1)|+2|f(0)|+|f(1)|\le400$$
So we have $|a|\le200$
But how to find an upper bound for $b$?
Due to the symmetry, I make a guess for the maxima of $|a|+|b|+|c|$ occurring when $b=0$, then we have
$$y=200x^2-100~~~\text{or}~~~y=-200x^2+100$$
But is there a rigorous way to prove it?
| Let $|ax^2+bx+c|\le100$ for all $|x|\le 1$
Firstly if $x=0, |c|\le 100(1)$
And if $x = \dfrac{1}{2}, then \left| \dfrac{a}{4}+\dfrac{b}{2}+c \right| < 100\Longrightarrow| a + 2b + 4c | < 400$
$| a | = | 2a + 2b + 2c - (a + 2b + 4c) + 2c | < 2 | a + b + c | + | a + 2b + 4c | + 2 | c | < 200+ 400 + 200< 800$ (if $x=1$ then $|a+b+c|\le 100$)\ then $|a|\le 800(2)$
$ | b | = | a + 2b + 4c - (a + b + c) - 3c | < | a + 2b + 4c | + | a + b + c | + 3 | c | < 400 + 100 + 300< 800$\ then $| b | < 800 (3)$
Finally $| a | + | b | + | c | \le 100+800+800\le 1700$ i.e $| a | + | b | + | c | \le1700$
| {
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ and
$\frac{1}{x}=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$
How can I find $x^{2022}+\frac{1}{x^{2022}}$?
| Your answers for $x=(1\pm i)/\sqrt2$, $1/x=(1\mp i)/\sqrt2$ are correct, now notice that
$$
{1\over x}={\bar x},
$$
the "complex conjugate", and
$$
x^2+{\bar x}^2={1+2i-1+1-2i-1\over 2}=0
$$
and $x^4={\bar x}^4=-1$,
so
$$
\begin{align}
x^{2022}+1/x^{2022}&=x^{4\times505}x^2+{\bar x}^{4\times505}{\bar x}^2\\
&=(-1)^{505}(x^2+{\bar x}^2)\\
&=0
\end{align}
$$.
| {
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"source": "stackexchange",
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Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) \ \ \text{ none of } (\text{a})-(\text{c}).\end{equation*}
$$$$
I have done the following :
(a) It cannot be true since $2\cdot \left (2^{2022}-1\right )$ exceeds the sum of the series, which is a subseries of the infinite geometric series $\left \{\frac{1}{2^n}\right \}$ which converges to
$\frac{1}{1-\frac{1}{2}}=2$.
So we have that $2\cdot \left (2^{2022}-1\right )\gg2>S$.
(b) $\frac{2^{2022}-1}{2}=2^{2021}-\frac{1}{2}\gg2>S$
Therefore it cannot be true.
(c) We want to check if $S=\frac{2^{2022}-1}{2^{2021}}=2-\frac{1}{2^{2021}}$.
Let $\hat{S}$ be the sum of the infinite series : \begin{equation*}\hat{S}=1+\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2^n}=2 \text{ when } n\to \infty\end{equation*}
We therefore want to check if $S=\hat{S}-\frac{1}{2^{2021}}$ or $\hat{S}-S=\frac{1}{2^{2021}}$ ?
We know that $\hat{S}-S=\frac{1}{2^{2023}}+\frac{1}{2^{2024}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty \ \ \ (\star)$
We can rewrite $\frac{1}{2^{2021}}=\frac{2}{2^{2022}}=\frac{1}{2^{2022}}+\frac{1}{2^{2022}}=\frac{1}{2^{2022}}+\frac{2}{2^{2023}} \Rightarrow$ etc $\Rightarrow \frac{1}{2^{2021}}=\frac{1}{2^{2022}}+\frac{1}{2^{2023}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty$
Comparison with (⋆) gives that $\frac{1}{2^{2021}}>\hat{S}-S \Rightarrow $ c) cannot be true.
So (d) must be the correct answer.
$$$$
Is everything correct ?
| If $S = 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2021}} + \frac{1}{2^{2022}}$, then just add $\color{blue}{\frac{1}{2^{2022}}}$ to both sides to get the result:
\begin{align}
S + \color{blue}{\frac{1}{2^{2022}}} &= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \frac{1}{2^{2021}} + \underbrace{\frac{1}{2^{2022}} + \color{blue}{\frac{1}{2^{2022}}}}\\
&= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \underbrace{\frac{1}{2^{2021}} + \ \ \ \ \ \ \ \frac{1}{2^{2021}}}\\
&= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2^{2020}}\\
&= \cdots\\
&= 1 + 1\\
&= 2\\
\ \\
\implies S &= 2 - \frac{1}{2^{2022}}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying indefinite integral (possibly in terms of elliptic integrals) I would like to simplify the following integral
$$\int ds \frac{s^n}{\sqrt{1-s^2}}\frac{1}{\sqrt{-(s-s_1)(s-s_2)(s-s_3)}}$$.
for $n=0, 1$, and $2$. Here, $s_1$, $s_2$, and $s_3$ are constants (they come from the roots of a complicated cubic polynomial). I'm pretty sure that this integral can be rewritten in terms of elliptic integrals for each $n$, but I'm not sure how to proceed. Any help would be greatly appreciated!
| I will write here a (very) partial answer to your problem.
First of all, I am going to consider only one case, that is $n = 0$.
Even in this case, it seems like no expression exists for a "primitive" of that integral.
However if we make a deal (a bad one) we can manage to find an expression. The deal is: one of the three roots must be zero. If that happens, then we have a very nasty expression for the integral.
I will use $a, b, c$ instead of $s_1, s_2, s_3$ to avoid confusion and to write clear.
In the first case, we need to assume $c = 0$ to get an expression for the integral, as said.
$$\int \frac{\text{d}s}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}} = -\frac{\color{blue}{2 (s-1) s (b-s) \sqrt{\frac{(s+1) (a-b)}{(b+1) (a-s)}} \left(b F\left(\sin ^{-1}\left(\sqrt{\frac{(a-1) (b-s)}{(b-1) (a-s)}}\right)|\frac{(a+1) (b-1)}{(a-1) (b+1)}\right)+(a-b) \Pi \left(\frac{a (b-1)}{(a-1) b};\sin ^{-1}\left(\sqrt{\frac{(a-1) (b-s)}{(b-1) (a-s)}}\right)|\frac{(a+1) (b-1)}{(a-1) (b+1)}\right)\right)}}{\color{red}{a (b-1) b \sqrt{1-s^2} \sqrt{\frac{(a-1) (s-1) (a-b) (b-s)}{(b-1)^2 (a-s)^2}} \sqrt{s^2 (a-s) (s-b)}}}$$
Where $F$ denotes the complete Elliptic integral of the first kind and $\Pi$ denotes the complete elliptic integral of the third kind.
Second case: $n = 1$.
Here we get, assuming $c = 0$ as before.
$$\int \frac{s}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}}\text{d}s = \frac{\color{blue}{2 \sqrt{\frac{(s-1) (a-b)}{(b-1) (a-s)}} \sqrt{\frac{(s+1) (a-b)}{(b+1) (a-s)}} \sqrt{-s^2 (s-a) (s-b)} F\left(\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)}}{\color{red}{s \sqrt{1-s^2} (a-b) \sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}}}$$
With the same meaning of $F$ as before.
Again I did not find any expression for $c\neq 0$.
Third Case: $n = 2$. Always assuming $c = 0$ we have
$$\int \frac{s^2}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}}\text{d}s = -\frac{\color{blue}{2 s (s+1) (b-s) \sqrt{\frac{(s-1) (a-b)}{(b-1) (a-s)}} \left(a F\left(\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)+(b-a) \Pi \left(\frac{b+1}{a+1};\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)\right)}}{\color{red}{(b+1) \sqrt{1-s^2} \sqrt{\frac{(a+1) (s+1) (a-b) (b-s)}{(b+1)^2 (a-s)^2}} \sqrt{s^2 (a-s) (s-b)}}}$$
Same meaning of $F$ and $\Pi$.
Again, no expression found for $c\neq 0$.
I apologise for thise answer which does not cover the general cases you presented here. However this could be a start.
Otherwise, you need numerical integration.
Postface
I coloured the terms because the writing is rather nasty, and confusing too. In each of the expressions, the results come into a form of fraction: the $\color{blue}{BLUE}$ part is the numerator and the $\color{red}{RED}$ part is the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| If we take mod 3 and $y=2z+1$ then $2^y\equiv -1$ but $$x^2+7x+4 \equiv 0,1$$ a contradiction, so $y=2z$. Let $2^z = t$, then we have
*
*If $x>0$ then
$$(x+2)^2\leq \underbrace{x^2+7x+4}_{t^2} \leq (x+4)^2 \implies t = x+3\implies x=5$$
*
*If $x<0$ then $x=-n$ with positive $n$. So $$(n-4)^2\leq \underbrace{n^2-7n+4}_{t^2} \leq (n-2)^2 \implies t = n-3\implies n=-5$$so no solution.
*If $x=0$ then $y= 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as follow
$z = r{e^{i\theta }}$& $\overline z = r{e^{ - i\theta }}$
$r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = i\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right) \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = {e^{i\frac{\pi }{2}}}\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right)$
$ \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = \left( {r{e^{ - i\left( {\theta - \frac{\pi }{2}} \right)}} + {r^2}{e^{i\left( {2\theta + \frac{\pi }{2}} \right)}}} \right)$
$ \Rightarrow r\left( {\cos \theta - i\sin \theta } \right) - {r^2}\left( {\cos 2\theta + i\sin 2\theta } \right) = \left( {r\left( {\sin \theta + i\cos \theta } \right) + {r^2}\left( { - \sin 2\theta + i\cos 2\theta } \right)} \right)$
$ \Rightarrow r\cos \theta - {r^2}\cos 2\theta - r\sin \theta + {r^2}\sin 2\theta - i\left( {r\sin \theta - {r^2}\sin 2\theta - r\cos \theta - {r^2}\cos 2\theta } \right) = 0$
Not able to proceed further
| First of all, clearly $z = 0$ works, so let's just assume $z \neq 0$.
Let $z = r e^{i \theta}$. We have:
\begin{align}
\bar{z} - z^2 &= i(\bar{z} + z^2) \\
(1 - i)\bar{z} &= (1 + i)z^2 \\
\frac{1 - i}{1 + i} &= \frac{z^2}{\bar{z}} \\
e^{-i \pi/2} &= r e^{3i\theta}
\end{align}
This shows that $r = 1$ and $\theta = -\frac{\pi}{6} + \frac{2 \pi}{3} n$ for some $n \in \mathbb{Z}$.
In summary, the solutions are $0, e^{-i\pi/6}, e^{i 3\pi / 6}$ and $e^{i 7\pi / 6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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