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Proving the square root of 2 exists(Possible typo in Zorich's Mathematical Analysis)
In the picture above, I don't understand how if $$s^2 \lt 2$$ then $$\left (s + \frac{2-s^2}{3s}\right)^2 \lt 2$$ holds. What insures that $(s + \frac{2-s^2}{3s}) \in X = \{x : x^2 \lt 2\}$ where $x$ is a positive real number ??
And when the identity is expanded should it not be $$\left(s+\frac\Delta{3s}\right)^2=s^2+2\cdot s \cdot\frac\Delta{3s}+\left(\frac\Delta{3s}\right)^2$$ instead of $$\left(s+\frac\Delta{3s}\right)^2=s^2+ 2 \cdot\frac\Delta{3s}+\left(\frac\Delta{3s}\right)^2$$
Thanks in advance!
P.S. The book is Zorich's Mathematical Analysis I in case this is an error.
| We know $1 < s^2 < 2$ so $0 < 2- s^2 = \triangle < 1$. So $ \frac{\triangle}{3s} < \frac {\triangle}3 < \triangle < 1$ so $(\frac{\triangle}{3s})^2 < \frac{\triangle}{3s}$.
That's the groundwork.
So $(s + \frac {2-s^2}{3s})^2 = (s + \frac {\triangle}{3s})^2$
$= s^2 + 2s*\frac {\triangle}{3s} + (\frac {\triangle}{3s})^2$
$< s^2 + 2*\frac {\triangle}3 + \frac {\triangle}3$
$= s^2 + 3\frac {\triangle}3 = s^2 + \triangle$
$= s^2 + (2 - s^2) = 2$.
So $s^2 < (2+ \frac {2-s^2}{3s})^2 < 2$.
And that's that.
(the book had a typo but they didn't make any conclusions based on it. They forgot to cancel out the lower $s$ and left it in for a few steps, and then, simply had it disappear. It was probably a type-setting error and not an actual error of the author.)
Or if you want to do it without the gordang triangle.
$(s + \frac {2-s^2}{3s})^2 = s^2 + 2s*\frac {2-s^2}{3s} + \frac {(2-s^2)^2}{9s^2}$
$=s^2 + \frac {4 - 2s^2}3 + \frac {4 - 4s^2 +s^2}{9s^2}$
$=s^2 +\frac 43 - \frac {2s^2}3 + \frac 4{9s^2} -\frac 49 + \frac {s^2}9$
$= \frac {4s^2}9 + \frac 89 + \frac 4{9s^2}$
$=\frac 49(s^2 + \frac 1{s^2} + 2)$
$= \frac 49(2 -h + \frac 1{2-h} + 2)$
$< \frac 49(2 + 1/2 + 2) = \frac {18}9 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Indentify the plane defined by $Re(z\overline{z})=\frac{lm(\overline{z})}{2}$ I tried:
$$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 = \frac{-y}{2} \Leftrightarrow \\
-2x^2 = y$$
But my book states:
Circunference of center $(0,-\frac{1}{4})$ and radius $\frac{1}{4}$.
What went wrong?
| Let $z=x+iy$ then $z\bar{z} = (x^2 + y^2) + 0i$ .
So $\Re(z\bar{z}) = x^2 + y^2$. Similarly $\bar{z} = x-iy$ so $\Im(\bar{z})=-y$. Hence the required plane is $$x^2 + y^2 = -\frac{y}{2} \iff x^2 + \left(y + \frac{1}{4}\right)^2 = \left(\frac{1}{4}\right)^2$$ which is a circle centered at $(0, -1/4)$ and radius $1/4$.
Recall that whilst $\Re(z) = x$, it need not be the case that $\Re(z\bar{z}) = x^2$. In fact, if you let $z\bar{z} = u+iv$ you see that $u = x^2 + y^2$ and $v=0$. Then clearly $\Re(z\bar{z}) = u = x^2 + y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sin x = \sin 2x$ The equation is: $\sin x = \sin 2x$
I recognize that $\sin 2x$ is a double angle and use the double angle identity, so it becomes:
$$\begin{array}{rrcl}
&\sin x &=& 2 \sin x \cos x \\
\implies& \sin x - 2 \sin x \cos x &=& 0
\end{array}$$
Then I am stuck... Not sure how to proceed.
| \begin{align}
\sin x & = \sin 2x \\
&= 2 \sin x \cos x
\end{align}
Then
\begin{align}
\sin x - 2 \sin x \cos x = 0
\end{align}
So
$$\sin x (1 - 2\cos x ) = 0$$
This imples that $\sin x = 0$ or $(1-2\cos x) = 0$.
$\sin x = 0$ implies that $x = n \pi$. Also $(1 - 2\cos x ) = 0$ implies that $\cos x = \frac{1}{2}$ and so $x = 2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}$. So that the solution set is $$\left \{n \pi,2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}: n \in\mathbb{Z}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangular series perfect square formula 8n+1 derivation In triangular series
$$1 $$
$$1+2 = 3$$
$$1+2+3 = 6$$
$$1+2+3+4 =10$$
$$\ldots$$
Triangular number in 8n+1 always form perfect square .
Like $8\cdot 1+1 = 9 , 8\cdot 3+1 = 25$ .
How this formula is derived ?
| $$\sum_{i=1}^ki=\sum_{i=1}^k\frac{(i+1)^2-i^2-1}{2}=\frac{(k+1)^2-1^2-k}{2}=\frac{k(k+1)}{2}$$
The $k$th triangular number is $\frac{k(k+1)}{2}$.
$$8\left[\frac{k(k+1)}{2}\right] +1=4k^2+4k+1=(2k+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve integral : $\int\frac{\sqrt{x^2 + 1}}{x^2 + 2}dx$ How to solve integral : $\int\frac{\sqrt{x^2 + 1}}{x^2 + 2}dx$
Integration by parts or substitution?
| Let $x = \tan u$, $dx = \sec^2u ~du$. Substitute in the integral to get
$$\int \frac{\sec^3u}{2+\tan^2u}du.$$
Now
\begin{align}
I = \int \frac{\sec^3u}{2+\tan^2u}du &= \int \frac{\frac{1}{\cos^3u}}{2+\frac{\sin^2u}{\cos^2u}}du\\
&=\int \frac{\frac{1}{\cos^3u}}{\frac{2\cos^2u+\sin^2u}{\cos^2u}}du\\
&= \int \frac{1}{\cos u(2\cos^2u+\sin^2u)}du\\
&= \int \frac{\sec u}{(1+\cos^2u)}du\\
&= \int \frac{\sec u}{(2-\sin^2u)}du\\
&= \int \frac{\cos u}{\cos^2u(2-\sin^2u)}du\\
&= \int \frac{\cos u}{(1-\sin^2u)(2-\sin^2u)}du\\
\end{align}
Now, Let $v = \sin u$, $dv = \cos u~ du$, then
\begin{align}
I &= \int \frac{1}{(1-v^2)(2-v^2)}dv = \int \frac{1}{(1-v)(1+v)(\sqrt{2}-v)(\sqrt{2}+v)}dv\\
&= \int \frac{-\frac{1}{2}}{(1-v)} + \frac{\frac{1}{2}}{(1+v)} + \frac{\frac{1}{\sqrt{8}}}{(\sqrt{2}-v)}+ \frac{-\frac{1}{\sqrt{8}}}{(\sqrt{2}+v)}dv\\
&= \frac{1}{2} \ln(1-v) - \frac{1}{2} \ln(1+v) -\frac{1}{\sqrt{8}} \ln(\sqrt{2}-v) + \frac{1}{\sqrt{8}} \ln(\sqrt{2}+v)\\
&= \frac{1}{2} \ln(1-\sin u) - \frac{1}{2} \ln(1+\sin u) -\frac{1}{\sqrt{8}} \ln(\sqrt{2}-\sin u) + \frac{1}{\sqrt{8}} \ln(\sqrt{2}+\sin u)
\end{align}
Therefore
\begin{align}
I &= \frac{1}{2} \ln(1-\frac{x}{\sqrt{1+x^2}}) - \frac{1}{2} \ln(1+\frac{x}{\sqrt{1+x^2}}) -\frac{1}{\sqrt{8}} ~~\ln(\sqrt{2}-\frac{x}{\sqrt{1+x^2}}) \\
&+ \frac{1}{\sqrt{8}} \ln(\sqrt{2}+\frac{x}{\sqrt{1+x^2}}) + C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do i find this limit WITHOUT L'hôpitals rule How do I find
$$\lim_{x\to 0}\frac{(x+4)^{3/2}+e^x-9}{x}$$
without l'hôpital rule? I know from l'hôpital the answer is 4.
| Less elegant than zhw'x answer.
For $(x+4)^{3/2}$, use the generalized binomial theorem of Taylor expansion
$$(x+4)^{3/2}=8+3 x+\frac{3 x^2}{16}+O\left(x^3\right)$$ Using the standard Taylor expnasion of $e^x$, we then have $$(x+4)^{3/2}+e^x-9=\left(8+3 x+\frac{3 x^2}{16}+O\left(x^3\right) \right)+\left( 1+x+\frac{x^2}{2}+O\left(x^3\right)\right)-9$$ $$(x+4)^{3/2}+e^x-9=4 x+\frac{11 }{16}x^2+O\left(x^3\right)$$
$$\frac{(x+4)^{3/2}+e^x-9}{x}=4 +\frac{11 }{16}x+O\left(x^2\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Guess the formula for $\sum\frac 1{(4n-3)(4n+1)}$ and prove by induction
For $n \ge 1$, let
$$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
Guess a simple explicit formula for $a_n$ and prove it by induction.
So I have guessed the formula as : $$\frac{n}{4n+1}$$
But wasn't sure how to prove it by induction
| Base case:
$$a_1=\frac1{4(1)+1}\color{green}\checkmark$$
Now the inductive step:
$$\begin{align}a_{n+1}&=a_n+\frac1{(4n+1)(4n+5)}\\&=\frac n{4n+1}+\frac1{(4n+1)(4n+5)}\\&=\frac{n(4n+5)}{(4n+1)(4n+5)}+\frac1{(4n+1)(4n+5)}\\&=\frac{4n^2+5n+1}{(4n+1)(4n+5)}\\&=\frac{(4n+1)(n+1)}{(4n+1)(4n+5)}\\&=\frac{n+1}{4(n+1)+1}\color{green}\checkmark\end{align}$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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fine the limits :$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$ fine the limits-without-lhopital rule and Taylor series :
$$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$$
i know that :
$$\lim_{x \to 0} \frac{\sin x}{x}=1=\lim_{x \to 0}\frac{\tan x}{x}$$
But I can not answer please help .
| Here is a different way.
You can use standard Taylor series expansions, as $x \to 0$, to get $$\sin 2x-2x\cos x=-\frac{x^3}3+o(x^4) $$ $$\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))=-18x+o(x^2) $$ $$x\sin x \tan x\sin 2x=2x^4+o(x^5)$$ then $$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=\lim_{x \to 0}\frac{\frac{18}{3}x^4+o(x^5)}{2x^4+o(x^5)}=3. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
Someone told me that it has $(29+1)/2=15$ solutions. How come? Any other method to solve this?
| First, factorise :
$$x^2-y^2=(x-y)(x+y)=5^{29}.$$
Then by the Fundamental Theorem of Arithmetics, both factors must be powers of $5$, i.e.
$$\left\{\begin{array}{}x+y & = & 5^a \\ x-y & = & 5^b,\end{array}\right.$$with $a+b=29$. Since $x+y>x-y$, we must have $a>b$; hence you will have one system for every $15\leq a\leq 29$. Since every system as above has a unique integer solution $$x=\frac{5^{a}+5^b}{2},\quad y=\frac{5^{a}-5^b}{2},$$
we obtain the $15$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relationship between two rotated ellipses centered at the origin Let there be two ellipses centered at the origin, and rotated angles $\alpha$ and $\alpha'$ from the x-coordinate axis. $a, b$ are the semiaxes of the first, $a',b'$ the semiaxes of the second, and the ratio $ab/a'b'$ is called $f$.
The equations for both ellipses can be written:
$$
\frac{(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac{(y\cos\alpha-x\sin\alpha)^2}{b^2}=1
$$
and
$$
\frac{(x\cos\alpha+fy\sin\alpha)^2}{a^2}+\frac{(fy\cos\alpha-x\sin\alpha)^2}{b^2}=1
$$
Where the equation for the second ellipse is written for convenience as a function of the parameters $a,b, \alpha$.
Now in the text I am following I find that the semiaxes of the second ellipse, $a', b'$, can be obtained from the second equation above, using the relation:
$$
\frac{1}{a'^2}+\frac{1}{b'^2}=\frac{\cos^2\alpha+f^2\sin^2\alpha}{a^2}+\frac{\sin^2\alpha+f^2\cos^2\alpha}{b^2}
$$
However I don't have any clue were this last relation comes from. Any help is welcome!
| The equation of an ellipse in standard position with semi-axis lengths $a'$ and $b'$ can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\frac1{a'^2}&0\\0&\frac1{b'^2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1.\tag1$$ The trace of the central matrix—the sum of its main diagonal elements—in this equation is equal to $\frac1{a'^2}+\frac1{b'^2}$. Rotating this ellipse amounts to a similarity transformation of the coefficient matrix, which leaves its trace unchanged. Expanding the equation of the rotated ellipse given in the question and collecting terms gives $$\left({\cos^2\alpha\over a^2} + {\sin^2\alpha\over b^2}\right)x^2+2f\cos\alpha\sin\alpha\left(\frac1{a^2} - \frac1{b^2}\right)xy + \left({f^2\sin^2\alpha\over a^2} + {f^2\cos^2\alpha\over b^2} \right)y^2=1.\tag2$$ In matrix form, the diagonal elements are the coefficients of $x^2$ and $y^2$, so take these two coefficients from equation (2) and rearrange their sum to get $$\frac1{a'^2}+\frac1{b'^2}={\cos^2\alpha+f^2\sin^2\alpha\over a^2}+{\sin^2\alpha+f^2\cos^2\alpha\over b^2}$$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286423",
"timestamp": "2023-03-29T00:00:00",
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Partial sum using telescopic series The summation of $(n^2+1)\cdot n!$ using to $N$ using telescopic series.
Question: Find a closed expression in $n$ for:
\begin{gather}
\text{Let } u_n = (n^2+1)\cdot n! \\
\sum_{n=1}^{N}u_n = \sum_{n=1}^{N}(n^2+1)\cdot n!\\
\end{gather}
My Attempt
I tried to split $(n^2+1)\cdot n!$ into two different terms that would telescope.
First try
\begin{gather}
(n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\
= (n+1)\cdot (n+1)! - 2n\cdot n!\\
\sum_{n=1}^{N}(n+1)\cdot (n+1)! - 2n\cdot n!\\
(2)\cdot(2)! - 2(1)\cdot(1)!\\
(3)\cdot(3)! - 2(2)\cdot(2)! \\
\vdots \\
(n+1)\cdot (n+1)! - 2(n)\cdot (n)!\\
\end{gather}
The $2$ on the second term is quite annoying and needless to say it does not work.
Second try
\begin{gather}
(n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\
= (n+1)\cdot (n+1)! - 2n\cdot n! = (n+1)\cdot(n+1)! -2[(n+1)! - (n)!]\\
\end{gather}
Didn't work.
Try #3
\begin{gather}
(n+1)\cdot(n+1)! -2[(n+1)! - (n)!] = (n+1)!\cdot(n-1)+2n!\\
\end{gather}
This one didn't help at all.
Could anyone please suggest a hint to arrive at the right telescoping expression?
| The key is that $$(n^2+1)\cdot n! = n\cdot(n+1)! - (n-1)\cdot n!.$$
Thus $$\sum_{n=1}^Nu_n=(N\cdot(N+1)!-(N-1)\cdot N!) + ((N-1)\cdot N! - (N-2)\cdot (N-1)!) +\\+ \cdots + (1\cdot2!-0\cdot1!) = N\cdot(N+1)!.$$
| {
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"url": "https://math.stackexchange.com/questions/2289308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve equation with absolute value of complex numbers Find all complex numbers $z$ such that $|z| + \overline{z}=9-3i$.
Thinking about using with the fact that $z=a+bi$ and $|z| = \sqrt{a^2+b^2}$ in some way. Not sure how.
| Write $\;z=x+iy\;$ , so that your equation is
$$\sqrt{x^2+y^2}+x-iy=9-3i\stackrel{\text{compare real and imag. parts}}\implies\begin{cases}\sqrt{x^2+y^2}+x=9\\{}\\y=3\end{cases}\;\;\implies$$
$$\sqrt{x^2+3^2}+x=9\implies x^2+9=81-18x+x^2\implies18x=72\implies x=4$$
and the unique solution is $\;4+3i\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f,g,h,i\in \{0,1,2\}$
$$M^{T}M= \begin{pmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{pmatrix}\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix} $$.
sum of diagonal entries $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2 = 5$$
How can I form different cases?
| Since the square of each number can only equal $0$, $1$ or $4$, it comes down to selecting either five numbers which equal $1$ (with the four remaining ones being $0$), or select one which equals $2$ and one which equals $1$ (with the seven remaining ones being $0$). As such, the number of valid matrices equals:
$${9 \choose 5} + {9 \choose 1} {8 \choose 1} = 126 + 72 = 198$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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completing square for a circle In the following question:
I don't understand how we can get from the original equation to the final equation using completing the square.
Any thoughts as how to get to the final equation?
| One way of doing this is to avoid fractions completely until the last step.
$$\begin{align}
\frac{x+y}{x^2+y^2+1}&=c\\
x+y&=c(x^2+y^2+1)\\
x+y&=cx^2+cy^2+c\\
cx^2-x+cy^2-y+c&=0\\
cx^2-x+cy^2-y&=-c\\
4c(cx^2-x+cy^2-y)&=4c(-c)\\
4c^2x^2-4cx+4c^2y^2-4cy&=-4c^2\\
(2cx)^2-2(2cx)+(2cy)^2-2(2cy)&=-4c^2\\
(2cx)^2-2(2cx)+1+(2cy)^2-2(2cy)+1&=2-4c^2\\
(2cx-1)^2 + (2cy-1)^2 &= 2-4c^2\\
\frac{(2cx-1)^2 + (2cy-1)^2}{4c^2} &= \frac{2-4c^2}{4c^2}\\
\bigg(x-\frac{1}{2c}\bigg)^2 + \bigg(y-\frac{1}{2c}\bigg)^2 &= \frac{1}{2c^2}-1\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the solution to the nonlinear PDE: $U - XUx - (1/2)*(Uy)^2 + X^2 = 0$ with $U(X,0) = X^2 - (1/6)*(x^4), 0Step 1: Rewrite the PDE
$P= Ux$ and $Q= Uy$
$$F = U - XP - (1/2)Q^2 + x^2$$
Step 2: Charpits Equations
$dx/dτ = Fp = -x$
$dy/dτ = Fq = -2Q$
$dp/dτ = -Fx-P*Fu = -2x$
$dq/dτ = -Fy-Q*Fu = -Q$
$du/dτ = PFp+QFq = -xp - 2Q^2$
Step 3: Integrate
At this point I am confused as to where to start?
Next Steps: Parameterise Initial data
$Xo = S, Uo = S^2 - (1/6)*S^4, Yo = 0$
$dUo/dS = Po*dxo/ds + Qo*dyo/ds$
$Po = 2S -(2/3)*S^3$
| Hint:
Let $U=X^2-V$ ,
Then $U_X=2X-V_X$
$V_Y=-U_Y$
$\therefore 2X^2-V-X(2X-V_X)-\dfrac{(-V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
$XV_X-V-\dfrac{(V_Y)^2}{2}=0$ with $V(X,0)=\dfrac{X^4}{6}$
Let $V=XW$ ,
Then $V_X=XW_X+W$
$V_Y=XW_Y$
$\therefore X(XW_X+W)-XW-\dfrac{(XW_Y)^2}{2}=0$ with $W(X,0)=\dfrac{X^3}{6}$
$X^2W_X=\dfrac{X^2(W_Y)^2}{2}$ with $W(X,0)=\dfrac{X^3}{6}$
$(W_Y)^2=2W_X$ with $W(X,0)=\dfrac{X^3}{6}$
$W_Y=\pm\sqrt{2W_X}$ with $W(X,0)=\dfrac{X^3}{6}$
$W_{XY}=\pm\dfrac{W_{XX}}{\sqrt{2W_X}}$ with $W(X,0)=\dfrac{X^3}{6}$
Let $Z=W_X$ ,
Then $Z_Y=\pm\dfrac{Z_X}{\sqrt{2Z}}$ with $Z(X,0)=\dfrac{X^2}{2}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dY}{dT}=1$ , letting $Y(0)=0$ , we have $Y=T$
$\dfrac{dZ}{dT}=0$ , letting $Z(0)=Z_0$ , we have $Z=Z_0$
$\dfrac{dX}{dT}=\mp\dfrac{1}{\sqrt{2Z}}=\mp\dfrac{1}{\sqrt{2Z_0}}$ , letting $X(0)=f(Z_0)$ , we have $X=f(Z_0)\mp\dfrac{T}{\sqrt{2Z_0}}=f(Z)\mp\dfrac{Y}{\sqrt{2Z}}$ i.e. $Z=F\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)$
$Z(X,0)=\dfrac{X^2}{2}$ :
$F(X)=\dfrac{X^2}{2}$
$\therefore Z=\dfrac{\left(X\pm\dfrac{Y}{\sqrt{2Z}}\right)^2}{2}$
$2Z=X^2\pm\dfrac{\sqrt2XY}{\sqrt{Z}}+\dfrac{Y^2}{2Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $g(x) = \sqrt{1 + x^2}$ is continuous Show that $g(x) = \sqrt{1 + x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon - \delta$ argument.
This is what I was thinking...
Let $p \in \mathbb{R}$. Then $g(x)-g(p) = \sqrt{1 + x^2} - \sqrt{1 + p^2}$
$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{x^2-p^2}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} = \frac{(x+p)(x-P)}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$.
Thus, $g(x) - g(p)| = \frac{|x+p||x-p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$
$ \quad \quad \quad \quad \quad \quad \leq \frac{|x| + |p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} |x -p|$ $= (\frac{|x|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} + \frac{|p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} ) |x-p| $
$ \quad \quad \quad \quad \quad \quad \leq (\frac{|x|}{\sqrt{1 + x^2}} + \frac{|p|}{\sqrt{1 + p^2}} ) |x-p|$
$ \quad \quad \quad \quad \quad \quad \leq (1+1) |x-p| = 2|x-p|$
Choose $\epsilon > 0$. Then $|g(x)-g(p)| < \epsilon$ whenever $|x-p| < \frac{\epsilon}{2}$. Let $\delta = \frac{\epsilon}{2}> 0$. Then for $\epsilon > 0$ there exists $\delta > 0$ such that $|g(x)-g(p)| < \epsilon$ whenever $|x-p|< \delta$. Hence, $g$ is continuous at $p \in \mathbb{R}$. Since $p$ is an arbitrary real number, $g$ is continuous on $\mathbb{R}$.
Is this correct?
| You can also follow this slightly shorter argument:
Fix an epsilon $\epsilon$.
$$|g(x)-g(p)|=|\sqrt{1+x^2}-\sqrt{1+p^2}|\leq \sqrt{|1+x^2-1-p^2|}\leq \sqrt{|x^2-p^2|}=\sqrt{|x+p||x-p|}$$
You can bound $|x+p|$ by making sure $|x-p|<1$ or $|x+p|<1+2p$. Also, you make sure that $|x-p|<\frac{\epsilon ^2}{1+2p}$ So that you have when $|x-p|<\min(1,\frac{\epsilon ^2}{1+2p})$
$$\sqrt{|x+p||x-p|}<\epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Sum of all possible values of $\gcd(a-1,a^2+a+1)$
Find the sum of all possible values of
$$\gcd(a-1,a^2+a+1)$$
where $a$ is a positive integer.
| \begin{align}
\gcd(a-1,a^2+a+1)&=\gcd(a-1,a^2+a+1-(a+2)(a-1))\\
&=\gcd(a-1,3)\\
&=\begin{cases} 3 & \textrm{if }a\equiv 1\quad(\textrm{mod }3) \\ 1 & \textrm{if }a\not\equiv 1\quad(\textrm{mod }3) \end{cases}
\end{align}
The sum is $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to write $\frac{x^3+4x}{x^2-4}$ as a serie at $x_0=1$? I did this, but what to do next?
$f(x)=x+\frac{4}{x-2}+\frac{4}{x+2}$
| In this case the $n$-th derivative of $f$ is explicitly calculable:
$$f(x) = x + 4\,\frac{1}{x - 2} + 4\,\frac{1}{x + 2}$$
$$f'(x) = 1 - 4\,\frac{1}{(x - 2)^2} - 4\,\frac{1}{(x + 2)^2}$$
$$f''(x) = 4\,\frac{2}{(x - 2)^3} + 4\,\frac{2}{(x + 2)^3}$$
$$\cdots$$
$$f^{(n)}(x) = (-1)^n4\,\frac{n!}{(x - 2)^{n+1}} + (-1)^n4\,\frac{n!}{(x + 2)^{n+1}}$$
Now, you can apply Taylor's theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$
Prove that $\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]=6$
My idea was to find the minimal polynomial of $\sqrt[3]{5}+\sqrt{2}$ over $\mathbb{Q}$ and to show that $\deg p(x)=6$
Attempt:
Let $u:=\sqrt[3]{5}+\sqrt{2}\\
u-\sqrt[3]{5}=\sqrt 2\\
(u-\sqrt[3]{5})^2=2\\
u^2-2\sqrt[3]{5}u+5^{2/3}-2=0\\
u^2-2-5^{2/3}=2\sqrt[3]{5}u\\
(u^2-2-5^{2/3})^3=2^3\cdot 5 \cdot u$
I'm stuck here
Here Wolfram's result
My previous question over $\mathbb{Q}(\sqrt[3]{5})$
| There is no need to compute minimal polynomials.
Let $\alpha = \sqrt[3]{5}+\sqrt{2}$. Then $(\alpha-\sqrt{2})^3=5$ and so $\alpha^3 + 6 \alpha -5 =\sqrt{2} (3\alpha^2 + 2)$, which gives
$(\alpha^3 + 6 \alpha -5)^2 =2 (3\alpha^2 + 2)^2$.
Therefore,
$\alpha$ is a root of a polynomial of degree $6$ and so
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right] \le 6$.
On the other hand,
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right]$
is a multiple of $6$ because
$\left[\mathbb{Q}(\sqrt[3]{5}):\mathbb Q\right]=3$
and
$\left[\mathbb{Q}(\sqrt{2}):\mathbb Q\right]=2$.
Therefore,
$\left[\mathbb{Q}(\sqrt[3]{5}+\sqrt{2}):\mathbb Q\right] = 6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find this limit of an integral Find $$\lim_{n\to\infty}n\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)$$
it is obvious
$$\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}})dx=B\left(\dfrac{n+1}{2n},\dfrac{2n+1}{4n}\right)$$
| You are missing a crucial factor $2$. You want to compute
$$ \lim_{n\to +\infty}n^2\left[\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)\,\Gamma\left(\frac{1}{2}+\frac{1}{4n}\right)}{2\,\Gamma\left(1+\frac{3}{4n}\right)}-\frac{\Gamma\left(\frac{1}{2}\right)^2}{2}\right] $$
hence it is enough to exploit the Taylor series of $\Gamma(z)$ in neighbourhoods of $z=\frac{1}{2}$ and $z=1$:
$$ \Gamma(1+x)=1-\gamma x + \frac{\gamma^2+\zeta(2)}{2} x^2+O(x^3) $$
$$ \Gamma\left(\frac{1}{2}+x\right)=\sqrt{\pi}-\sqrt{\pi}(\gamma+\log 4)x + \sqrt{\pi}\frac{\pi^2+2(\gamma+\log 4)^2}{4} x^2+O(x^3) $$
to get that the previous limit is just $-\infty$. However, if $n^2$ is replaced by $n$ the limit becomes $-\frac{3\pi\log 2}{4}$. The previous Taylor series are derived from $\Gamma'(x)=\psi(x)\Gamma(x)$, $\Gamma''(x)=\psi'(x)\Gamma(x)+\psi(x)^2\Gamma(x)$ and the well-known values of $\psi(1)$ and $\psi\left(\frac{1}{2}\right)$.
The first terms of the asymptotic expansion of the integral are given by:
$$ \int_{0}^{\pi/2}(\sin x)^{\frac{1}{n}}(\cos x)^{\frac{1}{2n}}\,dx = \frac{\pi}{2}-\frac{3\pi\log 2}{4n}+\frac{\pi^3+36 \pi\log(2)^2}{64 n^2}+O\left(\frac{1}{n^3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Analyze convergence of the series $\sum\limits_{n=1}^\infty\frac{1}{n(n^\frac14-n^\frac13+n^\frac12)}$ $$\sum_{n=1}^\infty\frac{1}{n(n^\frac{1}{4}-n^\frac{1}{3}+n^\frac{1}{2})}$$
It is possible to use the ratio test or the root test or any of them without give opposite answers ? There is a case in which a method give an answer and another other answer? Which of them is a better strategy for this one? Does this one converge or diverge? How I recognize which method I ought use?
Sorry for asking so many questions.
| Just for the fun.
Consider $$A=\frac{1}{x(x^\frac{1}{4}-x^\frac{1}{3}+x^\frac{1}{2})}$$ and make $x=y^{12}$ to get $$A=\frac{1}{y^{15} \left(y^3-y+1\right)}=\frac{1}{y^{18} \left(1-\frac 1{y^2}+\frac 1 {y^3}\right)}$$ Now, make the long division (or use Taylor series) to get $$\frac{1}{ 1-\frac 1{y^2}+\frac 1 {y^3}}=1+\frac 1{y^2}-\frac 1 {y^3}+\frac 1 {y^3}+\frac 1 {y^4}+O\left(\frac 1 {y^5} \right)$$ making $$A=\frac 1 {y^{18}}+\frac 1 {y^{20}}-\frac 1 {y^{21}}+\frac 1 {y^{22}}+O\left(\frac 1 {y^{23}} \right)$$ Back to $x$ $$A=\frac 1 {x^{3/2}}+\frac 1 {x^{5/3}}-\frac 1 {x^{7/4}}+\frac 1 {x^{11/6}}+\cdots$$ Replace $x$ by $n$ and prepare the summations : you face convergent series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral of $\frac{x^9}{(1+x^3)^{1/3}}$, recurrence relation I need to find $$\int \frac{x^9}{(1+x^3)^{1/3}}$$
I have recurrence relation:
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$$a(m+1+np)J_{m, p} = x^{m+1-n}(ax^n + b)^{p+1} - b(m+1-n) J_{m-n,p}$$
But when I tried to find the integral, I had problems with $$J_{0, p}$$, because I got
$$0 \cdot J_{0,p} = x^{-2}(x^3+1)^{\frac{2}{3}} + 2J_{-3, p}$$
| $$a = 1, n = 3, b = 1, p = -\frac{1}{3}$$
$$J_{m, p} = \int x^m (ax^n + b)^p dx$$
$m = 9$
$$9 J_{9, p} = x^{7}(x^3 + 1)^{\frac{2}{3}} - 7 J_{6,p}$$
$m = 6$
$$6 J_{6, p} = x^{4}(x^3 + 1)^{\frac{2}{3}} - 4 J_{3,p}$$
$m = 3$
$$3 J_{3, p} = x(x^3 + 1)^{\frac{2}{3}} - J_{0,p}$$
I think we have to stop the reccurrence at this stage and calculate the integral:
$$J_{0, p} = \int \frac{1}{(1+x^3)^{\frac{1}{3}}}dx =\dfrac{\ln\left(\left|\left(\frac{1}{x^3+1}-1\right)^\frac{2}{3}+\sqrt[3]{\frac{1}{-x^3-1}+1}+1\right|\right)-2\ln\left(\left|\sqrt[3]{\frac{1}{x^3+1}-1}+1\right|\right)}{6}-\dfrac{\arctan\left(\frac{2\sqrt[3]{\frac{1}{x^3+1}-1}-1}{\sqrt{3}}\right)}{\sqrt{3}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is Differentiation of total variation function $V(f, [0,x])$ possible in this condition? I am having difficulty of determining whether differentiation of total variation function $V(f, [0,x])$ is possible on $x = 0$ when $f:[0,1]→R$ is defined as:
$ f(x) = 0$ for $x = 0$ and $f(x) = x^2sin(1/x)$ for $x≠0$
I think it would not be possible intuitively, but I cannot come up with precise proof.
| To compute the derivative of the total variation of $f$ at $x = 0$ as a definitional limit, take a partition with subintervals
$\left[((k+1)\pi )^{-1},(k\pi)^{-1}\right].$ Since $f(x) = x^2 \sin(1/x)$ vanishes at the endpoints, there is an extremum at some point $\xi_k$ where
$$\frac{1}{(k+1)\pi} \leqslant \xi_k \leqslant \frac{1}{k\pi}.$$
The total variation of $f$ on the subinterval is
$$V\left[(k\pi)^{-1}\right] - V\left[((k+1)\pi)^{-1}\right] = 2|f(\xi_k)|,$$
since $f$ is alternately monotone increasing and decreasing in the subintervals on either side of $\xi_k$.
Observing where $f$ takes the extreme value in the interval we find
$$\frac{1}{((k+1/2)\pi)^2} = |f(\, ((k+1/2)\pi)^{-1})| < |f(\xi_k)| < \xi_k^2 < \frac{1}{(k\pi)^2}.$$
Hence,
$$\frac{2}{\pi^2}\frac{1}{(k+1/2)^2} \leqslant V\left[(k\pi)^{-1}\right] - V\left[((k+1)\pi)^{-1}\right] \leqslant \frac{2}{\pi^2}\frac{1}{k^2}$$
Summing from $k = n$ to $\infty$ we get
$$\frac{2}{\pi^2}\sum_{k=n}^{\infty}\frac{1}{(k+1/2)^2} < V[(n\pi)^{-1}] < \frac{2}{\pi^2}\sum_{k=n}^{\infty}\frac{1}{k^2}.$$
Using the following bounds for the sums,
$$\sum_{k=n}^{\infty}\frac{1}{k^2} < \int_{n-1}^{\infty}\frac{dx}{x^2}= \frac1{n-1},\\\sum_{k=n}^{\infty}\frac{1}{(k+1/2)^2} > \int_{n+1/2}^{\infty}\frac{dx}{x^2}= \frac1{n+ 1/2},$$
it follows that,
$$\frac{2}{\pi^2}\frac1{n + 1/2} < V[(n\pi)^{-1}] < \frac{2}{\pi^2}\frac1{n-1}.$$
Take $h$ such that $(n\pi)^{-1} < h < ((n-1)\pi)^{-1},$ and note that $h \to 0$ as $n \to \infty$.
Since $V(x)$ is increasing, we have
$$\frac{2}{\pi^2}\frac{1}{n + 1/2}<V[(n\pi)^{-1}] < V(h) < V[((n-1)\pi)^{-1}]< \frac{2}{\pi^2}\frac{1}{n-2}.$$
Since, $ (n-1)\pi < 1/h < n\pi,$ we have
$$ \frac{2}{\pi} \frac{1 - 1/n}{1 + 1/(2n)} = \frac{2}{\pi^2}\frac{(n-1)\pi}{n + 1/2} < \frac{V(h)}{h} < \frac{2}{\pi^2}\frac{n\pi}{n-2} = \frac{2}{\pi} \frac{1}{1 - 2/n}
$$
Taking limits as $n \rightarrow \infty$ and applying the squeeze theorem, we find the right-hand derivative of the total variation
$$V'(0) = \lim_{h \rightarrow 0} \frac{V(h)}{h } = \frac{2}{\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1)4^2+\dots+(4-1)4^{z-2}+(4-1)4^{z-1}$$
Which becomes:
$$4^z=1+4^1-4^0+4^2-4^1+4^3-4^2+\dots+4^{z-1}-4^{z-2}+4^z-4^{z-1}$$
After canceling everything out you are left with:
$$4^z=4^z$$
More generally:
$$a^z =1 + (a-1)\sum_{k=0}^{z-1} a^k$$
Therefore: $(2^x-1)/3$ will always be an integer when $x$ is even.
My question is: will there ever be an integer solution to $(2^x-1)/3$ when $x$ is odd?
| As you noted: $y =\frac {2^{2k} - 1}3$ is always an integer. So $2y = 2\frac {2^{2k} - 1}3 = 2\frac {2^{2k+1} - 2}{3} = \frac {2^{2k+1} - 1}3 - \frac 13$ is integer so $\frac {2^{2k+1} -1}3$ is not an integer.
So no.
If know modulo arithmetic it's easier.
$y = \frac {2^{2k} -1}3$ being an integer is the same as $3y +1 = 2^{2k}$ is the same as saying $2^{2k}$ has remainder $1$ when divided by $3$ which we write as $2^{2k} \equiv 1 \mod 3$. ($a \equiv b \mod n$ means $a$ and $b$ have the same remainder when divided by $n$).
So $2*2^{2k} = 2^{2k + 1}$ will have remainder $2*1 = 1$ when divided by $3$ or $2^{2k+1} = 2*2^{2k} \equiv 2*1 = 2 \mod 3$ and $2^{2k+1} \not \equiv 1 \mod 3$ So $\frac {2^{2k+1} - 2}3$ is an integer but $\frac {2^{2k+1} - 1}3$ is not.
In general though $\frac {2^{2k} -1 }3$ is an integer; $\frac {2^{2k+1} + 1}3 $ is an integer. And we can generalize that as $\frac {2^{x} + (-1)^x}3$ is an integer.
Or in modulo notation:
$(2)^x \equiv (-1)^2 \mod 3$ which makes sense as $2 \equiv -1 \mod 3$ [$2 = 0*3 + 2$ with $2$ remainder and $-1 = -1*3 + 2$ with $2$ remainder.] So $2^x = (3-1)^x = 3^x - x*3^{x-1}......+ x*3*(-1)^{x-1} + (-1)^x$ will have the same remainder as $(-1)^x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}{4}} &=
\frac{\cos(2x)}{\frac{4x-\pi}{4}} \\&=
\frac{4\cos(2x)}{4x-\pi} = \,\,???\end{align}$$
What do I do next?
| $$\cos(2x)=\sin\left(\frac{\pi}2-2x\right)\\\frac{\sin(\frac{\pi}2-2x)}{x-\frac{\pi}4}=\frac{2\sin(\frac{\pi}2-2x)}{2x-\frac{\pi}2}$$
Now it should be easy.
| {
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"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Help with this integral: $\int_{0}^{1}\frac{1}{1+x^{2}}dx$ (Riemann) I'm stuck when I try to solve this integral:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx$$
I try this:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}\frac{1}{n}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}$$
Can someone help with the next step? I'm stuck.
| We can proceed as follows. First, we expand the term $\frac{n}{k^2+n^2}$ in the geometric series
$$\frac{n}{k^2+n^2}=\sum_{\ell =0}^N (-1)^\ell \frac{k^{2\ell}}{n^{2\ell +1}}+(-1)^{N+1}\frac{k^{2N+2}}{n^{2N+1}(k^2+n^2)}\tag 1$$
Then, summing $(1)$ reveals
$$\begin{align}
\sum_{k=1}^n \frac{n}{k^2+n^2}&=\sum_{k=1}^n\sum_{\ell =0}^N (-1)^\ell \frac{k^{2\ell}}{n^{2\ell +1}}+\frac{(-1)^{N+1}}{n^{2N+1}}\sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\\\\
&=\sum_{\ell =0}^N \frac{(-1)^\ell}{n^{2\ell +1}} \sum_{k=1}^n k^{2\ell}+\frac{(-1)^{N+1}}{n^{2N+1}}\sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\tag2
\end{align}$$
Note that we have the relationships
$$\sum_{k=1}^n k^{2\ell}=\frac{n^{2\ell +1}}{2\ell +1}+O(n^{2\ell}) \tag3$$
and
$$\frac{n^{2N+1}}{2(2N+3)}\le \sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\le \frac{n^{2N+1}}{2N+3}\tag 4$$
Using $(3)$ and $(4)$ in $(2)$ and letting $N\to \infty$ yields
$$\sum_{k=1}^n \frac{n}{k^2+n^2}=\sum_{\ell =0}^\infty \frac{(-1)^\ell}{2\ell +1}+O\left(\frac1n\right)$$
whereupon letting $n\to \infty$ we obtain
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{n}{k^2+n^2}=\sum_{\ell =0}^\infty \frac{(-1)^\ell}{2\ell +1}$$
Finally, recalling that the Taylor series for $\arctan(x)=\sum_{\ell=0}^\infty \frac{(-1)^\ell\,x^{2\ell +1}}{2\ell +1}$, we see that
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{n}{k^2+n^2}=\pi/4$$
as expected!
| {
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} |
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
I've tried different substitutions like put $Q(x)$ $=$ $k$ for some $k$ and getting the equation $k(x^2-6x+8)$ $=$ $(k-2)(x^2-6x)$ $<=>$ $k$=$(6x-x^2)/4$, but that doesn't give equlity. Thank you for your help
| $$\begin{align*}
(x^2-6x+8) Q(x) &\equiv (x^2 -6x)Q(x-2)\\
(x^2 - 6x)[Q(x)-Q(x-2)] &\equiv -8Q(x)
\end{align*}$$
If $Q(x)$ has degree $n > 0$, $Q(x) - Q(x-2)$ has degree $n-1$, and so the left hand side has degree $n+1$. However, the right hand side has degree $n$.
So $n\not>0$, and $Q(x)$ can only be a constant function with degree $0$. Let $Q(x) \equiv q$,
$$\begin{align*}
(x^2-6x+8)q &\equiv (x^2 -6x)q\\
8q &\equiv 0\\
q &\equiv 0\\
Q(x) &\equiv 0
\end{align*}$$
Notes regarding that $Q(x) - Q(x-2)$ has degree $n-1$: Let
$$Q(x) \equiv ax^n + bx^{n-1} + \cdots$$
where $a \ne 0$ and all lower degree terms are omitted. Then $Q(x-2)$ is
$$\begin{align*}
Q(x-2) &\equiv a(x^n - 2nx^{n-1} + \cdots) + b(x^{n-1} - \cdots) + \cdots\\
&\equiv ax^n -2anx^{n-1} + bx^{n-1} + \cdots\\
Q(x)-Q(x-2)&\equiv 2anx^{n-1} + \cdots
\end{align*}$$
Since $a\ne 0$ and $n> 0$, the coefficient of $x^{n-1}$ is non-zero, and so the right hand side has degree $n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Growth series for the lamplighter group This paper gives the growth series of the lamplighter group as
$$ \frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2+x^3)^2(1-x-x^2)} = 1 + 3x + 6x^2 + 8x^3 + 10x^4 + \ldots $$
The first 3 coefficients seem to be correct. However, I get 12 group elements with a word norm of 3. They are the following:
\begin{array}{l l}
ttt & at^{-1}a \\
tta & at^{-1}t^{-1} \\
tat & t^{-1}at \\
tat^{-1} & t^{-1}at^{-1} \\
att & t^{-1}t^{-1}a \\
ata & t^{-1}t^{-1}t^{-1}
\end{array}
What is the reason for this discrepancy?
| You are right. On page 6 of that paper, equation $(3.5)$ is given as:
$$f_{G\wr\mathbb{Z}}(x)=\frac{f_G(x)(1-x^2)^2(1+xf_G(x))}{(1-x^2f_G(x))^2(1-xf_G(x))}$$
On substituting $(1+x)$ for $f_G(x)$, I get:
$$f_{{\cal{L}}_2}(x)=\frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2-x^3)^2(1-x-x^2)}$$
And when expanded, the result seems to agree with you:
$$1 + 3 x + 6 x^2 + 12 x^3 + 22 x^4 + 40 x^5 + 71 x^6 + 123 x^7 + 212 x^8 + 360 x^9 + 607 x^{10} +...$$
In the working at the bottom of the same page, a $^2$ goes missing, and then $(-x^2)(+x)$ becomes $+x^3$ in the denominator. It's just a mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do whole number solutions exist for $3x^2 = y^2$? I am investigating properties of square numbers and would like to find whole number solutions for this equation
$3x^2 = y^2$
or $\sqrt{3x^2} = y$
How do I prove that whole number solutions do not exist or how do I identify them?
| We can rearrange this to get
\begin{align}3x^2&=y^2\\
y&=\pm\sqrt{3x^2}\\
&=\pm\sqrt{3}x\end{align}
If we assume $x$ is a whole number, then $y$ can only be a whole number if $x=y=0$
Equally:
\begin{align}3x^2&=y^2\\
x^2&=\frac {y^2}{3}\\
x&=\pm\sqrt{\frac{y^2}{3}}\\
&=\pm\frac{y}{\sqrt{3}}\end{align}
Again, if we assume $y$ is a whole number, then $x$ can only be a whole number if $x=y=0$
Therefore the only solution for $x$ and $y$ whole numbers is $x=y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving of exponential equation If $3^a=21^b$ and $7^c=21^b$
prove that $b=\frac{ac}{(a+c)}$
Can someone please help me prove this ?
Already tried:
$3^a=3^b \times 7^b$ and $7^c=3^b \times 7^b$
Also, $3^a=7^b$.
| from the first, You have
$$3=(21)^{\frac {b}{a}} $$
and from the second,
$$7=(21 )^{\frac {b}{c} }$$
thus by product,
$$21=(21)^{\frac {b}{a}+\frac {b}{c}} $$
hence
$$1=\frac {b}{a}+\frac {b}{c} $$
and
$$\frac {1}{b}=\frac {1}{a}+\frac {1}{c} $$
$$\implies b=\frac {ac}{a+c} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Completing the square but in different situation
Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$
My attempt,
I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$
But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x^2+4x+(\frac{4}{2})^2=45+(\frac{4}{2})^2$. But in this question is different. Could someone give me some hints for it? Thanks in advance.
| Well, we have:
$$x^2+4\cdot\left(\frac{x}{x-2}\right)^2=45\tag1$$
Bring together using a common denominator:
$$\frac{x^2\cdot\left(x^2-4x+8\right)}{\left(x-2\right)^2}=45\tag2$$
Multiply both sides by $\left(x-2\right)^2$:
$$x^2\cdot\left(x^2-4x+8\right)=45\cdot\left(x-2\right)^2\tag3$$
Expand out terms of the right hand side:
$$x^2\cdot\left(x^2-4x+8\right)=45x^2-180x+180\tag4$$
Subtract $45x^2-180x+180$ from both sides:
$$x^2\cdot\left(x^2-4x+8\right)-45x^2+180x-180=0\tag5$$
The left hand side factors into a product with three terms:
$$\left(x-6\right)\cdot\left(x-3\right)\cdot\left(x^2+5x-10\right)=0\tag6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $c+d$ when $a,b,c,d\in\mathbb N$, $ab=2(c+d)$, $cd=2(a+b)$ Question:
\begin{align}
&a,b,c,d\in\mathbb N\\
&ab=2(c+d)\\
&cd=2(a+b)\\
&a+b\ge c+d\\
&\text{(Four numbers don't need to be all different.)}\\\\
&\text{Find all possible values for }c+d
\end{align}
What I tried was
\begin{align}
&x^2-(a+b)x+2(c+d)=0\\
&x^2-(c+d)x+2(a+b)=0\\
&(a+b)^2\ge8(c+d)\\
&(c+d)^2\ge8(a+b)\\
&\rightarrow a+b\ge8,\space c+d\ge8\\
\end{align}
which didn't help.
| Hint
$$0=ab+cd-2(c+d)-2(a+b)\\
8=(a-2)(b-2)+(c-2)(d-2)\\
$$
This leads to only few possibilities to check where all numbers are at least $2$.
Hint 2: If $a=1$ (or any other of the numbers) you have
$$b=2(c+d)\\
cd=2(b+1)$$
hence
$$cd=2(b+1)=2(2(c+d)+1)=4c+4d+2\\
(c-4)(d-4)=18$$
which is easy to solve by factoriations.
The case where one of the numbers is 0 is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Wieferich's criterion for Fermat's Last Theorem I have found the following way to prove some(Wieferich's) criterion for Fermat's Last Theorem and am wondering what would be wrong. My point of doubt is calculation of the Fermat-quotients of $y,z$ being $-1$, since I found these rules on Wikipedia.
Also, should I split this in parts? I can imagine people don't feel like going through too much text.
Anyway, have fun!
Theorem:
Let:
$ \quad \quad \quad \quad p$ be an odd prime,
$ \quad \quad \quad \quad \gcd(x,y,z) = 1$,
$ \quad \quad \quad \quad xyz \not \equiv 0 \pmod p$
If:
$ \quad \quad \quad \quad x^p = y^p + z^p$,
then $p$ is Wieferich-prime.
Proof:
Consider the following congruence:
$ \quad \quad \quad \quad (x^n - y^n)/(x - y) \equiv nx^{n - 1} \pmod {x - y}$
which we can prove by induction on $n$ and in which we divide first.
Let $n = p$.
Since:
$ \quad \quad \quad \quad \gcd(x - y,(x^p - y^p)/(x - y))$
$ \quad \quad \quad \quad = \gcd(x - y,px^{p - 1})$
$ \quad \quad \quad \quad = \gcd(x - y,p)$
$ \quad \quad \quad \quad = \gcd(x - z,p)$
$ \quad \quad \quad \quad = 1$,
it follows that:
$ \quad \quad \quad \quad x - y = r^p$,
$ \quad \quad \quad \quad (x^p - y^p)/(x - y) = s^p$,
$ \quad \quad \quad \quad x - z = t^p$,
$ \quad \quad \quad \quad (x^p - z^p)/(x - z) = u^p$,
$ \quad \quad \quad \quad rs = z$,
$ \quad \quad \quad \quad tu = y$,
for some $r,s,t,u$ with $\gcd(r,s) = \gcd(t,u) = 1$.
The following also holds for $x - z,t,u$:
$ \quad \quad \quad \quad s \equiv 1 \pmod p \implies s^p \equiv 1 \pmod {p^2}$
Now let:
$ \quad \quad \quad \quad s^p = px^{p - 1} \pmod {x - y}$
$ \quad \quad \quad \quad \implies s^p = px^{p - 1} + ar^p \equiv 1 \pmod {p^2}$, for some $a$
$ \quad \quad \quad \quad \implies s \equiv ar \equiv 1 \pmod p \implies s^p \equiv (ar)^p \equiv 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies ar^p \equiv 1/a^{p - 1} \pmod {p^2}$
$ \quad \quad \quad \quad \implies s^p = px^{p - 1} + ar^p \equiv px^{p - 1} + 1/a^{p - 1} \equiv 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies px^{p - 1} \equiv 1 - 1/a^{p - 1} \pmod {p^2}$
$ \quad \quad \quad \quad \implies p(ax)^{p - 1} \equiv a^{p - 1} - 1 \pmod {p^2}$
$ \quad \quad \quad \quad \implies q_p(a) \equiv 1 \pmod p$,
where $q_p(a)$ denotes the Fermat-quotient for $a$ modulo $p$.
So it follows that:
$ \quad \quad \quad \quad q_p(r) \equiv q_p(1/a) \equiv -q_p(a) \equiv -1 \pmod p$
Because of $q_p(s) \equiv 0 \pmod p$:
$ \quad \quad \quad \quad q_p(z) \equiv q_p(rs) \equiv q_p(r) + q_p(s) \equiv -1 + 0 \equiv -1 \pmod p$
Since the same holds for $x - z,t,u$, we now have:
$ \quad \quad \quad \quad q_p(y) \equiv q_p(z) \equiv -1 \pmod p$
From which it follows that:
$ \quad \quad \quad \quad y^{p - 1} \equiv 1 - p \pmod {p^2}$
$ \quad \quad \quad \quad z^{p - 1} \equiv 1 - p \pmod {p^2}$
$ \quad \quad \quad \quad \implies y^p \equiv y(1 - p) \pmod { p^2 }$
$ \quad \quad \quad \quad \implies z^p \equiv z(1 - p) \pmod { p^2 }$
We also note:
$ \quad \quad \quad \quad y^p \equiv (tu)^p \equiv t^p \equiv x - z \pmod {p^2}$
$ \quad \quad \quad \quad z^p \equiv (rs)^p \equiv r^p \equiv x - y \pmod {p^2}$
So we can set:
$ \quad \quad \quad \quad y(1 - p) \equiv x - z \pmod {p^2}$
$ \quad \quad \quad \quad z(1 - p) \equiv x - y \pmod {p^2}$
$ \quad \quad \quad \quad \implies (x - z)/y \equiv (x - y)/z \implies z(x - z) \equiv y(x - y) \pmod {p^2}$
$ \quad \quad \quad \quad \implies y^2 - z^2 \equiv (y + z)(y - z) \equiv x(y - z) \pmod {p^2}$
So either:
$ \quad \quad \quad \quad \implies x \equiv y + z \pmod {p^2}$
or:
$ \quad \quad \quad \quad \implies p | y - z$
Suppose $x \equiv y + z \pmod {p^2}$:
$ \quad \quad \quad \quad \implies (x - y)^p \equiv r^{p^2} \equiv r^p \equiv x - y \implies z^p \equiv z \pmod {p^2}$, contradicting:
$ \quad \quad \quad \quad z^p \equiv z(1 - p) \pmod { p^2 }$
So now we know $y \equiv z \pmod {p}$
But then:
$ \quad \quad \quad \quad y^p \equiv z^p \pmod {p^2}$
$ \quad \quad \quad \quad \implies x^p \equiv y^p + z^p \equiv 2z^p \pmod {p^2}$
Also:
$ \quad \quad \quad \quad x \equiv y + z \implies x \equiv z + z \equiv 2z \pmod p$
$ \quad \quad \quad \quad \implies x^p \equiv (2z)^p \pmod {p^2}$
We conclude:
$ \quad \quad \quad \quad x^p \equiv (2z)^p \equiv 2z^p \pmod {p^2}$
$ \quad \quad \quad \quad \implies (2z)^p - 2z^p \equiv 0 \pmod {p^2}$
$ \quad \quad \quad \quad \implies z^p(2^p - 2) \equiv 0 \pmod {p^2}$,
from which we can see $p$ must be a Wieferich-prime.
| A mistake they made is assuming $q_p(s) \equiv q_p(u) \equiv 0 \pmod p$, since $s^{p - 1} \equiv 1/s \pmod {p^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\
&= \cos^{-1}(0)=90° \\
\alpha_2 &= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\
&= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\
\alpha_3 &= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\
&= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5°
\end {aligned}$$
As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts?
| $180°-122.5°=57.5°$, and the angles sum to be $180°$, nothing wrong here.
The angles you find has nothing to do with whether there is such a triangle with the three points as vertices. The triangle is already formed by the three points before you measure the angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ and the straight lines $OP,OQ$ make angles $\alpha,\beta$ with the $x$-axis where $O$ is the origin then $\frac{\tan\alpha}{\tan\beta}=$
$(A)-1$
$(B)-2$
$(C)2$
$(D)\sqrt2$
Let $P$ be $(x_1,y_1)$ and $Q$ be $(x_2,y_2)$.
Equation of tangent is $\frac{y-y_1}{x-x_1}=\frac{3x_1^2}{2y_1}$
It passes through $(x_2,y_2)$,so $\frac{y_2-y_1}{x_2-x_1}=\frac{3x_1^2}{2y_1}$
I need to find out $\frac{\tan\alpha}{\tan\beta}=\frac{\frac{y_1}{x_1}}{\frac{y_2}{x_2}}=\frac{y_1x_2}{x_1y_2}$
I am stuck here.
| We have
\begin{align}
\frac{y_2-y_1}{x_2-x_1}&=\frac{3x_1^2}{2y_1}\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3x_1^3\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3y_1^2\\
2y_1y_2+y_1^2&=3x_1^2x_2\\
y_1^3(2y_2+y_1)^3&=27x_1^6x_2^3\\
y_1^3(2y_2+y_1)^3&=27y_1^4y_2^2\\
(2y_2+y_1)^3&=27y_1y_2^2\\
8y_2^3-15y_2^2y_1+6y_2y_1^2+y_1^3&=0\\
(y_2-y_1)^2(8y_2+y_1)&=0\\
8y_2&=-y_1\\
64y_2^2&=y_1^2\\
64x_2^3&=x_1^3\\
4x_2&=x_1
\end{align}
Therefore,
\begin{align}
\frac{\tan\alpha}{\tan\beta}&=\frac{y_1x_2}{x_1y_2}\\
&=\frac{-8y_2x_2}{4x_2y_2}\\
&=-2
\end{align}
This method works as when we solve the equation of the tangent with the curve, the $y$-coordinate (or $x$-coordinate if we rewrite it as an equation in $x$) of $P$ is a double root of the cubic equation. It is easy to find the third root, which gives $Q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Witty functional equation
Let
$$f(x) = 2/(4^x + 2)$$
for real numbers $x$. Evaluate
$$f(1/2001) + f(2/2001) + f(3/2001) + \cdots + f(2001/2001)$$
Any idea?
| We have $\displaystyle f(1-x)=\frac{2}{4^{1-x}+2}=\frac{2\cdot 4^x}{4+2\cdot 4^x}=\frac{4^x}{2+ 4^x}$
$\displaystyle f(x)+f(1-x)=\frac{2}{4^x+2}+\frac{4^x}{2+ 4^x}=1$
The required sum is
$$\frac{2000}{2}\times 1+f(1)=1000+\frac{2}{4+2}=\frac{3001}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 1,
"answer_id": 0
} |
Give me some hints in calculation this limit. $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ \tan (x^{2}-y^{2}) - y^{6}}{\sin(x^{6}-y^{6}) - x^{5}y +xy^{5} + \arctan(x^{2}y -xy^{2})}$$
I used a fact that $$\tan \alpha \sim \alpha \\ \arctan \alpha \sim \alpha \\ \sin\alpha \sim \alpha$$
Since now we have $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}$$
Then $$\lim_{x\to2\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}=\\ \\
=\lim_{x\to2\ \\ y\to2}\frac{(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)(x-y)} {(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x^{2}-y^{2})(x^{2}+y^{2}) - xy(x-y)}=\\=\lim_{x\to2\ \\y\to2}\frac{(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)} {(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x+y)(x^{2}+y^{2}) - xy} $$
And what to do next what multipliers to group? Help please.
| $\lim_\limits{x\to2\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}\\
\lim_\limits{x\to2\\\ y\to2}\frac{(x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5) + (x-y)(x+y)}{(x-y)(x^5+x^4y+x^3y2+x^2y^3+xy^5+y^5) - (x-y)(xy)(x^3+x^2y+xy^2+y^3) + xy(x-y)}\\
\lim_\limits{x\to2\\\ y\to2}\frac{(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5) + (x+y)}{(x^5+x^4y+x^3y2+x^2y^3+xy^5+y^5) - (xy)(x^3+x^2y+xy^2+y^3) + xy}\\
$
And now let x = y = 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get:
$$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$
$$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$
$$\sum_{r=-2}^{0}{r^3} = -9 = e$$
$$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$
$$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$
After some algebra we now know $a= -\frac12, b = -1, c = 1, d = \frac32, e = -9$:
$$\therefore \sum^{n}_{r=-2}{r^3}= -\frac12n^4 -n^3 + n^2 +\frac32n -9$$
We define $\sum^{n}_{r=0}{r^3}$:
$$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=-\frac12n^4 -n^3 + n^2 +\frac32n -9 -(-9)$$
$$-\frac12n^4 -n^3 + n^2 +\frac32n = \frac14(-2n^4-4n^3+4n^2 + 6n)$$
$$\frac14n^2\biggl(-2n^2-4n+4+\frac6n\biggl) = \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
$$ \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
My problem:
I got $d = \frac32$ and as far as i can see since it need to end up with $d = 0$ as the smallest term that can be left is a $n^2$ term as $\frac14n^2(n+1)^2$ is multiplied by $n^2$ meaning $e$ has to cancel out (which it does) but i don't know how to get rid of $d$, if you could explain it to me it would be much appreciated
| for $n \ge k$ let the sum $\sum_{r=k}^n r^3 $ be $S_{k}(n)$ so for $n \ge 0$
$$
S_{-2}(n) = S_0(n) -9
$$
since $S_0(0)=0$ this gives $S_{-2}(0)=e=-9$ and
$$
S_0(n)= an^4+bn^3+cn^2+dn
$$
now $S_0(n)-S_0(n-1) = n^3$
so
$$
\begin{align}
a(4n^3-6n^2+4n-&1) + \\
b(3n^2-3n+&1)+\\
c(2n-&1)+\\
&d= n^3
\end{align}
$$
comparing coefficients of $n^3$ gives $a=\frac14$.
the $n^2$ term gives $b=2a$ so $b=\frac12$.
the term in $n$ gives $4a-3b+2c=0$ so $c-\frac14$. and for the constant term: $a-b+c-d=0$ so $d=0$.
altogether this gives
$$
S_0(n)=\frac14 n^4 + \frac12 n^3 + \frac14 n^2 = \bigg(\frac{n(n+1)}2 \bigg)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we get
$$A^2B^4=(AB^2)A=BA^2$$ hence
$$A^2B^4=BA^2 \tag{2}$$ Now post multiplying with $B^4$ and using $(2)$we get
$$A^2B^8=B(A^2B^4)=B^2A^2$$ hence
$$A^2B^8=B^2A^2 \tag{3}$$
Now Pre Multiply with $B^2$ and use $(3)$ we get
$$B^2A^2B^8=B^4A^2$$ $\implies$
$$A^2B^8B^8=B^4A^2$$
$$A^2B^{16}=B^4A^2$$
Now pre multiply with $A^2$ and use $(2)$we get
$$A^4B^{16}=A^2B^4A^2$$ $\implies$
$$B^{16}=BA^4=B$$
is there any other approach to solve this?
| $A^4=I$ implies that $A$ is invertible. Hence $B^2=A^{-1}BA$ . Repeatedly squaring and using the previous step we get $B^{16}=A^{-4}BA^{4}$ which gives $B^{16}=B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 4
} |
Is this correct ${2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}?$ Given this infinite product
$$\lim_{n \to \infty}{2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}\tag1$$
Where $n\ge0$
Experimental on the infinite product calculator show $(1)=\sqrt{2-\sqrt{2}}$
How can we show that?
| In the same spirit as Meet Taraviya's answer, write
$$\prod_{n=0}^{\infty}\frac{4n+2}{4n+2+(-1)^n}=\prod_{k=0}^{\infty}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}$$ Considering the partial product
$$P_m=\prod_{k=0}^{m}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}=\sqrt{\pi }\frac{ \sec \left(\frac{\pi }{8}\right)\, \Gamma \left(2
m+\frac{5}{2}\right)}{4^{m+1}\,\Gamma \left(m+\frac{11}{8}\right)\, \Gamma
\left(m+\frac{13}{8}\right)}$$ Now, using Stirling approximation for large values of $m$
$$\log \left(\frac{\Gamma \left(2 m+\frac{5}{2}\right)}{\Gamma
\left(m+\frac{11}{8}\right) \Gamma \left(m+\frac{13}{8}\right)}\right)=2 m \log (2)+\log \left(2 \sqrt{\frac{2}{\pi }}\right)+\frac{3}{64
m}+O\left(\frac{1}{m^2}\right)$$ Truncating to $O\left(\frac{1}{m}\right)$ we then have $$\frac{\Gamma \left(2 m+\frac{5}{2}\right)}{\Gamma \left(m+\frac{11}{8}\right)
\Gamma \left(m+\frac{13}{8}\right)}\approx 4^m\times 2 \sqrt{\frac{2}{\pi }}$$ making $$P_\infty=\frac{\sec \left(\frac{\pi }{8}\right)}{\sqrt{2}}=2 \sin \left(\frac{\pi }{8}\right)=\sqrt{2-\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Surface area of an ellipse I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!
Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid.
$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$
| $\frac {x^2}{9} + \frac {y^2}{4} = 1\\
\frac {d}{dx}(\frac {x^2}{9} + \frac {y^2}{4} = 1)\\
\frac {2x}{9} + \frac {2y}{4}\frac {dy}{dx} = 0\\
\frac {dy}{dx} = - \frac {4 x}{9 y}\\
2\pi \int_{-3}^{3} y\sqrt {1+(\frac {dy}{dx})^2}\ dx\\
2\pi \int_{-3}^{3} \sqrt {y^2+ \frac {16}{81}x^2}\ dx$
Note you have the wrong limits of integration. Now you can substitute $y^2 = 4 - \frac 49 x^2$
$2\pi \int_{-3}^{3} \sqrt {4 - \frac {20}{81}x^2}\ dx\\
2\pi \int_{-3}^{3} \frac {2}{9}\sqrt {81 - 5 x^2}\ dx$
Which gets you back where you were. Now you need to make a trig substitution
$x = \frac 9{\sqrt 5} \sin \theta$
$
2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {2}{9}\sqrt {81 - 81\sin^2\theta} (\frac 9{\sqrt 5} \cos \theta\ d\theta)\\
2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {18}{\sqrt 5} \cos^2\theta\ d\theta\\
2\pi \frac {9}{\sqrt 5}(\theta + \sin\theta\cos\theta)|_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3}\\
\frac {36\pi}{\sqrt 5}\arcsin \frac{\sqrt 5}3 + 8\pi
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Diagonalizing symmetric real bilinear form I am given the following symmetric matrix:
$$
A=\begin{pmatrix}
1 & 2 & 0 & 1\\
2 & 0 & 3 & 0\\
0 & 3 & -1 & 1\\
1 & 0 & 1 & 4\\
\end{pmatrix}\in M_4(\Bbb R)
$$
Let $f\in Bil(V), f(u,v)=u^tAv.$
I want to find a base $B \subset \Bbb R^4$ such that the matrix representing $f $ in respect to $B$ is diagonal.
I took $
v_1=\begin{pmatrix}
1 \\
0 \\
0 \\
0
\end{pmatrix}
$ and found a vector space $V_2=\{v\in\Bbb R^4$| $f(v,v_1)=0\}$, and got $V_2 =sp\{ \begin{pmatrix}
-2 \\
1 \\
0 \\
0
\end{pmatrix},\begin{pmatrix}
0 \\
0 \\
1 \\
0
\end{pmatrix},\begin{pmatrix}
-1 \\
0 \\
0 \\
1
\end{pmatrix}\} =sp\{v_2,v_3,v_4\}$
Now, the matrix in repect to $B=\{v_1,v_2,v_3,v_4\}$ looks like this:$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -4 & 3 & -2\\
0 & 3 & -1 & 1\\
0 & -2 & 1 & 3\\
\end{pmatrix}
$$
I want to continue inductively, but I'm not sure how to procceed.
| The method of repeated completing squares also leads, by nature, to rational entries in this case. However, that is not the end of the story, as the given form is $SL_4 \mathbb Z$ equivalent to a diagonal form. In brief,
$$ (w + 2x+z)^2 + (x-2y+z)^2 - (9x-9y+13z)^2 + 19 (2x-2y+3z)^2 = w^2 - y^2 + 4z^2 + 4wx +6xy +2wz +2yz $$
$$
A =
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
2 & 0 & 3 & 0 \\
0 & 3 & -1 & 1 \\
1 & 0 & 1 & 4
\end{array}
\right)
$$
$$
Q =
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
0 & 1 & -2 & 1 \\
0 & 9 & -9 & 13 \\
0 & 2 & -2 & 3
\end{array}
\right)
$$
and
$$
D =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 19
\end{array}
\right)
$$
The one-line formula above is just
$$ Q^T D Q = A $$
As far as the order the question was asked, we take
$$
P = Q^{-1} =
\left(
\begin{array}{rrrr}
1 & 2 & -6 & 25 \\
0 & -1 & 4 & -17 \\
0 & -1 & 1 & -4 \\
0 & 0 & -2 & 9
\end{array}
\right)
$$
to get
$$ P^T A P = D. $$ Indeed,
$$
P^T =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & -1 & -1 & 0 \\
-6 & 4 & 1 & -2 \\
25 & -17 & -4 & 9
\end{array}
\right)
$$ and
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & -1 & -1 & 0 \\
-6 & 4 & 1 & -2 \\
25 & -17 & -4 & 9
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
2 & 0 & 3 & 0 \\
0 & 3 & -1 & 1 \\
1 & 0 & 1 & 4
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & -6 & 25 \\
0 & -1 & 4 & -17 \\
0 & -1 & 1 & -4 \\
0 & 0 & -2 & 9
\end{array}
\right) =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 19
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
| You know that $1-\cos x=2\sin^2\dfrac{x}2$ and $1+\cos x=2\cos^2\dfrac{x}{2}$, so $$\frac{1-\cos x}{1+\cos x}=\frac{2}{2}\left(\frac{\sin\dfrac{x}2}{\cos\dfrac{x}2}\right)^2=\tan^2\frac{x}2.$$
If you couldn't understand, please comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Note that :suppose $a=z+\frac 1z $so
$$z+\frac 1z=a\\z^2+\frac{1}{z^2}=(z+\frac 1z)^2-2z\frac1z=a^2-2\\
z^4+\frac{1}{z^4}=(z^2+\frac{1}{z^2})^2-2=(a^2-2)^2-2\\
z^3+\frac{1}{z^3}=(z+\frac 1z)^3-3z\frac1z(z+\frac 1z)=a^3-3a$$
what you have now is $$(a^2-2)^2-2=47 \to\\(a^2-2)^2=49 \\\begin{cases}a^2-2=7 & a^2=9\\\to a^2-2=-7 & a^2=-5 \end{cases}
\\a^2-2= 7 \\\to a^2=9 \\\to a=\pm 3 $$now
you want $$z^3+\frac{1}{z^3}=a^3-3a=\\(\pm 3)^3-3(\pm 3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!
| Let $a=\tan(\frac{\pi}{5})$ and $b= \tan(\frac{2 \pi}{5})$. Also let $s=a^2+b^2$ and $p=(a b)^2$. We have
\begin{eqnarray*}
a^2+b^2 &=&(ab)^2+5 \\
s&=&p+5.
\end{eqnarray*}
Now $a(1-b^2)=-2b$ and $b(1-a^2)=2a$ square these equations and add them together
\begin{eqnarray*}
3(a^2+b^2)+4(ab)^2&=&(ab)^2(a^2+b^2) \\
3s+4p&=&sp.
\end{eqnarray*}
Now eliminate $p$ and we have the quadratic $s^2-12s-20=0$. This has roots $2$ and $\color{red}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
find matrix $A$ in other basis Original matrix is:
$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$
original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$
basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$
Where:
$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} \\f_3=\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3}$$
Therefore:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1f_1+y_2f_2+y_3f_3$$
Expanding that we get:
$$x_1\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3}=y_1(2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3})+y_2(3\mathbf{e_1}+4\mathbf{e_2}+\mathbf{e_3} )+y_3(\mathbf{e_1}+2\mathbf{e_2}+2\mathbf{e_3})$$
How should I group things at right side around unit vectors? for example for first coordinate $2\mathbf{e_1} \, !=\, 3\mathbf{e_1}$, how could I group coordinates?
${\Large \textbf {Solution:}}$
Opening brackets:
$$x\mathbf{e_1}+x_2\mathbf{e_2}+x_3\mathbf{e_3} = 2y_1\mathbf{e_1}+3y_1\mathbf{e_2}+y_1\mathbf{e_3}+3y_2\mathbf{e_1}+4y_2\mathbf{e_2}+y_2\mathbf{e_3}+y_3\mathbf{e_1}+2y_3\mathbf{e_2}+2y_3\mathbf{e_3}$$
Sorting by unit vectors:
$$x_1\mathbf{e_2}+x_2\mathbf{e_3}+x_3\mathbf{e_3} = (2y_1+3y_2+y_3)\mathbf{e_1}+(3y_1+4y_2+2y_3)\mathbf{e_2}+(y_1+y_2+2y_3)\mathbf{e_3} \tag{1}$$
Based on fotmula:
$$A'=T^{-1}AT$$
Matrix $T$ is based on right side of equation $(1)$, and we have:
$$T=\begin{pmatrix} 2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$
The rest is more less technical.
$$T^{-1}=\begin{pmatrix} -6&5&-2\\4&-3&1\\1&-1&1 \end{pmatrix}$$
| Call the first and second basis $E$ and $F$ respectively. You are given $$A_E=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$ and the change of coordinates matrix that changes $F$ coordinates to $E$ coordinates is $$Q=\begin{pmatrix}2&3&1\\3&4&2\\1&1&2 \end{pmatrix}$$ Then $$A_F=Q^{-1}A_EQ.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Continued fraction in 8th root------ any simpler approach? It seems to be a problem from the Putnam Exam. The problem asked to find the exact value of
$$x=\sqrt[8]{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}}$$
And express as $\dfrac{a+b\sqrt{c}}{d}$, in terms of some integers $a,b,c,d$.
My approach
I've tried to treat it as normal continued fractions. It is assumed by the question that it is positive and real (per the question, $x\in\mathbb Q(\sqrt c)$ for some integer $c$), it's straight-forward to have:
$$
x=\sqrt[8]{2207-\frac1{x^8}}
$$$$
x^{16}-2207x^8+1=0
$$$$
x^8=\frac{2207\pm\sqrt{2207^2-4}}{2}
$$
Let $x^4=\sqrt \alpha\pm\sqrt \beta$, as
$$\alpha+\beta=\frac{2207}{2}$$
$$4\alpha\beta=\sqrt{\frac{2207^2-4}{4}}$$
Solve and reject inappropriate (as $x^4$ is positive by our assumption) solution to get
$$x^4=\sqrt{\frac{2207}{4}+\frac12}\pm\sqrt{\frac{2207}{4}-\frac12}$$
$$=\frac{47}{2}\pm\sqrt{\frac{2205}{4}}$$
And let $x^2=\sqrt{\gamma}\pm\sqrt{\delta}$, using the same approach to get
$$x^2=\frac72\pm\sqrt{\frac{45}{4}}$$
And hence
$$x=\frac{3\pm\sqrt 5}{2}$$
But as
$$x=\frac{3-\sqrt 5}{2}\lt\frac12\lt 1$$
When we take the fraction to the second evolution, a fallacy occurred. Thus,
$$x=\frac{3+\sqrt 5}{2}$$
I'm wondering whether my approach is valid. Also, this method seems to be a bit tedious, is there any easier one?
Thanks in advance.
| Let $x=a- \frac{1}{x}$ ; this satisfies $x^2-ax+1=0$ ... & $x^2$ satisfies
\begin{eqnarray*}
(x^2+1)^2=(ax)^2 \\
(\color{blue}{x^2})^2-(a^2-2)\color{blue}{x^2}+1=0
\end{eqnarray*}
So to "square root a continued fraction" we need to solve $a^2-2=b$ ... eighth root so we need to do this three times
\begin{eqnarray*}
a^2-2 &=&2207 \; \; \; &a&=&47 \\
b^2-2 &=&47 \; \; \; &b&=&7 \\
c^2-2 &=&7 \; \; \; &c&=&3 \\
\end{eqnarray*}
So we have $\color{red}{x=\frac{3 +\sqrt{5}}{2}}$. (Justify why the positive root has been chosen ... ?)
EDIT : There is often an ambiguity given by exactly how we define the convergents of a continued fraction ... see Continued fraction fallacy: $1=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
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If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!
| A non power/taylor series method using Leibniz' generalize product rule:
$$ (fg)^{n}(x) = \sum_{r = 0}^{n} \binom{n}{r} f^{n-r}(x) g^{r}(x)$$
Let $f(x) = \dfrac{1}{x^2 + x + 1}$ and $g(x) = x^2 + x +1 $
Then $f(x) g(x) = 1$. We apply the rule to $fg$.
If $a_k = f^{k}(0)$ the we get the recurrence relation
$$a_k + ka_{k-1} + k(k-1)a_{k-2} = 0$$
With $a_{0} = 1$ and $a_1 = -1$
Now let $b_k = \dfrac{a_k}{k!}$, then we get
$$b_k + b_{k-1} + b_{k-2} = 0$$
with $b_{0} = 1$ and $b_{1} = -1$.
Thus we get that the $b_i$ are
$$1, -1, 0, 1, -1, 0, 1, \dots$$
We see that $b_{3k} = 1$, $b_{3k+1} = -1$ and $b_{3k+2} = 0$.
Thus $$a_{36} = b_{36}\cdot 36! = 36!$$
| {
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Conjectured primality test II This question is closely related to my previous question .
Can you provide a proof or a counterexample for the following claim :
Let $n$ be a natural number , $n>1$ and $n \not\in \{4,8,9\}$ . Then $n$ is prime if and only if
$\displaystyle\sum_{k=1}^{n}\left(2^k+1\right)^{n-1} \equiv n \pmod{2^n-1}$
You can run this test here .
I was searching for a counterexample using the following two PARI/GP programs :
CE1(n1,n2)=
{
forcomposite(n=n1,n2,
s=sum(k=1,n,lift(Mod(2^k+1,2^n-1)^(n-1)));
if((Mod(s,2^n-1)==n),print("n="n)))
}
CE2(n1,n2)=
{
forprime(n=n1,n2,
s=sum(k=1,n,lift(Mod(2^k+1,2^n-1)^(n-1)));
if(!(Mod(s,2^n-1)==n),print("n="n)))
}
I've tested this claim up to 10000 and there were no counterexamples .
Remark
More generally we can formulate the following criterion :
Let $b$ and $n$ be a natural numbers , $b\geq 2$ , $n>1$ and $n \not\in \{4,8,9\}$ . Then $n$ is prime if and only if
$\displaystyle\sum_{k=1}^{n}\left(b^k+1\right)^{n-1} \equiv n \pmod{\frac{b^n-1}{b-1}}$
| This is a partial answer.
This answer proves that if $n$ is a prime and $b\ge 2$ is an integer, then
$$\displaystyle\sum_{k=1}^{n}\left(b^k+1\right)^{n-1} \equiv n \pmod{\frac{b^n-1}{b-1}}$$
Proof : (This proof is similar to the one in this answer to your previous question.)
For $n=2$, we have
$$\displaystyle\sum_{k=1}^{2}\left(b^k+1\right)\equiv (b+1)+(b^2+1)\equiv (b+1)^2-2(b+1)+2 \equiv 2 \pmod{b+1}$$
In the following, $n$ is an odd prime.
Let $N:=\frac{b^n-1}{b-1}$.
By the binomial theorem,
$$\begin{align}\displaystyle\sum_{k=1}^{n}\left(b^k+1\right)^{n-1}&=\sum_{k=1}^n\sum_{j=0}^{n-1}\binom{n-1}{j}(b^k)^{j}\cdot 1^{n-1-j}\\\\&=\sum_{j=0}^{n-1}\sum_{k=1}^n\binom{n-1}{j}(b^k)^{j}\\\\&=\sum_{k=1}^n\binom{n-1}{0}(b^k)^{0}+\sum_{j=1}^{n-1}\sum_{k=1}^n\binom{n-1}{j}(b^k)^{j}\\\\&=n+\sum_{j=1}^{n-1}\sum_{k=1}^n\binom{n-1}{j}(b^k)^{j}\\\\&=n+\sum_{j=1}^{n-1}\binom{n-1}{j}\sum_{k=1}^n(b^j)^{k}\tag1\end{align}$$
By the way,
$$\sum_{k=1}^n(b^j)^{k}=\frac{(b^{n+1})^j-b^j}{b^j-1}=\frac{(b(b-1)N+b)^j-b^j}{b^j-1}\tag2$$
By the binomial theorem,
$$\begin{align}(2)&=\frac{\left(\displaystyle\sum_{m=0}^{j}\binom{j}{m}(b(b-1)N)^{m}b^{j-m}\right)-b^j}{b^j-1}\\\\&=\frac{\left(\displaystyle\sum_{m=1}^{j}\binom{j}{m}(b(b-1)N)^{m}b^{j-m}\right)+b^j-b^j}{b^j-1}\\\\&=\frac{b^jN}{b^j-1}\sum_{m=1}^{j}\binom jm(b-1)^mN^{m-1}\tag3\end{align}$$
Note here that $\gcd(b^j,b^{j}-1)=1$.
Now, we use that if $n$ is an odd prime, then either $\gcd(N,b^j-1)=n$ and $b\equiv 1\pmod n$ or $\gcd(N,b^j-1)=1$ holds for $1\le j\le n-1$. (The proof can be seen at the end of this answer.)
Case 1 : When $\gcd(N,b^j-1)=n$ with $b\equiv 1\pmod n$,
$$\begin{align}(3)&=\frac{b^jN}{(b-1)(b^{j-1}+b^{j-2}+\cdots +b+1)}\sum_{m=1}^{j}\binom jm(b-1)^mN^{m-1}\\\\&=\frac{b^jN}{b^{j-1}+b^{j-2}+\cdots +b+1}\sum_{m=1}^{j}\binom jm(b-1)^{m-1}N^{m-1}\end{align}$$
Now, since $b^{j-1}+b^{j-2}+\cdots +b+1\equiv j\not\equiv 0\pmod n$, we have that $$\frac{1}{b^j-1}\sum_{m=1}^{j}\binom jm(b-1)^mN^{m-1}=\frac{1}{b^{j-1}+b^{j-2}+\cdots +b+1}\sum_{m=1}^{j}\binom jm(b-1)^{m-1}N^{m-1}$$
is an integer.
Case 2 : When $\gcd(N,b^j-1)=1$, we have that $$\frac{1}{b^j-1}\sum_{m=1}^{j}\binom jm(b-1)^mN^{m-1}$$ is an integer.
So, in either case, we have
$$(3)\equiv 0\pmod N\tag4$$
Therefore, from $(1)(2)(3)(4)$, we have
$$\begin{align}(1)&\equiv n+\sum_{j=1}^{n-1}\binom{n-1}{j}\cdot 0\qquad\pmod N\\\\&\equiv n\qquad \pmod N\qquad \blacksquare\end{align}$$
| {
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Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| $(a+b)^2 + (b+c)^2 + (a+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=2(a^2+b^2+c^2)+2(ab+bc+ac)=2(a^2+b^2+c^2)+2 \cdot 1=2((a+b+c)^2-2(ab+bc+ac))=2 \cdot (2)+2=6$
| {
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Differential equation via Laplace transform $$xy''+(2x+3)y'+(x+3)y=3e^{-x}$$
$$L[xy'']+L[(2x+3)y']+L[(x+3)y]=L[3e^{-x}]$$
$$-\frac{d}{dp}(p^2Y)-\frac{d}{dp}(2pY)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2)-\frac{dY}{dp}(2p)-\frac{dY}{dp}=\frac{3}{p+1}$$
$$-\frac{dY}{dp}(p^2+2p+1)=\frac{3}{p+1}$$
$$-dY=-\frac{3dp}{p^3+3p^2+3p+1}$$
$$\int-dY=\int\frac{3dp}{p^3+3p^2+3p+1}$$
$$-Y=\frac{-3}{2(x+1)^2}+C$$
$$Y=\frac{3}{2(x+1)^2}+C$$
This is the answer which I got, however, at the end of the book it says that the answer is $$Y=xe^{-x}$$
Can someone please tell me if my answer is correct, and if so how I can make it look like $Y=xe^{-x} \text{?}$
| Consider the equation
$$t \, y'' + (2 \, t + 3) \, y' + (t+3) \, y = a \, e^{-t}$$
then by the standard Laplace transform this becomes
\begin{align}
- \partial_{s}(s^2 \, \overline{y}) + y(0) - 2 \, \partial_{s}(s \, \overline{y}) + 3 \, s \, \overline{y} - 3 \, y(0) - \overline{y}' + 3 \overline{y} &= \frac{a}{s+1}
\end{align}
\begin{align}
- (s+1)^2 \, \overline{y}' + (s+1) \, \overline{y} &= \frac{a}{s+1} + 2 \, y(0) \\
\overline{y}' - \frac{1}{s+1} \, \overline{y} &= - \frac{a}{(s+1)^3} - \frac{2 \, y(0)}{(s+1)^2} \\
\frac{d}{ds} \left( \frac{\overline{y}}{s+1} \right) &= \frac{a}{3} \, \frac{d}{ds} \left( \frac{1}{(s+1)^3} \right) + 2 \, y(0) \, \frac{d}{ds} \left( \frac{1}{(s+1)^2} \right) \\
\frac{\overline{y}}{s+1} &= \frac{a}{3} \, \frac{1}{(s+1)^3} + 2 \, y(0) \, \frac{1}{(s+1)^2} + c_{0} \\
\overline{y} &= \frac{a}{3} \, \frac{1}{(s+1)^2} + \frac{2 \, y(0)}{s+1} + c_{0} \, (s+1).
\end{align}
Inversion of the transform leads to
$$y(t) = \frac{a}{3} \, t \, e^{-t} + 2 \, y(0) \, e^{-t} + c_{0} \, (\delta(t) + \delta'(t))$$.
For this case supposing $y(0) = 0$, $a=3$, and since $c_{0}$ is arbitrary, or that other conditions apply, then
$$y(t) = t \, e^{-t}.$$
| {
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} |
Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure
| Let $R>3$ be the circumradius of $ABC$ and $a,b,c$ its side lengths. We have
$$ a=2\sqrt{R^2-(R-2)^2},\qquad b=2\sqrt{R^2-(R-3)^2},\qquad c=2\sqrt{R^2-(R-1)^2} $$
hence
$$ a^2=16R-16,\qquad b^2=24R-36,\qquad c^2 = 8R-4 $$
and we also have
$$ R^2 = \frac{a^2 b^2 c^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} $$
so $R$ is the root of a cubic polynomial, and
$$ R = \color{red}{2+2\cos\frac{\pi}{9}}=4\cos^2\frac{\pi}{18}\approx 3.87938524.$$
It follows that the area of the given triangle equals
$$ \Delta=4\sqrt{6+10 \cos\left(\frac{\pi }{9}\right)+4 \cos\left(\frac{2 \pi }{9}\right)} \approx \color{red}{17.1865547} $$
| {
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Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$
$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$
The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?
| This would be one way of going about it but using a different fourier series:
$$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$
Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just use odd numbers instead with $(2n+1)$.
So we can make a new formula with only odds and we can do this by letting $n =(2n+1)$, and because we don't want the even results we can resolve the numerator to be equal to $2$.
$$f(x)\ =\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\cos\left(n\pi x\ +\frac{\pi x}{2}\right)$$
Then setting $x = 0$ gives:
$$0=\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$
After rearranging:
$$\frac{\pi^{2}}{8}=\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$
| {
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Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an other way, I've got an another result :
$f(x)=\dfrac{x^2}{x^4-1} =\dfrac{x^2-1+1}{(x^2-1)(x^2+1)} =\dfrac{x^2-1}{(x^2-1)(x^2+1)}+\dfrac{1}{(x^2-1)(x^2+1)} = \\\dfrac{1}{x^2+1} +\dfrac{1}{x^4-1}= \dfrac{1}{x^2+1} -\dfrac{1}{1-(x^2)^2}$
Thus $\boxed {F(x)= \arctan{x} -\operatorname{argth}x^2 +C} $
I think that might be wrong, because that doesn't match with the first result if I plot both on geogebra. I noticed as well their domains are different.
My question is why this two different results? Is one of them wrong?
| There is nothing strange in having different expressions for an antiderivative: an antiderivative of $1/x$ over $(0,\infty)$ is $\log x$ as well as $\log(3x)$, because antiderivatives are determined (over an interval) up to an additive constant. However, your second computation is wrong.
An antiderivative of $1/(1-x^2)$ is $\operatorname{artanh}x$ (or, with your notation, $\operatorname{argth}x$), but then it's false that $\operatorname{artanh}(x^2)$ is an antiderivative of $1/(1-x^4)$: indeed
$$
\frac{d}{dx}\operatorname{artanh}(x^2)=
\frac{1}{1-(x^2)^2}\cdot 2x
$$
by the chain rule.
The mistake would be the same as saying that, since an antiderivative of $x$ is $x^2/2$, an antiderivative of $x^4=(x^2)^2$ is $(x^2)^2/2$, which is clearly wrong.
| {
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let $f$ be a differentiable function. Compute $\frac{d}{dx}g(2)$, where $g(x) = \frac{f(2x)}{x}$. let $f$ be a differentiable function and $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$
Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative
$$\dfrac{d}{dx}g(2)=?$$
I know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$
and :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}=a\in\mathbb{R}$$
so :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)+7+f(4)}{x-4}=\dfrac{-3}{2}$$
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}+\dfrac{f(4)+7}{x-4}=\dfrac{-3}{2}$$
now what ?
| You have an extra $4$ in the numerator here:
i know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$
If $g(x) = \dfrac{f(2x)}x$, then
\begin{align*}
\frac d{dx} g(x) &= \frac d{dx} \left(\frac{f(2x)}x\right)\\[0.3cm]
&= \frac{x \frac d{dx} f(2x) - f(2x) \frac d{dx}x}{x^2}\\[0.3cm]
&= \frac{2x f'(2x) - f(2x)}{x^2}\\[0.3cm]
\end{align*}
By $\dfrac d{dx} g(2)$ I think you mean $\dfrac d{dx} g(x)\bigg|_{x=2}$. Or, more compactly, $g'(2)$.
So $g'(2) = \dfrac{4f'(4) - f(4)}4$.
We're given $\displaystyle \lim_{x\to 4} \frac{f(x) + 7}{x-4} = -\frac32$. Since $\lim_{x\to4} (x-4) = 4-4 = 0$, then $\displaystyle\lim_{x\to4} (f(x) + 7)$ must also be zero because this is the only way the limit of the quotient can be a finite number. Also, since $f$ is continuous (because it is differentiable) then we can safely say $\displaystyle\lim_{x\to 4}(f(x) + 7) = f(4) + 7$. Thus, $f(4) + 7 = 0$, and so we have $f(4) = -7$.
It remains to find $f'(4)$. By the definition of the derivative (sometimes called the alternate definition of the derivative), we have:
$$ f'(4) = \lim_{x \to 4} \frac{f(x) - f(4)}{x-4} = \lim_{x \to 4} \frac{f(x) + 7}{x - 4}$$
And we were given that this is $-\dfrac32$.
Now plug all the numbers into where they need to go, and simplify.
| {
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Find the value of this indefinite integral. $$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
$$
where both $a$ and $l$ are constants.
I've tried simplifying :
$$
\int \frac{dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(l^2+a^2+x^2-a^2-x^2)dx}{(a^2+x^2)\sqrt{l^2+a^2+x^2}}
\\
=\frac{1}{l^2}\int \frac{(\sqrt{l^2+a^2+x^2})dx}{(a^2+x^2)}-\frac{1}{l^2}\int \frac{dx}{\sqrt{a^2+x^2+l^2}}
$$
Now, I know how to simplify the second integral but I have no idea how to solve the first one.
Note : I need the value of this integral for a physics problem.
| Take $l^2+a^2 = r^2$ and $x=r\tan (\theta)$ so that the indefinite integral becomes $$\int \frac{r \sec^2(\theta) d\theta}{(r \sec(\theta))(r^2 \tan^2(\theta) + a^2)} = \int \frac{\sec(\theta) d\theta}{r^2 \sec^2(\theta) - l^2} = \int \frac{\cos(\theta) d\theta}{r^2-l^2 \cos^2(\theta)} =$$ $$\int \frac{\cos (\theta) d\theta}{(r^2-l^2)+l^2\sin^2(\theta)} = \int \frac{\cos(\theta) d\theta}{a^2+l^2 \sin^2(\theta)}$$Now take $\sin(\theta)=y$ so that the integral becomes
$$\int \frac{dy}{a^2+l^2y^2}$$Now take $y=\frac{a}{l}t$ and the integral becomes $$\frac al \int \frac{dt}{a^2(1+t^2)}=\frac 1{al}\int \frac{dt}{1+t^2} = \frac{1}{al} \tan^{-1}(t)=\frac{1}{al} \tan^{-1} \left( \frac la \left( \sin(\theta) \right) \right)$$ $$=\frac{1}{al} \tan^{-1} \left( \frac la \left( \frac{x}{\sqrt{x^2+r^2}} \right) \right)=\frac{1}{al} \tan^{-1} \left( \frac la \left( \frac{x}{\sqrt{x^2+a^2+l^2}} \right) \right)$$ as @innerproduct mentioned in his answer.
| {
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How to get the $\phi$ from $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$? $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,
where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.
I want to know how to get to this result.
I'm able to derive $c$ by taking the derivative of the equation, then squaring both and adding them together, and back-substituting the cosine of a double angle.
But how does one get to the expression for $\tan(\phi)$?
| For fun, here's a trigonograph, which leads to a counterpart cosine identity:
$$\begin{align}
a \sin x + b \sin(\theta+x) &= c\sin(\phi+x) \\
a \cos x + b \cos(\theta+x) &= c\cos(\phi+x)
\end{align}$$
where
$$c^2 = a^2 + b^2 - 2 a b \cos(180^\circ-\theta) = a^2 + b^2 + 2 a b \cos \theta \qquad\text{and}\qquad \tan\phi = \frac{b\sin\theta}{a+b\cos\theta}$$
| {
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Find the volume of the region above the cone $ z=\sqrt{x^2+y^2} $ and below the paraboloid $ z=2-x^2-y^2 $ . Find the volume of the region above the Cone $ z=\sqrt{x^2+y^2} $ and below the Paraboloid $ z=2-x^2-y^2 $ .
The options are (i) $\frac{13 \pi}{6}$, (ii) $\frac{7 \pi}{3}$, (iii) $\frac{13 \pi}{3}$, (iv) $4 \pi$
Answers:
The intersections of the above two surface is given by
$\sqrt{x^2+y^2}=2-x^2-y^2$
or, $ z=2-z^2 $,
or, $z=1$, since $z \geq 0$
Let , $x=r \cos \theta$ and $y=r \sin \theta$.
Then the volume is
$ V=\int_{0}^{2 \pi} \int_{r=0}^{1} \int_{z=r}^{2-r^2} r dz dr d \theta = \frac{5 \pi}{6} $ ,
which does not any of the given options . Am I right ? Any help is there ?
Does it matter '' above '' cone and '' below '' paraboloid ?
| Converting to cylindrical coordinates, the paraboloid is given by the equation $z = 2-r^2$ and the cone is given by the equation $z=r$. These curves intersect when $$2-r^2 = r \implies r=1\, \text{ or } r=-2$$ Since $r\geq0$, $r=1$. The volume is computed to be
\begin{align}
V &= \int_0^{2\pi}\int_0^1\int_r^{2-r^2}r\,dz\,dr\,d\theta \\
&=\int_0^{2\pi}\int_0^1 (2r-r^3-r^2)\,dr\,d\theta \\
&=\int_0^{2\pi}\frac{5}{12}\,d\theta \\
&=\frac{5\pi}{6}
\end{align}
It appears there must be an error in the options.
| {
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"url": "https://math.stackexchange.com/questions/2352056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is evident that $-1$ is not a root of the second equation, but how can I find out whether the other two roots are common or not?
| $$x^2+x+1=0$$
Let's just consider$$x=\exp\left(\frac{i\pi}{3}\right).$$
$$1990 \equiv 1989+1 \equiv 1 \mod 3$$
$$200 \equiv 198+2 \equiv 2 \mod 3$$
Hence $$x^{1990}+x^{200}+1 = x+x^2+1=0$$
$\exp\left(-\frac{i\pi}{3} \right)$ being the conjugate must be another solution as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational We have to show that roots of $$abc^2x^2 + (3a^2 c + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational.
This can be possible if the discriminant is a perfect square. SO I tried converting it into perfect square but failed:
$$\text{Discriminant}=(3a^2c+b^2c)^2-4abc^2(2b^2-6a^2-ab)\\
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)$$
I cannot proceed please help! Thanks!
| The product of the roots is
$$
\frac{-6a^2 -ab +2b^2}{abc^2}
= \frac{(b - 2 a) (3 a + 2 b)}{a b c^2}
$$
which immediately suggests what the roots are.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F(a)=F(b)$ implies $2a^2=b^2+ab$, and thus
$F(2)= 4+2a+b $
What Now?
Any help would be appreciated , thank you
| $$a^{ 2 }+{ a }^{ 2 }+b={ b }^{ 2 }+ab+b\\ { a }^{ 2 }-{ b }^{ 2 }=ab-{ a }^{ 2 }\\ \left( a-b \right) \left( a+b \right) =a\left( b-a \right) \\ \left( a-b \right) \left( a+b \right) +a\left( a-b \right) =0\\ \left( a-b \right) \left( 2a+b \right) =0\\ a\neq b,b=-2a\\ $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $n^a+mn^b+1$ divides $n^{a+b}-1$, prove $m=1$ and $b=2a$ Let $n,m,a,b $ be positive integers such that: $n\geq 2$, $a\leq b $ and $n^a+mn^b+1$ divides $n^{a+b}-1$. Prove that $m=1$ and $b=2a $.
We could factor the divisor to proceed by $n^{a+b}-1=(n-1)(1+n+n^2+...+n^{a+b-1})$ but I cannot see another way we could take this on from here.
| Suppose $k(n^a + mn^b + 1) = n^{a+b} - 1$. Since $k$ is the ratio of two positive integers (using $n\ge 2$), it is certainly positive, and also clearly $k < n^{a+b}/(mn^b) = n^a/m$. In particular $0 < k < n^a$.
On the other hand, looking at this equation modulo $n^a$ (using the fact that $a \le b$ and hence $n^a \mid n^b$) we see that $k \equiv -1 \pmod{n^a}$. Combining this with the known range of $k$, this uniquely identifies $k = n^a-1$. It also forces $m=1$ since otherwise $n^a/m$ would be too small.
We therefore have $(n^a - 1)(n^a + n^b + 1) = n^{a+b} - 1$ which easily simplifies to $n^{2a} = n^b$, hence $b = 2a$ (since $n > 1$).
| {
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Find all the natural numbers $n$ such that $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime How many natural numbers $n$ exist for which $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime?
I thought a lot but I don't how to calculate the correct part.
| So it's pretty clear that $n = 3k$ only works for $k=1$ i.e. $n=3$.
For $n = 3k+1$ we have that $n^2 = 9k^2 + 6k + 1$, so $\lfloor \frac{n^2}{3}\rfloor = 3k^2 + 2k$. If $k$ is even then this number is divisible by $2$. If $k$ is odd, then this number is divisible by $k$, which means that we still have a solution when $k=1$ i.e. $n=4$.
For $n=3k-1$ we can apply a similar technique and get $n^2 = 9k^2 - 6k + 1$ so $\lfloor \frac{n^2}{3}\rfloor = 3k^2 - 2k$. In this case $k=1$ gives us $1$ which is not a prime.
| {
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Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies \frac{x^2}{1+x^2}=\frac{y^2}{1+y^2}$$
$$\implies x^2+x^2y^2=y^2+y^2x^2$$
$$\implies x^2=y^2$$
$$\implies \pm x=\pm y$$
So $g(x)$ is not injective
| Let $x \neq 0$
$$f\left(\frac1x\right)=\frac{1/x}{1+1/x^2}=\frac{x}{1+x^2}=f(x)$$
Thus, $f$ is not injective. Your mistake was $x+xy^2=y+yx^2 \implies x=y$. This is certainly not true.
$$g(-2)=\frac{(-2)^2}{1+(-2)^2}=\frac{4}{1+4}=g(2)$$
Hence, $g$ is not injective.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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arctan series multisection by roots of unity I'm trying to multisect the series for $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots$ using the method of roots of unity, as described in the paper linked in one of the comments in How to express a power series in closed form. I want to take every second term of this series. But I think that I'm doing something wrong, because I get some weird results :q
Here's what I do:
First, since I'll be taking every 2nd term, I need square roots of unity:
$\omega = e^{i\frac{2\pi}{2}} = e^{i\pi} = -1 \\
\omega^0 = (-1)^0 = 1 \\
\omega^1 = (-1)^1 = -1 \\
\omega^{-1} = (-1)^{-1} = \frac{1}{-1} = -1$
I assume the step $k=2$ and the initial offset $r = 0$, and the function $f(x) = \arctan(x)$.
Now I use the formula for the closed form of the multisected series:
$$F(x) = \frac{1}{k}\sum_{n=0}^{k-1}\omega^{-nr}f(\omega^nx)$$
After substituting my numbers and the roots of unity to this formula I get:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot0}\arctan(\omega^nx) \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot0}\arctan(\omega^0x) + \omega^{-1\cdot0}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^0\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(-x) \right\} $$
Now since $\arctan(-x) = -\arctan(x)$, I get:
$$F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(x) \right\} = \frac{1}{2}\!\cdot\!0 = 0$$
:-/
I thought that I perhaps should have used $r=1$ as the offset, so I tried this one too:
$$F(x) = \frac{1}{2}\sum_{n=0}^{1}\omega^{-n\cdot1}\arctan(\omega^nx) \\
F(x) = \frac{1}{2}\left\{ \omega^{-0\cdot1}\arctan(\omega^0x) + \omega^{-1\cdot1}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{ \omega^0\arctan(\omega^0x) + \omega^{-1}\arctan(\omega^1x) \right\} \\
F(x) = \frac{1}{2}\left\{1\cdot\arctan(1\!\cdot\!x) + (-1)\!\cdot\!\arctan(-1\!\cdot\!x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) - \arctan(-x) \right\} \\
F(x) = \frac{1}{2}\left\{ \arctan(x) + \arctan(x) \right\} \\
F(x) = \frac{2\!\cdot\!\arctan(x)}{2} = \arctan(x)$$
So again, something seems to be wrong here: how come I obtained the same original function from taking only every second term of its power series?
Where did I make a mistake?
| We have
\begin{eqnarray*}
\tan^{-1}(x)= x- \frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
and also
\begin{eqnarray*}
\tanh^{-1}(x)= x+ \frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\frac{x^9}{9}+\cdots
\end{eqnarray*}
Now just add these to obtain
\begin{eqnarray*}
\frac{\tan^{-1}(x)+\tanh^{-1}(x)}{2} = x+ \frac{x^5}{5}+\frac{x^9}{9}+\frac{x^{13}}{13}+\frac{x^{17}}{17}+\cdots
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \theta}{4}+\frac{ir \sin \theta}{4}+\frac{6 \cos \theta}{r}-\frac{6 i \sin \theta}{r} \right|$$
So
$$\left|\frac{z}{4}+\frac{6}{z}\right|=\sqrt{\frac{r^2}{16}+\frac{36}{r^2}+3 \cos (2\theta)}$$
any clue from here?
| $|\frac {z}{4} + \frac {6}{z}| \ge 0$
I say that there exist a $z$ such that $1\le|z|\le 7$ and $\frac {z}{4} + \frac {6}{z} = 0$
and any such $z$ must minimize the objective.
$z = i 2\sqrt {6} $
To maximize the objective
$|\frac {z}{4} + \frac {6}{z}| \le |\frac {z}{4}| + |\frac {6}{z}|$
if $z$ is real then:
$|\frac {z}{4} + \frac {6}{z}| = |\frac {z}{4}| + |\frac {6}{z}|$
$z = 7$ maximises $|\frac {z}{4}| + |\frac {6}{z}|$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$
Compute $\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}$
As I see it there are two possibilities:
*
*$g(z)=\frac{\log(z+4)}{(z+i)^2}$ is holomorphic on $D_1(i)$, so we have
$\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}=\oint_{|z-i|=1}\frac{g(z)}{(z-i)^2}=2\pi i\frac{d}{dz}\left(g(z)\right)\mid_{z=i}$
*Also we see that $z=i$ is an isolated singularity in $D_1(i)$ with winding number 1, so we have
$\displaystyle\int_{|z-i|=1}\frac{\log(z+4)}{(z^2+1)^2}=2\pi i\text{Res}_i\left(\frac{\log(z+4)}{(z^2+1)^2}\right)$
and since $z=i$ is a pole of order 2 we can obtain the residue via
$\displaystyle \frac{d}{dz}\left((z-i)^2f(z)\right)\mid_{z\to i}=\frac{2\log(z+4)}{(z-i)(z+i)^2}\mid_{z\to i}+\frac{\log(z+4)}{(z+i)^2}\mid _{z\to i}$
Now I see that I must have done something wrong since the first summand is problematic.
| Both forms are indeed equivalent. Simply note that
$$\begin{align}
\frac{d}{dz}\left((z-i)^2f(z)\right)&=2(z-i)f(z)+(z-i)^2f'(z)\\\\
&=2(z-i)\frac{\log(z+4)}{(z^2+1)^2}+(z-i)^2\left(\frac{1}{(z+4)(z^2+1)^2}-\frac{4z\log(z+4)}{(z^2+1)^3}\right)\\\\
&=\frac{1}{(z+4)(z+i)^2}+\frac{2\log(z+4)}{(z+i)^3}\left(\frac{z+i}{z-i}-\frac{2z}{z-i}\right)\\\\
&=\frac{1}{(z+4)(z+i)^2}-\frac{2\log(z+4)}{(z+i)^3}\\\\
&=\frac{d}{dz}\left(\frac{\log(z+4)}{(z+i)^2}\right)
\end{align}$$
as was to be shown!
| {
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
| Yes Possible.
Evaluate
$$\lim_{ x\to \infty} \left( tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
$$L=\lim_{y \to 0}\frac{\left( tan^{-1}\left(\frac{1+y}{1+4y}\right)-\tan^{-1}{1}\right)}{y}$$
using the following formula, we can acheive, $$tan^{-1}A-tan^{-1}B = \frac{A-B}{1+AB}$$ we get
$$L=\lim_{y \to 0} \frac{1}{y} \times \frac{-3y}{(2+5y)}$$
$$L=\lim_{y \to 0} \frac{-3}{(2+5y)}=\frac{-3}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction of $\sum_{k=2}^n (k-1)(k)\binom{n}{k} = n(n-1)2^{n-2}$ I've been struggling with this sum for an while, plugging $n+1$ instead of $n$, knowing that $\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$ and after some manipulation I've found this sum.
$$2\sum_{k=1}^{n} k^2\binom{n}{k}$$
I couldn't see any way I could get out of here and I don't know how to start this proof without the property of the sum of binomial coefficients.
Thanks in advance.
| Applying Pascal's identity gives you
$$ \sum_k k(k - 1) \binom{n + 1}{k} = \sum_k k(k - 1) \binom{n}{k} + \sum_k k(k - 1) \binom{n}{k - 1}. $$
If we call the sum $f(n)$ then this says that
$$ f(n + 1) = f(n) + \text{ a sum that looks very much like } f(n). \tag{1}$$
We can deal with the second sum by changing variables:
$$ \sum_k k(k - 1) \binom{n}{k - 1} = \sum_k (k + 1)k \binom{n}{k} = f(n) + 2\sum_k k \binom{n}{k}. \tag{2}$$
Let $g(n) = \sum_k k \binom{n}{k}$. Then similarly we find that
$$ g(n + 1) = g(n) + \sum_k (k + 1) \binom{n}{k} = 2g(n) + \sum_k \binom{n}{k} = 2g(n) + 2^n. $$
Therefore
\begin{align}
g(n + 1) &= 2g(n) + 2^n \\
&= 2\left[2g(n - 1) + 2^{n - 1} \right] + 2^n \\
&= 4g(n - 1) + 2 \cdot 2^n \\
&= 8g(n - 2) + 3 \cdot 2^n \\
\end{align}
and by induction
$$g(n + 1) = 2^{n + 1}g(0) + (n + 1)2^n = (n + 1)2^n. \tag{3}$$
Substituting $(2)$ and $(3)$ into $(1)$ gives us
$$ f(n + 1) = 2f(n) + 2g(n) = 2f(n) + n2^n. $$
Hence induction gives
$$ f(n + 1) = n2^n + (n - 1)2^n + (n - 2)2^n + \dots + 1 \cdot 2^n = n(n + 1)2^{n - 1}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Wrong solution using Cramer's rule? I have tried to solve the following systems of equations:
$$
\left\{
\begin{array}{rrrcl}
x & + y & + z &=& 1 \\
2x & + 5y & + 2z &=& 4 \\
4x & + 2y & + 3z &=& 5
\end{array}
\right.
$$
The determinant of the matrix of the coefficients is $3$ and the other determinants are $-8$, $-2$, $7$. So the solutions are
$$
x = - \frac{8}{3} \,,
\quad
y = - \frac{2}{3} \,,
\quad
z = \frac{7}{3} \,.
$$
However, these values doesn't satisfy the system. Furthermore, Wolfram Alpha says that the solutions are $x = 8/3, y = 2/3, z = -7/3$. The roots are the same in numerical value, but with opposite sign. What is the error here?
| At least$$\Delta=1\cdot5\cdot3+1\cdot2\cdot2+4\cdot1\cdot2-4\cdot5\cdot1-1\cdot2\cdot2-3\cdot2\cdot1=-3$$
$$\Delta_x=1\cdot5\cdot3+1\cdot2\cdot5+4\cdot1\cdot2-5\cdot5\cdot1-1\cdot2\cdot2-3\cdot4\cdot1=-8$$
$$\Delta_y=1\cdot4\cdot3+1\cdot4\cdot2+5\cdot1\cdot2-4\cdot4\cdot1-1\cdot2\cdot5-3\cdot2\cdot1=-2$$ and
$$\Delta_z=1\cdot5\cdot5+1\cdot2\cdot2+4\cdot1\cdot4-4\cdot5\cdot1-1\cdot2\cdot4-5\cdot2\cdot1=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371900",
"timestamp": "2023-03-29T00:00:00",
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Primality test using Chebyshev and Legendre polynomials Inspired by Theorem 5 in this paper I have formulated the following claim :
Let $n$ be an odd number and $n>1$ . Let $T_n(x)$ be Chebyshev polynomial of the first kind and let $P_n(x)$ be Legendre polynomial , then $n$ is a prime number if and only if the following congruences hold simultaneously
$\bullet \: T_n(3) \equiv 3 \pmod n$
$\bullet \: P_n(3) \equiv 3 \pmod n$
You can run this test here .
I was searching for pseudoprimes using the following PARI/GP program :
CL(lb,ub)=
{
forstep(n=lb,ub,[2],
if(!ispseudoprime(n),
if((Mod(polchebyshev(n,1,3),n)==3),
if((Mod(pollegendre(n,3),n)==3),print(n)))))
}
I have tested this claim up to $1.4 \cdot 10^6$ and there were no counterexamples .
Question : Can you provide a proof or a counterexample for the claim given above ?
| This is a partial answer.
This answer proves that if $n$ is an odd prime, then $P_n(3)\equiv 3\pmod n$.
Using that $\binom nk\equiv 0\pmod n$ for $1\le k\le n-1$, we have
$$\begin{align}P_n(3)&=\frac{1}{2^n}\sum_{k=0}^{n}\binom nk^2(3-1)^{n-k}(3+1)^k\\\\&=\frac{1}{2^n}\sum_{k=0}^{n}\binom nk^2\cdot 2^{n-k}\cdot 2^{2k}\\\\&=\sum_{k=0}^{n}\binom nk^2\cdot 2^k\\\\&\equiv \binom n0^2\cdot 2^0+\binom nn^2\cdot 2^n\quad\pmod n\\\\&\equiv 1+2^n\quad\pmod n\end{align}$$
Now, since $\frac{n^2-1}{4}$ is even when $n$ is odd, we have
$$\begin{align}P_n(3)&\equiv 1+2^n\equiv 1+2\cdot\left(2^{\frac{n-1}{2}}\right)^2\equiv 1+2\cdot \left((-1)^{\frac{n^2-1}{8}}\right)^2\equiv 1+2\cdot (-1)^{\frac{n^2-1}{4}}\\\\&\equiv 3\pmod n\qquad\blacksquare\end{align}$$
According to Theorem 5 in the paper you showed, we can say that if $n$ is an odd prime, then $T_n(3)\equiv 3\pmod n$.
Therefore, we can say that if $n$ is an odd prime, then $T_n(3)\equiv P_n(3)\equiv 3\pmod n$.
| {
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$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Multiply $(2.1)$ through by $3$:
$$\frac{3(y + x)}{xy} = 4 \tag{2.2}$$
Let $y = mx$
Substitute $y = mx$ into $(1)$ and $(2.2)$:
$$x^2 + m^2x^2 = 10 \tag{1.1}$$
$$\frac{3(mx + x)}{mx^2} = 4$$
Factorise:
$$\frac{3x(m + 1)}{x(mx)} = 4$$
$$\frac{3(m + 1)}{(mx)} = 4$$
Cross multiply:
$$3(m+1) = 4mx$$
Collect like terms:
$$4mx - 3m = 3$$
Factorise:
$$m(4x - 3) = 3$$
Divide through by $(4x - 3)$:
$$m = \frac{3}{4x - 3} \tag{3}$$
Substitute $(3)$ into $(1.1)$
$$x^2 + \left(\frac{3}{4x - 3}\right)^2x^2 = 10$$
$$x^2 + \frac{9x^2}{16x^2 - 24x + 9} = 10$$
Multiply through by ${16x^2 - 24x + 9}$:
$$16x^2 - 24x^3 + 9x^2 + 9x^2 = 10(16x^2 - 24x + 9)$$
Divide through by $2$:
$$8x^4 - 12x^3 + 9x^2 = 80x^2 - 120x + 45$$
$$8x^4 - 12x^3 - 71x^2 +120x - 45 = 0$$.
$x = 0$ is not a solution of the above equation, so divide through by $x^2$:
$$8x^2 - 12x - 71 + \frac{120}{x} - \frac{45}{x^2} = 0 \tag{1.2}$$
Let $v = x - \frac{k}{x}$.
$$v^2 = x^2 - 2k + \frac{k^2}{x^2}$$
Rewriting $(1.2)$:
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12x + \frac{120}{x}\right)$$
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12(x - \frac{10}{x}\right) \tag{1.3}$$
Putting $k = 10$, works for $v$, but not for $v^2$.
I don't know any other approach to solving the equation from $(1.3)$.
| help from the graph :
$$x^2+y^2=10 \to \text {Circle}$$
$$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{3}\to y=\dfrac{3x}{4x-3}\to \text{Hemographic
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of the bounded decreasing sequence
Find the limit of the bounded decreasing sequence defined by
$a_1=3 \\a_{n+1}= \frac{1}{4-a_n}+1$
Can anyone teach me how to do this question? Thanks.
The answer is $\frac{5-\sqrt5}{2}$.
| We'll prove that $a_n\leq3$.
Indeed, $a_1\leq3$ and $a_{n+1}=\frac{1}{4-a_n}+1\leq\frac{1}{4-3}+1=2<3$.
In another hand $a_1>\frac{5-\sqrt5}{2}$ and
$$a_{n+1}-\frac{5-\sqrt{5}}{2}=\frac{1}{4-a_n}-\frac{3-\sqrt5}{2}=\frac{1}{4-a_n}-\frac{2}{3+\sqrt5}=$$
$$= \frac{1}{4-a_n}-\frac{1}{4-\frac{5-\sqrt5}{2}}=\frac{a_n-\frac{5-\sqrt5}{2}}{(4-a_n)\left(\frac{3}{2}+\frac{\sqrt5}{2}\right)},$$
which gives $a_n>\frac{5-\sqrt{5}}{2}$ by induction and
$$a_n-\frac{5-\sqrt{5}}{2}<\left(\frac{2}{3+\sqrt5}\right)^1\left(a_{n-1}-\frac{5-\sqrt{5}}{2}\right)<$$
$$<\left(\frac{2}{3+\sqrt5}\right)^2\left(a_{n-2}-\frac{5-\sqrt{5}}{2}\right)<...<\left(\frac{2}{3+\sqrt5}\right)^{n-1}\left(a_1-\frac{5-\sqrt{5}}{2}\right)$$
and since $\frac{2}{3+\sqrt5}<1$ we obtain
$$\lim_{n\rightarrow+\infty}a_n=\frac{5-\sqrt5}{2}$$
by the limit definition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the smallest value of $n$ such that $19 \,|\, 10^n+1$ A riddle I'm working on reduces to this question. I don't have a number theory background and don't really know how to approach this kind of problem.
In fact I'm not sure such an $n$ exists. I feel like it does, but my computer has been chugging on an R script for almost 20 minutes to find one.
Any suggestions for resources/additional reading about evaluating divisibility appreciated!
| The pattern
$\begin{array}{l}
10^3 + 1 = 1001 = 999+2 \\
10^4 + 1 = 10001 = 9999+2\\
10^5 + 1 = 100001 = 99999+2\\
\dots \\
10^9 + 1 = 999999999+2
\end{array}$
Multiples of $19$:
\begin{array}{r|ccccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
19n & 19 & 38 & 57 & 76 & 95 & 114 & 133 & 152 & 171 \\
\end{array}
$$\begin{array}{cccccccccc}
& & & 5 & 2 & 6 & 3 & 1 & 5 & 7 & 9\\
& & - & - & - & - & - & - & - & - & -\\
19 & ) & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\
& & 9 & 5 & & \\
& & - & -\\
& & & 4 & 9\\
& & & 3 & 8 \\
& & & - & -\\
& & & 1 & 1 & 9 \\
& & & 1 & 1 & 4 \\
& & & - & - & -\\
& & & & & 5 & 9 \\
& & & & & 5 & 7 \\
& & & & & & - & - \\
& & & & & & 2 & 9\\
& & & & & & 1 & 9\\
& & & & & & - & - \\
& & & & & & 1 & 0 & 9\\
& & & & & & & 9 & 5\\
& & & & & & & - & - & -\\
& & & & & & & 1 & 4 & 9\\
& & & & & & & 1 & 3 & 3\\
& & & & & & & - & - & -\\
& & & & & & & & 1 & 6 & 9\\
& & & & & & & & 1 & 7 & 1\\
& & & & & & & & - & - & -\\
& & & & & & & & & & -2\\
\end{array}
$$
Hence
$999999999 = 19 \times 52631579 - 2$
$10^9 + 1 = 19 \times 52631579$
$19 \mid 10^9+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| Let $x=at$ and by Partial-Fraction Decomposition we get
\begin{align*}\int_0^a \frac{x^4}{(x^2+a^2)^4}dx&=\frac{1}{a^3}\int_0^1 \frac{t^4}{(t^2+1)^4}dt\\
&=\frac{1}{a^3}\int_0^1\left(\frac{1}{(t^2+1)^2}-\frac{2}{(t^2+1)^3}+\frac{1}{(t^2+1)^4}\right)\,dt
\\&=\frac{1}{16a^3}\left[\arctan(t)
+\frac{t}{t^2+1}-\frac{(14/3)t}{(t^2+1)^2}+\frac{(8/3)t}{(t^2+1)^3}\right]_0^1\\&=\frac{3\pi-4}{192a^3}.
\end{align*}
P.S. Note that for $p\geq 2$,
$$(2p-2)\int\frac{dt}{(t^2+1)^p}=\frac{t}{(t^2+1)^{p-1}}
+(2p-3)\int\frac{dt}{(t^2+1)^{p-1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$
$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$
$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$
$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$
Could some help me to solve it, thanks
| Use $\sin{x}=\frac{2t}{1+t^2}$ and $\cos{x}=\frac{1-t^2}{1+t^2}$, where $t=\tan\frac{x}{2}$.
One of roots is $\sqrt2-1$ and the second is very ugly.
For $\tan\frac{x}{2}=\sqrt2-1$ we obtain $\frac{x}{2}=22.5^{\circ}+180^{\circ}k,$ where $k\in\mathbb Z$,
which gives $x=\frac{\pi}{4}$.
The second root does not give solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$
Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$
My attempts:
Using the given condition and double angle formula,
$\dfrac{\sin2x}{\sin2y} = \dfrac{3}{2}$
Now I am struggling to find the value of $\dfrac{\cos2x}{\cos2y}$
The simplest form I could reach was:
$\dfrac{2\cos^2x-1}{8\cos^2x-1}$
How do I continue from here? Simpler methods to solve the problem are welcome.
PS: The answer is $\dfrac{49}{58}$
| $\sin{y}=3\sin{x}$ and $\cos{y}=2\cos{x}$.
Thus,
$$1=\sin^2y+\cos^2y=9\sin^2x+4\cos^2x=4+5\sin^2x,$$
which is impossible.
If $\sin{y}=\frac{1}{3}\sin{x}$ we obtain:
$$1=\sin^2y+\cos^2y=\frac{1}{9}\sin^2x+4\cos^2x=\frac{1}{9}+\frac{35}{9}\sin^2x.$$
Now, you can get $\sin^2x$, $\sin^2y$ and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximize linear function over disk of radius $2$
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.
$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5), 4/\sqrt(5))$ gives a larger max.
How do I approach this to get this point?
| $$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 \leq 4\end{array}$$
Since the objective function is linear and nonzero, its gradient never vanishes. Thus, the maximum is attained at the boundary of the feasible region, i.e., on the circle of radius $2$ centered at the origin
$$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 = 4\end{array}$$
Using polar coordinates instead, the objective function is
$$2 \cos (\theta) + 4 \sin (\theta) = 2 \sqrt{5} \left( \dfrac{2}{2 \sqrt{5}} \, \cos (\theta) + \dfrac{4}{2 \sqrt{5}} \, \sin (\theta) \right) = 2 \sqrt{5} \, \cos \left( \theta - \arctan (2) \right)$$
Thus, the maximum is $\color{blue}{2 \sqrt{5}}$, which is attained at $\theta = \arctan (2)$, i.e., at
$$x = 2\,\cos \left( \arctan (2) \right) = \dfrac{4}{2 \sqrt{5}} = \color{blue}{\dfrac{2}{\sqrt{5}}}$$
$$y = 2\,\sin \left( \arctan (2) \right) = \dfrac{8}{2 \sqrt{5}} = \color{blue}{\dfrac{4}{\sqrt{5}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Sequence defined recursively: $\sum_{k=1}^n x_k = \frac{1}{\sqrt{x_{n+1}}}$ Let $(x_n)_{n \ge 1}$ defined as follows:
$$x_1 \gt 0, x_1+x_2+\dots+x_n=\frac {1}{\sqrt {x_{n+1}}}.$$ Compute the limit $\lim _ {n \to \infty} n^2x_n^3.$
MY TRY: I thought about using Stolz-Cesaro lemma, but I couldn't get to an appropriate form that leads to an easier limit.
| This answer will use the Stolz-Cesaro lemma.
$$L=\lim_{n \to \infty} n^2x_n^3$$
$$S_n=\sum_{k=1}^n{x_k} \quad\quad x_{n+1}=S_n^{-2}$$
We will try to find a recurrence relation that is easier to manipulate:
$$\frac{x_{n+1}}{x_n}=\frac{S_n^{-2}}{S_{n-1}^{-2}}=\frac{S_{n-1}^{2}}{S_n^{2}}=\left (\frac{S_{n-1}}{S_n}\right)^2$$
$$S_{n-1}=S_n-x_n$$
$$\frac{x_{n+1}}{x_n}=\left(\frac{S_n-x_n}{S_n}\right)^2$$
$$\frac{x_{n+1}}{x_n}=\frac{1}{x_nS_n^2}=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$
$$\left(\frac{S_n-x_n}{S_n}\right)^2=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$
$$\frac{S_n-x_n}{S_n}=\frac{1}{\sqrt{x_n}S_n}$$
$$S_n-x_n=\frac{1}{\sqrt{x_n}}$$
$$S_n=x_n+\frac{1}{\sqrt{x_n}}=x_n+x_n^{-\frac{1}{2}}$$
$$x_{n+1}=\frac{1}{\left(x_n+x_n^{-\frac{1}{2}}\right)^2}$$
$$x_{n+1}=\frac{x_n}{\left(x_n^{\frac{3}{2}}+1\right)^2} \quad n>1$$
Because of the limit we want to evaluate, we make the substitution: $a_n=n^2x_n^3 \quad x_n=a_n^{\frac{1}{3}}n^{-\frac{2}{3}} \quad L=a_{\infty}$
$$a_{n+1}^{\frac{1}{3}}(n+1)^{-\frac{2}{3}}=\frac{a_n^{\frac{1}{3}}n^{-\frac{2}{3}}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^2}$$
$$a_{n+1}(n+1)^{-2}=\frac{a_nn^{-2}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$
$$a_{n+1}=\frac{a_n(n+1)^{2}}{n^2\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$
To get rid of the square root we make the substitution: $a_n=b_n^2 \quad b_n=\sqrt{a_n} \quad L=b_{\infty}^2$
$$b_{n+1}^2=\frac{b_n^2(n+1)^{2}}{n^2\left(\frac {b_n}{n}+1\right)^6}$$
$$b_{n+1}=\frac{b_n(n+1)}{n\left(\frac {b_n}{n}+1\right)^3}$$
$$\frac{b_{n+1}}{n+1}=\frac{b_n}{n\left(\frac {b_n}{n}+1\right)^3}$$
This next substitution is somewhat obvious: $c_n=\frac{b_n}{n} \quad b_n= nc_n \quad L=\lim_{n \to \infty} (nc_n)^2$
$$c_{n+1}=\frac{c_n}{\left(c_n+1\right)^3}$$
Because we want the limit in a usable form for the Stolz-Cesaro lemma, we make the substitution: $c_n=\frac{1}{d_n} \quad d_n=\frac{1}{c_n}$
$$L=\lim_{n \to \infty} \left(\frac{n}{d_n}\right)^2=\left( \lim_{n \to \infty} \frac{n}{d_n}\right)^2$$
$$L=\left( \lim_{n \to \infty} \frac{n+1-n}{d_{n+1}-d_n}\right)^2$$
$$L=\lim_{n \to \infty} \frac{1}{(d_{n+1}-d_n)^2}$$
$$\frac{1}{d_{n+1}}=\frac{1}{d_n\left(\frac{1}{d_n}+1\right)^3}$$
$$d_{n+1}=d_n\left(\frac{1}{d_n}+1\right)^3=\frac{(d_n+1)^3}{d_n^2}$$
$$d_{n+1}=\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}$$
$$d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}$$
$$\lim_{n \to \infty} \left(d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}\right)$$
$$\lim_{n \to \infty} \left(d_{n+1}=d_n+3\right)$$
$$\lim_{n \to \infty} d_{n+1}-d_n=3$$
$$L=\frac{1}{3^2}=\frac{1}{9}$$
All of this is assuming that $d_n$ diverges to $+\infty$. Since $x_1>0$, $d_1>0$ (substitutions considered) is the only case that needs to be analyzed. We will prove that $d_{n+1}>d_n$ for $d_n>0$.
$$\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}>d_n$$
$$d_n^3+3d_n^2+3d_n+1>d_n^3$$
$$3d_n^2+3d_n+1>0$$
This is clearly true for $d_n>0$. Note that this also shows there are no solutions to $d_{n+1}=d_n$ for $d_n>0$. Therefore, we can conclude $L=\frac{1}{9}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving System of quadratic equations If $b²-4ac=0$ ($a \neq 0$, and $a, b, c \in \mathbb {R}$) and $x, y $ satisfy the system $$ax²+(b+3)x+c=3y$$ and $$ay²+(b+3)y+c=3x$$ then the value of $x/y$ is...?
| Assuming $x,y \in \mathbb{R}$, then $x/y = 1$.
Your equations give
$$
ax^2+bx+c = 3(y-x) = -3(x-y) = -(ay^2+by+c),
$$
so
$$ ax^2+bx+c = -(ay^2+by+c). $$
The assumption $b^2-4ac=0$, $a\neq 0$, means that one of the following must be true:
*
*$ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2$,
*$ax^2+bx+c = (\sqrt{a}x-\sqrt{c})^2$,
*$ax^2+bx+c = -(\sqrt{-a}x+\sqrt{-c})^2$, or
*$ax^2+bx+c = -(\sqrt{-a}x-\sqrt{-c})^2$.
For simplicity say we are in the first case. Then
$$
(\sqrt{a}x+\sqrt{c})^2 = -(\sqrt{a}y+\sqrt{c})^2 .
$$
This can only happen if $\sqrt{a}x+\sqrt{c} = \sqrt{a}y+\sqrt{c} = 0$, and then $x=y=-\sqrt{c}/\sqrt{a}$. The other three cases are similar.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality : $(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$ Find the largest integer $k$ such that
$$(a^3+3)(b^3+6)(c^3+12)>k(a+b+c)^3$$
for all positive real numbers $a, b, c$.
My attempt :
By Holder inequality, $(a^3+1+2)(2+b^3+4)(4+8+c^3)\geq(\sqrt[3]{8}a+\sqrt[3]{8}b+\sqrt[3]{8}c)^3$
$\Leftrightarrow(a^3+3)(b^3+6)(c^3+12)\geq8(a+b+c)^3$
There is no equality hold.
so $(a^3+3)(b^3+6)(c^3+12)(a+b+c)^3>8(a+b+c)^3$
| The equality should be for $(a,1,\sqrt[3]2)||(\sqrt[3]2,b,\sqrt[4]4)||(\sqrt[3]4,2,c),$ which is impossible.
Let $a_1+a_2=3$, $b_1+b_2=6$, $c_1+c_2=12$, $b_1c_1=a_1c_2=a_2b_2$, where $a_i\geq0$, $b_i\geq0$ and $c_i\geq0$.
Thus, for the equality occurring after using Holder we need
$$\left(a,\sqrt[3]{a_1},\sqrt[3]{a_2}\right)||\left(\sqrt[3]{b_1},b,\sqrt[3]{b_2}\right)||\left(\sqrt[3]{c_1},\sqrt[3]{c_2},c\right),$$
which is very ugly:
$k_{max}=8.093...$ for $(a,b,c)=(0.904...,1.368...,2.297...)$,
but it says $8$ is a maximum possible integer value, for which our inequality is true
and $(0.904,1.368,2.297)$ is a counterexample for all integer $k>8$.
By the way, there is a counterexample, which is a bit of better: $$(a,b,c)=(1,2,2).$$
In the exam you apply your proof for $k=8$ and for $k\geq9$ you write:
The equality $$(a^3+3)(b^3+6)(c^3+12)\geq k(a+b+c)^3$$
is wrong. $(1,2,2)$ is the counterexample.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rationalize $\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$ I am having trouble rationalizing the denominator of $$\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$$
I tried grouping the denominator as $(2 + 2\sqrt[6]{2}) + 2\sqrt[3]{2}$ and multiplying top and bottom by $(2 + 2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2$ to obtain $$\frac{(2+2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2}{(2+2\sqrt[6]{2})^3+16}$$
However, expanding the denominator produces $24 + 24\sqrt[6]{2}+24\sqrt[3]{2}+8\sqrt{2}$ which doesn't look any better than the original.
So what should I do differently or is there another way to approach this problem? Any advice would be appreciated. Thanks.
| Let $\sqrt[6]2=x\implies\sqrt[3]2=x^2$
We have $$\dfrac1{2+2x+2x^2}=\dfrac{1-x}{2(1-x^3)}=\dfrac{(1-x)(1+x^3)}{2(1-x^6)}$$
| {
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"url": "https://math.stackexchange.com/questions/2383577",
"timestamp": "2023-03-29T00:00:00",
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Transforming $-1+ \cos x +\sin x$ into $-2\sin^2 (x/2) +2\sin(x/2)\cos(x/2)$ This trigonometrical part was included in a complex variable problem. I tried to find how
$$ -1+ \cos (x) +\sin (x)$$
becomes
$$-2\sin^2 (x/2) +2\sin(x/2)\cos(x/2)$$
Can I let $1 = \sin^2 x +\cos^2x$ ?
Any help'll be appreciated.
| The identity $ \sin(2 \theta) = 2 \sin( \theta ) \cos( \theta ) $ give us that
$$
\sin(x) = \sin(2 \cdot x/2) = 2 \sin(x/2) \cos(x/2)
$$
Similary, we use that $ cos(2 \theta) = \cos^2( \theta ) - \sin^2( \theta ) $ to obtain
$$
-1 + \cos(x) = -1 + \cos(2 \cdot x/2) = -1 + ( \cos^2(x/2) - \sin^2(x/2) )
$$
Finally from the identity $ cos^2( \theta ) + \sin^2( \theta ) = 1 $ we deduce that $ -1 + \cos^2( \theta ) = - \sin^2( \theta ) $:
$$
-1 + ( \cos^2(x/2) - \sin^2(x/2) ) = ( -1 + \cos^2(x/2) ) - \sin^2(x/2)) = -\sin^2(x/2) - \sin^2(x/2) = -2 \sin^2(x/2)
$$
Adding the two equlities:
$$
-1 + \cos(x) + \sin(x) = -2 \sin^2(x/2) + 2 \sin(x/2) \cos(x/2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\lim_{x\to 0} \frac{e^x-1-x}{x^2}$ without L'Hôpital's Rule. I want to solve this limit without the use of L'Hôpital's rule:
$$\lim_{x\to 0} \frac{e^x-1-x}{x^2}.$$
| I would Taylor Expand
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
and plug in to find
$$\lim_{x\to 0}\frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x}{x^2}=\lim_{x\to 0}\frac{\sum_{n=2}^\infty \frac{x^n}{n!}}{x^2}=\lim_{x\to 0}\frac{\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots}{x^2}=\lim_{x\to 0} \frac{1}{2}+\frac{1}{6}x+\frac{1}{24}x^2+\cdots=\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\begin{align}
& =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt]
& =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}
Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right)
\end{align}
I can't think of what to do next..
| Write $t=\tan\frac{\theta}{2}$ so $\sin\theta=\frac{2t}{1+t^2},\,\cos\theta=\frac{1-t^2}{1+t^2}$. Hence $$\frac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1+t^2+2t-1+t^2}{1+t^2+2t+1-t^2}=\frac{2t+2t^2}{2+2t}=t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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An inequality with $\cos$ and triangle sides Here is the problem:
Let $ABC$ be a triangle with sides $a, b, c$. Show that $\dfrac{\cos A}{a^3}+\dfrac{\cos B}{b^3}+\dfrac{\cos C}{c^3}\geq\dfrac{3}{2abc}.$
Here's my attempt:
By the cosine formula, we have $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ etc, which the left hand side can be transformed into:
\begin{equation*}
\dfrac{a^2+b^2-c^2}{2abc^3}+\dfrac{a^2+c^2-b^2}{2ab^3c}+\dfrac{b^2+c^2-a^2}{2a^3bc}
\end{equation*}
And then I'm stuck. Can someone help me? Thanks.
| $$2abc\sum_{cyc}\frac {a^2+b^2-c^2}{2abc^3}=$$ $$=\sum_{cyc}(\frac {a^2}{c^2}+\frac {b^2}{c^2}-1)=$$ $$=-3+\sum_{cyc}(\frac {a^2}{b^2}+\frac {b^2}{a^2})=$$ $$=-3+\sum_{cyc}(2+(\frac {a}{b}-\frac {b}{a})^2\;)=$$ $$=+3+\sum_{cyc}(\frac {a}{b}-\frac {b}{a})^2\geq 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Given $m$, find three numbers $a,b,c$ such that $c-b=m(b-a)$ and $a+b,a+c,b+c$ are all squares I've worked with the next problem
*
*Given an integer $m$, find three integers $a,b,c$, with $b\neq a$, such that $c-b=m(b-a)$ and
$$a+b=x^2, \ \ a+c=y^2, \ \ b+c=z^2,$$
where $x,y, z \in \mathbb{Z}$.
I've not been able to give a result that guarantees the existence of solutions of this problem for some $m$ values.
Can you help me, please?
| Let
\begin{eqnarray*}
a&=&2((mp^2+2mpq-q^2)^2+(mp^2+q^2)^2-(-mp^2+2pq+q^2)^2) \\
b&=&2((mp^2+2mpq-q^2)^2-(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
c&=&2(-(mp^2+2mpq-q^2)^2+(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
\end{eqnarray*}
One can easily verify that
\begin{eqnarray*}
a+b&=& 4(mp^2+2mpq-q^2)^2 \\
b+c&=&4(-mp^2+2pq+q^2)^2 \\
c+a&=&4(mp^2+q^2)^2 \\
c-b&=&-16(pm-q)(p+q) pqm \\
b-a&=&-16(pm-q)(p+q) pq.
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/2385958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrt[3]x\left(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\right)$,
Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt[3]{x(x^3+2)^2}$,
LHS :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2}
{\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$
RHS :
$\sqrt[3]{3x(x^3+2)}\leq \sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{9x^2}\leq \sqrt[3]{(x^3+2)^2}$
so $\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\leq 3\sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{x}(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2})\leq 3\sqrt[3]{x(x^3+2)^2}$
Thus,
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
| Denote $u = \frac{x^3+y^3+z^3}{3xyz}$ and $v = \frac{xy+yz+zx}{x^2+y^2+z^2}$.
Then $u\ge 1$ and $0 \le v \le 1$.
We need to prove that $\sqrt[3]{3u} + \sqrt{v} \ge 1 + \sqrt[3]{3}$ or
$\sqrt[3]{3u} - \sqrt[3]{3} \ge 1 - \sqrt{v}$ or
$$\frac{3u - 3}{3^{2/3}(u^{2/3} + u^{1/3} + 1)} \ge \frac{1 - v}{1 + \sqrt{v}}.$$
By using $(w^3+1) - (w^2+w) = (w+1)(w-1)^2\ge 0$ for $w\ge 0$, we have
$u^{2/3} + u^{1/3} \le u + 1$ (simply letting $w = u^{1/3}$). Also, $3^{2/3} < 3$ and $\frac{1 - v}{1 + \sqrt{v}} \le 1 - v$.
It suffices to prove that
$\frac{3u-3}{3(u+2)} \ge 1 - v$
or $\frac{u-1}{u+2}\ge 1 - v$ or
$$\frac{x^3+y^3+z^3 - 3xyz}{x^3+y^3+z^3+6xyz} \ge \frac{x^2+y^2+z^2-xy-yz-zx}{x^2+y^2+z^2}.$$
Since $x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx)$, the above inequality is written as
$$\frac{(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2 - 6xyz)(x^2+y^2+z^2-xy-yz-zx)}{(x^3+y^3+z^3+6xyz)(x^2+y^2+z^2)} \ge 0.$$
By AM-GM, this inequality is true. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Maclaurin series with zero denominator when evaluating derivative I have following function:
$\displaystyle f(x) = \frac{\ln(1+x^2) - x^2}{\sqrt{1+x^4} - 1}$
As you can see, when doing the quotient rule for the denominator in your head, the derivative of this function results to $0$ in the denominator when evaluating for $x = 0$. My question is, how is it possible that there exists a Maclaurin series according to Wolfram Alpha?
Do I overlook something?
| $$\eqalign{\ln(1+x^2) &= - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4}+ \ldots\cr
\sqrt{1+x^4} - 1 &= \frac{x^4}{2} - \frac{x^8}{8} + \ldots\cr
\frac{\ln(1+x^2)}{\sqrt{1+x^4}-1} &= \frac{x^4 \left(-\dfrac{1}{2} + \dfrac{x^2}{3} + \ldots\right)}{x^4 \left(\dfrac{1}{2} - \dfrac{x^4}{8} + \ldots\right)}\cr
&= -1 + \frac{2}{3} x^2 + \ldots}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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First four nonzero terms of the McLaurin expansion of $\frac{xe^x}{\sin x}$ at $x_0=0$
Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series.
$$ \frac{xe^x}{\sin x}$$
Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x$
We have the following MacLaurin series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x +a_2 x^2 + a_3 x^3 +a_4 x^4 + ...$$
$$g(x) = \sum_{n=0}^{\infty} b_n x^n = 0 + x+ 0 -\frac{1}{6} x^3 + 0 + \frac{1}{120} x^5 +...$$
$$h(x) = xe^x = \sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} \frac{n}{(n)!}x^n = 0 + x + x^2 + \frac{1}{2} x^3 + \frac{1}{6} x^4 + \frac{1}{24} x^5$$
.
The Maclaurin coefficients of $f(x)g(x)$
$$ c_0 = a_0b_0 = 0$$
$$ c_1 = a_0b_1 + a_1b_0 = a_0$$
$$ c_2 = a_0b_2 + a_1b_1 +a_2b_0 = a_1 $$
$$ c_3 = a_0b_3 +\cdots a_3b_0 = - \frac{1}{6} a_0 + a_2$$
$$ c_4 = a_0b_4 +\cdots + a_4b_0 = - \frac{1}{6} a_1 + a_3 $$
$$ \implies f(x)g(x) = a_0 x^1 + a_1 x^2 + \left( - \frac{1}{6} a_0 + a_2\right)x^3 + \left(- \frac{1}{6} a_1 + a_3\right) x^4 + \cdots$$
.
Equating coefficients of $f(x)g(x)$ with coefficients of $h(x)$
$$ c_0 = 0$$
$$ c_1 = a_0 = 1 $$
$$ c_2 = a_1 = 1 $$
$$ c_3 = - \frac{1}{6} a_0 + a_2 = 1/2 \implies a_2 = 2/3 $$
$$ c_4 = - \frac{1}{6} a_1 + a_3 = 1/6 \implies a_3 = 1/3 $$
.
it follows that the first 4 nonzero terms are:
$$f(x) = \frac{x e^x}{\sin(x)} = 1 + x + \frac{2}{3} x^2 + \frac{1}{3} x^3$$
.
I would like to know if I am going in the right direction?
My main question regards the original question "Define if necessary the given function so as to be continuous at $x_0=0$" . I am not sure how to handle this question, if a definition is needed how? and why? if a definition is not needed how? and why?
Thx for your input/help.
| I think that your expansion is partially incorrect. After setting $f(0)=1$, note that for $x\not=0$,
\begin{align*}
f(x)=\frac{x e^x}{\sin(x)}&=x \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(x-\frac{x^3}{6}+o(x^4)\right)^{-1}\\
&=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1-\frac{x^2}{6}+o(x^3)\right)^{-1}\\
&=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)
\cdot\left(1+\frac{x^2}{6}+o(x^3)\right)\\
&=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1+\frac{x^2}{6}+o(x^3)\right)\\
&=1+x+\frac{x^2}{2}+\frac{x^3}{6}
+\frac{x^2}{6}+\frac{x^3}{6}+o(x^3)\\
&=1+x+\frac{2x^2}{3}+\frac{x^3}{3}+o(x^3)
\end{align*}
where we used the fact that
$$(1-z)^{-1}=1+z+z^2+o(z^2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$
Prove that:
$$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$$
I computed the indefinite integral:
$$\int\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}(\ln 2)^2}+C$$
How can I continue from here?
| $$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}>\int\limits_{1}^{+\infty}\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}\ln^2 2}|_1^{+\infty}=\frac{1}{4\ln^22}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| Partial answer, complete for $n$ even.
Suppose $n\geq2$. Clearly, $k>n$.
Then $k!$ is even, so there is an even number of odd numbers on the left. So $n\equiv0,3\pmod4$.
Suppose $n\geq4$ is even. Odd squares are $1$ mod $8$, and $8\mid k!$, so $n/2$ is divisible by $8$. Odd $16$th powers are $1$ mod $32$ and $32\mid k!$, so $n/2$ is divisible by $32$. [...] Odd $4^m$th powers are $1$ mod $2^{2m+1}$ (Euler) so $2^{2m+1}\mid n/2$... etc, contradiction. (Boring details left out.)
Suppose $n\geq3$ is odd. Odd numbers have odd powers that are congruent to themselves mod $8$ and $8\mid k!$, so $8$ divides $1+3+\cdots+n=(n+1)^2/4$, so $n\equiv-1\pmod{8}$. Then odd numbers have $n$th powers that are congruent to their inverse modulo $32$ and $32\mid k!$, so $32$ divides $(n+1)^2/4$, so $n\equiv-1\pmod{16}$. From now this argument does not give anything new: $64\mid(n+1)^2/4$ only implies $n\equiv-1\pmod{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.
| $$a=\sqrt{6-\sqrt{20}}$$
$$\implies a^2=6-\sqrt{20}$$
$$\implies(a^2-6)^2=20$$
$$\implies a^4-12a^2+16=0$$
$$\implies (a^2-2a-4)(a^2+2a-4)=0$$
so it will be one of the four possibilities $\pm\sqrt{5}\pm1$, and since $\sqrt{6-\sqrt{25}} \lt \sqrt{6-\sqrt{20}} \lt \sqrt{6-\sqrt{4}}$, you want the one which is in $[1,2]$
| {
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"timestamp": "2023-03-29T00:00:00",
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A quicker approach to the integral $\int{dx\over{(x^3+1)^3}}$ Source: A question bank on challenging integral problems for high school students.
Problem:
Evaluate the indefinite integral
$$\int{dx\over{(x^3+1)^3}}$$
Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fractions but it would be VERY time consuming and altogether boring. I am unable to find an alternate solution other than partials. Can anyone lead me towards a quicker approach, because the exam I'm preparing for is time bound and I can't afford to spend much time on a single problem. Thanks!
Edit 1:
I looked upon reduction formulas. I guess we can generalise this by using:
$$I_n = \int {dx\over{(x^3+1)^n}}$$
$$I_n = \int{(x^3+1)^{-n}dx}$$
Will try and solve it. Maybe we can reduce it to a simpler integral!
Edit 2:
Ok so I got the reduction formula. Can anybody verify if its right, like if you've solved it?
$$I_{n+1} = \frac {x}{3n(x^3+1)^n} + \frac{(3n-1)}{3n}I_n$$
Edit 3: Now I have reached the solution nearly
take $n=2$
$$I_3 = \frac{x}{6(x^3+1)^2}+\frac{5}{6}I_2$$
now take $n=1$
$$I_2 = \frac{x}{3(x^3+1)}+\frac{2}{3}I_1 $$
now we see that $I_1$ is nothing but
$$I_1= \int{\frac{dx}{x^3+1}}$$
which simplifies to
$$I_1 = \int{\frac{dx}{(x+1)(x^2+1-x)}}$$
Now this integral is cake, solve using Partial Fractions
$$\frac{1}{(x+1)(x^2+1-x)} \equiv \frac{A}{x+1} + \frac{Bx+C}{(x^2+1-x)}$$
And eventually $I_1$ is:
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
This way we get $I_1$. Putting in equation above we get $I_2$
Substitute $I_2$ in the equation above and obtain an expression for $I_3$ !
I'll verify ASAP :)
Final edit: Yeas! Reached the answer. Matches term to term!
Final answer is :
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
Use the above approach or any alternatives that are more quick are welcome!
| We have:
$$\int\frac{dx}{(x^3 + 1)^3} = \int \frac1{x^2}\frac{x^2}{(x^3 + 1)^3}dx$$
with
$$u = \frac1{x^2}, dv = \frac{x^2}{(x^3 + 1)^3}dx$$
this becomes:
$$\text{something } - \frac13\int\frac{1}{x^3(x^3 + 1)^2}dx$$
then:
$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \int\frac{x^3 + 1 - x^3}{x^3(x^3 + 1)^2}dx = \int\frac{dx}{x^3(x^3 + 1)} - \int\frac{dx}{(x^3 + 1)^2}$$
do same trick for $\int \frac{dx}{(x^3 + 1)^2}$ to get:
$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \text{something}' + \frac53 \int\frac{dx}{x^3(x^3 + 1)}$$
so this reduces to finding
$$\int\frac{dx}{x^3 + 1} $$
this is:
$$\int\frac{1 + x^3 - x^3}{x^3 + 1} dx = x - \int x \frac{x^2}{x^3 + 1}dx$$
integrate by parts and voila!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Positive definiteness of difference of inverse matrices Let $A$ and $B$ be two $n \times n$ symmetric and positive definite matrices.
If $A \prec B$, then is it true that $B^{-1} \prec A^{-1}$?
Here, $A \prec B$ means that $B-A$ is positive definite.
| For ease of notation I'll use $\ge$ instead of $\succeq$ and so on.
Lemma 1: $A \le B \Rightarrow C^T A C \le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x \ge 0$ for any conformable vector $x$ so that $C^T(B-A)C \ge 0.$
Lemma 2: $I \preceq B \Rightarrow$ $B$ is invertible and $B^{-1} \preceq I.$
Proof.
*
*Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G \le G^T B G = D.$ Thus all eigenvalues of $B$ are $\ge 1 >0,$ hence $B$ is invertible.
*Now write $B^{-1} = B^{-1/2}IB^{-1/2} \le B^{-1/2}BB^{-1/2} = I,$ where $B^\alpha := GD^{\alpha}G^T.$
Proposition: $0 < A \le B \Rightarrow$ $B$ invertible and $B^{-1} \le A^{-1}.$
Proof.
\begin{align*}
A\le B &\Rightarrow B-A \ge 0 \\
&\Rightarrow A^{-1/2}(B-A)A^{-1/2} \ge 0\\
&\Rightarrow A^{-1/2}BA^{-1/2} \ge I\\
&\Rightarrow A^{1/2}B^{-1}A^{1/2} \le I,\\
\end{align*}
hence
\begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}B^{-1}A^{1/2}A^{-1/2}\\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\\
&\le A^{-1/2}IA^{-1/2}\\
&= A^{-1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Infinite trigonometry Summation: $ \sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right) $ I would like to evaluate
$$
\sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)
$$
Summation image please view before solving
I saw a pattern and realized the answer will converge and the final summation will be 1-(1/√2) but I am going wrong somewhere .Answer given is 2-(1/√2)
Plz help
| This may be seen as a telescoping series, one may write for $N\ge1$,
$$
\small{\sum_{k=1}^N\! \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)\!=\sum_{k=1}^N \!\left(\! \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+1)}\! \right)\!+\!\sum_{k=1}^N\! \left(\! \cos \frac{\pi}{2(k+1)}-\cos \frac{\pi}{2(k+2)} \!\right)}
$$ giving
$$
\small{\sum_{k=1}^N \! \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)=\!\!\left(\cos \frac{\pi}{2}-\cos \frac{\pi}{2(N+1)}\right)\!+\!\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2(N+2)}\right)}
$$ then by letting $N \to \infty$, one gets
$$
\sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)=(0-1)+\left(\cos \frac{\pi}{4}-1\right)
$$I think you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $z\in\mathbb{C}$ such that $|z|=2i(\bar{z}+1)$ I was trying to solve the following problem:
Find $z\in\mathbb{C}$ such that $$|z|=2i(\bar{z}+1).$$
I tried to solve the problem by letting $z=x+iy$ and found out that the solutions are $z=-1+i\frac{1}{\sqrt{3}},-1-i\frac{1}{\sqrt{3}}$. But as tried to justify the solutions $z=-1-i\frac{1}{\sqrt{3}}$ does not satisfy the equation.
I couldn't find any error in my calculation. Any help would be highly appreciated. Thanks in advance.
| The equation you should get is
$$\sqrt{x^2+y^2} = 2i(x-iy + 1)\\
\sqrt{x^2 + y^2} = y + 2(x+1)i$$
which means that
*
*$2y=\sqrt{x^2+y^2}$
*$2(x+1)=0$
From the second equation, you get $x=-1$, and the first equation becomes
$$2y=\sqrt{y^2+1}$$
Now, you SQUARE the equation and get $$4y^2=y^2+1$$
and get $$y^2=\frac{1}{3}$$
Now, the equation $y^2=\frac13$ has two solutions, but since you got that equation by squaring some other equation, you should know that some of the solutions of the second equation may not solve the first equation. In your case, $y=\frac{1}{\sqrt{3}}$ solves the equation, because $$2\cdot \frac1{\sqrt{3}}=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+1}$$
but $y=-\frac{1}{\sqrt{3}}$ does not solve the equation, because $$2\cdot\left(- \frac1{\sqrt{3}}\right)\neq 2\cdot\frac1{\sqrt{3}}=\sqrt{\left(-\frac{1}{\sqrt{3}}\right)^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint!
| Use $\lfloor x \rfloor + \lfloor x + 1/2 \rfloor = \lfloor 2x \rfloor$ identity . So $$\lfloor \frac{4a-10}{15} \rfloor = a-4 $$ Try $a = 0 , 1,2,3, \dots$ The only solution is $a = 4$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that there are exactly two solutions to the equation $x^3 = x^2$. This is Velleman's exercise 3.6.12.c:
Prove that there are exactly two solutions to the equation $x^3 = x^2$.
Here's my proof of it:
Proof.
Existence. Let $x = 0$ then clearly $x^3 = x^2$ and let $x = 1$ then clearly $x^3 = x^2$.
Uniqueness. We choose an arbitrary $z$ such that $z^3 = z^2$. Now we consider two exhaustive cases:
Case 1. $z = 0 = x$, then clearly $0 = 0$.
Case 2. $z \neq 0$, then $z^2 \neq 0$ and dividing the equation $z^3 = z^2$ by $z^2$, we get $z = 1 = x$.
Is my proof correct?
Thanks.
| Yes, your proof is correct.
But here's a more standard proof . . .
\begin{align*}
&x^3=x^2\\[4pt]
\iff\;&x^3-x^2=0\\[4pt]
\iff\;&x^2(x-1)=0\\[4pt]
\iff\;&x^2=0\;\;\text{or}\;\;x-1=0\\[4pt]
\iff\;&x=0\;\;\text{or}\;\;x=1\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| The problem is that
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 \ne 3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = (397)_9. $$
What we are trying to do is solve
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 = a \cdot 9^2 + b \cdot 9^1 + c \cdot 9^0 = (abc)_9 $$
To find $a$ we take as many copies of $81$ from $397$ as we can:
$$ 397 - 4 \cdot 81 = 73. $$
Thus $a = 4$. Next, we take as many copies of $9$ from $73$ as we can:
$$ 73 - 8 \cdot 9 = 1, $$
so $b = 1$. Lastly, we take as many copies of $1$ from $1$ as we can:
$$ 1 - 1 \cdot 1 = 0. $$
Hence $c = 1$. Therefore the answer is $$(397)_{10} = (481)_9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
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