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Solving $ \frac{dy}{dx} = y^2 - 9$ $ \frac{dy}{dx} = y^2 - 9$
This is separable so I rewrite it as $ \frac{1}{(y^2 - 9)}dy = dx$ then I get
$$\int \frac{1}{(y^2 - 9)}dy = \int dx = \int 1dx = x + c,$$ for some $c \in \mathbb{R} $
The left hand side:
$$\int \frac{1}{(y^2 - 9)}dy = \int \frac{1}{(y+3)(y-3)}dy = \int \left( \frac{A}{y+3} + \frac{B}{ y-3}\right) \,dy $$
We have $A(y-3) + B(y+3) = 1 \implies A =-\frac{1}{6}, B = \frac{1}{6}$
So we get
$$ \int \left(\frac{A}{y+3} + \frac{B}{ y-3}\right) \, dy = \int \left (\frac{-\frac{1}{6}}{(y+3)} + \frac{\frac{1}{6}}{ (y-3)}\right)dy = \frac{1}{6}\int \left (\frac{1}{ (y-3)} - \frac{1}{ (y+3)}\right) dy = \frac{1}{6} \left(\int \frac{1}{ (y-3)}dy - \int \frac{1}{ (y+3)} dy\right ) = \frac{1}{6}\left ( \ln \left \vert y-3 \right\vert - \ln \left\vert y+3 \right\vert \right ) = \frac{1}{6}\ln \left \vert\frac{y-3}{y+3} \right \vert $$
So I have $\ln \left\vert\frac{y-3}{y+3}\right\vert = 6x + 6c \implies e^{\ln\left\vert\frac{y-3}{y+3}\right\vert} = \left\vert\frac{y-3}{y+3}\right\vert =e^{6x + 6c} $
Is this correct to far? If yes how do I solve this for $y$?
| Simplifying from where you stopped:
$\displaystyle \ln|\frac{y-3}{y+3}| = 6x + \underbrace{6c}_{=\,c}\Longrightarrow|\frac{y-3}{y+3}|=e^{6x+c}=e^{6x}\cdot \underbrace{e^c}_{=\,c}=Ce^{6x}$
$$|\frac{y-3}{y+3}|=Ce^{6x} \Longrightarrow \frac{y-3}{y+3}=-Ce^{6x} \Longrightarrow y-3=-Ce^{6x}(y+3) \Longrightarrow y(1+Ce^{6x})=-3Ce^{6x}+3 \Longrightarrow \boxed{\displaystyle y=\frac{-3Ce^{6x}+3}{Ce^{6x}+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/921589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$ I have tried parts which ended up into a more complex form, any suggestions?
$$\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
| Edit: I decided that the previous solution admits a minor but nice generalization, so here goes.
Let $$t = \dfrac{x}{\sqrt{x^4+1}}, \quad dt = \frac{1-x^4}{(x^4+1)^{3/2}} \, dx.$$ Then $$\frac{1}{1 \pm 2t^2} = \frac{x^4+1}{x^4 + 1 \pm 2x^2} = \frac{x^4+1}{(1 \pm x^2)^2},$$ so that
$$ \begin{align*} \int \frac{x^2 \mp 1}{(x^2 \pm 1)\sqrt{x^4+1}} \, dx &= \int \frac{(x^2 \mp 1)(x^2 \pm 1)(x^4+1)}{(x^2 \pm 1)^2 (x^4+1)^{3/2}} \, dx \\ &= \int \frac{1}{1 \pm 2t^2 } \cdot \frac{x^4-1}{(x^4+1)^{3/2}} \, dx \\ &= - \int \frac{dt}{1 \pm 2t^2}.\end{align*}$$ In the positive case, we get $$- \frac{1}{\sqrt{2}} \tan^{-1} \sqrt{2}t + C = - \frac{1}{\sqrt{2}} \tan^{-1} \frac{x \sqrt{2}}{\sqrt{x^4+1}} + C.$$ In the negative case, we get $$\frac{1}{2\sqrt{2}} \log \left| \frac{1 - \sqrt{2}t}{1 + \sqrt{2}t} \right| + C = \frac{1}{2 \sqrt{2}} \log \left| \frac{\sqrt{x^4+1} - x \sqrt{2}}{\sqrt{x^4+1} + x \sqrt{2}} \right| + C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The minimum of $x^2+y^2$ under the constraints $x+y=a$ and $xy=a+3$ I solved the following problem: If $x,y,a \in \mathbb{R}$ such that $x+y=a$ and $xy=a+3$, find the minimum of $x^2+y^2$
Here is my solution. $x^2+y^2=(x+y)^2 -2xy= a^2-2a-6$. The minimum value is obtained when $a=1$ and it is $-7$. Where did I go wrong?
The answer says that the minimum is $2$ when $x=y=-1$
| Another way:
$$(x+y)^2 \ge 2xy \implies a^2 \ge 2(a+3) \implies a \in (\infty, -2] \cup [3,\infty)$$
$$(x^2+y^2)(1+1)\ge(x+y)^2 \implies x^2+y^2 \ge \tfrac12 a^2 \ge 2$$
Equality is achieved in both inequalities when $x=y=-1$, so this is the minimum.
| {
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"url": "https://math.stackexchange.com/questions/925301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating Trig Functions How you I solve the following integral?
$\int_{-\pi}^\pi \cos(5x)\cos(nx)dx$
I know I need to use the ${1\over 2}(\cos(u-v)+\cos(u+v))$ but I keep getting zero.
| Firstly, $\cos(nx)\cos(5x)=\frac{1}{2}(\cos((n-5)x)+\cos((n+5)x))$.
If $n\neq\pm5$, we split the integral up as follows:
\begin{align*}
\int_{-\pi}^{\pi} \cos(nx)\cos(5x)dx&=\frac{1}{2}\left(\int_{-\pi}^{\pi}\cos((n-5)x)dx+\int_{-\pi}^{\pi}\cos((n+5)x)\right) \\
&= \frac{1}{2}\left[\frac{1}{n-5}\sin((n-5)x)+\frac{1}{n+5}\sin((n+5)x)\right]_{-\pi}^{\pi} \\
&= \frac{1}{2}\left(\left(\frac{1}{n-5}\sin((n-5)\pi)+\frac{1}{n+5}\sin((n+5)\pi)\right)-\left(\frac{1}{n-5}\sin(-(n-5)\pi)+\frac{1}{n+5}\sin(-(n+5)\pi)\right)\right) \\
&= \frac{1}{n-5}\sin((n-5)\pi)+\frac{1}{n+5}\sin((n+5)\pi)\:.
\end{align*}
Now if we are assuming that $n$ is an integer, then because $\sin$ is zero for integer multiples of $\pi$, this integral becomes 0.
If instead $n$ were equal to $\pm5$, using the evenness of the cosine function, we see:
\begin{align*}
\int_{-\pi}^{\pi} \cos(\pm5x)\cos(5x)dx &= \int_{-\pi}^{\pi} \cos^2(5x)dx \\
&= \frac{1}{2}\int_{-\pi}^{\pi}(\cos(10x)+1)dx \\
&= \frac{1}{2}\left[\frac{1}{10}\sin(10x)+x\right]_{-\pi}^{\pi} \\
&= \frac{1}{2}\left(\left(\frac{1}{10}\sin(10\pi)+\pi\right)-\left(\frac{1}{10}\sin(-10\pi)-\pi\right)\right) \\
&= \pi\:.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int_{1}^{\infty} \frac{\log^3 x}{x(x-1)} dx$ How do I arrive at the closed form expression of the integral $$\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx$$
Most probably the closed form is $\dfrac{\pi^4}{15}$
| Using a CAS, $$I_1=\displaystyle\int\dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(x)+3 \text{Li}_2(x) \log ^2(x)-6 \text{Li}_3(x) \log (x)-\frac{1}{4}
\log ^4(x)+\log (1-x) \log ^3(x)$$ $$I_2=\displaystyle\int_{1}^{a} \dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(a)+3 \text{Li}_2(a) \log ^2(a)-6 \text{Li}_3(a) \log (a)-\frac{1}{4}
\log ^4(a)+\log (1-a) \log ^3(a)-\frac{\pi ^4}{15}$$and, going to limit $$I_3=\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx=\frac{\pi ^4}{15}$$
For large values of $a$ $$I_2\simeq \frac{\pi ^4}{15}-\frac{\log ^3(a)+3 \log ^2(a)+6 \log (a)+6}{a}$$ For $a=10^3$, the value of the integral is $5.97352$ while the approximation leads to $5.97372$. For $a=10^4$, the value of the integral is $6.38423$ while the approximation leads to $6.38423$.
| {
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Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.
| Using Gerry Myerson's clever inequality observation, we have two possibilities.
Case 1: $y=x+1$. Then by substitution and simplification, we must solve
$$5x^2-9x+7=0,$$
which has no integer solutions, as seen by using the quadratic equation.
Case 2: $y=x+2$. This yields the [factored] quadratic
$$2x(x-9)=0.$$
Evidently, the only two integer solutions are $x=0$ and $x=9$, yielding the solutions $(x,y)=(0,2)$ and $(x,y)=(9,11)$.
There are no other solutions.
| {
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Can anybody help me solve this combinatorial identity? While trying to derive some physical equation, I noticed that the following identity was needed:
$\sum^{4a \leq 2k}_{a=0}{2k \choose 4a} + \sum^{4a+1 \leq 2k}_{a=0} {2k \choose 4a+1} = \left\{ \begin{array}{ll} \frac{2^k(2^k +1)}{2} & (k=4l+1, 4l+4)\\
\frac{2^k(2^k -1)}{2} & (k=4l+2, 4l+3) \end{array} \right.$
(where $l,a$ are positive integers)
My strategy was to change the $2^k$'s of RHS into $\sum_{a=0}^{a=k}{k \choose a}$ and use ${n \choose k}={n-1 \choose k} +{n-1 \choose k-1}$, but obtained only incomplete and partial relations. Can anybody solve this identity explicitly, or at least suggest another strategy that might work better?
| A tool: By the Binomial Theorem, we have
$$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\binom{n}{5}x^5+\cdots. \tag{1}.$$ Putting $x=1$ we get a familiar identity, and putting $x=-1$ we get something almost as familiar.
Using addition and subtraction, we get the sum of the binomial coefficients of the shape $\binom{n}{k}$, where $k$ ranges over the even numbers, and also the sum where $k$ ranges over the odd numbers.
Now comes the interesting part. Put $x=i$. We get
$$(1+i)^n=1+\binom{n}{1}i-\binom{n}{2}^2-\binom{n}{3}i+\binom{n}{4}+\binom{n}{5}i+\cdots \tag{2}.$$
We get a similar result by using $x=-i$. It can also be obtained from (2) by conjugation.
We connect Equation (2) with powers of $2$. Note that $1+i=\frac{1}{\sqrt{2}}\left(\cos(\pi/4)+i\sin(\pi/4)\right)$. Taking the $n$-th power, we get
$(1+i)^n=2^{n/2}\left(\cos(n\pi/4)+i\sin(n\pi/4)\right)$. Do the same with the substitution $x=1-i$. By taking real and imaginary parts, we get explicit formulas for sums of certain types of binomial coefficients.
By playing with the ideas above, we can get explicit formulas for $\sum \binom{n}{k}$, where (i) $k$ ranges over multiples of $4$; (ii) $k$ ranges over the numbers that have remainder $1$ on division by $4$; $k$ ranges over the numbers with remainder $2$ on division by $4$; (iv) $k$ ranges over the numbers with remainder $3$.
For remainder $0$ or $1$ (your problem) one may not need remainder $0$ and remainder $1$ separately.
| {
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"url": "https://math.stackexchange.com/questions/929354",
"timestamp": "2023-03-29T00:00:00",
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How to find $n$-th value in a series Let $(x_n, y_n, z_n) = (3, 1, 0)$ for $n=0$
For $n \ge 1$,
$$\begin{align}
x_n &= x_{n-1} +3 z_{n-1}\\
y_n &= x_{n-1} +2 z_{n-1}\\
z_n &= 5 y_{n-1}
\end{align}$$
Please let me know the formula to find $x_n,y_n,z_n$ values of any integer $n$.
The following would be the series:
$$\begin{array}{c:l}
n & x_n,y_n,z_n\\
\hline
0 & 3,1,0\\
1 & 3,3,5\\
2 & 18,13,15\\
3 & 63,48,65\\
\end{array}$$
| As pointed out by GerryMyerson, you can rewrite your problem as
\begin{align}
X_n=A\cdot X_{n-1} =\ldots=A^nX_0
\end{align}
With $X_n =(x_n,y_n,z_n)^T$ and $X_0=(3,1,0)$, we only need the appropriate matrix $A$.
Your recurrence relation is
\begin{align}
x_n &=x_{n-1}+3z_{n-1}\\
y_n &=x_{n-1}+2z_{n-1}\\
z_n &= 5 y_{n-1}
\end{align}
Or as matrix-vector product
\begin{align}
\begin{pmatrix}x_{n-1} \\ y_{n-1}\\ z_{n-1} \end{pmatrix}
=
\underbrace{\begin{pmatrix}1 & 0 & 3\\ 1 & 0 & 2 \\ 0 & 5 &0 \end{pmatrix}}_{=:A}
\begin{pmatrix}x_{n-1} \\y_{n-1}\\z_{n-1} \end{pmatrix}
\end{align}
Now we have to diagonalize $A$, i.e. finding matrices, such that $A=V D V^T$
Do you know how to proceed?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to Solve $ \int \frac{dx}{x^3-1} $ I am having quite a difficult time integrating
$$
\int \frac{\mathrm{d}x}{x^3-1}
$$
My first approach was to apply a partial fraction decomposition
$$
\int \frac{\mathrm{d}x}{x^3-1} = \int \frac{\mathrm{d}x}{(x-1)(x^2+x+1)} = \frac{1}{3} \int \frac{\mathrm{d}x}{x-1} - \frac{1}{3} \int \frac{x+2}{x^2+x+1}\mathrm{d}x$$
The first term is simple enough to integrate but I can't figure out the second. I've tried each of the integration tricks in my arsenal, integration of parts gives a more complex equation, a trigonometric substitution doesn't seem to get me anywhere, and lastly I'm not sure what to do with a u-substitution since if $ u = x^2+x+1 $ then $ \mathrm{d}u = (2x+1)\mathrm{d}x \neq (x+2)\mathrm{d}x $. And further breaking up the formula further doesn't seem to eliminate these problems.
My last effort was to get a solution from Wolfram and try to reverse engineer it by taking the derivative
$$
\frac{\mathrm{d}}{\mathrm{d}x} \left( - \frac{1}{6} \ln (x^2+x+1) + \frac{1}{3} \ln (1-x) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) \right)
$$
from which I got
$$
\frac{1}{3(1-x)} - \frac{1}{2(x^2 + x + 1)} - \frac{2x+1}{6x^2 + 6x + 6}
$$
However I can't seem to figure out the method or motivation for the fraction decomposition. I'm I missing something really obvious or am I missing a tool of integration? Thank you very much for your help!
From the first answer suggesting completing the square I believe I was able to figure the rest out. One then gets a difference of squares and can then use a trigonometric substitution where
$$ x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan \theta $$
the answer I got is
$$
\frac{1}{3} \ln |x-1| - \frac{1}{6} \ln(x^2+x+1) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) + C
$$
My last confusion is the discrepancy in the solutions. Wolfram's second term seems to be undefined over $ (1, \infty) $, whereas I believe that $$
\int \frac{\mathrm{d}x}{x^3-1}
$$ should be defined over that interval.
| Hint: Complete the square in the denominator: $x^2+x+1 = (x + 1/2)^2 + 3/4$. Now write the fraction as a linear combination of two fractions whose integral you can do.
| {
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Special values $\psi \left(\frac12\right)$ and $\psi \left(\frac13\right)$ I wonder if it is easy to prove that
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac13\right) & = -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3,
\end{align}
$$
where $\psi$ is the digamma function-the logarithmic derivative of $\Gamma$ function- and $\gamma$ is Euler's constant.
I started with $\psi \left(\dfrac12\right)=\dfrac{\Gamma'}{\Gamma}\left(\dfrac12\right)$ which is not easy to handle.
Thank you for your help.
| You might start from the sum form:
$$ \psi(z) = -\gamma + \sum_{k=1}^\infty \left( \dfrac{1}{k} - \dfrac{1}{k+z-1}\right)$$
For $z=1/2$ the partial sum up to $k=n$ (let's say for convenience that $n$ is even) is
$$ \eqalign{\sum_{k=1}^n \left( \dfrac{1}{k} - \dfrac{2}{2k-1} \right) &=
\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \ldots + \dfrac{1}{n} - \dfrac{2}{1} - \dfrac{2}{3} - \ldots - \dfrac{2}{2n-1}\cr
&= \sum_{k=1}^n (-1)^k \dfrac{1}{k} - \sum_{k=n/2+1}^{n} \dfrac{2}{2k-1}\cr}$$
As $n \to \infty$, the first sum approaches $\displaystyle \sum_{k=1}^\infty (-1)^k \dfrac{1}{k} = -\ln 2$, while the second is approximated by
$$ \int_{n/2}^n \dfrac{2 \; dx}{2x-1} = \ln \left( \dfrac{2n-1}{n-1} \right) \to \ln 2$$
so the result is
$$ \psi(z) = - \gamma - \ln 2 - \ln 2 = -\gamma - 2 \ln 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial fraction for $\displaystyle \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}$
Now Let $\cos x= y\;,$ and Let $\displaystyle \frac{1}{(1-y)(1+y)(5+4y)} = \frac{A}{1+y}+\frac{B}{1-y}+\frac{C}{5+4y}$
after solving We Get $\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{18}$ and $\displaystyle C = -\frac{16}{9}$
So $\displaystyle \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx = \frac{1}{2}\int \frac{1}{1+\cos x}dx - \frac{1}{18}\int\frac{1}{1-\cos x}dx - \frac{16}{9}\int \frac{1}{5+4\cos x}dx$
And after that we can solve easily Like for
$\displaystyle \int\frac{1}{1+\cos x}dx = \int\frac{1-\cos x}{\sin^2 x}dx = \int \left(\csc^2 x-\csc x\cdot \cot x\right)dx = -\cot x +\csc x+\mathcal{C}$
My Question is , Is there is any other method by which we can solbe the above question.
OR without using partial fraction,
Thanks
| Decompose the integrand as follows\begin{align}\int\frac{1}{\sin^2 x\left(5+4\cos x\right)}dx
=& \ \frac19 \int 5\csc^2x-\frac{4\cos x}{\sin^2 x}- \frac{16}{5+4\cos x}\ dx\\
= &\ \frac19\bigg(-5\cot x+\frac4{\sin x} -\frac{32}3\tan^{-1}\frac{\tan\frac x2}3 \bigg)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.
| In method 1, you have a mistake. If $$u=2x^2$$ $$x=\frac{\sqrt{u}}{\sqrt{2}}$$ $$dx=\frac{1}{2 \sqrt{2} \sqrt{u}}$$ and $$\int \frac{x}{\sqrt{1 + 2x^2}} dx=\frac{1}{4} \int \frac{du}{\sqrt{u+1}}=\frac{\sqrt{u+1}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Inclusion/Exclusion to solve $x_1+x_2+x_3=15$ with $x_1,x_2\leq 5$ and $x_3\leq 7$ for non negative integers $x_1,x_2,x_3$ I want to solve for:
Number of integers solutions to the equation $x_1+x_2+x_3=15$ with $x_1,x_2\leq 5$ and $x_3\leq 7$ for non negative integers $x_1,x_2,x_3$
How can this be done using the principle of inclusion/exclusion?
| If $N$ denotes the total number of ways $x_1+x_2+x_3 =15$ without any restrictions (of course, non-negative integers) and $M$ is the number of ways with desired constraints, then
$$
M = N - \left(N_{x_1>5} + N_{x_2>5} + N_{x_3>7}\right) + \left(N_{x_1>5\ and\ x_2>5} + N_{x_1>5\ and\ x_3>7} + N_{x_2>5\ and\ x_3>7}\right) - N_{x_1>5\ and\ x_2>5\ and\ x_3>7}
$$
where the $N$ with a subscript denotes the number of ways with the constraints in the subscript.
We have $$N_{x_1>5} = N_{x_2>5} = {11\choose 2},\ N_{x_3>7} = {9\choose2},$$
$$N_{x_1>5\ and\ x_2>5} = {5\choose 2},\ N_{x_1>5\ and\ x_3>7} = N_{x_2>5\ and\ x_3>7} = {3\choose 2},$$ and $$N_{x_1>5\ and\ x_2>5\ and\ x_3>7} = 0.$$
And the total number of ways without any constraint is ${17\choose 2}$. So,
$$
M = {17\choose 2} - \left( 2{11\choose 2} + {9\choose2}\right) + \left({5\choose 2} + 2{3\choose 2}\right) = 6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
parametric integral relating to hyperbolic function Suppose that $a$ is real number such that $0<a<1$, how can we calculate
$$ I(a)=\int_0^\infty \big(1-\frac{\tanh ax}{\tanh x}\big)dx .$$
As for some speical cases, I can work out $I(1/2)=1$. Any suggestion to the integral above? Tks.
| If you are interested in particular cases:
$$\begin{align}
I\left( \frac15 \right) & = \sqrt{2-\frac{2}{\sqrt{5}}} \, \, \pi \\
I\left( \frac14 \right) & = \frac{\pi}{2} + 1 \\
I\left( \frac13 \right) & = \frac{\sqrt{3}}{3} \, \pi \\
I\left( \frac25 \right) & = \left(\sqrt{2+\frac{2}{\sqrt{5}}} - \frac54 \right) \pi \\
I\left( \frac12 \right) & = 1 \\
I\left( \frac35 \right) & = \frac{10\pi}{3\sqrt3} + \sqrt{1+\frac{2}{\sqrt5}} \, \pi- \sqrt{5+2\sqrt5} \, \pi \\
I\left( \frac23 \right) & = \frac{3\pi}{4} - \frac{\sqrt{3} \, \pi}{3} \\
I\left( \frac34 \right) & = \frac{8\sqrt{3} \,\pi}{27} - \frac{\pi}{2} + \frac13 \\
\end{align}$$
There are also some closed-form for $a \geq 1$:
$$\begin{align}
I(1) & = 0 \\
I(2) & = -\frac{\pi}{4} \\
I(3) & = -\frac{2\sqrt{3} \, \pi}{9} \\
I(4) & = -\frac{\sqrt{2} \, \pi}{4} - \frac{\pi}{8} \\
I(5) & = -\frac{2}{5}\sqrt{1+ \frac{2}{\sqrt5}} \, \pi \\
I(6) & = -\frac{5 \pi}{12} - \frac{\sqrt{3} \, \pi}{9}
\end{align}$$
I've got them by using CAS and some manual simplification.
@spanferkel
$$\begin{align}
I\left(\frac 45 \right) & = -\frac{1}{40} \left(25+48 \sqrt{5-2 \sqrt{5}}-2 \sqrt{2050-352 \sqrt{5}-400 \sqrt{25-11 \sqrt{5}}}\right) \pi \\
I\left(\frac 16 \right) & = \frac{5 \pi }{3 \sqrt{3}}+1
\end{align}$$
The expressions are simplifications of sums with $\arctan$ terms.
@spanferkel
$$\begin{align}
I\left( \frac 18 \right) & = 1+\sqrt{2}\,\pi \\
I\left( \frac 28 \right) & = \frac{\pi}{2} +1\\
I\left( \frac 38 \right) & = -{\frac {16}{27}}\,\sqrt {3}\,\pi+\sqrt {2}\,\pi+\frac 13 \\
I\left( \frac 48 \right) & = 1\\
I\left( \frac 58 \right) & = -\sqrt {2}\,\pi +{\frac {32}{125}}\,\sqrt {25+10\,\sqrt {5}}\,\pi -{\frac {16}{125}}\,\sqrt {25-10\,\sqrt {5}}\,\pi+\frac 15 \\
I\left( \frac 68 \right) & = {\frac {8}{27}}\,\sqrt {3}\,\pi -\frac{\pi}{2} + \frac 13
\end{align}$$
I couldn't find a closed-form of $I(7/8)$. The expressions are coming again from sums in term of $\arctan$.
| {
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"answer_count": 3,
"answer_id": 1
} |
Prove by induction $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n}\ge\frac{2}{3}n\sqrt{n}$ for all positive integers Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$
Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$
I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
| For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
2k\sqrt{k} \geq (2k-1)\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 = 4k^3 - 3k + 1
$$
The rest should follow since $k \geq \frac1{3}$, since the induction is over the positive integers.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$ \begin{align}
& \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \\
& \text{ordering} \\
& a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)\cdot 1}{2\cdot 4\cdot 6...(2n-2)(2n)(2n+2)}= \\
& \underbrace{\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{4} \right)...\left( 1-\frac{1}{2n} \right)}_{n\text{ times}}\frac{1}{(2n+2)} \\
& \text{clearly} \\
& \left( 1-\frac{1}{2} \right)^{n}\frac{1}{(2n+2)}\le a_{n}\le \left( 1-\frac{1}{2n} \right)^{n}\frac{1}{(2n+2)} \\
& \text{Root test} \\
& \sqrt[n]{\left( 1-\frac{1}{2} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}}\le \sqrt[n]{a_{n}}\le \sqrt[n]{\left( 1-\frac{1}{2n} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}} \\
& n\to \infty \\
& \frac{1}{2}\le \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{a_{n}}\le 1 \\
& \text{nothing :(} \\
& \text{Ratio test} \\
& \frac{a_{n}}{a_{n-1}}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)}{2\cdot 4\cdot 6...(2n)(2n+2)}\centerdot \frac{2\cdot 4\cdot 6...(2(n-1)+2)}{1\cdot 3\cdot 5...(2(n-1)-1)}=\frac{2n-1}{2n+2} \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=1 \\
& \text{nothing again} \\
\end{align}
Any suggestions?
| You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2)}\\
& \sim \frac{1}{\sqrt{\pi n}\cdot (2n+2)}\\
& \sim \frac{1}{2\sqrt{\pi}\cdot n^{\Large\frac 32}}, \quad \text{for} \, n \, \text{great}
\end{align}
$$
where we have use Stirling's approximation, then you easily conclude to the convergence of the series.
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I find 2x2 orthonormal diagonalizing matrices using only trigonometry? I have a matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where all values are known), and I eventually want to diagonalize it into:
$$
A=UDV^T
$$
for orthonormal U and V. If I represent U and V as:
$$
U=\begin{bmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{bmatrix}
,
V=\begin{bmatrix} cos(\phi) & sin(\phi) \\ -sin(\phi) & cos(\phi) \end{bmatrix}
$$
Is there a geometric way to quickly find $\theta$ and $\phi$ using $a$, $b$, $c$, and $d$?
I realize that I could find the eigenvalues and work from there algebraically, but I'm wondering if there's a clean visual way of "seeing" $\theta$ and $\phi$, and then going from there.
I was thinking they would be related to the angles of triangles made up of the values of $A$, but I can't seem to figure out which angles they are.
Thanks!
EDIT: I realized I miswrote $U$ and $V$. $U$ and $V$ need to be rotation matrices for this to work. Any ideas?
| Problem statement and restrictions
The problem you propose is this
$$
\begin{align}
\mathbf{A} &= \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \\
% A
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]
%
&=
%
% U
\left[
\begin{array}{cr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{array}
\right]
% S
\left[
\begin{array}{cc}
\sigma_{1} & 0 \\
0 & \sigma_{2}\\
\end{array}
\right]
% V*
\left[
\begin{array}{rr}
\cos \phi & \sin \phi \\
-\sin \phi & \cos \phi \\
\end{array}
\right] \\[3pt]
%
&=
% svd
\left[
\begin{array}{cc}
\sigma_{1}\cos \theta \cos \phi+\sigma_{2}\sin \theta \sin \phi & \sigma_{1}\cos \theta \sin \phi-\sigma_{2}\cos \phi
\sin \theta \\
\sigma_{1}\cos \phi \sin \theta-\sigma_{2} \cos \theta \sin \phi & \sigma_{2}\cos \theta \cos \phi+\sigma_{1}\sin \theta
\sin \phi \\
\end{array}
\right]
%
\end{align}
$$
The task is to connect the Roman alphabetics to the Greeks. Note this restricts us to class of problems where the domain matrices are rotation matrices. This excludes reflection matrices and matrices which are compositions with reflections.
To proceed, require
*
*rank $\left( \mathbf{A} \right)$ = 2 $ \quad \Rightarrow \quad$ $\det \mathbf{A} \ne 0$.
*The variables $a$, $b$, $c$, and $d$ are real.
Linear regression example
Consider a closely related example which has been solved before. The problem is from linear regression
$$
y(x) = a_{0} + a_{1} x
$$
Given a sequence of $m$ measurements of the form $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, one can assemble to systems matrix $\mathbf{A}\in\mathbb{C}^{m\times 2}_{2}$ and the data vector $y\in\mathbb{C}^{m}$ which is not in the null space: $y\notin\mathcal{N}\left( \mathbf{A}^{*} \right)$.
The linear system is
$$
\begin{align}
\mathbf{A} a & = y\\
% A
\left[
\begin{array}{cc}
1 & x_{1} \\
\vdots & \vdots \\
1 & x_{m} \\
\end{array}
\right]
% a
\left[
\begin{array}{c}
a_{0} \\
a_{1}
\end{array}
\right]
%
& =
%
\left[
\begin{array}{c}
y_{1} \\
\vdots \\
y_{m}
\end{array}
\right]
%
\end{align}
$$
The dimensions of this matrix $\mathbf{A}$ here do not match those of your example, there are too many rows. But this may not be a problem, because once we compute the matrices $\mathbf{S}$ and $\mathbf{V}$, we can construct the matrix $\mathbf{U}$ via
$$
\mathbf{S}^{-1} \mathbf{A} \, \mathbf{V} = \mathbf{U}.
$$
Eigensystem
The first step in decomposition is to construct the product matrix $\mathbf{A}^{*}\, \mathbf{A}$ and resolve the eigensystem.
$$
\mathbf{A}^{*}\, \mathbf{A} =
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array}
\right]
$$
The characteristic polynomial is
$$
\begin{align}
p(\lambda) &= \lambda^{2} - \lambda \, \text{tr}
\left( \mathbf{A}^{*} \mathbf{A}\right) + \det
\left( \mathbf{A}^{*} \mathbf{A}\right) \\
\end{align}
$$
The roots $p\left( \lambda_{\pm} \right)$ are the eigenvalues. The square root of the eigenvalues produces the singular values
$$
\sigma = \sqrt{\frac{1}{2}
\left( \mathbf{1}\cdot\mathbf{1} + x \cdot x \pm \sqrt{4
\left(
\mathbf{1}\cdot x
\right)^{2} -
\left( \mathbf{1}\cdot\mathbf{1} - x \cdot x
\right)^{2}}
\right)}
$$
Introduce rotation matrix
Onto computing the domain matrix, and this is where your insight will help. Use the relationship
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} &= \mathbf{V} \, \mathbf{S}^{2} \, \mathbf{V}^{*} \\
%
&=
% V
\left[
\begin{array}{cr}
\cos \phi & \sin -\phi \\
\sin \phi & \cos \phi \\
\end{array}
\right]
%
\mathbf{S}^{2}
% V*
\left[
\begin{array}{rc}
\cos \phi & \sin \phi \\
-\sin \phi & \cos \phi \\
\end{array}
\right] \\[3pt]
%
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array}
\right]
%
&=
%
\left[
\begin{array}{cc}
\sigma_{1}^{2} \cos^{2} \phi + \sigma_{2}^{2} \sin^{2} \phi &
\left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi \\
\left(\sigma_{1}^{2} - \sigma_{2}^{2} \right) \cos \phi \sin \phi &
\sigma_{2}^{2} \cos^{2} \phi + \sigma_{1}^{2} \sin^{2} \phi
\end{array}
\right]
%
\end{align}
$$
Solution
We find solutions like
$$
\cos \phi = \sqrt{\frac{
\left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{1}^{2} -
\left( x \cdot x \right) \sigma_{2}^{2}}
{\sigma_{1}^{4} - \sigma_{2}^{4}}}, \qquad
%
\sin \phi = \sqrt{\frac{
\left(\mathbf{1}\cdot \mathbf{1}\right) \sigma_{2}^{2} -
\left( x \cdot x \right) \sigma_{1}^{2}}
{\sigma_{1}^{4} - \sigma_{2}^{4}}}
$$
This constructs the angles in the rotation matrix for the domain interms of the data.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/943254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to rationalize this root form? Suppose that we have a equation like this:
$$\sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a+b-2\sqrt{ab}}$$
In order to rationalize it, we can apply the formula:
$$\sqrt{a} + \sqrt{b} = \sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a} - \sqrt{b} = \sqrt{a+b -2\sqrt{ab}}$$, where $a>b$
My question is:
Is it possible to rationalize the form like this:
$$\sqrt{10+2(\sqrt{15}+\sqrt{10}+\sqrt{6})}$$ ?
Thanks
| Notice that there are $3$ terms with $2$ so probably it will be of form $(a+b+c)^2$
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\\a^2+b^2+c^2=10\\ab=\sqrt{15}\\ac=\sqrt{10}\\bc=\sqrt{6}\\b^2+c^2=10-a^2\\a^2b^2+a^2c^2=25\\a^2(b^2+c^2)=25\\a^2(10-a^2)=25\\10a^2-a^4-25=0\\a^4-10a^2+25=0\\(a^2-5)^2=0\\a=\sqrt{5}\\ab=\sqrt{15}\implies b=\sqrt{3}\\bc=\sqrt{6}\implies c=\sqrt{2}$$
So the answer is $\sqrt{2}+\sqrt{3}+\sqrt{5}$
| {
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"answer_count": 1,
"answer_id": 0
} |
What exactly happens in the algebraic steps here? $$ \frac{n(n+1)}{2} + (n+1) = (n+1)(\frac{n}{2} + 1) = \frac{(n+1)(n+2)}{2} $$
I don't understand what happens from the first to the second and from the second to the third one.
| Note that $$\frac{n(n+1)}{2} = \frac{n}{2}(n+1)$$ so in the original expression, we have the sum of two terms both of which contain a factor of $(n+1)$. Taking this common factor out we obtain
$$\frac{n(n+1)}{2} + (n+1) = \frac{n}{2}(n+1) + 1(n+1) = (n + 1)\left(\frac{n}{2}+1\right)$$
which is the second expression. As for the third, note that we can rewrite the second factor by putting both terms over the same denominator as such
$$\frac{n}{2} + 1 = \frac{n}{2} + \frac{2}{2} = \frac{n+2}{2}$$
which gives
$$(n + 1)\left(\frac{n}{2}+1\right) = (n+1)\left(\frac{n+2}{2}\right) = \frac{(n+1)(n+2)}{2}.$$
| {
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"answer_count": 3,
"answer_id": 0
} |
Solving equation involving binomial function Solve for $x$ in terms of $i$ and $j$:
$$
\binom{x}{i} = j
$$
where $x$ is Real; $i$ and $j$ are Integers: $x \geqslant i$, $i \geqslant1$, $j \geqslant 0$.
I came across this problem while trying to unrank combinations $\binom{n}{k}$, pp. 51, Algorithm 2.12. I tried the following:
$$
Minimize \left | \binom{x}{i}-j \right |
$$
$$
s.t. x\geqslant i
$$
A plot of $|\binom{x}{i}-j|$ for $i=20$ and $j=10$ is the following:
I ignore if there is a closed-form for the real solutions of $x$ in terms of $i$ and $j$.
| As a partial answer I generated some solutions using CAS.
For $i=1$: $x_1 = j.$
For $i=2$:
$$x_1 = 1/2+1/2\,\sqrt {1+8\,j},$$
$$x_2 = 1/2-1/2\,\sqrt {1+8\,j}.$$
For $i=3$:
$$x_1 = 1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}+{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1,$$
$$x_2 = -1/6\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-1/2\,{\frac {1}{
\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1+1/2\,i\sqrt {3} \left(
1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}} \right),$$
$$x_3 = -1/6\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-1/2\,{\frac {1}{
\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1-1/2\,i\sqrt {3} \left(
1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}} \right).$$
For $i=4$:
$$x_1 = 3/2+1/2\,\sqrt {5+4\,\sqrt {1+24\,j}},$$
$$x_2 = 3/2-1/2\,\sqrt {5+4\,\sqrt {1+24\,j}},$$
$$x_3 = 3/2+1/2\,\sqrt {5-4\,\sqrt {1+24\,j}},$$
$$x_4 = 3/2-1/2\,\sqrt {5-4\,\sqrt {1+24\,j}}.$$
For $i=5$ I didn't find closed-form.
And the solutions for $i=6$ are here.
I'm also very interested in a general solution.
| {
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} |
How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$.
For variance of $Y$, I need to find $E(Y^2)$ first.
This is what I have with U-substitution: $E(Y^2) = \frac{n\theta_{2} - n \theta_{1} + 2n \theta_{1} \theta_{2} - n\theta_{1}^{2} + \theta_{1}^{2}}{(n+1)}$.
But with this the $Var(Y)= E(Y^2) - \left [E(Y) \right ]^{2} \neq n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$
My mistake is definitely the $E(Y^2)$ part, so I would appreciate it if someone here can go over it with me.
Edit:
$$\begin{eqnarray}
\mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$ from below and also my own work.
But now how do I simplify $$Var(Y) = \mathbb{E}\left(Y^2\right) - \left [ \mathbb{E}\left(Y\right) \right ]^{2}$$ to $$Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}?$$
| The conceptual way to do this is to recognize that $Y$ is a location-scale transformation of a beta distribution. If $f_Y(y) \propto (y-\theta_1)^{n-1}$ for $y \in [\theta_1, \theta_2]$, then this suggests the linear, order-preserving transformation $$X = g(Y) = \frac{Y-\theta_1}{\theta_2 - \theta_1}.$$ Then $Y = g^{-1}(X) = (\theta_2 - \theta_1)X + \theta_1$, and the PDF of $X$ is given by $$f_X(x) = f_Y(g^{-1}(x)) \left| \frac{dg^{-1}}{dx} \right| = nx^{n-1}, \quad x \in [0,1].$$ We immediately recognize that $X \sim \mathrm{Beta}(a = n, b = 1)$. Hence $$\begin{align*} \mathrm{E}[X] &= \frac{a}{a+b} = \frac{n}{n+1}, \\ \mathrm{Var}[X] &= \frac{ab}{(a+b)^2(a+b+1)} = \frac{n}{(n+1)^2(n+2)}. \end{align*}$$ Since $Y = (\theta_2 - \theta_1)X + \theta_1$, it immediately follows that $\mathrm{Var}[Y] = (\theta_2-\theta_1)^2 \mathrm{Var}[X]$ and the result is proven.
In the event that we wish to derive the variance of $X$ directly from its PDF, the computation is a trivial exercise in integration: $$\mathrm{E}[X^k] = \int_{x=0}^1 x^k nx^{n-1} \, dx = n \int_{x=0}^1 x^{n+k-1} \, dx = n \left[\frac{x^{n+k}}{n+k} \right]_{x=0}^1 = \frac{n}{n+k}.$$ Thus $$\mathrm{Var}[X] = \mathrm{E}[X^2] - \mathrm{E}[X]^2 = \frac{n}{n+2} - \Bigl(\frac{n}{n+1}\Bigr)^2 = \frac{n}{(n+1)^2(n+2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
summation algebra for $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1}$ Why does $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1} = 1 + 2\sum_{n=1}^\infty x^n$? Shouldn't this be $1 + x + 2\sum_{n=1}^\infty x^n$ because of the $n+1$ in the second summation?
| This may convince you.
\begin{array}\
\displaystyle\sum_{n=0}^{\infty} x^n \hphantom{+ \displaystyle\sum_{n=0}^{\infty} x^{n+1}} &= 1 + \hphantom{2}x + \hphantom{2}x^2 + \dots\\
&\\
\hphantom{\displaystyle\sum_{n=0}^{\infty} x^n +} \displaystyle\sum_{n=0}^{\infty} x^{n+1} &= \hphantom{1 + 2}x + \hphantom{2}x^2 + \dots\\
&\\
\displaystyle\sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} x^{n+1} &= 1 + 2x + 2x^2 + \dots
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series
$$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$
converges?
My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n!
$$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \pi} \sqrt{2n \pi} (\frac{n}{e})^{2n}} =\frac{(2)^{2n}}{4^n \sqrt{n \pi}}$$
The series of the secomd term diverges. It is correct to conclude thatthe series diverges?
Another ideas are welcome!
Thanks
| Another approach: Note that $\frac{(2n)!}{n!n!}$ equals the binomial coefficient $\binom{2n}{n}$. The binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have sum $2^{2n} = 4^n$, thus their average is $\frac{4^n}{2n+1}$. Since $\binom{2n}{n}$ is the largest of these binomial coefficients, it is surely "above average": this means that $\binom{2n}{n} \geq \frac{4^n}{2n+1}$ from which it follows that
$$\frac{(2n)!}{n!n!} \frac{1}{4^n} = \binom{2n}{n} \frac{1}{4^n} \geq \frac{1}{2n+1}.$$ Since the series $\sum_{n \geq 1} \frac{1}{2n+1}$ diverges, the same holds for the series $\sum_{n \geq 1} \binom{2n}{n} \frac{1}{4^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=\lfloor\frac{\sqrt{8x+1}-1}{2}\rfloor$ Question:
let $x\ge 0$, show that
$$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=\lfloor\dfrac{\sqrt{8x+1}-1}{2}\rfloor$$
My idea: let $\lfloor \sqrt{2x}\rfloor =m$
then
$$\sqrt{2x}-1<m\le \sqrt{2x}$$
so
$$m^2\le 2x<(m+1)^2$$
so
$$2x-\lfloor\sqrt{2x}\rfloor\in [m^2-m,m^2+m+1)$$
so
$$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=m $$or
$$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=m-1$$
But Right is eigher is $m$ or $m-1$,then How prove LHS=RHS?
so follow I can't,can you help?Thank you
| Note that $\frac{\sqrt{8x+1}-1}{2}=\sqrt{2x+\frac 14}-\frac 12$.
We know that the LHS equals $m$ or $m-1$. Suppose that the LHS equals $m$. Then
$$m\le \sqrt{2x-m}< m+1$$
Thus
$$m^2+m\le 2x < (m+1)^2+m$$
So the RHS satisfies
$$\sqrt{m^2+m+\frac 14}-\frac 12\le \sqrt{2x+\frac 14}-\frac 12<\sqrt{(m+1)^2+m+\frac 14}-\frac 12$$
Note that $\sqrt{m^2+m+\frac 14}-\frac 12=\sqrt{(m+\frac 12)^2}-\frac 12=m$. We also have
$$\sqrt{(m+1)^2+m+\frac 14}-\frac 12< m+1$$
as can be checked by adding $\frac 12$ and squaring on both sides. After cancelling we are then left with $0<1$, a true statement.
Thus the RHS also equals $m$. An analogous calculation shows that the RHS is $m-1$ if the LHS is $m-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.
How can I proceed?
| Homage a lab
$2\cos A\cos B\cos C=(\cos(A+B)+\cos(A-B))\cos C=(\cos(A-B)-\cos C)\cos C=\frac{1}{4}\cos^2(A-B)-(\cos C-\frac{1}{2}\cos(A-B))^2\le\frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of possible eight digit number divisible by 9 An eight digit number divisible by 9 is o be formed by using 8 digits out of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. Find the number of ways in which it can be done.
I know divisible rule of 9 is sum of all the digits should be multiple of nine. But don't know how to use it in permutation.
| Eight digit no divisible by $9$ i.e. sum of digits divisible by $9$
$(1.)$ Total no formed by $(1,2,3,4,5,6,7,8) = 8!$
$(2.)$ Total no formed by $(0,2,3,4,5,6,7,9) = 7\times7!$
$(3.)$ Total no formed by $(1,0,3,4,5,6,9,8) = 7×7!$
$(4.)$ Total no formed by $(1,2,0,4,5,9,7,8) = 7×7!$
$(5.)$ Total no formed by $(1,2,3,0,5,6,7,8) = 7×7!\, 8! + 28 × 7 ! = 36 × 7 !$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dividing the function by the largest power of x to rationalize(infinity minus infinity)
Calculating the limits of the above when x approaches infinity, why can't we just divide the function by the largest power of x and then say inside of the square root should be equal to "1" when x is at infinity level?
This seems so right to me but I know it is dead wrong. Could somebody tell me what I am missing?
| Since $\lim_{x\to \infty}|x|\sqrt{1\pm\frac 2x}\to \infty$, we have $+\infty-\infty$. We cannot find the limit in this form.
Multiplying
$$\sqrt{x^2+2x}-\sqrt{x^2-2x}$$
by
$$\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\ (=1)$$
gives you
$$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-2x})(\sqrt{x^2+2x}+\sqrt{x^2-2x})}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$
$$=\lim_{x\to \infty}\frac{(x^2+2x)-(x^2-2x)}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}=\lim_{x\to \infty}\frac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$$$=\lim_{x\to\infty}\frac{\frac{4x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-2x}}{x}}=\lim_{x\to\infty}\frac{4}{\sqrt{\frac{x^2+2x}{x^2}}+\sqrt{\frac{x^2-2x}{x^2}}}$$$$=\lim_{x\to \infty}\frac{4}{\sqrt{1+\frac 2x}+\sqrt{1-\frac 2x}}=\frac{4}{\sqrt 1+\sqrt 1}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
If $x^2+y^2+z^2=3/2$, then $\sum\frac{x^2}{x(4x-3)+z^2+y^2} \le1+\frac{\sqrt2}2$ If $x$, $y$, and $z$ are real numbers satisfying
$$
x^2 +y^2 +z^2 = 3/2
$$
then prove that $1+\frac{\sqrt{2}}{2} \geq$ the cyclic sum of
$$
\frac{x^2}{x(4x-3)+z^2+y^2} .
$$
I've tried Cauchy-Schwarz, but I don't know how to restructure the cyclic sum to relate it to $1+\frac{\sqrt{2}}{2}$. Is there a way to do it using just basic inequalities such as Cauchy-Schwarz?
| I think your constant should be $1+\frac{\sqrt2}{2} \lt 1+\frac{2}{\sqrt2}$ which is stronger and still true. Since the denominator $4x^2+y^2+z^2-3x$ is bigger if $x$ is negative it is enough to prove that the inequality holds for positive real numbers $x,y,z$. Therefore, to make it a bit more beautiful, you can make the substitutions $x^2=\frac{3}{2}a$, $y^2=\frac{3}{2}b$ and $z^2=\frac{3}{2}c$. Then, the inequality changes to:
$$
\sum_{cyc} \frac{a}{4a+b+c-\sqrt{6a}} \le 1+\frac{\sqrt2}{2}
$$
For positive reals $a,b,c$ with $a+b+c=1$. Now, with AM-GM we have $\sqrt{6a}=\sqrt{(\sqrt{2})\cdot(3\sqrt{2}a)}\le\frac{1}{2}(\sqrt{2}+3\sqrt{2}a) \iff -\frac{1}{2}(\sqrt{2}+3\sqrt{2}a)\le-\sqrt{6a}$
And therefore:
$$
\sum_{cyc} \frac{a}{4a+b+c-\sqrt{6a}} \le \sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(\sqrt{2}+3\sqrt{2}a)} = \sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(\sqrt{2}(a+b+c)+3\sqrt{2} a)}=\sum_{cyc} \frac{a}{4a+b+c-\frac{1}{2}(4\sqrt{2}a+\sqrt{2}b +\sqrt{2} c)}=\sum_{cyc} \frac{a}{(1-\frac{\sqrt{2}}{2})(4a+b+c)}
$$
So it remains to prove that:
$$
\sum_{cyc} \frac{a}{(1-\frac{\sqrt{2}}{2})(4a+b+c)} \le 1+\frac{\sqrt2}{2} \iff \sum_{cyc} \frac{a}{4a+b+c} \le \frac{1}{2}
$$
If you multiply the whole inequality with $1-\frac{\sqrt{2}}{2}$. Now you can modify it as follows:
$$
\sum_{cyc} \frac{a}{4a+b+c} = \sum_{cyc} \frac{a}{2a+(a+b)+(a+c)} \le \sum_{cyc} \frac{a}{2a+2\sqrt{ab}+2\sqrt{ac}}=\frac{1}{2}\sum_{cyc} \frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{1}{2}
$$
Which follows again from AM-GM, with equality only if $a=b=c=\frac{1}{3}$ so in the original inequality, if $x=y=z=\frac{1}{\sqrt{2}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integrating $\sin^2(x) \cos x $ become $\frac {1}{3}\sin^3(x) + C$? What mathematical process is being done? Why does the $\cos x$ disappear?
| If you let $u = \sin x$, then $du = \cos x \ dx,$ and
$\int \sin^2x \cos x \ dx $ becomes $$\int u^2 du = \frac{1}{3} u^3 + C.$$
Then you substitute $u=\sin x$ back in.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this series.
| Observe that, by absolute convergence:
$$\begin{align}
\sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)& =\sum_{k=1}^{\infty} \dfrac{1+(-1)^k}{2}\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right)\\\\
&=\frac{1}{2}\sum_{k=1}^{\infty} \!\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\!\right)\right)\!+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right) \\\\
&=\frac{\gamma}{2}+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right).
\end{align}
$$
Then, as $N$ is great, write
$$
\begin{align}
\sum_{k=1}^{N} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)& =\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\sum_{k=1}^{N} (-1)^{k-1}\log\!\left(1+\dfrac{1}{k}\!\right)\\\\
&=\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\log \left(\prod_{k=1}^{N} \left(1+\dfrac{1}{k}\!\right)^{(-1)^{k-1}}\right) \tag2
\end{align}
$$ giving
$$
\begin{align}
\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)=\log 2-\log \left(\frac{\pi}{2}\right)
\end{align}
$$
where we have used Wallis' product formula for $\pi$.
Finally we obtain
$$
\sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)=\frac{\gamma}{2}-\frac{1}{2}\log \left(\frac{4}{\pi}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$ I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
| Using the series of $\ln^2(1-x)$,
\begin{align}
\int^1_0\ln^2{x}\ln^2(1-x) \ {\rm d}x
&=\sum^\infty_{n=1}\frac{2H_n}{n+1}\int^1_0x^{n+1}\ln^2{x}\ {\rm d}x\\
&=\sum^\infty_{n=1}\frac{4H_n}{(n+1)(n+2)^3}
\end{align}
Then integrate $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{(z+1)(z+2)^3}$ along an infinitely large square. The integral vanishes which implies the sum of its residues is zero. At the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
&=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{(z+1)(z+2)^3(z-n)^2}+\frac{2H_n}{(z+1)(z+2)^3(z-n)}\right]\\
&=\sum^\infty_{n=1}\frac{2H_n}{(n+1)(n+2)^3}-\sum^\infty_{n=1}\frac{4n+5}{(n+1)^2(n+2)^4}\\
&=\sum^\infty_{n=1}\frac{2H_n}{(n+1)(n+2)^3}+\frac{\pi^4}{30}+2\zeta(3)-\frac{91}{16}
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)
&=\operatorname*{Res}_{z=0}\frac{1}{z^2(z+1)(z+2)^3}\\
&=-\frac{5}{16}
\end{align}
At $z=-1$,
\begin{align}
{\rm Res}(f,-1)
&=0
\end{align}
At $z=-2$,
\begin{align}
{\rm Res}(f,-2)
&=\frac{1}{2}\lim_{z\to-2}\frac{{\rm d}^2}{{\rm d}z^2}\frac{(\gamma+\psi_0(-z))^2}{z+1}\\
&=-\frac{\pi^4}{36}+2\zeta(3)+\frac{2\pi^2}{3}-6
\end{align}
Therefore,
\begin{align}
\int^1_0\ln^2{x}\ln^2(1-x)\ {\rm d}x
&=2\left[-\frac{\pi^4}{30}-2\zeta(3)+\frac{91}{16}+\frac{5}{16}+\frac{\pi^4}{36}-2\zeta(3)-\frac{2\pi^2}{3}+6\right]\\
&=-\frac{\pi^4}{90}-8\zeta(3)-\frac{4\pi^2}{3}+24
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of the intersection of four circles of equal radius
This picture basically shows a rearrangement of four quarters of a circle of radius 1. It asks for the shaded area.
I got the answer to be $\frac{2\pi + 6}{13}$. But then it is incorrect.
The way I did it is by rearranging the four pieces into respectively a circle a square and a larger square with a inscribed circle. And naming each different pieces in the picture as x y and z.
| Note that we could use the first quadrant and then get the final answer by symmetry. We have three quarter-circles in this region with the equations$$\begin{array}{l}{C_1}:{\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\\{C_2}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = 1\\{C_3}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\end{array}$$we want to obtain the area beneath ${C_3}$ (to the $x$-axis) but to obtain the endpoint, we have to find the cross point of ${C_2}$ and ${C_3}$, which is $$\begin{array}{l}x = \frac{{\sqrt 3 - 1}}{2}\\y = 0\end{array}$$now, the final answer is $$I = 4\int\limits_0^{\frac{{\sqrt 3 - 1}}{2}} {\left( { - \frac{1}{2} + \sqrt {1 - {{\left( {x + \frac{1}{2}} \right)}^2}} } \right)dx} = 1 - \sqrt 3 + \frac{\pi }{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Interesting use of remainder theorem I am asking this question in reference to this post Write an Efficient Method to Check if a Number is Multiple of 3
In the proof of the method the author writes that any 2 digit number(AB) can be written in the form $$AB=11A-A+B$$Any 3 digit number(ABC) can be written as $$ABC = 99A + A + 11B – B + C = (99A + 11B) + (A + C – B)$$ And a 4 digit number(ABCD) as $$ 1001A + D + 11C – C + 999B + B – A$$
I have tried a few test cases and it works. But how did he come up with such an observation. And how can I represent a 5 digit number in this way?
Thanks in advance
| Considering congruences modulo $11$ is easier:
\begin{align}
10^0 & \equiv 1 \pmod{11} \\
10^1 & \equiv 10 \equiv -1\\
10^2 & \equiv -1\cdot 10\equiv -1\cdot(-1)\equiv 1\\
10^3 & \equiv 1\cdot 10\equiv 1\cdot(-1)\equiv -1\\
&\dots\\
10^{2k} & \equiv -1\cdot 10\equiv -1\cdot(-1)\equiv 1\\
10^{2k+1} & \equiv 1\cdot 10\equiv 1\cdot(-1)\equiv -1
\end{align}
Thus every power of $10$ is either congruent to $1$ (for even exponents) or to $-1$ (for odd exponents). This means that when you write a number as
$$
a_0+a_1\cdot 10^1+a_2\cdot 10^2+\dots+a_n\cdot 10^n
$$
you can substitute $10^{2k}$ with $1$ and $10^{2k+1}$ with $-1$ and this will make the new number differ from the old one by a multiple of $11$. In particular, checking whether the old number is divisible by $11$ is the same as checking the new number. In the case of five digits, the number is
$$
a_0+a_1\cdot 10+a_2\cdot 10^2+a_3\cdot 10^3+a_4\cdot 10^4
$$
and the number you can consider is
$$
a_0-a_1+a_2-a_3+a_4
$$
For two digits, $a_0+a_1\cdot 10$ becomes $a_0-a_1$, and so on.
| {
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Maximise the volume of an open triangular prism An open container is to be constructed out of 200 square centimeters of cardboard. The two end pieces are equilateral triangles. The open top is a horizontal rectangle. Find the lengths of the sides of the triangle for maximum volume of the container.
So effectively we have an equilateral triangular prism of length $L$.
I first found an equation for $L$: $$L=\frac{200-\sqrt{3}x^2}{2x}$$
I then used $V = \frac{1}{2}x^2Lsin60$ to get the volume, substituting for $L$ to get the volume in terms of $x$.
Differentiating with respect to $x$, setting equal to zero, and solving for $x$ gave me value of $x=6.2$
$$\frac{dv}{dx} = 50\sqrt{0.75} - \frac{9x^2}{8} = 0$$
x=6.204
But this answer was marked wrong by my teacher. Where had I gone wrong?
| Let's let our variable x represent the side of the equilateral triangle. In that case, the area of the equilateral triangle is $$\frac{x^2\sqrt{3}}{4}$$
Therefore, the remaining area for the cardboard to be used by the three rectangles is $$200-
\Big(\frac{x^2\sqrt{3}}{4}\Big)2$$ Because we have three sides, the area taken up by a single rectangle will be $$\frac{200-
\Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3}$$
Because one of the dimensions of the rectangle is x, we can divide by this to find the length, giving$$\frac{200-
\Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3x}$$
The total formula for the volume, therefore, is $$V(x)=\frac{x^2\sqrt{3}}{4} \cdot\frac{200-
\Big(\frac{x^2\sqrt{3}}{4}\Big)2}{3x}$$
$$\frac{x^2\sqrt{3}}{4} \cdot\frac{200-
\Big(x^2\sqrt{3}\Big)2}{12x}$$
$$\frac{200x^2\sqrt{3}-
2\Big(x^2\sqrt{3}\Big)^2}{48x}$$
$$\frac{200x^2\sqrt{3}-
6x^4}{48x}$$
$$\frac{100x\sqrt{3}-
3x^3}{24}$$
$$-8x^3+\frac{25\sqrt{3}}{6}x$$
The first derivative is
$$V'(x)=-24x^2+\frac{25\sqrt{3}}{6}$$
The roots of this equation are approximately -0.548 and 0.548. These are the critical points. The second derivative is
$$V''(x)=-48x$$
I will let you try and figure out which one is the minimum and which is the maximum, and what the final answer is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/963287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 + \log_6 256\\
x + \log_6 4 &= x\log_64 + \log_6 256\\
x-x\log_6 4 &= \log_6 256 - \log_6 4\\
x(1-\log_6 4) &= \log_6 256 - \log_6 4\\
x &= \dfrac{\log_6 256 - \log_6 4}{1-\log_6 4}\\
x &= 10.257}$$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..
| $$2^{2x}3^x=4^{3x+1},$$
$$(e^{\log2})^{2x}(e^{\log3})^x=(e^{\log4})^{3x+1},$$
$$e^{2x\log2}e^{x\log3}=e^{(3x+1)\log4},$$
$$e^{2x\log2+x\log3}=e^{(3x+1)\log4}.$$
Take the log,
$$2x\log2+x\log3=(3x+1)\log4,$$and
$$x=\frac{\log4}{2\log2+\log3-3\log4}=\frac1{\frac{\log3}{\log4}-2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that for any integer a, a^2 + 5 is not divisible by 4. My solution is:
Assume by contradiction that there is at least one number a such that $a^2$ + 5 is divisible by 4. Then a is either odd or even.
Consider the case when a is odd. Then a= 2k+1 for some integer k. Then $a^2$ = $4k^2$ + 4k + 1 so $a^2$ is odd. Then $a^2$ + 5 is an even number. In order for $a^2$ + 5 to be divisible by 4, $a^2$ + 5 = 4p for p ≥ 1, 4p-5 must be a perfect square. Since $a^2$ is odd, therefore 4p-5 must be an odd value thus it is a square of another odd integer.
4p-5 = $(2n+1)^2$ = ($4n^2$ + $4n$ + 1) Thus,$4p$ = $4n^2$ + $4n$ + $6$ => $p$= $n^2$ + $n$ + $1.5$ which shows that p is not an integer therefore 4p-5 cannot be a perfect square. Therefore, $a^2$ + 5 cannot be divisible by 4.
Consider the case when a is even. $a$ = $2k$ for some integer k. Then $a^2$ = $4k^2$. Clearly, $a^2$ is divisible by 4. But $a^2$ + 5 is odd value and clearly it is not divisible by 4.
Thus, in all cases we reach a contradiction. Therefore, $a^2$ + 5 is not divisible by 4 for any integers a.
Am i correct?
| Your idea is fine. More compactly, any number is congruent to $0,1,2,3$ modulo $4$. Squaring gives $0,1,0,1$. But $5=-1$ modulo $4$, but it cannot be the case $a^2=-1\mod 4$ for any $a$, by the above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Continuous Functions such that $f(0) = 1$ and $f(3x) - f(x) = x$ Just had a midterm with the following problem:
Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.
I was just curious how this would end up.
During the exam I tried setting $f(1) = c$, and end up getting $f(3) = 2c, f(9) = 4c, \ldots$ and from that $c = 3/2$ with the $f(0) = 1$ condition, but I was unable to make more progress on it.
| Notice that:
\begin{align*}
f(3x) - f(x) &= x \\
f(9x) - f(3x) &= 3x \\
f(27x) - f(9x) &= 9x \\
&~~\vdots \\
f(3^kx) - f(3^{k-1}x) &= 3^{k-1}x
\end{align*}
Summing them together, the LHS telescopes, giving us:
$$
f(3^kx) - f(x) = x\sum_{i=0}^{k-1}3^i = x \cdot \frac{3^k - 1}{2}
$$
Now consider what happens when we take the limit as $k \to -\infty$. By the continuity of $f$, we can move the limit inside the function, giving us:
\begin{align*}
f\left( \left[\lim_{k\to-\infty}3^k\right]x \right) - f(x) &= x \cdot \frac{\left[\lim\limits_{k\to-\infty}3^k\right] - 1}{2} \\
f(0) - f(x) &= x \cdot \frac{0 - 1}{2} \\
1 - f(x) &= \frac{-x}{2} \\
f(x) &= 1 + \frac{x}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Integrals: Partial Fractions $$ \int \frac{x^2-x+12}{x^3+3x} $$
I factored the denominator to get $ x(x^2+3) $. I then seperated the x and the $x^2+3$ into the partials $\frac{A}{x}$ and $\frac{Bx+C}{x^2+3}$.
After combining the two, I came up with $$ \frac{A(x^2+3)+ x(Bx+C)}{x(x^2+3)}$$
This is where I'm stuck. I'm assuming that A would equal 1 because it is where $x^2$ is in the original problem. The same would go for b equaling -1 and $C = 12$. I'm not so sure though.
| Note that $x^2-x+12=12$ when $x=0$.
Note also that $A(x^2+3)+x(Bx+C)=3A$ when $x=0$. Thus $A=\frac{12}{3}=4$.
The coefficient of $x^2$ in $x^2-x+12$ is $1$. In $A(x^2+3)+x(Bx+C)$ it is $A+B$. Since $A=4$, we have $B=-3$.
We leave finding $C$ to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
| There have already been enough answers, but none addresses why your proposed answer is incorrect. Mine will do just that.
The most basic mistake is that you use what you want to prove as if it was true. This is a logical fallacy called circular reasoning. There is also a number of other mistakes:
*
*From $b^2=4-a^2$ it follows that $b= \pm \sqrt{4-a^2}$ and not necessarily that $b=\sqrt{4-a^2}$.
*Similarly, from $(a+ib)^3=8$ it doesn't follow that $a+ib = 2$. What follows is that $a+ib \in \{2,2 \angle \frac{2\pi}{3}, 2 \angle \frac{4\pi}{3}\}$.
*In the equation $a+ib=2$ (which you mistakenly derived), you substituted $ib$ with what you (mistakenly due to the logical fallacy and (1)) thought $b$ was equal to. That is, you forgot to multiply $\sqrt{4-a^2}$ by $i$.
*In the final step you seem to suggest that $\sqrt{x^2-y^2}=x-y$, but this doesn't hold in general.
Finally, the tag linear-algebra doesn't apply to your question.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
| First Proof
$a+b=x$ , $b+c=y$ , $c+a=z$
$\therefore 2 \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-6\\=\underbrace{\dfrac{x}{y}+\dfrac{y}{x}}_{\geq 2}+\underbrace{\dfrac{y}{z}+\dfrac{z}{y}}_{\geq 2}+\underbrace{\dfrac{z}{x}+\dfrac{x}{z}}_{\geq 2}-3 \geq 3$
Second Proof
$f(a,b,c)=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right) \implies f(ta,tb,tc)=f(a,b,c)$ $\forall t\in \mathbb{R}\setminus\{0\}$
$a+b+c=1 \implies f(a,b,c)=\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3 \geq \dfrac{9}{2}-3=\dfrac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
$$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$
But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$
Where did I make a error in my calculation?
Thanks!
| $$
\frac{3}{\left(\frac 3 2\right)} = 3\cdot\frac 2 3 \ne 3\cdot \frac 3 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Inverse Trigonometric Integrals How to calculate the value of the integrals
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx
$$
and
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$
| Here is a simple and a nice way to evaluate the first and the second integral.
Evaluation of $1^{\mbox{st}}$ Integral :
Making substitution $x=\tan\theta\,$ followed by integration by parts, we get
\begin{align}
\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx&=\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\\
&=-\theta^2\cot\theta\bigg|_0^{\Large\frac{\pi}{4}}+2\int_0^{\Large\frac{\pi}{4}}\theta\cot\theta\,d\theta\tag{1} \\
&=-\frac{\pi^2}{16}+2\theta\ln(\sin\theta)\bigg|_0^{\Large\frac{\pi}{4}}-2\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta\tag{2}\\
&=-\frac{\pi^2}{16}-\frac{\pi}{4}\ln2+G+\frac{\pi}{2}\ln2\tag{3}\\
&=\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2}
\end{align}
Evaluation of $2^{\mbox{nd}}$ Integral :
Again making substitution $x=\tan\theta\,$ followed by integration by parts, we get
\begin{align}
\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx&=\int_0^{\Large\frac{\pi}{4}}\frac{\theta^3\cos\theta}{\sin^3\theta}\,d\theta\\
&=-\left.\frac{\theta^3}{2\sin^2\theta}\right|_0^{\Large\frac{\pi}{4}}+\frac{3}{2}\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\tag{4}\\
&=-\frac{\pi^3}{64}++\frac{3}{2}\left[\color{red}{G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2}\right]\\
&=\color{blue}{\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2}
\end{align}
Explanation :
$(1)$ Integration by parts, $u=\theta^2\,\mbox{ and }\,dv=\dfrac{d\theta}{\sin^2\theta}$
$(2)$ Integration by parts, $u=\theta\,\mbox{ and }\,dv=\cot\theta\,d\theta$
$(3)$ The evaluation of $\displaystyle\int_0^{\Large\frac{\pi}{4}}\ln(\sin\theta)\,d\theta$. See Mr. Tunk-Fey's answer, his answer is the best!
$(4)$ Integration by parts, $u=\theta^3\,\mbox{ and }\,dv=\dfrac{\cos\theta}{\sin^3\theta}\,d\theta$
Done! $\,$ (>‿◠)✌
| {
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"url": "https://math.stackexchange.com/questions/984026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Closed-form of $\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx$ Inspired by this question, is there a closed-form of
$$\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx\,?$$
Here $n \in \mathbb{N_+}$. In the answers to the question above we could find proofs of cases $n=2,3$.
I state here some specific cases.
$$\begin{align}
\int_0^1\frac{\arctan x}{x}\,dx & = G, \\
\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx & =G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2,\\
\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx & = \frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2.\\
\end{align}$$
Furtheremore I've evaluated $n=4,5$ cases.
$$\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^4dx$$
equals to
$$2G-\frac{3\pi^4}{256}-\frac{\pi^3}{48}-\frac{\pi^2}{8}-\frac{\pi^2G}{8}+\frac{3\pi}{64}\zeta(3)-\frac{\pi^3}{96} \ln2+\frac{\pi}{2} \ln2+\frac{1}{768}\psi_3\left(\frac{1}{4}\right),$$
and
$$\int_{0}^{1}\left(\frac{\arctan(x)}{x}\right)^5dx$$
equals to
$$\frac{5G}{2}-\frac{25\pi^4}{512}-\frac{5\pi^3}{192}-\frac{5\pi^2}{32}-\frac{5\pi^2 G}{8}+\frac{15\pi}{64}\zeta(3)-\frac{5\pi^3}{96}\ln 2+\frac{5\pi}{8}\ln 2 + \frac{5}{768}\psi_3\left(\frac{1}{4}\right).$$
Here $G$ is Catalan's constant, $\zeta$ is the Riemann zeta function, $\psi_3$ is the polygamma function of order $3$, and $\pi$ is also a famous constant.
Note that the problem is related to Dirichlet beta function, since
$$\begin{align}
\beta(2) & = G \\
\beta(3) & = \frac{\pi^3}{32} \\
\beta(4) & = \frac{1}{768}\left(\psi_3\left(\frac{1}{4}\right)-8\pi^4\right).
\end{align}$$
| Using substitution and integration by parts,
\begin{align}&\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\int_0^{\pi/4}\frac{x^n}{\tan^n x}\sec^2 x\,dx\\&=\left[\frac{x^n}{\tan^n x}\tan x\right]_0^{\pi/4}-\int_0^{\pi/4}\frac{nx^{n-1}}{\tan^{n-1}x}-\frac{nx^n}{\tan^n x}\sec^2x\,dx\end{align}
So we have $$\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\frac{n}{n-1}\int_0^{\pi/4}\frac{x^{n-1}}{\tan^{n-1}x}\,dx-\frac{1}{n-1}\left(\frac{\pi}{4}\right)^n$$
Now define $\displaystyle I(m,n)=\int_0^{\pi/4}x^m\cot^nx\,dx$ for $m\ge n$. Using integration by parts, we get a recurrence formula.
\begin{align}I(m,n)&=\int_0^{\pi/4}x^{m}\cot^{n}x\,dx\\&=\left[\frac{1}{m+1}x^{m+1}\cot^n x\right]_0^{\pi/4}+\frac{n}{m+1}\int_0^{\pi/4}x^{m+1}\cot^{n-1}x(\cot^2x+1)\,dx\end{align}
So we have $\displaystyle I(m,n)=-I(m,n-2)+\frac{m}{n-1}I(m-1,n-1)-\frac{1}{n-1}\left(\frac\pi4\right)^m$. We now need to determine $I(m,0)$ and $I(m,1)$ to find $I(n-1,n-1)$. It is easily seen $$I(m,0)=\frac{1}{m+1}\left(\frac\pi4\right)^{m+1}$$ $I(m,1)$ is a bit of a problem. This is where things start to get messy.(All zetas, catalans, and polygammas appear here)
\begin{align}I(m,1)&=\int_0^{\pi/4}x^m\cot x\,dx\\&=\left[x^m\log\sin x\right]_0^{\pi/4}-m\int_0^{\pi/4}x^{m-1}\log\sin x\,dx\\&=\frac{\log 2}{2}\left(\frac{\pi}{4}\right)^m+\sum_{n=1}^\infty\frac mn\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx\end{align} Now using integration by parts,$$\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx=\frac{1}{2n}\left(\frac\pi4\right)^{m-1}\sin\frac{n\pi}2+\frac{m-1}{(2n)^2}\left(\frac\pi4\right)^{m-2}\cos\frac{n\pi}2-\frac{(m-1)(m-2)}{(2n)^2}\int_0^{\pi/4}x^{m-3}\cos 2nx\,dx$$
This yields
\begin{align}\int_0^{\pi/4}x^{m-1}\cos 2nx\,dx=&\sin\frac{n\pi}{2}\sum_{k=1}^{[\frac{m+1}2]}\frac{(-1)^{k-1}}{(2n)^{2k-1}}\left(\frac\pi4\right)^{m-2k+1}\frac{(m-1)!}{(m-2k+1)!}\\&+\cos\frac{n\pi}{2}\sum_{k=1}^{[\frac m2]}\frac{(-1)^{k-1}}{(2n)^{2k}}\left(\frac\pi4\right)^{m-2k}\frac{(m-1)!}{(m-2k)!}+\left(\frac{(m-1)!}{(2n)^m}(-1)^{m/2}\right)_{2|m}\end{align}
where the last term $\displaystyle\left(\frac{(m-1)!}{(2n)^m}(-1)^{m/2}\right)$ is present iff $m$ is even. Now plug this up on the above summation and we get the $I(m,1)$.
\begin{align}I(m,1)=&\sum_{k=1}^{[\frac{m+1}2]}\frac{(-1)^{k-1}\pi^{m+1-2k}}{2^{2m+1+4k}}\binom{m}{2k-1}\left(\psi_{2k-1}\left(\frac14\right)-\psi_{2k-1}\left(\frac34\right)\right)\\&+\sum_{k=1}^{[\frac m2]}\frac{(-1)^{k-1}\pi^{m-2k}(2^{2k}-1)}{2^{2m+2k+1}}\frac{m!}{(m-2k)!}\zeta(2k+1)+\left(\frac{(-1)^{m/2}m!}{2^m}\zeta(m+1)\right)_{2|m}\end{align}
| {
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"answer_id": 0
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Finding $\sum \frac{1}{n^2+7n+9}$ How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$
I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.
| $$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+7n+9}
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}-\frac1{n+\frac72+\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}+\frac1{-n-\frac72-\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\sum_{n\in\mathbb{Z}}\frac1{n+\frac{1}{2}-\frac{\sqrt{13}}{2}}\\
&-\frac1{\sqrt{13}}\left(\frac1{\frac52-\frac{\sqrt{13}}{2}}+\frac1{\frac32-\frac{\sqrt{13}}{2}}+\frac1{\frac12-\frac{\sqrt{13}}{2}}\right)\\
&-\frac1{\sqrt{13}}\left(\frac1{-\frac52-\frac{\sqrt{13}}{2}}+\frac1{-\frac32-\frac{\sqrt{13}}{2}}+\frac1{-\frac12-\frac{\sqrt{13}}{2}}\right)\\
&=\frac\pi{\sqrt{13}}\cot\left(\frac\pi2-\frac{\pi\sqrt{13}}{2}\right)-\frac1{\sqrt{13}}\left(\frac{\sqrt{13}}{3}-\frac{\sqrt{13}}{1}-\frac{\sqrt{13}}{3}\right)\\[3pt]
&=\frac\pi{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)+1
\end{align}
$$
where we use $(7)$ from this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Square root and principal square root confusion A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the concept with the following two questions...
Question 1:
$\sqrt{12x^2} = \sqrt{4\cdot3x^2} = 2|x|\sqrt{3}$
Why do we only have to keep in mind that $x$ could be negative, yet we just factorize $\sqrt{12}$ to $\sqrt{3\cdot4}$ and put the 2 right in front of the radical sign?
$\sqrt{12}$ could be factorized as $\sqrt{-3 \cdot -4}$ as well.
Question 2:
I've been told to take the square root of both sides in the following equation, but the square root doesn't have its own symbol right? Only the principal root does... So if you'd use the radical sign you're using the principal root and therefore you're missing out on a solution in the following scenario:
$(x + 8)² = 1 \iff \sqrt{(x + 8)^2} = \sqrt1 \iff x + 8 = 1 \iff x = -7$
Does that mean that whenever we want to take the square root of something (not the principal) we just use a $\pm$ symbol instead? For example:
$(x + 8)^2 = 1 \iff x + 8 = \pm 1 \iff x = -8\pm1$
| *
*We have to keep in mind that $x$ might be negative because we don't know what $x$ is and it might by negative.
But we don't have to keep anything in mind about $12$ because we know exactly what $12$ is.
But perhaps we need to understand why we are allowed to split $\sqrt {ab}=\sqrt{a}\sqrt{b}$ in the first place.
Real Number Answer (Assumes Since $m^2 \ge 0$ for all $m \in \mathbb R$ then $x^2 = $a negative number is impossible)
If $k=\sqrt{a}\sqrt{b}>0$ (presuming $a,b> 0$) then $k^2 = (\sqrt{a}\sqrt{b})^2 = \sqrt{a}^2 \sqrt{b}^2 = ab$ so if we ask "What is $\sqrt {ab}$, that is what is the unique non-negative number so $x^2=ab$". Well, we just demonstrated that $k = \sqrt{a}\sqrt{b}$ is such a number.
So if we have $M > 0$ and $M$ could be written as $M = a*b$ where $a,b\ge 0$ we would have $\sqrt M = \sqrt{a}\sqrt {b}$.
Now you are correct that $M = (-a)*(-b)$. But how does that help us in anyway. We explained why $M =a*b$ works but $M=(-a)*(-b)$ is as useful for find square roots as noting that $M = j+k$ for some $j$ and $k$. It's just not relevant.
We need $\sqrt{a}$ so that $\sqrt{a}^2 =a$ as well as a $\sqrt{b}$ so that $\sqrt{b}^2 = b$ for this to work and if we tried it with $-a*-b$ we'd need $\sqrt{-a}$ so that $\sqrt{-a}^2 = -a$ and there's just no such thing.
Unless we use complex numbers...
Complex number: There exist $i$ and $-i = (-1)*i$ so that $i^2=-1$. (And $(-i)^2 = i^2 = -1$ as well.
So if $k^2 = 12$ we could do $(\sqrt{3}2)^2 = 3*4 = 12$ so $k=\sqrt 3\sqrt 4=2$ is an acceptable answer. So is $(\sqrt{3}i*2i)^2 = (\sqrt\3i)^2*(2i)^2 = (-3)*(-4) = 12$ but $\sqrt{3}i*2i$ is just another way of writing $\sqrt{3}i*2i = (\sqrt{3}2)*(i^2)= -2\sqrt 3$.
So the two values of $k$ are $2\sqrt 3$ and $-2\sqrt 3$ just as we wanted.
But we can do $\sqrt{12} = \sqrt{-3}\sqrt{-4} = \sqrt 3i* 2i=-2\sqrt 3$.
(I suppose I should point out we tend not to worry about primary square roots in complex numbers because it just opens a can of worms.)
*"I've been told to take the square root of both sides in the following equation"
Well, short answer, since we do not know whether $x+8$ is positive or negative we don't know that $\sqrt{(x+8)^2} = x+8$. If $x = -17$ that is false.
$\sqrt{(-17+8)^2} = \sqrt{(-9)^2} =\sqrt{81}= 9\ne -17+8$.
Instead all we know is that $\sqrt{(x+8^2} = |x+8|$
$\sqrt{(-17+8)^2}=\sqrt{81} = 9 $ which does $|-9| = |-17+8|$.
So to solve
$(x + 8)^2 = 1 \implies \sqrt{(x+8)^2} = \sqrt 1\implies |x+8| = 1$
$\implies -(x+8) = 1$ OR $x+8 = 1$ $\implies$
$x+8 = -1$ ** OR ** $x+8 = 1$
And we can write that last line as
$x+8 = \pm 1$.
Or we can take a short cut and go directly to:
$(x+8)^2 = 1$
$\sqrt{(8+x)^2} = \sqrt 1$
$8+ x = \pm 1$.
.... and from here we can go
$x = \pm 1 -8$ so $x= -1 -8=-9$ OR $x=1-8=-7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $ab-h^2 =0 $ represent a parabola,
$ab-h^2>1$ represent an ellipse
$ab-h^2 <0$ represent a hyperbola
Here $ab-h^2 <0$ therefore equation (1) represents hyperbola and equation (2) is parabola. ( where a,b, h are coefficients of $x^2,y^2,xy$ respectively.
Now how to get the minimum distance between the two curves please suggest .. thanks.
| $y^2-xy-2x^2=0$, factorize it we get $(y-2x)(y+x)=0$, so the graph is two straight line $-2x+y=0$ or $x+y=0$.
Points on $y^2=x-2$ can be written as $(t^2+2,t)$, the distance from $(t^2+2,t)$ to $-2x+y=0$ is $\frac{|-2(t^2+2)+t|}{\sqrt{(-2)^2+1^2}}= \frac{|2t^2-t+4|}{\sqrt{5}}=\frac{|2(t-1/4)^2+31/8|}{\sqrt{5}}$.
The distance from $(t^2+2,t)$ to $x+y=0$ is $\frac{|t^2+2+t|}{\sqrt{(1)^2+1^2}}= \frac{|(t+1/2)^2+7/4|}{\sqrt{2}}$.
$\frac{7}{4\sqrt2}<\frac{31}{8\sqrt5}$ so $\frac{7}{4\sqrt2}=\frac{7\sqrt{2}}{8}$ is minimum distance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.
Is the answer
$$
\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?
$$
I understand the concept of Matrix Transformation, I don't think I'm doing it right.
| There is a very simple method to solve the problem described in "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely."
Consider the formula
$$
\vec{L}(\vec{p}) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{x} & \vec{y} & \vec{z} \\
p_1 & a_1 & b_1 & c_1 \\
p_2 & a_2 & b_2 & c_2 \\
p_3 & a_3 & b_3 & c_3 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{pmatrix}
},
$$
where $\vec{L}$ is linear transformation acting on arbitrary point $\vec{p}$. $\vec{L}$ has the property
$$
\vec{L}(\vec{a}) = \vec{x};\quad
\vec{L}(\vec{b}) = \vec{y};\quad
\vec{L}(\vec{c}) = \vec{z}.
$$
Indices designate components of the corresponding vector.
Let's consider your case.
We need such $\vec{L}$ that
$$
\vec{L}: \begin{pmatrix}1\\ 1\\ 1\end{pmatrix} \mapsto
\begin{pmatrix}1\\ 1\\ 1\end{pmatrix};~
\vec{L}: \begin{pmatrix}0\\ 1\\ 0\end{pmatrix} \mapsto
\begin{pmatrix}0\\ 1\\ 0\end{pmatrix};~
\vec{L}: \begin{pmatrix}1\\ 0\\ 2\end{pmatrix} \mapsto
\begin{pmatrix}1\\ 0\\ 1\end{pmatrix}.
$$
Now I plug them into the general expression
$$
\vec{L}(\vec{p}) =
(-1)
\frac{
\det
\begin{pmatrix}
0 & (1,1,1)^T & (0,1,0)^T & (1,0,1)^T \\
\begin{matrix}
p_{1} \\
p_{2} \\
p_{3} \\
\end{matrix} &
%
\begin{matrix}1\\ 1\\ 1\end{matrix} &
%
\begin{matrix}0\\ 1\\ 0\end{matrix} &
%
\begin{matrix}1\\ 0\\ 2\end{matrix}
\end{pmatrix}
}{
\det
\begin{pmatrix}
\begin{matrix}1\\ 1\\ 1\end{matrix} &
%
\begin{matrix}0\\ 1\\ 0\end{matrix} &
%
\begin{matrix}1\\ 0\\ 2\end{matrix}
\end{pmatrix}
}.
$$
Doing determinants I get
$$
= \left[
2\begin{pmatrix}1\\ 1\\ 1\end{pmatrix} -
2\begin{pmatrix}0\\ 1\\ 0\end{pmatrix} -
\begin{pmatrix}1\\ 0\\ 1\end{pmatrix}
\right] p_1 -
\begin{pmatrix}0\\ 1\\ 0\end{pmatrix} p_2 -
\left[
\begin{pmatrix}1\\ 1\\ 1\end{pmatrix} -
\begin{pmatrix}0\\ 1\\ 0\end{pmatrix} -
\begin{pmatrix}1\\ 0\\ 1\end{pmatrix}
\right] p_3
$$
or simplified
$$
\vec{L}(\vec{p}) =
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} p_1 +
\begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix} p_2 +
\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix} p_3 .
$$
Of course, you can write that in a vector form
$$
\vec{L}(\vec{p}) =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix}.
$$
Now you can easily check
$$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}1\\ 1\\ 1\end{pmatrix} =
\begin{pmatrix}1\\ 1\\ 1\end{pmatrix};~
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}0\\ 1\\ 0\end{pmatrix} =
\begin{pmatrix}0\\ 1\\ 0\end{pmatrix};~
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} =
\begin{pmatrix}1\\ 0\\ 1\end{pmatrix};
$$
For more details on the methods used, you can always refer to "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely". The latter contains many problems similar to this one as explained by the authors of the method presented.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/996217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Tricky Decomposition Decompose the following polynomial:
$A=x^4(y^2+z^2) + y^4(z^2+x^2) + z^4(x^2+y^2) +2x^2 y^2 z^2 $
(by the way I'm not quite sure about the tags and the title feel free to edit them if you wish)
| $$
A = x^4(y^2+z^2) + x^2(y^4+2y^2 z^2 +z^4) +y^2z^2(y^2+z^2) \\
= (y^2+z^2)\left(x^4+(y^2+z^2)x^2 +y^2z^2\right) \\
=(y^2+z^2)(x^2+y^2)(z^2+x^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational Are there infinitely many pairs of different positive rational numbers $x,y$ such that $\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational?
Consider such a pair. Then we have $x^2+y^3=a^2$ and $x^3+y^2=b^2$ for some rationals $a,b$. So $(a^2-y^3)^3=x^6=(b^2-y^2)^2$. Does this lead to something useful?
| Yes, there are infinitely many solutions in integers.
In particular, we have a solution with $y = 2x$ whenever $x + 4$ and $8 x + 1$ are both squares (i.e. $x^2 + (2x)^3 = x^2 (1 + 8 x)$ and $(2x)^2 + x^3 = x^2 (4 + x)$). If $x + 4 = u^2$ and $8 x + 1 = v^2$, eliminating $x$ gives us the
equation $8 u^2 - 31 = v^2$, which is similar to a Pell equation. One solution
is $u = 2, v = 1$, and whenever $\pmatrix{u\cr v}$ is a solution so is
$ \pmatrix{3 & 1\cr 8 & 3\cr} \pmatrix{u\cr v}$. This gives the sequence
$$ \pmatrix{u\cr v\cr} = \pmatrix{2\cr 1\cr}, \pmatrix{7\cr 19\cr}, \pmatrix{40\cr 113\cr}, \ldots $$
corresponding to
$$ \pmatrix{x\cr y\cr} = \pmatrix{0\cr 0\cr}, \pmatrix{45\cr 90\cr}, \pmatrix{1596\cr 3192\cr}, \ldots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible? How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible on $\mathbb Q[X]$?
I can try to find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X^2+aX+b)(X^2+cX+d)$$
and prove that $a,b,c,d\notin\mathbb Q$ and find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X-a)(X^3+bX^2+cX+d)$$
and prove that $a,b,c,d\notin \mathbb Q$, but it's really long and I would like to know if there is an other method.
| Let $P(X)=A(X)B(X)$ a decomposition of your polynomial into quadratics. Observe that
$$
P(X)\equiv X(X^3+3X^2+X+2) \bmod 5
$$
and the deg 3 factor is irreducible mod 5 because it has no roots mod 5. Thus one of the quadratic factors $A, B$ has only one root mod 5, but this is not possible. Hence there's no factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find basis for subspace given by $x_1/3=x_2/4=x_3/2$ Find a basis for the subspace W: $$
W = \left\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} \in \mathbb{R}^3 \left| \dfrac{x_{1}}{3}=\dfrac{x_{2}}{4}=\dfrac{x_{3}}{2} \right.\right\}
$$
The two equation signs are really throwing me off. If it was a linear equation with one equation sign then I could easily find the basis by putting the equation in a matrix and solve for each variable but I cannot do this for this equation. Can anyone help me out? Thanks.
| What you have here are two equation:
$$
\frac{x_{1}}{3}=\frac{x_{2}}{4}
$$
$$
\frac{x_{2}}{4}=\frac{x_{3}}{2}
$$
from the first one you get $x_{2}=\frac{4}{3}x_{1}$ and from the
second one you get $x_{3}=\frac{1}{2}x_{2}$, now using $x_{2}=\frac{4}{3}x_{1}$
we get $x_{3}=\frac{1}{2}\cdot\frac{4}{3}x_{1}=\frac{2}{3}x_{1}$
so
$$
W=\{\begin{pmatrix}x_{1}\\
\frac{4}{3}x_{1}\\
\frac{2}{3}x_{1}
\end{pmatrix}\mid x_{1}\in\mathbb{R}\}=\{\begin{pmatrix}1\\
\frac{4}{3}\\
\frac{2}{3}
\end{pmatrix}x_{1}\mid x_{1}\in\mathbb{R}\}=sp\{\begin{pmatrix}1\\
\frac{4}{3}\\
\frac{2}{3}
\end{pmatrix}\}
$$
A small thing to note: we have two independent equations and so the
dimension of $W$ is $3-2=1$ as expected
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$
| $$\frac{ky^2}{2}+cy=\frac{kx^2}{2}-cx \iff y^2+\frac{2cy}{k}=\frac2k \left (\frac{kx^2}{2}-ck \right ) \iff y^2+\frac{2cy}{k}+\frac{c^2}{k^2}=\frac2k \left (\frac{kx^2}{2}-ck \right )+\frac{c^2}{k^2} \iff \left (y+\frac{c}{k}\right )^2 =\frac{(c-kx)^2}{k^2}$$
I think you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule.
I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having trouble differentiating $f(x) = (\frac{x+1}{x-1})^{\sqrt{x^2-1}}$. I've used Mathematica, but I won't understand it unless I see the step-by-step process.
| Note that:
$$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = \left(1 + \frac{2}{x-1}\right)^{\sqrt{x^2-1}} = e^{\sqrt{x^2-1} \log \left(1 + \frac{2}{x-1}\right)}$$
Now we work with the exponent:
$$\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \log \left(1 + \frac{2}{x-1}\right)
\overset{(1)}{=}
\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \frac{2}{x-1}
\overset{(2)}{=}
2$$
Therefore:
$$\lim_{x\rightarrow \infty} \left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = e^2$$
Observations.
$(1)$ Note that $\log(1+z) \sim z$ when $z$ goes to zero. Here "$z = \frac{2}{x-1}$".
$(2)$ Clearly $\lim_{x\rightarrow \infty} \frac{\sqrt{x^2-1}}{x-1}=
\lim_{x\rightarrow \infty}\sqrt{\frac{(x-1)(x+1)}{(x-1)^2}}=
\lim_{x\rightarrow \infty} \sqrt{\frac{x+1}{x-1}}=
\lim_{x\rightarrow \infty} \sqrt{1+\frac{2}{x-1}}
=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
A limit-determinant question
Let $d_n$ be the determinant of the $n\times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos1,\cos2,\ldots,\cos n^2$. (For example,
$$d_3=
\begin{vmatrix}
\cos1&\cos2&\cos3\\
\cos4&\cos5&\cos6\\
\cos7&\cos8&\cos9\\
\end{vmatrix}$$
The argument of $\cos$ is always in radians, not degrees.) Evaluate $\lim_{n\to\infty}d_n$.
Interesting question, I don't know where to start.
| $$\begin{array}{l} \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\ =\left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 3\cos 1 - \sin 3\sin 1 & \cos 3\cos 2 - \sin 3\sin 2 & \cos 3\cos 3 - \sin 3\sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\= \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ - \sin 3\sin 1 & - \sin 3\sin 2 & - \sin 3\sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right|\\= -\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \sin 1 & \sin 2 & \sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\=-\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \sin 1 & \sin 2 & \sin 3 \\ \cos 6\cos 1-\sin 6\sin 1 & \cos 6\cos 2-\sin 6\sin 2 & \cos 6\cos 3-\sin 6\sin 3 \end{array}\right| \\=-\sin 3 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \sin 1 & \sin 2 & \sin 3 \\ -\sin 6\sin 1 & -\sin 6\sin 2 & -\sin 6\sin 3 \end{array}\right| \\=\sin 3 \sin 6 \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \sin 1 & \sin 2 & \sin 3 \\ \sin 1 & \sin 2 & \sin 3 \end{array}\right| =0. \end{array}$$
Can you get the value of $d_n$ following the same idea?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving Complex Quadratic equations After working a few exercises on the topic, the questions become progressively harder. In this particular exercise I was asked to solve the equations. However I can't quite seem to break this problem as I did previously and I'll explain why.
The equation is as follows: $$z^2 + (1-i)z + (-6 + 2i) = 0$$
My thought was to use the general formula to get the value of $z$, however it doesn't seem to be the case, or isn't as straight-forward as previous quadratics I've worked with. Please do see my attempt and correct my mistakes whether it be on this question are any other thing in particular that I'm doing wrong, self-teaching is somewhat harder.
My attempt:
Use the formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
In my case, $a = 1, b = (1-i), c = (-6 + 2i)$
$$\therefore z = -(1-i) \pm \frac{\sqrt{(1-i)^2 - 4(1)(-6+2i)}}{2(1)}$$
Which boils down to: $$z = -1+i \pm{\sqrt{12-4i}}$$
From here on I'm confused as to how to continue. Am I going completely off track?
| $$
z^2 + (1-i)z + (-6 + 2i) = 0\\
\begin{align}
\implies z &= \frac{-(1-i) \pm \sqrt{(1-i)^2 - 4(-6 + 2i)}}{2}\\
&= \frac{1}{2}\left( i - 1 \,\pm \sqrt{1+ 2i^2 - i^2 +24 - 8i}\right)\\
&= \frac{1}{2}\left(i-1 \pm 2i\sqrt{2i - 6}\right)\\
&= \left(\frac{1}{2}\pm\sqrt{2(i - 3)} \right)i - \frac{1}{2}
\end{align}
$$
Always remember that the quadratic equation is just a result of "Completing the Square" for a general quadratic:
$$ az^2 + bz + c = 0\\
\implies z^2 + \frac{b}{a}z + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 +\frac{c}{a} = \left(z + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\cdot\frac{4a}{4a} = 0\\
\implies z + \frac{b}{2a} = \sqrt{\frac{b^2 + 4ac}{4a^2}}\\
\implies z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sin^2(x), \cos^2(x),$ and $\sin^4(x)$ and linear dependence Since
$$
\sin^{4}x=\sin^{2}x\sin^{2}x = \sin^{2}x(1-\cos^{2}x) = \sin^{2}x-\sin^{2}x\cos^{2}x
$$
does this mean that
$$
\sin^{2}x,\cos^{2}x, \text{ and } \sin^{4}x
$$
are linearly dependent?
| You can once again use the Wronskian: $$W(x) = \begin{vmatrix} \sin^2x & \cos^2x & \sin^4x \\ \sin(2x) & -\sin(2x) & 4\sin^3x \cos x \\ 2\cos(2x) & -2\cos(2x) & 12\sin^2x \cos^2x + -4\sin^4x\end{vmatrix},$$ and see if it is identically zero, or not. But that's quite a pain here, so you can take a leap of faith: $$W(\pi/4) = \begin{vmatrix} 1/2 & 1/2 & 1/4 \\ 1 & -1 & 1 \\ 0 & 0 & 2\end{vmatrix} = -2 \neq 0,$$ and so they are linearly independent. Or you can use the definition of linear dependence. Consider: $$a\sin^2x + b\cos^2x + c\sin^4x = 0$$
This has to be true for all $x$, so, taking $x = 0$ gives $b = 0$, leaving us with $a\sin^2x + c \sin^4x = 0$, that is, $(\sin^2x) (a + c \sin^2 x) = 0$. Now, if we don't take values of $x$ that make $\sin x = 0$, we can look only at $$a + c\sin^2x = 0.$$
Taking $x = \pi/2$ gives us $a = -c$, and $x = \pi/4$ gives $a = -c/2$. Hence $a = c = 0$ too, and the functions are linearly dependent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2$ This question came up while I was revising the Gamma function.
$$
\prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2
$$
Please prove me prove this infinite product, and explain the steps involved. I'm really having trouble with this one. Many thanks.
| Let $P_m = \displaystyle\prod_{n=1}^{m}\left[1+\dfrac{(-1)^{n+1}}{2n-1}\right]$. Then, $P_{2m} = \dfrac{2}{1}\dfrac{2}{3}\dfrac{6}{5}\dfrac{6}{7} \cdots \dfrac{4m-2}{4m-3}\dfrac{4m-2}{4m-1}$.
This can be rewritten as $P_{2m} = \dfrac{2^{2m}(2m)!^3}{(4m)!(m!)^2} = \dfrac{2^{2m}\binom{2m}{m}}{\binom{4m}{2m}}$.
The Central Binomial Coefficient satisfies $\dbinom{2n}{n} \sim \dfrac{4^n}{\sqrt{\pi n}}$ for large $n$.
Thus, $P_{2m} = \dfrac{2^{2m}\binom{2m}{m}}{\binom{4m}{2m}} \to \dfrac{2^{2m}\frac{4^m}{\sqrt{\pi m}}}{\frac{4^{2m}}{\sqrt{2\pi m}}} = \sqrt{2}$. Also, $P_{2m-1} = \dfrac{4m-2}{4m-1}P_{2m} \to 1 \cdot \sqrt{2} = \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral of $\frac{\arctan x}{x^2+1}$ EDIT: I was studying from a site that uses really ambiguous notation so I misread $\arctan\ (x)^2$ as $\arctan\ (x^2)$. Now I can see why the integral is actually $\frac{1}{2} \arctan^2\ x + c $. Thanks to everyone who answered and corrected me!
Why is $$\int\frac{\arctan\ x}{x^2+1} dx=\frac{1}{2}\arctan(x^2)+c$$
instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\begin{align} &\int\frac{\arctan\ x}{x^2+1} dx=\int \arctan\ x\ dx \\
&x\ \arctan\ x-\int \frac{x}{x^2+1}\ dx \\
&x\ \arctan\ x -\frac{1}{2}\ln(x^2+1)+c\end{align}$$
Is there something I'm not getting? I'd like to see the steps for the correct calculation so I can better understand it. Thanks!
| Your notation may be a bit off?
Doing the $u$-substitution gives
$$\int \frac{\arctan x}{1+x^2}dx = \int u\text{ }du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C.$$
Note that $(\arctan x)^2 \neq \arctan \left(x^2\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Help with Mathematical Induction $$1^3+2^3+\cdots+n^3=\left[\frac{n(n+1)}2\right]^2$$
so far I have..
$$1^3+2^3+\cdots+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$
then..
$$\left[\frac{k(k+1)}2\right]^2+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$
where do I go from here so that the rhs equals the lhs?
| Inductive step:
$$\begin{align*}\underbrace{1^3+2^3+\ldots+n^3}_{=\left(\dfrac{n(n+1)}{2}\right)^2}+(n+1)^3&=\left(\dfrac{n(n+1)}{2}\right)^2+(n+1)^3=(n+1)^2\cdot\left(\frac{n^2}{4}+(n+1)\right)=\\&=(n+1)^2\cdot\left(\frac{n^2+4n+4}{4}\right)=(n+1)^2\cdot\left(\frac{(n+2)^2}{2^2}\right)=\\&=\left(\frac{(n+1)(n+2)}{2}\right)^2\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lipschitz continuity of a function Show that $g(x) = \frac{1}{x^{2} +1}$ is Lipschitz conitnuous.
From the definition, we must show that $\forall x,y \in \mathbb{R}$, $|f(x)-f(y)| \leq K|x-y|$, for some real constant $K$.
First, I tried to find:
$|f(x)-f(y)|=|\displaystyle\frac{1}{x^{2} +1} - \frac{1}{y^{2} +1}|=|\frac{x^2-y^2}{(1+x^2)(1+y^2)}| \leq \frac{(|x-y|)(|x|+|y|)}{(1+x^2)(1+y^2)}$
Using the triangle inequality. However, I hit a dead end here. What would be the best way to approach this problem? Thank you for the help.
| \begin{align}
\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|&=\left|\frac{x+y}{(1+x^2)(1+y^2)}\right| |x-y|\\
&\le \max\left(\frac{x+y}{(1+x^2)(1+y^2)}\right)|x-y|\\
&={3\sqrt{3}\over 8}\,|x-y|
\end{align}
where the last equality follows from a little calculus.
Of course, we do not need such a sharp $K$, so we can of course estimate $$\left|\frac{x+y}{(1+x^2)(1+y^2)}\right|$$ instead as outlined in other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
| Rewrite it as
$$4b^2(c^2-d^2) = a^2(4d^2-c^2) = a^2(4d^2-4c_0^2)$$
where $c_0 = c/2$. Then
$$b^2(c^2-d^2) = a^2(d^2-c_0^2)$$
We can assign: $b^2 = d^2-c_0^2$ and $4d^2-c^2 = a^2$.
For the first equation we use the famous Pythagorean triple: $b = 3, d = 5, c_0 = 4 \rightarrow c = 8$.
Then it brings us to the last variable: $a = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find the determinant of a 4x4 matrix when the diagonal is made up of variables? Evaluate: $\det(A)$, where $A=
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a\end{bmatrix}$
| When computing the determinant, you are allowed to add any linear combination of the other rows (or columns) to any particular row (or column). Also recall that the determinant is linear in the rows (or columns), and this allows you to break out common factors.
One way of doing it might look like
$$
\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a
\end{vmatrix}=
\begin{vmatrix}
a+3 & 1 & 1 & 1 \\
a+3 & a & 1 & 1 \\
a+3 & 1 & a & 1 \\
a+3 & 1 & 1 & a
\end{vmatrix}=
(a+3)
\begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a
\end{vmatrix}=
(a+3)
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & a-1 & 0 & 0 \\
0 & 0 & a-1 & 0 \\
0 & 0 & 0 & a-1
\end{vmatrix}=
(a+3)
\begin{vmatrix}
a-1 & 0 & 0 \\
0 & a-1 & 0 \\
0 & 0 & a-1
\end{vmatrix}=
(a+3)(a-1)^3.
$$
In the first step, the three rightmost columns are added to the first (leftmost). The second step breaks out the common factor $(a+3)$ from the first column. Step three subtracts the first row from each of the other rows, and the fourth step is expansion along the first column (which has only one non-zero term).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral (close form?) I'm struggling to evaluate the following integral:
$\displaystyle \int_{-1}^{1}\frac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{\left ( 1+x^2 \right )\left ( 1+x^4 \right )\left ( 1+x^6 \right )}}\,dx$.
I see that the integrand is even, thus the integral can be re-written as: $\displaystyle \int_{-1}^{1}=2\int_{0}^{1}$, which does not help at all. Applying the sub $u=1-x$ is useless because it would result in a giant mess.
Do you see a pattern or an alternate approach?
| I'll play and see if
I can find anything interesting.
$I
=\int_{0}^{1}\dfrac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{( 1+x^2 ) ( 1+x^4 ) ( 1+x^6 )}}\,dx$
Let $y=x^2$.
Then
$dy = 2x\, dx
$
or
$dx
=\frac{dy}{2x}
=\frac{dy}{2\sqrt{y}}
$.
Then
$\begin{array}\\
I
&=\frac12\int_{0}^{1}\dfrac{1+2y+3y^2+4y^3+5y^4+6y^5+7y^6}{\sqrt{y( 1+y ) ( 1+y^2 ) ( 1+y^3 )}}
dy\\
&=\frac12\int_{0}^{1}\dfrac{g(y)dy}{\sqrt{f(y)}}
\\
\end{array}
$
where
$g(y)
=1+2y+3y^2+4y^3+5y^4+6y^5+7y^6
$
and
$\begin{array}\\
f(y)
&=y( 1+y ) ( 1+y^2 ) ( 1+y^3 )\\
&=y\left( ( 1+y^2 ) ( 1+y^3 )+y ( 1+y^2 ) ( 1+y^3 )\right)\\
&=y\left( 1+y^2+y^3+y^5+y(1+y^2+y^3+y^5)\right)\\
&=y\left( 1+y^2+y^3+y^5+y+y^3+y^4+y^6\right)\\
&=y\left(1+y+y^2+2y^3+y^4+y^5+y^6\right)\\
&=y+y^2+y^3+2y^4+y^5+y^6+y^7\\
\end{array}
$
Noting that
$\begin{array}\\
f'(y)
&=1+2y+3y^2+8y^3+5y^4+6y^5+7y^6\\
&=(1+2y+3y^2+4y^3+5y^4+6y^5+7y^6)+4y^3\\
&=g(y)+4y^3\\
\end{array}
$
$\begin{array}\\
I
&=\frac12\int_0^1 \dfrac{g(y)dy}{\sqrt{f(y)}}\\
&=\frac12\int_0^1 \dfrac{(f'(y)-4y^3)dy}{\sqrt{f(y)}}\\
&=\frac12\int_0^1 \dfrac{f'(y)dy}{\sqrt{f(y)}}
-2\int_0^1 \dfrac{y^3\,dy}{\sqrt{f(y)}}\\
\end{array}
$
I'll stop here;
this seems like a good start.
Any errors here
should be reported
to the appropriate committee
for immediate remediation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How find this sum $\sum_{k=1}^{\infty}\frac{1}{1+a_{k}}$
Let $\{a_{n}\}$ be the sequence of real numbers defined by $a_{1}=3$ and for all $n\ge 1$,
$$a_{n+1}=\dfrac{1}{2}(a^2_{n}+1)$$ Evaluate
$$\sum_{k=1}^{\infty}\dfrac{1}{1+a_{k}}$$
My idea 1:
since
$$2a_{n+1}=a^2_{n}+1$$
so we have
$$2(a_{n+1}-1)=(a_{n}+1)(a_{n}-1)\Longrightarrow \dfrac{1}{1+a_{n}}=2\cdot\dfrac{a_{n}-1}{a_{n+1}-1}$$
so we must find this sum
$$\sum_{n=1}^{\infty}\dfrac{1}{1+a_{n}}=\sum_{n=1}^{\infty}\dfrac{2(a_{n}-1)}{a_{n+1}-1}$$
then I can't find this sum
other idea: maybe we can find this $a_{n}$ closed form?
$$2a_{n+1}-1=(a_{n})^2$$ I want let
$a_{n}=\cos^2{b_{n}}$,so
$$\cos{2b_{n+1}}=(\cos{b_{n}})^4$$
then I can't follow works.
| First, observe that
$$
a_{n+1} -1= \frac{a_n^2-1}{2} = \frac{(a_n+1)(a_n-1)}{2},
$$
and consequently, (one can easily show that $a_n >1$ for all $n\ge 1$)
$$
\frac{1}{a_{n+1}-1}=\frac{1}{a_n-1} - \frac{1}{a_n+1}.
$$
Rewriting this as
$$
\frac{1}{a_n+1} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}
$$
Summing up this identity from $n=1$ to $N$ we get
$$
\sum_{n=1}^N \frac{1}{a_n+1} = \frac{1}{a_1-1} - \frac{1}{a_{N+1}-1}
$$
Finally, observe that $a_n \ge n$ for all $n\ge 1$ (since $\frac{n^2+1}{2}\ge n+1$ as soon as $n\ge 2$) and conclude that the sum is $\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A sequence is given by $a_1 = 1$, $a_{n+1} = \sqrt{1+2a_n}$ for $n=1,2,3,4...$ Show that the sequence is increasing and bounded by 3.
Find out if it converges, and find its limit.
So far i think the limit is
$$
1+\sqrt{2},
$$
and that i should use induction to find that its bounded by 3.
I have solved everything except the convergence part. Can someone show me, and relieve my headache?
thanks for any help!
| Assume that, as $n\to\infty$, the limit of $a_n$ is $u$, hence $a_{n+1}=a_n=u$.
$$\begin{align}
a_{n+1}&=\sqrt{1+2a_n}\\
u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$
NB:
$$\begin{align}
a_1&=1\\
a_2&=\sqrt{1+2}=\sqrt{3}\\
a_3&=\sqrt{1+2\sqrt{3}}\\
a_4&=\sqrt{1+2\sqrt{1+2\sqrt{3}}}\\
a_5&=\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{3}}}}\\
\vdots \\
u=\lim_{n\to\infty}{a_n}&=\sqrt{1+2\underbrace{\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\cdots}}}}}_{u}}\\u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{n}{(n+r)^2}$ $$
\lim_{n\ \to\ \infty}\left[\,%
{n \over \left(\, n + 1\,\right)^{2}}+
{n \over \left(\, n + 2\,\right)^{2}}+\cdots +
{n \over \left(\, 2n\,\right)^{2}}\,\right]
$$
How can I deal with this limit in a reasonable way ?.
EDIT: I am STILL looking for a solution without use of integration
|
I am looking for a solution without use of calculus
I'm interpreting that as "without integration", since limits are calculus.
Write
$$\begin{aligned}
\frac{1}{(n+r)^2} &= \frac{1}{(n+r)(n+r+1)} + \left(\frac{1}{(n+r)^2}-\frac{1}{(n+r)(n+r+1)}\right)\\
&= \left(\frac{1}{n+r} - \frac{1}{n+r+1}\right) + \frac{1}{(n+1)^2(n+r+1)}.
\end{aligned}$$
Then you get
\begin{align}
\sum_{r=1}^n \frac{n}{(n+r)^2} &= n\sum_{r=1}^n \left(\frac{1}{n+r}-\frac{1}{n+r+1}\right) + \underbrace{\sum_{r=1}^n \frac{n}{(n+r)^2(n+r+1)}}_{R_n}\\
&= \underbrace{n\left(\frac{1}{n+1} - \frac{1}{n+n+1}\right)}_{C_n} + R_n.
\end{align}
It is easily seen that $C_n \to \frac{1}{2}$, and $R_n\to 0$ follows since it is the sum of $n$ positive terms each of which is smaller than $\frac{1}{n^2}$, so $0 < R_n < \frac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$? is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$?
I know that $f(x)$ cannot be expressed as a product of degree 1 polynomial and degree 3 polynomial since it has no roots in $R$ but what about two degree 2 polynimials?
| If $f=(ax^2+bx+c)(dx^2+ex+f)$ with $a,d \neq 0$ mod $7$ one can divide the first factor by $a$ to make it monic, then remultiply by $a$ to compensate. So resetting the coefficients at this point $f=(x^2+bx+c)(dx^2+ex+f).$ [these are "reset" $d,e,f$] Then the second factor is also monic to make the coefficient of $x^4$ in the product match that of $f.$ So we have
$$f=(x^2+bx+c)(x^2+ex+f).$$
Now we have $cf=1$ so that $c,f$ make one of the pairs (in either order) $(1,1),\ (2,4),\ (3,5),\ (6,6).$ We also have $b+e=1$ so that $b,e$ make up one of the pairs $(0,1),\ (2,6),\ (3,5),\ (4,4).$ Now the coefficient of $x^2$ in the product is $(c+f)+(be).$ The values of $c+f$ are $1,2,5,6,$ while the values of $be$ are $0,1,2,5.$ So to get the coefficient of $x^2$ to be $1$ we need a sum of $1$ mod $7$ using one term from the first list and one from the second. The only choice is $c+f=6,\ be=2.$ This leads to $(c,f)=(2,4)$ [without loss] and to $(b,e)=(4,4).$ But then the product of the two factors becomes $x^4+8x^3+22x^2+24x+8,$ which has all coeffients $1$ mod $7$ except that of $x^1$ which happens to come out $3$ mod $7.$
Added note to explain a comment below: The expansion of $(x^2+bx+c)(x^2+ex+f)$ is
$$x^4+(b+e)x^3+(c+f+be)x^2+(bf+ce)x+cf.$$
So although true that $bf+ce=1,$ it is also true that the simpler relation $b+e=1$ must hold, in order that this product should match $f$ which has all coefficients $1$ mod $7.$
The above argument works by assuming the cubic and constant coefficients are $1$ to get relations separately between $b,e$ versus $c,f$ and then putting these into the $x^2$ coefficient, since that has a simple expression in terms of sum/products from $b,e$ and $c,f.$ The last thing one would try to match would be the coefficient $bf+ce$ of the power $x^1,$ and that gives what goes wrong in this case since it cannot be made $1$ after the other coefficients of $f$ have been matched.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$ Let $a,b,c,d,e$ be integers such that $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$. Prove that $a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$.
I'm reminded of the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. But for $5$th degree, how can I find a factorization for $a^5+b^5+c^5+d^5+e^5-5abcde$?
| If you use the following notations:
$$s_1=a+b+c+d+e$$
$$s_2=ab+ac+ad+ae+bc+bd+be+cd+ce+de$$
$$s_3=abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde$$
$$s_4=abcd+abce+abde+acde+bcde$$
$$s_5=abcde$$
Then we have:
$$a^5+b^5+c^5+d^5+e^5=\left(\left(s_1\left(s_1^2-2s_2\right)-s_1s_2+3s_3\right)s_1-s_2\left(s_1^2-2s_2\right)+s_3s_1-4s_4\right)s_1-s_2\left(s_1\left(s_1^2-2s_2\right)-s_1s_2+3s_3\right)+s_3\left(s_1^2-2s_2\right)-s_4s_1+5s_5$$
The given condition is in this notation:
$$s_2=0$$
From that it follows that:
$$a^5+b^5+c^5+d^5+e^5-5abcde=s_1\left(s_1^4+5s_1s_3-5s_4\right)$$
And the desired result follows.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Product identity for $n^n$ I came across the rather nice identity
\begin{align}
&&\frac{(-n)^{n-1} \Gamma (n+1)}{(1-n)_{n-1}}&&\tag{1}&\\
\\
&=&\prod _{k=1}^{n-1} \frac{(k+1) n^2}{n^2-k n}&&\tag{2}&\\
\\
&=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n}
\cdots\frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2&&\tag{3}&\\
\\
&=&\underbrace{n\cdot n\cdot n\cdots n\cdot n\cdot n}_{n\text{ times}}&&\tag{4}&\\
\\
&=&n^{n}&&\tag{5}&\\
\end{align}
where ${(1-n)_{n-1}}$ in $(1) $ is the Pochhammer symbol, where $(x)_{n}\equiv\dfrac{\Gamma(x+n)}{\Gamma(x)}$.
Interestingly, Mathematica doesn't simplify the expression to steps $(4)$ & $(5).$ How is this identity proven, and for what $n$ is this not the case?
| Steps (1) to (2)
\begin{align}
f(n)
&= \frac{(-n)^{n-1} \Gamma(n+1)}{(1-n)_{n-1}} \\
&= \frac{(-n)^{n-1} n!}
{\underbrace{(1-n)(1-n+1)(1-n+2)\cdots -2 \cdot -1}_{n-1 \tiny \mbox{ factors}}} \\
&= \frac{(-n)^{n-1} n!}
{(1-n)(2-n)(3-n)\cdots -2 \cdot -1} \\
&= \frac{n^{n-1} n!}
{(n-1)(n-2)(n-3)\cdots 2 \cdot 1} \\
\end{align}
here we could continue directly with
\begin{align}
f(n) = \frac{n^{n-1} n!}{(n-1)!} = n^{n-1} n = n^n
\end{align}
or complicate things with multiplying nominator and denominator with $n^{n-1}$
\begin{align}
f(n)
&= \frac{n^{2n-2}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{(n^2)^{n-1}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{n^2\cdot 2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \frac{2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n^3}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\
&= \prod_{k=1}^{n-1} \frac{(k+1) n^2}{n^2-kn}
\end{align}
Steps $(2)$ to $(5)$
For even $n=2m+2$ and $m\ge 0$ one gets $2m+1$ factors:
\begin{align}
f(n)
=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n}
\cdots
\frac{n^3-3n^2}{4n}
\cdot
\frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2 \\
=&\underbrace{\frac{2 n^2}{n^2- n}\cdot
\underbrace{\frac{3 n^2}{n^2-2 n}\cdot
\underbrace{\frac{4 n^2}{n^2-3 n}
\cdots
\frac{n^2- 3n}{4}}
\cdot
\frac{n^2- 2n}{3}}\cdot\frac{n^2- n}{2}}_{\tiny \mbox{left factor} \times \mbox{right factor}}\cdot n^2 \\
=&(n^2)^m n^2 \\
=&n^{2m+2} \\
=&n^n
\end{align}
Note: The underbraces mean that $m$ pairs consisting of a left and a right side factor each, from the outside to the inside, multiply, resulting into $n^2$ each.
For odd $n=2m+3$ and $m\ge 0$ one gets $2m+1+1$ factors:
\begin{align}
f(n)
&=(n^2)^m \cdot \frac{(m+2)n^2}{n^2-(m+1)n} \cdot n^2 \\
&=(n^2)^m \frac{m+2}{(2m+3)-(m+1)} n^3 \\
&=n^{2m+3} \\
&=n^n
\end{align}
So this seems to hold for $n\ge 2$.
| {
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Determining the asymptotics of the Summatory function of an Arithmetic Function We define the arithmetic function: $\displaystyle f(n) = \max\limits_{p^{\alpha} || n} \alpha$, that is if $\displaystyle n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (prime factorization of $n$) then $f(n) = \max\limits_{1 \le i \le k} \alpha _i$.
How do we determine the asymptotic growth of the summatory function of $f$, that is $$\displaystyle S(x) = \sum\limits_{1 \le n \le x} f(n)$$
Is it possible to express the Dirichlet series corresponding to $f$, that is $\displaystyle \sum\limits_{1 \le n} \frac{f(n)}{n^s}$ in terms of Dirichlet series of some known arithmetic functions ?
| If we group the numbers by value of $f(n)$, we have
$$S(x) = \sum_{k \leqslant \log_2 x} k\cdot\operatorname{card} \{ n \leqslant x : f(n) = k\} = \sum_{k \leqslant \log_2 x} \operatorname{card} \{ n \leqslant x : f(n) \geqslant k\}.$$
If $p(k,x)$ denotes the count of numbers $\leqslant x$ that are multiples of a $k$-th power of a prime, that becomes
$$S(x) = \sum_{k} p(k,x).$$
Since
$$\lfloor x\rfloor - p(k,x) = \sum_{ n\leqslant x} \mu(n)\left\lfloor\frac{x}{n^k} \right\rfloor$$
for $k\geqslant 2$, we have, omitting the upper bound on the sums, since all later terms are $0$,
\begin{align}
S(x) &= \sum_{k=1}^\infty p(k,x)\\
&= \lfloor x\rfloor - 1 + \sum_{k=2}^\infty p(k,x)\\
&= \lfloor x\rfloor - 1 - \sum_{k=2}^\infty \left(\sum_{m=2}^\infty \mu(m)\left\lfloor \frac{x}{m^k}\right\rfloor\right)\\
&= \lfloor x\rfloor -1 - \sum_{k=2}^\infty \left(\sum_{m=2}^\infty \mu(m)\frac{x}{m^k} - \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\right)\\
&= x\biggl(1 - \sum_{k=2}^\infty \sum_{m=2}^\infty \frac{\mu(m)}{m^k}\biggr)
- (1+x-\lfloor x\rfloor) + \sum_{k=2}^\infty \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor \frac{x}{m^k}\right\rfloor\biggr)\\
&= \Biggl(\sum_{k=1}^\infty \biggl(1 - \frac{1}{\zeta(k)}\biggr)\Biggr)\cdot x + O(1) + O(\sqrt{x}). \tag{1}
\end{align}
It remains to prove the $O(\sqrt{x})$ estimate for the last sum. We have
\begin{align}
\left\lvert \sum_{k=2}^\infty \sum_{m=2}^\infty \mu(m)\biggl(\frac{x}{m^k} - \left\lfloor \frac{x}{m^k}\right\rfloor\biggr)\right\rvert
&\leqslant \sum_{k=2}^\infty \sum_{m=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\\
&= \sum_{m=2}^\infty \sum_{k=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr)\\
&= \sum_{m=2}^{\lfloor \sqrt{x}\rfloor}\sum_{k=2}^\infty \biggl(\frac{x}{m^k} - \left\lfloor\frac{x}{m^k}\right\rfloor\biggr) + \sum_{m=\lfloor\sqrt{x}\rfloor+1}^\infty\sum_{k=2}^\infty \frac{x}{m^k}.
\end{align}
The last sum can be explicitly evaluated,
$$\sum_{m= a+1}^\infty \sum_{k=2}^\infty \frac{1}{m^k} = \sum_{m=a+1}^\infty \frac{1}{m(m-1)} = \frac{1}{a},$$
and so
$$\sum_{m=\lfloor \sqrt{x}\rfloor+1}^\infty\sum_{k=2}^\infty \frac{x}{m^k} = \frac{x}{\lfloor\sqrt{x}\rfloor} \in O(\sqrt{x}).$$
For the other sum, we split the inner sum at $\frac{\log x}{\log m}$. For $k \leqslant \frac{\log x}{\log m}$, we use $0 \leqslant z-\lfloor z\rfloor < 1$ to obtain the estimate
$$\sum_{m=2}^{\lfloor \sqrt{x}\rfloor} \frac{\log x}{\log m},$$
which with
$$\sum_{m=2}^a \frac{1}{\log m} = \frac{a}{\log a} + O\left(\frac{a}{(\log a)^2}\right)$$
yields a bound
$$\sum_{m=2}^{\lfloor\sqrt{x}\rfloor}\sum_{k=2}^{\left\lfloor\frac{\log x}{\log m}\right\rfloor}\biggl(\frac{x}{m^k}-\left\lfloor\frac{x}{m^k}\right\rfloor\biggr) \leqslant 2\sqrt{x} + O\left(\frac{\sqrt{x}}{\log x}\right)\in O(\sqrt{x}),$$
and for the remaining part we use
$$\sum_{k=a}^\infty \frac{1}{m^k} = \frac{1}{m^a(1-m^{-1})}$$
to obtain
$$\sum_{m=2}^{\lfloor\sqrt{x}\rfloor}\sum_{k=\left\lfloor\frac{\log x}{\log m}\right\rfloor+1}^\infty \frac{x}{m^k} \leqslant \sum_{m=2}^{\lfloor\sqrt{x}\rfloor} \frac{x}{m^{\left\lfloor\frac{\log x}{\log m}\right\rfloor+1}(1-m^{-1})}\leqslant \sum_{m=2}^{\lfloor\sqrt{x}\rfloor} \frac{1}{1-m^{-1}} \leqslant 2\lfloor\sqrt{x}\rfloor.$$
Thus the asymptotic $(1)$ is proved.
For the Dirichlet series, we can - for $\operatorname{Re} s > 1$ - write
$$\sum_{n=1}^\infty \frac{f(n)}{n^s} = \sum_{k=1}^\infty \underbrace{\sum_{n=1}^\infty\frac{[f(n) \geqslant k]}{n^s}}_{\sigma_k(s)},$$
where $[\,\cdot\,]$ is the Iverson bracket. Since $[f(n)\geqslant 1] = 1$ for all $n\geqslant 2$ and $f(1) = 0$, we have $\sigma_1(s) = \zeta(s) - 1$, and for $k\geqslant 2$ we note that
\begin{align}
\sigma_k(s) &= \sum_{m=2}^\infty (-\mu(m))\sum_{n=1}^\infty \frac{1}{(m^kn)^s}\\
&= \sum_{m=2}^\infty \frac{-\mu(m)}{m^{ks}}\zeta(s)\\
&= \biggl(1 - \sum_{m=1}^\infty \frac{\mu(m)}{m^{ks}}\biggr)\zeta(s)\\
&= \biggl(1 - \frac{1}{\zeta(ks)}\biggr)\zeta(s),
\end{align}
hence
$$\sum_{n=1}^\infty \frac{f(n)}{n^s} = \zeta(s)\cdot\sum_{k=1}^\infty\Bigl(1-\frac{1}{\zeta(ks)}\Bigr) = \zeta(s)\cdot \sum_{k=1}^\infty \frac{\zeta(ks)-1}{\zeta(ks)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031380",
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"source": "stackexchange",
"question_score": "4",
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Find the area of a triangle if its two sides measure $6 in.$ and $9 in.$, and the bisector of the angle between the sides is $4\sqrt{3}$ in. Find the area of a triangle if its two sides measure $6$ in. and $9$ in., and the bisector of the angle between the sides is $4\sqrt{3}$ in. I'm thinking of using the formula $A$=$\frac{1}{2}bh$ I can't find the base or height, I used the Angle bisector formula which is $l=$$\frac{\sqrt{ab[(a+b)^2-c^2]}}{a+b}$ So i found out C which i think is the base should i multiply it by $2$? because I think it's the half. From here I'm lost
| If you know the lengths of two sides $a$ and $b$, and the angle $\theta$ between them, then the area of the triangle is $\frac{1}{2}ab\sin \theta$. Let's label the sides $a = 6$, $b = 4\sqrt 3$, and $c = 9$, and the half-angle as $\theta$. The sum of each sub triangle area must equal the total triangle area:
$$A = \frac{1}{2}ab\sin\theta + \frac{1}{2}bc\sin\theta = \frac{1}{2}ac\sin2\theta $$
Solving this,
$$ ab\sin\theta + bc\sin\theta = ac\sin2\theta = 2ac\sin\theta\cos\theta$$
$$ \cos\theta = \frac{b(a+c)}{2ac} = \frac{5}{3\sqrt 3}$$
Then,
$$\sin\theta = \sqrt{1-\cos^2\theta} = \frac{\sqrt{2}}{3\sqrt{3}} \qquad \sin2\theta = \frac{10\sqrt 2}{27}$$
And so the area is
$$ A = \frac{1}{2}ac\sin 2\theta = 10\sqrt 2 \text{ sq. in.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033054",
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"source": "stackexchange",
"question_score": "2",
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Show that $\sum_{i=1}^{r} i^2 = \binom{r+1}{3} + \binom{r+2}{3}$ by finding generating function Find the generating function for the sequence $c_r$ where $c_0 = 0$ and $ c_r = \sum_{i=1}^{r} i^2 $ for $r \in \mathbb N$. Hence show that $\sum_{i=1}^{r} i^2 = \binom{r+1}{3} + \binom{r+2}{3}$
EDIT:
Sorry, I forgot to mention what I got so far.
Since
$c_0 = 0$
$c_1 = 1$
$c_2 = 1 + 4$
Hence, the generating function for $c_r$ is
$C(x) = 0 + 1x + (1+4)x^2 + (1+4+9)x^3 + ...$
| Let $C(x)$ be the GF for the sequence $\langle c_n \rangle_{n\ge 0}$, where $c_n$ is the $n^\text{th}$ partial sum of the sequence of squares. If $f(x)$ is the GF for the sequence of squares, then $\frac{f(x)}{1-x}$ is the GF for the sequence of the partial sums of squares.
So, we first find the GF for $\langle n^2 \rangle_{n\ge 0}$. Starting from $\frac{1}{1-x} = \sum_{k=0}^\infty x^k$, we differentiate, multiply by an $x$, and do those two moves once more. Then we have
$f(x) = \frac{x(x+1)}{(1-x)^3}.$
So the GF for the sequence of the partial sums of squares is $$C(x) =\frac{x(x+1)}{(1-x)^4} = \dfrac{x^2}{(1-x)^4} + \dfrac{x}{(1-x)^4}.$$
By Newton's binomial theorem, these two terms are
$$C(x) = \sum_{k=0}^\infty {{4+k-1} \choose {k}}x^{k+2} + \sum_{k=0}^\infty {{4+k-1}\choose k}x^{k+1}.$$
After playing around with our indices, we have
$$\sum\limits_{k=0}^\infty c_k x^k = 1+\sum_{k=2}^\infty \left[ {{k+1} \choose {k-2}} + {{k+2}\choose {k-1}} \right]x^{k}.$$
Comparing coefficients and using the identity ${n \choose k} = {n \choose {n-k}}$, we have
\begin{align*}
c_0 &= 0 \\
c_1 &= 1 \\
c_n &= {{n+1} \choose 3} + {{n+2}\choose 3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034529",
"timestamp": "2023-03-29T00:00:00",
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Prove, using the method of mathematical induction that the following holds true For natural numbers $n\ge1$ show the following inequality using induction.
$$n^{1/n}\le 1+\sqrt{\frac{2}{n}}$$
| To solve this problem, merely binomial theorem will be enough.
The base case $n = 1$ is obviously true. Then we assume that $$n^{\frac1n}\le 1+\sqrt{\frac{2}{n}} \iff n \le \left(1+\sqrt{\frac2n}\right)^n,$$ and we try to prove $$(n+1)^{\frac1{n+1}}\le 1+\sqrt{\frac{2}{n+1}} \iff n+1 \le \left(1+\sqrt{\frac2{n+1}}\right)^{n+1}.$$
$\mathbf{Claim} \quad \left(1+\sqrt{\dfrac2n}\right)^n + 1 \le \left(1+\sqrt{\dfrac2{n+1}}\right)^{n+1}.$
With this claim, it's clear that $n+1 \le \left(1+\sqrt{\dfrac2n}\right)^n+1 \le \left(1+\sqrt{\dfrac2{n+1}}\right)^{n+1}$, so once the claim is proved, we're done.
Proof of the claim
Let $a_n = \left(1+\sqrt{\dfrac2n}\right)^n$. Then we want $a_n + 1\le a_{n+1} \,\forall n \in \Bbb{N}$. This is a bit tricky if we use binomial theorem and condense the higher power terms with a summation sign.
\begin{split}
a_n + 1&= \left(1+\sqrt{\frac2n}\right)^n \\
&= 1 + 1 + \sqrt{2n} + \frac{n(n-1)}{2} \frac2n + \sum_{k=3}^n \binom{n}{k} \left(\sqrt{\frac2n}\right)^k \quad \mbox{(Here we assume $n \ge 3$)}\\
&= 2 + \sqrt{2n} + (n-1) + \sum_{k=3}^n \binom{n}{k} \frac{(\sqrt{2n})^k}{n^k} \\
&= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} \\
&= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n} \\
&= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} 1 \left( 1-\frac{1}{n} \right) \cdots \left( 1-\frac{k-1}{n} \right) \\
&< 1 + \sqrt{2(n+1)} + n + \sum_{k=3}^n \frac{(\sqrt{2(n+1)})^k}{k!} 1 \left( 1-\frac{1}{n+1} \right) \cdots \left( 1-\frac{k-1}{n+1} \right) \\
&= 1 + \sqrt{2(n+1)} + \binom{n+1}{2} \frac2{n+1} + \sum_{k=3}^n \frac{(\sqrt{2(n+1)})^k}{k!} \frac{n+1}{n+1} \frac{n}{n+1} \cdots \frac{n-k+2}{n+1} \\
&= 1 + \sqrt{2(n+1)} + \binom{n+1}{2} \frac2{n+1} + \sum_{k=3}^n \binom{n+1}{k} \left(\sqrt{\frac2{n+1}}\right)^k \\
&= \sum_{k=0}^n \binom{n+1}{k} \left(\sqrt{\frac2{n+1}}\right)^k \\
&= \left(1+\sqrt{\frac2n}\right)^n \\
&= a_{n+1}
\end{split}
What is left is the base cases for $n = 1$ and $n = 2$.
\begin{split}
a_1 + 1 &= 1 + 1 + \sqrt2 < 2 + 2 = 4 = \left( 1 + \sqrt{\frac22} \right)^2 = a_2\\
a_2 + 1 &= 1 + \left( 1 + \sqrt{\frac22} \right)^2 \\
&= 1 + 1 + \sqrt2 \cdot \sqrt2 + \frac22 \\
&= 1 + \sqrt2 \cdot \sqrt2 + \left( \frac22 + 1 \right) \\
&< 1 + \sqrt2 \cdot \sqrt3 + 2 \\
&< 1 + \sqrt2 \cdot \sqrt3 + \binom{3}{2} \frac23 + \left(\sqrt\frac23\right)^3\\
&= \left(1+\sqrt\frac23\right)^3 \\
&= a_3
\end{split}
| {
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Prove that $13\vert(3^{n+1} +3^{n} +3^{n-1})$
Prove that $3^{n+1} +3^{n} +3^{n-1}$ is divisible by $13$ for all positive integral values of $n$
| Besides the obvious factorization approach, you can reason by induction.
$$3^{(n+1)+1}+3^{(n+1)}+3^{(n+1-1)}=3\cdot 3^{n+1}+3\cdot 3^n+3\cdot 3^{n-1}=3 \cdot (3^{n+1}+3^n+3^{n-1}),$$
and
$$3^{1+1}+3^{1}+3^{1-1}=13.$$
Hence
$$S_{n+1}=3\cdot S_n\text{, and }S_1=13.$$
| {
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"source": "stackexchange",
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Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
| Note that
$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b)$$
Now put $a=(x-5)$ and $b=2$
$$(x-5)^2-4=(x-5)^2-2^2=((x-5)+2)((x-5)-2)=(x-5+2)(x-5-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038224",
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"source": "stackexchange",
"question_score": "2",
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Proving $x^2 - y^2 + z^2 \gt (x - y + z)^2$ Prove that
$$x^2 - y^2 + z^2 > (x - y + z)^2$$
where: $x < y <z$ for all natural numbers.
Thank for help.
| I think the formula $A^2 − B^2 = (A− B)(A+ B)$, will be perfect.
$x^2 − y^2 > (x − y + z)^2 − z^2$
$(x − y)(x + y) > (x − y + z − z)(x − y + z + z)$,
$(x − y)(x + y) > (x − y)(x − y + 2z)$.
| {
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Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| $$
f'(x)= \frac{-1}{1+{(\frac{1-x}{1+x}})^2}.\frac{d}{dx}(\frac{1-x}{1+x})
$$
$$f'(x)=\frac{-(1+x)^2}{2(1+x^2)}.\frac{-1(1+x)-(1-x)}{(1+x)^2}$$
$$f'(x)=\frac{-1(-1-x-1+x)}{2(1+x^2)}$$
$$f'(x)=\frac{1}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Use the definition of infinite limit to prove $\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$
Prove
$$\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$$
And I was given the solution like this: but I could not understand how it removes the complicated terms.
Let $\delta=\min(0.5,\frac{1}{5M})$.
$$\frac{x}{x^2-1}=\frac{x}{(x+1)(x-1)}
\geq\frac{0.5}{\left(\frac{1}{5M}\right)(1.5+1)}=5M\times0.2=M$$
I understand the definition of $M-\delta$, but what I don't understand is what is the solution doing, I mean the process of estimation to get rid of the complicated terms. Anyone can enlighten me? thanks!
| We can first use partial fraction decomposition to $\frac{x}{x^2-1},$ which leads to
\begin{gather*}
\frac{x}{x^2-1}=\frac{1}{2(x+1)}+\frac{1}{2(x-1)},
\end{gather*}
which implies that
\begin{gather*}\tag{1}
\frac{x}{x^2-1}>\frac{1}{2(x-1)}, \qquad \text{provided } x>-1.
\end{gather*}
Let $M>0.$ Set
$$\delta=\frac{1}{2M}.$$
For every $x>1,$ we have
\begin{align*}
0<x-1<\delta\implies \frac{x}{x^2-1}&>\frac{1}{2(x-1)}\qquad\qquad \text{by }(1)\\
&>\frac{1}{2\delta}=\frac{1}{2\frac{1}{2M}}=M,
\end{align*}
which, by definition of limit, completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Surface integrals: Find the area of the portion of the cone $x^2+y^2=z^2$ above the $xy$ plane and inside the cylinder $x^2+y^2=ax$ I need to find the area of the portion of the cone $x^2+y^2=z^2$ above the $xy$ plane and inside the cylinder $x^2+y^2=ax$ .
For this, I used cylindrical coordinates to parametrize the region:
$$x=r\cos \theta$$
$$y=r\sin \theta$$
$$z=z$$
Since $x^2+y^2=z^2$ then $r=z$. And since $x^2+y^2=ax$ then $r=a\cos \theta$. So the final parameterization, for $-\pi/2<\theta<\pi/2$ and $0<r<a\cos \theta$, is:
$$x=r\cos \theta$$
$$y=r\sin \theta$$
$$z=r$$
And so, $T_r =(\cos\theta,\sin\theta,1)$ and $ T_{\theta}=(-r\sin\theta ,r \cos\theta, 0)$. Which gives $T_r \times T_{\theta}=(-r\cos\theta, -rsin\theta, r^2)$
And so the area is given by:
$$\int_{-\pi/2}^{\pi/2} \int_0^{a \cos\theta}r(1+r^2)^{1/2} dr d\theta$$
Which seems simple, but gives a hell-difficult integral when integrating respect to $\theta$. So I wanted to ask you: is my approach correct? Is there a simpler way to parametrize this surface?
| The final double-integral you have doesn't evaluate to an elementary function, but it can be recognized as a combination of Elliptic Integrals.
Take the inner integral first. By putting $u=1+r^2$, It is easy to show that
$$
\int_{0}^{a\cos\theta} r\sqrt{1+r^2}dr = \frac{1}{3} \sqrt{a^2 \cos ^2\theta +1} \left(a^2 \cos ^2 \theta + 1 \right) - \frac{1}{3}
$$
So, your outer integral is
$$
\int_{-\pi/2}^{\pi/2} \frac{1}{3} \sqrt{a^2 \cos ^2\theta +1} \left(a^2 \cos ^2 \theta + 1 \right) - \frac{1}{3} d\theta
$$
For $a \neq 1$, it is well known that one gets elliptic integrals of various kinds. The limits $-\pi/2, \pi/2$ suggest that we should be able to get by with complete elliptic integrals. From the definition of $K(k)$ and $E(k)$, we have
$$
K(k) = \int_{0}^{\pi/2} \frac{1}{\sqrt{1-k^2 \sin^2 \theta}} d\theta \\
E(k) = \int_{0}^{\pi/2} \sqrt{1-k^2 \sin^2 \theta} d\theta
$$
You can use these to evaluate the integral above. For instance, the even integral
$$
\int_{-\pi/2}^{\pi/2} \sqrt{a^2 \cos ^2\theta +1} \; d\theta \\
= 2 \int_{0}^{\pi/2} \sqrt{1 + a^2 - a^2 \sin ^2\theta} \; d\theta \\
= 2\sqrt{1+a^2} \int_{0}^{\pi/2} \sqrt{1 - \frac{a^2}{1+a^2} \sin ^2\theta} \; d\theta \\
= 2\sqrt{1+a^2} E \left( \frac{a^2}{1+a^2} \right)
$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluting $\tan15°$ using difference formula
Evalute $\tan15°$ using difference formula
Steps I took:
$$\begin{align}
\tan(45-30)&=\frac { \tan(45)-\tan(30) }{ 1+\tan(45)\tan(30) }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \cdot \frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1-\frac { \sqrt3 }{ 3 } }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } } \\
\end{align}$$
Where do I go from here to arrive at the correct answer? And what mistakes did I make, if any?
|
Hint: Go on
$$\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } }=\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } }\cdot\frac33=
\frac {3-2\sqrt3+1 }{3-1}=
\frac {4-2\sqrt3}{2}=2-\sqrt3$$
$$\large\tan15^{\circ}=2-\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.
Therefore the nth term of given series is :
$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)
By using partial fraction :
and solving for A,B and C we get A = 1/2, B = -1, C =1/2
Putting the values of A,B and C in (1) we get :
$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$
But by putting $r =1,2,3, \cdots$ I am not getting the answer. Please guide how to solve this problem . Thanks.
| The $r$th term is
$$\frac12 \left (\frac1{r}-\frac1{r+1} \right )-\frac12 \left (\frac1{r+1}-\frac1{r+2} \right )$$
so the sum telescopes. The result is
$$\sum_{r=1}^n \frac{2 r-1}{r (r+1) (r+2)} = \frac12\left (1-\frac1{2} \right )-\frac12 \left (\frac1{n+1}-\frac1{n+2} \right ) = \frac14 - \frac1{2 (n+1) (n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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For fixed $m$, find all positive integer solutions to $a^m+b^m = p^n$ where $p$ is prime The problem:
For fixed $m$,
find all solutions to
$a^m+b^m = p^n$
where $p$ is prime,
all variables are positive integers,
$a \le b$,
and
$m \ge 2$.
This is a generalization
of this question:
Solve $a^{2013}+b^{2013}=p^n$ for all positive integers a,b,n and prime numbers p
My answer showed that
if $m$ is odd
and divisible by 3
then the only solution
(aside from the trivial one of
$a=b=1, p=2, n=1$)
is
$a=1, b=2, m=3, p=3, n=2$
(i.e.,
$1+8=9$).
I am now asking about
the solutions for
other values of $m$.
Some observations:
If $a = b$,
the equation is
$2a^m = p^n$,
so $p=2$,
$a=2^k$,
and
$n = km+1$.
If $m$ is even,
then $p^n$ is the
sum of two squares,
so the results there
can be used.
In particular, if $p$
is of the form
$4k+3$,
then $n$ must be even.
For what other classes of $m$,
or even particular values of $m$,
can all solutions be found?
| The only positive integer solutions of $p^n=a^m+b^m$ with $p$ prime, $m\ge 3$ not a power of $2$ are $(p,n,a,b,m)=(3,2,2,1,3),(2,km+1,2^k,2^k,m),\forall k\ge 1$.
If $a=b$, then $(p,n,a,b,m)=(2,km+1,2^k,2^k,m), k\ge 1$. If $(a,b,m)=(2,1,3)$, then $(p,n,a,b,m)=(3,2,2,1,3)$. Otherwise: let $m=2^kt$ with $t\ge 2$ odd. Then by Zsigmondy's Theorem:
$$a^m+b^m=\left(a^{2^k}+b^{2^k}\right)\left(a^{m-2^k}-a^{m-2^k-1}b\pm\cdots+b^{m-2^k}\right)$$
has a prime divisor that does not divide $a^{2^k}+b^{2^k}$, so $a^m+b^m$ has at least two different prime divisors.
However, this is not so simple if $m$ is a power of $2$. Some partial results:
If $m=2$, then $a^2+b^2=p^n$ has infinitely many positive integer solutions with $p$ prime. E.g., if $n=1$, then there exists a unique solution if $p\equiv 1\pmod{4}$; no solutions if $p\equiv 3\pmod{4}$. See Fermat's theorem on sums of two squares (or Fermat's Christmas Theorem). Uniqueness isn't mentioned in the Wikipedia page, but it's easy to prove; see e.g. here (there's a small mistake; it should be $(ad+bc)^2+(ac-bd)^2$).
If $m=2, n=2$, then $a^2+b^2=p^2$ has infinitely many positive integer solutions. It's well-known (see here) that we must have $p=k(m^2+n^2)$ for some positive integers $k,m,n$ (wlog $m>n$). Clearly $k=1$. If we can find such $m,n$, then $\{a,b\}=\{2mn,m^2-n^2\}$ solves the equation. So $a^2+b^2=p^2$ is solvable iff $p\equiv 1\pmod{4}$.
If $m=4$, in $1961$ it was unknown whether there are inifinitely many primes of the form $a^4+b^4$ (see here). If $m=2^t$ with $t\ge 2$, $n$ must be odd (see here).
Also see Generalized Fermat Numbers. In particular, it is open whether there are infinitely many primes of the form $a^{2^n}+1$, $n\ge 1$, in particular if there are infinitely many primes of the form $n^2+1$ (see here).
If $\gcd(b,p)=1$, we must have $p\equiv 1\pmod{4}$, because $-1$ is a quadratic residue mod $p$ iff $p\equiv 1\pmod{4}$ (Quadratic Reciprocity).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of solutions to equation, range restrictions per variable Find the number of solutions of the equation $x_1+x_2+x_3+x_4=15$ where variables are constrained as follows:
(a) Each $x_i \geq 2.$
(b) $1 \leq x_1 \leq 3$ , $0 \leq x_2$ , $3 \leq x_3 \leq 5$, $2 \leq x_4 \leq 6$
I believe I understand part A. I can fix the values at greater than two by adding two to each term, as in:
$(x_1 + 2) + (x_2 + 2) + (x_3+2) +(x_4+2) = 15$
$x_1+x_2+x_3+x_4=7$
So, $C(7+4-1,7) = {10! \over 7!3!} = 120$
For part B, i really have no idea where to start. Thanks for any insight!
| Assume $a$ as a real number, such that $0<a <1$
The number of integral solutions should be the same as:
The coefficient of $a ^ {15}$ in
$ \left(a + a^{2} + a^{3} \right)\left(1+a + a^{2} + a^{3} + ... \right)\left(a^{3} + a^{4} + a^{5} \right)\left(a^{2} + a^{3} + a^{4} +a^{5} + a^{6}\right) $
This is same as
the coefficient of $a ^ {9}$ in
$ \left(1 + a + a^{2} \right)^{2}\left(1+a + a^{2} + a^{3} + ... \right)\left(1+a+a^{2} + a^{3} + a^{4}\right) $
This is same as
the coefficient of $a ^ {9}$ in
$ \left(1 -a^{3} \right)^{2}\left(1-a^{5}\right)\left(1-a\right)^{-4} $
Which is same as
the coefficient of $a ^ {9}$ in
$ \left(1 -2a^{3} + a^6 \right)\left(1-a^{5}\right)\left(1+4a+\frac{4.5}{2!}a^2+\frac{4.5.6}{3!}a^3+\frac{4.5.6.7}{4!}a^4+...\right) $
Which is same as
the coefficient of $a ^ {9}$ in
$ \left(1 -2a^{3} + a^6 - a^5+ 2 a^8 \right)\left(1+4a+\frac{4.5}{2!}a^2+\frac{4.5.6}{3!}a^3+\frac{4.5.6.7}{4!}a^4+...\right) $
Which is
$\binom{12}{3} - 2\binom{9}{3} + \binom{6}{3} - \binom{7}{3} + 2\binom{4}{3} = 45$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How prove this inequality $\prod_{1\le i
let $z_{1},z_{2},z_{3},z_{4},z_{5}$ are complex numbers,and such
$$|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2+|z_{5}|^2=5$$
$$A=\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix}$$
show that
$$\det{(AA^{*})}\le 5^5$$
my idea:
we must prove
$$|z_{1}-z_{2}|^2|z_{1}-z_{3}|^2|z_{1}-z_{4}|^2|z_{1}-z_{5}|^2|z_{2}-z_{3}|^2|z_{2}-z_{4}|^2|z_{2}-z_{5}|^2|z_{3}-z_{4}|^2|z_{3}-z_{5}|^2|z_{4}-z_{5}|^2\le 5^5$$
then let $$LHS=f(z_{1},z_{2},z_{3},z_{4},z_{5})$$
maybe can use this indentity
$$\sum_{i=1}^{n}|z_{i}-z|^2=n\sum_{i=1}^{n}|z_{i}|^2-|\sum_{i=1}^{n}z_{i}|^2$$
where
$$z=\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}$$
But following I can't
| The required inequality is a straightforward corollary of Hadamard’s inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
} |
Prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent How do I prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent for every x in real numbers except for $x=0$?
I tried using the ratio test, but it doesn't seem to be conclusive.
| Assuming you want to know whether $\sum_{n = 1}^\infty\frac{1}{n^2x^2 + 1}$ converges.
Two cases:
*
*$|x| \geq 1$. In this case, we have $$\frac{1}{n^2x^2 + 1} \leq \frac{1}{n^2 + 1} \leq \frac{1}{n^2}$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges.
*$0<|x| < 1$. In this case, we have
$$
\frac{1}{n^2 x^2 + 1} \leq \frac{1}{n^2x^2 + x^2} = \frac{1}{x^2}\frac{1}{n^2 + 1} \leq \frac{1}{x^2}\cdot\frac{1}{n^2}
$$
and $\sum_{n = 1}^\infty\frac{1}{x^2}\cdot\frac{1}{n^2} = \frac{1}{x^2}\sum_{n = 1}^\infty\frac{1}{n^2}$ converges.
Therefore, the sum $\sum_{n = 1}^\infty\frac{1}{n^2x^2 + 1}$ converges as long as $x \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. I then set $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$. However, I afterwards realized that I had squared an inequality to get this. The following is what my proof looked like before I realized this mistake:
Let $\varepsilon > 0$. Choose $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$ and let $n > N$. Then $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. Thus $2n\varepsilon > 1 - \varepsilon^2$, $\varepsilon^2 + 2n\varepsilon > 1$, $\varepsilon^2 + 2n\varepsilon + n^2 = (\varepsilon + n)^2 > 1 + n^2$, $\varepsilon + n > \sqrt{1 + n^2}$, and $\varepsilon > \sqrt{n^2 + 1} - n$. Hence $| \sqrt{n^2 + 1} - n - 0 | = \sqrt{n^2 + 1} - n <
\varepsilon.$ Therefore, $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$.
How should I find $N$? Also, are there any other errors in this proof?
| $$\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) =\lim _{n\to \infty} \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n}\\ = \lim _{n\to \infty} \frac{n^2 + 1-n^2}{\sqrt{n^2 + 1} + n}=\lim _{n\to \infty} \frac{1}{\sqrt{n^2 + 1} + n}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
| Hint: $\left(\dfrac{x^2+2}{x^2-4}\right)^{9x^2} = \left(\left(1+\dfrac{6}{x^2-4}\right)^{x^2-4}\right)^9\cdot \left(1+\dfrac{6}{x^2-4}\right)^{36}$
| {
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"url": "https://math.stackexchange.com/questions/1055428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Multiply out the right hand side and confirm that the two expressions are equal.
$$\frac{x^6-1}{x-1}=(1+x+x^2+x^3+x^4+x^5)\iff x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 8
} |
Double substitution when integrating. I need to integrate
$$f(x) = \cos(\sin x)$$
my first thought was substituting so that $u = \cos(x)$, but that doesn't seem to do the trick. Is there any way to do a double substitution on this?
Any ways on how to proceed would be appreciated.
| You can express in terms of Incomplete Bessel Functions:
Consider $$\begin{align}J_0(1,w)&=\dfrac{2}{\pi}\int_0^w\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\cos\left(x-\dfrac{\pi}{2}\right)~d\left(x-\dfrac{\pi}{2}\right)
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\sin x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\sin x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cos\sin\left(\dfrac{\pi}{2}-x\right)~d\left(\dfrac{\pi}{2}-x\right)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cos\cos(\pi-x)~d(\pi-x)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\pi\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-J_0(1)\end{align}
$$
Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx=J_0(1)+J_0(1,w)$
$\therefore\int\cos\sin x~dx=\dfrac{\pi}{2}\left(J_0(1)+J_0\left(1,x-\dfrac{\pi}{2}\right)\right)+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Weak principle of induction for $5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$
Show that
$$5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$$
Proving the base case $n(1)$:
$5(1)= \frac{5(1)(1+1)}{2}$
$5 = \frac{5(2)}{2}$
$5 = 5$
Induction hypothesis:
$n = k$
$5+10+15+\ldots+5k = \frac{5k(k+1)}{2}$
Induction step (adding $k+1$):
$5+10+15+\ldots+5k+5k+1 = \frac{5k+1(k+1+1)}{2}$
Substituting $\frac{5k(k+1)}{2}$ for $5k$:
$\frac{5k(k+1)}{2}+5k+1 = \frac{(5k+1)(k+1+1)}{2}$
Simplifying:
$\frac{5k(k+1)+2(5k+1)}{2} = \frac{(5k+1)(k+2)}{2}$
$\frac{5k^2+5k+10k+2}{2} = \frac{5k^2+10k+k+2}{2}$
These aren't equal, so what did I do wrong here?
|
Since you have your answer now! Let's see it with a different approach
$$S=\color{red}{5}+\color{blue}{10}+\cdots+\color{green}{5(n-1)}+\color{orange}{5n}$$
$$S=\color{red}{5n}+\color{blue}{5(n-1)}+\cdots+\color{green}{10}+\cdots+\color{orange}{5}$$
$$2S=\underbrace{\color{red}{5(n+1)}+\color{blue}{5(n+1)}+\cdots+\color{green}{5(n+1)}+\color{orange}{5(n+1)}}_{\text{n terms}}$$
$$S=\frac{5n(n+1)}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3^2=5\cdot 2^1$$
$$2^2+4^2=5\cdot 2^2$$
$$2^2+6^2=5\cdot 2^3$$
| Note that $5$ is a sum of two squares, for $5=2^2+1^2$.
If $n$ is even, say $n=2k$, then $5\cdot 2^n=(2^k\cdot 2)^2+(2^k\cdot 1)^2$.
Also, $10=3^2+1^2$. So if $n$ is odd, say $n=2k+1$, then $5\cdot 2^{n}=10\cdot 2^{2k}=(2^k\cdot 3)^2+(2^k\cdot 1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Help with $\int \frac{1}{(\sin x + \cos x)}$ Kindly solve this question
$$\int \frac{1}{(\sin x + \cos x)} dx$$
I reached up to
$$\frac{(1+\tan^2x)}{1-\tan^2x + 2\tan x}$$
| HINT:
$$\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right)$$ or $$\sin x+\cos x=\sqrt2\cos\left(x-\frac\pi4\right)$$
More generally, set $a=r\cos\phi,b=r\sin\phi$ to find
$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin\left(x+\arctan2(b,a)\right)$
Similarly, set $a=r\sin\psi,b=r\cos\psi$ to find
$a\sin x+b\cos x=\sqrt{a^2+b^2}\cos\left(x-\arctan2(a,b)\right)$
arctan2 has been defined here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Just a proof of algebra If $a+b+c=0$,
Show that
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this problem but can't find a solution. Please help me.
| If you expand the expression and factor you get $$\frac{(a^3+b^3+c^3-ab^2-a^2b-ac^2-a^2c-bc^2-b^2c+3abc)(b-c)(c-a)(a-b)}{abc (b-c)(c-a)(a-b)}$$ and if you then use $c=-a-b$ you get $$\frac{-9a^2b-9ab^2}{-a^2b-ab^2}=9$$ providing you have not divided by zero (so the $a,b,c$ must be distinct and non-zero as Suzu Hirose said in the comments).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x$ $\forall x\neq -1$ Given function $y=f(x)$
such that
$$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$
find $f(x)$ and $f(2007)$.
| With
$$f(x) = \frac{x+3}{x-1}-\frac{3}{2}$$
we have
$$a := f\left(\frac{x-3}{x+1}\right) = -x-\frac{3}{2}$$
and
$$b := f\left(\frac{x+3}{x+1}\right)=2x+\frac{3}{2}$$
Therefore the sum is $a+b=x$ and $f(2007)=-\frac{999}{2006}.$
Edit regarding the comments: My function $f(x)\,$ is a solution of the functional equation (FE). The road to it was as follows: If the FE is valid for all $x\ne -1\;$ and you plug in large $x$ into the FE, you see that there should be a singularity at $x=1.\;$ Making the ansatz $f(x) = \frac{g(x)}{x-1}+\cdots$ and playing arround with Maple arrives at the given function.
If the OP has a different function solving the FE, this only shows that there are at least two solutions.
| {
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"url": "https://math.stackexchange.com/questions/1060737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
| This is the same as proving that $2^{6n-1}+5\cdot3^{2n}$ is divisible by $11$ for all $n$. The case $n=1$ is obvious.
By induction hypothesis, you can assume $2^{6n-1}+5\cdot3^{2n}=11k$, which can be written
$$
2^{6n-1}=11k-5\cdot3^{2n}
$$
Now
\begin{align}
2^{6(n+1)-1}+5\cdot3^{2(n+1)}
&=2^6\cdot2^{6n-1}+45\cdot3^{2n}\\
&=2^6(11k-5\cdot3^{2n})+45\cdot3^{2n}\\
&=11\cdot 2^6k+3^{2n}(45-5\cdot64)
\end{align}
and you're done because $45-5\cdot64=-275=-11\cdot16$.
| {
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"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 4
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ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
then total no. of Subest of $X = 2^5 = 32$
(Which also contain $\phi$)
$\bullet\; $ If $A = \{1\}\;,$ Then $B=\left (\{2\}\;,\{3\}\;,\{4\}\;,\{5\}\;,\{2,3\}\;,\{2,4\}\;,\{2,5\}\;,\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{2,3,4\}\;,\{2,3,5\}\;,\{3,4,5\}\;,\{2,4,5\}\;,\{2,3,4,5\}\right )$
Similarly for $A=\{2\}\;,A=\{3\}\;,A=\{4\}$ and $A=\{5\}\;,$ we get $15$ ordered pair for each single
element ed set $A$
$\bullet\; $ If $A = \{1,2\}\;,$ Then $B=\left(\{3\}\;,\{4\}\;,\{5\},\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{3,4,5\}\right)$
Similarly for $A=\{1,3\}\;,A=\{1,4\}\;,A=\{1,5\}\;,A=\{2,3\}\;,A=\{2,4\}\;,A=\{2,5\}\;,A=\{3,4\}\;,A=\{3,5\}\;,A=\{4,5\}$ we get $7$ ordered pair for each double elemented set $A$
$\bullet\; $ If $A=\{1,2,3\}\;,$ Then $B=\left(\{4\}\;,\{5\}\;,\{4,5\}\right)$
So for $10$ ordered pair of $A\;,$ we get $3$ ordered pair of $B$
$\bullet \;$ If $A = \{1,2,3,4\}\;,$ Then $B = \{5\}$
So for $5$ ordered pair of $A\;,$ we get $1$ ordered pair of $B$
So Total ordered pair of $\left(A,B\right)$ is $ = \left(5\times 15\right)+\left(10 \times 7\right)+\left(10 \times 3\right)+\left(5 \times 1\right) = 75+70+30+5 = 180$
If Question is How can we solve using Combination (selection) way.
plz explain me, Thanks
| First let us try to get the number of ordered pairs $(A,B)$ such that $A\cup B = \phi$, without any restrictions on $A$ and $B$. Clearly, each element can EITHER be in only $A$ OR only $B$ OR neither. So, that is $3^5$. Now, remove the cases when $A = \phi$. This is $2^5-1$. Similarly for $B$. So, the total is $3^5 - 2^5 - 2^5 + 1 = 180$ (the +1 is because we have counted the pair $(A,B) = (\phi,\phi)$ twice).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$.
Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$.
How can I prove this?
This is what I have so far:
$0 < x < y$ (definition of positive)
$0 < 1 < \frac{y}{x}$ (division by $x$)
$0 < \frac{1}{y} < \frac{1}{x}$ (division by $y$)
But I think it seems too simple.
| $y>x$ means $y=x+M$ for some $M>0$.
$1/y>0$ is obvious: it's a ratio between positive numbers.
Then
$$
\frac1{y}=\frac1{x+M}=\frac1{x(1+\frac{M}{x})}=\frac1x\cdot\frac{1}{1+\frac{M}{x}}
$$
Next observe that $M,x>0$ hence $M/x>0$, thus $1+\frac{M}{x}>1$.
Now, by axioms of real numbers, the sign of an inequality is preserved is you multiply both sides by a positive quantity $c>0$. Let's multiply both sides of $1+\frac{M}{x}>1$ by the positive quantity $c=\frac{1}{1+\frac{M}{x}}$. We obtain then
$$
\frac{1}{1+\frac{M}{x}}<1.
$$
Hence you know that multiplying a positive quantity by a factor in $]0,1[$ you'll obtain something smaller (exercise). Thus
$$
\frac1x\cdot\frac{1}{1+\frac{M}{x}}<\frac1x
$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Probabilistic Proof of $\prod\limits_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$ Using probability methods (characteristic function?) prove
$$\prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$$
I know what is characteristic function but I have no idea how use it in this task. I will grateful for help.
| I don't know if it answers your question since I don't use characteristic function. By using
$$ \sin(2\theta)=2\sin\theta\cos\theta $$
one has
\begin{eqnarray}
\prod_{i=1}^n\cos\left(\frac t{2^i}\right)&=&\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\cos\left(\frac t{2^n}\right)\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\cos\left(\frac t{2^n}\right)\sin\left(\frac t{2^n}\right)}{\sin\left(\frac t{2^n}\right)}\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)2\cos\left(\frac t{2^n}\right)\sin\left(\frac t{2^n}\right)}{2\sin\left(\frac t{2^n}\right)}\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\sin\left(\frac t{2^{n-1}}\right)}{2\sin\left(\frac t{2^n}\right)}
\end{eqnarray}
and continue doing this, one has
\begin{eqnarray}
\prod_{i=1}^n\cos\left(\frac t{2^i}\right)
&=&\cdots\\
&=&\frac{\cos\left(\frac t{2}\right)\sin\left(\frac t{2}\right)}{2^{n-1}\sin\left(\frac t{2^n}\right)}
&=&\frac{\sin t}{2^{n}\sin\left(\frac t{2^n}\right)}.
\end{eqnarray}
Thus
$$ \prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\prod_{i=1}^n\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\frac{\sin t}{2^{n}\sin\left(\frac t{2^n}\right)}=\frac{\sin t}{t}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Simplifying Permutations Could someone explain the process of simplifying the following permutation in $S_6$
(1,3,5)(2,4,5)(2,3,6)
An explanation on how you arrived at the simplified form would also be greatly appreciated.
Thanks
| In the book I read this from they started at the right. Having this in mind we will now simplify $(1,3,5)(2,4,5)(2,3,6)$. To do this we start with $1$. Where does $1$ end up after this permutation? to see this look for the first appearence of $1$ (starting from the right), it only appears once and it goes to $3$, so write $(1,3$ . We now wish to know where $3$ goes to. This time $3$ appears in the first cycle (right to left), so $3$ goes to $6$. And the other cycles don't move $6$, so we write $(1,3,6$ we know want to go where $6$ goes to. Notice $6$ goes to $2$, and then $2$ goes to $4$ in the next cycle, and two is not moved by the last cycle, so $6$ goes to $4$, we proceed to write $(1,3,6,4$. We know wish to know where $4$ goes. $4$ is not moved by the first cycle, but it is moved by the second cycle, notice $4$ goes to $5$, and in the last cycle $5$ is sent to $1$, so we write $(1,3,6,4)$. We have closed the first cycle, however notice we have not inspected what happens to elements $2$ and $5$ yet. since $2$ is the smallest numbers missing we write $(1,3,6,4)(2$ and proceed to see where $2$ is mapped under the permutation. Notice $2$ is sent to $3$ by the first cycle and $3$ not moved by the second cycle, it is however moved by the last cycle, it is sent to $5$, so $2$ goes to $5$, we proceed to write $(1,3,6,4)(2,5$ Where does $5$ go to? well all the other elements are taken so $5$ must go to $2$ and we conclude:
$(1,3,5)(2,4,5)(2,3,6)=(1,3,6,4)(2,5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
| Noting that
$$ \int_0^1x^n\ln x\,dx=-\frac{1}{(n+1)^2} $$
we have
\begin{eqnarray}
\int_0^\infty\frac{(x^2-1)\ln x}{1+x^4}dx&=&2\int_0^1\frac{(x^2-1)\ln x}{1+x^4}\,dx\\
&=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^2-1)x^{4n}\ln x\,dx\\
&=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n+2}-x^{4n})\ln x\,dx\\
&=&2\sum_{n=0}^\infty(-1)^n\left(\frac1{(4n+1)^2}-\frac1{(4n+3)^2}\right)\\
&=&2\sum_{n=-\infty}^\infty(-1)^n\frac1{(4n+1)^2}\\
&=&\frac{1}{32}\left(\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{8})^2}-\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{8})^2}\right)\\
&=&\lim_{b\to0}\frac{1}{32}\left(\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{1}{8}}-\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{3}{8}}\right)\\
&=&\frac{\pi^2}{4\sqrt2}.
\end{eqnarray}
Here we use this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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