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|---|---|---|
Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$
| Observe that the solution set of $x^2=x$ is $\{0,1\}$ whereas the solution set of $x=1$ is $\{1\}$. Hence dividing by $x$ is not an equivalence transformation. Instead of dividing by $x$ factor it out: $x^2-x=0\iff x(x-1)=0$ to get $x=0$ or $x=1$.
In your first case we will get $(2x-3)(x+5-9)=0\iff 2x-3=0$ or $x-4=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find k with little oh notation? For which values of $k$ are the following true (as $x \to 0$)?
(a) $\sqrt{1+x^2} = 1 + o(x^k)$
(b) $\sqrt[3]{1+x^2} = 1 + o(x^k)$
(c) $1 - \cos(x^2) = o(x^k)$
(d) $1 - \cos^2 x = o(x^k)$
How would you find $k$ for problems like these?
| The expression
\begin{align*}
f(x)=g(x)+o(x^k)\qquad\qquad x\rightarrow 0
\end{align*}
is just another notation for
\begin{align*}
\lim_{x\rightarrow 0}\frac{f(x)-g(x)}{x^k}=0\tag{1}
\end{align*}
In case $\sqrt{1+x^2} = 1 + o(x^k)$ we take the Taylor series expansion
\begin{align*}
\sqrt{1+x^2}&=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n}x^{2n}\\
&=1+\frac{1}{2}x^2+\frac{1}{2}\left(-\frac{1}{2}\right)x^4+\cdots\tag{2}
\end{align*}
We consider according to (1) and (2)
\begin{align*}
\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-1}{x^k}
&=\lim_{x\rightarrow 0}\frac{1+\frac{1}{2}x^2-\frac{1}{4}x^4+\cdots-1}{x^k}\\
&=\lim_{x\rightarrow 0}\frac{\frac{1}{2}x^2-\frac{1}{4}x^4+\cdots}{x^k}\\
&=0\tag{3}
\end{align*}
and conclude (3) is correct for each $k<2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-2)\gt0}$
${3-x^2-2x\gt0}$
${-x^2-2x+3\gt0}$
${-1(x^2+2x-3)\gt0}$
$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$
${(x-1)(x+3)\lt0}$
At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.
| Note that if $x>3\implies x-2>1$ then $\dfrac1x<\dfrac13$ and $\dfrac1{x-2}<1$ so $$\frac3{x(x-2)}<\frac3{3(3-2)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding points on a line that are closest
Find the points that give the shortest distance between the lines$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-t\\-1+2t\\-1+t\end{pmatrix}\\\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5+3s\\0\\2-s\end{pmatrix}$$
So I subtracted the second line from the first to get these two equations:
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}=0$
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}3\\ 0\\ -1\end{pmatrix}=0$
I know I am supposed to rearrange them together two get a system of equations but I am not sure how.
Any help please?
| Two lines in $\mathbb R^3$ either intersect (distance = 0) or not intersect (distance = the distance between two parallel planes on which the two lines lie accordingly). Classify it first, then use $u\times v$, where $u, v$ indicate the direction of the two lines, to find the normal for the plane. Hope that helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\pi (r^2-x^2)$ We are told to integrate $\pi (r^2-x^2)$ from $-r$ to $r$
$$V=\int_{-r}^{r}\pi(r^2-x^2)dx=2\pi\int_0^r(r^2-x^2)dx=2\pi\left[r^2x-\frac{x^3}{3}\right]_0^r=2\pi\left(r^3-\frac{r^3}{3}\right)=\frac{4}{3}\pi r^3$$
why does the integral of $r^2$ equal $r^2 x$?
| We have $$\begin{align}V&=\int_{-r}^r\pi(r^2-x^2)\,dx=\int_{-r}^r(\pi r^2-\pi x^2)\,dx=\int_{-r}^r\pi \color{red}r^2\,d\color{blue}x\end{align}-\int_{-r}^r\pi \color{blue}x^2\,d\color{blue}x$$ and since we are integrating with respect to $x$ (in blue) we treat $r$ (in red) as a constant.
Hence $$\begin{align}V&=[\pi \color{red}r^2\color{blue}x]_{-r}^r-\left[\frac\pi3x^3\right]_{-r}^r\\&=\pi r^3-\pi r^2(-r)-\left(\frac\pi3r^3-\frac\pi3(-r)^3\right)\\&=2\pi r^3-\frac23\pi r^3=\frac43\pi r^3\end{align}$$ as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing $\cos^{10}x\le 1-x^2$ for $x$ in $[0,0.5]$ I have to prove this inequality:
$$\cos^{10}x\le 1-x^2 \quad \text{for all}\quad x \in [0,0.5]$$
Is there any easy and/or elegant way to do this? I can do this with Taylor, but it's really a mess. :(
Thanks in advance
| Note that
$$\cos(x)^{10}\le 1-x^2\iff 10\log\cos x\le\log (1-x^2)$$
and
*
*$10\log\cos x\le 10 \log
(1-\frac12x^2+\frac1{24}x^4)\le-5x^2+\frac5{12}x^4\quad$ by $\log(1-x)<x$
*$\log (1-x^2)\ge-x^2-x^4$ to be proved
and since
$$-5x^2+\frac5{12}x^4\le -x^2-x^4 \iff4x^2-\frac{17}{12}x^4\ge 0\iff x^2(4-\frac{17}{12} x^2)\ge 0\\\iff -\sqrt{\frac{48}{17}}\le x\le \sqrt{\frac{48}{17}}$$
the inequality is proved.
To prove
*
*$\log (1-x^2)\ge-x^2-x^4$
let consider
*
*$g(x)=\log (1-x^2)\ge+x^2+x^4$
and note that
*
*$g(0)=0$
*$g'(x)=\frac{2x^3-4x^5}{1-x^2}$ and $g'(x)>0$ for $x\in(0,1/2)$
and thus $g(x)\ge0$ for $x\in[0,1/2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Integrating function of log $$
\int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x-2)\right)~{\rm d}x
$$
In the above given question, I am unable to integrate the function inside log. I am substituting $(x-2) = t$ and then reducing the question into a simpler form with only $t$ as a variable, but i am getting a function of $t$ inside log which I am unable to open or integrate. Please provide a proper approach to my way or any suggestions are well appreciated.
| First, make the substitution you've already mentioned:
$$\int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x - 2)\right)~{\rm d}x$$
$$t=x-2$$
$${\rm d}t={\rm d}x$$
$$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t$$
Now, we solve using integration by parts where $u = \ln\left(\sqrt{t^2+4} + t\right)$ and ${\rm d}v = {\rm d}t$. Calculating further, we get that $v=t$ and
$${\rm d}u = \frac{1}{\sqrt{t^2+4} + t} \cdot \frac{\sqrt{t^2+4} + t}{\sqrt{t^2+4}}~{\rm d}t = \frac{{\rm d}t}{\sqrt{t^2+4}}$$
This gives us
$$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = \left.t \ln\left(\sqrt{t^2+4} + t\right)\right\rvert_{-2}^2 - \int_{-2}^2 \frac{t}{\sqrt{t^2+4}}~{\rm d}t$$
Now, we may make the substitutions $w=t^2+4$ and ${\rm d}w = 2t~{\rm d}t$ (which also means $\frac{1}{2}{\rm d}w=t~{\rm d}t$), as well as evaluating the solved part of the integral.
$$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = \left[2\ln(2\sqrt{2} + 2) + 2\ln(2\sqrt{2} - 2)\right] - \int_8^8 \frac{{\rm d}w}{\sqrt{w}}$$
Because the limits of integration on our integral are now equal, we can eliminate it altogether and continue to evaluate.
$$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = 2\left[\ln(2\sqrt{2} + 2) + \ln(2\sqrt{2} - 2)\right]$$
$$= 2\ln[(2\sqrt{2} + 2)(2\sqrt{2} - 2)]$$
$$= 2\ln(8 - 4) = 2\ln4 = \ln\left(4^2\right) = \ln 16 \approx 2.773$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{dx} &= 2x^{2-1} \\
&= 2x = \text{Slope at P.}
\end{align}
$$
Now, the equation for any straight line is also satisfied for the tangent:
$$
\begin{align}
y - y_0 &= m(x - x_0) \\
\implies y - y_0 &= 2x (x - x_0) \\
\text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\
\implies y - 4 &= 2x(x+2)\\
\implies y - 4 &= 2x^2 + 4x\\
\implies y &= 2x^2 + 4x +4\\
\end{align}
$$
This is where the problem occurs. If I were to try to solve for $y$ using:
$$
y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
I'd get:
$$
\begin{align}
y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\
x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\
x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\
x &= \frac{-4 \pm 4i}{4} \\
x &= -1 \pm i
\end{align}
$$
Is this the correct direction, or did I do something wrong?
| You need a point and a slope. The point is $(2,4)$ and the slope is $y'(2)=4.$
Thus the equation is $$ y-4=4(x-2)$$ or $$ y=4x-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Help to solve a system equation: $x-y+xy=-4$; $xy(x-y)=-21$. I need to solve a system equation. Here's how it looks:
$x-y+xy=-4$
$xy(x-y)=-21$
I tried to substitute $x-y$ with $w$ and $xy$ with $t$ to simplify everything. After that I got this system equation:
$w+t=-4$
$tw=-21$
I solved this new system equation and got these results:
$t_1=-7, w_1=3$
$t_2=3, w_2=-7$
After all these steps I ended up with 2 new system equations:
1.
$xy=-7 (\leftarrow t)$
$x-y=3 (\leftarrow w)$
2.
$xy=3 (\leftarrow t)$
$x-y=-7 (\leftarrow w)$
Looks like the first one doesn't have any solutions. And I can't solve the second one. Am I doing wrong steps? Please help me to solve this system equation. Thanks.
| \begin{align}
x-y+xy&=-4
,\\
xy(x-y)&=-21
.
\end{align}
We can consider $u,v$
\begin{align}
u&=x-y
,\\
v&=xy
\end{align}
as two roots of the quadratic equation,
\begin{align}
z^2+az+b&=0
\tag{1}\label{1}
,\\
a&=-(u+v)=-(x-y+xy)=4
,\\
b&=uv=xy(x-y)=-21
,
\end{align}
so \eqref{1} is
\begin{align}
z^2+4z-21&=0
\end{align}
which has two real solutions
\begin{align}
z_{1,2}&=3,-7
.
\end{align}
Since we can not distinguish
which of the two is $u$ and which is $v$,
we need to consider both cases:
\begin{align}
\text{Case 1.}\quad
&
\begin{cases}
u&=x-y=3
,\\
v&=xy=-7
.
\end{cases}
\end{align}
And
\begin{align}
\text{Case 2.}\quad
&
\begin{cases}
u&=x-y=-7
,\\
v&=xy=3
.
\end{cases}
\end{align}
We can exploit the quadratic equation again (twice),
considering two roots, $x$ and $-y$,
since we can get the values of the sum and product of named roots
as
\begin{align}
\text{case 1.}\quad
&
\begin{cases}
x+(-y)=3
,\\
x(-y)=-xy=7
,
\end{cases}
\end{align}
and
\begin{align}
\text{case 2.}\quad
&
\begin{cases}
x+(-y)=-7
,\\
x(-y)=-xy=-3
.
\end{cases}
\end{align}
The two quadratic equations are then
\begin{align}
\text{case 1:}\quad
t^2-3t+7&=0
,\\
\text{case 2:}\quad
t^2+7t-3&=0
.
\end{align}
The first case does not have real roots,
but the second one provides two real roots,
\begin{align}
t_{1,2}&=
-\tfrac72\pm\tfrac12\,\sqrt{61}
.
\end{align}
And as before, we have to consider two options,
\begin{align}
\begin{cases}
x&=-\tfrac72+\tfrac12\,\sqrt{61}
,\\
y&=-(-\tfrac72-\tfrac12\,\sqrt{61})
.
\end{cases}
\end{align}
and
\begin{align}
\begin{cases}
x&=-\tfrac72-\tfrac12\,\sqrt{61}
,\\
y&=-(-\tfrac72+\tfrac12\,\sqrt{61})
.
\end{cases}
\end{align}
The final substitution into original pair of equations
confirms that the two solutions are indeed
\begin{align}
\begin{cases}
x&=-\tfrac72+\tfrac12\,\sqrt{61}
\\
y&=\phantom{-}\tfrac72+\tfrac12\,\sqrt{61}
\end{cases}
\end{align}
and
\begin{align}
\begin{cases}
x&=-\tfrac72-\tfrac12\,\sqrt{61}
,\\
y&=\phantom{-}\tfrac72-\tfrac12\,\sqrt{61}
.
\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.
In addition, this identity is supposed to be true:
$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$
| First we can use the identities
$$ (x+y+z)^2=x^2+y^2+z^2+2(yz+zx+xy) $$
and
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)$$
As $x+y+z=0$, we have $x^2+y^2+z^2=-2(yz+zx+xy)$ and $x^3+y^3+z^3=3xyz$.
So, R.H.S is $-xyz(yz+zx+xy)$.
From the above results, we have
\begin{align*}
(x^3+y^3+z^3)(x^2+y^2+z^2)&=x^5+y^5+z^5+x^3(y^2+z^2)+y^3(z^2+x^2)+z^3(x^2+y^2)\\
(3xyz)(-2yz-2zx-2xy)&=x^5+y^5+z^5+x^3[(y+z)^2-2yz]\\
&\quad +y^3[(z+x)^2-2zx]+z^3[(x+y)^2-2xy]\\
-6xyz(yz+zx+xy)&=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\\
-6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)-2xyz(x^2+y^2+z^2)\\
-6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)+4xyz(yz+zx+xy)\\
x^5+y^5+z^5&=-5xyz(yz+zx+xy)
\end{align*}
Hence, L.H.S. is also $-xyz(yz+zx+xy)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solving $x^2+y^2=9$, $\arctan\frac{y+2}{x+2} + \arctan\frac{y-2}{x+4} =2\arctan\frac{y}{x}$ without graphing? The system of equations:
$$\begin{align}
x^2 + y^2 &= 9 \\[6pt]
\operatorname{arctan}\frac{y+2}{x+2} + \operatorname{arctan}\frac{y-2}{x+4} &=2\,\operatorname{arctan} \frac{y}{x}
\end{align}$$
I tried to interpret the second equation by setting $x$ and $y$ as the legs of a right triangle, but I'm still unable to solve this system of of equations geometrically. Any tips?
| Essentially, the equations say that the origin is at distance $3$ from $(x,y)$ and on the angular bisector of the triangle with vertices $(x,y)$, $(x+2,y+2)$, $(x+4,y-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum length of the hypotenuse
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.
My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h(x)=x+y=x+\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h'(x)=1+\frac{\sqrt{x^2-a^2}.b-xb.\frac{x}{\sqrt{x^2-a^2}}}{x^2-a^2}=1+\frac{x^2b-a^2b-x^2b}{(x^2-a^2)^{3/2}}\\
=1+\frac{-a^2b}{(x^2-a^2)^{3/2}}=\frac{(x^2-a^2)^{3/2}-a^2b}{(x^2-a^2)^{3/2}}
$$
$$
h'(x)=0\implies (x^2-a^2)^{3/2}=a^2b\implies (x^2-a^2)^{3}=a^4b^2\\
\implies x^6-3x^4a^2+3x^2a^4-a^6=a^4b^2\implies x^6-3x^4a^2+3x^2a^4-a^6-a^4b^2=0\\
$$
How do I proceed further and find $h_{min}$ without using trigonometry ? Or is there anything wrong with my calculation ?
| Let $\angle PCB=\angle ABN=\theta$. Then $\displaystyle \cos \theta =\frac{a}{x}$ and $\displaystyle \sin \theta=\frac{b}{y}$.
Then we have to minimize $$x+y=\frac{a}{\cos \theta}+\frac{b}{\sin \theta}$$
So using Holder,s Inequality
$$(\cos^2\theta+\sin^2\theta)\cdot \bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\cdot \bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\geq \bigg(a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg)^3$$
So $$\bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bigg)\geq \bigg(a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg)^{\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$ So this is the problem..
$$
\frac{25s}{(s^2+16)(s-3)(s+3)}
$$
So what I did was...
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {A}{s^2+16}+\frac {B}{s-3}+\frac{C}{s+3}
$$
then...
$$\begin{align}
25s &= A(s-3)(s+3)+B(s^2+16)(s+3)+C(s^2+16)(s-3) \\
25s &= A(s^2-9)+B(s^3+3s^2+16s+48)+C(s^3-3s^2+16s-48) \\
25s &= As^2-9A+Bs^3+3Bs^+16Bs+48B+Cs^3-3Cs^2+16Cs-48C
\end{align}$$
rearranged to get...
$$
25s = Bs^3+Cs^3+As^2+3Bs^2-3Cs^2+16Bs+16Cs-9A+48B-48C
$$
factoring out...
$$
25s = (B+C)s^3+(A+3B-3C)s^2+(16B+16C)s-9A+48B-48C
$$
Now, this is the part where U messed up. I've been trapped with this question for more than an hour. I don't know if I made any mistakes as I've looked at it over and over again.
The answer is supposed to be:
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {-s}{s^2+16}+\frac {1}{2(s-3)}+\frac{1}{2(s+3)}
$$
Please be detailed. Thank you!
| You must write
$$\frac{25s}{(s^2+16)(s-3)(s+3)}=\frac{As+B}{s^2+16}+\frac{C}{s-3}+\frac{D}{s+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
" Let $A$ be a symmetric $2 \times 2$ matrix with the property $A^{-1} = A$. Find all possible trace values of $\operatorname{tr}A$" I need some help solving this.
I have tried:
$$
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
=\frac{1}{\operatorname{det}A}\cdot \begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}$$
I ended up with $$a=\frac{d}{\operatorname{det}A},$$
and
$$d=\frac{a}{\operatorname{det}A}.$$
Then
$$\operatorname{tr}(A)=a+d=\frac{a+d}{\operatorname{det}A},$$
but I don't really think it works.
| You have $A^{-1} = A \implies A^2 = I$. So we just calculate that for the matrix you have $$\begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c& d \end{pmatrix}\cdot\begin{pmatrix} a & b \\ c& d \end{pmatrix} = \begin{pmatrix} a^2 + bc & (a+d)\cdot b \\ (a+d)\cdot c & bc + d^2 \end{pmatrix}$$
Therefore, you get $(a+d)\cdot b = (a+d)\cdot c = 0$.
Case 1: $a+d = 0$, we're done.
Case 2: Let's assume $b$ or $c$ are $0$ then $a^2 = d^2 = 1 \implies a^2 - d^2 = 0 \implies a = \pm d$.
If $a = -d \implies a+d = 0$
If $a = d$, then $A = aI$, and $A^2 = I \implies a = \pm 1 \implies A = \pm I$
Therefore, for $A^2 = I$, we have $\text{tr} A = 0, 2 , -2$
Continuing from your chain of thought
In your calculations, you obtained the following,
$$a+d = \frac{a+d}{\det A}$$
Now, as we know if $A^2 = I$, then $\det A = \pm1$.
Therefore, for $\det A = -1$. You get $$a+d = -(a+d) \implies \text{tr} A = 0$$
For $\det A = 1$, you get $$\begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} d & -b\\ -c & a \end{pmatrix}$$
$\implies b = c = 0, a = d$. Therefore $A = aI$, and $\text{tr} A = 2a$
$$A^2 = I \implies (aI)^2 = I \implies a^2 = 1 \implies a = \pm 1$$
Hence, $\text{tr} A = \pm 2$
Comment: For a more 'clean' answer, please refer to Omnomnomnom's answer/response. I provided an elementary answer to demonstrate one exists, but the better and faster way, in my opinion, is still using eigenvalues.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Divisibility of $n^2+9$ by $n+3$ How would one find all integers $n$ such that $n+3 \vert n^2 +9$? I assume it is important that $(n+3)^2 - 6n = n+3$, but I am struggling to see how you can find all $n$, and confirm an upper bound such that there are no more such $n$.
| Alternatively:
$n+3|n^2+9 \iff$
$n+3|n^2 + 9 - n(n+3)\iff$
$n+3|-3n + 9\iff$
$n+3|-3n + 9 + 3(n+3)\iff$
$n+3|18$
$n+3 = \pm 1; \pm 2;\pm 3;\pm 6;\pm 9;\pm 18$
$-21,-12,-9,-6,-5,-4,-2,-1,0,3,6,15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2c^2(a^2 + b^2)$
At this point if we take the $LHS - RHS$ we can write the expression as the sum of squares proving the inequality.
I was wondering, is it possible to divide both sides by $c^2(a^2 + b^2)$ and show somehow that
$((a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4)/(c^2(a^2 + b^2)) ≥ 2$
I tried but was not able to.
| We need to prove that
$$(a^2+b^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or
$$c^4-2(a^2+b^2)c^2+2(a^4+b^4)\geq0$$ or
$$(c^2-a^2-b^2)^2+(a^2-b^2)^2\geq0.$$
Yes, you can prove this inequality by the dividing.
Indeed, if $c^2(a^2+b^2)=0$ then the inequality is obvious.
Let $c^2(a^2+b^2)\neq0$.
Thus, by AM-GM and Cauchy-Schwarz
$$\frac{c^4+2(a^4+b^4)}{c^2(a^2+b^2)}=\frac{c^2}{a^2+b^2}+\frac{2(a^4+b^4)}{c^2(a^2+b^2}\geq$$
$$\geq2\sqrt{\frac{c^2}{a^2+b^2}\cdot\frac{2(a^4+b^4)}{c^2(a^2+b^2)}}=2\sqrt{\frac{2(a^4+b^4)}{(a^2+b^2)^2}}=$$
$$=2\sqrt{\frac{(1+1)(a^4+b^4)}{(a^2+b^2)^2}}\geq2\sqrt{\frac{(a^2+b^2)^2}{(a^2+b^2)^2}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Edge length of an equilateral triangle if distances from a point $P$ to its vertices is given A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.
The answer is $\sqrt{169 + 60\sqrt(3)}$.
If $s$ is the edge length of the triangle, and if $x$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 12, and if $y$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 13, according to the Law of Cosines,
\begin{equation*}
s^{2} = 169 - 120\cos(x)
\end{equation*}
\begin{equation*}
s^{2} = 194 - 130\cos(y)
\end{equation*}
and since $\cos(360 - (x + y)) = \cos(x + y)$,
\begin{equation*}
s^{2} = 313 - 312\cos(x+y)
\end{equation*}
So,
\begin{align*}
\cos{x} = - \frac{s^{2} - 169}{120} , \\
\cos{y} = - \frac{s^{2} - 194}{130} , \\
\cos(x+y) = - \frac{s^{2} - 313}{312} .
\end{align*}
For any real numbers $\theta$ and $\phi$,
\begin{equation*}
\bigl[\cos{\theta}\cos{\phi} - \cos(\theta + \phi)\bigr]^{2}
= \sin^{2}\theta\sin^{2}\phi
= \bigl(1 - \cos^{2}\theta\bigr)\bigl(1 - \cos^{2}\phi\bigr) .
\end{equation*}
So,
\begin{align*}
&\left[\frac{s^{2} - 169}{120} \cdot \frac{s^{2} - 194}{130} + \frac{s^{2} - 313}{312}\right]^{2} \\
&\qquad \qquad = \left(1 - \left(\frac{s^{2} - 169}{120}\right)^{2}\right)\left(1 - \left(\frac{s^{2} - 194}{130}\right)^{2}\right)
\end{align*}
or equivalently, by multiplying by $(120^{2})(130^{2})312$,
\begin{align*}
&\Bigl[312(s^{2} - 169)(s^{2} - 194) + (120^{2})(130^{2})(s^{2} - 313)\Bigr]^{2} \\
&\qquad \qquad
= 312 \Bigl((120^{2})(130^{2}) - 130^{2}(s^{2} - 194)\Bigr)
\Bigl((120^{2})(130^{2}) - 120^{2}(s^{2} - 194)\Bigr) .
\end{align*}
How is the quartic equation in the variable $s$ to be solved?
|
Let $ABC$ be an equilateral triangle. $P$ is a point inside $\triangle ABC$ such that $PA=5$, $PB=12$ and $PC=13$. Rotate $C$ and $P$ about $A$ through $60^\circ$ to $B$ and a point $X$. Rotate $A$ and $P$ about $B$ through $60^\circ$ to $C$ and a point $Y$. Rotate $B$ and $P$ through $60^\circ$ to $A$ and a point $Z$.
Since $\triangle ABX\cong\triangle ACP$, $\triangle BCY\cong\triangle BAP$ and $\triangle CAZ\cong\triangle CBP$,
$$[AXBYCZ]=2[\triangle ABC]$$
Note that $\triangle APX$, $\triangle BPY$ and $\triangle CPZ$ are equilateral triangles of sides $5$, $12$ and $13$ respectively. Also, $\triangle PBX$, $\triangle YPC$ and $\triangle AZP$ are right-angled triangles of sides $5$-$12$-$13$.
\begin{align*}
[\triangle ABC]&=\frac{1}{2}\left[\frac{1}{2}(5)^2\sin60^\circ+\frac{1}{2}(12)^2\sin60^\circ+\frac{1}{2}(13)^2\sin60^\circ+3\times\frac{1}{2}(5)(12)\right]\\
&=\frac{169\sqrt{3}}{4}+45
\end{align*}
So,
\begin{align*}
\frac{1}{2}(AB)^2\sin60^\circ&=\frac{169\sqrt{3}}{4}+45\\
AB^2&=169+60\sqrt{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Uniform Convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$
Uniform convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$
Using Dirichlet Test, it can be shown that the it uniformly converges for $\mathbb{R}$ \ $\{0\}$.
In $x=0$ there is obviously a pointwise convergence to 0.
However, I'm struggling to make a conclusion for $\mathbb{R}$.
If you could help me find a conclusion or provide another direction, it would be appreciated.
| Note
\begin{eqnarray}
S_N(x)&=&\sum_{n=1}^{N}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}=\frac{x^2}{1+x^2}=\sum_{n=1}^{N}(-1)^{n-1}\frac{1}{(1+x^2)^{n-1}}\\
&=&\frac{x^2}{1+x^2}\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{1+\frac{1}{1+x^2}}=x^2\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{x^2+2}\\
\end{eqnarray}
and hence
\begin{eqnarray}
\left|S_N(x)-\frac{x^2}{x^2+2}\right|&=&\frac{x^2}{(1+x^2)^{N}(x^2+2)}.
\end{eqnarray}
Let
$$ f_N(x)=\frac{x}{(1+x)^{N}(x+2)}, x\ge0. $$
Solveing $f'(x)=0$ gives $x=x_N\equiv\frac{\sqrt{N^2+1}-N+1}{N}$ and noting $f_N''(x_N)<0$. Therefore $f_N(x)$ attains the max at $x=x_N$. Thus
\begin{eqnarray}
&&\max\left|S_N(x)-\frac{x^2}{x^2+2}\right|=\max f_N(x^2)=\max_{x\ge0}f_N(x)\\
&=&f_N(x_N)=\frac{\sqrt{N^2+1}-N+1}{\sqrt{N^2+1}+N+1}\left(\frac{N}{\sqrt{N^2+1}+1}\right)^N\\
&\le&\frac{\sqrt{N^2+1}-N+1}{\sqrt{N^2+1}+N+1}\\
&\le&\frac{2}{\sqrt{N^2+1}+N+1}\\
&\to&0
\end{eqnarray}
as $N\to\infty$ and thence $\{S_N(x)\}$ converges uniformly to $\frac{x^2}{x^2+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ iff $$\frac{x}{2}=\frac{y}{4}$$, then $$20\ge(2x+4y)^2$$
$$-4.47\le2x+4y\le4.47$$
| The distance from the circle's center ($ x^2+y^2 = \frac{1}{20}$) to the line $2x + 4y = 1$ is given by
$$
d = \frac{1}{\sqrt{2^2+4^2}} = \sqrt{\frac{1}{20}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Commutator Group of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$
Let $\operatorname{GL}_2(\mathbb{R})$ be the general linear group of $2\times2$ matrices and $\operatorname{SL}_2(\mathbb{R})$ the special linear group of $2 \times 2$ matrices. Show that the commutator subgroup of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$.
I can show that the commutator subgroup is contained in $\operatorname{SL}_2(\mathbb{R})$ as if $A,B \in \operatorname{GL}_2(\mathbb{R})$ then
$$
\det(ABA^{-1}B^{-1}) = \det(A)\det(B)\det(A^{-1})\det(B^{-1}) = 1.
$$
But how can I show the reverse inclusion? That is, that $\operatorname{SL}_2(\mathbb{R})$ is contained in the commutator subgroup of $\operatorname{GL}_2(\mathbb{R})$.
Any help will be appreciated.
| If $x\in\mathbb R$, then$$\begin{pmatrix}1&x\\0&1\end{pmatrix}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}^{-1}\begin{pmatrix}1&x\\0&1\end{pmatrix}^{-1}$$and therefore $\left(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\right)$ is a product of commutators. For the same reason $\left(\begin{smallmatrix}1&0\\x&1\end{smallmatrix}\right)$ is a product of commutators. On the other hand, if $x\in\mathbb{R}\setminus\{0\}$,$$\begin{pmatrix}x&0\\0&x^{-1}\end{pmatrix}=\begin{pmatrix}x&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}x&0\\0&1\end{pmatrix}^{-1}\begin{pmatrix}0&1\\1&0\end{pmatrix}^{-1}.$$
Now, let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in SL_2(\mathbb{R})$. If $a\neq0$, then$$\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&0\\\frac ca&1\end{pmatrix}\begin{pmatrix}1&ab\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix},$$which is a product of commutators. Otherwise, $b\neq0$ and$$\begin{pmatrix}0&b\\c&d\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&-\frac db\\0&1\end{pmatrix}\begin{pmatrix}b^{-1}&0\\0&b\end{pmatrix}.$$Finally, $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ is a commutator, since it is equal to$$\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}-1&0\\1&2\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}^{-1}\begin{pmatrix}-1&0\\1&2\end{pmatrix}^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding all complex entries? Find all complex triples $(x, y, z) $such that the following matrix is diagonalizable
$A = \begin{bmatrix}1&x&y \\ 0&2 & z \\0&0&1\end{bmatrix}$
my attempts :
matrix A is an upper triangular matrix.
so the the eigenvalues of A are diagonal entries $1,2,1$
This implies that $A$ is diagonalizable.
Case 1: if $x=y= z= 0$
case 2 : if $x\neq y \neq z \neq 0$
Now Im confused that How can i find all complex triples $(x, y, z) $such that the following matrix is diagonalizable
| Diagonalizable means the minimal polynomial is squarefree. The full characteristic polynomial is $(\lambda - 1)^2 (\lambda - 2).$
Diagonalizable if and only if
$$ (A - I)(A - 2I) = 0. $$
This gives a restriction on the triple $x,y,z.$ And, when we do have $y = xz,$ we have
$$
\left(
\begin{array}{ccc}
1&-x&-xz \\
0&1&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&xz \\
0&2&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&0 \\
0&1&-z \\
0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
1&0&0 \\
0&2&0 \\
0&0&1 \\
\end{array}
\right)
$$
On the other hand, if $y = xz + t$ with $t \neq 0$ we have
$$
\frac{1}{t}
\left(
\begin{array}{ccc}
0&t&zt \\
1&-x&-xz \\
0&0&t \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&xz +t \\
0&2&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
x&t&0 \\
1&0&-z \\
0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
2&0&0 \\
0&1&1 \\
0&0&1 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Result comparison of a trigonometric equation The Problem: Solve the equation: $$ \cos x=\cos3x+ 2\sin2x
\\$$
The Result: $$ x=k\frac{\pi}{2},k\in \mathbb{Z}$$
My solution:$$ \cos x=4\cos^3x-3\cos{x}+4\sin{x}\cos{x}\\ 4\cos^3x-4\cos{x}+4\sin{x}\cos{x}=0\\ 4\cos{x}(\cos^2{x}+\sin{x}-1)=0\\
\cos{x}(1-\sin^2{x}+\sin{x}-1)=0\\
\cos{x}(\sin{x}-\sin^2{x})=0\\
\cos{x}\sin{x}(1-\sin{x})=0\\
x=\frac{\pi}{2}+l\pi \lor x=m\pi \lor x=\frac{\pi}{2}+2n\pi; \hspace{0.4cm}l,m,n\in \mathbb{Z}$$
Question: Is my solution equal to the result?
| My resolution:
$$\cos x= \cos \left(3x\right)+2\sin \left(2x\right)$$
$$-\cos \left(3x\right)-2\sin \left(2x\right)+\cos \left(x\right)=0$$
Using the following identity
$$-\cos \left(p\right)+\cos \left(q\right)=2\sin \left(\frac{p+q}{2}\right)\sin \left(\frac{p-q}{2}\right)$$
I obtaining,
$$-2\sin \left(2x\right)+2\sin \left(\frac{-x+3x}{2}\right)\sin \left(\frac{x+3x}{2}\right)=0$$
Hence
$$-2\sin \left(2x\right)+2\sin \left(2x\right)\sin \left(x\right)=0$$
Factorizing
$$2\sin \left(2x\right)\left(\sin \left(x\right)-1\right)=0$$
Solving each part separately:
$$ \sin \left(2x\right)=0\quad \mathrm{or}\quad \sin \left(x\right)-1=0$$
I have your solutions:
$$\sin \left(2x\right)=0\quad :\quad x=k\pi,\:x=\frac{\pi}{2}+k\pi$$
$$\sin \left(x\right)-1=0\quad :\quad x=\frac{\pi}{2}+2k\pi,\, k\in\mathbb Z$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
My attempt:
Consider $\frac{1}{1 - z}$. Let $z = x + iy$, we know that $|z| < 1 \implies x, y < 1$. $$\frac{1}{1 - z} = \frac{1}{1 - x - iy} = \frac{1 - x + iy}{(1 - x)^2 + y^2}.$$
$$\Re \left(\frac{1}{1 - z} \right) = \frac{1 - x}{(1 - x)^2 + y^2}.$$
I got $$\frac{1}{(x-1)^2 + y^2} > \frac{1}{2}.$$
How do I manage the numerator? Can you help me? I welcome the alternative approaches.
| You need to use more than just $x,y<1$ for example with $x=y=\frac{3}{4}$ you obtain $\Re\left(\frac{1}{1-(x+iy)} \right)=\frac{2}{5} <\frac{1}{2}$.
Hint:
Using your computations:
$$ \Re\left(\frac{1}{1-z}\right)=\frac{1-x}{(1-x)^2+y^2}=\frac{1-x}{x^2+y^2-2x+1}$$
but $x^2+y^2<1$ so:
$$\frac{1-x}{x^2+y^2-2x+1}>\frac{1-x}{1-2x+1}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0 & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ :
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.
| To find other values of $x$ besides $0$ for which
$$
\det \begin{bmatrix}
x & 0 & 0 & \cdots & n-1 \\
0 & x & 0 & \cdots & n-2 \\
0 & 0 & x & \cdots & n-3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \cdots & x
\end{bmatrix} = 0
$$
consider taking $\frac{n-1}{x}$ times the first row, plus $\frac{n-2}{x}$ times the second row, plus $\frac{n-3}{x}$ times the third, and so on, on the grounds that you will get a row vector which is equal to the last row in its first $n-1$ entries.
Its last entry is going to be $\frac{(n-1)^2 + (n-2)^2 + \dots + 2^2 + 1^2}{x} = \frac{n(n-1)(2n-1)}{6x}$. So if this happens to be equal to $x$, then the last row is a linear combination of the other rows, which means the determinant is $0$.
There are two values of $x$ for which this works: the two square roots of $\frac{n(n-1)(2n-1)}{6}$. This gives us the last two eigenvalues: the whole list is
$$
a, a, \dots, a, a - \sqrt{\frac{n(n-1)(2n-1)}{6}}, a + \sqrt{\frac{n(n-1)(2n-1)}{6}}
$$
and their product is
$$
a^{n-2}\left(a^2 - \frac{n(n-1)(2n-1)}{6}\right)
$$
as you conjectured.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sums of Nilpotent Matrices Let $A =$ diag$(a_1,a_2,…,a_n)$, where the sum of all $a_i$’s is zero.
Show that A is a sum of nilpotent matrices.
My idea:
$\begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}$
Where $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}^2$ = $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}^2$ = $0_2$
Extending the 2 $\times$ 2 matrix, we get
$\begin{bmatrix}
2 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 & 0 & 0 \\
-1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & -1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1
\end{bmatrix}$
Where $\begin{bmatrix}
1 & 1 & 0 & 0 \\
-1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & -1 & -1
\end{bmatrix}^2$ = $\begin{bmatrix}
1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1
\end{bmatrix}^2$ = $0_4$
Kindly correct me if I’m wrong, and also, please help me how to generalized this. Thank you in advance.
| The diagonal matrices of the form $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$ form a basis of the space of the matrices with null trace. In fact, a matrix with null trace is $\operatorname{diag}(a_1,\ldots,a_n)$ with $a_1+a_2+\cdots+a_n=0$, which is equal to$$a_1\operatorname{diag}(1,0,0,\ldots,0,-1)+a_2\operatorname{diag}(0,1,0,\ldots,0,-1)+\cdots+a_{n-1}\operatorname{diag}(0,0,\ldots,0,1,-1).$$Now, you can apply your idea to each matrix of the type $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding all values in $\mathbb R$ for quadratic absolute value equation The question:
Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$
So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution set, $|x^2+2x+2|\ge-1$ as indicated by the roots of the RHS equation, but this is where I get stuck.
Where am I going wrong?
| Since $|x| = x$ or $-x$, $$|x^2+2x+2| = |x^2-3x-4|$$ if and only if
$$x^2+2x+2 = x^2-3x-4$$ or
$$-(x^2+2x+2) = x^2-3x-4$$ or
$$x^2+2x+2 = -(x^2-3x-4)$$ or
$$-(x^2+2x+2) = -(x^2-3x-4)$$
Some cases are equivalent (can you guess which and why?), so the computations are easier than they look.
| {
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Find the convergence domain of the series $\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$
Exercise: Find the convergence domain of the series $$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$$
To solve the series I used the comparison test then the Leibniz criteria:
$$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}\leqslant\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{(x-5)^n}$$
Now applying Leibniz criteria the series converge for $x\neq 5$.
Solution(book): $x\geqslant5\frac{1}{3},x<4\frac{2}{3}$.
Question:
The solution surprised me. What am I doing wrong? Why is it not the same?
Thanks in advance!
| We have
$$
\sum_{k=0}^{\infty}(-1)^k y^k = \lim_{n\rightarrow \infty}\frac{y^n+1}{y+1}
$$
here $\vert y\vert < 1 \Rightarrow$ convergence
Now
$$
\sum_{k=0}^{\infty}(-1)^k \frac{y^k}{k} = \frac{1}{y}\int_0^y \sum_{k=0}^{\infty}(-1)^k y^k dy = \frac{1}{y}\int_0^y\frac{dy}{y+1} = \frac{\ln(y+1)}{y}
$$
for $y \ne 0$ and $\vert y\vert< 1$ so the convergence is linked to
$\vert\frac{1}{3(x-x)}\vert < 1 \Rightarrow 5\frac{1}{3} < x$ and $x < 4\frac{2}{3}$
| {
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"source": "stackexchange",
"question_score": "3",
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Continuity of functions of two variables Find the points where the function $f(x,y)$ is continuous where
$$f(x,y)=\left\{ \begin{array}{ll}
\frac{x^2\sin^2y}{x^2+2y^2} & \mbox{if $(x,y)\not=(0,0)$};\\
0 & \mbox{if $(x,y)=(0,0)$}.\end{array} \right.$$
What I attempted: Here $f(x,y)$ is continuous at all the points $(x,y)\not=(0,0)$
We will check continuity at the point $(x,y)=(0,0)$
$f(x,y)=0$ is well defined. Now,
$$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}$$
Here $0\le \frac{x^2\sin^2y}{x^2+2y^2}\le \frac{x^2y^2}{x^2+2y^2}$ (As $\sin^2y \le y$) [Am I correct here?]
If we let $x=r\cos\theta$, $y=r\sin\theta$, then $$\frac{x^2y^2}{2x^2+y^2}=\frac{r^2 \cos^2\theta r^2 \sin^2\theta}{r^2+r^2 \sin^2\theta}=\frac{r^2 \sin^2\theta \cos^2\theta }{1+\sin^2\theta}\le r^2$$ (As $\frac{\sin^2\theta}{1+\sin^2\theta}\le1$, $\cos^2\theta\le 1$ )
As $(x,y)\to (0,0)$, $r\to 0$. Therefore using Sandwich Theorem we should have $$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}=0$$
So, it is continuous at $(0,0)$.
I am not sure whether it is correct or not.The question was asked in an interview and I used the above technique. However, they did not pointed my mistake but just gave an another problem of an similar kind where $f(x,y)=\frac{x^2+\sin^2y}{2x^2+y^2}$ for all $(x,y) \not= (0,0)$ and $0$ for the origin.I tried the same method but failed. Then they asked me to leave.
| Here is the answer for the second question: $\frac {x^{2}+\sin ^{2} (y)} {2x^{2}+y^{2}}$ is $\frac 1 2$ when $y=0$ and $\frac {\sin ^{2} (y)} {y^{2}}$ when $x=0$. The last expression approaches 1 as $y \to 0$. Hence the functioin does not even have a limit as $(x,y) \to 0$ and the function is not continuous.
| {
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Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$
I have noticed that
$$\dfrac{1}{3}< 1-\dfrac{1}{2^2}+\dfrac{1}{3};\\
\dfrac{1}{3}+\dfrac{1}{5}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5};\\
\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}\\
$$
which is true in general because
$$1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{36}-\dfrac{1}{64}-\dots=1-\sum_{n=1}^\infty \dfrac{1}{(2n)^2}=1-\dfrac{1}{4}\sum_{n=1}^\infty\dfrac{1}{n^2}=1-\dfrac{\pi^2}{24}>0.
$$
Thus
$$
\sum_{n=1}^\infty \dfrac{1}{2n+1}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots
$$
Finally, as the series $\sum_{n=1}^\infty \dfrac{1}{2n+1}$ diverges, so does the series requested.
My two questions are:
I) Is this reasoning correct?
II) Can this exercise be done without using that $\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$? I would like to find a solution with more elementary tools.
Thank you.
| $${1\over2n-1}-{1\over(2n)^2}={4n^2-2n+1\over4n^2(2n-1)}\gt{4n^2-4n+1\over4n^2(2n-1)}={2n-1\over4n^2}\ge{1\over4n}$$
so
$$\left(1-{1\over4}\right)+\left({1\over3}-{1\over16}\right)+\left({1\over5}-{1\over36}\right)+\cdots\gt{1\over4}+{1\over8}+{1\over12}+\cdots={1\over4}\left(1+{1\over2}+{1\over3}+\cdots\right)$$
| {
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Show $x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent. I want to show that the equation
$x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent.
This question has been solved with infinite descent by showing that $x,y,z$ are all even and we can infinitely find smaller solutions:
Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Can it be shown instead by assuming you have the least positive solution and then finding a smaller solution.
I let ($a,b,c$) be the least positive solution with gcd(a,b,c) = 1, and I want to find a smaller solution.
I have that exactly one of $a,b,c$ must be even because if two were even, then the left hand side would be odd while the right hand side is even. And if all three are even, then gcd($a,b,c$) $\ge$ 2.
So we can assume $a$ is even. I've tried considering the equation mod $a,b,c,2$ but couldn't seem to find a way to construct a smaller solution.
| $x^2+y^2+z^2=2xyz...........................(1)$
in this equation $RHS$ is even .then, $LHS$ is also even.
w.l.o.g, $z$ is even $\exists z_1 $ such that $z= 2 z_1$ and $x,y$ are in same parity.
substituting $z= 2 z_1$ in given equation
$x^2+y^2+4z_1^2=4xyz_1 ........................(2)$
$x^2+y^2=4xyz_1-4z_1^2$
$RHS$ is multiple of 4 .then ,$LHS$ must be multiple of 4
therefore $x$ and $y$ are even.
let $x=2x_1$ and $y=2y_1$
substituting $x=2x_1$ and $y=2y_1$ in $(2)$ we get,
$4x_1^2+4y_1^2+4z_1^2=16x_1y_1z_1 $
$\implies x_1^2+y_1^2+z_1^2=4x_1y_1z_1 $
$LHS$ is multiple of $4$.then,$RHS$ is also multile of $4$
w.l.o.g,$z_1$ is even, let $z_1=2z_2$
$\implies x_1^2+y_1^2+4z_2^2=8x_1y_1z_2 $
in similar argument .We can find three sequence of infinite integers
$$x_1 \gt x_2 \gt x_3 \dots \gt x_n$$
$$y_1 \gt y_2 \gt y_3 \dots \gt y_n$$
$$z_1 \gt z_2 \gt z_3 \dots \gt z_n$$
that can solve the equation $x^2+y^2+z^2=2xyz$.
by using infinite descent we can prove
$x^2+y^2+z^2=2xyz$ has no solution greater than $0$.
| {
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Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$ Show that $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}$$
This is a 9th grade problem.
I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only $2n+1$ terms. Right?
After that I have no idea. Thx!
| There are $2n+1$ terms in the sum, you just need to pair up the terms
symmetrically from both ends, take average and compare with the term in the middle.
$$\begin{align}\sum_{k=n+1}^{3n+1} \frac{1}{k}
&= \sum_{k=-n}^n\frac{1}{2n+1+k}
= \frac12\sum_{k=-n}^n\left(\frac{1}{2n+1+k} + \frac{1}{2n+1-k}\right)\\
&= \sum_{k=-n}^n \frac{2n+1}{(2n+1)^2-k^2}
\stackrel{\color{blue}{\text{ assume } n> 0}}{>} \sum_{k=-n}^n \frac{1}{2n+1} = \frac{2n+1}{2n+1} = 1\end{align}$$
| {
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USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):
Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$.
Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$
This question is on symmetric polynomials. I recall as many facts as I can think of regarding inequalities. (This test was closed book).
*
*AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.
*quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.
*The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.
*I am still looking for other inequalities that might work.
It's tempting to race through this problem with the first solution that comes to mind. I'm especially interested in some kind of organizing principle.
| $ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$=$ 4(ab+bc+ca)+ 4\min(a^2,b^2,c^2)-2(ab+bc+ca)$
Now, suppose $a≤b≤c$ without losing generality.
Hence,
$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$
=$ 4(ab+bc+ca)+ 4a^2-2(ab+bc+ca)$
= $4((a+b+c)a+bc)-2(ab+bc+ca)$
=$4(4a(abc)^{1/3})+bc)-2(ab+bc+ca)≥ 16(abc)^{2/3}-2(ab+bc+ca)
=(a+b+c)^2-2(ab+bc+ca)=a^2+b^2+c^2$
| {
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How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
| $$\sin(2x)=\frac{24}{25}$$
The identities you want are:
$$\sin(2x)=2\sin(x)\cos(x) \space [I]$$
and
$$\sin^2(x)+\cos^2(x)=1\space[II]$$
Square $[I]$ to get:
$$4\sin^2(x)\cos^2(x)=\frac{576}{625}\to\sin^2x(1-\sin^2x)=\frac{144}{625}$$
$$\to\sin^4(x)-\sin^2x+\frac{144}{625}=0$$
Let $\sin^2(x)=\sigma$. Your equation becomes:
$\sigma^2-\sigma+\frac{144}{625}=0$
Solve via Quadratic Formula:
$$\sigma=\frac{1\pm\sqrt{1-\frac{576}{625}}}{2}=\frac 12\pm\frac{\sqrt{\frac{49}{625}}}{2}=\frac{9}{25}, \frac{16}{25}.$$
Since $\sin^2(x)=\frac{9}{25}, \frac{16}{25}, \sin(x)=\pm \frac 35, \pm \frac 45$
Check these values with the function $f(S)=\sin(2\cdot\sin^{-1}(S))$, (where $S$ is the solutions i.e. $\pm\frac 35, \pm \frac 45$) to see the ones which provide the correct solution, and note the values which work are $\frac 35, \frac 45$.
We can see them graphed here:
| {
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Maximizing $3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. What went wrong?
Let $f(x) = 3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. For some $x \in \left[0,\frac{\pi}{2}\right]$, $f$ attains its maximum value, $m$. Compute $m + 100 \cos^2 x$.
What I did was rewrite the equation as $f(x)=6\cos^2x+8\sin x\cos x+3$. Then I let $\mathbf{a}=\left<6\cos x,8\cos x\right>$ and $\mathbf{b}=\left<\cos x,\sin x\right>$.
Using Cauchy-Schwarz, I got that the maximum occurs when $\tan x=\frac{4}{3}$, and that the maximum value is $10\cos x$. However, that produces a maximum of $9$ for $f(x)$, instead of the actual answer of $11$.
What did I do wrong, and how do I go about finding the second part? Thanks!
| Let's ignore the condition that $x\in[0,\pi/2]$ for the moment.
We are maximising $3u^2+8uv+9v^2$ subject to the constraint $u^2+v^2=1$.
For a given $a$, $3u^2+8uv+9v^2=a$ is soluble under this constraint
iff $3u^2+8uv+9v^2=a(u^2+v^2)$ is. This means that $Q(u,v)=(3-a)u^2+8uv+(9-a)v^2=0$. If this has a solution $\ne(0,0)$ it can be scaled to one with
$u^2+v^2=1$. This is the case if the quadratic form $Q$ factors over the
reals, that is if $8^2-4(3-a)(9-a)\ge0$. This gives $a^2-12a+11\le0$,
equivalently $(a-1)(a-11)\le0$. The possible values of $a$ form
the interval $[1,11]$.
Does $11$ occur as a maximum for some acute angle $x$? When $a=11$,
$Q(u,v)=-8u^2+8uv-2v^2=-2(2u-v)^2$, so at a maximum, $v=2u$. The
solutions of this when $u^2+v^2=1$ are $(1/\sqrt5,2/\sqrt5)$
and $(-1/\sqrt5,-2/\sqrt5)$ and the first comes from the acute
angle $x=\tan^{-1}2$.
| {
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Finding the number of councils with at least two girls There are 12 people, 7 girls and 5 boys.
You are supposed to choose a council of 4 people so that there are at least 2 girls.
I know you can solve this by taking the total amount of choices and subtracting the scenarios with one or no girls.
12C4 - (5C4) - (7C1)(5C3) = 420
But I'm pretty sure you can obtain the answer through other considerations.
If you assume that the first two seats must be girls, then the final two spots would have 10 and 9 people to choose from. So I reasoned that the answer must be
7*6*10*9 = (7C2)(10C2) = 420*9
So I'm a factor of 9 off. My hunch is that my answer is double counting somehow but I'm having trouble identifying my mistake. Help?
| As lulu suggests in the comments, a way to count the number of councils with at least two girls is to add the number of ways of selecting a council with exactly two, exactly three, or exactly four girls.
The number of ways of selecting exactly $k$ girls and $4 - k$ boys from seven girls and five boys is $$\binom{7}{k}\binom{5}{4 - k}$$ Hence, the number of ways of selecting a council with at least two girls is $$\binom{7}{2}\binom{5}{2} + \binom{7}{3}\binom{5}{1} + \binom{7}{4}\binom{5}{0} = 21 \cdot 10 + 35 \cdot 5 + 35 \cdot 1 = 210 + 175 + 35 = 420$$ as you found by subtracting the number of cases in which fewer than two girls are selected from the number of ways of selecting four of the twelve available people.
Where did you go wrong?
If you assume that the first two seats must be girls, then the final two spots would have $10$ and $9$ people to choose from. So I reasoned that the answer must be $$7 \cdot 6 \cdot 10 \cdot 9 = \binom{7}{2}\binom{10}{2}$$
What matters here is who is selected, not the order in which they are selected. Your count $7 \cdot 6 \cdot 10 \cdot 9$ is the number of ways of selecting a girl, a second girl, selecting one of the ten remaining people, and then selecting one of the remaining nine people in that order.
Since the number of ways of selecting a subset of $k$ objects from a set of $n$ objects is
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
the number of ways of selecting two objects from a set with $n$ objects is
$$\binom{n}{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)(n - 2)!}{2 \cdot 1 \cdot (n - 2)!} = \frac{n(n - 1)}{2}$$
In particular,
$$\binom{7}{2} = \frac{7 \cdot 6}{2} = 6 \cdot 3 = 21$$
and
$$\binom{10}{2} = \frac{10 \cdot 9}{2} = 5 \cdot 9 = 45$$
Hence, the statement
$$7 \cdot 6 \cdot 10 \cdot 9 = \binom{7}{2}\binom{10}{2}$$
is false since
$$7 \cdot 6 \cdot 10 \cdot 9 = 42 \cdot 90 > 21 \cdot 45 = \binom{7}{2}\binom{10}{2}$$
The answer
$$\binom{7}{2}\binom{10}{2}$$
is still too large. To see why, suppose Anne, Belinda, Claire, and Edward are selected for the council. You count this selection three times, once for each of the $\binom{3}{2}$ ways of designating two of the three women as the designated women.
$$
\begin{array}{l l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Belinda} & \text{Claire, Edward}\\
\text{Anne, Claire} & \text{Belinda, Edward}\\
\text{Belinda, Claire} & \text{Anne, Edward}
\end{array}
$$
Suppose Anne, Belinda, Claire, and Debra are selected for the council. You count this selection six times, once for each of the $\binom{4}{2}$ ways of designating two of the four women as the designated women.
$$
\begin{array}{l l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Belinda} & \text{Claire, Debra}\\
\text{Anne, Claire} & \text{Belinda, Debra}\\
\text{Anne, Debra} & \text{Belinda, Claire}\\
\text{Belinda, Claire} & \text{Anne, Debra}\\
\text{Belinda, Debra} & \text{Anne, Claire}\\
\text{Claire, Debra} & \text{Anne, Belinda}
\end{array}
$$
Notice that
$$\binom{2}{2}\binom{7}{2}\binom{5}{2} + \color{red}{\binom{3}{2}}\binom{7}{3}\binom{5}{1} + \color{red}{\binom{4}{2}}\binom{7}{4}\binom{5}{0} = \binom{7}{2}\binom{10}{2}$$
The reason you got too large an answer is that the group of seven girls and the group of ten additional people are not disjoint.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Without-loss-of-generality question Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.
Here's the statement of the problem:
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that
$$
\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq\frac{3}{16}.
$$
Here's how they start in the solution:
We prove the homogenized inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
for all positive real numbers $a,b,c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a,b,c>0$, fulfilling this condition, the inequality
$$
\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\leq\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).
$$
And so on... If someone's interested in the rest, it's problem A2, solution 2 in this pdf: https://www.imo-official.org/problems/IMO2009SL.pdf
Why is there no loss of generality in such a choice of $a,b,c$?
| Suppose that we know that if $a+b+c=1$ the homogenized inequality is true. Then if we have arbitrary $A$, $B$, $C$, with $A+B+C=x$, we can define $a=A/x$ and so on. When we plug this into the inequality, the $x$'s drop out, so that $A$, $B$, and $C$ also satisfy the inequality.
| {
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Compute limit of double exponent expression. Compute $$\large\lim_{n\to \infty}\left(\left(\prod_{i=1}^{n+1}{i^{i^p}}\right)^{1/(n+1)^{p+1}} - \left(\prod_{i=1}^{n}{i^{i^p}}\right)^{1/n^{p+1}}\right),$$
where $p$ is some nonnegative real number.
I found this problem while working through the Problems in Real Analysis: Advanced Calculus on the Real Axis by Teodora-Liliana Radulescu, Titu Andreescu, and Vicenţiu D. Rădulescu, but I don't know how to do it. Can someone help me?
| The calculations got a bit messy so there could be errors, let me know if you find some.
Let the sequence $a_n$ be defined by $a_n = \prod\limits_{i=1}^n i^{i^p}$.
Then by Abel's summation formula we have $$\log \left(a_n^{1/n^{p+1}}\right) = \frac{1}{n^{p+1}}\sum\limits_{i=1}^ni^p\log i = \frac{1}{n^{p+1}}\left(n^pS(n) - p\int\limits_1^nx^{p-1}S(x) \text{d}x\right)$$
where $S(x) = \sum\limits_{1 \leq i \leq x}\log i = x \log x - x + \mathcal{O}\left(\log x\right)$.
The expression becomes messy: $$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\int\limits_1^nx^p\left(\log x - 1 + \mathcal{O}\left(\frac{\log x}{x} \right)\right)\text{d}x = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\int\limits_1^nx^p\left(\log x - 1\right)\text{d}x + \mathcal{O}\left(\frac{p}{n^{p+1}}\int\limits_1^nx^{p-1}\log x \text{d}x\right) = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) - \frac{p}{n^{p+1}}\left(\frac{n^{p+1}\left((p+1)\log n - 1\right)+1}{(p+1)^2} - \frac{n^{p+1}-1}{p+1}\right) + \mathcal{O}\left(\frac{\log n}{n}\right) = $$
$$\log n - 1 + \mathcal{O}\left(\frac{\log n}{n}\right) -\frac{p}{p+1}\log n + \frac{p}{(p+1)^2} + \frac{p}{p+1} + \mathcal{O}\left(\frac{\log n}{n}\right) = $$
$$\frac{\log n}{p+1}-\frac{1}{(p+1)^2}+\mathcal{O}\left(\frac{\log n}{n}\right)$$
So we get $$a_n^{1/n^{p+1}} = \left(1+\mathcal{O}\left(\frac{\log n}{n}\right)\right)e^{-1/(p+1)^2}n^{1/(p+1)}$$
Then we have $$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = e^{-1/(p+1)^2}\left((n+1)^{1/(p+1)}-n^{1/(p+1)}\right) + \mathcal{O}\left(n^{-p/(p+1)}\log n\right)$$
So we finally get
$$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \mathcal{O}\left(n^{-p/(p+1)}\log n\right)$$
which converges to $0$.
I just realized that the last statement holds only for $p \gt 0$. If $p = 0$, the bound is not strong enough, and we need a better approximation.
When $p = 0$, the formula for $\log\left(a_n^{1/n^{p+1}}\right)$ becomes:
$$\log\left(a_n^{1/n^{p+1}}\right) = \frac{1}{n}\sum\limits_{i=1}^n\log i = \frac{1}{n}S(n)$$
and we use the better estimate $$S(n) = \sum\limits_{i=1}^n\log i = n\log n - n + \frac{1}{2}\log n + C + \mathcal{O}\left(\frac{1}{n}\right)$$ coming from Euler-Maclaurin summation (the constant value is $C = \log\sqrt{2\pi}$ but this is not necessary here).
Then we get $$\log\left(a_n^{1/n^{p+1}}\right) = \log n - 1 + \frac{1}{2}\frac{\log n}{n} + \frac{C}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)$$
and so $$a_n^{1/n^{p+1}} = \frac{n}{e}\left(1 + \frac{\log n}{2n} + \mathcal{O}\left(\frac{\log^2 n}{n^2}\right)\right)\left(1 + \frac{C}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)\right)\left(1 + \mathcal{O}\left(\frac{1}{n^2}\right)\right)$$
$$a_n^{1/n^{p+1}} = \frac{1}{e}\left(n + \frac{1}{2}\log n + C\right) + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
Then we have $$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \frac{1}{e}\left(1 + \frac{1}{2}\log\left(1+\frac{1}{n}\right)\right) + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
$$a_{n+1}^{1/(n+1)^{p+1}} - a_n^{1/n^{p+1}} = \frac{1}{e} + \mathcal{O}\left(\frac{\log^2 n}{n}\right)$$
so the limit is $\frac{1}{e}$ in the case where $p = 0$.
| {
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Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positive integer), we get
$\frac{x^2-1}{xy+1}=0$ which is a non-negative integer, regardless of $y$. So, one solution is $(x,y)=(1,k)$ where $k$ is a positive integer.
Now, let's say $y=1$. Then $\frac{x^2-1}{xy+1}=\frac{x^2-1}{x+1}=x-1$ which is always a non-negative integer so $(x,y)=(k,1)$ is also a solution. However, I don't know how to find the other or prove that those are the only ones.
| $x+1=xy+1 ⇒ y=1$ then x can get any value greater than 0.
B: $x-1=xy+1 ⇒ x(y-1)=-2 ⇒ x=2, y=0$ are only possible solutions.
| {
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Finding the sum of $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, $\cos\frac{5π}{7}$ by first finding a polynomial with those roots
Without using tables, find the value of $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$$
This is a very common high school trigonometric problem, and the usual way to solve this is by repeated application of trigonometric identities. But I thought of a bit different approach.
Somehow, if we can find a polynomial whose roots are the three terms of the above expression, then we can apply Vieta's formula to find the value.
So please help me with it. (Any hint will be appreciated.)
| Consider numbers: $$\omega_k=e^{i\frac{(2k-1)\pi}{7} }=\cos\left(\frac{(2k-1)\pi}{7}\right) + i \sin\left(\frac{(2k-1)\pi k}{7}\right), \qquad k=1,...,7.$$
Easy to see (De Moivre's formula) that $$w_k^7=e^{i(2k-1)\pi}=e^{i\pi}=-1, \qquad k=1,...,7.$$
So, all $\omega_k$ are solutions of equation
$$
\omega^7-1 = 0.\tag{1}
$$
Denote unknown value $\omega$ as $\omega=\cos\alpha+i\sin\alpha$ (simply, $\omega = c+is$).
So:
$$
c^7 + i 7c^6s- 21 c^5s^2- i 35c^4s^3 + 35 c^3s^4 +i 21c^2s^5-7cs^6-is^7-1=0.
$$
Focusing on real part of the equation, we have
$$
c^7 - 21 c^5s^2 + 35 c^3s^4 -7cs^6+1=0.\tag{2}
$$
Now replace each $s^2$ with $(1-c^2)$:
$$
64c^7-112c^5+56c^3-7c+1=0.\tag{3}
$$
Knowing that $c=-1$ is one of solutions of equation $(3)$ and that $\cos\left(\dfrac{(2k-1)\pi}{7}\right)$ are double-solutions of eq. $(3)$ (I mean: $\cos \frac{\pi}{7}=\cos\frac{13\pi}{7}$, $\cos \frac{3\pi}{7}=\cos\frac{11\pi}{7}$, $\cos \frac{5\pi}{7}=\cos\frac{9\pi}{7}$), we can factor its LHS:
$$
\left(8c^3-4c^2-4c+1\right)^2(c+1)=0.
$$
Therefore, the equation for $\cos\left(\frac{\pi}{7}\right)$, $\cos\left(\frac{3\pi}{7}\right)$,$\cos\left(\frac{5\pi}{7}\right)$ has form
$$
8c^3-4c^2-4c+1=0.
$$
| {
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Partitions of numbers Let $p_{2}(n)$ denote the number of partitions of the positive integer n into at most 2 parts. Prove for each $n \ge 4$ that $p_{2}(n) + p_{2}(n-3) = n$
I understand that I need to split this into odd and even cases - I would just like my reasoning validating. Any comments/feedback on my proof or notation are very welcome.
Starting with the even numbers:
The number of partitions of size 2 of an even number $n$ is given by $\frac{n}{2} + 1$
The number of partitions of size 2 of an even number $n-3$ is given by $\frac{n-3}{2} + 1$ Since $n-3$ is odd, we only take the integer part of $\frac{n-3}{2}$ which is $\frac{n}{2} -2$. Adding on the 1 gives $\frac{n}{2} -1$.
Now for the even case, we do $p_{2}(n) + p_{2}(n-3) = \frac{n}{2} +1 + \frac{n}{2} -1 =n$
Odd Numbers
The number of partitions of size 2 of an odd number $n$ is given by $\frac{n}{2} + 1$. Since $n$ is odd, we only need the integer part of $\frac{n}{2}$ which is $\frac{n-1}{2}$. So we have $\frac{n-1}{2} +1$.
The number of partitions of size 2 of an odd number $n-3$ is given by $\frac{n-3}{2} + 1$. Since $n-3$ is even, we get $\frac{n-1}{2} -1 +1 = \frac{n-1}{2}$.
Now for the odd case, we do $p_{2}(n) + p_{2}(n-3) = \frac{n-1}{2} +1 + \frac{n-1}{2}=n$
| It is simpler to have a "compact view" of the arguments. From the post, the formula
$$p_2(n)=\left[\frac n2\right]+1$$
is known, and was shown. Now we can cover all cases in one breath:
$$
\begin{aligned}
p_2(n)+p_2(n-3)
&=
\left[\frac n2\right]+1
+
\left[\frac {n-3}2\right]+1
\\
&=
\left[\frac n2\right]
+
\left[\frac {n-3}2+1+1\right]
\\
&=
\left[\frac n2\right]
+
\left[\frac {n+1}2\right]
\\
&=
n\ .
\end{aligned}
$$
We have used some properties of the floor function $[x+1]=[x]+1$ and
$[x]
+
\left[x+\frac 12\right]=
[2x]$, where $x\in \Bbb R$, which are simpler to check in the case $x$ is half an integer.
| {
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Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked the numbers from $1$ to $1000$, and the results are:
$1.$ The numbers on the left are the last digit of each digit squared.
$2.$ The numbers on the right are the number of times that the last digit is repeated.
$$
\begin{array}{cc}
0: &100, \\
1: &200, \\
4: &200, \\
5: &100, \\
6: &200, \\
9: &200
\end{array}
$$
So, why does this happen? What is the property that all integers have?
| These numbers are the squares modulo 10. Notice that the square of the number $10n+k$ is
$$ (10n+k)^2 = 10(10n^2+2nk)+k^2, $$
so the last digit of the square is determined by only the last digit of the original number. In particular, we find
$$ 0^2=0 \quad 1^2=1 \quad 2^2 = 4 \quad 3^2 = 9 \quad 4^2 = 10+6 \\
5^2 = 20+5 \quad 6^2 = 30+6 \quad 7^2 = 40+9 \quad 8^2 = 60+4 \quad 9^2 = 80+1, $$
or writing "$\equiv$" to mean that they have the same last digit,
$$ 0^2 \equiv 0 \\
1^2 \equiv 1 \equiv 9^2 \\
2^2 \equiv 4 \equiv 8^2 \\
3^2 \equiv 9 \equiv 7^2 \\
4^2 \equiv 6 \equiv 6^2 \\
5^2 \equiv 5, $$
so every last digit except $0$ and $5$ is the last digit of two squares out of a block of 10 consecutive numbers, while $0$ and $5$ are the last digit of only one each.
| {
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?
| This is in fact a composition problem. We want to find all compositions of $x$ and $y$ such that $z$ takes values in $0,2,4,6,8,10$.
Then we count All 2-Compositions of $0,2,4,6,8,10$, that is how can we add 2 numbers to get these values.
Let $C(n,k,a,b)$ be all k-compositions of n using the numbers from a to b.
We want:
$\sum_{i \in (0,2,4,6,8,10)} C(i,2,0,i)$ = 1+3+5+7+9+11=36.
We know $C(n,k,0,n)=$ $\sum_{i=2}^{10} [{i-1 \choose 1}+2]+1=66$ since we need 2 extra to account for 0 and 1 at the end to account for $x=0$ and $y=0$.
So we get $\frac {36}{66}$ as required.
| {
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If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.
| An approach using factoring expressions is the following: define
$ \quad x := e^{i\alpha}, \quad y := e^{i\theta}, \quad $ and
$$ a := \frac{\cos(\alpha -3\theta)}{\cos^3 \theta},
\, b := \frac{\sin(\alpha -3\theta)}{\sin^3 \theta},
\, c := \cos \alpha - \frac{2-a^2}a,
\, d := \cos \alpha - \frac{2-b^2}b, \, e := a-b, $$
$$ F_ 1 \!:=\! 3 x^2 - y^4 + x^2 y^4 - 3 y^8,
\, F_ 2 \!:=\! 3 x^2 - y^2 + x^2 y^2 - 3 y^4,
\, F_ 3 \!:=\! 3 x^2 + y^2 - x^2 y^2 - 3 y^4. $$
Now $\, e = 8 F_ 1 y^2/(x (1-y^4)^3) \,$ and thus if $\, F_ 1=0 \,$
then $\,a=b=m.\,$ Factorization gives
$$ c = F_1F_2/(2 x (1+y^2)^3(x^2+y^6) \quad \text{and}
\quad d = F_1F_3/(2 x (1-y^2)^3(x^2-y^6))$$ so we have $\, c=d=0\,$
and we are done.
| {
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Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this? Using induction, as above I've shown the base case is true. Then if we assume that
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward. Perhaps its not true? Is there a counterexample?
| I know this is an old question but I wanted to add my two cents. What you wronte has a closed for. $\sum_{3j=0}^n\binom{n}{3j} = \frac{2^n+m}{3}$ where m = 2, 1, −1, −2, −1, 1, when n is congruent to
0, 1, 2, 3, 4, 5 (mod 6). according to https://math.hmc.edu/benjamin/wp-content/uploads/sites/5/2019/06/Sums-of-Evenly-Spaced-Binomial-Coefficients.pdf
That said, let us write your special case
$\sum_{j=1}^{n-1} \binom{3n}{3j} = -2 + \sum_{j=1}^{n-1} \binom{3n}{3j}$
Obvoiusly $3n\equiv 0,or 3 mod{6}$. Then m = 2 or -2. Therefore
$\sum_{j=1}^{n-1} \binom{3n}{3j} = -2 +\frac{2^{3n}\pm2}{3} = \frac{2^{3n}\pm2-6}{3} = \frac{2^{3n}-4}{3} or \frac{2^{3n}-8}{3}$
Since $gcd(3,4)=1$ and $4|2^{3n}-4$ and $4|2^{3n}-8$ Then $4|\sum_{j=1}^{n-1} \binom{3n}{3j} |$ .
This gives us even a stronger version of your conclusion
$4|\sum_{j=1}^{n-1} \binom{3n}{3j}$ if n is even and $8|\sum_{j=1}^{n-1} \binom{n}{3j}$ if n is odd I hope this helped
| {
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Cominatorics (sticks and plus, generating function) If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.
My solution with in-ex (Wrong):
$$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} = 25$$ $,1\le x_{i} \ge 4 $
$$x_{i}= 1 +z_{i} $$
$$z_{1}+1 + z_{2}+1 + z_{3}+1 + z_{4}+1 + z_{5} +1+ z_{6}+1 + z_{7}+1 +z_{8} +1+ z_{9}+1 + z_{10}+1 = 25$$
$$z_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$
${sticks+plus} \choose sticks $$=$${24} \choose 15$. That is all cominations when everyone gets at least one.
The forbidden combinations, $z_{i} \ge 4 $,:$$z_{i}=4+u_{i}$$ $$4+u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$
$$u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 11$$
${sticks+plus} \choose sticks $$=$${20} \choose 11$, and this can be done in ten ways. So my answer is: ${24} \choose 15$$-10$${20} \choose 11$
According to the book this is wrong.
When doing it with GF I get: $X^{10}\frac{(1-x^{4})^{10}}{(1-x)^{10}}$. And then I don'tknow what to do. How to now which coefficient to find?
| You have $[ x^{10} ]\cdot\frac{(1-x^4)^{10}}{(1-x)^{10}}$, where $[ x^{k} ]$ denote the coefficient of operator.
$\frac{1}{(1-x)^{10}}$ can be written as $\sum_{j=0}^\infty \binom{-10}{j}(-x)^j$ (Binomial series).
$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{-10}{j}(-x)^j$$
Now we use that $\binom{-r}{s}=\binom{r+s-1}{r-1}\cdot (-1)^s$
$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$
The coefficient of operator $[ x^{10} ] $can be combined with $x^k, 0\leq k\leq 10$ as $[ x^{10} ]\cdot x^k=[ x^{10-k} ]$
$$\left([ x^{10} ]\cdot \binom{10}{0} -[ x^{6} ]\cdot \binom{10}{1}+[ x^{2} ]\cdot \binom{10}{2}-\ldots \right)\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$
Finally we choose the corresponding coefficients
$${19 \choose 9 }-10\cdot {15 \choose 9} +45\cdot {11 \choose 9}=44,803$$
| {
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"url": "https://math.stackexchange.com/questions/2795862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing concentric super ellipses to equal area slices How the concentric super ellipses as shown in the figure can be divided into parts containing equal area such that the total area of the superellipse A = A1 + A2 + ... An where n = 60 in the shown figure given its semi-major axis a and semi-minor axis b. The origin of the sectors lies at the center of the superellipse. Any help is appreciated.
| Following the same idea presented here Dividing an ellipse into equal area, it suffices consider the affinity by $X=ax$ and $Y=by$ such that
$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$
and divide the unit circle in 60 equals parts by the lines
*
*$y=0$
*$y=\pm\frac{\sqrt 3}{3}x$
*$y=\pm{\sqrt 3}x$
*$x=0$
and by the circles
*
*$x^2+y^2=\frac{\sqrt 5}5$
*$x^2+y^2=\frac{\sqrt {10}}5$
*$x^2+y^2=\frac{\sqrt {15}}5$
*$x^2+y^2=\frac{2\sqrt {5}}5$
thus the equations for the lines and ellipses dividing the main ellipse into $60$ sectors of equal area $A=\frac{\pi ab}{60}$ are
*
*$Y=0$
*$Y=\pm\frac{\sqrt 3}{3}\frac b a X$
*$y=\pm{\sqrt 3}\frac b a X$
*$X=0$
for the lines and
*
*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt 5}5$
*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {10}}5$
*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {15}}5$
*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{2\sqrt {5}}5$
| {
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Indefinite integral. Help $\int\frac{2x+3}{x^2-4}dx=\int\frac{2x+3}{(x+2)(x-2)}dx$
So...
$A=\frac{1}{4}$
$B=\frac{7}{4}$
$\frac{1}{4}\log_{10}|x+2|+\frac{7}{4}\log_{10}|x-2|+c$
The result should be: $\log_{10}|x^2-4|+\frac{3}{4}\log_{10}|\frac{x-2}{x+2}|+c$
What did I do wrong?
| Your computation is correct (apart from using $\log_{10}$ instead of $\log$, with implied base $e$, or $\ln$).
Probably the book chose a different approach:
$$
\frac{2x+3}{x^2-4}=\frac{2x}{x^2-4}+\frac{3}{x^2-4}=
\frac{2x}{x^2-4}+\frac{3}{4}\frac{1}{x-2}-\frac{3}{4}\frac{1}{x+2}
$$
so integration gives
$$
\log\lvert x^2-4\rvert+\frac{3}{4}\log\left|\frac{x-2}{x+2}\right|+c
$$
However this is in no way “better” than your form for the antiderivative.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor series of $\sqrt{1+x}$ I have to give the Taylor series of $\sqrt{1+x}$ with development point $0$.
But I am not quite sure what to do.
Normally you have to give the Taylor series up to a certain order.
Well $T_f(x,0)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$
I calculated the derivative of $f(x)=\sqrt{1+x}$ up to order 5 and locked for a regularity.
$f^{(n)}(0)=(-1)^n\cdot\frac{\prod_{k\leq n\,\text{with}\, k\,\text{odd}}k}{2^{n}}$
Then the Taylor series is given by:
$T_{f}(x,0)=\sum_{n=0}^\infty \frac{(-1)^n}{\prod_{1\leq k\leq n\,\text{with}\, k\,\text{even}}k\cdot 2^n}x^n$
Which is pretty ugly to work with.
Can you confirm this solution?
Thanks in advance.
| Almost correct:
$$\begin{align*}
f^{(0)}(x) & = (1+x)^{\frac{1}{2}} \\[1ex]
f^{(1)}(x) & = \frac{1}{2} \cdot (1+x)^{-\frac{1}{2}} \\[1ex]
f^{(2)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot (1+x)^{-\frac{3}{2}} \\[1ex]
\vdots \\
f^{(n)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot \ldots \cdot \left( - \frac{2n-3}{2} \right) \cdot (1+x)^{-\frac{2n-1}{2}} \\[1ex]
f^{(n)}(0) & = \frac{(-1)^{\color{red}{n-1}}}{2^n} \cdot \prod_{\substack{k \leqslant \color{red}{2n-3} \\ k \text{ odd}}} k \cdot (1+x)^{-\frac{2n-1}{2}}
\end{align*}$$
Usually a convenient notation is used: for $\mu \in \mathbb{R}$ and $n \in \mathbb{N}$ let
$$\binom{\mu}{n} = \frac{\mu \cdot (\mu-1) \cdot \ldots \cdot (\mu - n + 1)}{n!}.$$
Then the Taylor series expresses nicely as
$$T_f(x, 0) = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
closed formula for Sum of this series Is there any closed form of this series?
$$\frac{x!}{1!} + \frac{(x+1)!}{2!} + \frac{(x+2)!}{3!} +\cdots + \frac{(x+n-1)!}{n!}$$
I tried to manipulate the expansion of $(1+x)^n$ but can't seem to get the term $(1+x)!$ from $x!$. Need help on this one.
| By the Pascal formula we get:
$$\begin{align}
\frac{x!}{1!}+\dots+\frac{(x+n-1)!}{n!}&=(x-1)!\left(\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}\right)\\\\
&=(x-1)!\left(\binom{x}{x}+\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}-1\right)\\\\
&=(x-1)!\left(\binom{x+n}{x}-1\right)\\\\
&=x\cdot\frac{(x+n)!}{n!}-(x-1)!
\end{align}$$
Where I added and subtracted a term $\binom{x}{x}$ so I can use the Pascal formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}}$
Using the substitution $t=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ the equation can be written as
$(t)^x+(\frac{1}{t})^x=2^x$.
And from this i wrote:
$t^{2x}-(2t)^x-1=0$.
Now i don't how to proceed. Any suggestions?
Thanks!
| Hint:
$$\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}=\cos(45^\circ+30^\circ)=\cos75^\circ$$
Similarly, $$\dfrac{\sqrt{2+\sqrt3}}2=\sin75^\circ$$
Now for $0<A<90^\circ,(\cos A)^x,(\sin A)^x$ is deceasing fucntion
| {
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Finding $\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}$.
Find an expression for
$$\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}.$$
This was given in the chapter for APs. However, I do not see how this relates to them. I tried using telescopic sums, but I am not proficient in them and was thererfore unable to solve this question.
I tried calculating the sum for a few consecutive values, and tried constructing a polynomial that covered them all, but it proved to be difficult.
Any hints on how to solve this question?
Please help.
| A telescoping series:
$\sum_{n=1}^{k} f(n) - f(n+1) = f(1) - f(k+1)$
How do we get our series into that form?
$(n^4 + n^2 + 1) = (n^2+n +1)(n^2-n+1)\\
\frac {n}{(n^4 + n^2 + 1)} = \frac {1}{2(n^2 - n + 1)} - \frac {1}{2(n^2 + n + 1)}$
Here is the really tricky bit:
$\frac {1}{(n^2 + n + 1)} = \frac {1}{(n+1)^2 - (n+1) + 1}$
$\sum_{n=1}^{k}\frac {n}{n^4 + n^2 + 1} = \frac 12 - \frac {1}{2(k^2+k+1)}$
| {
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Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. As the question states:
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related question, and understand the steps to be:
*
*find the first few terms of the Taylor polynomial.
*Generalize the terms by making use of an infinite sum to represent the function as the Taylor series.
*use the infinite sum in the ratio test to find the radius of convergence.
Progress so far:
*
*The first 6 terms (n = 0 to n = 5) of the Taylor polynomial I have calculated to be:
$x^3 \cdot \ln{(\sqrt{x})} + \frac{1}{2}(x-a) + \frac{5}{4}(x-a)^2 + \frac{11}{12}(x-a)^3 + \frac{1}{8}(x-a)^4 - \frac{1}{40}(x-a)^5$
It is at this point however that I fall over. It is not intuitive to me how I can write the c-terms as a function without utilizing some sort of online maths engine for fitting the data to a curve.
Is there some sort of first-year-student-friendly technique for modelling these data points systematically? Alternatively, does someone have an intuition they would be willing to share for solving this problem?
| Write $x=1+t$, so you need to find the Taylor series of
$$
\frac{1}{2}(1+t)^3\ln(1+t)
$$
at $t=0$. Since
$$
\ln(1+t)=\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}
$$
your Taylor series is
$$
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+1}}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+2}}{n}+
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+3}}{n}
$$
that can be rewritten as
$$
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+
\frac{3}{2}\sum_{n>1}\frac{(-1)^{n}t^{n}}{n-1}+
\frac{3}{2}\sum_{n>2}\frac{(-1)^{n-1}t^{n}}{n-2}+
\frac{1}{2}\sum_{n>3}\frac{(-1)^{n-2}t^{n}}{n-3}
$$
Now isolate the only term for $n=1$, the two terms for $n=2$ and the three terms for $n=3$:
$$
\frac{t}{2}+\frac{1}{2}\left(-\frac{t^2}{2}+3t^2\right)+
\frac{1}{2}\left(\frac{t^3}{3}-\frac{3t^3}{2}+3t^3\right)+\\
\frac{1}{2}\sum_{n>3}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n3}{n-1}+\frac{(-1)^{n-1}3}{n-2}+\frac{(-1)^{n-2}}{n-3}\right)t^n
$$
The last summation can be rewritten as
$$
\frac{1}{2}\sum_{n>3}\left(-\frac{1}{n}+\frac{3}{n-1}-\frac{3}{n-2}+\frac{1}{n-3}\right)(-1)^nt^n=
\sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)}
$$
so the end result is
$$
\frac{t}{2}+\frac{5}{4}t^2+\frac{11}{12}t^3+\sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)}
$$
and you can check that the coefficient for $t^4$ is indeed $1/8$; the coefficient for $t^5$ is $-1/40$ as you computed.
The Taylor series for $f(x)=x^3\ln\sqrt{x}$ around $1$ is obtained by substituting $t$ with $x-1$. The radius of convergence is the same as for $\ln(1+t)$, that is, $1$. You can also compute it with the ratio test, as further check.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which integers can be written as $x^2+2y^2-3z^2\ $?
For which integers $n$ has the diophantine equation $$x^2+2y^2-3z^2=n$$ solutions ?
These theorems
https://en.wikipedia.org/wiki/15_and_290_theorems
do not apply because the given quadratic form is not positive (or negative) definite. It seems that the quadratic form is universal (for every integer $n$ a solution exists) , but I have no idea how this can be proven.
| The jpegs did not come out well.... in Modern Elementary Theory of Numbers by Leonard Eugene Dickson, (1939) , on the top of page 161, homework exercise 2 is this: given that $C$ is a positive integer, $$ x^2 + 2 y^2 - C z^2 $$ is universal if and only if $C$ is odd and every prime factor of $C$ is $\; \equiv 1 \; \mbox{or} \; 3 \pmod 8$
Dickson and his students Ross and Oppenheim found all universal indefinite ternary quadratic forms, collected into four types (up to $SL_3 \mathbb Z$ equivalence of forms). Take $M$ any integer and $N$ any odd integer, not necessarily positive in either case. In all four cases it it obvious that they are universal, just experiment a bit with each. In the first, take $(x,1,0)$ for example.
$$ xy - M z^2 $$
$$ 2xy - N z^2 $$
$$ 2xy + y^2- N z^2 $$
$$ 2xy + y^2- 2 N z^2 $$
In the fourth form, (I) for odd numbers $(x,1,0);$ (II) for numbers $2 \pmod 4$ take $(x,2,1);$ (III) for numbers $0 \pmod 4$ take $(x,2,0).$
Notice there is no $x^2$ term in any of these, so that the form evaluated at $(1,0,0)$ comes out to $0. \;$ A fundamental part of this is that a universal (ternary) form must non-trivially represent $0.$ Not true for quaternaries, such as $w^2 - 2 x^2 - 3 y^2 + 6 z^2$
Meanwhile, your $x^2 + 2 y^2 - 3 z^2$ is equivalent to the fourth type listed above, namely $2xy+y^2 + 6 z^2.$ For quadratic forms, we take (half) the Hessian matrix of second partials; forms with such Gram matrices $G,H$ are equivalent when there is an integer matrix $P$ with $\det P = 1$ and $P^T GP = H.$ Notice that this means $Q = P^{-1}$ is also of integers with $\det Q = 1,$ while $Q^T H Q = G.$ I have a program that finds me such matrices $P.$ Equivalent forms integrally represent exactly the same values.
The first matrix identity below reads $$ (x+y)^2 + 2 (x+3z)^2 - 3(x+2z)^2 = 2xy+y^2 + 6 z^2 $$
$$ $$
$$
\left(
\begin{array}{ccc}
1&1&1 \\
1&0&0 \\
0&3&2 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&0&0 \\
0&2&0 \\
0&0&-3 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&1&0 \\
1&0&3 \\
1&0&2 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
0&1&0 \\
1&1&0 \\
0&0&6 \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
0&1&0 \\
-2&2&1 \\
3&-3&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
0&1&0 \\
1&1&0 \\
0&0&6 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
0&-2&3 \\
1&2&-3 \\
0&1&-1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&0&0 \\
0&2&0 \\
0&0&-3 \\
\end{array}
\right)
$$
For the second matrix identity, take $u = -2y+3z, \; v = x + 2 y - 3 z, \; w = y - z, \;$ giving
$$ 2uv+v^2 + 6 w^2 = x^2 + 2 y^2 - 3 z^2 $$
| {
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If $\frac 1{a+b+c}= \frac 1a+ \frac 1b+ \frac 1c$, then show for odd $n$, $ \frac 1{a^n+b^n+c^n}= \frac 1{a^n}+ \frac 1{b^n}+ \frac 1{c^n}$ The stated duplicate question does not address solution in terms of $p, q, r,$ i.e. not in terms of the Viete's formula. My approach is based on the viete's formula by finding the relation between the coefficients of a polynomial to sums and products of its roots.
If $\frac 1{a+b+c}= \frac 1a+ \frac 1b+ \frac 1c$, then show for odd $n$, $ \frac 1{a^n+b^n+c^n}= \frac 1{a^n}+ \frac 1{b^n}+ \frac 1{c^n}$
Cannot establish the significance of odd $n$, i.e., why not applicable for even $n$.
Also, a simple approach is: $\frac 1{a+b+c}= \frac{ab+bc+ca}{abc}\implies abc = (ab+bc+ca)(a+b+c)$.
The two product terms are usually associated to a polynomial (as shown below) with $3$ roots: $a,b,c,$ having $-p=$ sum of roots $=a+b+c$, and $q=$ sum of product of roots two at a time$=ab+bc+ca$, and the single term for all $3$ roots $=-r= abc.$
So, $-r = -pq\implies r = pq$.
Derivation of polynomial with $3$ roots:
Let $a,b,c$ be the roots of $x^3 + px^2 + qx + r = 0$. Then, writing the polynomial in terms of roots gives $(x - a)(x - b)(x -c)$.
$\therefore$ $x^3 + px^2 + qx + r = (x - a)(x - b)(x - c)$.
$= (x^2 - (a + b)x + ab)(x - c)$
$= x^3 - (a + b)\cdot x^2 + (ab)\cdot x - c\cdot x^2 + (a+b)c\cdot x - abc$
$= x^3 - (a + b +c) x^2 + (ab + bc + ca) x - abc$
$\therefore$ equating coefficients
(a) $a + b + c = -p$.
(b) $ab+ bc + ca = q$.
(c) $abc = -r$.
Unable to proceed further.
I want to add that the source has cryptic answer that has next step as : The required polynomial is : $f(x)= x^3 +px^2+qx+pq$. <-- This step stucks me. The next one ones are easy: This implies $(x+p)(x^2+q)$. As one root is $=-p$, so $a=-p= a+b+c\implies c=-b$. So, $\frac 1{a^n+b^n+c^n} = \frac 1{a^n}=\frac 1{a^n+b^n-b^n}=\frac 1{a^n+b^n+c^n}$.
The last part uses odd $n$ property, but the stucking step is issue.
| The polynomial is $$f(x)=x^3+px^2+qx+r=(x-a)(x-b)(x-c)$$
You have proven that $r=pq$ from the condition that is given.
$$(x+p)(x^2+q)=(x-a)(x-b)(x-c)$$
$$(x+p)(x-i\sqrt{q)}(x+i\sqrt{q})=(x-a)(x-b)(x-c)$$
WLOG, if $a=-p$, then $b=-c$ and $\frac1{b}=-\frac1{c}$.
If $n$ is odd, then we have $b^n=(-1)^nc^n=-c^n$ and $\frac1{b^n}=(-1)^n\frac1{c^n}=-\frac1{c^n}$, that is we have $b^n+c^n=0$ and $\frac1{b^n}+\frac1{c^n}=0$. Both sides of the equation that we want to show become $\frac1{a^n}$.
However, if $n$ is even, then we have $b^n=c^n$ and $\frac1{b^n}=\frac1{c^n}$ and the LHS of what we want to show become $\frac1{a^n+2b^n}$ but the RHS becomes $\frac1{a^n}+\frac2{b^n}$.
| {
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Summation of product of integers with consecutive integers as first $m$ factors It is well known that
$$\sum_{r=1}^n r^\overline{m}=\frac {n^{\overline{m+1}}}{m+1}\\
\text{i.e.}
\sum_{r=1}^n \scriptsize(r+1)(r+2)\cdots (r+m-1)=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\\
\text{which can also be written as }\\
\qquad \scriptsize m!\sum_{r=1}^n\binom {r+m-1}m=\frac 1{m+1}\binom {n+m}{m+1}$$.
Can it be shown that
$$\sum_{r=1}^n \left[r^\overline{m}\prod_{k=1}^q (r+p_k)\right]
=\frac {n^{\overline{m+1}}}{m+1}P_q(n)\\
\text{i.e. }\tiny \sum_{r=1}^n r(r+1)(r+2)\cdots (r+m-1)(r+p_1)(r+p_2)\cdots (r+p_q)
=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\cdot \left(a_qn^q+a_{q-1}n^{q-1}+\cdots+a_1 q+a_0\right)$$
where $p_k$ are positive integers greater than $m$, and $P_q(n)$ is a polynomial of degree $q$ in $n$?
A few examples (with solutions from Wolframalpha):
1. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)}(r+5)=\frac 1{4}\boxed{n(n+1)(n+2)}(n+7)$
2. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)}(r+4)(r+8)=\frac 1{12}\boxed{n(n+1)(n+2)(n+3)}(2n^2+30n+103)$
3.$\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)(r+3)}(r+5)(r+8)=\frac 1{105}\boxed{n(n+1)(n+2)(n+3)(n+4)}(15n^2+235n+884)$
|
The assumption is correct (and in fact can be generalized) since we have
\begin{align*}
\sum_{r=1}^nr^{\overline{m+1}}&=\frac{n^{\overline{m+2}}}{m+2}\\
&=\frac{n^{\overline{m+1}}}{m+1}\cdot\frac{m+1}{m+2}(n+m+1)\\
&=\frac{m+1}{m+2}(n+m+1)\sum_{r=1}^n r^{\overline{m}}\tag{1}
\end{align*}
as well as
\begin{align*}
\sum_{r=1}^nr^{\overline{m}} r&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)\\
&=\sum_{r=1}^nr^{\overline{m+1}}-m\sum_{r=1}^nr^{\overline{m}}\tag{2}\\
\end{align*}
and for positive integer $k$
\begin{align*}
\sum_{r=1}^nr^{\overline{m}} r^k&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)r^{k-1}\\
&=\sum_{r=1}^nr^{\overline{m+1}}r^{k-1}-m\sum_{r=1}^nr^{\overline{m}}r^{k-1}\\
\end{align*}
Let's look at OP's first example with
\begin{align*}
\sum_{r=1}^nr(r+1)&=\sum_{r=1}^nr^{\overline{2}}=\frac{1}{3}n(n+1)(n+2)\tag{3}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{r=1}^nr^{\overline{2}}(r+5)}
&=\sum_{r=1}^nr^\overline{2}r+5\sum_{r=1}^nr^{\overline{2}}\\
&=\left(\sum_{r=1}^nr^{\overline{3}}-2\sum_{r=1}^nr^{\overline{2}}\right)+5\sum_{r=1}^nr^{\overline{2}}\tag{apply 2}\\
&=\frac{3}{4}(n+3)\sum_{r=1}^nr^{\overline{2}}+3\sum_{r=1}^nr^{\overline{2}}\tag{apply 1}\\
&=\frac{3}{4}(n+7)\sum_{r=1}^nr^{\overline{2}}\\
&\,\,\color{blue}{=\frac{1}{4}n(n+1)(n+2)(n+7)}\tag{apply 3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$
Compute $I=\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$
My 2 attempts:
First:
We observe that $\frac {1}{x^2+2x+2}=\frac {1}{(x+1)^2+1}$ and $\frac 1{(x+1)^2+1}=(\arctan{(x+1)})^{'}.$ Then:
$$\int_0^2\frac{\arctan{x}}{x^2+2x+2}dx=\int_0^2\arctan{x}(\arctan{(x+1)})^{'}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{1\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{(1+x^2-x^2)\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\arctan{(x+1)}dx-\int_0^2\frac{x^2\arctan{(x+1)}}{1+x^2}dx$$
Where the second integral is pretty easy to solve using integration by parts but the second one is not very pleasant so... I thought I should try something else.
Second attempt:
Let $u=\arctan(x+1)$ then $x=\tan u-1$ then $$I=\int_{\arctan1}^{arctan3}\arctan{(\tan u -1)du}$$
then I could let $w=\tan(u)-1$ then $dw=((w+1)^2+1)du$... but it's a lot of work and I really think you can do this more easily...
Any hints?
| Some thoughts here:
\begin{align}
I&=\int\frac{\arctan x dx}{(x+1)^2+1}\\
&=\int\frac{\arctan (u-1) du}{u^2+1}\\
&=\int\frac{\arctan(\tan w-1)\sec^2w dw}{\sec^2w}\\
&=\int\arctan(\tan w-1)dw
\end{align}
Using similar substitutions, consider
$$J=\int\frac{\text{arccot} x dx}{(x+1)^2+1}=\int\text{arccot}(\tan w-1)dw$$
Notice
$$\arctan x+\text{arccot}x=\frac\pi 2$$
So
$$I+J=\frac{\pi}{2}\int^2_0\frac{dx}{(x+1)^2+1}=\frac{\pi}2(\arctan3-\frac{\pi}4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Gauss' law and a half-cylinder The question is:
A half cylinder with the square part on the $xy$-plane, and the length $h$ parallel to the $x$-axis. The position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$.
$S_1$ is the curved portion of the half-cylinder $z=(r^2-y^2)^{1/2}$ of length $h$.
$S_2$ and $S_3$ are the two semicircular plane end pieces.
$S_4$ is the rectangular portion of the $xy$-plane
Gauss' law:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}$$
$\mathbf E$ is the electric field $\left(\frac{\text{Newton}}{\text{Coulomb}}\right)$.
$\mathbf{\hat{n}}$ is the unit normal vector.
$dS$ is an increment of the surface area $\left(\text{meter}^2\right)$.
$q$ is the total charge enclosed by the half-cylinder $\left(\text{Coulomb}\right)$.
$\epsilon_0$ is the permitivity of free space, a constant equal to $8.854\times10^{-12}\,\frac{\text{Coulomb}^2}{\text{Newton}\,\text{meter}^2}$.
The electrostatic field is:
$$\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$$
where $\lambda$ is a constant.
Use this formula to calculate the part of the total charge $q$ for the curved portion $S_1$ of the half-cylinder:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}=\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$
The goal is to find the total charge $q$ enclosed by the half-cylinder, expressed in terms of $\lambda$, $r$ and $h$.
The solution should be:
$$\pi r^2\lambda h\epsilon_0$$
This is what I've tried:
First calculate Gauss' law for $S_1$:
\begin{align}
f(x,y)&=z=(r^2-y^2)^{1/2}=\sqrt{(r^2-y^2)} \\
\frac{\partial f}{\partial x}&=\frac12(r^2-y^2)^{-\frac12}\cdot 0=0 \\
\frac{\partial f}{\partial y}&=\frac12(r^2-y^2)^{-\frac12}\cdot -2y=-\frac{y}{\sqrt{(r^2-y^2)}}=-\frac yz \\
\\
\mathbf{E}&=\lambda(x\mathbf{i}+y\mathbf{j}) \\
E_x[x,y,f(x,y)]&=\lambda x \\
E_y[x,y,f(x,y)]&=\lambda y \\
E_z[x,y,f(x,y)]&=0 \\
\\
\text{length}&=h \\
\end{align}
Using the formula
$$\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$
we get:
\begin{align}
&\iint_R\left\{-\lambda x\cdot 0-\lambda y\cdot -\frac{y}{z} + 0\right\}\,dx\,dy \\
&=\iint_R\frac{\lambda y^2}{z}\,dx\,dy \\
&=\lambda\iint_R\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
\end{align}
Since the length is $h$ and the length is parallel to the $x$-axis:
\begin{align}
&\lambda \int_R\int_0^h\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
&=\lambda\int_R\left[\frac{y^2x}{\sqrt{r^2-y^2}}\right]_0^h\,dy \\
&=\lambda\int_R\frac{y^2h}{\sqrt{r^2-y^2}}\,dy \\
\end{align}
Subsitute:
\begin{align}
y&=r\sin\theta \\
\theta&=\arcsin\left(\frac1r y\right) \\
\frac{dy}{d\theta}&=\frac{d}{d\theta}\left(r\sin\theta\right)=r\cos\theta \\
dy&=r\cos(\theta)\,d\theta \\
\\
&\lambda\int\frac{hr^2\sin^2\theta}{\sqrt{r^2-r^2\sin^2\theta}}\cdot r\cos(\theta)\,d\theta \\
&=\lambda h\int\frac{r^3\sin^2\theta\cos\theta}{r\sqrt{1-\sin^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\cos\theta}\,d\theta \\
&=\lambda hr^2\int\sin^2\theta\,d\theta \\
&=\lambda hr^2\int\frac{1-\cos2\theta}{2}\,d\theta \\
&=\frac12\lambda hr^2\int1-\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\int1\,d\theta-\int\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\left[\theta-\frac12\sin2\theta\right] \\
\\
\text{substitute back } \theta=\arcsin\left(\frac1r y\right)\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac1r y\right)-\frac12\sin\left(2\arcsin\left(\frac1r y\right)\right)\right] \\
\text{the boundaries of }y\text{ are }-r\text{ and }r\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac{y}{r}\right)-\frac12\sin\left(2\arcsin\left(\frac{y}{r}\right)\right)\right]_{-r}^r \\
&=\frac12\lambda hr^2\left(\frac{\pi}{2}-0\right) - \left(-\frac{\pi}{2}-0\right) \\
&=\frac12\pi\lambda hr^2
\end{align}
Calculate Gauss' law for $S_2$ and $S_3$:
The surfaces of $S_2$ and $S_3$ are equal.
Since:
$\bullet$ the position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$, the direction of the electrostatic field at both surfaces is opposite: $(\lambda x \mathbf{i})$,
$\bullet$ and the unit normal vectors are in opposite direction,
the addition of the result of Gauss' law will not be equal to 0.
The surface of each of the surfaces is $\frac12 \pi r^2$.
The electric field in the $x$-direction is $\lambda x\mathbf{i}$.
$x$ for $S_2$ = $\frac12 h$.
$x$ for $S_3$ = $-\frac12 h$.
$\mathbf{\hat{n}}$ for $S_2$ = $\mathbf{i}$.
$\mathbf{\hat{n}}$ for $S_3$ = $-\mathbf{i}$.
Therefore for $S_2$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda \frac12 h\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}
And for $S_3$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda (-\frac12 h)\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}
Calculate Gauss' law for $S_4$:
Since $S_4$ lies in the $xy$-plane, the electrostatic field $\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$ lies parallel to the surface, thus the result of Gauss' law is $0$.
The net result is:
\begin{align}
\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS&=\frac{q}{\epsilon_0} \\
&=\frac12 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 +0 \\
&=\pi\lambda hr^2 \\
\end{align}
The total charge $q$ enclosed in the half-cylinder is thus:
$$q=\pi\lambda hr^2 \epsilon_0$$.
This solves the problem I was having.
| The lower and upper boundaries for $y$ in $S_1$ should be $-r$ and $r$ instead of $0$ and $r$. I was thinking in polar coordinates instead of in Cartesian coordinates...
I'll edit the original post accordingly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proof of divergence of harmonic series with Cauchy criteria I have a proof where the divergence harmonic series is shown via the Cauchy criteria.
$\epsilon := 1/2$ and $m := 2n$.
$$
a_m - a_n = (1 + \frac{1}{2} + \ldots + \frac{1}{m}) - (1+\frac{1}{2} + \ldots + \frac{1}{n}) \\
= (1 + \frac{1}{2} + \ldots + \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{2n}) - (1+\frac{1}{2} + \ldots + \frac{1}{n}) \\
= (\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n}) \geq \frac{1}{2n} + \frac{1}{2n} + \ldots + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2} = \epsilon)
$$
I'm trying to understand where the $\geq$ and the $\frac{1}{2n} + \frac{1}{2n} + \ldots + \frac{1}{2n}$ comes from in the last line. Also, why should $\frac{1}{n+1}$ be larger than $\frac{1}{2n}$ if in the second line the series is progressing from $\frac{1}{n+1}$ to $\frac{1}{2n}$. In my understanding, the value should then be smaller.
Thanks for your help!
| Observe that if $1 \le k \le n \implies n+k \le 2n \implies \dfrac{1}{n+k} \ge \dfrac{1}{2n}$. Let $k$ varies from $1$ to $n$ and add them up. Also the number of inequalities is $2n-(n+1)+1 = n$ which explains the numerator $n$ of the fraction $\dfrac{n}{2n}$ that is $\dfrac{1}{2}$ after simplifying .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A couple of big-O questions about a problem from number theory I am going through official solutions for IMO'09 problems. Currently stuck at solution 2 for N7. I am copying parts that confuse me below, but if anyone is interested in the rest, this is a link to full pdf.
Now, here are my questions:
*
*Letting $s_n=(ab)^n-\gamma_1\left(\frac{a}{b}\right)^n-\ldots-\gamma_k\left(\frac{a}{b^{2k-1}}\right)^n$, I don not quite get how they derive $z_n=s_n+O\left(\frac{b}{a}\right)^n$. Will the following explanation be good enough?
If we let $\alpha_n=z_n-s_n$, then $(s_n+\alpha_n)^2=z_n^2\equiv z_n^2+0$. From the solution $z_n^2+O(B^n)=\{s_n+O\left(\frac{b}{a}\right)^n\}^2$, and since $0=O(B^n)$, we necessarily have $\alpha_n=O\left(\frac{b}{a}\right)^n$. Is that it?
*When coefficients $\delta_i$ are introduced, $\delta_0$ is determined to be greater than $0$ because $z_n>0$. How? Is it because for large values of $n$ term $(ab)^n$ will dominate, so its coefficient needs to be positive?
*After the last centered formula they derive $a=b^{k-1}$ from asymptotic calculations. What are these calculations and why does $a=b^{k-1}$ follow from them? Something like the following?
If we assign $\epsilon_n$ to be $O\left(\frac{A}{B^k}\right)^n$, then $\epsilon_n\to0$, because $A<B^k$. If in the last centered formula we now move everything, except $\epsilon_n$, to the left side, we will have, in particular, $A^nB^n(1-\delta_0^2)$ there. If $1-\delta_0^2\neq0$ this term is dominating for large $n$ and thus the left side will never tend to $0$, but the right side tends to $0$, so $1-\delta_0^2=0$. And so on for $\delta_1$ and the rest of them…
Is that it?
*Immediately after what is discussed in 3. they infer that there should be some $P\in\mathbb{Q}[x]$ for which $(X-1)(X^{k-1}-1)=P(X)^2$. Why? As far as I can see, if we let $X=B^n$, then $A^n=X^{k-1}$ and we have $(A^n-1)(B^n-1)=(X-1)(X^{k-1}-1)$. If we now view it as polynomial in variable $X$, from the assumption that $(A^n-1)(B^n-1)$ is a perfect square for all $n$ this polynomial maps countably infinite number of integers to perfect squares. How do we now derive that it necessarily needs to be a perfect square of some polynomial in $X$?
Thank you.
| $\def\peq{\mathrel{\phantom{=}}{}}$First, there is a typo in the definition of $\{γ_n\}$. It should be$$
2γ_1 = 1, \quad γ_n = \sum_{j = 1}^{n - 1} γ_j γ_{n - j}. \quad (n \geqslant 2)
$$
For the first question, there is some abuse of notations in the official answer. In fact, since $0 < γ_n < 1$ for all $n \geqslant 1$ and $B^{k - 1} \leqslant A < B^k$, then\begin{align*}
s_n^2 &= \left( (ab)^n - \sum_{j = 1}^k γ_j \left( \frac{a}{b^{2j - 1}} \right)^n \right)^2\\
&= ((ab)^n)^2 - 2(ab)^n · \sum_{j = 1}^k γ_j \left( \frac{a}{b^{2j - 1}} \right)^n + \left( \sum_{j = 1}^k γ_j \left( \frac{a}{b^{2k - 1}} \right)^n \right)^2\\
&= (AB)^n - \sum_{j = 1}^k 2γ_j \left( \frac{A}{B^{j - 1}} \right)^n + \sum_{j = 2}^k \sum_{l = 1}^{j - 1} γ_l γ_{j - l} \left( \frac{a}{b^{2l - 1}} \right)^n \left( \frac{a}{b^{2(j - l) - 1}} \right)^n\\
&\peq + \sum_{\substack{1 \leqslant j, l \leqslant k\\j + l \geqslant k + 1}} γ_j γ_l \left( \frac{a}{b^{2j - 1}} \right)^n \left( \frac{a}{b^{2l - 1}} \right)^n\\
&= (AB)^n - \sum_{j = 1}^k 2γ_j \left( \frac{A}{B^{j - 1}} \right)^n + \sum_{j = 2}^k \sum_{l = 1}^{j - 1} γ_l γ_{j - l} \left( \frac{A}{B^{j - 1}} \right)^n + O\left( \left( \frac{A}{B^k} \right)^n \right)\\
&= (AB)^n - 2γ_1 A^n - \sum_{j = 2}^k \left( 2γ_j - \sum_{l = 1}^{j - 1} γ_l γ_{j - l} \right) \left( \frac{A}{B^{j - 1}} \right)^n + O\left( \left( \frac{A}{B^k} \right)^n \right)\\
&= (AB)^n - A^n + O\left( \left( \frac{A}{B^k} \right)^n \right) = z_n^2 + B^n - 1 + O(B^n). \tag{1}
\end{align*}
Note that $z_n \sim (ab)^n\ (n → ∞)$ and $a > b$, then (1) implies $s_n \sim (ab)^n\ (n → ∞)$ and$$
|z_n - s_n| = \frac{|z_n^2 - s_n^2|}{z_n + s_n} \sim \frac{B^n}{2(ab)^n} = 2\left( \frac{b}{a} \right)^n, \quad n → ∞
$$
which implies $z_n = s_n + O\left( \left( \dfrac{b}{a} \right)^n \right)$.
For the second question, your reasoning is correct.
For the third question, it can be derived analogously to (1) that for some $N_0 \geqslant 1$ and any $n \geqslant N_0$,\begin{align*}
&\peq A^n B^n - A^n - B^n + 1 = z_n^2 = \left( δ_0 (ab)^n - \sum_{j = 1}^k δ_j \left( \frac{a}{b^{2j - 1}} \right)^n \right)^2\\
&= δ_0^2 (AB)^n - 2δ_0 δ_1 A^n - \sum_{j = 2}^k \left( 2δ_0 δ_j - \sum_{l = 1}^{j - 1} δ_l δ_{j - l} \right) \left( \frac{A}{B^{j - 1}} \right)^n + O\left( \left( \frac{A}{B^k} \right)^n \right). \tag{2}
\end{align*}
Note that $A > B$. First, dividing by $(AB)^n$ on both sides of (2) and making $n → ∞$ yields $δ_0^2 = 1$, thus $δ_0 = 1$. Next, subtracting $(AB)^n$ from both sides of (2), then dividing by $A^n$ and making $n → ∞$ yields $2δ_0 δ_1 = 1$, thus $δ_1 = \dfrac{1}{2}$. Now, for $m = 2, \cdots, k - 2$, each time subtracting$$(AB)^n - A^n - \sum_{j = 2}^{m - 1} \left( 2δ_j - \sum_{l = 1}^{j - 1} δ_l δ_{j - l} \right) \left( \frac{A}{B^{j - 1}} \right)^n
$$
from both sides of (2), then dividing by $\left( \dfrac{A}{B^{m - 1}} \right)^n$ and making $n → ∞$ yields $2δ_m = \sum\limits_{j = 1}^{m - 1} δ_j δ_{m - j}$. Now, (2) reduces to\begin{align*}
-B^n + 1 &= - \left( 2δ_0 δ_{k - 1} - \sum_{l = 1}^{k - 2} δ_l δ_{k - 1 - l} \right) \left( \frac{A}{B^{k - 2}} \right)^n\\
&\peq - \left( 2δ_0 δ_k - \sum_{l = 1}^{k - 1} δ_l δ_{k - l} \right) \left( \frac{A}{B^{k - 1}} \right)^n + O\left( \left( \frac{A}{B^k} \right)^n \right). \tag*{(3)}
\end{align*}
Suppose $a > b^{k - 1}$, then $A > B^{k - 1}$ and analogously $2δ_{k - 1} = \sum\limits_{j = 1}^{k - 2} δ_j δ_{k - 1 - j}$. Then (3) becomes$$
-B^n + 1 = -\left( 2δ_0 δ_k - \sum_{l = 1}^{k - 1} δ_l δ_{k - l} \right) \left( \frac{A}{B^{k - 1}} \right)^n + O\left( \left( \frac{A}{B^k} \right)^n \right),
$$
and divinding by $B^n$ and making $n → ∞$ leads to contradiction (Note that $A < B^k$). Thus, $a = b^{k - 1}$ and $k \geqslant 3$.
For the last question, I do not understand the logic within that step, but here is a way using similar ideas: Since $a = b^{k - 1}$, then$$
(b^{2n(k - 1)} - 1)(b^{2n} - 1) = z_n^2 = \left( δ_0 b^{nk} - \sum_{j = 1}^k δ_j b^{n(k - 2j)} \right)^2\\
\Longrightarrow ((b^n)^{2(k - 1)} - 1)((b^n)^2 - 1)(b^n)^{2k} = \left( δ_0 (b^n)^{2k} - \sum_{j = 1}^k δ_j (b^n)^{2(k - j)} \right)^2.
$$
Thus,$$
(X^{2(k - 1)} - 1)(X^2 - 1)X^{2k} = \left( δ_0 X^{2k} - \sum_{j = 1}^k δ_j X^{2(k - j)} \right)^2 = (Q(X))^2,
$$
which leads to contradiction since $k \geqslant 3$ and $\dfrac{X^{2(k - 1)} - 1}{X^2 - 1}$ has no multiple zeros.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tricky question on polynomials
For any real numbers $x$ and $y$ satisfying $x^2y + 6y = xy^3 +5x^2 +2x$, it is known that $$(x^2 + 2xy + 3y^2) \, f(x,y) = (4x^2 + 5xy + 6y^2) \, g(x,y)$$
Given that $g(0,0) = 6$, find the value of $f(0,0)$.
I have tried expressing $f(x,y)$ in terms of $g(x,y)$. But seems that some tricks have to been done to further on the question. Can anyone figure out the expression?
| The given equation implies $$\frac{y}{x} = \frac{y^2 + 5x + 2}{x^2 + 6}$$ so we can rewrite the expression for $f/g$ as $$\frac{4 + 5(y/x) + 6(y/x)^2}{1+2(y/x) + 3(y/x)^2} = \frac{4(x^2 + 6)^2 + 5(y^2 + 5x+2)(x^2 + 6) + 6(y^2 + 5x+2)^2}{(x^2+6)^2 + 2(y^2 + 5x+2)(x^2 + 6) + 3(y^2 + 5x+2)^2}$$
Assuming $f$ and $g$ are supposed to be continuous, we see that (taking the limit along the curve defined by the equation) this rewritten expression gives $$\frac{f(0,0)}{6}=\frac{f(0,0)}{g(0,0)} = \frac{36\cdot 4 + 5\cdot 12 + 24}{36+24+12} = \frac{228}{72} = \frac{19}{6} \implies f(0,0) = 19$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0}^{1} \frac{\ln(1-x)}{x}dx = -\int_{0}^{1}\frac1x \cdot (x+x^2/2 +x^3/3 \ldots)dx =- \int_{0}^{1}(1+x/2+x^2/3 \dots )dx = -( 1 + \frac1{2^2} + \frac1{3^3} \dots) = -\frac{\pi^2}{6}$$
Using these integral, I evaluate $$ \int_{0}^{1} \frac{\ln(x)}{1-x} $$
by letting $1-x=t$
$$ \int_{0}^{1} \frac{\ln(x)}{1-x} =\int_{0}^{1} \frac{\ln(1-t)}{t} dx = -\frac{\pi^2}{6}$$
However, I'm facing problem while integrating
$$\int_{0}^{1} \frac{\ln(x)}{1+x}$$ using the same above way.
| Hint: integrate by parts.
Let $u = \ln(x)\implies u' = \dfrac1x$ and $v' = \dfrac1{1 + x}\implies v = \ln(1 + x)$. Therefore,
$$\int\dfrac{\ln(x)}{1 + x}\,\mathrm dx = \ln(x)\ln(1 + x) - \int\dfrac{\ln(1 + x)}x\,\mathrm dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x_3) = x^3+px+q = 0$$
Now find the coefficient near the powers of $x$:
$$
x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 = x^3+px+q
$$
That means that I can write a system of equations:
$$
\begin{cases}
-(x_1 + x_2 + x_3) = 0 \\
x_1x_2 + x_1x_3 + x_2x_3 = p \\
- x_1x_2x_3 = q
\end{cases}
$$
At this point I got stuck. I've tried to raise $x_1 + x_2 + x_3$ to 3 power and expand the terms, but that didn't give me any insights. It feels like I have to play with the system of equations in some way but not sure what exact.
| Every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials, in this case $s_1=x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$ and $s_3=x_1x_2x_3$. Since $x_1^3+x_2^3+x^3$ is homogeneous, we can find $a$, $b$ and $c$ such that
$$
x_1^3+x_2^3+x_3^3=as_1^3+bs_1s_2+cs_3
$$
*
*For $x_1=1$, $x_2=0$, $x_3=0$: $1=a$
*For $x_1=1$, $x_2=1$, $x_3=0$: $2=8a+2b$
*For $x_1=1$, $x_2=1$, $x_3=1$: $3=27a+9b+c$
Therefore $a=1$, $b=-3$, $c=3$ and finally
$$
x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3
$$
This is a general result.
In your case, by Viète’s formulas
$$
x_1+x_2+x_3=0,\qquad
$$
so in the end
$$
x_1^3+x_2^3+x_3^3=3x_1x_2x_3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Lagrange multiplier when decisions variables are not in the same set
Find the maximum of $2x+y$ over the constraint set $$S = \left\{ (x,y) \in \mathbb R^2 : 2x^2 + y^2 \leq 1, \; x \leq 0 \right\}$$
I want to use Lagrange multipliers to find the optimal solution. However, Lagrange requires $\vec x \in A$. In our case $R$, however $x$ can only be negative or zero. How can I get rid of this constraint? My idea is to do $x=w-z, w-z \le 0, w,z \in R$, but I am not sure if this is the right way to do it.
| You can solve this problem introducing convenient slack variables in order to eliminate inequalities. The inequalities
$$
2x^2+y^2 \le 1 \rightarrow 2x^2+y^2-1+\epsilon_1^2 = 0\\
x \le 0 \rightarrow x+\epsilon_2^2 = 0
$$
and the lagrangian thus is
$$
L(x,y,\lambda_1,\lambda_2,\epsilon_1,\epsilon_2) = 2x+y+\lambda_1(2x^2+y^2-1+\epsilon_1^2)+\lambda_2(x+\epsilon_2^2)
$$
the stationary condition $\nabla L = 0$ gives the conditions
$$
\left\{
\begin{array}{rcl}
\lambda_2+4 \lambda_1 x+2& = & 0 \\
2 \lambda_1 y+1& = & 0 \\
\epsilon_1^2+2 x^2+y^2-1& = & 0 \\
\epsilon_2^2+x& =& 0 \\
\epsilon_1 \lambda_1& =& 0 \\
\epsilon_2 \lambda_2& =& 0
\end{array}
\right.
$$
after solving we have
$$
\begin{array}{ccccccc}
x & y & \lambda_1 & \lambda_2 & \epsilon_1 & \epsilon_2 & 2x+y \\
0 & -1 & \frac{1}{2} & -2 & 0 & 0 & -1 \\
0 & 1 & -\frac{1}{2} & -2 & 0 & 0 & 1 \\
-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{\sqrt{3}}{2} & 0 & 0 & -\frac{1}{\sqrt[4]{3}} & -\sqrt{3} \\
\end{array}
$$
Here when $\epsilon_i = 0$ it means that the corresponding restriction is active.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
A urn contains blue balls and red balls. I need to find probabiltiy of drawing more blue balls than red balls
A urn contains 5 identical blue balls and 4 identical red balls. Taking 5 balls at random from the urn what is the probability that the number of blue balls be greater than the number of red balls?
My first guess was setting the ways that I can draw the balls.
It was:
$\color{blue} {BBBBB}$ ; $\color{blue} {BBBB}\color{red}{R}$; $\color{blue} {BBB}\color{red}{RR}$; $\color{blue} {BB}\color{red}{RRR}$; $\color{blue} {B}\color{red}{RRRR}$
Only $3$ cases have the number of blue balls greater than the number of red balls. Then the odds must be $\displaystyle{\frac{3}{5}}$.
But this answer sounds strange for me. I think that it is wrong?
Could anyone help me how to figure out this question?
| Hint
There are $31$ permutations for selecting $5$ balls:
$$\left(
\begin{array}{ccccc}
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} \\
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} \\
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} \\
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{red}{R} \\
\color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} \\
\color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{red}{R} \\
\color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{blue}{B} \\
\color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{red}{R} \\
\color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} \\
\color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} \\
\color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} \\
\color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{red}{R} \\
\color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} \\
\color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{red}{R} \\
\color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{red}{R} & \color{blue}{B} \\
\color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{red}{R} & \color{red}{R} \\
\color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} \\
\color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} \\
\color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} \\
\color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} & \color{red}{R} \\
\color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} \\
\color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} & \color{red}{R} \\
\color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{blue}{B} \\
\color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{red}{R} & \color{red}{R} \\
\color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} \\
\color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} \\
\color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{blue}{B} \\
\color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{red}{R} & \color{red}{R} \\
\color{red}{R} & \color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{blue}{B} \\
\color{red}{R} & \color{red}{R} & \color{red}{R} & \color{blue}{B} & \color{red}{R} \\
\color{red}{R} & \color{red}{R} & \color{red}{R} & \color{red}{R} & \color{blue}{B} \\
\end {array}
\right)$$
(Was too long to post as a comment, my apologies)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$
We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$
Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$
$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$
$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.
| Continuing from where you stopped.
Multiply both sides by $xyz$ to clear out the denominators. Then multiple the first six terms by $x+y+z=1$, while the RHS with $(x+y+z)^2=1$ to get:
$$\sum_{\text{sym}}x^4y + \sum_{\text{sym}}x^3y^2 \ge 2\sum_{\text{cyc}}x^3yz + 2\sum_{\text{cyc}}x^2y^2z$$
This inequality is true, as we have $\sum_{\text{sym}}x^4y \ge 2\sum_{\text{cyc}}x^2y^2z$ and $\sum_{\text{sym}}x^3y^2 \ge 2\sum_{\text{cyc}}x^3yz$. These follow by summing the cyclic AM-GM inequalities:
$$x^4y + z^4y \ge 2x^2z^2y$$
$$\text{and}$$
$$x^3y^2 + x^3z^2 \ge 2x^3yz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Help with finding Limit What is the limit of $$\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}}$$
Does any one can help, I am not sure how to solve this.
| $$
\begin{aligned}
\lim _{n\to \infty }\left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}} &= \lim _{n\to \infty}\exp\left(\frac{3n^3+4}{4n^4-1}\ln\left(\frac{n!}{n^n}\right)\right)\\ &\approx
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}}{e^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\left(\frac{1}{2}\ln \left(2\pi n\right)-n\right)\right)\\ &=
\color{red}{e^{-\frac{3}{4}}}
\end{aligned}$$
Solved with Stirling approximation
$$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x, \text{ for } x \to \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825213",
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"source": "stackexchange",
"question_score": "1",
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Rearrangement of polynomial under square root In this page, the following equation
$$\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a \label{a} \tag{1}$$
is suggested to be arranged in this way:
$$\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2}$$
and then it is squared in both sides. What if instead the square of $\ref{a}$ is directly taken? $x, y$ are real variables and $a,c$ are real constants.
$$(x + c)^2 + y^2 + 2\sqrt{(x + c)^2 + y^2}\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2 = 4a^2 \label{b} \tag{2}$$
Focusing on the remaining square root only:
$$\sqrt{(x + c)^2 + y^2}\sqrt{(x - c)^2 + y^2} = \\
= \sqrt{(x^2 + c^2 + y^2 + 2cx)(x^2 + c^2 + y^2 - 2cx)} = \\
= \sqrt{(x^2 + c^2 + y^2)^2 - 4c^2x^2} = \\
= \sqrt{x^4 + c^4 + y^4 + 2c^2x^2 + 2 c^2y^2 + 2x^2y^2 - 4c^2x^2} = \\
= \sqrt{x^4 + c^4 + y^4 - 2c^2x^2 + 2 c^2y^2 + 2x^2y^2}$$
It could still seem the square of $(\pm x^2 \pm c^2 \pm y^2)$, but there is no combination of signs of $x^2,c^2,y^2$ which generates only one negative $- 2c^2x^2$ term and two positive ones as above. How to proceed, then, to simplify it?
Equation $\ref{b}$ should become:
$$(x + c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2} - (x - c)^2 - y^2 = 4a^2$$
| I think the following way is better.
Since $a>0$ we obtain that our equation is equal to
$$\left(\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}\right)^2=4a^2$$ or
$$\sqrt{(x^2+y^2+z^2)^2-4x^2c^2}=2a^2-x^2-y^2-c^2$$ and since $a>c$ and by the triangle inequality
$$\sqrt{x^2+y^2}<\frac{d_1+d_2}{2}=a,$$ we obtain:
$$2a^2-x^2-y^2-c^2>2a^2-a^2-c^2=a^2-c^2>0,$$ which says that our equation is equivalent to
$$(x^2+y^2+z^2)^2-4x^2c^2=(2a^2-x^2-y^2-c^2)^2$$ and the rest is smooth.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| Wihout induction, it is simply a consequence of lil' Fermat:
As $3$ is prime, Fermat asserts that for any $n$, $\;n^3\equiv n\mod 3$, so
$$n^4-n^2=n^3\cdot n-n^2\equiv n^2-n^2\mod 3. $$
| {
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Solve the recurrence $T(n) = 2 T(n/2) + 2$ T(2) = 1
T(1) = 0
Ans is (3/2)* n - 2
My solution is :
T(n) = 2 T(n/2) + 2
T(n) = 4 T(n/4) + 4
T(n) = 8 T(n/8) + 6
T(n)=(2^k)T(n/2^k) + 2k
where k = log(n) ..... in base 2
as n/(2^k) = 1 for T(1)
I don't how to solve this type of question to get in terms of n only. Can anyone help me out?
| I am a bit confused because as pointed out in the comments, the intial conditions don't exactly make sense.
However, just taking T(2) = 1, we can obtain our result:
Since, $T(n) = 2*T(n/2) + 2 => T(n) = 2*(2*T(n/4) + 2) + 2 => T(n) = 2^2*T(n/2^2) + 2^2 + 2$
Proceeding further we get $T(n) = 2^2 * (2*T(n/2^3)+2) + 2^2 + 2 => T(n) = 2^3*T(n/2^3) + 2^3 + 2^2 + 2$
As you can see, the relation we get now is $T(n) = 2^k T(n/2^k) + 2 + 2^2 +...+2^k$
Now because I want to use the initial condition, let $n = 2^{k+1}$
Therefore, $T(n) = \frac{2^{k+1}}{2}T(\frac{2^{k+1}}{2^k}) + 2 + 2^2 +...+ 2^k$
Therefore, $T(n) = \frac{n}{2}T(2) + 2 + 2^2 +...+ 2^k$
Since, $2 + 2^2 +...+ 2^k$ is a geometric progression, using the formula for a geometric progression sum, we get $2 + 2^2 +..+2^k = \frac{2(1-2^k)}{1-2} = 2(2^k-1) = 2(\frac{n}{2} - 1)$
Using this result and the fact that T(2) = 1, we get $T(n) = \frac{n}{2}(1) + 2(\frac{n}{2}-1) = \frac{n}{2} + n - 2 = \frac{3n}{2} - 2$.
| {
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} |
What about the definite integral $\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy$? I would like to know if it is possible to justify the calculation of next definite integral
$$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy.$$
Question.(Corrected, see the comments) My reasonings and expriments with Wolfram Alpha online calculator, suggest me that the following identity holds $$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy=-\frac{1}{32}\left(16C-\pi^2+4\pi\log 2\right),$$
where $C$ denotes the Catalan's constant. Am I right? Do you know it or is it possible to justify? Many thanks.
My motivation was to compute an example of a reduction formula for integrals that I've known from a preprint by M.L. Glasser (Universidad de Valladolid). I've deduced the conjeture in the Question from my subsequent calculations for the limits of integration to deduce the closed-form of our doble integral (to me seem that were difficult calculations and the justification should be tedious, this is why I am asking here) with the mentioned CAS.
| As already pointed out in the comments, we have
$$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$
Using
$$ \int \limits_0^x \operatorname{artanh} (t) \, \mathrm{d} t = x \operatorname{artanh}(x) + \frac{1}{2} \log (1 - x^2) = \frac{1}{2} \left[x \log \left(\frac{1+x}{1-x}\right) + \log (1 - x^2) \right] $$
for $x \in (-1,1)$, we can integrate by parts to get
\begin{align}
I &= - \frac{\pi}{4} \log(2) + \int \limits_0^1 \frac{t \operatorname{artanh}(t) + \frac{1}{2} \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\
&= - \frac{\pi}{4} \log(2) + \frac{1}{2} \int \limits_0^1 \frac{2 t \log(1+t) - t \log(1-t^2) + \log(1-t^2)}{1+t^2} \, \mathrm{d} t \\
&\equiv - \frac{\pi}{4} \log(2) + I_1 + I_2 + I_3 \, .
\end{align}
We obtain
$$ I_1 = \int \limits_0^1 \frac{t \log(1+t)}{1+t^2} \, \mathrm{d} t = \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} $$
from this question and we find
$$ I_2 = \frac{1}{2} \int \limits_0^1 \frac{- t \log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{4} \int \limits_0^1 \frac{- \log(1-s)}{1+s} \, \mathrm{d} s = \frac{\pi^2}{48} - \frac{\log^2 (2)}{8}$$
using the substitution $t^2 = s$ and this question. A well-known representation of Catalan's constant and the change of variables $t = \tan(s)$ yield
\begin{align}
I_3 &= \frac{1}{2} \int \limits_0^1 \frac{\log(1-t^2)}{1+t^2} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^1 \frac{\log(t)}{1+t^2} \, \mathrm{d} t + \frac{1}{2} \int \limits_0^1 \frac{\log \left(\frac{1}{t} - t\right)}{1+t^2} \, \mathrm{d} t \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{\cos(s)}{\sin(s)} - \frac{\sin(s)}{\cos(s)}\right) \, \mathrm{d} s \\ &= - \frac{\mathrm{C}}{2} + \frac{1}{2} \int \limits_0^{\pi/4} \log \left(\frac{2 \cos(2s)}{\sin(2 s)}\right) \, \mathrm{d} s \\
&= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) - \frac{1}{4} \int \limits_0^{\pi/2} \log(\tan(r)) \, \mathrm{d} r \\
&= - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \, ,
\end{align}
where the last integral vanishes by symmetry (use $r \rightarrow \frac{\pi}{2} - r$).
Putting everything together, we confirm the conjectured result:
\begin{align}
I &= - \frac{\pi}{4} \log(2) + \frac{\pi^2}{96} + \frac{\log^2 (2)}{8} + \frac{\pi^2}{48} - \frac{\log^2 (2)}{8} - \frac{\mathrm{C}}{2} + \frac{\pi}{8} \log(2) \\
&= - \frac{\mathrm{C}}{2} - \frac{\pi}{8} \log(2) + \frac{\pi^2}{32} \, .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the range of convergence of $\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{1}{2n+3}.$ I was fiddling with the integral
$$\int_0^1 x^2\sqrt{1-x^2} \ dx $$
and I expanded the term under square root using a binomial series. Integrating, I got the result
$$\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{x^{2n+3}}{2n+3}\Biggr|_0^1.$$
I would like to know if evaluating this series at the upper limit 1 would make it converge, since binomial series has a convergence of $|x|<1$? Also if it does converge what is the range of convergence?
| Since
$\begin{array}\\
\dfrac{\binom{\frac12}{n+1}}{\binom{\frac12}{n}}
&=\dfrac{\frac{\prod_{k=0}^{n+1}(\frac12-k)}{(n+1)!}}{\frac{\prod_{k=0}^{n}(\frac12-k)}{n!}}\\
&=\dfrac{\frac12-(n+1)}{(n+1)}\\
&=-(1-\dfrac{1}{2(n+1)})\\
\end{array}
$
$(-1)^n\binom{\frac12}{n}$
decreases,
as does
$(-1)^n\binom{\frac12}{n}\dfrac1{2n+3}
$.
Therefore the sum converges for
$|x| < 1$
and conditionally converges
for $x = -1$.
For $x=1$,
we can do,
for $n \ge 1$,
$\begin{array}\\
\binom{\frac12}{n}
&=\dfrac{\prod_{k=0}^{n}(\frac12-k)}{n!}\\
&=\dfrac{\prod_{k=0}^{n}(1-2k)}{2^nn!}\\
&=\dfrac{\prod_{k=1}^{n}(1-2k)}{2^nn!}\\
&=\dfrac{(-1)^n\prod_{k=1}^{n}(2k-1)}{2^nn!}\\
&=\dfrac{(-1)^n\prod_{k=1}^{n}(2k-1)(2k)}{2^{2n}(n!)^2}\\
&=\dfrac{(-1)^n(2n)!}{2^{2n}(n!)^2}\\
&=\dfrac{(-1)^n}{2^{2n}}\binom{2n}{n}\\
&\approx\dfrac{(-1)^n}{2^{2n}}\dfrac{4^n}{\sqrt{\pi n}}\\
&=\dfrac{(-1)^n}{\sqrt{\pi n}}\\
\end{array}
$
so
$(-1)^n\binom{\frac12}{n}\dfrac{x^{2n+3}}{2n+3}
\approx \dfrac{1}{\sqrt{\pi n}}\dfrac{x^{2n+3}}{2n+3}
$
and the sum of these
converges at
$x = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a+b\sqrt{3}=\sqrt{21-12\sqrt{3}}, a,b \in \mathbb {Z}$ Find a+b So far I've reasoned that $\mathbf{a}$ and $\mathbf{b}$ can't be both negative, because $\sqrt{21-12\sqrt{3}}$ cannot be negative.
Also $\mathbf{a}$ and $\mathbf{b}$ can't be both positive, because $\sqrt{21-12\sqrt{3}}$ is from 0 to 1, thus there is no positive whole numbers which could satisfy that $\mathbf{a}$ plus $\mathbf{b}*\sqrt{3}$ is close to 0 and 1.
At this stage, I don't know what to do. I appreciate any help.
| Write
$$\left(a+b\sqrt{3}\right)^2=a^2+3b^2+2ab\sqrt{3}=21-12\sqrt{3}$$
This means
$$\begin{cases}
a^2+3b^2&=21\\
2ab&=-12
\end{cases}$$
Now substitute $a=-6/b$. You get the following biquadratic
$$b^4-7b^2+12=0$$
Has two solution for $b^2$, $3$ to be discarded because $b$ is an integer and $4$ which means that
$$b=\pm 2\\a=\pm 3$$
The solution to retain is the non negative one ie $-3+2\sqrt{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$
Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ :
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$
I tried using AM-GM or AM-HM but I can't figure it out.
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\root3\of{\Big(a+\frac{1}{b}\Big)^2\Big(b+\frac{1}{c}\Big)^2\Big(c+\frac{1}{a}\Big)^2}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(\Big(a+\frac{1}{b}\Big)\Big(b+\frac{1}{c}\Big)\Big(c+\frac{1}{a}\Big)\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(abc+a+b+c+\frac1a+\frac1b+\frac1c+\frac{1}{abc}\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(2+a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2/3}$$
| Let us show stronger, with an intermediate step:
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^2 +
\left(b+\frac{1}{c}\right)^2 +
\left(c+\frac{1}{a}\right)^2
&\geq 4(a+b+c)
\\
&\geq 3(a+b+c+1)
\ .
\end{aligned}
$$
The last inequality, compactly written $a+b+c\ge3\sqrt[3]{abc}=3$ is simple, so let us show the first one. We write
$$
a=A^3\ ,\qquad
b=B^3\ ,\qquad
c=C^3\ ,\qquad A,B,C>0\ ABC=1\
$$
and the given inequality, written first as $\sum(a+ac)^2\ge \dots$, becomes the homogeneous inequality:
$$
A^6(C^3+ABC)^2+
B^6(A^3+ABC)^2+
C^6(B^3+ABC)^2
\ge
4(A^3+B^3+C^3)A^3B^3C^3\ .
$$
At this point we can forget the constraint $ABC=1$.
Let us see how to dominate with the terms on the left side the ones on the right side.
First, let us expand $(C^3+ABC)^2 = C^6 + 2\,C^3\cdot ABC + (ABC)^2$. Then:
$$
\begin{aligned}
A^6C^6+B^6A^6+C^6B^6
&\ge (A^3+B^3+C^3)A^3B^3C^3\ ,
\\
&\qquad\text{which follows from}
\\
&\qquad\frac 12A^6C^6+\frac 12B^6A^6\ge A^6B^3C^3
\\
&\qquad\text{and the cycled versions of it.}
\\[2mm]
2(A^6C^3+B^6A^3+C^6B^3)ABC
&\ge 2(A^3+B^3+C^3)A^3B^3C^3\ ,
\\
&\qquad\text{which follows from}
\\
&\qquad\frac 23A^6C^3+\frac 13B^6A^3\ge (A^6C^3)^{2/3}(B^6A^3)^{1/3}=A^5B^2C^2
\\
&\qquad\text{and the cycled versions of it. Finally:}
\\
(A^6+B^6+C^6)A^2B^2C^2
&\ge
(A^3+B^3+C^3)A^3B^3C^3
\\
&\qquad\text{follows from}
\\
&\qquad\frac 46A^6+\frac 16B^6+\frac 16C^6\ge A^4BC
\\
&\qquad\text{and the cycled versions of it.}
\end{aligned}
$$
We have used the generalized form of the AM-GM inequality with more general weights $\lambda_1,\lambda_2,\dots,\lambda_n\ge 0$, total weight being one, $$\sum \lambda_kx_k\ge \prod x_k^{\lambda_k}\ .$$
(This is the Jensen inequality for the concave function $\ln$.)
Note: The idea was simple, it is the standard way to show such inequalities. Typing all the stuff...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Check the proof that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic
I have to prove that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic
If the function is periodic then:
$$
f(x) = f(x+T)\\
\cos(x)\cdot\cos(\sqrt{3}x) = \cos(x+T)\cdot\cos(\sqrt{3}(x+T))
$$
Consider the function at $0$:
$$
\cos(T)\cdot\cos(\sqrt{3}T) = 1
$$
But this equation has only one solution at $T=0$ which contradicts the initial assumption that there exists a positive period.
Or the other way:
$$
\cos(T)\cdot\cos(\sqrt{3}T) = 1
$$
Let $\cos(T) = 1$ and $\cos(\sqrt{3}T) = 1$, hence
$$
T = 2 \pi m \\
\sqrt{3}T=2\pi n
$$
Substituting $T$ in the second equation gives:
$$
2\pi m\sqrt{3} = 2\pi n \\
\sqrt{3} = \frac{n}{m}
$$
But $m,n \in \mathbb N$ and $\sqrt{3} \in \{\mathbb R \setminus \mathbb Q\}$ which gives a contradiction.
Is the prove above valid?
Update:
For $\cos{T} = -1$ and $\cos\sqrt{3}T = -1$:
$$
T = \pi+ 2 \pi m \\
\sqrt{3}T=\pi + 2\pi n \\
\sqrt{3} = \frac{2n + 1}{2m+1}
$$
Which is a contradiction.
| As an alternative, we have that
$$\cos({x})\cdot \cos({\sqrt{3}x})=\frac12 \cos(x(\sqrt3-1))+\frac12\cos(x(\sqrt 3+1))$$
and $\cos(x(\sqrt3-1))$ has period $\frac{2\pi}{\sqrt 3-1} $ while $\cos(x(\sqrt3+1)+$ has period $\frac{2\pi}{\sqrt 3+1}$
and $\not\exists k\in\mathbb{Z}$ such that
$$\frac{2\pi}{\sqrt 3-1}=k\frac{2\pi}{\sqrt 3+1}\iff \frac{\sqrt 3+1}{\sqrt 3-1}=k\iff (\sqrt 3+1)^2=2k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Number of different sums with k numbers from {1, 5, 10, 50}
Say we have $k$ numbers, each of which belongs to the set $S = \{1, 5, 10, 50\}$
How many different sums can be created by adding these numbers?
If $k = 1$, the are four different sums.
Also, if $k = 2$, there are ten:
$$\begin{align} 1 + 1 = 2 \quad 1 + 5 &= 6 \quad 1 + 10 = 11 \\ 1 + 50 = 51 \quad 5 + 5 &= 10 \quad 5 + 10 = 15 \\ 5 + 50 = 55 \quad 10 + 10 &= 20 \quad 10 + 50 = 60 \\ 50 + 50 &= 100 \quad \end{align}$$
| At least for small values of $k$, this answer will be tractable. Your answer is the number of terms in the expansion of $(x+x^{5}+x^{10}+x^{50})^k$ which can be found using a computer algebra system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:
$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$
and
$$B=\frac {A^3-I_2}{A-I_2}-I_2$$
$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
But I get stuck here.
| The matrix $C=A^3$ satisfies its characteristic equation, that is: $det(A^3-xI)=0$, which is $(4-x)(-2-x)+9=0$.
See https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem.
This gives $C^2-2C+I=0$, thus $C$ has the eigenvalue 1. The eigenvalues of $A^3$ are the cubes of the eigenvalues of $A$. If the eigenvalues of $A$ are the two complex cube roots of unity, then we'd get $A^2+A=-I$, but then $A^3-I$ is non-zero, thus the eigenvalues of $A$ are also both 1, which means that $A$ satisfies $A^2=2A-I$.
This gives $A^3=2A^2+A=2(2A-I)+A=3A-2I$. Thus, $A^2+A=3A-I=A^3+I$.
Remark: The conclusion that the eigenvalues of $A$ must both be 1 is wrong, as they could be any of the other two cube roots of unity, occuring twice. This is shown in the other answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Domain of Trigonometric Inequality The problem: Find values of $x$ on the interval $[0,\pi]$ for which $\cos(x)< \sin(2x)$
My answer came out to be $0.52< x<1.57$ or $2.62< x\leq π$. However, the textbook's answer is slightly different: $0.52< x<1.57$ or $2.62< x<π$. At $x=π$, $\cos(x)< \sin(2x)$, and $x$ can be $[0,π]$ so why is $π$ left out of the solution?
| $\cos(x)-2\sin(x)\cos(x)<0$ $\leftrightarrow$
$\cos(x)[1-2\sin(x)]<0$
*
*$\cos(x)<0$ if and only if $ \frac{\pi}{2}+2k\pi<x<\frac{3\pi}{2}+2k\pi$
while
*$\sin(x)>\frac{1}{2}$ if and only if
$\frac{\pi}{6}+2k\pi<x<\frac{5\pi}{6}+2k\pi$
So the solution is
$\frac{\pi}{6}+2k\pi<x<\frac{\pi}{2}+2k\pi$
or
$\frac{5\pi}{6}+2k\pi<x<\frac{3\pi}{2}+2k\pi$
So $\pi$ is oviously a solution of your problem
(You can observe that $\frac{\pi}{6}\sim 0,52$ , $\frac{\pi}{2}\sim 1,57$, $\frac{5\pi}{6}\sim 2,62 $ and $\frac{3\pi}{2}\sim 4,71$ )
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed after this......
| The reciprocal quartic should be transformed as:
$$\left( x+\frac{1}{x} \right)^2+k\left( x+\frac{1}{x} \right)+\color{red}{9}=0$$
Now,
\begin{align}
x+\frac{1}{x} &= \frac{-k \color{red}{\pm} \sqrt{k^2-36}}{2} \\
x &=
\frac{-k\color{red}{\pm} \sqrt{k^2-36}}{2(2)} \color{blue}{\pm}
\frac{1}{2} \sqrt{
\left( \frac{-k\color{red}{\pm} \sqrt{k^2-36}}{2} \right)^2-4} \\
&=
\frac{-k \color{red}{\pm} \sqrt{k^2-36}}{4} \color{blue}{\pm}
\frac{\sqrt{k^2 \color{red}{\mp} k\sqrt{k^2-36}-26}}{2\sqrt{2}}
\end{align}
For four distinct real roots,
*
*$k^2-36>0$ and
*$k^2 \pm k\sqrt{k^2-36}-26>0$
$\implies k^2-26>|k|\sqrt{k^2-36}>0$
$\implies (k^2-26)^2 > k^2(k^2-36)$
$\implies k^4-4\times 13k^2+4\times 13^2 > k^4-4\times 9k^2$
$\implies 13^2 > 4k^2$
And the roots are positive, $\text{sum of roots}=-k>0$
Hence,
$$\fbox{$-\frac{13}{2}<k<-6$}$$
| {
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"question_score": "2",
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} |
Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $
Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$
My Attempt:
I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$
This reduced $f(x)$ to,
$$f(x)=\frac{1-3\tan^2{x}}{3-\tan^2{x}}$$
I don't know how to solve any further. I thought of using derivative, but the function is dicontinuous at times. How do I solve it? Any hints would be helpful.
Thanks.
| Note that $\tan(x)^2 \ge 0$ and
$$f(x)= \frac{1-3\tan(x)^2}{3-\tan(x)^2} = 3+\frac{8}{\tan(x)^2-3}.$$
Clearly, for $\tan(x)^2 > 3$, $\frac{8}{\tan(x)^2-3} > 0$, and $f(x) >3$.
For $0 \le \tan(x)^2 < 3$, we have $-3 \le \tan(x)^2-3 < 0 \implies \frac{8}{\tan(x)^2-3}\le-\frac{8}{3} \implies f(x) = 3+\frac{8}{\tan(x)^2-3} \le 3-\frac{8}{3} = \frac{1}{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all complex numbers satisfying $z\cdot\bar{z}=41$, for which $|z-9|+|z-9i|$ has the minimum value My first attempt was to express $z$ as $x+iy$ and minimize the expression $\sqrt{(x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}$ where $x^2+y^2=41$.
That said, it seems to me that using the geometric interpretation could be easier. As far as I understand, I need to find points on the circle for which the sum of distances to the points $(9,0)$ and $(0,9)$ is lowest. This interpretation, however, doesn't help with regard to calculations.
Is there some simple trick or idea I'm missing?
Thank you!
| Hint: Show that $$\sqrt{(8x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}\geq 9\sqrt{2}$$ and the equal sign holds if $$x=4,y=5$$
ok we will prove the inequality above:
squaring all we get
$$2\sqrt{(8x-9)^2+y^2}\sqrt{(x^2+(y-9)^2}\geq 162-(8x-9)^2-(y-9)^2-x^2-y^2$$,
now we use that $x^2+y^2=41$:
we get
$$2\sqrt{(8x-9)^2+41-x^2}\sqrt{41-y^2+(y-9)^2}\geq 121-(8x-9)^2-(y-9)^2$$
squaring again and factorizing we get
$$- \left( 64\,{x}^{2}+{y}^{2}-144\,x-18\,y+41 \right) \left( 64\,{x}^{
2}+{y}^{2}-144\,x-18\,y+42 \right)
\geq 0$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What does is mean $Y=\min \{X_1,X_2\}$? $Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?
Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent.
After some calculations:
*
*$P(X_1=0)=\frac{1}{4}$
*$P(X_1=1)=\frac{1}{2}$
*$P(X_1=2)=\frac{1}{4}$
* $P(X_2=1)=P(X_2=2)=P(X_2=3)=\frac{1}{3}$
*
*I know how to work with $X_1, X_2$ but I have no clue how to do anything with $Y, Z$. Why $Y\sim X_1$ and $Z\sim X_2$?
*How do I approach the question: Calculate $P(Y=y, Z=z)$?
| Assume that $X_1$ and $X_2$ are independent. The joint probability function of $X_1,X_2$ is given by
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 1/12 & 1/6 & 1/12 \\
2 & 1/12 & 1/6 & 1/12 \\
3 & 1/12 & 1/6 & 1/12
\end{array}
$$
The table for $Y=\min\{X_1,X_2\}$ is
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 0 & 1 & 1 \\
2 & 0 & 1 & 2 \\
3 & 0 & 1 & 2
\end{array}
$$
So $P(Y=0)=1/4$, $P(Y=1)=7/12$, $P(Y=2)=1/6$. The table for $Z=\max\{X_1,X_2\}$ is
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 1 & 1 & 2 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3
\end{array}
$$
So $P(Z=1)=1/4$, $P(Z=2)=5/12$ and $P(Z=3)=1/3$.
For the joint probability of $Y,Z$,$P(Y=y,Z=z)$, where $y\leq z$, analyze by cases:
$$Y=0,Z=1\leftrightarrow X_1=0,X_2=1$$
$$Y=0,Z=2\leftrightarrow X_1=0,X_2=2$$
$$Y=0,Z=3\leftrightarrow X_1=0,X_2=3$$
$$Y=1,Z=1\leftrightarrow X_1=1,X_2=1$$
$$Y=1,Z=2\leftrightarrow X_1=1,X_2=2\;\text{or}\;X_1=2,X_2=1$$
$$\vdots$$
Using independence yo get the joint probability
$$
\begin{array}{c|lcr}
Z/Y & 0 & 1 & 2 \\
\hline
1 & 1/12 & 1/6 & 0 \\
2 & 1/12 & 1/4 & 1/12 \\
3 & 1/12 & 1/6 & 1/12
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Exercise about the form of elements of $\mathbb{Q}(u)$
Let $u \in \mathbb{C}$ be a root of $x^3-x^2+x+2$. Consider $f = (u^2+u+1)(u^2-u)$ and $g= (u-1)^{-1}$ in $\mathbb{Q}(u)$ and express them in the form $au^2+bu+c$ with coefficients in $\mathbb{Q}$.
We have $f = (u^2+u+1)(u-1)u=(u^3-1)u=u^4-u = u^4 -u(u^3-u^2+u+2)-u=u^3-(u^3-u^2+u+2)-u^2-3u=-4u-2$.
While $g=-(u^2+1)/3$ (I found it solving $(u-1)(au^2+bu+c)$ for $a,b,c$).
Is this correct? Are there smarter ways to solve this exercise?
| Here is a solution using Euclidean division.
For $f$, we get
$$
f=(u^2+u+1)(u^2-u) = (u + 1)(u^3-u^2+u+2) + (-4 u - 2) = -4 u - 2
$$
For $g$, we get
$$
0 = u^3-u^2+u+2 = (u^2 + 1)(u-1) + 3
$$
and so the inverse of $u-1$ is $-(u^2 + 1)/3$.
Note that this works for $g$ only because $u-1$ has degree $1$ and so the remainder is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to understand $2$'s complement Using $n$-bit integers we can usually represent anything from $0$ to $2^n-1$. If we're interpreting a number with $2$'s complement, then the first bit is the sign bit, and we represent $-x$ with $2^n-x$. So for example, normally:
$n=4$:
\begin{array}{|c|c|}
\hline
\text{Decimal} & \text{Binary}\\ \hline
0 & 0000 \\ \hline
1 & 0001 \\ \hline
2 & 0010 \\ \hline
3 & 0011 \\ \hline
4 & 0100 \\ \hline
5 & 0101 \\ \hline
6 & 0110 \\ \hline
7 & 0111 \\ \hline
8 & 1000 \\ \hline
9 & 1001 \\ \hline
10 & 1010 \\ \hline
11 & 1011 \\ \hline
12 & 1100 \\ \hline
13 & 1101 \\ \hline
14 & 1110 \\ \hline
15 & 1111 \\ \hline
\end{array}
But using $2$'s complement:
\begin{array}{|c|c|}
\hline
\text{Decimal} & \text{Binary}\\ \hline
0 & 0000 \\ \hline
1 & 0001 \\ \hline
2 & 0010 \\ \hline
3 & 0011 \\ \hline
4 & 0100 \\ \hline
5 & 0101 \\ \hline
6 & 0110 \\ \hline
7 & 0111 \\ \hline
-8 & 1000 \\ \hline
-7 & 1001 \\ \hline
-6 & 1010 \\ \hline
-5 & 1011 \\ \hline
-4 & 1100 \\ \hline
-3 & 1101 \\ \hline
-2 & 1110 \\ \hline
-1 & 1111 \\ \hline
\end{array}
Now apparently if we add two numbers using $2$'s complement we still get the right answer even if we have to truncate the leftmost bit, and this is bending my brain at the moment.
If I have two negative numbers $-x$ and $-y$, then adding them gives us $2^{n+1} - (x + y)$ which after truncating gives us $-(x+y)$.
If I have a positive number $x$ and a negative number $-y$, then adding them gives us $x + (2^n - y) = 2^n - (-x + y)$ which is the representation for $-(-x+y)$ or $x-y$.
If I have a negative number $-x$ and a positive number $y$ then adding them gives us $2^n-x + y = 2^n-(x-y)$ which is the representation for $-(x-y)$ or $y-x$.
If I have two positive numbers $x$ and $y$ then adding them gives us $x+y$.
My question:
Looking at the four cases mentioned earlier, do we normally have to handle them separately, or is it really just doing arithmetic mod $2^n$ and everything will just work itself out naturally as long as everything ($x$, $y$, as well as $x+y$) is confined to $-2^{n-1}$ through $2^{n-1}-1$? For example using $4$ bits we could not be able to represent $(-8) + (-7) = -15$ since the sum falls outside the scope?
| It's the latter – arithmetic mod $2^n$ and everything works itself out naturally. And yes, we can't represent $-15$ with $4$ bits because it's outside the scope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient except simple division.
| $$x^6+x^3+1=(x+1)(x^5-x^4+x^3)+1.$$
You may apply so-call long division method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| We have: $\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=(a+b)^2+2(a+b)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2=1+2ab+4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{1+2ab}{(ab)^2}= 5+2ab+\dfrac{2}{ab}+\dfrac{1}{(ab)^2}+\dfrac{2}{ab}= 5+2ab+\dfrac{4}{ab}+\dfrac{1}{(ab)^2}= 1+\left(4+\dfrac{1}{(ab)^2}\right)+ \left(2ab+\dfrac{4}{ab}\right)\ge 2\sqrt{4\cdot\dfrac{1}{(ab)^2}}+(a+b)^2+\dfrac{4}{\dfrac{(a+b)^2}{4}}=\dfrac{4}{ab}+(a+b)^2+\dfrac{16}{(a+b)^2}\ge \dfrac{4}{\dfrac{a^2+b^2}{2}}+\dfrac{((a+b)^2-8)((a+b)^2-2)}{(a+b)^2}+10= 8+10+\dfrac{(a+b)^2-8)((a+b)^2-2)}{(a+b)^2}\ge 18$ because $0<(a+b)^2\le 2(a^2+b^2) = 2\implies a+b+\dfrac{1}{a}+\dfrac{1}{b}\ge \sqrt{18}=3\sqrt{2}$ , with $=$ occurs at $a = b = \dfrac{1}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 5
} |
Write $\omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.
At each point $p ∈ \mathbb R^3$, define a bilinear function $\omega_p$
on $T_p(\mathbb R^3)$ by
$$\omega_p(\textbf{a},\textbf{b})=\omega_p(\begin{bmatrix}
a^1 \\
a^2 \\
a^3 \\ \end{bmatrix}, \begin{bmatrix}
b^1 \\
b^2 \\
b^3 \\ \end{bmatrix})= p^3\det \begin{bmatrix}
a^1&a^2 \\
b^1 &b^2 \\ \end{bmatrix} $$ , for tangent vectors $\textbf{a},\textbf{b} ∈ T_p(\mathbb R^3)$, where $p^3$ is the third component of $p = (p^1,
p^2, p^3)$. Since $\omega_p$ is an alternating bilinear function on
$T_p(\mathbb R^3)$, $\omega$ is a $2-$form on $\mathbb R^3$. Write
$\omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.
I know that $\omega=a_{12}dx^1 ∧dx^2+a_{13}dx^1 ∧dx^3+a_{23}dx^2 ∧dx^3,$ $\textbf{a}=a^1\frac{\partial}{\partial x^1}+a^2\frac{\partial}{\partial x^2}+a^3\frac{\partial}{\partial x^3},\textbf{b}=b^1\frac{\partial}{\partial x^1}+b^2\frac{\partial}{\partial x^2}+b^3\frac{\partial}{\partial x^3}.$
$\omega(\textbf{a},\textbf{b})=\omega(a^1\frac{\partial}{\partial x^1}+a^2\frac{\partial}{\partial x^2}+a^3\frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+b^2\frac{\partial}{\partial x^2}+b^3\frac{\partial}{\partial x^3})$ $$=a_{12}dx^1 ∧dx^2+a_{13}dx^1 ∧dx^3+a_{23}dx^2 ∧dx^3 (a^1\frac{\partial}{\partial x^1}+a^2\frac{\partial}{\partial x^2}+a^3\frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+b^2\frac{\partial}{\partial x^2}+b^3\frac{\partial}{\partial x^3})=a_{12}a^1b^2+a_{13}a^1b^3+a_{23}a^2b^3.$$
How does this answer related to $$p^3\det \begin{bmatrix}
a^1&a^2 \\
b^1 &b^2 \\ \end{bmatrix}. $$ Please help me.
| The general formula for $\omega_p=a_{12}dx_1\wedge dx_2+a_{13}dx_1\wedge dx_3+a_{23}dx_2\wedge dx_3$ is $$\omega_p(a,b)=a_{12}\det \begin{pmatrix}a^1&a^2\\b^1&b^2\end{pmatrix}+a_{13}\det\begin{pmatrix}a^1&a^3\\b^1&b^3\end{pmatrix}+a_{23}\det\begin{pmatrix}a^2&a^3\\b^2&b^3\end{pmatrix}.$$
Choose a special $\omega_p=p^3dx^1\wedge dx^2$,
$\omega_p(a,b)=\omega_p(a^1\frac{\partial}{\partial x^1}+a^2\frac{\partial}{\partial x^2}+a^3\frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+b^2\frac{\partial}{\partial x^2}+b^3\frac{\partial}{\partial x^3})=p^3\det \begin{pmatrix}a^1&a^2\\b^1&b^2\end{pmatrix}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find out the subsequential limits of a sequence
Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.
We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
| Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$
Let's look at
$$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$
We know that
$\dfrac4{n^2}$ is decreasing and $2$ is constant.
The first few terms of $\dfrac{3^n}{n!}$ are
$$\dfrac{3}{1},\,
\dfrac{3\cdot3}{2\cdot1},\,
\dfrac{3\cdot3\cdot3}{3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3}{4\cdot3\cdot2\cdot1},\,
\dfrac{3\cdot3\cdot3\cdot3\cdot3}{5\cdot4\cdot3\cdot2\cdot1},\cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $\lvert x_n\rvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $(1+i)^{(1-2i)}$ Find all the values of $(1+i)^{(1-2i)}$ and show that there are small values as we wish (else from $0$) and big values as we wish
\begin{align}
(1+i)^{(1-2i)}&=e^{\ln(1+i)^{(1-2i)}}
=e^{(1-2i)\ln(1+i)}
\\&=e^{(\ln\sqrt{2}+i(\frac{\pi}{2}+2\pi k))(1-2i)}
\\&=e^{\ln\sqrt{2}-2\ln\sqrt{2}*i+i(\frac{\pi}{2}+2\pi k)+2(\frac{\pi}{2}+2\pi k))}
\\&=\sqrt{2}*e^{-2\ln\sqrt{2}*i}*i*e^{2(\frac{\pi}{2}+2\pi k)}
\\&=e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}
\\&=e^{-i(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=e^{-i}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*2e^{i(\pi+4\pi k)}+i\sqrt{2}
\\&=\frac{-3}{2}+i\sqrt{2}
\end{align}
What can I conclude from it on the values?
Or should I stop here $e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}$ and $e^{\pi+4 \pi k}cis(-2\ln\sqrt{2}+\frac{\pi}{2})$
| Hint. Note that $|e^z|=|e^{\text{Re}(z)+i\text{Im}(z)}|=e^{\text{Re}(z)}$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $\pi/4+2\pi k$)
$$\text{Re}(\ln\sqrt{2}+i(\frac{\pi}{4}+2\pi k))(1-2i))=\ln\sqrt{2}+\pi\left(4k+\frac{1}{2}\right)$$
where $k\in\mathbb{Z}$. Now consider the limits as $k\to \pm \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $(2^2-1)(3^2-1)(4^2-1)....(300^2-1)$ divisible by $7^{95}$? Is $(2^2-1)(3^2-1)(4^2-1)....(300^2-1)$ divisible by $7^{95}$?
And what about $7^{100}$?
This is taken out of one of the TAU entry tests.
I seem to always have a struggle with these exercises, probably because I don't enough experience.
What I've tried is to see if I can notice some pattern in the sequence but that really didn't add much except that I know for sure that it's divisible by 7.
Another thing that I tried is to set the sequence like this:
$((7-5)^2-1)((7-4)^2-1)((7-3)^2-1)...((7+293)^2-1)$
But unfortunately, I can't really see how that helps me either, I still can't extract a 7 out of the expression.
| Note that $x^2-1=(x-1)(x+1)$. Then
$$
(2^2-1)(3^2-1)\cdots (300^2-1)=(2-1)(2+1)(3-1)(3+1)\cdots(300-1)(300+1).
$$
Since $49=7^2 < 300 < 7^3=343$, we will check the number of multiples of $7$ and $49$. Since $7\cdot 43=301$ and $7\cdot 44=308$, the number of multiples of $7$ less than $300$ and $302$ is $42$ and $43$, respectively. Also, since $49\cdot 6=294$ and $49\cdot 7=343$, the number of multiples of $49$ less than $300$ and $302$ is $6$ and $6$, respectively. $42+43+6+6=97$ and $n=97$ is the maximum such that $7^n$ divides $(2^2-1)(3^2-1)\cdots (300^2-1)$. Therefore $(2^2-1)(3^2-1)\cdots (300^2-1)$ is divisible by $7^{95}$ but not divisible by $7^{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856585",
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
| HINT: $$\left[\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)+\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right)\right]^2$$ is equal to $$2+2\sin\frac{\pi}{20}\cos\frac{\pi}{20}-2\sin\frac{3\pi}{20}\cos\frac{3\pi}{20}+2\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right).$$ The double-angle and addition formulas for $\sin$ might come in handy...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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My solution to inhomogeneous $\frac{d^2y}{dx^2} + y = \sin{x}$ does not conform to my book's solution! I need help with the solution of this particular equation:
$$\frac{d^2y}{dx^2} + y = \sin{x}$$
Due to me having to go to work, I cannot display all my work in mathjax, my shift starts in 5 min...but my solution is: $$y= \frac{-x}{2} + \frac{(\cos{2x}\sin{x})}{4} + C_1\cos{x} + C_2\sin{x}$$
Whereas my book's solution is: $$y= \frac{-x}{2} +C_1\cos{x} + C_2\sin{x}$$
I have used the method of variation of prameters, cramer's rule and basic antidifferentiation to solve resulting system.
Thank you all!
| $$y'' + y = \sin{x}$$
$$y''\cos(x) + y\cos(x) = \frac 12 \sin{(2x)}$$
$$y''\cos(x) -y'\sin(x)+y'\sin(x)+ y\cos(x) = \frac 12 \sin{(2x)}$$
$$(y'\cos(x))'+(y\sin(x))'= \frac 12 \sin{(2x)}$$
First integration
$$(y'\cos(x))+(y\sin(x))= -\frac 14 \cos{(2x)}+K_1$$
$$(\frac y {\cos(x)})'= -\frac 14 \frac {\cos{(2x)}+K_1}{\cos^2(x)}$$
Second integration
$$(\frac y {\cos(x)})= -\frac 14 \int \frac {\cos{(2x)}+K_1}{\cos^2(x)}dx+K_2$$
$$(\frac y {\cos(x)})= -\frac 14 \int \frac {2\cos^2(x)+K_1}{\cos^2(x)}dx+K_2$$
$$ y = -\frac 12x\cos(x)+ K_1\cos(x)\int \frac {dx}{\cos^2(x)}+K_2\cos(x)$$
Finally
$$ \boxed{y = -\frac 12x\cos(x)+ K_1\sin(x)+K_2\cos(x)}$$
| {
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"source": "stackexchange",
"question_score": "1",
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What is $\log(n+1)-\log(n)$? What is gap $\log(n+1)-\log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
| Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{2}{2 n+1}$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{6 n+3}{6 n^2+6n+1}$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{60 n^2+60 n+11 } {60 n^3+90 n^2+36 n+3 }$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{420 n^3+630 n^2+260 n+25 }{420 n^4+840 n^3+540 n^2+120 n+6 }$$
These are respectively equivalent to Taylor series to $O\left(\frac{1}{n^3}\right)$, $O\left(\frac{1}{n^5}\right)$, $O\left(\frac{1}{n^7}\right)$ and $O\left(\frac{1}{n^9}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
How to get the shaded region of the rectangle? I have this problem:
So my development was:
Denote side of rectangle with: $2a, 2b$.
So, $4ab= 64, ab = 16$
Denote shaded region with $S$
Denote area of triangle $DGH = A_1$ and triangle $FBE = A_2$.
So, $A_1 + A_2 + S = 64$
$S = 64 - A_1 - A_2$
The triangles $A_1, A_2$ are congruent because $LAL$ congruence criterion.
The area of $A_1$ and $A_2$, is the same and i got it with this way:
Since, the $\angle{GDH} = 90$ and the median from this angle to the base $HG$, that is the altitude of the triangle $DGH$, will measure the half of the $HG$ side.
And the $HG$ side by Pythagorean theorem, will be $\sqrt{a^2 + b^2}$, that will be the base of the triangle.
And the altitude will be: $\frac{\sqrt{a^2 + b^2}}{2} $,
So the Area of $A_1 = \frac{a^2 + b^2}{4}$
So, $A_1 + A_2 = \frac{a^2 + b^2}{2}$
Then, $64 - (\frac{a^2 + b^2}{2}) = S$
And, $-(a^2 - 8ab + b^2) = 2S$
And I have not been able to continue from here, what should I do? Thanks in advance.
| Hint:
*
*there are eight small red triangles all with the same area
*six of them are shaded
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
How to determine if a set of five $2\times2$ matrices is independent $$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right],
\left[\begin{matrix}1&4\\0&3\end{matrix}\right]\bigg\}$$
How can I determine if a set of five $2\times2$ matrices are independent?
| Stretch out the matrices to complete the rows of the following matrix
$$\newcommand{\adj}{\operatorname{adj}}
M(v)=\begin{bmatrix}
v_1&1&2&2&1\\
v_2&2&1&-1&2\\
v_3&0&1&1&2\\
v_4&1&0&1&1\\
v_5&1&4&0&3
\end{bmatrix}\tag1
$$
Note that the top row of the adjugate of $M(v)$
$$
\begin{bmatrix}
-14&-14&-14&28&14
\end{bmatrix}\tag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $\adj M(v)$. By Laplace's Formula,
$$
\det M(v)=u\cdot v\tag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $\det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-\frac1{14}u$ with the fixed columns of $M(v)$ as
$$
1\ \overbrace{\begin{bmatrix}
1&2\\2&1
\end{bmatrix}}^\text{row $1$}
+
1\ \overbrace{\begin{bmatrix}
2&1\\-1&2
\end{bmatrix}}^\text{row $2$}
+
1\ \overbrace{\begin{bmatrix}
0&1\\1&2
\end{bmatrix}}^\text{row $3$}
-2\ \overbrace{\begin{bmatrix}
1&0\\1&1
\end{bmatrix}}^\text{row $4$}
-
1\ \overbrace{\begin{bmatrix}
1&4\\0&3
\end{bmatrix}}^\text{row $5$}
=
\begin{bmatrix}
0&0\\0&0
\end{bmatrix}\tag4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as,
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
After further simplification I got,
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$
Taking $-2^{99}$ common I got,
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So,
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
| Here, you forgot red:
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1\cdot \color{red}{2}}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2\color{red}{2^2}}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}\color{red}{2^{99}}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 = \begin{bmatrix}
1 & 6\\
0 & 1
\end{bmatrix} $
......
$A^n = \begin{bmatrix}
1 & 2n\\
0 & 1
\end{bmatrix} $
This was very basic approach. I want to know if there is any other way a smart trick or something to solve this problem ?
| What you did was the smart approach. That is, you computed the first few terms of the sequence $(A^n)_{n\in\mathbb N}$ and you noticed a patern. The only thing that remains to be done is to prove that the pattern is real, but that's easy. Obviously,$$A^1=A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$$and$$A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\implies A^{n+1}=A.\begin{pmatrix}1&2n\\0&1\end{pmatrix}=\begin{pmatrix}1&2(n+1)\\0&1\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A special Vandermonde system with integer coefficients Consider the Vandermonde system
$$\left(\begin{matrix} 1&1&1&\dots&\dots&1\\
1&2&3&4&\dots&n\\
1&2^2&3^2&4^2&\dots&n^{2}\\
1&2^3&3^3&4^3&\dots&n^{3}\\
\vdots&&&&&\vdots\\
1&2^{n-1}&3^{n-1}&4^{n-1}&\dots&n^{n-1}\\
\end{matrix}
\right)~
\left(\begin{matrix} x_{1,n}\\ x_{2,n}\\\dots\\\vdots\\ \dots \\ x_{n,n}
\end{matrix}\right)
=\left(\begin{matrix} 1\\-1\\1\\-1\\ \vdots\\ (-1)^{n-1}
\end{matrix}
\right)
$$
The solution for $n=6$ is $(21, -70, 105, -84, 35, -6)$.
Is there a formula for the general solution involving 2 parameters $j$ and $n $ and, may be, binomial coefficients?
| With help from OEIS A127717 we conjecture that
for $0\le p\lt n$
$$\sum_{k=1}^n k^p (-1)^{k+1}
\sum_{q=k}^n {q-1\choose k-1} q = (-1)^p.$$
This is
$$\sum_{q=1}^n q \sum_{k=1}^q {q-1\choose k-1}
(-1)^{k+1} k^p
= \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} (k+1)^p
\\ = \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} p! [z^p] \exp((k+1)z)
\\ = p! [z^p] \exp(z)
\sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} \exp(kz)
\\ = p! [z^p] \exp(z)
\sum_{q=1}^n q (1-\exp(z))^{q-1}.$$
Now with $(1-\exp(z))^{q-1} = (-1)^{q-1} z^{q-1} + \cdots$ and $n\gt p$
we may certainly extend $q$ to infinity without contributing to $[z^p]$
and we find
$$p! [z^p] \exp(z)
\sum_{q\ge 1} q (1-\exp(z))^{q-1}
\\ = p! [z^p] \exp(z) \frac{1}{(1-(1-\exp(z)))^2}
\\ = p! [z^p] \exp(z) \exp(-2z) = p! [z^p] \exp(-z) = (-1)^p.$$
This is the claim. We see that the desired coefficient is given by
$$\bbox[5px,border:2px solid #00A000]{
x_{k,n} = (-1)^{k+1}
\sum_{q=k}^n {q-1\choose k-1} q.}$$
Remark. A careful examination of the OEIS entry shows that we
can simplify $x_{k,n}.$ We get for the sum
$$\sum_{q=k}^n {q-1\choose k-1} q
= k \sum_{q=k}^n {q\choose k}
= k \sum_{q=0}^{n-k} {q+k\choose k}
\\ = k \sum_{q=0}^{n-k} [z^k] (1+z)^{q+k}
= k [z^k] (1+z)^k \sum_{q=0}^{n-k} (1+z)^q
\\ = k [z^k] (1+z)^k \frac{(1+z)^{n-k+1}-1}{1+z-1}
\\ = k [z^{k+1}] (1+z)^k ((1+z)^{n-k+1}-1)
\\ = k [z^{k+1}] (1+z)^k (1+z)^{n-k+1}
= k [z^{k+1}] (1+z)^{n+1}.$$
We thus have
$$\bbox[5px,border:2px solid #00A000]{
x_{k,n} = (-1)^{k+1}
k {n+1\choose k+1}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The product of three consecutive integers is ...? Odd? Divisible by $4$? by $5$? by $6$? by $12$? If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
*
*My answer is correct
*What is the formal proof of what I said in $ i) $
| One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2\;$mod $3$.
*
*If $n\equiv 0\;(\text{mod}\;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\\[4pt]$
*If $n\equiv 1\;(\text{mod}\;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\\[4pt]$
*If $n\equiv 2\;(\text{mod}\;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n} \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{1\cdot2\cdot3\cdots(n-1)} -\frac{1}{2(1\cdot2\cdot3\cdots n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{2}\sum_ {n=2}^\infty \frac{1}{n!}\right) \\[10pt]
& = e - \frac{1}{2} (e- 1)= \frac{1}{2}(e+1)
\end{align}
Is it correct ???
if not correct then any hints/solution will be appreciated..
thanks in advance
| You made a mistake after this line
$$\sum_ {n=1}^{\infty}(\frac{2n}{2.4.6........2n} -\frac{1}{2.4.6........2n})$$
You simplified numerator and denominator by $2n$ but then
$$\sum_ {n=1}^{\infty}(\frac{1}{2.4.6........2(n-1)} -\frac{1}{2.4.6........2n})$$
You have at the denominator
$$2^nn!$$
So that
$$\sum_ {n=1}^{\infty}\frac{2n}{2^nn!}=\sum_ {n=1}^{\infty}\frac{1}{2^{n-1}(n-1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\iint_D x \sin (y -x^2) \,dA$. Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral
$$\iint_D x \sin (y -x^2) \,dA.$$
I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!
| I'm not sure Kishores answer is precisely the same, but I try to be more illustrative and short. First of all we transform the integral by defining \begin{align}
u &= y-x^2 \\
v &= x^2
\end{align}
which leads to
$$
\int_{\rm D} x\sin(y-x^2) \, {\rm d}x \, {\rm d}y
=\frac{1}{2}\int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v \, .
$$
In the $(x,y)$-plane the area of integration looks like this
which in the $(u,v)$-plane becomes
so that the $v$ integral can be done trivially
\begin{align}
\frac{1}{2} \int_{\rm D} \sin u \, {\rm d}u \, {\rm d}v &= \frac{1}{2} \int_0^1 {\rm d}u \sin u \int_{\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u}^{\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u} {\rm d}v \\
&=\frac{1}{2} \int_0^1 {\rm d}u \sin u \left\{1+\frac{\sqrt{5-4u}}{2} - \frac{\sqrt{9-4u}}{2} \right\} \, .
\end{align}
Integrals of this form lead to Fresnel-Functions
\begin{align}
&\int \sin u \, \sqrt{1-au} \, {\rm d}u \\ = &\sqrt {\frac{\pi \,a}{2}} \left\{ \cos \left( \frac{1}{a} \right) \, {\rm C} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) + \sin \left( \frac{1}{a} \right) \, {\rm S} \left( \sqrt {{\frac {1-au}{\pi \,a/2}}} \right) \right\} -\cos (u) \, \sqrt {1-au}
\end{align}
where $0\leq a \leq 1$. The numerical value of the integral is $\approx 0.1141123$.
In the $(u,v)$-plane
\begin{align}
&{\rm 1)} \quad 2-x \text{ is mapped on } v=\frac{5}{2} - \frac{\sqrt{9-4u}}{2} - u \\
&{\rm 2)} \quad 1-x \text{ is mapped on } v=\frac{3}{2} - \frac{\sqrt{5-4u}}{2} - u \\
&{\rm 3)} \quad x^2 \text{ is mapped on } u=0 \\
&{\rm 4)} \quad x^2+1 \text{ is mapped on } u=1 \, .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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