Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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why $x^2 - 25y^2 = 1$ cannot have integer solutions other than $( \pm 1,0)$ pell equations must have n to be nonsquared,so $x^2 - 25y^2 = 1$ is not a pell equation, but the thing is how can i show $x^2 - 25y^2 = 1$ have no integer solutions? or it has integer solutions.
Solve : $x^2-25y^2=1$
| Given $x^2-25y^2=1$
$(x-5y)(x+5y)=1$
$1=-1 \cdot -1$ and $1=1 \cdot 1$ are the only ways to write $1$ as the product of two integers.
Hence we must have $x-5y=x+5y=1$ or $x-5y=x+5y=-1$ and so it gives the only solutions as $x= \pm 1$ and $y=0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Argument and modules of complex numbers Can you help me to find the argument and modules of this complex number please?
$$z=\frac{2-i}{-\sqrt {3}-2i}$$
| $$z=\frac{2-i}{-\sqrt {3}-2i}\cdot\frac{-\sqrt {3}+2i}{-\sqrt {3}+2i}=\frac{-2\sqrt3+2i-i\sqrt3-2i^2}{3+4}=$$
$$=\frac{2-2\sqrt3+(2-\sqrt3)i}{7}=\frac{2-2\sqrt3}{7}+\frac{2-\sqrt3}{7}i=a+bi$$
then
$$|z|=\sqrt{a^2+b^2}$$
$$\arg z=\arctan\frac{a}{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent? Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent ?
$\sum_{n=2}^{\infty} |\frac{(-1)^n }{n^2+(-1)^n}| =\frac{1}{|n^2+(-1)^n|}$
now $|n^2+(-1)^n| > n^2 - 2 $
=>$\frac{1}{|n^2+(-1)^n|}<1/(n^2 - 2)$
$1/(n^2 - 2)$ is convergent by Limit comparison test with $1/n^2$
therefore,$\frac{1}{|n^2+(-1)^n|}$ is convergent by comparison test
Therefore, case of absolute convergence?
is this correct ?
| $\sum\limits_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}=\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}-\sum\limits_{n=1}^{\infty} \frac{1}{(2n+1)^2-1}$
with
$\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}\leq \frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{24}$
and
$\sum\limits_{n=1}^{\infty} \frac{1}{(2n+1)^2-1}=\frac{1}{4}\sum\limits_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1})=\frac{1}{4}$
Therefore
$|\sum\limits_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}|\leq \sum\limits_{n=2}^{\infty} \frac{1}{n^2+(-1)^n}\leq \frac{\pi^2}{24}+\frac{1}{4}$
=> absolute convergence
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.
| Note that for $n>1$,
$$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$
$$\sum_{n=2}^{2015}\frac{1}{n^3-1}=\frac{1}{2(1)}-\frac{1}{2(2)}-\frac{1}{2(2015)}+\frac{1}{2(2016)}=\frac{2031119}{8124480}<\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int_0^{\pi/2} \frac{\sin x}{x(x+1)} \, \mathrm{d}x < \ln\frac{\pi+2}{2}$ How to prove that $$\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{x(x+1)}dx < \ln\frac{\pi+2}{2}$$
How prove tasks like this?
| We have $\sin x < x$ for $x > 0$ hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x < \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \ln (x+1)\big]_0^{\pi/2} = \ln \left(\frac{\pi}{2} + 1\right).$$
Also $\sin x > \frac{2}{\pi}x$ for $x > 0$ (and $x < \pi/2$) hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x > \frac{2}{\pi} \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \frac{2}{\pi}\ln (x+1)\big]_0^{\pi/2} = \frac{2}{\pi}\ln \left(\frac{\pi}{2} + 1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex eigenvectors... I can't get the right answer even though I'm using software I'm using software to calculate my eigenvectors, and I can't get the correct answers...
I have $$A = \begin{pmatrix} 1 & 2+i \\ 2+i & 1 \end{pmatrix}.$$ I solved the eigenvalues to be $3+i, -1-i$ (which is correct according to my software). Then I compute $\ker(A - (3+i)I)$ and $\ker(A-(-1-i)I)$, which gives me eigenvectors of $(1,)^T, (-1,0)^T$. But, plugging them in, they don't satisfy the requirement of an eigenvector. Can someone show me how my method is wrong and how to arrive at the correct answer?
| We all agree that given
$$
\mathbf{A} = \left[
\begin{array}{cc}
1 & 2+i \\
2+i & 1 \\
\end{array}
\right]
$$
the eigenvalues are
$$
\lambda \left( \mathbf{A} \right) = \left\{ 3 + i, -1 - i \right\}
$$
The eigenvectors are
$$
v_{1} =
\color{blue}{\left[
\begin{array}{c}
1 \\
1 \\
\end{array}
\right]}, \qquad
v_{2} =
\color{red}{\left[
\begin{array}{r}
-1 \\
1 \\
\end{array}
\right]}
$$
The eigenvector equations are
$$
\begin{align}
%
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) \cdot v_{1}
&= \mathbf{0} \\[5pt]
%
\left[
\begin{array}{rr}
-2-i & 2+i \\
2+i & -2-i \\
\end{array}
\right]
\color{blue}{\left[
\begin{array}{r}
1 \\
1 \\
\end{array}
\right]}
&=
\left[
\begin{array}{r}
0 \\
0 \\
\end{array}
\right] \\[10pt]
\end{align}
$$
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) \cdot v_{2}
&= \mathbf{0} \\[5pt]
%
\left[
\begin{array}{rr}
2+i & 2+i \\
2+i & 2+i \\
\end{array}
\right]
\color{red}{\left[
\begin{array}{r}
-1 \\
1 \\
\end{array}
\right]}
&=
\left[
\begin{array}{r}
0 \\
0 \\
\end{array}
\right] \\[10pt]
%
\end{align}
$$
We don't know what software package the OP used. So to advance the discussion, here is Mathematica output.
Of course we can solve the eigenvalue equation directly.
$$
\mathbf{A} v = \lambda v
$$
The answer is the same.
| {
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Find partial sum of this series: $\sum_{n=1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$? $\sum_{1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$
How to find partial sum, sum and prove convergence by definition? Thanks a lot.
| $$\sum_{n=1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$$
Resolve it into partial fractions,
$$=\frac{1}{2}\sum_{n=1}^{\infty }[\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the locus of $|z-2i|=3|z+3|$ I got as far as
$$|z-2i|=3|z+3| \Leftrightarrow \\
(\ldots) \Leftrightarrow \\
x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\
x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\
-8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\
8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\
y^2+\frac{1}{2}y-x^2-\frac{27}{4}x=\frac{85}{8}$$
Then I tried to complete the square:
$$y^2+\frac{1}{2}y = 0\Leftrightarrow (y+\frac{1}{4})^2-\frac{1}{16}$$
$$-x^2-\frac{27}{4}x=0 \Leftrightarrow -(x+\frac{27}{8})^2+\frac{729}{64} = 0 \Leftrightarrow (x+\frac{27}{8})^2-\frac{729}{64} = 0$$
And so the equation becomes:
$$(y+\frac{1}{4})^2-\frac{1}{16}+(x+\frac{27}{8})^2-\frac{729}{64}=\frac{85}{8} \Leftrightarrow \\
(y+\frac{1}{4})^2+(x+\frac{27}{8})^2=\frac{85}{8}+\frac{1}{16}+\frac{729}{64} $$
According to my book the center is $-\frac{27}{8}-\frac{1}{4}i$ which I got right but the radius is $\frac{\sqrt{117}}{8}$.
What went wrong?
| Problem is way up at the top. You wrote $x^2 - y^2 + 4y-4$, when it should have been $x^2 + y^2 - 4y+4$. Similarly, $9x^2 - 9y^2 + 54 x + 81$ should have been $9x^2 + 9y^2 + 54 x + 81$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL? I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?
| Put it on a cartesian plane $A = (0,0); B= (1,0); C=(1,1); D=(0,1)$ and $K = (1,k)$ and $L=(j, 1)$.
Then ... it all comes out in the wash.
$AK = \sqrt{k^2 + 1} = KL = \sqrt{(1-k)^2 + (1-j)^2} = AL = \sqrt{1 + j^2} ; 1>k > 0; 1>j> 0$
So $k^2 + 1 = k^2 + j^2 - 2(k+j) + 2 = 1+ j^2$
So $j = k$ and $k^2 - 4k + 1 = 0$ so $k = 2 \pm \frac{\sqrt{12}}{2}$ and (as $k < 1$) $k = 2 - \sqrt{3}$
So $AK = KL = AL = \sqrt{8 - 4\sqrt{3}}$
So area of triangle is $\frac 12 b*h = \frac 12 * AK * \frac {\sqrt{3}}2*AK = \frac {\sqrt{3}}4*AK^2 = \frac {\sqrt 3}4*(8 - 4\sqrt{3}) = 2\sqrt{3} - 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding solutions to polynomial modular equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$ I'm more or less okay with linear modulo equations, and was wondering how to solve polynomial modulo equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$.
The possible solutions are 2, 7, 9, 11 and 16 (found on an online calculator). What is the general method to solving such equations? A method better than just guess for all $x \leq 19$, since then it would be difficult if the modulo is big. Thanks!
| there are six sixth roots of one $\pmod {19},$ those being the three cube roots of $1$ and the three cube roots of $-1.$ For each of these, solve $\frac{x+1}{x} \equiv t \pmod {19}$ where $t^6 \equiv 1 \pmod {19}.$ Or, for $t \neq 1,$ you see
$$ \frac{1}{x} \equiv t - 1 \pmod {19}, $$
$$ x \equiv \frac{1}{t-1} \pmod{19} $$
The cube roots of one are $1,7,11 \pmod {19}.$ For instance, $18 \cdot 19 = 342$ and $7^3 = 343.$ Then $7 \cdot 11 = 77 = 1 + 4 \cdot 19.$
The cube roots of $-1$ are therefore $18, 12, 8 \pmod {19}.$
multiplicative inverses
$$
\begin{array}{cccccccccccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
1 & 10 & 13 & 5 & 4 & 16 & 11 & 12 & 17 & 2 & 8 & 9 & 3 & 15 & 14 & 6 & 9 & 18
\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$ uniformly continuous or not Find out if function $$f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$$ is uniformly continous or not in area $D=\{0<x^2+y^2<1\}$.
I found out that we have no $\lim\limits_{x,y\to 0}f(x,y)$, because $$\lim\limits_{x,y\to 0}\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\rho^3\sin^3\alpha+\rho^3\cos^3\alpha}}{\sqrt[5]{\rho^5\sin^5\alpha+\rho^5\cos^5\alpha}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\sin^3\alpha+\cos^3\alpha}}{\sqrt[5]{\sin^5\alpha+\cos^5\alpha}}$$
hence we can't use The Uniform Continuity Theorem as we can't determ $f(0,0)$. Function doesn't have bounded partial derivatives, so I think it's not uniformly continous, but I don't know how to show that
| It's not uniformly continuous. Take $x = \rho\cos{\alpha},y= \rho\sin{\alpha}$. Let $\alpha =\pi/4,$, and let $\rho \rightarrow 0^+$, then $f(x,y) \rightarrow \frac{2^{1/3}}{2^{1/5}}$, but take $\alpha =0$, then $f(x,y) \rightarrow 1$.
So let $\epsilon = \frac{1}{2}|\frac{2^{1/3}}{2^{1/5}} - 1|$, then $\forall \delta > 0$, we can find very small $\rho_1, \rho_2>0$, such that $f(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4})$ is very close to $\frac{2^{1/3}}{2^{1/5}}$, and $f(\rho_2 \cos{0},\rho_2\sin{0})$ is very close to $1$, and $(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}), (\rho_2 \cos{0},\rho_2\sin{0})$ are very close to each other.
Then $||(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}) - (\rho_2 \cos{0},\rho_2\sin{0})|| < \delta$, $|f(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}) - f(\rho_2 \cos{0},\rho_2\sin{0})| > \epsilon$.
Thus $f(x, y)$ is not uniformly continuous in any neighborhood of $(0,0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| By substituting $x\mapsto\frac1x$, we get
$$
\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x=\int_1^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
$$
Therefore, using formula $(6)$ from this answer, we get
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
&=2\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x\\
&=2\int_0^1\left(1+x^2-x^4-x^6+x^8+x^{10}-x^{12}-x^{14}+\cdots\right)\,\mathrm{d}x\\
&=2\left(1+\frac13-\frac15-\frac17+\frac19+\frac1{11}-\frac1{13}-\frac1{15}+\cdots\right)\\
&=2\sum_{k=1}^\infty\left(\frac1{8k-7}+\frac1{8k-5}-\frac1{8k-3}-\frac1{8k-1}\right)\\
&=\frac14\sum_{k=1}^\infty\left[\left(\tfrac1k-\tfrac1{k-\frac18}\right)+\left(\tfrac1k-\tfrac1{k-\frac38}\right)-\left(\tfrac1k-\tfrac1{k-\frac58}\right)-\left(\tfrac1k-\tfrac1{k-\frac78}\right)\right]\\
&=\frac14\left(\color{#C00}{H_{-\frac18}}+\color{#090}{H_{-\frac38}-H_{-\frac58}}\color{#C00}{-H_{-\frac78}}\right)\\[4pt]
&=\frac\pi4\left(\color{#C00}{\cot\left(\frac\pi8\right)}+\color{#090}{\cot\left(\frac{3\pi}8\right)}\right)\\[5pt]
&=\frac\pi{\sqrt2}
\end{align}
$$
Since $\cot\left(\frac\pi8\right)=\sqrt2+1$ and $\cot\left(\frac{3\pi}8\right)=\sqrt2-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Large powers of sine appear Gaussian -- why? As part of approximating an integral, I have noticed that $\sin^k(x), x\in[0, \pi]$ look almost identical to $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ once $k$ is large enough (in practice, the two equations are visually identical for $k\geq 15$). As an example, see this picture of the two functions for $k=10$:
Can anybody explain why these functions practically are identical?
Edit: I should mention that the expression $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ is derived by a 2nd order Taylor expansion of $\log\sin^k(x)$ around the mode $x_0 = \frac{\pi}{2}$, i.e. a Laplace approximation.
| This is too long for a comment.
Semiclassical made a very good comment.
Let us use Taylor series around $x=\frac \pi 2$
$$\sin(x)=1-\frac{1}{2} \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} \left(x-\frac{\pi
}{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ $$\sin^k(x)=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} k (3 k-2)
\left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ On the other hand $$e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=1-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2+\frac{1}{8} k^2 \left(x-\frac{\pi
}{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ Computing the difference $$\sin^k(x)-e^{-\frac{1}{2} k \left(x-\frac{\pi }{2}\right)^2}=-\frac{1}{12} k \left(x-\frac{\pi }{2}\right)^4+O\left(\left(x-\frac{\pi
}{2}\right)^6\right)$$ AT least, they are very close in the area around the maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Note that if $z^3=1, z\neq 1$ so that $z^3-1=(z-1)(z^2+z+1)=0$ then $z^5+z+1=z^2+z+1=0$. A key to this observation is just seeing whether an appropriate root of unity may be a root.
This tells you that $x^2+x+1$ is a factor of $x^5+x+1$, and the question then is how you do the division. The factorisation method you have been shown is equivalent to doing the polynomial long division.
Another method, not suggested by others, is to use the fact that you know $x^2+x+1$ is a factor and multiply through by $x-1$ so that this factor becomes $x^3-1$.
So $(x-1)(x^5+x+1)=x^6-x^5+x^2-1=(x^3-1)(x^3+1)-x^2(x^3-1)=(x^3-1)(x^3-x^2+1)=(x-1)(x^2+x+1)(x^3-x^1+1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Why is $x^2+x$ the same as writing $(x+0.5)^2-0.25$? I find it extremely weird that a translation vertically by a variable should cause the graph to move both vertically and horizontally. Also, why 0.5 and 0.25?
| Okay so $0.5 = \frac{1}{2}$ and $0.25 = \frac{1}{4}$.
Let's start from $$\left ( x + \frac{1}{2} \right )^{2} - \frac{1}{4}$$
$$= \left (\frac{2x+1}{2} \right )^{2} - \frac{1}{4}$$
$$= \frac{(2x+1)^{2}}{4} - \frac{1}{4}$$
$$= \frac{(2x+1)^{2}-1}{4}$$
$$= \frac{4x^{2}+4x}{4}$$
Then factor out $4x$ in the numerator to obtain $$= \frac{4x(x+1)}{4}$$
Then you'll get $x^{2}+x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What can be the values of $m$ in $(m-3)x^2 + (m+2)x + 2m + 1$? What can be the values of $m$, if $(m-3)x^2 + (m+2)x + 2m + 1$ should be always greater than or equal to zero?
I think we should be using Delta, but I've got no idea how.
| Let,
\begin{align}
y&=ax^2+bx+c\\
&=a\left(x+\dfrac{b}{2a}\right)^2+\left(c-\dfrac{b^2}{4a}\right)\\
&=a\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b^2-4ac}{4a}\right)\\
&=a\left[\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a^2}\right]
\end{align}
From this we can conclude that, when $D\le0$ and $a\ge0$, then only $y\ge0$. Applying this to $y=(m-3)x^2 + (m+2)x + 2m + 1$:
\begin{align}
D&\le0\\
\implies(m+2)^2-4(m-3)(2m+1)&\leq0\\
\implies7m^2-24m-16&\ge0\\
\implies(7m+4)(m-4)&\ge0\\
\implies m\le-\dfrac47\quad\text{and,}\quad m&\ge4\tag{i}
\end{align}
Again, $a\ge0\implies m-3\ge0\implies m\ge3.\tag{ii}$
Therefore from (i) and (ii) we get, $m\ge4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Primes where $(p-1) ^ 2 + 1$ and $(p+1) ^ 2 + 1$ are also prime I'm following a math course about basic number theory. The course contains an Open Problems section with Landau’s conjecture, that states: "There are infinitely many primes of the form $n^2 + 1$.". Examples include:
$2 = 1^2 + 1$
$5 = 2^2 + 1$
$17 = 4^2 + 1$
$37 = 6^2 + 1$
I noticed that n in these examples are primes minus one. So I made a program that checks for the first x primes if $(p-1) ^ 2 + 1$ is also prime. This yields the primes:
$2, 3, 5, 7, 11, 17, 37, 41, 67, 127, 131, 151, 157, ..$
I also checked out $(p+1) ^ 2 + 1$, which yields:
$3, 5, 13, 19, 23, 53, 73, 83, 89, 109, 149, 179, 223, ..$
Both formula's seem to yield numbers at about the same rate.
However, what I find strange is the intersection of the sets. If I look for primes where both $(p-1) ^ 2 + 1$ and $(p+1) ^ 2 + 1$ are also prime, only 3 and 5 seem to suffice. From there the sets seem to go their seperate ways. Now I would like some insight into why this is, but I can't find the above formula's or the sets using searches. With my computer I tried the first 50 million primes but only 3 and 5 seem to have this these properties. What am I looking at here? Are there other primes known that satisfy these properties?
| Have you thought about comparing $(p - 1)^2 + 1$ to $(p + 1)^2 + 1$ for just a few values of $p$ and seeing what happens?
*
*$p = 2$ gives us $(p - 1)^2 + 1 = 2$ and $(p + 1)^2 + 1 = 10 = 2 \times 5$.
*$p = 3$ gives us $(p - 1)^2 + 1 = 5$ and $(p + 1)^2 + 1 = 17$.
*$p = 5$ gives us $(p - 1)^2 + 1 = 17$ and $(p + 1)^2 + 1 = 37$.
*$p = 7$ gives us $(p - 1)^2 + 1 = 37$ and $(p + 1)^2 + 1 = 65 = 5 \times 13$.
*$p = 11$ gives us $(p - 1)^2 + 1 = 101$ and $(p + 1)^2 + 1 = 145 = 5 \times 29$.
*$p = 13$ gives us $(p - 1)^2 + 1 = 145 = 5 \times 29$ and $(p + 1)^2 + 1 = 197$.
*$p = 17$ gives us $(p - 1)^2 + 1 = 257$ and $(p + 1)^2 + 1 = 325 = 5^2 \times 13$.
*$p = 19$ gives us $(p - 1)^2 + 1 = 325 = 5^2 \times 13$ and $(p + 1)^2 + 1 = 401$.
This suggests that one of $(p - 1)^2 + 1$ or $(p + 1)^2 + 1$ is prime and the other is a multiple of $5$ (though of course it's also possible both values fail to yield a prime, e.g., $p = 29$). Hopefully this enables you to make sense of Robert's tightly packed answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\frac{b}{3x-2}\right) dx=k$$
then
$$\lim_{N \to \infty} \left[\frac{2}{3}ln(x)+(a-\frac{2}{3})ln (x+3)-\frac{b}{3}ln(3x-2)\right]_3^N =k$$
By logarithm properties:
$$\lim_{N \to \infty} \left[ln\frac{(x)^\frac{2}{3}(x+3)^\left(a-\frac{2}{3}\right)}{(3x-2)\frac{b}{3}}\right]_3^N =k$$
Evaluating:
$$\lim_{N \to \infty} \left[ln\frac{(N)^\frac{2}{3}(N+3)^\left(a-\frac{2}{3}\right)}{(3N-2)^\frac{b}{3}}\right]-\left[ln\frac{(3)^\frac{2}{3}(6)^\left(a-\frac{2}{3}\right)}{(7)^\frac{b}{3}}\right] =k$$
But, I do not know what to do after this step. Any recommendation will be appreciated
| Hint: It converges when
$$ \lim _{ n\to \infty } \ln { \left[ \sqrt [ 3 ]{ \frac { { n }^{ 2 }{ \left(
n+3 \right) }^{ 3a-2 } }{ { \left( 3n-2 \right) }^{ b } } } \right] } \\ \\ 2+3a-2=b\quad \Rightarrow b=3a $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Complex Analysis - Laurent series of $\frac{z}{z^2+1}$ in $|z-3|>2$ I'm supposed to find the Laurent series for $\frac{z}{z^2+1}$ in $D$ : $|z-3|>2$
the nature of the domain $D$ tells me that the analytic part of the Laurent series expansion must be $0$
which means the series I'm looking for should look like this :
$$\frac{a_{-1}}{z-3}+\frac{a_{-2}}{(z-3)^2}+\frac{a_{-3}}{(z-3)^3}\dots\dots$$
I tried to manipulate the function to create a $\frac{1}{1-\frac{2}{z-3}}$ term but I keep failing.
please help me
thanks !
| Let $f(z)=\frac{z}{z^2+1}$. There are pole singularities of $f$ at $z=\pm i$. Hence, if we expand $f$ in a Laurent series around $z=3$, that series will have only a principal part for $|z-3|<\sqrt{10}$ and will have no principal part for $|z-3|>\sqrt{10}$.
For $|z-3|>\sqrt{10}$, we can write
$$\begin{align}
\frac{z}{z^2+1}&=\frac{1/2}{z-i}+\frac{1/2}{z+i}\\\\
&=\frac{1/2}{(z-3)+3-i}+\frac{1/2}{(z-3)+3+i}\\\\
&=\frac{1}{2(z-3)}\left(\frac{1}{1+\frac{3-i}{z-3}}+\frac{1}{1+\frac{3+i}{z-3}}\right)\\\\
&=\frac{1}{2(z-3)}\sum_{n=0}^\infty (-1)^n\left(\left(\frac{3-i}{z-3}\right)^n+\left(\frac{3+i}{z-3}\right)^n\right)\\\\
&=\sum_{n=0}^\infty (-1)^n(10)^{n/2}\cos(n\arctan(1/3))(z-3)^{-(n+1)}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Count permutations of [1, 1, 1, 2, 2, 3, 3, 4, 5] with length 6 How can I count number of permutations such as:
[1, 1, 1, 2, 2, 3]
[1, 2, 1, 3, 2, 4]
[1, 1, 1, 2, 2, 5]
...
So I have:
'1' x 3
'2' x 2
'3' x 2
'4' x 1
'5' x 1
So permutation valid if it contains $\le3$ of '1' and $\le2$ of '2' and and $\le2$ of '3' and $\le1$ of '4' and $\le1$ of '5' and it's length is 6.
I've counted them on Python with filtering and the answer is 3630. But how it works?
Thanks!
| I believe you are forced to sum mutually exclusive and exhaustive possibilities.
I suggest defining $N(a,b)$ as the number of permutations having $a$ "1"'s and $b$ other doubles. Then ...
$$N(a,b) = \binom 2 b \cdot \frac{6!}{a!(2!)^b}$$
and
$$ N_{tot} = N(0,2)+N(1,2)+N(1,1)+N(2,2)+N(2,1)+N(2,0)+N(3,1)+N(3,0) $$
* EDIT *
I my expression gave the number of permutations given a specific combination of "single" extra numbers (extra means not including 1 or any doubles )
there will be $4-b$ "single" numbers from which we must choose $6-a-2b$
So my formula above should be amended to read
$$N(a,b) =\binom{4-b}{6-a-2b}\binom 2 b \cdot \frac{6!}{a!(2!)^b}$$
Evaluating the formula ...
$$ \begin{eqnarray*}
N_{tot} =& \binom 2 2 \binom 2 2 \frac{6!}{(2!)^2}
+ \binom 2 1 \binom 2 2 \frac{6!}{(2!)^2}
+ \binom 3 3 \binom 2 1 \frac{6!}{ 2!}
\\+ &\binom 2 0 \binom 2 2 \frac{6!}{2! (2!)^2}
+ \binom 3 2 \binom 2 1 \frac{6!}{ 2! (2!)}
+ \binom 4 4 \binom 2 0 \frac{6!}{ 2! }
\\+ & \binom 3 1 \binom 2 1 \frac{6!}{ 3! (2!)}
+ \binom 4 3 \binom 2 0 \frac{6!}{ 3! }
\end{eqnarray*}$$
$$ =(1)(1)(180)+(2)(1)(180)+(1)(2)(360)+ (1)(1)(90) \\+ (3)(2)(180) + (1)(1)(360) + (3)(2)(60) +(4)(1)(120) $$
$$ =180+360+720+90+1080+360+360+480 = 3630$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Egyptian fraction for $\varphi- {F(2n+2) \over F(2n+1)}$ The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$.
https://math.stackexchange.com/a/746678/134791
This may be written as the following closed form for an Egyptian fraction.
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$
where $\varphi$ is the golden ratio
$$\varphi = \frac{1+\sqrt{5}}{2}$$
and $F(n)$ are the Fibonacci numbers as described by
$$F(n)=F(n-1)+F(n-2)$$
with $F(0)=0, F(1)=1$.
The result generalizes to other samplings.
$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$
https://math.stackexchange.com/a/2307929/134791
Is there a similar formula for $\varphi- \dfrac{F(2n+2)
}{F(2n+1)}$?
Related questions
Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions?
Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$
| Let $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2} = -\alpha^{-1}$.
In terms of $\alpha, \beta$, we have the Binet's formula:
$$F(n) = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$
For any odd integer $m$, this leads to
$$\varphi - \frac{F(m+1)}{F(m)} = \alpha - \frac{\alpha^{m+1} - \beta^{m+1}}{\alpha^m - \beta^m}
= \frac{\beta^{m-1} + \beta^{m+1}}{\alpha^m - \beta^m}
= \frac{(\beta-\alpha)\beta^m}{\alpha^m - \beta^m}
= \frac{\alpha-\beta}{\alpha^{2m} + 1}
$$
Notice $\displaystyle\;\frac{1}{\alpha^{2m}+1}\;$ can be rewritten as
$$\begin{align}
& \left(\frac{1}{\alpha^{2m}+1} + \frac{1}{\alpha^{4m}-1}\right)
-\left(\frac{1}{\alpha^{4m}-1} - \frac{1}{\alpha^{8m}-1}\right)
-\left(\frac{1}{\alpha^{8m}-1} - \frac{1}{\alpha^{16m}-1}\right)
-\cdots\\
= & \frac{\alpha^{2m}}{\alpha^{4m}-1}
- \frac{\alpha^{4m}}{\alpha^{8m}-1}
- \frac{\alpha^{8m}}{\alpha^{16m}-1}
- \cdots\\
= & \frac{1}{\alpha^{2m} - \beta^{2m}}
- \frac{1}{\alpha^{4m} - \beta^{4m}}
- \frac{1}{\alpha^{8m} - \beta^{8m}}
- \cdots
\end{align}
$$
Combine what is already known for even $m$, we obtain following formula for general $m$.
$$\varphi - \frac{F(m+1)}{F(m)} =
\begin{cases}
\displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 1 \pmod 2\\
\displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)}, & m \equiv 0 \pmod 2
\end{cases}
$$
One may wonder what happens if we replace Fibonacci numbers by Lucas numbers.
Using a similar approach, one can show that
$$\varphi - \frac{L(m+1)}{L(m)} =
\begin{cases}
\displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)},& m \equiv 1 \pmod 2\\
\displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 0\pmod 2
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Consider the following.
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{a} \\
(\sqrt{-3 - 2 i} + \sqrt{-3 + 2 i} )^2 &= a \\
-6 + \sqrt{(-3-2i)(-3+2i)} &= a \\
(-3-2i)(-3+2i) &= (a+6)^2 \\
(a+6)^2 - 13 &=0 \\
a^2 + 12 a + 23 &= 0 \\
\end{align}
from here it is determined that $a = -6 \pm \sqrt{13}$, which yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\frac{x}{1-x^2}=x+x^3+x^5+\cdots$$
$$\frac{x^2}{1-x^4}=x^2+x^6+x^{10}+\cdots$$
$$\cdots$$
Add them all and get the answer. Unfortunately, I havn't got a direct method to calculate it. Appreciate for your help!
| Let $x\in(0,1)$ and let $\varepsilon>0$. I shall prove that there is a natural number $p$ such that$$n\geqslant p\Longrightarrow\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$Take $p'\in\mathbb N$ such that $\left|\frac x{1-x}-(x+x^2+\cdots+x^{p'})\right|<\varepsilon$. Take $p\in\mathbb N$ such that, when you express any element of $\{1,2,\ldots,p'\}$ as the product of an odd number with a power of $2$, then the exponent of that power of $2$ is always smaller than or equal to $p$. Then $$n\geqslant p\Longrightarrow x+x^2+\cdots+x^{p'}\leqslant\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}<x+x^2+x^3+\cdots=\frac x{1-x}$$and therefore$$\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Regarding a number theory proof Before I begin I should mention that I have only recently started studying number theory.
It is required to prove that there are no positive integers $a,b,n>1$ such that
$$a^n - b^n\mid a^n+b^n$$
that is $k$ can never be an integer if
$$\frac{a^n+b^n}{a^n-b^n} = k$$
Now if $b|a$ then we can write
$$\frac{(a/b)^n+1}{(a/b)^n-1} = k$$
which is clearly not true for all $a/b$. However I have no idea how to prove the general result. Help will be appreciated.
| Notice if $a^n - b^n|a^n + b^n$ then $a^n - b^n|a^n + b^n + a^n - b^n = 2a^n$ and $a^n - b^n| a^n + b^n -(a^n- b^n) = 2b^n$.
So if $p|a^n - b^n$ and $p$ is prime either $p = 2$ or $p|a$ and $p|b$.
Let $d = \gcd (a,b)$ and let $a = a'd; b = b'd$ then $\frac{a^n - b^n}{a^n + b^n}= \frac {a'^n - b'^n}{a'^n + b'^n}$ and $a^n - b^n| a^n + b^n \iff a'^n - b'^n|a'^n + b'^n$.
So $a'^n - b'^n|2a'^n$ and $a'^n- b'^n|2b'^n$ and for any prime $p \ne 2$ if $p|a'^n - b'^n$ then $p|a'$ and $p|b'$ so $p|\gcd(a'b') =1$ which is impossible. So no prime other than $2$ divides $a'^n - b'^n$.
So $a'^n - b'^n = 2^m$. But only $2$ can't divide both $a'$ and $b'$ so $a'^n - b'^n = 2$ or $a'^n - b'^n = 1$.
So $(a'^n - b'^n) = (a'-b')(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})$.
Clearly $a'-b' < (a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})$ and $(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1}) > 0$ so
$a -b = 1$ and $(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})= 2$
$(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1}) = 2$ is only possible if $n= 1$. And indeed if $a-b =1$ then $1|a+b$ but... $n$ must be bigger than $1$ and that is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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3-Variable Constrained Inequality I had a chance to read a newly published problem-solving journal very recently and saw this inequality:
Statement: If $a, b, c \in [1,\infty)$ with $a+b+c = 6$, then show that:
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} < \dfrac{3\sqrt{3}abc}{2}$. Just curious about your techniques,etc...
| Let's look at $f(x)= \sqrt{x^2-1}$ for $x\gt1$
$f''(x)=\dfrac {-1}{(x^2-1)^{3/2}}\lt0$ so it's concave and by jensen
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} \leq 3 \sqrt{(\dfrac {a+b+c}{3})^2-1}=3\sqrt{3}$
By symmetry we can assume $a\geq2 \Rightarrow abc\geq2 \Rightarrow \dfrac{3\sqrt{3}abc}{2}\geq3\sqrt{3}$
(I didn't care equality cases which can be easily eliminated)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute the series $\sum_{k\geq 1}\frac{1}{k(k+m)}$ I tried to do it like this:
$$\sum_{k\geq 1} \frac{1}{k(k+m)}=$$ $$\ = \sum_{k\geq 1} \frac{1}{km}-\frac{1}{m(k+m)}=$$ $$=\sum_{k\geq 1} \frac{1}{m}(\frac{1}{k}-\frac{1}{k+m})=$$ $$=\frac{1}{m}\sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} $$
I do know that this sum should equal to$\ \frac{H_m}{m}$, but i don't know how to prove it from here...
| $$\require{cancel} \sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} = \left(\frac{1}{1} - \cancel{\frac{1}{m+1}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{m+2}}\right) + \ldots + \left(\cancel{\frac{1}{m+1}} - \frac{1}{2m+1}\right) + \left(\cancel{\frac{1}{m+2}} - \frac{1}{2m+2} \right) + ...$$
So, all the second terms will cancel out and all the first terms after $k\geq m+1$ will also cancel out leaving,
$$\sum\limits_{k=1}^{m}\frac{1}{k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Trigonometric integrals. Integrate $\int\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\mathrm{d}x$ I tried factoring the denominator but does not seem to suit the factorization of quadratic. Is there another way?
| \begin{align*}
J &= \int \frac{1}{\cos^4 - \cos^2 x \sin^2 x + \sin^4 x} dx \\
&= \int \frac{\sec^4}{1 - \tan^2 x + \tan^4 x} dx
\end{align*}
Let $u = \tan x$ and $\sec^2 x = \tan^2 x + 1$
$$ J = \int \frac{(1+\tan^2 x) \sec^2 x}{1- \tan^2 x +\tan^4 x} dx $$
again $ u = \tan x$
$$ J = \int \frac{u^2 + 1}{u^4 - u^2 + 1} du $$
Since $ \int \frac{u^2 + 1}{u^4 - u^2 + 1} du = \arctan(\frac{u}{1-u^2}) + c $
$$ J = -\arctan(2\cot(2x)) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| Green triangle has 2x area of red triangle. Heights of both triangles are equal because all sides of square are equal. so base of green triangle = 2x base of red triangle.So r=length of base of red triangle and 2r=base of green triangle. Let s= length of side of square also height of both triangles. for red triangle 2=1/2rs so rs=4. same relation hold for green triangle. 4=1/2(2rs) so rs=4 again. Referring to the yellow triangle bottom side=s-r and right side=s-2r. So the area of the triangle is 3= 1/2 (s-r)(s-2r) then
6=( s-r)( s-2r) so 6=s^2-3sr+2r^2. since rs=4. r=4/s. substitute in equation. then
6=S^2-3s(4/s)+2(16/s^2) 6=s^2-12+32/s^2 , 18=s^2+32/s^2, 18s^2=s^4+32, s^4-18s^2+32=0, Let x=s^2, then x^2-18x+32=0, Factor (x-16) (x-2)=0 x=16, x=2 disregard x=2 x=16=s^2
then 16-4-2-3=7 area of Blue triangle
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 11
} |
Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic) I tried using the following approach:
$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$
Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,
$$x=(2222^5+5555^2)k,\ k\in N$$
$$x=(1111^2)(32\cdot1111^3+25)k$$
The term in parentheses was found to be $54165190296027657$, which is divisible by $7$. Thus $7 | x$.
Another method (edit):
We can also write $$2222^{5555} = (7\times 317+3)^{5555} = 7p+3^{5555}$$
Similarly $$5555^{2222}=(793\times7+4)^{2222} = 7q+4^{2222}$$
So we have $$x=7(p+q)+(3^5+4^2)r$$
Here we have $3^5+4^2$ is divisible by $7$. So $x$ is also divisible by $7$.
But is there a simple way without multiplication of large numbers? Thank you!
| Because $2226$ divided by $7$, $5551$ divided by $7$,
$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1})$, for odd $n$ we have $a^n+b^n=(a+b)(a^{n-1}-...+b^{n-1})$ and
$$2222^{5555}+5555^{2222}=$$
$$=(2226-4)^{5555}+4^{5555}+$$
$$+(5551+4)^{2222}-4^{2222}+$$
$$-\left(4^{5555}-4^{2222}\right)$$
and $$4^{5555}-4^{2222}=4^{2222}\left(4^{3333}-1\right),$$
where the last expression divided by $4^3-1$, which divided by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325830",
"timestamp": "2023-03-29T00:00:00",
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\frac{c}{a}$
$α+δ+β+δ=\frac{-2B}{A}$
⇒ $α+β+2δ =\frac{-2B}{A}$
$(α+δ)(β+δ)=\frac{C}{A}$
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$
$(α+β)^2=\frac{4b^2}{a^2}$
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)
$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)
From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
| Instead doing all the hard work you did, you can notice that the difference of roots $(\vert x_1-x_2\vert )$ is same for both the equations. Hence :
$$|\alpha-\beta|=|(\alpha+\delta)-(\beta+\delta)|=\sqrt{(\alpha+\beta)^2-4\alpha \beta}=\sqrt{(\alpha+\delta +\beta+ \delta)^2-4(\alpha+\delta)( \beta+\delta)}$$
$$\implies \sqrt{\left(\frac {-2b}{a} \right)^2 -4\left(\frac ca \right)}= \sqrt{\left(\frac {-2B}{A} \right)^2 -4\left(\frac CA\right)}$$
$$\implies
\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation:
\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j , \ \ 0 \leq t \leq 2 \pi \end{align} But I am not sure . Any help is there ?
| $\mathbf u = (\frac {\sqrt 2}{2} \mathbf i - \frac {\sqrt 2}{2} \mathbf j)\\
\mathbf v = (\frac {\sqrt 6}{6} \mathbf i + \frac {\sqrt 6}{6} \mathbf j -\frac {\sqrt 6}{3} \mathbf k)$
$(x,y,z) = (1,1,-1) + 3\mathbf u \cos t + 3\mathbf v \sin t\\
x = 1 + 3\frac {\sqrt 2}{2} \cos t + \frac{\sqrt {6}}{2} \sin t\\
y = 1 - 3\frac {\sqrt 2}{2} \cos t + \frac {\sqrt {6}}{2} \sin t\\
z = -1 - \sqrt {6} \sin t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The volume of the water in the hemispherical bowl is given by $V=\frac{\pi}{3}y^2(3R-y)$ when the water is $y$ m deep. Water is flowing out at the rate of $6m^3$/min from a reservoir shaped like a hemispherical bowl of radius $R=13m.$The volume of the water in the hemispherical bowl is given by $V=\frac{\pi}{3}y^2(3R-y)$ when the water is $y$ m deep.Find
$(a)$At what rate is the water level changing when the water is $8m$ deep.
$(b)$At what rate is the radius of the water surface changing when the water is $8m$ deep.
$(a)V=\frac{\pi}{3}y^2(3R-y)$
$\frac{dV}{dt}=\frac{\pi}{3}[y^2\frac{d}{dt}(3R-y)+(3R-y)\frac{d}{dt}y^2]$
$\frac{dV}{dt}=\frac{\pi}{3}[y^2(-1)\frac{dy}{dt}+(3R-y)2y\frac{dy}{dt}]$
$6=\frac{\pi}{3}[8^2(-1)\frac{dy}{dt}+(3\times 13-8)(16)\frac{dy}{dt}]$
$\frac{dy}{dt}=\frac{1}{24\pi}$
This answer is correct but i am not able to solve the second part.
| Since your hemisphere is the lower half of a circle, you can express the radius as the x-coordinate (see image added in edit below),
$$ R^2 = x^2 + y^2 \implies x = \sqrt{R^2 - y^2} $$
now differentiate to find $\frac{dx}{dt}$,
$$ \frac{dx}{dt} = \frac{1}{2} (R^2-y^2)^{-\frac{1}{2}} (-2y) \frac{dy}{dt} = \frac{-y}{\sqrt{R^2-y^2}} \frac{dy}{dt} $$
and you know $y=8$, $R=13$, and $\frac{dy}{dt}=\frac{1}{24\pi}$, so you should be able to plug these values in and get $\frac{dx}{dt} = \frac{-1}{3\pi\sqrt{105}} \approx -0.0103546$
Edit:
| {
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Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{(x+4)}{\sqrt{6}})+C$
However the result is $\frac{arctan(\frac{(x+4)}{\sqrt{6}})}{\sqrt{6}}+C$
Why is my result wrong? I can't see any mistake.
| You can write it shortly as $$\int \frac { 1 }{ 6+\left( x+4 \right) ^{ 2 } } dx=\frac { 1 }{ 6 } \int \frac { \sqrt { 6 } d\left( \frac { x+4 }{ \sqrt { 6 } } \right) }{ 1+{ \left( \frac { x+4 }{ \sqrt { 6 } } \right) }^{ 2 } } =\frac { \sqrt { 6 } }{ 6 } \arctan { \left( \frac { x+4 }{ \sqrt { 6 } } \right) } +C$$
| {
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How to evaluate $\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$? How to evaluate $\displaystyle\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$?
I know that I can evaluate the integral using residue theorem, but which is the result with passages? Please
| With the substitution $z(\theta) = e^{i \theta}$ so that $d\theta = \frac{dz}{iz}$, we can rewrite the integral as the counterclockwise contour integral
$$
\oint_{|z| = 1} \frac{1}{(A + \frac 12 B(z + z^{-1}))^2} \frac 1{iz}\,dz =
\frac 4i \oint_{|z| = 1} \frac{z}{(Bz^2 + 2Az + B)^2}\,dz
$$
Evaluate this integral using the residue theorem.
Assume that $|A| > |B|$ so that we have a nice formula. With the quadratic formula, we find that
$$
Bz^2 + 2Az + B = 0 \implies z = \frac{-A \pm \sqrt{A^2 - B^2}}{B} =
-\frac{A}{B} \pm \sqrt{\frac{A^2}{B^2} - 1}
$$
Only the root $z_+ = \frac{-A + \sqrt{A^2 - B^2}}{B}$ satisfies $|z_+| < 1$. The residue theorem tells us that for $f(z) = \frac{z}{(Bz^2 + 2Az + B)^2}$, we have
$$
\oint_{|z| = 1} \frac{z}{(Bz^2 + 2Az + B)^2} = 2 \pi i \operatorname{Res}(f;z_+)
$$
Note that the pole at $z_+$ has order $2$. We calculate the residue
$$
\operatorname{Res}(f;z_+) = \lim_{z \to z_+} \frac{d}{dz}\left[\frac{z}{B^2(z - z_-)^2}\right] = \lim_{z \to z_+} \frac{(z-z_-)^2 + 2z(z-z_-)}{B^2[(z - z_-)^2]^2}
\\ =
\frac{(z_+-z_-)^2 + 2z_+(z_+-z_-)}{B^2[(z_+ - z_-)^2]^2}
$$
Simplify, noting that $z_+ - z_- = 2 \frac{\sqrt{A^2 - B^2}}{B}$.
| {
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$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$
Where $\beta(n)$ is Beta dirichlet function
$(1)$ becomes
$$ln\left({2\over \pi}\right)+{1\over n\Gamma(n)}\sum_{n=1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}{x^{n-1}\over e^x+e^{-x}}\mathrm dx\tag2$$
How can we show that the closed form for $(1)$ is correct?
| Here is a fairly elementary proof of this result. I derive a closed form equivalent to the OP's.
We begin with the following Lemma:
Lemma $1$: $\sum_{k=1}^p \ln(r*k+n) = \ln\left[r^n\left(\frac{n+r}{r}\right)_p\right] \quad \forall p\in\mathbb{N}$
Proof of Lemma $1$: $\sum _{k=1}^p\ln \left(rk+n\right) = \ln \left(\prod _{k=1}^p\left(rk+n\right)\right) = \ln \left(r^p\prod _{k=0}^{p-1}\left(k+\frac{n}{r}\right)\right) = \log\left[r^n\left(\frac{n+r}{r}\right)_p\right]$
We are now ready to prove the result
$$\begin{align} \sum_{n=1}^{p}(-1)^{n-1}&\left({\beta(n)\over n}-\ln{n+1\over n}\right)\\ &= \sum_{n=1}^{p}\left(-1\right)^{n-1}\left(\frac{1}{n}\sum _{k=0}^{\infty}\frac{\left(-1\right)^k}{\left(2k+1\right)^n}-\ln \left(\frac{n+1}{n}\right)\right)\\
&=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{n=1}^{p}(-1)^n\sum _{k=0}^{\infty}\frac{(-1)^k}{n\left(2k+1\right)^n}\\
&\stackrel{*}{=} \sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^k\sum _{n=1}^\infty\frac{\left(-1\right)^n}{n\left(2k+1\right)^n}\\
&\stackrel{**}{=}\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^n\ln \left(\frac{2k+1}{2k+2}\right)\\
&=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=1}^p\left(-1\right)^k\ln \left(\frac{2k+1}{2k+2}\right)+\ln(2)\\
&=\sum _{k=1}^p\left(-1\right)^k\left[\ln \left(\frac{k+1}{k}\right)-\ln \left(\frac{2k+1}{2k+2}\right)\right]+\ln(2)\\
&=\sum _{k=1}^p\left[\ln \left(\frac{2k+1}{2k}\right)-\ln \left(\frac{4k+1}{4k+2}\right)-\ln \left(\frac{2k}{2k-1}\right)+\ln \left(\frac{4k-1}{4k}\right)\right]+\ln(2)\\
&= \sum _{k=1}^p\left(\ln \left(2k+1\right)-2\ln \left(2k\right)-\ln \left(4k+1\right)+\ln \left(4k+2\right)+\ln \left(2k-1\right)+\ln \left(4k-1\right)-\ln \left(4k\right)\right)+\ln(2)\\
&=\ln \left(2\left(3/2\right)_p\right)-2\ln \left(\left(1\right)_p\right)-\ln \left(4\left(5/4\right)_p\right)+\ln \left(16\left(3/2\right)_p\right)+\ln \left(\frac{1}{2}\left(1/2\right)_p\right)+\ln \left(\frac{1}{4}\left(3/4\right)_p\right)-\ln \left(\left(1\right)_p\right)+\ln(2)\\
&= \ln \left(\frac{2\cdot \left(1/2\right)_p\left(3/4\right)_p\left(3/2\right)_p\left(3/2\right)_p}{\left(1\right)_p\left(1\right)_p\left(1\right)_p\left(5/4\right)_p}\right)\\
&\stackrel{***}{=} \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)+\ln \left(\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right)
\end{align}$$
Luckily, as can be checked using Stirling's Formula, the second logarithm goes to $0$ as $p$ goes to infinity, i.e.
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right) = \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right) + \ln \left(\lim_{p \to \infty}\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right) = \color{red}{\ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)}$$
Footnotes:
$*\;\;\;\;\;\;$ Interchange double summations
$**\;\;\;\,$ Taylor Series for Natural Logarithm
$***\;$ Gamma representation for Pochhammer Symbol
| {
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How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
I have tried many different substitutions to no avail. Any help is appreciated.
| $$I = \int \frac{1}{\sin x + \cos x - 1}\, dx$$
Apply integral Subtitution
$$u = \tan\left(\frac{x}{2}\right)$$
$$\sin(x) = \frac{2u}{1+u^2},\quad \cos(x) = \frac{1 - u^2}{1 + u^2}\, \quad dx = \frac{2}{1 + u^2}$$
It follows
$$\int\frac{1}{u(-u + 1)}\, du = -\int\frac{1}{u(u - 1)}\, du $$
by partial fraction
$$-\int\left(\frac{1}{(u - 1)}- \frac{1}{u}\right)\, du $$
$$-\left[\int\frac{1}{(u - 1)}\, du -\int \frac{1}{u}\, du\right] $$
$\int \frac{1}{u}\, du = \ln(u)+C $
$\int\frac{1}{(u - 1)}\, du = \ln(u - 1)+C$
$$-\left[\ln(u - 1) - \ln(u)\right]+C $$
$u = \tan\left(\frac{x}{2}\right)$
$$-\ln\big(\tan\left(\frac{x}{2}\right) - 1\big) + \ln\big(\tan\left(\frac{x}{2}\right)\big)+C $$
$$I= \ln\big(\tan\left(\frac{x}{2}\right)\big)-\ln\big(\tan\left(\frac{x}{2}\right) - 1\big) +C $$
| {
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"url": "https://math.stackexchange.com/questions/2338996",
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$1-x+x^2-x^3+..(-1)^nx^n$ I have the following sum:
$1-x+x^2-x^3+..(-1)^nx^n, x\neq -1$
So what I thought was separating it in two cases like this:
Case 1. n is even
$$
1+x^2+x^4+...+x^n-x(1+x^2+...+x^n)
$$
Which I can turn into $\frac{1-x^{n+2}}{1-x^2}-\frac{x(1-x^{n+2})}{1-x^2}=\frac{1-x^{n+2}}{1-x^2}(1-x)$
Case 2. n is odd
$$
1+x^2+x^4+...+x^{n-1}-x(1+x^2+...+x^{n-1})=\frac{1-x^{n+1}}{1-x^2}-x(\frac{1-x^{n+1}}{1-x^2})
$$
My question is: Assuming what I've written is correct, which I'm not entirely sure, how can I combine the two cases for n even and odd into one equation?
| Call $S(x)$ your sum and note that
$$
S(x)(1+x)=1+(-1)^{n}x^{n+1}.
$$
| {
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} |
$| \cos (x) -\cos (y)| < \frac{y-x}{2}$ if $0\le x < y \le \pi/6$ If $0\leq{x<y<\frac{\pi}{6}}$
I am trying to prove that
$$| \cos (x) - \cos (y)| < \frac{y-x}{2}$$
Using the mean value theorem, defining $g(x)=\frac{y-x}{2}$ and $f(x)=| \cos (x) - \cos (y)|$. Then if show that $h(x)=g(x)-f(x)>0$, I would have proved it.
| If $0<y<x<\frac {\pi}{6}$
then $\cos y>\cos x$ since cosine is a decreasing function in this interval.
$|\cos x - \cos y| = -\cos x + \cos y$
Proposition:
$\frac {(x-y)}{2} > - \cos x + \cos y\\
\frac 12 > \frac {- \cos x + \cos y}{x-y}$
Mean value theorem: If $f(x)$ is differentiable in the interval $[x,y]$ there exists a $c \in (x,y)$ such that $f'(c) = \frac {f(x) - f(y)}{x-y}$
$f(x) = -\cos x\\
f'(x) = \sin x$
$\frac 12 >\sin c> 0\\
\forall c\in(x,y), \frac 12 > \sin c> 0\\
\frac 12 > \frac {(-\cos x) - (-\cos y)}{x-y}>0$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate divergence of the given function? The vector function is:
$$\mathbf{v}=\frac{1}{r^2}\hat{\mathbf{r}}$$
$r$ is the magnitude of position vector and
$\hat{\mathbf{r}}$ is the unit vector along the position vector
Now divergence will be
$$\nabla \cdot \mathbf{v}={\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)}\cdot \mathbf{v}$$
How is this evaluated?
| Without switching coordinate systems, this is my favorite method, since it breaks down the identity into small pieces.
Let $\mathbf{r} = x\mathbf{i} + y \mathbf{j} + z \mathbf{k}$, and $r = \sqrt{x^2 + y^2 + z^2}$. Notice that
\begin{align*}
\mathbf{v} &= \frac{\mathbf{r}}{r^3}\\
\mathbf{r}\cdot\mathbf{r} &= r^2 \\
\nabla r &= \frac{\mathbf{r}}{r} \\
\nabla \cdot \mathbf{r} &= 3 \\
\end{align*}
We can use the product rule for the divergence, and the power rule for the gradient:
\begin{align*}
\nabla \cdot \mathbf{v} &= \nabla\cdot(r^{-3} \mathbf{r}) \\
&= (\nabla r^{-3}) \cdot \mathbf{r} + r^{-3} \nabla \cdot \mathbf{r} \\
&= (-3)r^{-4}\nabla r \cdot \mathbf{r} + 3 r^{-3} \\
&= (-3)r^{-4}(r^{-1}\mathbf{r})\cdot \mathbf{r} + 3 r^{-3} \\
&= (-3)r^{-5}\mathbf{r}\cdot\mathbf{r} + 3r^{-3} \\
&= (-3)r^{-5}r^2 + 3r^{-3} \\
&= (-3)r^{-3} + 3r^{-3} = 0
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is it possible factor out $(x - b)$ from $(x^n-b^n)$ when $n$ is a fraction? Factoring out $(x-b)$
But how to factor out $(x-1)$ when the polynomial has the power of a negative integer where $n=-4$
$x^{-4}-1$
And even worst how to perform the factorization when the polynomial is to the power of an fraction? $n=-4/7$
$x^{-4/7}-1$
| For your example :
You could call $x^{-1/7}=\frac{1}{x^{1/7}}=y$
So
$$y^4-1=(y-1)(y+1)(y^2+1)=\left(\frac{1}{x^{1/7}}-1\right)(y^3+y^2+y+1)$$
$$=-\frac{1}{x^{1/7}}\left(x^{1/7}-1\right)(y^3+y^2+y+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration by parts
I have done the 1st part of question and the answer I got is $\frac{5e^4 - 1}{32} $which I verified from the calculator too. But I am confused how to approach to the deducing part using previous result(since it is stated HENCE ). Any help is greatly appreciated.
| $$\int _{ 1 }^{ e }{ { x }^{ 3 }{ \left( \ln { x } \right) }^{ 2 } } dx=\frac { 1 }{ 4 } \int _{ 1 }^{ e }{ { \left( \ln { x } \right) }^{ 2 } } d\left( { x }^{ 4 } \right) ={ \frac { { x }^{ 4 }{ \left( \ln { x } \right) }^{ 2 } }{ 4 } }_{ 1 }^{ e }-\frac { 2 }{ 4 } \int _{ 1 }^{ e }{ { x }^{ 3 } } \ln { x } dx=\\ =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( \int _{ 1 }^{ e }{ \ln { x } } d\left( { x }^{ 4 } \right) \right) =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { { x }^{ 4 }\ln { x } }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ { x }^{ 3 }dx } \right) =\\ =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { e }^{ 4 }-{ \frac { { x }^{ 4 } }{ 4 } }_{ 1 }^{ e } \right) =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { e }^{ 4 }-\frac { { e }^{ 4 } }{ 4 } +\frac { 1 }{ 4 } \right) =\frac { 8{ e }^{ 4 }-4{ e }^{ 4 }+{ e }^{ 4 }-1 }{ 32 } =\frac { 5{ e }^{ 4 }-1 }{ 32 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Transform curve into canonical form and determine type I have the curve:
$$9x^2-6xy+y^2+6x-2y-3=0$$
I have to transform it into the nnormal form and determine its type.
Well, we all know that such a curves are given by equation:
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F = 0$$
Which is purely the case.
$\textbf{Thoughts:}$
1.) I determined if given curve is central or not:
$$\Delta=\begin{vmatrix} A&B\\B&C \end{vmatrix} = \begin{vmatrix}9&-3\\-3&1 \end{vmatrix} = 9 -9=0$$
Therefore, the curve is not central.
2.) Now have to find an angle of transformation, by the formula:
$$\tan^2(\varphi)+\frac{a_{11}-a_{22}}{a_{12}}\tan(\varphi)-1=0$$
Where:
$$\begin{array}{|c|c|c|}\hline a_{11}&a_{12}&a_{22}\\ \hline 9 & -3 & 1 \\ \hline \end{array}$$
Therefore:
$$\tan^2(\varphi)-\frac{8}{3}\tan(\varphi)-1=0$$
or
$$3\tan^2(\varphi)-8\tan(\varphi)-3=0$$
which is square equation and substitute $t=\tan(\varphi)$ we get:
$$3t^2-8t-1=0$$
And roots are $t_1=3$ and $t_2=-\frac{1}{3}$ (which is strange for tangent, in my opinion).
3.) Find transformation, and appropriate $\sin x$ and $\cos x$
It is well known that there are transformations:
$$\begin{cases} x = x_1\cos(\varphi)-y_1\sin(\varphi) \\ y= x_1\sin(\varphi)+y_1\cos(\varphi) \end{cases} \tag{1}$$
So for $\cos (\varphi)$: $$\cos(\varphi)=\frac{1}{\sqrt{1+\tan^2(\varphi)}}$$
And for $\sin (\varphi)$: $$\sin(\varphi) = \frac{\tan(\varphi)}{\sqrt{1+\tan^2(\varphi)}}$$
As it does not matter which root to apply here I take $3$ and get:
$$\cos(\varphi) = \frac{1}{\sqrt{10}}$$
and
$$\sin(\varphi) = \frac{3}{\sqrt{10}}$$
or, if substitute results into $(1)$ we get:
$$\begin{cases}x = x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}} \\ y = x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}} \end{cases} \tag{2}$$
4.) System $(2)$ shows us new coordinates, which we should substitute into original equarion, and have:
$$9\left(x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right)^2-6\left(x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right)\cdot \left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}} \right) + \left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}}\right)^2 + 6\left( x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right) - 2\left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}}\right) - 3 = 0 \tag{3}$$
And I am totally stuck here
| Hint:
Use Gauß' algorithm to write this polynomial as a sum of squares:
\begin{align}
9x^2-6xy+y^2+6x-2y-3&=(9x^2-6xy+6x)+y^2-2y-3\\
&=\bigl[(3x-y+1)^2-y^2+2y-1\bigr]+y^2-2y-3\\
&=(3x-y+1)^2-4\\
&=(3x-y-1)(3x-y+3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge? Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge ? If then find the limit . Also the order of convergence $ \eta $.
Answer: \begin{align} b_{0 }=2 \\ b_{1}=2+\frac{1}{2} \\ b_{2}=2+\frac{1}{2+\frac{1}{2}} \\ b_{3}=2+\frac{1}{2+ \frac{1}{2+\frac{1}{2}}}\\ ..... \end{align} $ Actually \ this \ sequence \ is \ a \ continued \ fraction. $ So the sequence is convergent.
Let $ \lim_{n \rightarrow \infty } b_{n}=b $ , then taking the limit of both sides of $ b_{n}=2+\frac{1}{b_{n-1}} $ , we get
$ a=2+\frac{1}{a} $
or, $ a^{2}-2a-1=0 $ ,
or, $ a=\frac{2 \pm 2 \sqrt 2}{2}=1 \pm \sqrt 2 $ . But $ a \nless 0 $ .
So $ a=1+\sqrt 2=2.414 $.
Am I true ? Any help is there ?
| It's obvious that $b_n>2$.
Thus, $$|b_n-1-\sqrt2|=\left|\frac{1}{b_{n-1}}-\frac{1}{1+\sqrt2}\right|=|\frac{|b_{n-1}-1-\sqrt2|}{(1+\sqrt2)b_{n-1}}<\frac{|b_{n-1}-1-\sqrt2|}{2(1+\sqrt2)},$$
which says that $\{b_n\}$ converges to $1+\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A double-asymptote function I am facing a calculus problem which returns me a function which looks like this:
It seems to me, that it behaves half like a logarithm, half like an exponential. It is, of course, neither of both. It is different.
Some conditions are:
*
*$f\left(\frac{1}{2e}\right) = \frac{1}{2e}$
*$f(0) = W\left(\frac{1}{e}\right)$ and, by symmetry, $f\left(W\left(\frac{1}{e}\right)\right) = 0$
Its asymptotes are established as $y = \frac{1}{e}$ and $x = \frac{1}{e}$, so it looks symmetrical in respect of $f(x) = x$. Does anyone know which algebraic expression $f(x)$ correspond to this graph?
Preferably not a piecewise function, but just a single algebraic expression.
| Here is a possible answer for your question.
As Lambert W function is used in the question, it inspired me to think of the following algebraic expression as the form:
$$(a-bx)^{a-cy}=(a-by)^{a-cx},\tag{1}$$
where $a>e$ and $b,c>0$.
The above is the graph of the expression by substituting $a=\frac 1 e+e$ and $b=c=e$. It is symmetrical. The asymptotes are $x\approx0.7$ and $y\approx0.7$.
The shape is similar to yours if the straight line $y=x$ is removed.
By substituting the correct values of $a$, $b$ and $c$ and assuming $x\neq y$ when $x$ and $y \neq \frac 1 {2e}$, the expression might be found.
UPDATE-1
The values I calculated are $a\approx 4.26480$, $b\approx 8.87465$ and $c\approx 9.14674$.
The graph satisfies all constraints.
UPDATE-2
Condition A:
$ \left\{
\begin{aligned}
x = & 0\\
y = & W\left(\frac 1 e\right) \\
\end{aligned}
\right.$
Condition B: $ \left\{
\begin{aligned}
x = \frac 1 {2e}\\
y = \frac 1 {2e} \\
\end{aligned}
\right.$
Condition C: $y\to \frac 1 e$ as $x\to -\infty$.
We try to use the above 3 conditions to find the values of $a$, $b$ and $c$.
First we use condition C to prove
$$b=e(a-1).\tag{2}$$
Since $y\to \frac 1 e$ as $x\to -\infty$, we assume $\lim_{x\to -\infty}y=\frac 1 e$.
Claim:
$$y=\frac a b+\frac {W_k\left(-Ke^{\left(\frac {ab} c -a\right)K}\right)}{bK},\tag{3}$$
where $K=\frac{c\log(a-bx)}{b(a-cx)}.$
Take $\log$ on both sides in $(1)$,
\begin{align*}
(a-cy)\log(a-bx)&=(a-cx)\log(a-by)\\
(\frac {ab} c-by)\cdot \frac c b \log(a-bx)&=(a-cx)\log(a-by)\\
(\frac {ab} c-by)\cdot \frac{c\log(a-bx)}{b(a-cx)}&=\log(a-by).
\end{align*}
Let $K=\frac{c\log(a-bx)}{b(a-cx)}$,
\begin{align*}
\left(\frac {ab} c-by\right)K&=\log(a-by)\\
e^{\left(\frac {ab} c-by\right)K}&=a-by\tag{$\text{i}$}\\
e^{\frac {ab} c \cdot K}\cdot e^{-Kby}&=a-by\\
-Ke^{\frac {ab} c \cdot K}&=-K(a-by)e^{Kby}\tag{$\text{ii}$}\\
-Ke^{\frac {ab} c \cdot K}\cdot e^{-Ka}&=(Kby-Ka)e^{Kby-Ka}\tag{$\text{iii}$}\\
\therefore y&=\frac a b+\frac {W_k\left(-Ke^{\left(\frac {ab} c -a\right)K}\right)}{bK}.\tag{$\text{iv}$}
\end{align*}
Comment:
*
*In $(\text{i})$ on the R.H.S. we use the formula $e^{\log X}=X$.
*In $(\text{ii})$ we times $-Ke^{Kby}$ on both sides.
*In $(\text{iii})$ we times $e^{-Ka}$ on both sides.
*In $(\text{iv})$ we use the definition of W-function $Xe^X=Y\Leftrightarrow X=W_k(Y)$.
Note that
$$\lim_{x\to -\infty} K=\lim_{x\to -\infty} \frac{c\log(a-bx)}{b(a-cx)}=0,$$
Therefore, take the principle branch of $W$, set $k=0$,
$$\lim_{x\to -\infty} y=\lim_{K\to 0}\frac a b+\frac {W_0\left(-Ke^{\left(\frac {ab} c -a\right)K}\right)}{bK}=\frac{a-1}b=\frac 1 e.\tag{$\text{v}$}$$
$$\therefore b=e(a-1).$$
Comment:
*
*In $(\text{v})$ we use $e^{\left(\frac {ab} c -a\right)K}\sim 1$ and $W_0(-K)\sim -K$ as $K\to 0$.
Next we use condition B to prove
$$ c=\frac{2ea}{1+\frac{a+1}{a-1}\log\left(\frac {a+1} {2}\right)}.\tag{4}$$
Since it is symmetrical in respect of $y=x$, it's obvious that $y'|_{x=y=\frac 1 {2e}}=-1$, that is, the slope of tangent at $\left(\frac 1 {2e},\frac 1 {2e}\right)=-1$.
For $(1)$, differentiate both sides with respect to $x$,
$\left(\frac {-b(a-cy)}{a-bx}-cy'\log(a-bx)\right)(a-bx)^{a-cy}=\left(\frac {-b(a-cx)}{a-by}y'-c\log(a-by)\right)(a-by)^{a-cx}.\tag{5}$
By putting $b=e(a-1)$, $x=y=\frac 1 {2e}$ and $y'|_{x=y=\frac 1 {2e}}=-1$ into $(5)$,
$\require{cancel} \left(\frac {-e(a-1)(a-\frac c {2e})}{a-\frac {e(a-1)} {2e}}+c\log(a-\frac {e(a-1)} {2e})\right)\bcancel{(a-\frac {e(a-1)} {2e})^{a-\frac c {2e}}}=\left(-\frac {-e(a-1)(a-\frac c {2e})}{a-\frac {e(a-1)} {2e}}-c\log(a-\frac {e(a-1)} {2e})\right)\bcancel{(a-\frac {e(a-1)} {2e})^{a-\frac c {2e}}},$
$$\frac {-(a-1)(2e\cdot a- c )}{a+1}+c\log\left(\frac {a+1} {2}\right)=\frac {(a-1)(2e\cdot a- c)}{a+1}-c\log\left(\frac {a+1} {2}\right)$$
$$\therefore c=\frac{2ea}{1+\frac{a+1}{a-1}\log\left(\frac {a+1} {2}\right)}.$$
Then we use condition A to prove
$$c=\frac 1 {W\left(\frac 1e \right)}\left(a-\frac{a\log\left(a-e(a-1)W\left(\frac 1e \right)\right)}{\log a}\right).\tag{6}$$
Substitute $x=0$, $y=W\left(\frac 1 e\right)$ and $(2)$ into $(1)$, we get
$$a^{a-c W\left(\frac 1 e\right)}=\left(a-e(a-1)W\left(\frac 1 e\right)\right)^a$$
$$\left(a-c W\left(\frac 1 e\right)\right)\log a=a\log\left(a-e(a-1)W\left(\frac 1 e\right)\right)\tag{$\text{vi}$}$$
$$a-c W\left(\frac 1 e\right)=\frac {a\log\left(a-e(a-1)W\left(\frac 1 e\right)\right)}{\log a}$$
$$\therefore c=\frac 1 {W\left(\frac 1e \right)}\left(a-\frac{a\log\left(a-e(a-1)W\left(\frac 1e \right)\right)}{\log a}\right).$$
Comment:
*
*In $(\text{vi})$ we take $\log$ on both sides.
Finally, put $(4)$ and $(6)$ together,
$$\frac{2ea}{1+\frac{a+1}{a-1}\log\left(\frac {a+1} {2}\right)}=\frac 1 {W\left(\frac 1e \right)}\left(a-\frac{a\log\left(a-e(a-1)W\left(\frac 1e \right)\right)}{\log a}\right).\tag{7}$$
Solve $a$, $a\approx 4.26480$,
$b\approx 8.87465$ and
$c\approx 9.14674$.
| {
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"url": "https://math.stackexchange.com/questions/2345224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\sum\limits_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \Delta D$? In a book on economics, I have read that $\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \frac{r\Delta D}{r} = \Delta D$.
$t$ is the index of the current time period; $r$ represents the interest rate; $\Delta D$ is the change of debts.
Why can this equation be solved like this? Let’s say $r = 2$. Then, $\sum_{t = 1}^{\infty} \frac{2 \Delta D}{(1 + 2)^t}$ should result in something like $\frac{2\Delta D}{3}$, shouldn’t it?
What am I getting wrong here? Why does $\sum_{t = 1}^{\infty} \frac{r}{(1 + r)^t} = \frac{r}{r}$?
| The main theme here is the geometric series expansion
\begin{align*}
\sum_{t=0}^\infty q^t=\frac{1}{1-q}\qquad\qquad |q|<1
\end{align*}
Since OPs series starts with index $t=1$ we consider
\begin{align*}
\sum_{t=1}^\infty q^t&=\left(\sum_{t=0}^\infty q^t\right)-1\\
&=\frac{1}{1-q}-1\\
&=\frac{q}{1-q}\qquad\qquad |q|<1\tag{1}
\end{align*}
Using (1) we can transform OPs series
\begin{align*}
\sum_{t=1}^\infty\frac{1}{(1+r)^t}&=\sum_{t=1}^\infty\left(\frac{1}{1+r}\right)^t
=\frac{\frac{1}{1+r}}{1-\frac{1}{1+r}}\\
&=\frac{1}{r}\tag{2}
\end{align*}
valid for $\left|\frac{1}{1+r}\right|<1$.
Putting all together we conclude from (2) the following is valid for
$\left|\frac{1}{1+r}\right|<1$:
\begin{align*}
\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t}&=r\Delta D\cdot\sum_{t=1}^\infty \frac{1}{(1+r)^t}\\
&=r\Delta D\cdot\frac{1}{r}\\
&=\Delta D\tag{3}
\end{align*}
With respect to the example $r=2$: Since $\left|\frac{1}{1+2}\right|=\frac{1}{3}<1$ we obtain according to (2)
\begin{align*}
\sum_{t = 1}^{\infty} \frac{2\Delta D}{(1 + 2)^t}&=2\Delta D\cdot\sum_{t = 1}^{\infty}\left(\frac{1}{3}\right)^t\\
&=2\Delta D\cdot\frac{1}{2}\\
&=\Delta D
\end{align*}
in accordance with (3).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{x+12}{x-3} \gt |x-1| + 1$ Solve $\frac{x+12}{x-3} \gt |x-1| + 1$
I finally got $\frac{(x-6)(x+2)}{x-3} \lt 0$
That means the solutions are $3\lt x \le 6$ and $x\lt -2$ Am I right?
| at first we note that $$x\ne 3$$ and then we do case work:
$$x\geq 1$$ then we have $$\frac{x+12}{x-3}>x$$ then we have to solve
$$\frac{-x^2+4x+12}{x-3}>0$$
solving this we obatin $$3<x<6$$
in the second case we assume
2) $$x<1$$ and this is equivalent to
$$\frac{x+12}{x-3}>2-x$$
can you finish?
the last inequation is equivalent to
$$\frac{x^2-4x+18}{x-3}>0$$
this gives a contradiction
| {
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"url": "https://math.stackexchange.com/questions/2346157",
"timestamp": "2023-03-29T00:00:00",
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Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}
\qquad
g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}
$
has six elements. List them, then write the table of this group:
$\varepsilon = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 3 & 4 & 5\end{pmatrix}$
$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}$
$g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}$
$h = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \, \end{pmatrix}$
$k = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$l = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$
\begin{array}{c|ccc}
\circ & \varepsilon & f & g & h & k & l \\
\hline
\varepsilon \\ f \\ g \\ h \\ k \\ l
\end{array}
$
| You don't need to use a computer program for the computation. The two given maps act nontrivially on disjoint subsets of $\{1, \dots, 5\}$, so the map $\langle{f, g\rangle} \to \mathbb{Z}_2 \oplus \mathbb{Z}_3$ given by $f \to (1,0)$ and $g\to (0, 1)$ is a well-defined isomorphism.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding the tangent line that intersects a quartic at two points. I'm trying to find a straightforward calc solution to this question. I found an algebraic solution but I don't think it's the quickest way to do it. Thanks in advance!
Find the linear function $g(x)=mx+b$ whose graph is tangent to the graph of $f(x)=x^4-6x^3+13x^2-10x+7$ at two points.
Here's my algebraic solution: Let $(x_1,f(x_1))$ and $(x_2,f(x_2))$ be the points where $g(x)$ is tangent to $f(x)$. Let's form the function $h(x)=f(x)−g(x)$, so $h(x)=x^4−6x^3+13x^2−(10+m)x+7−b$. The zeros of $h(x)$ are $x=x_1$ and $x=x_2$and we also know that $h(x)≥0$ for all $x$. What's nice is that $h(x)$ can also be written as $h(x)=(x−x_1)^2(x−x_2)^2$. We can expand this to get $h(x)=x^4−2(x_1+x_2)x^3+(x^2_1+4x_1x_2+x^2_2)x^2−2(x^2_1x_2+x_1x^2_2)x+x^2_1x^2_2$. By comparing the two forms of $h(x)$ we can get that $x_1=1$ and $x_2=2$ (or vice versa) and therefore $m=2$ and $b=3$. I want to see a calc solution since this question appeared on a calc test.
| Let $h(x)=f(x)-g(x)$ be the distance between the functions. Note that $h(x)$ is a monic quartic polynomial. If the tangent line $g(x)$ intersects the graph of $f(x)$ at $x=c$ and $x=d$, then $h(x)$ must vanish $x=c$ and $x=d$, so the monomials $x-c$ and $x-d$ divide $h(x)$.
Additionally, since the derivatives of $f(x)$ and $g(x)$ match at $x=c$ and $x=d$, the derivative of $h'(x)$ must vanish there as well. This means $h(x)$ must be divisible by at least $(x-c)^2$ and $(x-d)^2$. The only such monic quartic is $h(x)=(x-c)^2(x-d)^2$.
Thus we are seeking $c$ and $d$ such that
$$(x-c)^2(x-d)^2=x^4-2(c+d)x^3+(c^2+d^2-4cd)x^2-2(cd^2+c^2d)x+c^2d^2\\=x^4-6x^3+13x^2-10x+7-(mx+b)$$
So from the cubic term we have $c+d=3$, and the quadratic term gives $c^2+d^2-4cd=13.$ Substitution yields a quadratic with solutions $c=1,2$. Then $12=m+10$ so $m=2$, and $7-b=4$ so $b=3.$ Thus $g(x) = 2x+3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of roots we have $$\sum_{cyc}(a-b)^2 = 2\left(\sum_{cyc} a^2-\sum_{cyc}{ab}\right)\\
= 2\left(\sum_{cyc} a\right)^2-6\left(\sum_{cyc} ab\right)\\
= -6p$$
This is coefficient of $x^2$ in the sought cubic.
But now computing coefficient of $x^2$ requires simplifying factors like $(a-b)^2\cdot(b-c)^2$ which becomes very lengthy.
Is there a shorter way around?
Thanks!
| HINT.-$$a+b+c=0\\ab+ac+bc=p\\abc=r$$ This gives
$$(a-b)^2+(b-c)^2+(c-a)^2=(c^2+4bc+4b^2)+(a^2+4ac+4c^2)+(b^2+4ab+4b^2)
=5(a^2+b^2+c^2)+4p$$
Since $a^2+b^2+c^2+2(ab+ac+bc)=0$ we have $$(a-b)^2+(b-c)^2+(c-a)^2=-10p+4p=-6p$$ Let $A$ and $B$ the other Vieta coefficients we need so$$X^3+6pX^2+AX+B=0$$ Taking into account what we desire, we have
$$(a-b)^6+6p(a-b)^4+A(a-b)^2+C=0\\(b-c)^6+6p(b-c)^4+A(b-c)^2+C=0$$ This a linear system in $A$ and $B$ whose solution is
$$A=\frac{(a-b)^6+6p(a-b)^4-(b-c)^6-6p(b-c)^4}{(a-b)^2-(b-c)^2}\\C=\frac{(b-c)^2[(a-b)^6+6p((a-b)^4]-(a-b)^2[(b-c)^6+6p(b-c)^4]}{(a-b)^2-(b-c)^2}$$ What remains to be simplified.The denominator is equal to $(a-c)(a-2b+c)$. The two numerators are polynomials of degree six for $A$ and degree eight for $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y+z)=\frac{20}{x}+\frac{30}{y}$$
$$\implies\frac{20y+30x}{xy}$$
$$\implies\frac{20*\frac{3z}{5}+30*\frac{2z}{5}}{\frac{2z}{5}\frac{3z}{5}}$$
Which gives me
$$\frac{100}{z}$$
Back to the 3rd equation!
| Simple, follow the equations you get:
$${y(x+y+z)\over x(x+y+z)}={30\over20} \implies {y\over x}={3\over2}\implies y={3\over2}x\\{z(x+y+z)\over x(x+y+z)}={50\over20} \implies {z\over x}={5\over2}\implies z={5\over 2}x\\x(x+y+z)=20\implies x\left(x+{3\over2}x+{5\over2}x\right)=20\implies{10\over 2}x^2=20\implies x^2=4\\\implies x=\pm2\\\implies y=\pm3\\\implies z=\pm5$$ all have to be of same sign in this case and you get :
$$x+y+z=\pm10\implies2(x+y+z)=\boxed{\pm20}$$
| {
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"url": "https://math.stackexchange.com/questions/2353505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find 3 vectors in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$ Find $3$ vectors $\vec{a},\vec{b},\vec{c}$ in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$
I know that the dot-product of two vectors in $\mathbb{R}^3$ is represented like this $(\vec{a},\vec{b})=|\vec{a}||\vec{b}|\cos(\angle\vec{a},\vec{b})$ so I've used random vector $\vec{a}(1,0,0)$ and added the following three conditions:
1.$(\vec{a},\vec{b})=|\vec{a}||\vec{b}|\cos(\frac{\pi}{3}) \Rightarrow 2b_1=b_1^2+b_2^2+b_3^2$
2.$(\vec{a},\vec{c})=|\vec{a}||\vec{c}|\cos(\frac{\pi}{3}) \Rightarrow 2c_1=c_1^2+c_2^2+c_3^2$
3.$(\vec{b},\vec{c})=|\vec{b}||\vec{c}|\cos(\frac{\pi}{3}) \Rightarrow 2(c_1b_1+c_2b_2+c_3b_3)=(b_1^2+b_2^2+b_3^2)(c_1^2+c_2^2+c_3^2) \Rightarrow b_1c_1=c_2b_2+c_3b_3$
And this is the point where I stuck... Any ideas how to solve this problem?
| Let see if we can find the most symmetric triple of vectors all in the positive octant. Let the $3$ unit vectors be
\begin{eqnarray*}
( \alpha, \alpha , \beta) , ( \alpha,\beta ,\alpha , )( \beta,\alpha, \alpha )
\end{eqnarray*}
We require that they are of unit length & their dot products are $1/2$. So we have
\begin{eqnarray*}
2\alpha^2+\beta^2=1 \\
\alpha^2+2 \alpha \beta= \frac{1}{2}.
\end{eqnarray*}
These are easily solved to give $ \alpha=\frac{1}{3 \sqrt{2}}$,$\beta=\frac{2 \sqrt{2}}{3}$.
So the three vectors $\color{red}{(\frac{1}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}},\frac{2 \sqrt{2}}{3}),(\frac{1}{3 \sqrt{2}},,\frac{2 \sqrt{2}}{3}\frac{1}{3 \sqrt{2}}),(\frac{2 \sqrt{2}}{3},\frac{1}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}})}$ will do.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Rotate $xy=1$ by $\frac{\pi}{4}$ in a negative (clockwise) direction. I was studying hyporbolae for the first time and noticed that $y=\frac{1}{x}$ is a rotated hyperbola. I had seen equations like $y=\frac{1}{x}$ before but never noticed they where hyperbolae.
Anyway using geometry and the general form of a hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, I realised that $xy=1$ is $\frac{x^2}{2}-\frac{y^2}{2}=1$ which has been rotated clockwise by $\frac{\pi}{4}$.
I tried to rotate $xy=1$ using the rotation matrix: $ \left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right)$
$$\left( \begin{array}{cc} \cos\theta & \sin\theta\\ \sin\theta & \cos\theta\end{array} \right) \left( \begin{array}{cc} x \\ \frac{1}{x}\end{array} \right)=\left( \begin{array}{cc} x'\\ y' \end{array} \right)$$
$$\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=x', -\frac{x}{\sqrt{2}}+\frac{1}{\sqrt{2}x}=y'$$
I used two methods. The first worked and the second did not. I'd like to know where I'm going wrong with the second approach.
In the first approach I squared both sides of the above equations and then used simultaneous equations to remove the $x$. Then dividing across by $2$ gives the desired result.
Just out of curiosity I tried another method. I first multiplied across both equations by $\sqrt{2}x$ and rearranged to quadratic form: $$x^2-\sqrt{2}x'x+1=0, x^2-\sqrt{2}y'x-1=0$$
I equated the solutions of each to get: $$\frac{\sqrt{2}x' \pm\sqrt{2x'^2-4}}{2}=\frac{-\sqrt{2}y' \pm\sqrt{2y'^2+4}}{2}$$
The $\pm$s are ugly but I figured I could get rid of them through a series of squaring. I tried a few different ways starting with roots on opposite sides, roots on the same side etc. but I can't quite get it through. Should this work or am I going down a dead-end road ?
Ok I got it: $$\frac{\sqrt{2}x \pm\sqrt{2x^2-4}}{2}=\frac{-\sqrt{2}y \pm\sqrt{2y^2+4}}{2}$$ simplifies to $$\frac{x^2}{2}-\frac{y^2}{2}=1$$
| I think its easier if we keep the variable $y$ around. Then
$$
\begin{pmatrix}
X\\
Y
\end{pmatrix}
=
\begin{pmatrix}
\cos(\pi/4) & -\sin(\pi/4)\\
\sin(\pi/4) & \cos(\pi/4)
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}
$$
so $X = \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{2} y$ and $Y = \frac{\sqrt{2}}{2} x + \frac{\sqrt{2}}{2} y$. Inverting this system, we have
$$
x = \frac{\sqrt{2}}{2} X + \frac{\sqrt{2}}{2} Y \qquad y = \frac{\sqrt{2}}{2} X - \frac{\sqrt{2}}{2} Y
$$
and substituting these expressions into the equation $xy = 1$, we find
\begin{align*}
1 = xy = \left(\frac{\sqrt{2}}{2} X + \frac{\sqrt{2}}{2} Y\right)\left(\frac{\sqrt{2}}{2} X - \frac{\sqrt{2}}{2} Y\right) = \frac{X^2}{2} - \frac{Y^2}{2} \, .
\end{align*}
Just as a note, in general, squaring equations can introduce spurious solutions because the squaring map $a \mapsto a^2$ is not one-to-one. The degree $4$ formula you are looking for is actually the equation of a "doubled" hyperbola, since
$$
x^4+y^4-4x^2+4y^2-2x^2y^2+4 = (x^2 - y^2 - 2)^2 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| At $x=2$ we have
$$y=\frac{4 c}{(2-a) (2-b) (2-c)}+\frac{2 b}{(2-a) (2-b)}+\frac{a}{2-a}+1=-\frac{8}{(a-2) (b-2) (c-2)}$$
Which is
$$y=\frac{8}{-a b c-4 (a+b+c)+2 (a b+a c+b c)+8}$$
In the given equation $x^3 -3x^2 + 2x +4 = 0$ we know the roots $a,b,c$ so we can write $(x-a)(x-b)(x-c)=0$ that is $$x^3-x^2 (a+b+c)+x (a b+a c+b c)-a b c=0$$
Thus, comparing with the equation's coefficient, we get
$$a+b+c=3;\;a b+a c+b c=2;\;abc=-4$$
Substitute in the expression for $y$ and get
$$y=\frac{8}{4-4 \cdot 3+2 \cdot 2+8}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need help in limits that contain arithmetic progression Let $a_n$ be the $n$-th term of an arithmetic progression with the initial term $a_1=2$ and with the common difference 5. That is, $a_n=2+5(n-1).$
Evaluate $$\lim_{n\rightarrow\infty}n(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}).$$
Well I consider that arithmetic progression here is increasing from 2 to infinity which makes me think that first term in the bracket is greater than the second one. So I think that answer should be positive infinity. But it is not. Can someone help me with it.
| $$\sqrt{a_n^2+3}-\sqrt{a_n^2-3} = (\sqrt{a_n^2+3}-\sqrt{a_n^2-3})\cdot\frac{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}} = \\ = \frac{a_n^2+3 - (a_n^2-3)}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}$$
Simplify this expression, and then also use the fact that
$$\sqrt{an^2 + bn + c} =n\cdot\sqrt{1 + \frac bn +\frac c{n^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.
| We can use partial fraction decomposition to write this as \begin{align}\sum_{n=1}^\infty \frac{8n}{4n^4+1}&=\sum_{n=1}^\infty\left(\frac{2}{2 n^2 - 2 n + 1} - \frac{2}{2 n^2 + 2 n + 1}\right)\end{align}
If we start writing out terms of this sequence, we get \begin{array}{ccccccccccc}(n=1)&2&-&\frac{2}{5}\\
(n=2)&&+&\frac 25&-&\frac 2{13}\\
(n=3)&&&&+&\frac 2{13}&-&\frac 2{25}\\
&&&&&&+&\cdots\\
(n=\infty)&&&&&&&+&\frac{2}{2\infty^2-2\infty+1}&-&\frac{2}{2\infty^2+2\infty+1}\end{array}
We therefore have a telescoping sequence, so the sum is:
$$\sum_{n=0}^\infty \frac{8n}{4n^4+1}=2-\frac{2}{2\infty^2+2\infty+1}$$
We can see that, $$\lim_{n\to\infty}\dfrac{2}{2n^2+2n+1}=0$$
Therefore, $$\sum_{n=1}^\infty\frac{8n}{4n^4+1} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
An expression involving the roots of a quadratic polynomial I have the equation $$ 9x^2 - 11x + 1 = 0 $$ whose two roots are $ \alpha $ and $ \beta $ .
I need to evaluate $$ \frac 1 {(9\alpha-11)^2} + \frac{11\beta - 1} 9$$
What I've tried
*
*Expanded the denominator and add them, but nothing simplifies and I get even a complex expression.
*I found that $ \alpha + \beta = \frac{11}{9}
\implies \alpha = \frac{11}{9} - \beta $
How to evaluate the value of the expression in an easier way?
| Just to show another way:
$$
\eqalign{
& 9\alpha ^{\,2} - 11\alpha + 1 = 0\quad \Rightarrow \quad \alpha \left( {9\alpha - 11} \right) = - 1\quad \Rightarrow \quad \left( {9\alpha - 11} \right) = - 1/\alpha \cr
& 9\beta ^{\,2} - 11\beta + 1 = 0\quad \Rightarrow \quad 11\beta - 1 = 9\beta ^{\,2} \cr
& 9\left( {\alpha ^{\,2} + \beta ^{\,2} } \right) - 11\left( {\alpha + \beta } \right) + 2 = 0\quad \Rightarrow \quad 9\left( {\alpha ^{\,2} + \beta ^{\,2} } \right) = 11\left( {\alpha + \beta } \right) - 2 = 11\left( {{{11} \over 9}} \right) - 2 = {{103} \over 9} \cr
& {1 \over {\left( {9\alpha - 11} \right)^{\,2} }} + {{11\beta - 1} \over 9} = \left( { - \alpha } \right)^{\,2} + \beta ^{\,2} = 103/81 \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Transition from exponential form to logarithmic form in an inequality This is my first question on math stack exchange and I am happy to join the community!
I would like to ask if someone could explain me in detail, how this transition is made (because I could not find any reference in the book):
$$\left(1-\frac{2}{n}\right)^k\le\frac{1}{3}$$ to $$k\ge\frac{\ln\frac{1}{3}}{\ln\left(1-\frac{2}{n}\right)}$$
Thank you very much in advance.
| It may be more complicated, as there are counter-examples to your statement, such as
*
*$n=1,k=3$ in which case $\left(1-\frac{2}{n}\right)^k = -1 \lt \frac13$
*$n=2, k=2$ in which case $\left(1-\frac{2}{n}\right)^k = 0 \lt \frac13$
*$n=-1, k=-2$ in which case $\left(1-\frac{2}{n}\right)^k = \frac19 \lt \frac13$
where $k\ge\dfrac{\log(\frac{1}{3})}{\log(1-\frac{2}{n})}$ is either meaningless or wrong.
Instead try something like
*
*If $n = 2$ then $\left(1-\frac{2}{n}\right) = 0 $ and $0^k \le \frac13$ is true for all $k > 0$
*If $n= 0$ then $\frac{2}{n}$ is not a real number
Otherwise take logarithms to get
$$k\log\left(1-\frac{2}{n}\right)\le\log\left(\frac{1}{3}\right)$$
*
*If $n \lt 0$ then $\left(1-\frac{2}{n}\right) > 1 $ so $\log\left(1-\frac{2}{n}\right) > 0$ so dividing both sides by $\log\left(1-\frac{2}{n}\right)$ gives $k \le \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(1-\frac{2}{n}\right)}$
*If $n\gt 2$, and $0 \lt 1-\frac{2}{n} \lt 1$ so $\log\left(1-\frac{2}{n}\right)$ is negative so dividing both sides by $\log\left(1-\frac{2}{n}\right)$ gives
$k \ge \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(1-\frac{2}{n}\right)}$
That leaves the case where $0 < n <2$ so $\left(1-\frac{2}{n}\right) < 0 $ and you have issues with powers of a negative number.
*
*If $k$ is an odd integer and $0 < n <2$ then $\left(1-\frac{2}{n}\right)^k$ is negative so always less than $\frac13$
There might be also be a solution with $k$ an even integer (let's say $2m$) in which case you are also trying to solve $\left(\frac{2}{n} -1\right)^k \le \frac13$. It will give similar solutions, restricted to $k$ being an even integer:
*
*if $n=1$ then $(-1)^{2m} =1$ so never less than or equal to $\frac13$
*if $1 < n<2$ then $k \le \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(\frac{2}{n}-1\right)}$
*if $0 < n<1$ then $k \gt \dfrac{\log\left(\frac{1}{3}\right)}{\log\left(\frac{2}{n}-1\right)}$
Incidentally, $\log\left(\frac{1}{3}\right) = -\log\left(3\right)$, which might simplify the notation
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| This sure looks like a Lagrange multiplier problem to me. Let $f(x,y) = 3x^2-2xy$ and $g(x,y) = \frac{x^2}{4}-y^2.$ Then $\nabla f = (6x-2y,-2x)$ and $\nabla g = (\frac{x}{2}, -2y)$. We get the system of 3 equations in three variables:
$$6x-2y = \lambda \frac{x}{2}$$
$$-2x = \lambda (-2y)$$
$$\frac{x^2}{4}-y^2 = 1$$
After some algebraic juggling, there are $8$ solutions:
$$x= \pm\frac{1}{3}\sqrt{18\pm 6\sqrt{10}}, y= \mp \frac{1}{6}\sqrt{18\pm 6\sqrt{10}}(-3\pm\sqrt{10}),$$
where there are $4$ choices for $x$ and for each one, two corresponding choices for $y$.
If all $8$ points are plugged into $f$, there are $4$ results: $$ \pm 2\sqrt{10}\pm\frac{20}{3},$$
and the max and min are obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.
Where, what, and why is wrong with my solution?
| First the domain of the inequation is $(-\infty,-4/3]\cup[4/3,+\infty)$.
Next, you should know that, when $A\ge 0$,
$$\sqrt A>B\iff A >B^2\quad\text{or}\quad B<0.$$
In the present case, one gets
\begin{cases}
9x^2-16>(3x+1)^2\iff x<-\dfrac{17}6 \\
\quad \text{or}\\
x<-\dfrac13.
\end{cases}
Both conditions yield $x\in (-\infty,-1/3)$. Taking the domain of validity of the inequation into account, one obtains
$$x\le-\frac43.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of square root of $n$ primes as a nested square root $$\sqrt{2}+\sqrt{3}=\sqrt{5+\sqrt{24}}$$
$$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$
These are two examples of how sum of k square root of primes (not necessarily consecutive) can be represented as nested square roots.
Is there any way to find all the terms of the nested roots given a set of primes? How do I compute them, individually?
i.e given $$A = [2, 3, 5]$$ representing $$\sqrt{2}+\sqrt{3}+\sqrt{5}$$
how do I get $$B = [14,140, 4096, 8847360]$$ representing $$\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$
The above are just examples, how do I compute for an arbitrary number of primes? What if the primes are given as
$$A = [a_0, a_1, ..., a_n]$$ How can I compute the nested loop?
| The first it's just
$$\sqrt{p}+\sqrt{q}=\sqrt{p+q+2\sqrt{pq}}=\sqrt{p+q+\sqrt{4pq}}.$$
For the second and for the rest maybe the following would help.
$$\sqrt2+\sqrt3+\sqrt5=\sqrt{10+2(\sqrt{6}+\sqrt{10}+\sqrt{15})}=$$
$$=\sqrt{14+2(\sqrt{6}+\sqrt{10}+\sqrt{15}-2)}=$$
$$=\sqrt{14+2\sqrt{35+2\sqrt{10}+6\sqrt6}}=$$
$$=\sqrt{14+2\sqrt{35+2\sqrt{\left(\sqrt{10}+3\sqrt{6}\right)^2}}}=$$
$$=\sqrt{14+2\sqrt{35+4\sqrt{16+3\sqrt{15}}}}=$$
$$=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $ \sum_{i=1}^{b-1} \left\lfloor \frac{a}{b}i \right\rfloor = \sum_{j=1}^{a-1} \left\lfloor \frac{b}{a}j \right\rfloor $ I am trying to solve the following problem:
Let $a$ and $b$ be integers greater than one which have no common divisors. Prove that:
$$ \sum_{i=1}^{b-1} \left\lfloor \frac{a}{b}i \right\rfloor = \sum_{j=1}^{a-1} \left\lfloor \frac{b}{a}j \right\rfloor $$
and find the value of this common sum.
I would like to know if you have any tips about how to approach this problem.
Here is what I've tried until now:
By computing the sum for a few cases, I noticed that the equality seems to be valid even when $a$ and $b$ have common divisors. So, ignoring this constraint and without loss of generality, I consider $a>b$, $a = b + n$.
Then we can write:
$$ \sum_{i=1}^{b-1} \left\lfloor \frac{a}{b}i \right\rfloor = \frac{b(b-1)}{2} + \sum_{i=1}^{b-1} \left\lfloor \frac{a-b}{b}i \right\rfloor = \frac{b(b-1)}{2} + \sum_{i=1}^{b-1} \left\lfloor \frac{n}{b}i \right\rfloor$$
and
$$ \sum_{j=1}^{a-1} \left\lfloor \frac{b}{a}j \right\rfloor = \frac{a(a-1)}{2} - \sum_{j=1}^{a-1} \left\lceil \frac{a-b}{a}j \right\rceil = \frac{a(a-1)}{2} - \sum_{j=1}^{a-1} \left\lceil \frac{n}{a}j \right\rceil$$
I am trying to use induction on $n$ to prove the equality, but I don't know if it is the best way to solve it.
| Since $a$ and $b$ are coprime, the numbers $0,a,2a,\ldots,(b-1)a$ form a complete set of residues modulo $b.$ Hence, if $\{ia/b\}$ denotes the fractional part of $ia/b,$ and if $i\neq j$ are any two distinct elments of $\{0,1,\ldots,b-1\},$ then $b^{-1}\{ia/b\}\neq b^{-1}\{ja/b\},$ which implies that each $b^{-1}\{ia/b\}$ is a unique element of the set $\{0,1/b,2/b,\ldots,(b-1)/b\}.$ Hence $$\sum_{k=0}^{b-1}\left\lfloor\dfrac{ka}{b}\right\rfloor=\sum_{k=0}^{b-1}\dfrac{ka}{b}-\sum_{k=0}^{b-1}\left\{\dfrac{ka}{b}\right\}=\dfrac{a(b-1)}{2}-\dfrac{b-1}{2}=\dfrac{(a-1)(b-1)}{2}$$ because $\sum_{k=0}^{b-1}(ka)/b=(a/b)\sum_{k=0}^{b-1}k$ and $\sum_{k=0}^{b-1}\{ka/b\}=b^{-1}\sum_{k=0}^{b-1}k.$ Thus, by symmetry $$\sum_{k=0}^{b-1}\left\lfloor\dfrac{ka}{b}\right\rfloor=\dfrac{(a-1)(b-1)}{2}=\sum_{k=0}^{a-1}\left\lfloor\dfrac{kb}{a}\right\rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A generalized Ahmed's integral Let $\vec{A}:=(A_1,A_2,A_3)$ be a vector where all its components are positive real numbers.
In the context of this question An integral involving error functions and a Gaussian we came across a following integral.
\begin{equation}
I(\vec{A}) := A_1\int\limits_0^{A_3} \frac{ \arctan\left(\frac{A_2}{\sqrt{1+A_1^2 +\xi^2}}\right)}{(1+\xi^2)\sqrt{1+A_1^2+\xi^2}} d\xi
\end{equation}
By performing the following transformations, firstly by substituting $\xi = \sqrt{1+A_1^2} \tan(\theta)$and then by substituting $t=\tan(\theta/2)$ we brought the quantity being sought for to the following form:
\begin{eqnarray}
&&I(\vec{A}) = \arctan(\frac{A_1 A_3}{\sqrt{1+A_1^2+A_3^2}}) \arctan(\frac{A_2}{\sqrt{1+A_1^2+A_3^2}})+\\
&&4 A_2\sqrt{1+A_1^2}\int\limits_0^{B} \arctan(\frac{t}{\sqrt{1+A_1^2}-A_1}) \cdot \frac{ t}{A_2^2 (1-t^2)^2 + (1+A_1^2) (1+t^2)^2} dt-\\
&&4 A_2\sqrt{1+A_1^2}\int\limits_0^{B} \arctan(\frac{t}{\sqrt{1+A_1^2}+A_1}) \cdot \frac{ t}{A_2^2 (1-t^2)^2 + (1+A_1^2) (1+t^2)^2} dt
\end{eqnarray}
where $B:= (-\sqrt{1+A_1^2} + \sqrt{1+A_1^2+A_3^2})/A_3$.
Now the question is how do we complete the calculation? Is the result expressed through elementary functions only and if not what kind of special functions enter the result?
| Define $\phi:= \arccos(A_2/\sqrt{1+A_1^2+A_2^2})$ and $\alpha := \sqrt{1+A_1^2}-A_1$ and $\beta:=\sqrt{1+A_1^2}+A_1$ and
\begin{eqnarray}
&&{\mathcal F}^{(a,b)}(t):=\int \arctan(\frac{t}{a}) \frac{1}{t-b} dt = \log(t-b) \arctan(\frac{t}{a})\\
&&-\frac{1}{2 \imath} \left( \log(t-b) \left[ \log(\frac{t-\imath a}{b-\imath a}) - \log(\frac{t+\imath a}{b+\imath a})\right] + Li_2(\frac{b-t}{b-\imath a}) - Li_2(\frac{b-t}{b+\imath a})\right)
\end{eqnarray}
By using the following partial fraction decomposition :
\begin{eqnarray}
\frac{t}{A_2^2(1-t^2)^2 + (1+A_1^2) (1+t^2)^2} =
\frac{I}{8 A_2 \sqrt{1+A_1^2}} \left( -\frac{1}{t-e^{\imath\phi}} + \frac{1}{t-e^{-\imath\phi}}+\frac{1}{t+e^{-\imath\phi}}-\frac{1}{t+e^{\imath\phi}}\right)
\end{eqnarray}
and by integrating each fraction over $t$ using the antiderivative above we arrive at the following result:
\begin{eqnarray}
&&I(\vec{A})= \arctan\left( \frac{A_1 A_3}{\sqrt{1+A_1^2+A_3^2}}\right) + \arctan\left( \frac{A_2}{\sqrt{1+A_1^2+A_3^2}}\right)+\\
&&\left.\frac{\imath}{2} \left(
{\mathcal F}^{\alpha,+\exp(-\imath \phi))}(t) +
{\mathcal F}^{\alpha,-\exp(-\imath \phi))}(t) -
{\mathcal F}^{\alpha,-\exp(+\imath \phi))}(t) -
{\mathcal F}^{\alpha,+\exp(+\imath \phi))}(t)
\right)\right|_0^B -\\
&&\left.\frac{\imath}{2} \left(
{\mathcal F}^{\beta,+\exp(-\imath \phi))}(t) +
{\mathcal F}^{\beta,-\exp(-\imath \phi))}(t) -
{\mathcal F}^{\beta,-\exp(+\imath \phi))}(t) -
{\mathcal F}^{\beta,+\exp(+\imath \phi))}(t)
\right)\right|_0^B
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Functional equation: $ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) $
Exercise: Find all of functions $f: \mathbb{R} \to \mathbb{R}$ such that $$ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) \text , \forall x , y , z \in \mathbb R \text . $$
In fact, I solved this question as follows:
*
*First step: Calculate $ f ( 0 ) $. In this step, I obtained that $ f ( 0 ) = 0 $ or $ f ( 0 ) = - 1 $.
*Second step: Consider two cases of the first step. I determinated the solutions are $ f ( x ) = 0 $, $ f ( x ) = - 1 $ and $ f ( x ) = x $.
The problem is that this way is very long. Who has another way to solve my exercise?
| The case $ f ( 0 ) = 0 $ is solved by @TobEnrack. For the case $ f ( 0 ) = -1 $, let $ x = y = 0 $ in
$$ f \big( x + y ^ 2 + z ^ 3 \big) = f ( x ) + f ( y ) ^ 2 + f ( z ) ^ 3 \tag 0 \label 0 $$
and you get
$$ f \big( z ^ 3 \big) = f ( z ) ^ 3 \tag 1 \label 1 $$
Again, letting $ x = z = 0 $ in \eqref{0} you have
$$ f \big( y ^ 2 \big) = f ( y ) ^ 2 - 2 \tag 2 \label 2 $$
Now combining \eqref{1} and \eqref{2} you get
$$ f ( x ) ^ 6 - 2 = \big( f ( x ) ^ 3 \big) ^ 2 - 2 = f \big( x ^ 3 \big) ^ 2 - 2 = f \Big( \big( x ^ 3 \big) ^ 2 \Big) = f \big( x ^ 6 \big) \\ = f \Big( \big( x ^ 2 \big) ^ 3 \Big) = f \big( x ^ 2 \big) ^ 3 = \big( f ( x ) ^ 2 - 2 \big) ^ 3 = f ( x ) ^ 6 - 6 f ( x ) ^ 4 + 12 f ( x ) ^ 2 - 8 $$
By simple algebra, it's easy to derive
$$ f ( x ) ^ 2 = 1 \tag 3 \label 3 $$
Finally, letting $ z = 0 $ in \eqref{0} and using \eqref{3} you get
$$ f \big( x + y ^ 2 \big) = f ( x ) \tag 4 \label 4 $$
Hence, if $ x \le y $, substituting $ \sqrt { y - x } $ for y in \eqref{4}, it's easy to see that
$$ f ( y ) = f \big( x + ( y - x ) \big) = f \Big( x + { \sqrt { y - x } } ^ 2 \Big) = f ( x ) $$
Since for every $ x $, either $ x \le 0 $ or $ 0 \le x $, therefore $ f ( x ) = -1 $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find angles in a triangle, with two similar triangles with scale factor $\sqrt{3}$
Triangle ABC has point D on BC which creates triangle ABD and ACD. They differ with the scale factor $\sqrt{3}$. What are the angles?
I know ADB and ADC cannot be right, as it shares the side AD and cannot have a scale factor of $\sqrt{3}$.
I have tried to approach it with $\frac{AD}{DC}=\frac{BD}{AD}=\sqrt{3}$ (or $\frac{AD}{AC}=\frac{BA}{AD}=\sqrt{3}$). Meaning that $DC=x$ $AD=\sqrt{3}x$ $BD=3x$. This is where I get stuck. Because I dont know if the triangle is right angled, and cannot use Pythagoras. I dont have any angles, hence cannot use $\frac{sin(a)}{angleA}$. How do I solve this?
| A solution is given by dropping an altitude to the hypotenuse of a 30-60-90 right triangle. Is this the only one? To find out we define the six quantities $a = BC$, $b = CA$, $c = AB$, $d = AD$, $m = BD$, and $n = DC$. There are four ways in which the triangles $ABD$ and $ACD$ could be similar:
\begin{align}
1)& \;\; \frac{b}{c} = \frac{d}{m} = \frac{n}{d} = \sqrt{3},\\
2)& \;\; \frac{b}{m} = \frac{d}{c} = \frac{n}{d} = \sqrt{3},\\
3)& \;\; \frac{b}{d} = \frac{d}{c} = \frac{n}{m} = \sqrt{3},\\
4)& \;\; \frac{b}{d} = \frac{d}{m} = \frac{n}{c} = \sqrt{3}.\\
\end{align}
(The other two permutations have $d/d = \sqrt{3}$, so we can rule them out immediately.) We need to solve for six quantities. Each of the cases above reduces this to three. The relationship $a = m+n$ reduces it two. Setting one of the lengths to 1 reduces it one. So we only need one more relationship. For this we use Stewart's Theorem:
$$
man + dad = bmb + cnc.
$$
Cases 1 through 3 each have 1 solution with positive side lengths, and case 4 has 2 solutions. However, only in case 1 do the three resulting triangles all satisfy the triangle inequality. This solution is the 30-60-90 right triangle
$$
a = 4, \;\; b = 2\sqrt{3}, \;\; c = 2, \;\; d = \sqrt{3}, \;\; m = 1, \;\; n = 3.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $z=e^{i\theta}$ right after:
$\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=-i\cdot\int_1^{-1}\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2}dz$
This now gives me 4 new integrals, for example
$\displaystyle-i\int_1^{-1}\frac{z^2}{2z^2+5z+2}dz$, $\displaystyle-i\int_1^{-1}\frac{1}{2z^4+5z^3+2z^2}dz$ and so on.
But since I haven't been able to solve any of the new integrals, I'm a little lost.
Edit: Can't I do a partial fractions decomposition of all the 4 integrals and solve them seperately?
| Looks like there should be a partial fraction decomposition:
$\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2} = Az + B + \frac C{z} + \frac D{z^2} + \frac E{z^3} + \frac {F}{z+2} + \frac {G}{2z + 1}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
| An alternative solution ...
We have $n^4-n^2+64=A^2$. Multiply this by $4$ and complete the square $(2n^2-1)^2+255=4A^2$. So
\begin{eqnarray*}
(2A+2n^2-1)(2A-2n^2+1)= 3 \times 5 \times 17.
\end{eqnarray*}
This gives
\begin{eqnarray*}
2A+2n^2-1 = x \\
2A-2n^2+1= y
\end{eqnarray*}
There are $4$ possible values for $(x,y)$ ... $(15,17),(17,15),(51,5),(85,3),(255,1)$. These lead to $(A,n^2)$ having the values $(8,0),(8,1),(14,12),(22,21),(64,64)$. The first second and last will give valid answers $\color{red}{n=0}$, $\color{red}{n= \pm 1}$ and $\color{red}{n= \pm 8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Peculiar Sum regarding the Reciprocal Binomial Coefficients Whilst playing around on Wolfram Alpha, I typed in the sum
$$\sum_{x=0}^\infty \frac{1}{\binom{2x}{x}}=\frac{2}{27}(18+\pi\sqrt 3)$$
I'm not sure how to derive the answer. My first instinct was to expand the binomial coefficient to get
$$\sum_{x=0}^\infty \frac{x!^2}{(2x)!}$$
and then to try using a Taylor Series to get the answer. I thought that if I could find a function $f(n)$ with
$$f(n)=\sum_{x=0}^\infty \frac{x!^2n^x}{(2x)!}$$
Then my sum would be equal to $f(1)$. How do I find such a function?
EDIT: I continued on this path and realized that I can use this to set up a recurrence relation for $f^{(x)}(0)$:
$$f^{(0)}(0)=1$$
$$f^{(x)}(0)=\frac{x^2}{2x(2x-1)}f^{(x-1)}(0)$$
However, I'm not sure how this helps me find $f(1)$...
Am I on the right track? Can somebody help me finish what I started, or point me towards a better method of calculating this sum?
Thanks!
| \begin{eqnarray*}
\binom{2n}{n} ^{-1} = \frac{2n+1}{2^{2n+1}} \int_{-1}^{1} (1-x^2)^n dx
\end{eqnarray*}
Substitute this for summand and inerchange the order of the integral and sum.
\begin{eqnarray*}
\int_{-1}^{1} \sum_{n=0}^{ \infty} \frac{2n+1}{2^{2n+1}} (1-x^2)^n dx &=& \frac{1}{2} \int_{-1}^{1} \left(2\frac{(\frac{1-x^2}{4})}{(1-(\frac{1-x^2}{4}))^2}+ \frac{1}{1-(\frac{1-x^2}{4})} \right) dx \\
= \int_{-1}^{1} \frac{16}{(3+x^2)^2} dx - \int_{-1}^{1} \frac{2}{(3+x^2)} dx
\end{eqnarray*}
Now use the standard integrals
\begin{eqnarray*}
\int_{-1}^{1} \frac{1}{(3+x^2)} dx = \frac{ \pi}{3 \sqrt{3}} \\ \int_{-1}^{1} \frac{1}{(3+x^2)^2} dx = \frac{ 1}{12} + \frac{ \pi}{18 \sqrt{3}} \\
\end{eqnarray*}
and the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $\Biggl(\Biggl(\tan^2\frac{7\pi}{24}-\tan^2\frac{\pi}{24}\Biggr):\Biggl(1-\tan^2\frac{7\pi}{24}\tan^2\frac{\pi}{24}\Biggr)\Biggr)^2$ This problem reminds me of the formula $\tan(x\pm y)=\frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}.$ But two minus signs in the problem interfere in applying the formula. I can't think of other ways. How to apply this formula?
| Hint: $A^{2}-B^{2} = (A-B)(A+B)$
Further hint:
Use the hint to rewrite the expression as $\left(\frac{\tan{(7\pi/24)}-\tan{(\pi/24)}}{1+\tan{(7\pi/24)}\tan{(\pi/24)}}\cdot\frac{\tan{(7\pi/24)}+\tan{(\pi/24)}}{1-\tan{(7\pi/24)}\tan{(\pi/24)}}\right)^{2}$
Answer:
The addition formula for $\tan$ then gives you the answer: $\left(\tan{\frac{6\pi}{24}}\tan{\frac{8\pi}{24}}\right)^{2} = \left(\tan{\frac{\pi}{4}}\tan{\frac{\pi}{3}}\right)^{2} = \left(1\cdot\sqrt{3}\right)^{2} = 3$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\;\log_m mn + \log_n mn = 49$, find $(\log_m n)^{3/4} + (\log_n m)^{3/4}.$ Given that $\log_m mn + \log_n mn = 49,\;$ find the value of $$(\log_m n)^{3/4} + (\log_n m)^{3/4}.$$
I have rewritten this expression but have hit a dead end.
| Let $\sqrt[4]{\log_nm}=x$. Hence, $x>0$ and $$ x^4+\frac{1}{x^4}=47 \iff
\left(x^2+\frac{1}{x^2}\right)^2=49$$
$$\iff x^2+\frac{1}{x^2}=7$$
$$\iff\left(x+\frac{1}{x}\right)^2=9$$
$$\iff x+\frac{1}{x}=3.$$
Thus, $$\left(\log_m n\right)^{3/4} + \left(\log_n m\right)^{3/4}=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=27-9=18.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Region of Convergence for Laurent Expansion Find the Laurent expansion of $\frac{z}{(z+1)(z+2)}$ about the singularity $z=-2$. Specify the region of convergence and the nature of singularity at $z = -2$.
The Laurent expansion I get is
$1+(z+2)+(z+2)^2+ \ldots + \frac{2}{z+2}$
The singularity is a simple pole.
But how to find the Region of Convergence.
| The Laurent series can be obtained as
$$\frac{z}{(z+1)(z+2)} = \frac{1}{z+2}\frac{z+2 - 2}{z+2 - 1} \\ = \left(\frac{2}{z+2} - 1\right)\frac{1}{1 - (z+2)} \\ = \left(\frac{2}{z+2} - 1\right)\sum_{n=0}^\infty (z+2)^n$$
and the geometric series on the RHS converges when $|z+2| < 1$.
| {
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"url": "https://math.stackexchange.com/questions/2376267",
"timestamp": "2023-03-29T00:00:00",
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Prove $3^n$ divides any number formed from $3^n$ identical digits I know this is a repost of this, however I had some trouble understanding the answers as I am new to number theory. I asked my doubt in a comment but got no reply.
One of the answers says:
A number whose digits are all equal and of length $3^n$ is thus of the
form $c = \sum\limits_{i=0}^{3^n-1} a \cdot 10^i = a \dfrac{10^{3^n}-1}{10-1}$ by the geometric series.
Since we have to account for the possibility that $a = 1$ we need to
show that $3^n \mid \dfrac{10^{3^n}-1}9$, i.e. $3^{n+2} \mid 10^{3^n}-1$.
This we can do by induction: Base case is trivial, suppose $3^{n+2}\mid 10^{3^n}-1$.
Then we have $10^{3^n} \equiv 1 \pmod {3^{n+2}}$, and so: $10^{3^n}\equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$. Now compute $10^{3^{n+1}} = \left(10^{3^n}\right)^3$ modulo $3^{n+3}$.
I can't understand how can we write $10^{3^n} \equiv 1 \pmod {3^{n+2}}$ as $10^{3^n} \equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$. And how to proceed further to prove $10^{3^{n+1}} \equiv 1$ (mod $3^{n+3})$. I figured we could cube both sides of $10^{3^n} \equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$, however it leads to a lot of extra terms on right hand side of the equation.
| We have
$$\frac{10^{3^{n+1}}-1}9=\frac{a^3-1}9=\frac{10^{3^n}-1}9(a^2+a+1)$$
where $a=10^{3^n}$. The induction step boils down to confirming that
$3$ divides $a^2+a+1$. If we could prove that $a\equiv1\pmod 3$ we could
confirm that.
| {
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How do we know that these roots are non-real? The roots, x and y, of a quadratic equation $$ax^2 + bx + c=0$$ satisfy the following inequality:
$$x^2 + y^2 < 0$$
What can you conclude about the nature of the roots of the quadratic equation?
$$x^2 + y^2 = (x+y)^2 - 2xy = \frac{b^2 - 2ac}{a^2}$$
since $$x^2 + y^2 <0$$, we can say $$b^2 - 2ac < a^2$$
therefore, $$b^2 - 4ac < a(a-2c)$$
therefore,$$ D < a(a-2c)$$
How do I proceed from here?
| Alternatively:
$$x^2+y^2=\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2=\frac{b^2-4ac}{a^2}<0 \Rightarrow$$
$$b^2-4ac<0 \Rightarrow x,y\in \mathbb{C}.$$
| {
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Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by contradiction:
Assume $c$ is divisible by $3$ and $a$ or $b$ is not divisible by $3$.
Since $c$ is divisible by $3$ we can write $c$ as $ \ c = 3m \implies c^{2} = 9m^{2} \implies 9 | c^{2}$.
Since $a$ and $b$ are not divisible by $3$, $\ a = 3k+1$ and $ \ b = 3n+1$ for some integers $\ k,n.$
Then,
$ a^{2} + b^{2} = (3k+1)^{2} + (3n+1)^{2} = 9k^{2} + 6k +9n^{2} + 6n + 2$.
I am stuck here. I can't find a contradiction. How can I show that $ a^{2} + b^{2} $ is not divisible by $9$.
| We have $a^2+b^2=c^2\equiv 0\pmod 3$. Since $a^2,b^2\equiv 0$ or $1\pmod 3$ we must to have $a\equiv b\equiv 0\pmod 3$.
| {
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"source": "stackexchange",
"question_score": "2",
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} |
What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$
What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$, where $x,y,z$ are integers.
My Attempt
$(x^2 y + y^2 z + z^2 x) +(x^2 y + y^2 z + z^2 x) =2186+2188=4374$.
From this we can derive $(x+y+z)(xy+yz+zx)-3xyz=4374$ . We can subtract the given equations to get $xy(x-y)+yz(y-z)+zx(z-x)=-2$. But after this , I can't figure out how to proceed.
Any help is appreciated. Thanks in advance.
| Subtracted the first equation from the second:
$$\left(x^2 z+x y^2+y z^2\right)-\left(x^2 y+x z^2+y^2 z\right)=2$$
Simplify and factor:
$$-x^2 y+x^2 z+x y^2-x z^2-y^2 z+y z^2=(x-z)(-x y+x z+y^2-y z)= (x - z)(y - x) (y - z)$$
Because $x$, $y$ and $z$ are integers, $x-z$, $x-y$ and $z-y$ must be integers too, and the only integers possible are $\pm 1$ and $\pm 2$, so $x$, $y$ and $z$ must also be consecutive integers.
Because the equations are symmetrical in interchange of $x$, $y$, and $z$, without loss of generality, we can choose $x$ to be the largest.
This gives us $y-x=-2,\;x-z=1,\;y-z=-1$ that is $y = x-2 , \;z =x -1 $
Plug these results in one of the two equations $x^2 y+x z^2+y^2 z=2186$ to get
$$x^2 (x-2)+(x-1)^2 x+(x-1) (x-2)^2-2186=0$$
Expand and simplify
$$x^3-3 x^2+3 x-730=0$$
$$(x-1)^3 - 729 = 0$$
Since $729=9^3$, the solution is $x=10$.
Substituting in the other equations gives $y=8$ and $z=9$.
$x^2+y^2+z^2=10^2+8^2+9^2=245$
| {
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Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$... Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$. Find the number of six-ball combinations if balls of the same color are considered identical.
Using the Inclusion-Exclusion Principle and $\binom{n}{m}_{R}=\binom{n+m-1}{m}$ where m is the number of balls and n is the number of boxes (or in this case the length of the combination), I should he able to solve this problem.
However, I have not idea how to solving this problems. Can someone please help me? Thank you
| Generating Function Approach
$$
\begin{align}
\left[x^6\right]\left(1+x+x^2\right)^6
&=\left[x^6\right]\left(\frac{1-x^3}{1-x}\right)^6\\
&=\left[x^6\right]\sum_{k=0}^6(-1)^k\binom{6}{k}x^{3k}\sum_{j=0}^\infty(-1)^j\binom{-6}{j}x^j\\
&=\sum_{k=0}^2(-1)^k\binom{6}{k}(-1)^{6-3k}\binom{-6}{6-3k}\\
&=\sum_{k=0}^2(-1)^k\binom{6}{k}\binom{11-3k}{5}\\
&=\binom{6}{0}\binom{11}{5}-\binom{6}{1}\binom{8}{5}+\binom{6}{2}\binom{5}{5}\\[9pt]
&=462-336+15\\[15pt]
&=141
\end{align}
$$
Casing on the Number of Pairs
Choose $k$ colors to have two balls of each of those colors and one of the rest
$$
\begin{align}
\sum_{k=0}^3\overbrace{\ \ \ \binom{6}{k}\ \ \ }^{\substack{\text{number of}\\\text{ways to choose}\\\text{paired colors}}}\overbrace{\binom{6-k}{6-2k}}^{\substack{\text{number of}\\\text{ways to choose}\\\text{singleton colors}}}
&=\binom{6}{0}\binom{6}{6}+\binom{6}{1}\binom{5}{4}+\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{3}{0}\\
&=1+30+90+20\\[9pt]
&=141
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$?
Calculate
$\displaystyle\lim_{n \to \infty}
\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.
$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) =
\infty - \infty$ We have an indeterminate form
So I proceeded to factorize $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]$$
taking the limit:
$$\lim\limits_{n \rightarrow \infty} n \left[ \sqrt{\frac{n+1}{n}}-1 \right]= \infty \cdot 0$$
indeterminate again
What am i missing? How is the way forward to proceed? Much appreciated
| Trick is to write $\sqrt{n^2 + n} - n$ in a nice form.Below we use rationalization
$\lim _{n\rightarrow \infty}\sqrt{n^2 + n} - n = \lim_{n \rightarrow \infty} \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n \rightarrow \infty} \sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}}) = \lim_{n \rightarrow \infty} \frac{\sqrt{n}(1)}{\sqrt{n}(\sqrt{1+\frac{1}{n}} + 1)} = \frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c)$ I am stuck with the following problem.
If $a>0$, $b>0$, $c >0$ and not all equal then prove that:
$$b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c).$$
Additional info:I'm looking for solutions using AM-GM .
I don't know how to progress .
I will be grateful if someone explains . Thanks in advance ..
| From C-S:
$$\left(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge \left(a+b+c\right)^2 \Rightarrow $$
$$\left(a^2b^2+b^2c^2+c^2a^2\right)^2\ge a^2b^2c^2(a+b+c)^2.$$
equality occurs when $a=b=c.$
Taking square root and $a\ne b\ne c$ will do.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$.
Then recall that in evaluating a discontinuous integrand, the integral is defined by the relations $$\int_a ^b f(x) = lim_{x->b^-} \int_a ^x f(x) dx $$
if $x = b$ is the discontinuous point or $$\int_a ^b f(x) = lim_{x->a^+} \int_x ^b f(x) dx $$
if $x = a$ is the discontinuous point.
With that in mind:
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$
$$\int _0 ^3\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} + \int _3 ^4\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} $$
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} + \int _3 ^4\frac{dx}{(x-1)^2 -(2)^2} $$
Remembering that $\int \frac{du}{u^2 +a^2} = \frac{1}{a} \arctan \left( \frac{u}{a}\right) + C, $
we get:
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} +\int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$
equals
$$lim_{x->3^+} \int_0 ^x \frac{dx}{(x-1)^2 -(2)^2} + lim_{x->3^-} \int_x ^4 \frac{dx}{(x-1)^2 -(2)^2} $$
equals
$$lim_{x->3^+} \left( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_3 ^x + lim_{x->3^-} \left(\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_x ^4$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{3-2}{2}\right) \right ) + \left ( \frac{1}{2} \arctan \left( \frac{4-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)- 0.2318 \right ) + \left ( \frac{\pi}{8}-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
which makes $$\int _0 ^4\frac{dx}{x^2 - 2x - 3} = 0.1609$$
But in my book, it said there is no value of $\int _0 ^4\frac{dx}{x^2 - 2x - 3}.$
How do you prove that the integral $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ has no value?
| $$I=\int _0 ^4\frac{dx}{x^2 - 2x - 3} =\frac{1}{4} \int _0 ^4\left(\frac{1}{x - 3}- \frac{1}{x+1}\right)dx$$
As already pointed out, there is a singularity at $x=3$.
So, at the common sens, the integral isn't convergent. But this is not true in the sens of Cauchy integration (Cauchy Principal Value : http://mathworld.wolfram.com/CauchyPrincipalValue.html )
$$I=\frac{1}{4}\lim_{\epsilon\to 0} \left( \int _0 ^{3-\epsilon}\frac{dx}{x - 3} +\int _{3+\epsilon}^4\frac{dx}{x - 3} - \int _0^4\frac{dx}{x+1}\right)$$
$\int _0 ^{3-\epsilon}\frac{dx}{x - 3} +\int _{3+\epsilon}^4\frac{dx}{x - 3} = \left[\ln|x-3| \right]_{x=0}^{x=3-\epsilon} +\left[\ln|x-3| \right]_{x=3+\epsilon}^{x=4} = \ln(\epsilon)-\ln(3)+\ln(1)-\ln(\epsilon) = -\ln(3)$
$\int _0^4\frac{dx}{x+1}=\left[\ln|x+1| \right]_{x=0}^4=\ln(4+1)-\ln(1)=\ln(5)$
$$\text{PV}\int _0 ^4\frac{dx}{x^2 - 2x - 3} = \frac{1}{4}\left(-\ln(3)-\ln(5)\right)=-\frac{\ln(15)}{4}$$
The symbol PV means that the integral is considered on the sens of a Cauchy integral. Indeed, the integral as a finite value, the so called Principal Value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $P(a)=0 \Rightarrow P(a+1)=1$ then $P(x)$ has no repeated roots.
Let $P(x) \in \mathbb{R}[x]$ be polynomial with all real roots and has the property that $P(a)=0 \Rightarrow P(a+1)=1$ for all $a \in \mathbb{R}$. Prove that $P(x)$ has a repeated root.
I think this problem statement is not true because if we suppose that $P(x)=x$ then $P(0)=0 \Rightarrow P(1)=1$, $\;P(x)$ has no repeated root.
Please suggest.
| For $n>1$ we have $P(x)=(x-a)^n$ then $P(a+1)=(a+1-a)^n=1$
If $P(x)$ has no repeated root then it doesn't happen in general
Anyway carat, the original poster, is right: for $n=1$ it is false. It even makes no sense talking of multiple root for $n=1$
Suppose the root are $3$ then we have
$P(x)=(x - a) (x - b) (x - c)$
Use the condition that $P(a+1)=1;\;P(b+1)=1;\;P(c+1)=1$
$$
\left\{ \begin{gathered}
\left( {a + 1 - a} \right)\left( {a + 1 - b} \right)\left( {a + 1 - c} \right) = 1 \hfill \\
\left( {b + 1 - a} \right)\left( {b + 1 - b} \right)\left( {b + 1 - c} \right) = 1 \hfill \\
\left( {c + 1 - a} \right)\left( {c + 1 - b} \right)\left( {c + 1 - c} \right) = 1 \hfill \\
\end{gathered} \right.
$$
which simplifies to
$$\left\{ \begin{gathered}
\left( {a - b + 1} \right)\left( {a - c + 1} \right) = 1 \hfill \\
\left( {b - a + 1} \right)\left( {b - c + 1} \right) = 1 \hfill \\
\left( {c - a + 1} \right)\left( {c - b + 1} \right) = 1 \hfill \\
\end{gathered} \right.$$
which is verified when $a=b=c$
Therefore $x=a$ is a triple root. $P(x)=(x-a)^3$
In a similar way it can be proved for any degree
| {
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"timestamp": "2023-03-29T00:00:00",
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is there any analytical way to konw if $\frac{1}{2x}+\frac{x}{2} >1$ for $(1,\infty)$ or $(0,\infty)$? I was solving a physics problem and I got
$$\frac{a^2+b^2}{2ab}$$
or equivalents
$$\frac{1+ \left( \frac{b}{a}\right)^2}{2\cdot\frac{b}{a}}.$$
If $x=\frac{b}{a}$, we obtain
$$\frac{1+x^2}{2x}$$ or $$\frac{1}{2x}+\frac{x}{2}.$$
I need to know if this expresions are $>1$ when $a<b$ (or $x>1$).
I know the answer is yes (I used geogebra to plot) but I'd like to know if there is any analytical way to get the answer. Thank you.
| For $x > 0$ (and $x$ must be greater than zero):
$\frac {1}{2x} + \frac x2 > 1 \iff$
$\frac 1x + x > 2 \iff$
$1 + x^2 > 2x \iff $
$x^2 - 2x + 1 > 0 \iff$
$(x - 1)^2 > 0 \iff$
$x - 1 \ne 0 \iff$
$x \ne 1$.
That's it. If $x = 1$ then $\frac 1{2x}+ \frac x2 = 1$ other wise $\frac 1{2x}+ \frac x2 > 1$
If $x < 0$ well .... $\frac 1{2x} + \frac x2 < 0 < 1$ so duh... but we can use the same argument to show that $\frac 1{2x} +\frac x2 < -1$ for $x < 0; x\ne -1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed-form for Floor Sum 1 Does a closed form exist for the following sum?
$$\sum_{k=0}^n \lfloor \sqrt{k} + \sqrt{k + n} \rfloor$$
If not, why is this sum so radically different than the sums below?
Closed forms do exist for the following sums*:
$$\sum_{k=0}^n \lfloor \sqrt{k + n} \rfloor$$
$$\sum_{k=0}^n \lfloor \sqrt{k} \rfloor$$
There is this floor functional identity:
$$\lfloor \sqrt{k} + \sqrt{k + 1} \rfloor = \lfloor\sqrt{4k+2}\rfloor$$
Don't know if this will help.
Thanks
*Existing closed forms
$$\sum_{k=0}^n \lfloor \sqrt{k} \rfloor=2\left(\sum_{k=0}^{\lfloor \sqrt{n} \rfloor-1}k^2\right)+\left(\sum_{k=0}^{\lfloor \sqrt{n} \rfloor-1}k\right)+\lfloor\sqrt{n}\rfloor\left(n-\lfloor\sqrt{n}\rfloor^2+1\right)$$
$$\left(\sum_{k=0}^n k^2\right)=\frac{2n^3+3n^2+n}{6}$$
$$\left(\sum_{k=0}^n k\right)=\frac{n^2+n}{2}$$
$$\sum_{k=1}^n \lfloor \sqrt{k+C} \rfloor=\sum_{k=C+1}^{C+n} \lfloor \sqrt{k} \rfloor=\sum_{k=0}^{C+n} \lfloor \sqrt{k} \rfloor-\sum_{k=0}^{C} \lfloor \sqrt{k} \rfloor$$
| Hint:
Such sums are made of runs of equal values, that are delimited by the indexes such that the general term crosses an integer.
$$\sqrt k+\sqrt{k+n}=m$$ when
$$k+2\sqrt k\sqrt{k+n}+k+n=m^2$$
or
$$4k(k+n)=(m^2-n-2k)^2$$
or
$$k=\frac{(m^2-n)^2}{m^2}=m^2-2n+\frac1{m^2}.$$
As $k$ is an integer, the final fraction can be ignored and a run of weight $m$ has length $(m+1)^2-2n-m^2+2n=2m+1$.
Hence for complete runs, i.e. up to some $k^*=(m+1)^2-2n$ (excluded), the sum is that of $(2m+1)m$.
For the last incomplete run, the length is $n-k^*$, for weight $m+1$.
| {
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Convergence of a series depending on a parameter. I was studying the convergence of a series with a parameter and I want to ask you if my conclusion is correct and if there is a better method to do it.
$$\sum_{n=1}^{\infty}\left(\frac{\pi}{2}-\arcsin\frac{n}{n+4} \right)^{\alpha}$$
I'm asked to say for which $\alpha$ the series is convergent. I tried to do this:
$$\sum_{n=1}^{\infty}\left(\frac{\pi}{2}-\arcsin\frac{n}{n+4} \right)^{\alpha}=
\sum_{n=1}^{\infty}\left(\arccos\frac{n}{n+4} \right)^{\alpha}=\sum_{n=1}^{\infty}\left(\arccos\left(1-\frac{4}{n+4}\right)\right)^{\alpha}$$
Now I tried to see the Taylor Series of $\arccos(1-y)$ as $y\rightarrow 0$, $y=\frac{4}{n+4}$.
By deriving $g(y)=\arccos(1-y)$ I obtain:
$$g'(y)=\frac{1}{\sqrt{1-(1-y)^2}}=(-x^2+2x)^{\frac{1}{2}}=(2x)^{\frac{1}{2}}(1-\frac{x}{2})^{\frac{1}{2}}=\sqrt{2y}+o(\sqrt{2y})$$
So I see that:
$$g(y)=\int\frac{1}{\sqrt{2y}}+o\left(\frac{1}{\sqrt{2x}}\right)dx=\sqrt{2x}+o(\sqrt{2x})$$
From this, I can conclude that my series is:
$$\sum_{n=1}^{\infty}\left(\arccos\left(1-\frac{4}{n+4}\right)\right)^{\alpha} \sim \left(\sqrt{\frac{8}{n+4}}\right)^{\alpha}$$
So, for asymptotic comparison to $\left(\frac{1}{n}\right)^{\frac{\alpha}{2}}$ the series converges if and only if $\alpha > 2$.
Am I right? If yes, may I ask you if there is a better method to do this computation? And, expecially, if there is a better method (if possible without involving integrals) to compute the Taylor series of inverse trigonometric functions!
Thanks in advance.
| $\sin(x) = \sin(\frac{\pi}{2}) + \cos\left(\frac{\pi}{2}\right)\left(x - \frac{\pi}{2}\right) - \frac{1}{2}\sin\left(\frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right) \\ = 1 - \frac{1}{2}\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right)$
Then
\begin{align}
\lim_{x\rightarrow \frac{\pi}{2}^-} \frac{1 - \sin(x)}{\left(\frac{\pi}{2} - x\right)^2} = \frac{1}{2}
\end{align}
Using $u = \sin(x)$ we have
\begin{align}
\lim_{u \rightarrow 1^-} \frac{1 - u}{\left(\frac{\pi}{2} - \arcsin(u)\right)^2} = \frac{1}{2}
\end{align}
Taking the root we have $\frac{\pi}{2} - \arcsin(u) \sim_{u\rightarrow 1} \sqrt{2(1-u)}$.
| {
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Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve in the xy-plane. Now if I could get y as a function of x, I could easily parametrize the curve but I am not able to do that.
Is there an easier way to solve the problem?
| HINT: it is $$y_{1,2}=-\frac{x}{2}\pm\sqrt{\frac{1}{2}-\frac{3}{4}x^2}$$ and with that $y$ we can compue $z$
| {
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The largest integer that divides $p^4-1$, in which p is a prime greater than $5$ The question is to find the largest integer that divides all $p^4-1$, where p is a prime greater than 5. Being asked this question, I just assume this number exists. Set $p = 7$, then $p^4-1=2400$. I don't have any background in number theory and not sure what to do next. Thank you for your help!
| Let $n$ be the largest integer that divides $p^4-1$ for all prime $p\geq 7.$
We have $11^4-1=14640$ and $7^4-1=2400.$ The $gcd$ of $14640$ and $2400$ is $240.$ So $$n\leq 240.$$ If $p$ is odd then modulo $16$ we have $p^4\in \{(\pm 1)^4, (\pm 3)^4,(\pm 5)^4,(\pm 7)^4\}=\{1^2,9^2, 25^2, 49^2\}=\{1^2,9^2,9^2,1^2\}=$
$=\{1,81,81,1\}=\{1\}.$
If $p$ is not divisible by $3$ then modulo $3$ we have $p^4\in \{(\pm 1)^4\}=\{1\}.$
If $p$ is not divisible by $5$ then modulo $5$ we have $p^4\in \{(\pm 1)^4,(\pm 2)^4\}=\{1,16\}=\{1\}.$
So for any integer $p$ that is not divisible by $2,3,$ or $5$ we have $p^4\equiv 1 \pmod {16}$ and $p^4 \equiv 1 \pmod 3$ and $p^4 \equiv 1 \pmod 5;$ and since $16,3,$ and 5 are pair-wise co-prime, therefore $p^4\equiv 1 \pmod {16\cdot 3\cdot 5}=240,$ so $$n\geq 240.$$
| {
"language": "en",
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"question_score": "5",
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Solve $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}+\frac{1}{z}$
Solve $$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}+\frac{1}{z},$$ where $x, y, z$ are positive integers.
So far I have:
If $x=y=z \Rightarrow(x, y, z) = (2, 2, 2)$.
If $x=z \Rightarrow (x, y, z) = (k, 2, k)$, where $k$ is natural.
If $y=z \Rightarrow (x, y, z) = (2, k, k)$, where $k$ is natural.
If $x=y \Rightarrow (x-4)(z+2) = -8 \Rightarrow x = 1, 2,3$. The first two cases yield already known solutions. The third yields $(x, y, z) = (3, 3, 6)$.
Now I tried to look at the case where $a \ne b \ne c$ and try to establish some bounds (e.g. $z > x, y$) but this has not yielded anything useful.
| By symmetry, we assume $x\leq y$ throughout this discussion.
Cases:
If $x=1$, then the equation simplifies to $\frac{1}{2}+\frac{1}{y}=\frac{1}{z}$. Then $z=1$ since otherwise the LHS is larger than the RHS. When $z=1$, we have $y=2$.
If $x=2$, then the equation simplifies to $\frac{1}{y}=\frac{1}{z}$, which can be solved for any $y$ and $z$.
If $x=3$, then we have $3\leq y<5$ as possible left-hand-sides (each of which can be checked).
If $x=4$, then $y\geq 4$ and the LHS is less than the RHS and there are no solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
| $\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{\sin^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{1-\cos^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{(1-\cos \pi x)(1+\cos \pi x)}=$
$=\lim_{x\to 1} \dfrac {\cos^2 \pi x}{1-\cos \pi x}=\dfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $f(x)=\frac{x^2}{1+x^2}$ bijective? $f: \mathbb{R} \to \mathbb{R}$ via $f(x) = \frac {x^2}{1+x^2}$
Injective:
Suppose f(a) = f(b)
$$ \implies \frac {a^2}{1+a^2} = \frac {b^2}{1+b^2} \implies a^2 +a^2b^2 = b^2 +b^2a^2$$
$$ \implies a^2 = b^2 \implies a \neq b$$
Surjective: If $x<0$, the image of $f$ is $[0, \infty)$
Hence $f$ is not surjective.
Therefore, $f$ is neither injective or surjective.
| $f $ is an even function
$$\implies f (1)=f (-1)=\frac {1}{2} $$
$\implies f $ is not injective
$\implies f $ is not bijective.
| {
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If one root of the equation $ax^2+bx+c=0$ be the square of the other. If $a \neq 0$ and if one root of the equation $ax^2+bx+c=0$ is the square of the other, prove that: $$b^3+a^2c+ac^2=3abc.$$
My Attempt:
Given:
$$ax^2 + bx + c=0$$
Let $\alpha $ and $\beta $ be the roots of the equation.
$$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
According to the question:
$$\alpha = \beta^2 $$
| $$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
Substitute
$$\alpha = \beta^2$$
into previous equations,
$$a(\beta^2+\beta) = a\beta(\beta+1) = - b\tag{1}$$
$$a\beta^3 = c\tag{2}$$
Cube equation $(1)$,
$$a^3 \beta^3 (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$
Using equation $(2)$,
$$a^2 c (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$
$$a c (a\beta^3 + 3a\beta( \beta + 1) + a) = -b^3$$
Using equation $(1)$ and $(2)$,
$$a c (c - 3b + a) = -b^3$$
$$ac^2-3abc+a^2c=-b^3$$
$$b^3+ac^2+a^2c = 3abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
solving $\cos z + \sin z = i$ I found a solution but when testing it, it does not work....?
$$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi$$
Upon testing it in W|A with $k=0$, I don't get the required result $(=i)$. I'm not sure where I went wrong and I did this a few times already.
| You may have applied the quadratic form incorrectly as I can see.
You have:
$$e^{2iz}(1+i)+2e^{iz}-1+i=0$$
with $a=1+i,b=2,c=-1+i$.
$$\implies e^{iz}=\frac{-2\pm \sqrt{2^2-4(i+1)(i-1)}}{2(1+i)}=\frac{-2\pm\sqrt{4-4(i^2-1)}}{2(i+1)}=\frac{-2\pm\sqrt{4-4(-2)}}{2(i+1)}$$
$$=\frac{-2\pm\sqrt{4+8}}{2(i+1)}=\frac{-2\pm\sqrt{12}}{2(i+1)}=\frac{-2\pm2\sqrt{3}}{2(i+1)}=\frac{-1\pm\sqrt{3}}{1+i}$$.
$$\therefore iz=\ln\left(\frac{-1\pm\sqrt{3}}{1+i}\right)\Rightarrow z=-i\ln\left(\frac{-1\pm\sqrt{3}}{1+i}\right)+2k\pi$$
which I believe can be further simplified with complex algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\triangle BAC$ is a right-angled isosceles triangle Suppose in the figure below, $AD\ \text{//} \ BC$, $BD=BC$, $CD=CE$,
and $ABCD$ is a trapezoid; $\measuredangle ABD=15°$
Prove that: $\triangle BAC$ is a right-angled isosceles triangle.
| Let $F$ the intersection of lines $BA$ and $CD$, $\angle BCD= x$ and $BC=L$.
Then
$$\angle BFC= x-\frac{\pi}{12}. \tag{1}$$
Applying some trig we get:
$$CD=2L\cos x \tag{2}$$
$$DE =4L(\cos x)^2 \tag{3} $$.
Applying sine rule in triangle $DAE$ we get:
$$DA=\frac{4L(\cos x)^2\sin x}{-\sin(3x)} \tag{4} $$
The similarity of triangles $FDA$ and $FCB$ produces:
$$FD=\frac{2L\cdot DA \cos x}{L -DA} \tag{5}$$
Applying sine rule in triangle $FDB$ we get:
$$FD=\frac{L \sin(\frac{ \pi}{12})}{ \sin( x-\frac{ \pi}{12})} \tag{6}$$
Using $(4)$ and making $(5)$ and $(6)$ equal, we get:
$$\sin(\frac{\pi}{12})(7-8(\sin x)^2)=-8(\cos x)^3\sin( x-\frac{ \pi}{12}) \tag{7}$$
After some trig relations, we get:
$$\tan(x-\frac{ \pi}{12})=\tan x (8(\sin x)^2-7)$$
Solving for $x$ we get:
$$x = \frac{5\pi}{12}$$
Thereafter it is easy to conclude that angle $BAC$ is a right angle and $\triangle ABC$ is isosceles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{11}$$
Then for some polynomial $Q(x)$ and $R(x)$; we have
$$P_1(x)=Q(x)\cdot P_2(x)+R(x)$$
Now, we put the values of $x$ such that $R(x)=0$ and form equations, but this method is way too long and solving the 11 set of equations for 11 variable (Since $R(x)$ a polynomial of at most 10 degree) is impossible to do for a competitive exam where the average time for solving a question is 3 minutes.
Another method is using the original long division method, and following the pattern, we can predict $Q(x)$ and $R(x)$, but it's also very hard and time taking.
I am searching for a simple solution to this problem since last a week and now I doubt even we have a simple solution to this question.
Can you please give me a hint/solution on how to proceed to solve this problem in time?
Thanks!
| \begin{eqnarray*}
p_2(x) &=& (x^6+1)(x^5+x^4+x^3+x^2+x+1)\\
&=& (x^6+1)(x^3+1)(x^2+x+1)\\
&=& (x^6+1)(x+1)(x^2-x+1)(x^2+x+1)\\
&=& (x+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
\begin{eqnarray*}
p_1(x) &=& (x^{12}+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\
&=& (x^{12}+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
Write $$p_1(x) = k(x)p_2(x) +r(x)$$ then $(x^6+1)(x^4+x^2+1)$ divides $r(x)$.
So $r(x) = (x^6+1)(x^4+x^2+1)s(x)$ where $s(x)$ is constant. So $$x^{12}+1 = k(x)(x+1)+ s(x)$$ Put $x=-1$ and we get: $s(-1)= 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer. Given
$$(1+x)^n= \binom {n}{0} + \binom{n}{1} x+ \binom{n}{2} x^2+ \cdots + \binom {n}{n} x^n.$$
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer.
I tried to use negative $x$ and even $i$ but could not eliminate $\binom{n}{2} $ term.
| Hint: What is $(1+1)^n+(1-1)^n+(1+i)^n+(1-i)^n$?
| {
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"source": "stackexchange",
"question_score": "10",
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
| hint AM-GM
$$(a+b)^2-a^2-b^2=2ab\le a^2+b^2$$
$$\implies 1-a^2-b^2\le (a^2+b^2) $$
$$\implies 1\le 2 (a^2+b^2). $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$.
So I know it's got something to do with Euler's totient function and the Euler-Fermat theorem, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance.
| Hint:
Note that, $a^3\equiv 0,1^3,2^3,3^3,4^3,5^3,6^3(\mod 7)\Rightarrow a^3\equiv 0,1,1,-1,1,-1,-1(\mod 7)$
More explanation :
Then $7\mid a\Rightarrow 7\mid a^3$ or $7\mid b\Rightarrow 7\mid b^3$.
If $7$ divides none of $a,b$, then either $a^3+b^3\equiv 0(\mod 7)$ or $a^3-b^3\equiv 0(\mod 7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be easy to implement in a computer program.
For instance, suppose
$$\begin{eqnarray}
F_0&=&0\\
F_1&=&1\\
F_{n+1}&=&F_n+F_{n-1},
\end{eqnarray}$$
so that $F_n$ is the $n^{th}$ term in the Fibonacci sequence. If we set
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
we see that
$$A^1=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix},$$
and it can be shown that
$$
A^n =
\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}.$$
This example is "interesting" in that it provides a novel way to compute the Fibonacci sequence. It is also relatively easy to implement a simple program to verify the above.
Other examples like this will be much appreciated.
| Here's the polynomial derivative matrix (where $n$ is the degree of the polynomial):
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}\newcommand{\i}{\mathrm{i}}\newcommand{\text}[1]{\mathrm{#1}}\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}\newcommand{\i}[0]{\mathrm{i}}\newcommand{\kume}[1]{\mathbb{#1}}\newcommand{\bold}[1]{\mathbf{#1}}\newcommand{\italic}[1]{\mathit{#1}}\newcommand{\kumedigerBETA}[1]{\rm #1\!#1}$
$$\large \mathbf {Derivative\s matrix\s}_{(n+1)\times (n+1)}=\begin{bmatrix}
0 & 1 & 0 & 0 & \cdots & 0 \\
0 & 0 & 2 & 0 & \cdots & 0 \\
0 & 0 & 0 & 3 & \cdots & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
Let me do a sample calculation:
Question: What's $$\derivative{}{}{x}\left(2x^3+5x^2+7\right)$$
Find the derivative matrix and multiply it with coefficients of the polynomial.
$$\mathbf {Answer}=\begin{bmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3\\
0 & 0 & 0 & 0\\
\end{bmatrix}
\begin{bmatrix}
7\\
0\\
5\\
2\\
\end{bmatrix}=\begin{bmatrix}
0\\
10\\
6\\
0\\
\end{bmatrix}\Rightarrow6x^2+10x
$$
And integral calculation (where $n$ is the degree of the polynomial, again)!
$$\large \mathbf {Integral\s matrix\s}_{(n+\bold{2})\times (n+\bold{2})}=\begin{bmatrix}
0 & 0 & 0 & 0 & \cdots & 0 \\
1 & 0 & 0 & 0 & \cdots & 0 \\
0 & \frac{1}{2} & 0 & 0 & \cdots & 0 \\
0 & 0 & \frac{1}{3} & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
Question: What's $$\int\left(2x^3+5x^2+7\right)\cdot\mathrm{d}x$$
$$\mathbf {Answer}=\begin{bmatrix}
0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
0 & \frac{1}{2} & 0 & 0 & 0\\
0 & 0 & \frac{1}{3} & 0 & 0\\
0 & 0 & 0 & \frac{1}{4} & 0\\
\end{bmatrix}
\begin{bmatrix}
7\\
0\\
5\\
2\\
0\\
\end{bmatrix}=\begin{bmatrix}
0\\
7\\
0\\
\frac{5}{3}\\
\frac{1}{2}\\
\end{bmatrix}\Rightarrow \frac{1}{2}x^4+\frac{5}{3}x^3+7x+\mathit{constant}
$$
Warning: Don't forget $+constant$.
Inspired by: $3$Blue$1$Brown
| {
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"url": "https://math.stackexchange.com/questions/2408108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 26,
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Factorization and GCD for Rijndael's finite field I'm factoring polynomials in $GF(2^8)$ with modulo polynomial $m = 2^8 + 2^4 + 2^3 + 2^1 + 2^0$
In particular, I factored
a = 0x49 = $$2^6 + 2^3 + 2^0 = 2^1 * (2^1 + 2^0)^6 * (2^2 + 2^1 + 2^0) * (2^4 + 2^1 + 2^0) * (2^3 + 2^1 + 2^0) \bmod{m}$$
b = 0x64 = $$2^6 + 2^5 + 2^2 = (2^1)^3 * (2^1 + 2^0) * (2^3 + 2^1 + 2^0) * (2^3 + 2^2 + 2^0) \bmod{m}$$
these numbers are multiplicative inverses and I can calculate directly using long division that $ab\bmod{m} = 2^0$.
I assume GCD should also be $1$, i.e. $2^0$.
Now I want to calculate GCD using just factored irreducible polynomials.
I know that for integers GCD equals the product of prime numbers (including their powers) present in both factorization, does it still hold here?
$gcd = 2^1 * (2^1 + 2^0) * (2^3 + 2^1 + 2^0) \bmod{m} = 2^5 + 2^4 + 2^3 + 2^1$
Can anyone please tell me where I am mistaken?
Thank you!
| A fundamental problem is that while there are primes in the ring of polynomials $GF(2)[x]$ (where you can also run the (extended) Euclidean algorithm, there are no primes in the field $GF(2^8)=GF(2)[x]/\langle m(x)\rangle$. The same holds in all fields. Technically, all the non-zero elements of a field are units, and hence divisible by each other.
This manifests in many ways (reverting to the notation that $2^k$ is the residue class of $x^k$ modulo $m(x)$):
*
*Because $ab=2^0$ the element $b$ divides all the "factors" you listed. For example $2^3+2^1+2^0=ab(2^3+2^1+2^0)$ is clearly "divisible" by $b$. Q: Which one of $b$ and $2^3+2^1+2^0$ is a factor of the other? A: Both!
*Because $2^8+2^4=2^4(2^4+2^0)=(2^1)^4(2^1+2^0)^4$, and also
$2^8+2^4=2^3+2^1+2^0$ (in $GF(2^8)$ "=" means the same thing as congruent modulo $m$), we see that
$$2^3+2^1+2^0=(2^1+2^0)^4(2^1)^4.$$ Therefore the factor $2^3+2^1+2^0$ is not a "prime" either.
*For some reason unknown to me here a non-primitive polynomial $m(x)$ is used in defining the field (I erred here initially in one of the comments). Anyway a computer check reveals that $x^{51}\equiv 1\pmod {m(x)}$, and that $51$ is the smallest positive exponent when this happens. Therefore the powers of $2^1$ cover one fifth of the elements of $GF(256)$. Another computer check reveals that $2^1+2^0$ is not among those $51$. But, again by computer, we do have the relation
$$(2^1+2^0)^5=(2^1)^{42}=(2^1)^{-9}.$$
This implies that any element $z\in GF(2^8), z\neq0,$ can be written uniquely in the form
$$z=(2^1)^k*(2^1+2^0)^\ell,$$ with $0\le k<51$, $0\le \ell<5$.
*Therefore you never need "primes" other than $2^1+2^0$ and $2^1$. And these two are also "divisible" by each other.
It may be useful to look at the analogy of the residue class ring $\Bbb{Z}_{17}$ of integers modulo the prime $17$:
*
*We have $2^4=-1$ in this ring, and therefore also $2^8=1$, so do we really want to call $2$ a "prime". It is a factor of $1$ after all!
*We have $3=20=2^2\cdot5$, so $3$ is not a "prime" but it doesn't really make sense to call $2^2\cdot 5$ a factorization of $3$ into "primes" either.
*Similarly $5^2=25=8=2^3$, and therefore also $5^6=(2^3)^3=2^9=2^8\cdot 2=2$. Actually all the non-zero element of $\Bbb{Z}_{17}$ are powers of $5$. But also $5=-12=-2^2\cdot3$ is "divisible" by both $2$ and $3$.
The analogy I wanted to make in comparing $GF(2^8)$ and $\Bbb{Z}_{17}$ is that when working in $GF(2^8)$ we may occasionally benefit from working inside the polynomial ring $GF(2)[x]$ where we have things like the extended Euclidean algoritm. Similarly in $\Bbb{Z}$ we have the extended Euclidean algorithm allowing us to calculate inverses in $\Bbb{Z}_{17}$. But, talking about primes in $GF(2^8)$ is equally futile as talking about primes in $\Bbb{Z}_{17}$.
I'm afraid that...
To clear up all the fog it may be necessary to read an introductory level textbook on abstract algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of real solutions of $f(f(x))$ Given $f(x)=x^3-12x+3$, find number of real solutions of $f(f(x))=0$.
My Try: Since $f(x)$ is having three real toots say $\alpha$, $\beta$ and $\gamma$, we have $$f(f(\alpha))=f(f(\beta))=f(f(\gamma)) =f(3).$$
Hence by Rolles theorem $\exists$ atleast one $c $ such that $f'(c)=0$
But how to count number of roots?
| $f(x)=x^3 - 12 x + 3$ has $3$ real solutions, as you stated: $\alpha,\beta,\gamma\in\mathbb{R}$
$f(f(x))=0 $ means $\left(x^3-12 x+3\right)^3-12 \left(x^3-12 x+3\right)+3=0$
translates in three $3-$rd degree equations
$\alpha ^3-12 \alpha +3=\alpha;\;\beta ^3-12 \beta +3=\beta;\;\gamma^3-12 \gamma +3=\gamma$
that is
$\alpha ^3-13 \alpha +3=0;\;\beta ^3-13 \beta +3=0;\;\gamma^3-13\gamma +3=0$
Each of which you can easily prove has three real roots, so the global number of real roots is $9$
Hope it is useful
| {
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How to solve a matrix system using Gauss elimination $$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)$$
I am struggling to solve this system $Ax=b$. I understand the basics of Gauss elimination but am not sure how to handle it with the alpha and beta. It needs to be solved giving conditions for alpha and beta for no solutions, one solution and infinitely many solutions.
I also need to compute the determinant of A and give a condition on $\alpha$ such that $A$ is invertible.
| $$\det\left(
\begin{array}{rrr}
-1 & 2 & 1 \\
3 & a & -2 \\
-1 & 5 & 2 \\
\end{array}
\right)=-3-\alpha$$
if $-3-\alpha\ne 0$ that is $\alpha\ne -3$ there exists one and only one solution
$$\left\{\frac{3 a-b+6}{a+3},\frac{b+3}{a+3},\frac{3 (2 a-b+3)}{a+3}\right\}$$
if $\alpha=-3$ the determinant of the matrix $A$ of coefficients is zero $\text{rank}(A)=2$, so we must consider the augmented matrix $A|B$
$$\left(
\begin{array}{rrr|r}
-1 & 2 & 1 & 3 \\
3 & -3 & -2 & b \\
-1 & 5 & 2 & 9 \\
\end{array}
\right)$$
if $\text{rank}(A|B)\ne \text{rank}(A)$ the system has no solutions, therefore to have infinite solutions we must have $\text{rank}(A|B)=2$ so all $3-$rd order determinants extracted from $A|B$ must be zero
$$
\det \left(
\begin{array}{rrr}
2 & 1 & 3 \\
-3 & -2 & \beta \\
5 & 2 & 9 \\
\end{array}
\right) =\beta+3
$$
if $\beta\ne -3$ the system is impossible
if $\beta+3=0\to \beta=-3$ the determinant is zero the rank of the augmented matrix is equal to the rank of $A$ and we have infinite solutions
$\{t,\;t,\;3-t\}$
| {
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Maclaurin Series Expansion $\ln(8+x^3)$.
Find Maclaurin Series Expansion of function $\ln(8+x^3)$. For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series?
$$ (\ln(8+x^3))'=\frac{3x^2}{8+x^3} =\frac{\frac{3}{8}x^2}{1-(-\frac{x^3}{8})}=$$
$$ =\sum_{n=1}^{\infty}\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}$$
and
$$ \sum_{n=1}^{\infty}\int\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}dx=$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\int x^{3n+2}dx=$$
$$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{x^{3n+3}}{3n+3}+C$$
for $x=0$
$$\ln(8+0^3)= \sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{0^{3n+3}}{3n+3}+C $$
$$ C=0 $$
$8+x^3>0 $
$x>-2 $
and
$|\frac{-x^3}{8}|<1$
$x<2$
I have to find if it converges only at $x=2$, $x=-2$ is out of domain? For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series? From Abel theorem $\forall$ x which converges $f(x)$=sum of series?
| HINT:
$$\ln(8+x^3)=\ln8+\ln\left(1+\dfrac{x^3}8\right)$$
See Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1<x \le 1$?
OR
Taylor series for $\log(1+x)$ and its convergence
OR
What is the correct radius of convergence for $\ln(1+x)$?
| {
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Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
| Scalar product:
Let $\vec A : = (a,b,c)$, $\vec B: = (b,c,a)$.
$|\vec A \cdot \vec B| \le |A| |B| $.
$\Rightarrow$ :
$2 \le ab +bc + ca \le a^2 +b^2 + c^2$.
Equality:
$a=b=c = \sqrt{\frac{2}{3}}$.
| {
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