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Functional expression for the sum of a certain power series Does anybody recognize the following power series together with a functional expression for the sum:
$$
\sum_{n = 0}^{\infty} \left( \begin{array}{c} 2n \\ n \end{array} \right) x^n
$$
| I wanted to elaborate on the alternate derivation Jack suggested. I'm sure there's a shorter way, but here it goes.
One can use induction to prove $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta.$$
The $n=0$ case is clear. For the inductive step note that
$$\binom{2n+2}{n+1} = 4\frac{2n+1}{2n+2}\binom {2n}n,\quad \text{and} \quad \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = \frac{2n+1}{2n+2}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta,$$
where the second is due to the following partial integration:
\begin{align}
\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = 0 + \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)\sin^2(\theta)\,d\theta &= \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)(1-\cos^2(\theta))\,d\theta.
\end{align}
These imply that
$$\frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{1}{4^n}\frac{2n+1}{2n+2}\binom {2n}n = \frac{2n+1}{2n+2}\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta.$$
Meaning that the inductive proof is complete, so let's use its result.
\begin{align}
\sum_{n\geq 0}\left ( 4^n\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \right )x^n
&= \tag{1}
\frac 2\pi \int_{0}^{\pi/2} \left (\sum_{n\geq 0} 4^n \cos^{2n}(\theta) x^n\right ) \,d\theta
\\&=
\frac 2\pi \int_{0}^{\pi/2} \frac{1}{1-4\cos^2(\theta)x} \,d\theta
\\&=
\frac 2\pi \int_{\arctan0}^{\arctan \infty} \frac{1}{1-4\cos^2(\theta)x} \,d\theta
\\&= \tag{2}
\frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\cos^2(\arctan u)x} \,du
\\&= \tag{3}
\frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\frac 1{1+u^2}x} \,du
\\&=
\frac 2\pi \int_0^\infty \frac 1 {(1-4x)+u^2} \,du
\\&=
\frac 2\pi \left[ \frac{\arctan \left( \frac u {\sqrt{1-4x}} \right)
}{\sqrt{1-4x}}\right]_{u=0}^{\infty}
\\&=
\frac 2\pi \frac{\frac \pi 2}{\sqrt{1-4x}}
\\&=
\frac 1 {\sqrt{1-4x}}
\end{align}
Further explanation:
(1)
Recognize that this is a geometric series, which converges for $|x| < \frac 14$, since $|x| < \frac 14 \Rightarrow |4\cos^2(\theta)x| < 1.$ By analyticity, we may exchange the order of integrations. It also has a nice closed form.
(2)
Substitute $\theta = \arctan u$, use $\arctan' = \frac 1 {1+\operatorname{id_{\mathbb R}}^2}. $
(3)
For example, you can use $\tan^2(\theta) = \frac 1 {\cos^2(\theta)} - 1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$
I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$
$$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.
If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.
Thank you!
| This inequality is something hard
$ a\geq b\geq\sqrt{3}\geq c>0$ such that $abc=a+b+c$ and constraints $(C)$ and $(2)$ and $(3)$ then we have :
$$\frac{1}{7b+a}+\frac{1}{7a+c}+\frac{1}{7c+b}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$
And :
$$\frac{1}{7b+a}\leq \left(\frac{1}{b}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (C)$$
$$\frac{1}{7c+b}+\frac{1}{7a+c}\leq \left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}\quad (2)$$
Simplify the condition $(C)$ is easy comparing the function :
$$f(x)=\frac{1}{7x+1}$$
Or:
$$f\Big(\frac{6+\frac{a}{b}}{7}\Big)\leq f\left(\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}\right)$$
For the constraint $(2)$ we know that :
$$f(a)=\frac{1}{7c+b}+\frac{1}{7a+c}-\left(\frac{1}{c}+\frac{1}{a}\right)\frac{1}{1+7\left(\frac{\frac{b}{a}+\frac{a}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3}{9}\right)^{\frac{1}{3}}}$$
is an increasing function (as the difference of a positive decreasing function and a negative increasing function) on $(\sqrt{3},\frac{\frac{3}{2}\sqrt{3}-\frac{1}{2}c+c}{c(\frac{3}{2}\sqrt{3}-\frac{1}{2}c)-1})$ with :
$$b=\frac{a+c}{ac-1}$$
So we have to find an approximation of the roots !
We can choose $b\geq\frac{3}{2}\sqrt{3}-\frac{1}{2}c\quad (3)$
A bit of algebra and we get a polynomial with a root equal to $\sqrt{3}$ and one other root .Now the magic key is,the inequality and the constraints $(C),(2)$ are homogenous so we can introduce a coefficient and play with it keeping in mind the others constraints.
I continue later...Thanks!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
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Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$ Can someone explain how is the RHS concluded? I did with sample numbers and it is all correct. but I can't figure out how C(12,6) comes to play.
$$
\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} = (2^{12} - \binom{12}{6}) / 2
$$
| Suppose we toss a fair coin for $12$ times. The number of combinations is $2^{12}$.
Now notice the number of outcomes $N(H>T)$ such that the total number of Head is more than Tail is same as that of Tail more than Head $N(H<T)$, by symmetry.
We know $N(H>T)$ is the total number of choosing less than $6$ Tails, which is
$$N=\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$$
We know that the no. of outcomes when the number of Head and Tail equals is ${12}\choose{6}$.
Since the total number of possible outcome is
$$2N+{{12}\choose{6}}=2^{12}$$
The claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Solve boundary value problem $x^2y'' + 3xy' + y = x^2$ Problem
$x^2y'' + 3xy' + y = x^2$
BC: $y(1) = 0, y(e) = 0$
I tried to first solve for the homogenous solution, but I get a really weird equation to solve for values of r. How do you solve the above equation?
My Work
$y=e^{rx} \\
x^2r^2 + 3xr + 1 = x^2 \\
x^2r^2 + 3xr + 1-x^2 = 0 \\
r = \frac{-3x \pm \sqrt{9x^2-4(1-x^2)x^2}}{2x^2}
$
| $$x^2\frac{d^2y}{dx^2} +3x\frac{dy}{dx} + y = x^2$$
This is known as an Euler Equation - differential equations where the degree of $x$ matches the order of the derivative. They can be solved using the change of variable $x= e^z$
So we get $$x\frac{dy}{dx} = \frac{dy}{dz}$$
and $$x^2\frac{d^2y}{dx^2}= \frac{d^2y}{dz^2} -\frac{dy}{dz}$$
Substituting this in, you should find that the coefficients of $x$ cancel out, and you get a more easily solved ODE.
You should end up with this ODE:
$$\frac{d^2y}{dz^2} -\frac{dy}{dz} + 3\frac{dy}{dz} + y = e^{2z}$$
$$\frac{d^2y}{dz^2} + 2\frac{dy}{dz} + y = e^{2z}$$
Auxilliary equation:
$$\lambda ^2 +2\lambda + 1 = 0$$
$$\implies \lambda = -1$$
So $$y_{CF} = c_{1}e^{-z}+c_{2}ze^{-z}$$
Then we try $$y_{PI} = Ae^{2z}$$
Substituting into our ODE:
$$\implies (4A+4A+A)e^{2z} = e^{2z} \implies A=\frac{1}{9}$$
$$y_{GS} = c_{1}e^{-z}+c_{2}ze^{-z} + \frac{1}{9}e^{2z}$$
Changing back in terms of $x$, we have $z = \ln x$
$$y_{GS} = c_{1}e^{\ln(x^{-1})}+c_{2}\ln (x)e^{\ln (x^{-1})} + \frac{1}{9}e^{\ln (x^{2})}$$
$$y_{GS} = \frac{c_{1}}{x} + \frac{c_{2}\ln x}{x} + \frac{x^2}{9}$$
$$y(1) = 0 \implies c_{1} + \frac{1}{9} = 0 \implies c_{1} = - \frac{1}{9}$$
$$y(e) = 0 \implies -\frac{1}{9e} +\frac{c_{2}}{e} +\frac{e^2}{9} = 0$$
$$c_{2} = \frac{1}{9}(1-e^3)$$
So $$y_{PS} = -\frac{1}{9x} + \frac{1}{9}(1-e^3)\cdot \frac{\ln x}{x} + \frac{x^2}{9}$$
$$y_{PS} = \frac{1}{9}(-\frac{1}{x} + \frac{(1-e^3)\ln x}{x} + x^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$.
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$ $\left|f_{n}\left(x\right)-f\left(x\right)\right| \le \varepsilon$
And this how i tried to continue :
$\left|f_{n}\left(x\right)-f\left(x\right)\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\left|x\right|\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|=\left|\frac{\left(\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right)\left(\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}\right)}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{x^{2}+\frac{1}{n}-x^{2}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|$
Which lead me to the same problem of $\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|$
any advice ?
| You can notice that
$$
\sqrt{x^2+\frac1n}\le\sqrt{x^2+2\frac{|x|}{\sqrt n}+\frac1n}=\sqrt{\left(|x|+\frac1{\sqrt n}\right)^2}=|x|+\frac1{\sqrt n}.
$$
So $0<f_n(x)-|x|\le\dfrac1{\sqrt n}$. Hence, the uniform convergence (on all $\mathbb R$, not only on $[-1,1]$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I need some help finding my mistake for solving a second order recurrence relation I have the recurrence relation: $$ a_n = a_{n-1} + 2a_{n-2} ; a_0 = 2, a_1=1$$
This is what I did:
$$ let \ \ g(x) = \sum a_n x^n \ \ then; \\
g(x) -a_0 -a_1x = \sum_{n=2}^\infty a_nx^n \\
\implies g(x) -2 -x = \sum_{n=2}^\infty (a_{n-1} + 2a_{n-2})x^n \\
\implies g(x)-2-x= \sum_{n=2}^\infty a_{n-1}x^n + 2\sum_{n=2}^\infty a_{n-2}x^n \\
\implies g(x)-2-x= x(g(x)-1) + 2x^2g(x)\\
\implies g(x)(1-x-2x^2) = 2\\
\implies g(x) = \frac{2}{1-x-2x^2}; \ \frac{2}{1-x-2x^2}= \frac{2}{3(1+x)} + \frac{4}{3(1-2x)} \\
\implies \frac{2}{3}\sum_{n=0}^\infty (-1)^nx^n + \frac{4}{3}\sum_{n=0}^\infty (2x)^n \\
$$
Giving $$ a_n = \frac{2}{3}(-1)^n + \frac{4}{3}(2^n) \implies a_n = \frac{2}{3}((-1)^n + 2^{n+1}) $$
But when I typed in the original recurrence relation into Wolfram Alpha and into matlab I would get the recurrence relation of: $$a_n = (-1)^n + 2^n $$ And I cannot find my mistake to get this recurrence relation. Any help would be appreciated.
| @WW1 says it all. You mistook $2$ for $1$ in what is supposed to be
$g(x)−2−x=x(g(x)−2)+2x^2g(x)$
Checking:
$g(x)=(2-x)/(1-x-2x^2)=(1-2x+1+x)/(1-x-2x^2)$
$g(x)=1/(1+x)+1/(1-2x)$
which you will see, gives the right answer when we work out the Taylor series and read off the coefficients for each power of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The $n$-th derivative of the reciprocal of a function and a binomial identity While I was looking for an answer to this MSE post in order to prove
\begin{align*}
\frac{d^n}{dx^n}\left(\frac{1}{1-e^{-x}}\right)=(-1)^n\sum_{j=1}^n{n\brace j}j!\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\tag{1}
\end{align*}
with the numbers ${n\brace j}$ denoting the Stirling numbers of the second kind I considered the following formula of the reciprocal of the $n$-th derivative of a function which might be interesting by itself.
With $D_x:=\frac{d}{dx}$ the following relationship is valid according to (3.63) in H.W. Goulds Binomial Identities, vol. I
\begin{align*}
D_x^n\left(\frac{1}{f(x)}\right)=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{\left(f(x)\right)^{j+1}}D_x^n\left(\left(f(x)\right)^j\right)
\end{align*}
$$ $$
Applying this formula to the function $f(x)=\frac{1}{1-e^{-x}}$ we obtain
\begin{align*}
D_x^n&\left(\frac{1}{1-e^{-x}}\right)\\
&=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{\left(1-e^{-x}\right)^{j+1}}D_x^n\left(\left(1-e^{-x}\right)^j\right)\\
&=\sum_{j=0}^n\frac{(-1)^j}{\left(1-e^{-x}\right)^{j+1}}\binom{n+1}{j+1}D_x^n\left(\sum_{k=0}^j\binom{j}{k}(-1)^ke^{-kx}\right)\\
&=(-1)^n\sum_{j=0}^n\frac{(-1)^j}{\left(1-e^{-x}\right)^{j+1}}\binom{n+1}{j+1}\sum_{k=0}^j\binom{j}{k}(-1)^kk^ne^{-kx}\\
&=\frac{(-1)^n}{(1-e^{-x})^{n+1}}\sum_{j=1}^n(-1)^j\left(1-e^{-x}\right)^{n-j}\binom{n+1}{j+1}\sum_{k=1}^j\binom{j}{k}(-1)^kk^ne^{-kx}\tag{2}
\end{align*}
On the other hand with the identity ${n\brace j}=\frac{1}{j!}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^n$
we obtain from (1)
\begin{align*}
D_x^n&\left(\frac{1}{1-e^{-x}}\right)\\
&=(-1)^n\sum_{j=1}^n{n\brace j}j!\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\\
&=(-1)^n\sum_{j=1}^n\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^n\\
&=\frac{(-1)^n}{(1-e^{-x})^{n+1}}\sum_{j=1}^n(-1)^je^{-jx}(1-e^{-x})^{n-j}\sum_{k=1}^j\binom{j}{k}(-1)^{k}k^n\tag{3}
\end{align*}
I have difficulties to prove the equality of (2) with (3). So putting $y=e^{-x}$ I would like to ask for a prove of the following relationship
Claim: The following is valid for $n\geq 1$ and $y\geq 0$.
\begin{align*}
\sum_{j=1}^n&(-1)^j\left(1-y\right)^{n-j}\binom{n+1}{j+1}\sum_{k=1}^j\binom{j}{k}(-1)^kk^ny^k\\
&=\sum_{j=1}^n(-1)^jy^j(1-y)^{n-j}\sum_{k=1}^j\binom{j}{k}(-1)^{k}k^n
\end{align*}
Please note I'm not interested in a proof by induction. I would like to see how to transform one side into the other, maybe with the help of generating functions.
Many thanks in advance.
| Extracting coefficients on $[y^m]$ where $0\le m\le n$ we see that we
have to prove that
$$\sum_{j=1}^n (-1)^j {n+1\choose j+1}
\sum_{k=1}^j {j\choose k} (-1)^k k^n {n-j\choose m-k} (-1)^{m-k}
\\ = \sum_{j=1}^n (-1)^j {n-j\choose m-j} (-1)^{m-j}
\sum_{k=1}^n (-1)^{k} {j\choose k} k^n$$
or alternatively
$$\sum_{j=1}^n (-1)^j {n+1\choose j+1}
\sum_{k=1}^j {j\choose k} k^n {n-j\choose m-k}
= \sum_{j=1}^n {n-j\choose m-j}
\sum_{k=1}^n (-1)^{k} {j\choose k} k^n.$$
Re-write this as follows:
$$\sum_{k=1}^n k^n \sum_{j=k}^n
(-1)^j {n+1\choose j+1} {j\choose k} {n-j\choose m-k}
= \sum_{k=1}^n k^n (-1)^k
\sum_{j=1}^n {n-j\choose m-j} {j\choose k}.$$
We have the claim if we can show that
$$ \sum_{j=k}^n
(-1)^j {n+1\choose j+1} {j\choose k} {n-j\choose m-k}
= (-1)^k \sum_{j=1}^n {n-j\choose m-j} {j\choose k}.$$
For the LHS we introduce
$${j\choose k} = {j\choose j-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j-k+1}} (1+z)^j \; dz$$
This has the property that it vanishes when $j\lt k$ including $j=-1$
so we may lower the index of the sum to $-1.$ We also introduce
$${n-j\choose m-k}
= \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n-j} \; dw$$
and obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
z^{k-1}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n}
\\ \times \sum_{j=-1}^n {n+1\choose j+1} (-1)^j
\frac{(1+z)^j}{z^j (1+w)^j}
\; dw\; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^{k}}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n+1}
\\ \times \sum_{j=-1}^n {n+1\choose j+1} (-1)^{j+1}
\frac{(1+z)^{j+1}}{z^{j+1} (1+w)^{j+1}}
\; dw\; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^{k}}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n+1}
\\ \times \left(1-\frac{1+z}{z(1+w)}\right)^{n+1}
\; dw\; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \frac{1}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(wz-1)^{n+1}
\; dw\; dz.$$
Extracting coefficients we find
$$- (-1)^{n+1-m+k} {n+1\choose m-k}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \frac{1}{1+z}
z^{m-k} \; dz
\\ = (-1)^k {n+1\choose m-k}.$$
Observe that we didn't use the differential in the integral which
means this also goes through using formal power series
only. Continuing with the RHS we find
$$(-1)^k \sum_{j=k}^n {n-j\choose m-j} {j\choose k}
= (-1)^k \sum_{j=0}^{n-k} {n-k-j\choose m-k-j} {j+k\choose k}.$$
We introduce
$${n-k-j\choose m-k-j}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k-j+1}} (1+z)^{n-k-j} \; dz$$
This vanishes when $j\gt m-k$ at some point at most at the upper index
(recall that $m\le n$). Hence we are justified in extending $j$ to
infinity and obtain
$$\frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n-k}
\sum_{j\ge 0} {j+k\choose k} \frac{z^j}{(1+z)^j}
\; dz
\\ = \frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n-k}
\frac{1}{(1-z/(1+z))^{k+1}}
\; dz
\\ = \frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n+1}
\frac{1}{(1+z-z)^{k+1}}
\; dz
= (-1)^k {n+1\choose m-k}.$$
We have shown that the coefficients on $k^n$ in the outer sum are
equal and hence so are the coefficients on $[y^m].$ This concludes the
argument.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there $n\in\mathbb N$ for which $3n+1$ and $2n+1$ are perfect squares and $5n+3$ is a prime?
Is there $n\in\mathbb N$ for which $3n+1$ and $2n+1$ are perfect squares and $5n+3$ is prime?
My trying:
We know that $n=3n+1-(2n+1)$ is not prime.
But how it can help for $5n+3$?
$\mathbb N=\{1,2,...\}$.
Thank you!
| Lets take $2n+1=a^2$ and $3n+1=b^2$,WLOG assume $a,b>0$ then $4a^2-b^2=(2a-b)(2a+b)=5n+3$ this is prime only if $2a-b=1$ and $2a+b$ is prime.
Take $b=2a-1$ then $$3n+1=4a^2-4a+1\\3n=4(2n+1)-4\sqrt{2n+1}\\5n+4=4\sqrt{2n+1}\\25n^2+40n+16=16(2n+1)\\25n^2+8n=0$$so either $n=0$ or $n=-\frac{8}{25}$,but $0,-\frac{8}{25}\not\in \Bbb{N}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help finding integer solutions of equation. There are infinitely many integer solutions for the equation $4x + 6y = 8 $
My work:
$2x+3y=4$ and $2x=4-3y$ so $x=2-(\frac 32)y$
Similarly $y=\frac 43 - \frac 23 x$ are integer solutions of equation.
But correct answer is $x = −4 + 3t, y = 4 − 2t$ for all integers $t$. Help me understand where I have gone wrong.
| There are infinitely many solutions to the equation $4x+6y=8$ since $(4,6) =2$ and $2\mid8$. We use Euclidean Algorithm to determine $m,n$ such that $4m+6n=2$. Here $m=-1,n=1$ and $8=2.4$.
Thus $x_0 = 4(-1)$ and $y_0=4.1=4$ is a particular solution.
The solutions are given by $x=−4+3t,y=4−2t$ for all integers $t$.
| {
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How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction.
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
| One question you seem to have is:
How do we know that $$ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $$ means
$$
\sum_{i=1}^n\frac{1}{n+i}?$$
The answer is that we don't for sure though experience suggests it does.
So if that IS the intended meaning then
here is no ambiguity to
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{1}{n+1}=\frac{n}{n+1}.$$
Since that is what was requested, it seems that the assumed definition is correct.
But maybe something like $1/2+1/3+...+1/17$ is the sum of $1/p$ for $p$ prime in some range?
Here the final term is even and whatever the rule is should depend only on $n.$
If we think that the first term is intended to be the largest (which is reasonable but not explicit ) then at least it is valid that
$$ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad \le \frac{1}{n+1}+\frac{1}{n+1}+...+\frac{1}{n+1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you derive the following two summations?
I tried using the geometric series formula, $$ \sum_{n=0}^{k-1} ar^{n}=a\left (\frac{1-r^{n}}{1-r} \right ) $$, but I got $1-2\left ( \frac{1}{2}^{n-1} \right )$ for the first one and $2-2\left ( \frac{1}{2}^{n-1} \right )$ for the second.
Edit: I derived the first answer by doing $1\cdot \frac{1-\left (\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}$, which results in $2\cdot\left(1-\left (\frac{1}{2}\right)^{n-1} \right)$ = $2-2\left (\frac{1}{2}\right)^{n-1}$. Since it starts out as n=1, I did $2-2\left (\frac{1}{2}\right)^{n-1}-1$, so the final answer I got is $1-2\left ( \frac{1}{2}^{n-1} \right )$
| Your calculations are correct. You just got which problems you were addressing muddled up.
Notice that $\sum\limits_{k=0}^{n-1}2^{-k} = \dfrac{1-2^{-n}}{1-2^{-1}} = (2-2^{n-1})$ answers the second problem, not the first.
And likewise $\sum\limits_{k=1}^{n-1}2^{-k} = \sum\limits_{k=0}^{n-1}2^{-k} -1 = (1-2^{n-1})$ answers the first problem, not the second.
| {
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"timestamp": "2023-03-29T00:00:00",
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show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$ Let $x,y,z\in R$,and such
$$xy+yz+xz=-1$$ show that
$$(x-3)^4+(y-3)^4+(z-3)^4\ge 193$$
it seem use Cauchy-Schwarz inequality to solve it?But I try sometime can't get this answer,even now I can't find this equality when $=$?
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Hence, the expression $\sum\limits_{cyc}(x-3)^4$ is a linear expression of $w^3$,
which says that it gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$. Thus, $z=-\frac{1+x^2}{2x}$ and
$$\min_{xy+xz+yz=-1}\sum_{cyc}(x-3)^4=\min\left(2(x-3)^4+\left(-\frac{1+x^2}{2x}-3\right)^4\right)=191.779...,$$
which occurs for $x_1=-0.12...$, where $x_1$ is a negative root of the following equation.
$$11x^6-90x^5+321x^4-180x^3-107x^2-18x-1=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $1 + \frac{n}{2} \leq 1+ \frac{1}{2} +\frac{1}{3} +\cdots + \frac{1}{2^n}$ for all natural numbers $n$ Definitions
*
*$H_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ for all $n \in \mathbb{N}$
The Question
*
*Prove $1 + \frac{n}{2} \leq H_{2^n}$ for all $n \in \mathbb{N}$
My Work
*
*
Base Case:
$1+\frac{1}{2} \leq 1+\frac{1}{2} = H_1$
*
Inductive Hypothesis: $1 + \frac{k}{2} \leq H_{2^k}$ for all $k \in \mathbb{N}$
*
Induction Step: $1+\frac{k+1}{2} = 1+\frac{k}{2} + \frac{1}{2} \leq H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} = H_{2^{k+1}} $
My Problem
*
*My problem is actually understanding the $H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}$ step. I think that's how the proof should finish, but I don't know why.
My Question
*
*Can someone explain why the inequality under the "My Problem" header is true? Or if it even is true, am I going about this proof the wrong way?
| If $n\leq 2^k$, then $$\frac{1}{2^k+n}\geq \frac1{2^k+2^k}=\frac1{2^{k+1}}$$ Doing this substitution, we now have a sum of $2^k$ identical fractions, which may then be simplified greatly.
| {
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How have I incorrectly computed $\int \frac{x^{2}+4}{x(x-1)^{2}} dx$? My work:
$\displaystyle \int \frac{x^{2}+4}{x(x-1)^{2}}dx$$ =\int \frac{A}{x} dx+\int \frac{B}{x-1}dx +\int \frac{C}{(x-1)^{2}}dx$
$x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $
$x^2 + 4 = x^{2}(A+B) + x(-2A+C) + (A-B)$
equating coefficients:
$1 = A + B $
$0 = - 2A + C $
$4 = A - B$
then
$1 - B = A$
so
$4 = 1 - B - B$
$\frac{-3}{2} = B$
and
$4 = A - \frac{-3}{2}$
$\frac{5}{2} = A$
then
$0 = - 2(\frac{5}{2}) + C$
$0 = - 5 + C$
$5 = C$
plugging A, B, and C back into the integral I got:
$= \frac{5}{2}ln|x| - \frac{3}{2}ln|x - 1| - 5ln|x - 1| + C$
wolfram state this is wrong. What have I done wrong here?
Thank you
| $x$ has a $-B$ coefficient as well, so $0 = - 2A - B - C$. And $4 = A-B$ should just be $4=A$. It looks like you mistakenly did $Bx(x-1) = Bx^2 - B$ instead of $Bx^2 - Bx$.
Here's a handy tip for dealing with partial fraction decompositions.
When you get to $x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $, you can exploit the fact that this is an identity in the variable $x$. In other words, this equation is true no matter what $x$ is. So, you can plug in "helpful" values of $x$ to determine the values of $A$, $B$, $C$.
For example, if $x=0$ then $0^2 + 4 = A(-1)^2$, which gives $A = 4$. Plugging in $x=1$ immediately gives you $C$. Then since you have $A$ and $C$ you can plug in any other value of $x$ to get $B$. No need to mess around with systems of equations.
| {
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Alternating sum of reciprocals Find the infinite sum: $\frac{1}{(1)(2)}+\frac{1}{(3)(4)}+\frac{1}{(5)(6)}+\cdots$.
I thought of expanding it as $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}- \cdots$, but it would yield nothing useful as this runs into the non-convergence issue.
| We have,
$$\sum_{n \geq 1, \text{odd}} \frac{1}{n(n+1)}$$
$$=\sum_{n \geq 1, \text{odd}} (\frac{1}{n}-\frac{1}{n+1} )$$
$$=\sum_{n \geq 1} \frac{1-(-1)^n}{2} (\frac{1}{n}-\frac{1}{n+1})$$
$$=\frac{1}{2}\sum_{n \geq 1} (\frac{1}{n}-\frac{1}{n+1})-\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n}+\frac{1}{2} \sum_{n \geq 1} \frac{(-1)^n}{n+1}$$
Now note,
$$\int_{0}^{1} x^{n-1} dx=\frac{1}{n}$$
So that,
$$\sum_{n \geq 1} \frac{(-1)^n}{n}$$
$$=\sum_{n \geq 1} (-1)^n \int_{0}^{1} x^{n-1} dx$$
$$=\int_{0}^{1} \sum_{n \geq 1} (-1)^n x^{n-1} dx$$
$$=-\int_{0}^{1} \sum_{n \geq 1} (-x)^{n-1} dx$$
$$=-\int_{0}^{1} \frac{1}{1+x} dx$$
$$=-\ln 2$$
And the sum clearly converges by the alternating series test as $\frac{1}{n} \to 0$ and $\frac{1}{n}$ is decreasing for $n>0$.
Utilizing this result we may continue as,
$$=\frac{1}{2}\left(1+\ln 2+(\ln 2-1) \right)$$
$$=\ln 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrate $e^{2x} \cdot \sin 3x dx$ using integration by parts technique I'm trying to integrate $e^{2x} \cdot \sin 3x dx$ using integration by parts technique. According to some websites my answer is incorrect.
My attempt was:
$\int e^{2x}\sin (3x) dx = [u=e^{2x}, du = 2e^{2x} , dv= \sin 3x, v= -1/3\cos 3x] = uv -\int vdu = -\frac{1}{3}e^{2x}\cos 3x + \frac {2}{3}\int e^{2x} \cos (3x) dx = [u=e^{2x}, du = 2e^{2x}, dv = \cos (3x), v = \frac{1}{3}\sin (3x)] = -\frac{1}{3}e^{2x} \cos3x + \frac{1}{3} e^{2x}\sin 3x -\frac{2}{3}\int e^2x \sin (3x) dx $
And so I get:
$\frac {5}{3}\int e^{2x} \sin (3x)dx = -\frac{1}{3}e^{2x}\cos 3x + \frac{1}{3}e^{2x}\sin 3x$
And finally:
$\int e^{2x} \sin (3x)dx = \frac{1}{5}e^{2x} (\sin(3x)-\cos(3x))+C$.
According to the websites I checked the answer should be $\frac{1}{13}e^{2x}(2\sin(3x)-3\cos(3x) + C$.
Where is my mistake?
| In your second integration by parts you forgot the factor of $\frac{2}{3}$ from the first one, so
\begin{align*}
... &= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x\,dx\right) \\
&= -\frac{4}{9}\int e^{2x}\sin 3x\,dx - \frac{1}{3}e^{2x}\cos 3x + \frac{2}{9}e^{2x}\sin 3x.
\end{align*}
Now simplify.
| {
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Solve $x^x\equiv1 \pmod {14}$ I know that the order $x$ must be relatively prime to $14$ in order to have a solution so do I just check $1,3,5,9,11,13$ and see which ones raised to themselves are congruent to $1$ mod $14$?
| This answer uses results mentioned in the comments above (specifically @RobertIsrael)
You are correct that you need only consider bases which are relatively prime to $14$. In other words, you only need to consider the classes of
$$
1,3,5,9,11,13.
$$
Next, we should compute the orders of each of these. Using $o(n)$ for the order of $n$ modulo $14$, we have
\begin{align*}
o(1)&=1&o(3)&=6\\
o(5)&=6&o(9)&=3\\
o(11)&=3&o(13)&=2.
\end{align*}
Now, what solves $x^x\equiv 1\pmod {14}$. Let's consider each case:
*
*If $x$ is equivalent to $1\pmod{14}$, then, since the order of $1$ is $1$, any power will do, so $\{1+14n:n\in\mathbb{Z}\}$ all satisfy $x^x\equiv 1\pmod{14}$.
*If $x$ is equivalent to $3\pmod{14}$, then, since the order of $3$ is $6$, to get $1$, you need the power to be a multiple of $6$. In other words, $x=3+14n$ and $3+14n$ is a multiple of $6$. This is impossible because $6$ and $14$ are even while $3$ is odd.
*If $x$ is equivalent to $5\pmod{14}$, then, since the order of $5$ is $6$, to get $1$, you need the power to be a multiple of $6$. In other words, $x=5+14n$ and $5+14n$ is a multiple of $6$. This is impossible because $6$ and $14$ are even while $5$ is odd.
*If $x$ is equivalent to $9\pmod{14}$, then, since the order of $9$ is $3$, to get $1$, you need the power to be a multiple of $3$. In other words, $x=9+14n$ and $9+14n$ is a multiple of $3$ since $9$ is already a multiple of $3$. This only happens when $n$ is a multiple of $3$, so $\{9+14\cdot 3n:n\in\mathbb{Z}\}$ are the only $x$'s equivalent to $9$ which satisfy the condition.
*If $x$ is equivalent to $11\pmod{14}$, then, since the order of $11$ is $3$, to get $1$, you need the power to be a multiple of $3$. In other words, $x=11+14n$ and $11+14n$ is a multiple of $3$. This only happens when $n$ is equivalent to $2\pmod 3$ because $11$ and $14$ are equivalent to $-1\pmod3$. So, $\{11+14\cdot (2+3n):n\in\mathbb{Z}\}$ are the only $x$'s equivalent to $11$ which satisfy the condition.
*Finally, if $x$ is equivalent to $13\pmod{14}$, since the order of $13$ is $2$, to get $1$, you need the power to be a multiple of $2$. In other words, $x=13+14n$ and $13+14n$ is a multiple of $2$. This is impossible because $2$ and $14$ are even while $13$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this definite integral? Here is the integral we have to evaluate:
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
After observing, i realise that i can't evaluate these integrals from area of circle, I say that $u=\sqrt{x^2+4}$. Then i can say $dx=\frac{udu}{\sqrt{u^2-4}}$. The first term would transform into:
$$\int_2^{2\sqrt{5}}\frac{u^2\,du}{\sqrt{u^2-4}}$$
Similarly i say $v=\sqrt{x^2-4}$. Then i can also say $dx=\frac{vdv}{\sqrt{v^2+4}}$ and the second term would transform into:
$$\int_0^4\frac{v^2\,dv}{\sqrt{v^2+4}}$$
But this integrals also doesn't seem easy to solve. It goes without saying that after substitution, the bounds of integrals interchanged. Maybe it can be helpful to solve the problem. Thank you for your effort!
| Hint -
1.) $\int \sqrt{x^2+a^2} = \frac 12 x \sqrt{x^2+a^2} - \frac{a^2}2 sinh^{-1}\frac xa + c$
Or
$= \frac 12 x \sqrt{x^2+a^2} + \frac{a^2}2 \ln|x + \sqrt{x^2+a^2}+c$
2.) $\int \sqrt{x^2-a^2} = \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 cosh^{-1}\frac xa + c$
Or
$= \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 \ln|x + \sqrt{x^2-a^2} + c$
For more formulas See this link.
After applying these formulas fill limits.
| {
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Show $\lim_{(x,y)\to(0,0)}\frac{x^3-xy^3}{x^2+y^2}=0$ I need to show if the following limit is exists (and exists $0$),
$$
f(x,y)=\frac{x^3-xy^3}{x^2+y^2},
$$
for $\vec x\to 0$. I tried out the following:
$$
\frac{x^3-xy^3}{x^2+y^2}\leq\frac{x^3-xy^3}{x^2}=1-\frac{y^3}{x^2},
$$
but obviously I'm stuk here. A similar approach by omitting $y^2$ gives the same problem. What could I do next?
| With polar coordinates we have $$x=r\cos { \theta ,y=r\sin { \theta } } \\ \lim _{ (x,y)\to (0,0) } \frac { x^{ 3 }-xy^{ 3 } }{ x^{ 2 }+y^{ 2 } } =\lim _{ r\rightarrow 0 } \frac { { r }^{ 3 }\cos ^{ 3 }{ \theta } -{ r }^{ 4 }\cos { \theta \sin ^{ 3 }{ \theta } } }{ { r }^{ 2 } } =0$$
| {
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Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}
| For any $a$ and $b$, finding $n$ and $k$ such that
$a=\dfrac{n}{k-a}$
$b=\dfrac{n}{k-b}$
"proves" that $a=b\;\;\forall a,b\in\mathbb{N}$.
| {
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Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\
&=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\
&=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6. -(eqn 1)
\end{align*}
$$
I'm not sure how to go on about rest of the problem.
[update]
Reading comments, I tried assuming $z = e^{i\theta}$
$2\cos\theta = e^{i\theta} + e^{-i\theta}$
$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$
$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$ (from eqn 1)
| Following your idea, you can suppose that $|z|=1$ and then $z=\text{cis}\theta$ and $z^{-1}=\text{cis}(-\theta)$
$$z+z^{-1}=2\cos\theta \to (z+z^{-1})^4=2^4\cos^4\theta$$
on the other hand:
$$z^2+z^{-2}=(z+z^{-1})^2-2=2^2\cos^2\theta-2=2\cos2\theta$$
$$z^4+z^{-4}=(z^2+z^{-2})^2-2=(2\cos2\theta)^2-2=2\cos4\theta$$
Replacing on your last equation you get:
$$2^4\cos^4\theta=2\cos4\theta+8\cos2\theta+6\\
\cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3)$$
| {
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Deriving formula from geometric relations If we know that
and we have this triangle
Then how can we derive this relation?
($\alpha$ and $\beta$ are assumed to be small, so that $\tan(\alpha) \approx \alpha$ and $\tan(\beta) \approx \beta$). We also know that $\theta = \alpha + \beta$, so
$$\alpha \approx \frac{4GM}{c^2b}-\tan(\beta)=\frac{4GM}{c^2b}-\frac{b}{D}$$
Also, $b=d\tan(\alpha)=D\tan(\beta)$. But what next? Where is the square root coming from?
| $$\tan \alpha=\frac{b}{d} \text{ and } \tan \beta=\frac{b}{D}$$
so,
$$\frac{D/d}{D+d}=\frac{1}{b}\frac{\tan \alpha/\tan \beta}{\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}}=\frac{1}{b}\frac{\tan^2 \alpha}{\tan \alpha+\tan \beta}$$
and,
$$\frac{4GM}{c^2}\frac{D/d}{D+d}=\theta\cdot b\cdot \frac{1}{b}\frac{\tan^2 \alpha}{\tan \alpha+\tan \beta}=\frac{\theta\cdot\tan^2 \alpha}{\tan \alpha+\tan \beta} \quad (*)$$
Using that $\tan\alpha \approx \alpha$ and $\tan\beta \approx \beta$ we get
$$\frac{\theta}{\tan \alpha+\tan \beta}\approx \frac{\theta}{ \alpha+ \beta}=1$$
and backing to $(*)$ we get
$$\alpha^2=\frac{4GM}{c^2}\frac{D/d}{D+d}\to \alpha=\sqrt{\frac{4GM}{c^2}\frac{D/d}{D+d}}$$
| {
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Can someone please check if my reasoning for this proof is valid or not? I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.
Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$
Obviously, the base case works for n=10
$1024=2^{10}\gt1000=10^3$
The induction hypothesis: Assume $P_n$ is true $\rightarrow$ $2^n\gt n^3$
I want to then prove that $2^{n+1}\gt (n+1)^3$
Now, using the induction hypothesis:
$2^n\gt n^3$
multiply both sides by 2
$2^{n+1}\gt 2n^3$
Using the fact that $n\ge10$ this implies that $n^3\ge10n^2$
$2n^3=n^3 +n^3\gt n^3 +10n^2=n^3 +3n^2 +7n^2$
Using the fact that $7\gt 1$ this implies that $7n\gt n$ since n is positive.
$n^3 +3n^2 +7n^2\gt n^3 +3n^2 +n^2$
Once again, using $n\ge 10$ this implies $n^2\ge 10n$
$n^3 +3n^2 +n^2\gt n^3 +3n^2 +10n=n^3 +3n^2 +3n+7n$
Again $7\gt 1$
$n^3 +3n^2 +3n+7n\gt n^3 +3n^2 +3n +n$
Using $n\ge 10$ one last time
$n^3 +3n^2 +3n +n\gt n^3 +3n^2 +3n+10$
Since $10\gt 1$
$n^3 +3n^2 +3n+10\gt n^3 +3n^2 +3n+1=(n+1)^3$
Thus, through the chain of inequalities, I have proved that $2^{n+1}\gt (n+1)^3$.
QED
Sorry if there are any errors in my reasoning. Thank you for reading and feedback.
| It's good but it'd be easier (both to do and follow) to go forward:
$(n+1)^3 = $$n^3 + 3n^2 + 3n + 1 $$< n^3 + 3n^2 + 3n^2 + 3n^2 = $$n^3 + 9n^2 < n^3 + n*n^2 = $$n^3 + n^3 =2n^3 $$< 2*2^n = 2^{n+1}$
| {
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Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
| Putting the parametric equation for both circles
$$x=2\sin a,y=2\cos a,z=3\sin b,t=3\cos b\\$$From this you get$$xt-yz=6\sin a\cos b+6\sin b\cos a=6\sin(a+b)=6\\\sin(a+b)=1\\a+b=\frac{\pi}{2}+2k
\pi$$
Which implies $x=2\sin(\pi/2-b)=2\cos b$ and $y=2\cos(\pi/2-b)=2\sin b$
From this we see that $(x,y,z,t)=(2\cos b,2\sin b,3\sin b,3\cos b)$ for each $b$
| {
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Finding Recurrence relations for combinatorics problems After you graduate you accept a job that promises a starting salary of $40,000$ and a raise at the end of each year equal to $5\%$ of your current salary plus $1000$. For example, your raise at the end of the first year is $3000$. Let $S_n$ be your salary after $n$ years, so that $S_0 = 40,000$.
A- Find a recurrence relation.
B- Determine how much you will be making after $2$ years, after $5$ years, after $10$ years.
I did part A. I don't know if I did it correctly, so it's $S_{n+1} = S_n \cdot 1.05+1000$ with $S_0=40,000$.
| The recurrence relation
\begin{align*}
S_0 & = 40,000\\
S_{n + 1} & = S_n \cdot 1.05 + 1000
\end{align*}
that you stated is correct.
Let's look at the first few terms of the sequence.
\begin{align*}
S_1 & = S_0 \cdot 1.05 + 1000\\
& = 40,000 \cdot 1.05 + 1000\\
S_2 & = S_1 \cdot 1.05 + 1000\\
& = (40,000 \cdot 1.05 + 1000) \cdot 1.05 + 1000\\
& = 40,000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000\\
S_3 & = S_2 \cdot 1.05 + 1000\\
& = (40,000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000) \cdot 1.05 + 1000\\
& = 40,000 \cdot 1.05^3 + 1000 \cdot 1.05^2 + 1000 \cdot 1.05 + 1000\\
& = 40,000 \cdot 1.05^3 + 1000(1.05^2 + 1.05 + 1)
\end{align*}
If we use the geometric series formula
$$\sum_{k = 0}^{n} r^{k - 1} = 1 + r + r^2 + \cdots r^{n - 1} = \frac{1 - r^n}{1 - r}$$
we can express $S_3$ in the form
$$S_3 = 40,000 \cdot 1.05^3 + 1000 \cdot \frac{1 - 1.05^3}{1 - 1.05}$$
Can you find a formula for $S_n$?
| {
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Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$ Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$
Attempt: i have break $2012$ in to sum of power of $2$
as $2012 = 2^{10}+2^{9}+2^{8}+2^7+2^6+2^4+2^3+2^2$
but wan,t be able to go further, could some help me , thanks
| What is the coefficient of $x^5$ in $(x+1)(x^2+2)(x^4+4)(x^8+8)(x^{16}+16)$?
$5$ in binary is $101_2$, or $5 = 2^0 + 2^2$
This means the coefficient of $x^5$ will be formed by looking at the term $x^5 = x^{2^0}x^{2^2} = x \cdot x^4$ and multiplying it by the constant term contributed from everything else in the expression.
$(x+1)(x^2+2)(x^4+4)(x^8+8)(x^{16}+16) \\= \left[(x+1)(x^4+4)\right](x^2+2)(x^8+8)(x^{16}+16) \\= \left[x^5 + 4x + x^4 + 4\right](x^2+2)(x^8+8)(x^{16}+16)$
Note that now the answer only depends on taking $x^5$ times the constant term from the righthand part, i.e. $2^1 \cdot 2^3 \cdot 2^4 = 2 \cdot 8 \cdot 16 = 256$
In other words: Multiply together the constant terms corresponding to the pieces that aren't involved in the binary representation of the exponent you want.
We know that $x^{2012} = x^{2^2}x^{2^3}x^{2^4}x^{2^6}x^{2^7}x^{2^8}x^{2^9}x^{2^{10}}$, therefore the coefficient of $x^{2012}$ is $2^0 \cdot 2^1 \cdot 2^5 = 64$.
| {
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Proving $\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$ for all $n \in N$ using mathematical induction I'm having a horrible time completing this since this is the first time I'm learning mathematical induction.
So far I have this:
Let $P(n)$ be:
$P(n):\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$
For the base case, $n=1$
$LHS = \sum_{i=1}^{0} {\frac{1}{4i^2-1}} = {\frac{1}{4(1^2)-1}} = {\frac{1}{4-1}} = {\frac{1}{3}}$
$RHS = \frac{n}{2n +1} = \frac{1}{2(1) +1} = \frac{1}{2 +1} = \frac{1}{3}$
Therefore $P(1)$ holds.
Assuming $P(k)$, $P(k)$ holds for $n=k$:
$P(k): \sum_{i=1}^{k} {\frac{1}{4i^2-1}} = \frac{k}{2k +1}$
Proving for $P(k+1)$:
$P(k+1):\\ \\\ \\ \sum_{i=1}^{k+1} {\frac{1}{4i^2-1}}$
$= \sum_{i=1}^{k} {\frac{1}{4i^2-1}} + \frac{1}{4(k+1)^2-1}$
$=\frac{k}{2k +1} + \frac{1}{4(k+1)^2-1}$
Can anyone help me from this point onward?
| for the sum $$\sum_{i=1}^{k+1}\frac{1}{4i^2-1}$$ we can write:
$$\frac{n}{2n+1}+\frac{1}{4(n+1)^2-1}$$
and this must be $$\frac{n+1}{2(n+1)+1}$$
can you proceed?
| {
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Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies that $\frac{dx}{d\theta}=\frac{\cos \theta}{2\sqrt{\sin \theta}}.$
So we have $\int \frac{1}{1-x^4}dx=\int\frac{1}{\cos^2\theta} \cdot \frac{\cos \theta}{2\sqrt{\sin \theta}} d\theta = \int\frac{1}{2\cos\theta \sqrt{\sin\theta}}d\theta.$ Then I stuck here.
Any hint would be appreciated.
| HINT:$$\frac{1}{1-x^4}=\frac{1}{2}\left(\frac{1}{1+x^2}+\frac{1}{1-x^2}\right)$$
You handle these. I guess?
| {
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How to show that $\mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ is finitely generated? We can say that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is finitely generated if minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$ is in $\mathbb{Z}[X]$. After some calculations it can be shown that $f(X) = X^2 - X - 1 \in \mathbb{Z}[X]$ is the minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$. I think that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is generated by $\left \{1, \cfrac{1 + \sqrt{5}}{2}\ \right \}$ so that any element of it can be written in the form $a + b \cfrac{1 + \sqrt 5}{2}$ or simply $\cfrac{c}{2} + \cfrac{d \sqrt{5}}{2}$ with $c, d \in \mathbb{Z}$.
By this claim, if I take some $f\in \mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ then I should be able to write it in the form above. Now let $f = a_n \left(\cfrac{1 + \sqrt{5}}{2} \right)^n + a_{n - 1} \left(\cfrac{1 + \sqrt{5}}{2} \right)^{n - 1} + \dots + a_1 \left(\cfrac{1 + \sqrt{5}}{2} \right) + a_0$, each $a_i \in \mathbb{Z}$. The $k^{th}$ term in the partial sum above is equal to $a_k.\cfrac{n!.5^{\frac{k}{2}}}{k!(n-k)!2^k}$ . I can not see that how these terms cancel out each other and in the end we have something like $\cfrac{c}{2}+\cfrac{d\sqrt{5}}{2}$. How do we show this?
| You might want to check that
$$\left(\frac{1+\sqrt{5}}2\right)^2=\frac{3+\sqrt5}2,$$
$$\left(\frac{1+\sqrt{5}}2\right)^3=2+\sqrt5.$$
$$\left(\frac{1+\sqrt{5}}2\right)^4=\frac{7+3\sqrt5}2$$
etc.
Indeed you can prove (by induction maybe) that
$$\left(\frac{1+\sqrt{5}}2\right)^n=\frac{a_n+b_n\sqrt5}2$$
where $a_n$ and $b_n$ are integers, moreover they are either both odd
integers or both even integers.
| {
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Find all $n$ for which $504$ doesn't divide $n^8-n^2$ I have seen a similar problem where someone showed that it does divide $n^9-n^3$ but here one has to show that it isn't the case for $n^8-n^2$ for all $n$.
| As I said, $n^8-n^2=n^2(n^6-1)$ and $504=2^3\times 3^2\times 7$.
If $n$ is not a multiple of $7$, then $n^6-1$ is divisible by $7$ by Fermat and on the other hand if $n$ is a multiple of $7$, $n^2$ is divisible by $7$.
As for the powers of $2$, if $n$ is even, $n^2$ is divisible by $8$ if $n \equiv 0(\textrm{mod}\hspace{3pt} 4)$.
If $n$ is odd, then $n^6-1$ is divisible by $8$. (We have $n^2\equiv 1(\textrm{mod}\hspace{3pt} 8)$ for $n$ odd.)
As for the powers of $3$, if $n$ is a multiple of 3 we are done as $n^2$ is divisible by $9$. Otherwise, Fermat-Euler tells us $n^6\equiv 1(\textrm{mod}\hspace{3pt} 9)$
So our only case that fails is if $n$ is even, and $n 2(\textrm{mod}\hspace{3pt} 4)$.(In other words, $4k+2$ where $k$ is an integer is the answer.)
| {
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$ \sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] $ find the sum of the series, no solution was provided.
$$ \sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} + \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
I've tried a few many ways, I'll add my attempts in but they weren't correct
| In this mathstackexchange post, an answer shows how a Fourier expansion of $x(1-x)$ on $[0,1]$ gives $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}=\dfrac{\pi^3}{32}.$$ On the other hand, we have $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\sum_{n=0}^\infty \frac{1}{n^2}-\sum_{n=0}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$ So, the two results put together gives $$\sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] =\dfrac{\pi^3}{32}+\dfrac{\pi^2}{8}.$$
| {
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Sum of all possible angles. If $$\tan\left(\frac{\pi}{12}-x\right) , \tan\left(\frac{\pi}{12}\right) , \tan \left(\frac{\pi}{12} +x\right)$$ in order are three consecutive terms of a GP then what is sum of all possible values of $x$.
I am not getting any start, can anybody provide me a hint?
| Clearly, $x=n\pi$ (where $n$ is any integer) is a trivial solution.
Otherwise
$$\dfrac{\tan\dfrac\pi{12}}{\tan\left(\dfrac\pi{12}-x\right)}=\dfrac{\tan\left(\dfrac\pi{12}+x\right)}{\tan\dfrac\pi{12}}$$
$$\iff\dfrac{\sin\dfrac\pi{12}\cos\left(\dfrac\pi{12}-x\right)}{\cos\dfrac\pi{12}\sin\left(\dfrac\pi{12}-x\right)}=\dfrac{\cos\dfrac\pi{12}\sin\left(\dfrac\pi{12}+x\right)}{\sin\dfrac\pi{12}\cos\left(\dfrac\pi{12}+x\right)}$$
Using Componendo & Dividendo and $\sin(A\pm B)$ formulae,
$$\dfrac{\sin\left(\dfrac\pi{12}+\dfrac\pi{12}-x\right)}{\sin\left\{\dfrac\pi{12}-\left(\dfrac\pi{12}-x\right\}\right)}=\dfrac{\sin\left(\dfrac\pi{12}+x+\dfrac\pi{12}\right)}{\sin\left\{\dfrac\pi{12}+x-\dfrac\pi{12}\right\}}$$
As $\sin x\ne0,$
$$\sin\left(\dfrac\pi6-x\right)=\sin\left(\dfrac\pi6+x\right)$$
Using Prosthaphaeresis Formulas, $$\sin\left(\dfrac\pi6+x\right)-\sin\left(\dfrac\pi6-x\right)=2\sin x\cos\dfrac\pi6$$
So, $\sin x$ has to be $0$ which is impossible as $x\ne n\pi$
| {
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Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Rearranging the identity $1 + \tan^2A = \sec^2A$ yields
$$\sec^2A - \tan^2A = 1$$
Factoring yields
$$(\sec A + \tan A)(\sec A - \tan A) = 1$$
Since $\sec A + \tan A = 4$, we have
$$4(\sec A - \tan A) = 1$$
which yields the system of equations
\begin{align*}
\sec A + \tan A & = 4\\
\sec A - \tan A & = \frac{1}{4}
\end{align*}
Solve the system for $\sec A$ and $\tan A$, then use the identities
\begin{align*}
\cos A & = \frac{1}{\sec A}\\
\sin A & = \tan A\cos A
\end{align*}
to solve for sine and cosine.
| {
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Irreducibility of $x^{11}+x^{10}+x^{9}+\dots+x+1$
Prove that$$g(x)=x^{11}+x^{10}+x^{9}+\cdots+x+1$$ is irreducible over $\mathbb{Q}$.
This is my (incorrect-see comments) attempt
$$g(x)=\frac{(x-1)(x^{11}+x^{10}+x^{9}+\cdots+x+1)}{x-1}=\frac{x^{12}-1}{x-1},$$
So$$g(x+1)=\frac{(x+1)^{12}-1}{x}=x^{11}+12x^{10}+66x^{9}+\cdots+12.$$ By Eisenstein criteria this polynomial is irreducible over $\mathbb{Q}[x]$ since $3$ divides $a_{i}$ where $i=0,1,2...10$, but $3^{2}$ does not divide $a_{0}$ and $3$ does not divide $a_{11}$.
Assume for contradiction $g(x)$ is reducible which means $\exists p(x),q(x)\in\mathbb{Q}$ both non-unit such that $g(x)=p(x)q(x)$. Hence $$g(x+1)=p(x+1)q(x+1)\Longrightarrow g(x+1)$$is reducible. A contradiction.
I am curious to know how to prove this result.
| Hint:
It is not true because:
$$
(x^{12}-1)=(x^6+1)(x^6-1)=(x^6+1)(x^3+1)(x^3-1)=
$$
$$
=(x^6+1)(x+1)(x^2-x+1)(x-1)(x^2+x+1)
$$
and
$$
(x^{12}-1)=(x-1)(x^{11}+x^{10}+ \cdots +x+1)
$$
| {
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Calculating $\int_{0}^{\frac{\pi}{6}} \cos{(x)} \sqrt{2\sin (x)+1} dx$ I am trying to calculate the value of the following
$$\int_{0}^{\frac{\pi}{6}} cosx \sqrt{2sin x+1} dx$$
I used a substitution method.
$$u = 2 \sin (x) + 1$$
$$\frac{du}{dx} = 2\cos (x)$$
$$\frac{u}{2 \cos (x)}du = dx$$
hence
$$\int_{0}^{\frac{\pi}{6}} \cos (x) \sqrt{2 \sin (x)+1} dx$$ =
$$\int_{1}^{2} \cos (x) \sqrt{u} \times\frac{u}{2 \cos (x)}du$$
= $$\int_{1}^{2} \frac{1}{2}u^\frac{3}{2}du$$
= $$\left[\frac{1}{5}u^\frac{5}{2}\right]_1^2$$
but I can't seem to move any further.
Can anyone help please?
Many thanks.
UPDATE
The third line is incorrect. It should be
$$\frac{du}{2 \cos (x)} = dx$$
hence
= $$\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$$
= $$\left[\frac{1}{3}u^\frac{3}{2}\right]_1^2$$
= $$\frac{1}{3}\times (2\sqrt3 - 1)$$
= $$\frac{2\sqrt3-1}{3}$$
| You had this:
$$\frac{du}{dx}=2\cos(x)$$
You should've moved the $dx$ to the other side to get
$$du=2\cos(x)\ dx$$
but you had an extra $u$. Follow this, and the rest of your work is fine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{\cos x}=2\cos x$$
$$\sin x + 1=2\cos^2 x$$
$$\sin x +1=2-2\sin^2 x$$
$$2\sin^2 x +\sin x - 1=0$$
$$2\sin^2 x +2\sin x -\sin x - 1=0$$
$$2\sin x (\sin x +1) -1(\sin x+1)=0$$
$$(\sin x +1)(2\sin x-1)=0$$
So, what's the next?
| Following on from $$(\sin x +1)(2\sin x-1)=0$$
Note it's useful to sketch the graph of $\sin x$ in the given interval:
We get that either:
$\sin x=-1\implies x=\frac{3\pi}{2},\quad$or
$\sin x =\frac{1}{2}\implies x=\frac{\pi}{6}\quad\text{or}\quad x=\frac{5\pi}{6}$
Now we must be careful since we had $\cos x$ in the denominator in the original equation, we require that $\cos x \neq 0$. This eliminates the solution $x=\frac{3\pi}{2}$. Therefore the total number of solutions is $2$ - so the answer is $(c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear transformation problem M2x2 to P2 Can you guys help me with this question?
Let $T:{M_{2\times2}} \to {P_2}$ be defined by $$T\begin{pmatrix} a&b\\c&d \end{pmatrix}=(a+b-c-d)t^2+ (c+d)t+ (a+b). $$ Find the matrix of T with respect to the standard bases for ${M_{2\times 2}}$ and ${P_2}$.
The standard bases for ${P_2}$ is $$\left\{ {1,t,{t^2}} \right\}$$ and for ${M_{2x2}}$ is $$\left\{ \begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} \right\}$$
| The matrix of a linear transformation comes from expressing each of the basis elements for the domain in terms of basis elements for the range upon applying the transformation.
For example,
$$T \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = 1\cdot 1 + 0\cdot t + 1 \cdot t^2.$$
Thus the first column of the matrix for $T$ with respect to these bases will be
$$\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.$$
Repeat for the three remaining basis elements of $M_{2 \times 2}$.
Solution:
$$Mat(T) = \begin{bmatrix} 1 & 1 &0 & 0 \\ 0 & 0 & 1 &1 \\ 1 & 1 & -1 & -1\end{bmatrix} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What minimum number (A) can be taken so that (A)^N is larger than the product of N numbers? Given a sequence of N numbers say 2,8,4,7,6,5. How can we calculate a minimum number say A such that AN is greater than the product of 2*8*4*7*6*5 = 13440?
So the minimum number satisfying the above condition is 5.
As 56 = 15625 which is greater than 13440.
But 46 = 4096 which is less than 13440.
| The geometric mean will do it: in your case we have
$\sqrt[6]{2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5} = (2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5)^{\frac{1}{6}} = 4.87603...$ so, as you noted, $5^6 > 2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5$ but $4^6 < 2 \cdot 8 \cdot 4 \cdot 7 \cdot 6 \cdot 5$.
In general, if we take $N$ numbers $a_1,..., a_N$ then setting $b =\sqrt[N]{a_1 \cdot a_2 \cdots a_N}$ we have
$$b^N = a_1 \cdot a_2 \cdots a_N$$ and so $x^N > a_1 \cdot a_2 \cdots a_N$ for any $x>b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Naive Bayes Classification Example Given the following data:
$$\begin{array}{c|c|c|c|c|}
\text{Instance} & \text{A} & \text{B} &\text{C} &\text{Class} \\ \hline
\text{1} & 1 & 2 & 1 & 1 \\ \hline
\text{2}& 0 & 0 & 1 & 1 \\ \hline
\text{3} & 2 & 1 & 2 & 2 \\ \hline
\text{4} & 1 & 2 & 1 & 2 \\ \hline
\text{5} & 0 & 1 & 2 & 1 \\ \hline
\text{6} & 2 & 2 & 2 & 2 \\ \hline
\text{7} & 1 & 0 & 1 & 1\\ \hline
\end{array}$$
Predit the class label for instance $(A=1, B=2, C=2)$ using naive Bayes classifcation.
Let $C_{1}$ be class $1$ and $C_{2}$ be class $2$.
I have so far that the class prior probabilties are:
$$P(C_{1})=\dfrac{4}{7}$$ $$P(C_{2})=\dfrac{3}{7}$$
Using Bayes Theorem: $$P(C{i}|X)=\dfrac{P(X|C{i})P(C_{i})}{P(X)}$$
I know that $P(X|C_{i})=\prod^{n}_{k=1}P(X_{k}|C_{i})$ but not sure how to calculate this.
Where do I go from here to go about answering the question?
| For $C_1$, by the assumption of Naive Bayesian Classifier, we have
$$
P(A = 1, B = 2, C=2 \mid C_1) = P(A = 1 \mid C_1) \cdot P(B = 2 \mid C_1) \cdot P(C = 2 \mid C_1)
$$
Take $P(A = 1 \mid C_1)$ as an example. There are $4$ training records of $C_1$, among which there are $2$ records with $A = 1$. Therefore, $P(A = 1 \mid C_1) = \frac{2}{4}$. Similarly, you can calculate $P(B = 2 \mid C_1)$ and $P(C = 2 \mid C_1)$.
It is similar to calculate $P(A = 1, B = 2, C = 2 \mid C_2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| The method does work. How did you deal with $2x^2$ anyway? You know that $x^2 \equiv 0, 1, 4 \pmod 5$. So you just double those, like this: $0, 2, 8$, rewrite as $0, 2, 3$.
Likewise with $3x$, you just have to triple $0, 1, 2, 3, 4$ to get $0, 3, 6, 9, 12$ which we rewrite as $0, 3, 1, 4, 2$.
And lastly $4 \equiv 4 \pmod 5$, obviously.
So the possibilities are:
*
*If $x \equiv 0 \pmod 5$, then $2x^2 + 3x + 4 \equiv 0 + 0 + 4 = 4$.
*If $x \equiv 1 \pmod 5$, then $2x^2 + 3x + 4 \equiv 2 + 3 + 4 = 9 \equiv 4$.
*If $x \equiv 2 \pmod 5$, then $2x^2 + 3x + 4 \equiv 8 + 6 + 4 = 18 \equiv 3$.
*If $x \equiv 3 \pmod 5$, then $2x^2 + 3x + 4 \equiv 18 + 9 + 4 = 31 \equiv 1$.
*If $x \equiv 4 \pmod 5$, then $2x^2 + 3x + 4 \equiv 32 + 12 + 4 = 48 \equiv 3$.
As we failed to find a $0$ this way, we conclude that $2x^2 + 3x + 4$ is never divisible by $5$. Not as elegant as completing the square, but easy enough for a child to do.
| {
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"url": "https://math.stackexchange.com/questions/2250718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of possible values of $n$ of an AP The sum of the first $n$ terms of an AP whose first term is a (not necessarily positive) integer and the common difference is $2$, is known to be 153. If $n>1$, then the number of possible values of $n$?
My approach :
$153= \frac{n}{2}[ 2a+ (n-1)\cdot2]$
from this we get;
$(a-1)n+2n^2-153=0$;
So shouldn't be the number of possible values of $n$ be $2$ as $D>0$?
| Some initial observations might help.
As the AP terms sum to $153$ and the common difference is $2$, the AP must consist of sequential odd numbers. Also, the $153= 3\cdot 3\cdot 17$ hence factors are $1,3,9,17,51,153$.
Consider an AP, $AP1$, with positive first term $2r+1$ and with $m$ terms.
Consider its "mirror AP", $AP2$ with the same last term ($2r+2m-1)$ as $AP1$ but first term $-(2r-1)$. Number of terms is $2r+m$. The first $r$ terms are negative values of the second $r$ terms so the first $2r$ terms sum to zero. This means $AP2$ has the same sum as $AP1$
Hence, for a given value of $m$ , possible AP lengths $n$ are $m$ and $m+2r$.
Last term is $2r+2m-1$. Sum of AP equals $153$ (given), i.e.
$$\begin{align}
153&=\frac m2\big[(2r+1)+(2r+2m-1)\big]\\
&=m(m+2r)\\
&=1\cdot 153\\
&=3\cdot 51\\
&=9\cdot 17
\end{align}$$
As $n>1$ therefore $\color{red}{n=3, 9, 17, 51, 153}$.
The possible APs are as follows (terms in blue sum to zero):
$$
\begin{array}
\hline
\hline
&n=m
&n=m+2r\\\\
\hline
m=1: \qquad
&\text{N/A}
&\color{blue}{-151,-149,\cdots 149,151}, 153\\
&&(n=153)\\\\
m=3:\qquad
&49,51,53
&\color{blue}{-47,-45,\cdots 45,47},49,51, 53\\
&(n=3)
&(n=51)\\\\
m=9:\qquad
&9,11,13,\cdots, 25
&\color{blue}{-7,-5,\cdots 5,7},9,11,13,\cdots, 25\\
&(n=9)
&(n=17)\\
\hline
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Permutations avoiding repeated elements in a row Assume a working set of: {A, B, C}
I'm creating a new set of 10 elements using the working set. The only restriction is that no single element can repeat more than 4 in a row.
For example:
Valid: AAABBCABAA
Invalid: AAAAABCABC
Valid: AAAABCABCA
Invalid: BABAACCCCC
How many permutations are possible with new set size of 10 using the working set without violating this rule?
How many permutations exist for variable ranges of set sizes using the working set? For example, how many combinations exist that do no have any element repeat from the working set for sizes 8 through 10?
| The following answer is long but introduces what I feel is a wonderfully elegant method which itself introduces the use of regular expressions to find the generating function that enumerates valid strings.
Let's imagine listing all the different valid strings and let's separate the strings with "+"s. We will call this list a regular expression $R$. The list will begin
$$\begin{align}
R=\epsilon &+ A + B + C &\\
&+ AA + AB + AC + BA + BB + BC + CA + CB + CC &\\
&+ AAA + AAB + AAC + ABA + ABB + ABC + ACA + ACB + ACC &\\
&+ BAA + BAB + BAC + BBA + BBB + BBC + BCA + BCB + BCC &\\
&+ CAA + CAB + CAC + CBA + CBB + CBC + CCA + CCB + CCC &\\
&+\cdots +AAAAB + AAAAC + \cdots &
\end{align}$$
Where $\epsilon$ is the empty string.
Now lets have some rules:
*
*We will allow ourselves to factorise regular expressions (whilst maintaining order) then, for example, we may write $AC+BC$ as $(A+B)C$ or we may write $CA+CB$ as $C(A+B)$ but $C(A+B)\ne (A+B)C$, also we cannot factor $AC+CA$.
*It doesn't matter in which order we write the list so for example $A+B=B+A$.
*We can define the removal of strings from the list with subtraction, so for example $A+B+AB +AA-AB =A+B+AA$.
*We will take powers of letters to mean multiple consecutive occurrences E.g. $A^2B^3C^2=AABBBCC$.
With these rules we can manipulate our regular expressions.
First, split up our regular expression $R$ into regular expressions $R_A$, $R_B$ and $R_C$ consisting of all of those valid strings ending with $A$, $B$ and $C$ respectively hence
$$R=\epsilon + R_A + R_B + R_C$$
Then any string in our regular expression $R_A$ can only be expressed as a valid string that ends with something other than $A$ followed by either $1$,$2$, $3$ or $4$ consecutive $A$s we may write this
$$R_A=(R-R_A)A+ (R-R_A)AA+(R-R_A)AAA+(R-R_A)AAAA=(R-R_A)(A+A^2+A^3+A^4)$$
similarly
$$R_B=(R-R_B)(B+B^2+B^3+B^4)$$
$$R_C=(R-R_C)(C+C^2+C^3+C^4)$$
As it stands this is the most that can be achieved with regular expressions but we must recognise what it is that we want to achieve: we only want to count the valid strings of length $k$.
In other words we don't care about the order of the letters or about distinguishing them, we may as well label them all $x$.
Now we know that order within a string doesn't matter we can treat our regular expressions exactly like equations in our variable $x$ so that
$$R=\epsilon + R_A + R_B + R_C \quad \text{becomes}\quad r(x)=1+r_a(x)+r_b(x)+r_c(x)\tag{1}\label{1}$$
and
$$R_A=(R-R_A)(A+A^2+A^3+A^4)\quad \text{becomes}\quad r_a(x)=(r(x)-r_a(x))(x+x^2+x^3+x^4)$$
$$R_B=(R-R_B)(B+B^2+B^3+B^4)\quad \text{becomes}\quad r_b(x)=(r(x)-r_b(x))(x+x^2+x^3+x^4)$$
$$R_C=(R-R_C)(C+C^2+C^3+C^4)\quad \text{becomes}\quad r_c(x)=(r(x)-r_c(x))(x+x^2+x^3+x^4)$$
these may be rearranged to give
$$r_a(x)=r_b(x)=r_c(x)=\frac{r(x)(x+x^2+x^3+x^4)}{1+x+x^2+x^3+x^4}$$
which can be substituted into $\eqref{1}$ to give
$$r(x)=1+3\frac{r(x)(x+x^2+x^3+x^4)}{1+x+x^2+x^3}$$
$$\implies r(x)=\frac{1+x+x^2+x^3+x^4}{1-2(x+x^2+x^3+x^4)}\tag{2}\label{2}$$
this is our generating function for valid strings and has the expanded form
$$r(x)=\sum_{k=0}^{\infty}r_kx^k$$
As you can see the exponent on the $x$ enumerates the valid string lengths and $r_k$ is the number of valid strings length $k$.
You have asked for the coefficients $r_8$, $r_9$ and $r_{10}$.
We can either use a computer algebra system such as sage to expand the generating function for us or we can manipulate it into a form that will give us a recurrence for $r_k$.
Using sage we simply input
r(x)=(1+x+x^2+x^3+x^4)/(1-2*(x+x^2+x^3+x^4))
show(taylor(r(x),(x,0),10))
which returns the Taylor expansion about $x=0$ of $r(x)$ up to the $x^{10}$ term:
$$55896 \, x^{10} + 18792 \, x^{9} + 6318 \, x^{8} + 2124 \, x^{7} + 714 \, x^{6} + 240 \, x^{5} + 81 \, x^{4} + 27 \, x^{3} + 9 \, x^{2} + 3 \, x + 1$$
As you can see
$$r_8=6318$$
$$r_9=18\,792$$
$$r_{10}=55\,896$$
Or, the recurrence which I mentioned above can be found by rearranging $\eqref{2}$
$$r(x)= 1+x+x^2+x^3 +x^4 + 2(x+x^2+x^3+x^4)r(x)$$
$$\implies\sum_{k=0}^{\infty}r_kx^k=1+x+x^2+x^3+x^4+\sum_{k=0}^{\infty}2(r_{k-1}+r_{k-2}+r_{k-3}+r_{k-4})x^k$$
$$\implies r_k=2(r_{k-1}+r_{k-2}+r_{k-3}+r_{k-4})+\sum_{j=0}^{4}\delta_{j,k}$$
where
$$\delta_{j,k}=\begin{cases}1 & j=k\\0 & j\ne k\end{cases}$$
A lovely tool for listing terms using a recurrence is Microsoft Excel because we can easily use formulae to relate cells in a column to previous cells in said column.
A third route is also open to us: we may factorise the denominator of $r(x)$ then express as partial fractions
$$\begin{align}r(x)&=\frac{1+x+x^2+x^3+x^4}{(1-\phi_1x)(1-\phi_2x)(1-\phi_3x)(1-\phi_4x)}\\&=(1+x+x^2+x^3+x^4)\left(\frac{A_1}{1-\phi_1x}+\frac{A_2}{1-\phi_2x}+\frac{A_3}{1-\phi_3x}+\frac{A_4}{1-\phi_4x}\right)\end{align}$$
which will eventually yield a closed for for $r_k$, however this requires us to solve a quartic which is, in general, a messy business so I will leave that to the reader.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \sqrt{ab}$.
Because $\gcd(a,b) = 1$, there exists integers $k$ and $l$ such that $ak + bl = 1$. This means that $\sqrt{a}(k\sqrt{}) + \sqrt{b}(l\sqrt{b}) = 1$.
Hence $\sqrt{a}$ and $\sqrt{b}$ are both positive integers and we can set $\sqrt{a} = c$ for some arbitrary integer $c$ and $\sqrt{b} = d$ for some arbitrary integer $d$. Therefore, $a = c^2$ and $b = d^2$.
| Consider the prime factorization of $ab=n^2$. Each prime factor appears an even number of times because $n^2$ is a square (each prime factor of $n$ is repeated twice in the factorization of $n^2$).
The condition $\gcd(a,b)=1$ means that $a$ and $b$ have no common prime factors.
That means each prime factor of $a$ itself appears an even number of times; ditto for $b$. This in turn means that $a$ is a square and $b$ is a square.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:
Attempt 1:
Write out the terms of $b_n$:
$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$
Adding up the terms you get:
$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$
But a dead end here.
Attempt 2:
For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$
Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges.
For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end.
| Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then
$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$
Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have
$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$
Note here that
*
*$\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.
*$\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.
So, taking limsup as $n\to\infty$ yields
$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$
Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$ I need to find out whether the following improper integral converges:
$$\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$$
I tried two comparison tests that failed, any ideas?
| Note that we can write
$$\begin{align}
\int_0^{N\pi/5}\frac{1}{1+x^2\sin^2(5x)}\,dx&=\frac15\int_0^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&=\frac15\int_0^{\pi/2}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\sum_{n=1}^{N-1}\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\int_{(N-1/2)\pi}^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\tag 1
\end{align}$$
The first on the right-hand side of $(1)$ creates no issue and the third integral is easily seen to converge as $N\to \infty$.
We focus attention on the second integral on the right-hand side of $(1)$ and write
$$\begin{align}
\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx&\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2\sin^2(x)}\,dx\\\\
&\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2 x^2}\,dx\\\\
&=\frac{20}{(2n+1)\pi}\arctan((n+1/2)\pi^2/10)\\\\
&\ge \frac{20}{(2n+1)\pi}\left(\frac{\pi}{2}-\frac{1}{(n+1/2)\pi^2/5}\right)
\end{align}$$
Inasmuch as the harmonic series diverges, the integral of interest diverges likewise.
| {
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Exact period of simple pendulum. Edit: Here is in depth derivation.
Suppose the pendulum is composed of a string of length $L$ and has a point mass of mass $m$ at the end of the string. Say we incline it at an angle $\theta_0 \in (0,\pi)$ counterclockwise from horizontal (counterclockwise counted positive and clockwise counted negative). Let the mass at the vertical position posses $0$ potential energy. Then it posses $mg(L-L\cos \theta_0)$ amount of Potential Energy at the signed angle of $\theta_0$. At any angle the mass posses a Kinetic energy of $\frac{1}{2}mv^2=\frac{1}{2}m \left(L\frac{d\theta}{dt}\right)^2$ and a potential energy of $mg(L-L\cos \theta)$. By conservation of mechanical energy,
$$\frac{1}{2}m\left(L\frac{d\theta}{dt}\right)^2+mgL(1-\cos \theta)=mgL(1-\cos \theta_0)$$
As the pendulum counterclockwise from an angle of $-\theta_0$ to $\theta_0$, $\frac{d\theta}{dt} \geq 0$ so,
$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}(\cos \theta-\cos \theta_0)}$$
This motion is half the cycle (to show this look at the equation counterclockwise motion from $\theta_0$ to $-\theta_0$), so it takes half the period to occur. From which we find,
$$T=2\sqrt{\frac{L}{2g}} \int_{-\theta_0}^{\theta_0} \frac{1}{\sqrt{\cos \theta-\cos \theta_0}} d\theta$$
As the integrand is even we get,
$$=4\sqrt{\frac{L}{2g}}\int_{0}^{\theta_0} \frac{1}{\sqrt{\cos \theta-\cos \theta_0}} d\theta$$
Now we make the substitution $\sin x=\dfrac{\sin \frac{\theta}{2}}{\sin \frac{\theta_0}{2}}$. $x \in [0,\frac{\pi}{2}]$ and $\theta \in [0,\theta_0]$ correspond together, so let $x \in \left[0,\frac{\pi}{2}\right]$. Then note the identities,
$$1-2\sin^2 \left(\frac{\theta}{2}\right)=\cos \theta$$
$$1-2\sin^2 \left(\frac{\theta_0}{2} \right)=\cos \theta_0$$
Give,
$$\sqrt{\cos \theta-\cos \theta_0}=\sqrt{2} \sin \frac{\theta_0}{2} \cos x$$
(If we let $\theta_0 \in (0,\pi]$ As $\cos x$ is nonnegative for $x \in \left[0,\frac{\pi}{2} \right]$). Also note the identity,
$$\cos \frac {\theta}{2}=\sqrt{1-\sin^2 \frac{\theta}{2}}$$
For $0 \leq \theta \leq \theta_0 \leq \pi$. The identities we found together convert the earlier expression we found for the period into,
$$T=4 \sqrt{\frac{L}{2g}} \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2 \sin^2 x}} dx$$
$$=4\sqrt{\frac{L}{g}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-k^2 \sin^2 x}} dx$$
Where $k=\sin (\frac{\theta_0}{2})$.
We also have the binomial series expansion,
$$(1-k^2\sin^2 x)^{-\frac{1}{2}}=\sum_{n=0}^{ \infty} {-\frac{1}{2} \choose n} (-1)^n k^{2n} \sin^{2n} x$$
A standard exercise in many books is to show for integers $n \geq 2$,
$$\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=\frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x dx$$
Hence showing for $n \geq 1$,
$$\int_{0}^{\frac{\pi}{2}} \sin^{2n} x dx=\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \frac{\pi}{2}$$
Using this gives,
$$T=2\pi \sqrt{\frac{L}{g}}\left(1+ \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} {-\frac{1}{2} \choose n} (-1)^n k^{2n} \right)$$
Also a famous result for $n \geq 1$ is,
$$(-1)^n {-\frac{1}{2} \choose n}=\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}$$
So the exact period is,
$$T=2\pi \sqrt{\frac{L}{g}}\left(1+ \sum_{n=1}^{\infty}\left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \right)^2 k^{2n} \right)$$
As claimed.
The equation that models a simple pendulum is,
$$-g\sin \theta=L \theta''$$
Where the derivative above is a time derivative. I read in my book that the period of of the pendulum starting from an angle of $\theta(0)=\theta_0$ is exactly,
$$T=2\pi\sqrt{\frac{L}{g}}\left[1+\left(\frac{1}{2}\right)^2 \sin^2 \left(\frac{\theta_0}{2}\right)+\left(\frac{1 \cdot 3}{2 \cdot 4} \right)^2 \sin^4 \left(\frac{\theta_0}{2}\right)+\cdots \right]$$
My question is how to get it?
Here's something I tried use $\sin (\theta)=\theta-\frac{\theta^3}{3}+\cdots$ to come up with a solution though I see if I include anything other than one other term I am lost. With one term I can get the first term in the period.
Here's another thing I tried to do, take Laplace transforms on both sides to get:
$$-g \int_{0}^{\infty} e^{-st} \sin (\theta(t))dt=L(s^2F(s)-s\theta(0)-\theta'(0))$$
But again it seems like there is no hope to solve that integral.
| Hint: Write $-g\sin \theta=L \theta''=L\dfrac{d^2\theta}{dt^2}$ and
$$-g\sin \theta\dfrac{d\theta}{dt}=L\dfrac{d^2\theta}{dt^2}\dfrac{d\theta}{dt}$$
$$-2g\sin \theta d\theta=L\times2\dfrac{d^2\theta}{dt^2}\dfrac{d\theta}{dt}dt=L\Big[\left(\dfrac{d\theta}{dt}\right)^2\Big]'dt$$
after integration
$$2g\cos\theta=L\left(\dfrac{d\theta}{dt}\right)^2+C$$
| {
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Write the function $\frac{1}{(z+1)(3-z)}$ as a Laurent series. $$f(z)=\frac{1}{(z+1)(3-z)}=\frac{1}{4z+4} + \frac{1}{12-4z}$$
$$\frac{1}{4z+4}=\frac{1}{4z}\frac{1}{1-\frac{-1}{z}}=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k$$
$$\frac{1}{12-4z}=\frac{1}{12}\frac{1}{1-\frac{z}{3}}=\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$
$$f(z)=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$
I can rewrite that as $$f(z)=\frac{1}{4z}\sum_{k=-\infty}^{0} (-1)^k z^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k z^k$$.
I need to move the $\frac{1}{4z}$ and $\frac{1}{12}$ into the sums but finding a series that will converge to each, but I have no idea what to use for either. Any suggestions? Am I taking a wrong approach or is there an obvious series to use for this?
Edit: The center is $0$ and the region in $1 \le |z| \le 3$.
I think I can use a geometric sequence to say $\frac{1}{4z}=\sum_{k=0}^{\infty}\frac{1}{2}(\frac{1}{2})^{k-1}\frac{z}{k}$ and $\frac{1}{12}=\sum_{k=0}^{\infty}\frac{1}{36}(\frac{2}{9})^{k-1}\frac{z}{k}$. I'm pretty sure that's true, but it seems like it makes the whole thing a complicated mess.
| I can write the expression as:
$$\frac{1}{(z+1)(-z+3)} = \frac{-1}{(z+1)(z-3)}$$
Use partial fractions:
$$\frac{-1}{(z+1)(z-3)} = \frac{A}{z+1}+\frac{B}{z-3}$$
$$-1 = A(z-3)+B(z+1)$$
If z=3, then
$$-1 = B(3+1)$$
$$-1 = B(4)$$
$$B=\frac{-1}{4}$$
If z=1, then
$$-1 = A(1-3)$$
$$-1 = A(-2)$$
$$1 = A(2)$$
$$A = 1/2$$
So, we have:
$$\frac{-1}{(z+1)(z-3)} = (\frac{1}{2})\frac{1}{z+1}+(\frac{-1}{4})\frac{1}{z-3}$$
The previous expression is a Laurent Series. Because the expression has no expressions of the form
$$\frac{1}{z^{n}}$$
The Laurent Series behaves as a Taylor Series.
If we suppose that in z=0 the expression is not analytic, then we can write it as
$$\frac{-1}{(z+1)(z-3)} = (\frac{1}{2})\frac{1}{z(1+\frac{1}{z})}+(\frac{-1}{4})\frac{1}{z(1-\frac{3}{z})}$$
| {
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Proving that $3^n+7^n+2$ is divisible by 12 for all $n\in\mathbb{N}$. Can someone help me prove this? :( I have tried it multiple times but still cannot get to the answer.
Prove by mathematical induction for $n$ an element of all positive integers that $3^n+7^n+2$ is divisible by 12.
| A general rule of thumb for induction is to try to get
$3^{n+1} + 7^{n+1}+2 = somethingtodo with(3^n + 7^n+2) = something to do with(multiple of 12) = multiple of 12$
So $3^{n+1} + 7^{n+1} +2 = $
$3*3^n + 7*7^n + 2=$
$3*3^n + 3*7n + 4*7^n + 2= $
$3(3^n + 7^n) + 4*7^n + 2 = $
$3(3^n + 7^n + 2) - 6 + 4*7^n + 2 = $
$3(multiple of 12) + 4*7^n -4 $
so now it's a matter of proving $4*7^n -4$ is a multiple of $12$.
$3(multiple of 12) + 4*7^n -4 = $
$3(multiple of 12) + 4(7^n - 1)=$
$3(multiple of 12) + 4*(7-1)(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + 4*6(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + 24*(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + (multiple of 12)*(7^{n-1} + .... + 7 + 1)=$
$multiple of 12$
....
So to formally put this together.
Base case:
$n = 1$. $3^1 + 7^1 + 2 =12$ is a multiple of $12$.
Inductive case:
Assume $3^n + 7^n + 2 = 12K$ is a multiple of $12$..
Then $3^{n+1}+7^{n+1} + 2 =$
$3*3^n + 3*7^n+ 6 + 4*7^n-4 =$
$3(3^n + 7^n + 2) + 4(7^n - 1)=$
$3(12K) + 4(7-1)(7^{n-1} + ... + 1)$
Let $M = (7^{n-1} + ... + 1)$. It's worth noting that if $n = 1$ then $7^{n-1} + ..... + 1 = 1$. It is important in a proof by induction to not make any assumptions that are not provable for the base $n=1$ case. In this case that $7^{n-1} + .... +1$ actually exists.)
$3(12K) + 4(7-1)(7^{n-1} + ... + 1)=$
$3(12K) + 4*6*M = $
$12[3K + 2M]$
is a multiple of $12$.
Conclusion: $3^n + 7^n + 2$ is a multiple of $12$ for all natural $n$.
| {
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Show this matrix is invertible Consider the matrix
\begin{equation}
S_n =
\begin{bmatrix}
\frac{1}{2!} & \frac{1}{3!} & \cdots & \frac{1}{(n+1)!} \\
\frac{1}{3!} & \frac{1}{4!} & \cdots & \frac{1}{(n+2)!} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{(n+1)!} & \frac{1}{(n+2)!} & \cdots & \frac{1}{2n!} \\
\end{bmatrix}.
\end{equation}
I am interested to know if $S_n$ is invertible for all $n$. I do not need the inverse explicitly, knowing it exists is sufficient (ie. non-zero determinant for all $n$ is enough).
If this matrix has a special name, I would appreciate if someone could bring it to my attention.
| For each $k\le n$, define a $k\times k$ matrix as follows:
$$
A_k=\begin{bmatrix}
\dfrac{(n+1)!}{(n-k+2)!} &\dfrac{(n+2)!}{(n-k+3)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+1)!}\\
\dfrac{(n+1)!}{(n-k+3)!} &\dfrac{(n+2)!}{(n-k+4)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+2)!}\\
\vdots &\vdots &\ddots &\vdots &\vdots\\
\dfrac{(n+1)!}{(n-1)!} &\dfrac{(n+2)!}{n!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+k-2)!}\\
\dfrac{(n+1)!}{n!} &\dfrac{(n+2)!}{(n+1)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+k-1)!}\\
1&1&\cdots&\cdots&1
\end{bmatrix}
$$
Since $A_n=S_n\operatorname{diag}\left((n+1)!,(n+2)!,\ldots,(2n)!\right)$, it suffices to show that $A_k$ is invertible for each $1\le k\le n$. We shall prove this by mathematical induction. The base case $k=1$ (with $A_1=[1]$) is trivial. For the inductive step, subtract every $j$-th column (with $j\ge2$) by the column on its left to obtain
$$
\left[\begin{array}{c|cccc}
\dfrac{(n+1)!}{(n-k+2)!} &\dfrac{(n+1)!}{(n-k+3)!}(k-1) &\cdots &\cdots &
\dfrac{(n+k-1)!}{(n+1)!}(k-1)\\
\dfrac{(n+1)!}{(n-k+3)!} & \dfrac{(n+1)!}{(n-k+4)!}(k-2) &\cdots &\cdots &\dfrac{(n+k-1)!}{(n+2)!}(k-2)\\
\vdots &\vdots &\ddots &\vdots &\vdots\\
\dfrac{(n+1)!}{(n-1)!} &\dfrac{(n+1)!}{n!}(2) &\cdots &\cdots &\dfrac{(n+k-1)!}{(n+k-2)!}(2)\\
n+1 &1 &\cdots &\cdots &1\\
\hline
1&0&\cdots&\cdots&0
\end{array}\right].
$$
Now we are done because the top right sub-block of this matrix is $\operatorname{diag}(k-1,k-2,\ldots,1)A_{k-1}$. This proof also shows that $|\det S_n|=\prod_{k=1}^n\frac{(k-1)!}{(n+k)!}$.
| {
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Infinite series of $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$ Find Value of $$S=\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$$
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
But here i cannot express above sum in form of $$\lim_{n \to \infty} \frac{1}{n} \times \sum_{r=1}^{n}f(\frac{r}{n})$$
| As you already wrote $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$,
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
When you take $n^2$ common out from the square root then your eq. becomes:
$$S=\frac{1}{n}\lim {n \to \infty}[1+ \frac{1}{\sqrt(1+ \frac{1}{n^2})}+......+\frac{1}{\sqrt(1+\frac{2}{n})^2}]$$
Can you solve from here?
| {
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Three quadratic equations have positive roots The three quadratic equations $ax^2-2bx+c=0$, $bx^2-2cx+a=0$ and $cx^2-2ax+b=0$ has both roots positive. Then which of the following is/are true
A) $a^2=bc$
B) $b^2=ac$
C) $c^2=ab$
D) $a=b=c$
I assumed $a>0$ so parabola is open upwards and hence it should definitely cut positive $Y $axis as it has both roots positive and hence $c \gt 0$.
Since $c \gt 0$ the third parabola is open upwards and so $b \gt 0$ and finally with same reasoning $a \gt 0$. Now each Discriminant is Non negative so
$$b^2 \ge ac$$
$$c^2 \ge ab$$
$$a^2 \ge bc$$
adding all
$$a^2+b^2+c^2-ab-bc-ca \ge 0$$ $\implies$
$$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0$$
if we take equality above i will get $a=b=c$ so fourth option is valid. But how to check whether other options a is/are true
| Since the equations have two roots, the leading coefficients must be non-$0\,$, so $abc \ne 0$. Assume WLOG that $a \gt 0\,$ (otherwise use the same argument for $a,b,c \mapsto -a,-b,-c\,$). Since both roots of each equation are positive, it follows by Vieta's relations that $b \gt 0 , c \gt 0\,$ as well, so in the end all three of $a,b,c$ are strictly positive numbers. Now, picking up from where the OP left off:
$$b^2 \ge ac$$
$$c^2 \ge ab$$
$$a^2 \ge bc$$
Those are all positive numbers, then multiplying together gives $\,a^2 b^2 c^2 \ge a^2 b^2 c^2\,$. But the latter is in fact an equality, which means that each of the previous three inequalities must be an equality as well, therefore $\,a^2=bc, b^2=ca, c^2=ab\,$. Furthermore, $a^2=bc=\sqrt{ac} \cdot c \implies a=c\,$, so in the end $a=b=c$.
| {
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Value of integral under certain condition If f is a real valued function and satisfies $f (x)+f (1-\frac {1}{x})=\arctan (x) $ for all $x $ except $0$ then find the least integer greater than or equal to $N=\int _0 ^1 f (x)dx $ . I havent done any progress . What I tried was putting $x $ as $\frac {x-1}{x} $ to get $f(\frac {x-1}{x})+f (\frac {1}{1-x})=\arctan (\frac {x-1}{x}) $. I think we have to find some relation between like $f (x)+f (1-x)=c $ where c is some constant and use $N=\int _0 ^1 \frac {1}{2}c $ to get the required value.
| Using your hints, we have, substituting $x$ with $\frac{x-1}{x}$ and then with $\frac{1}{1-x}$, $$ f(x) + f\left( 1 - \frac{1}{x}\right) = \arctan(x)$$ $$ f\left( \frac{x-1}{x}\right) + f\left(\frac{1}{1-x} \right) = \arctan\left(\frac{x-1}{x}\right)$$ and $$ f(x) + f \left( \frac{1}{1-x} \right) = \arctan\left( \frac{1}{1-x}\right) $$
We would like to isolate $f(x)$. Hence we add the first and third equations and subtracting the second gives : $$ 2f(x) = \arctan(x) + \arctan\left( \frac{1}{1-x}\right) - \arctan\left( \frac{x-1}{x}\right) $$
Now we want to use the relation $\arctan(x) + \arctan\left( \frac{1}{x}\right) = \pi/2$. To do so just compute \begin{align*} 2(f(x) + f(1-x)) & = (\arctan(x) + \arctan(1/x))\\ & + \left(\arctan(1-x) + \arctan \left( \frac{1}{1-x}\right) \right) \\ & - \left( \arctan\left( \frac{x-1}{x} \right) + \arctan \left( \frac{x}{x-1} \right)\right)\\ & = \frac{3\pi}{2} \end{align*}
Hence $$ 4 \int_0^1 f(x)dx = \int_0^1 2(f(x) + f(1-x))dx = \frac{3\pi}{2}$$ Finally $$ \int_0^1f(x)dx = \frac{3 \pi}{8}$$
| {
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$\lim_{n\rightarrow \infty } (\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{(n+1)^2}\cdot n^{n^2}})$ The given question is asking me to test the series for convergence or divergence.
$$\sum_{n=1}^{\infty } (\frac{n}{n+1})^{n^2}$$
My attempt, I'm using ratio test and I'm stuck at the limits part.
I don't know how to evaluate $\lim_{n\rightarrow \infty } (\frac{(n+1)^{2n^2+2n+1}}{(n+2)^{(n+1)^2}\cdot n^{n^2}})$
The answer given which is $\frac{1}{e}<1$, so it converges. Hope someone can provide the solution for the limit part, Thanks in advance.
| If you want to deal with the limit you posted, consider $$u_n= \left(\frac{n}{n+1}\right)^{n^2}\implies \log(u_n)=n^2 \log\left(\frac{n}{n+1}\right)=n^2 \log\left(1-\frac{1}{n+1}\right)$$ Now, use Taylor expansion and get $$\log(u_n)=-n+\frac{1}{2}-\frac{1}{3 n}+\frac{1}{4
n^2}+O\left(\frac{1}{n^3}\right)$$ Similarly $$u_{n+1}= \left(\frac{n+1}{n+2}\right)^{(n+1)^2}\implies \log(u_{n+1})=n^2 \left(1+\frac 1n\right)^2\log\left(1-\frac{1}{n+2}\right)$$ and Taylor again to get $$\log(u_{n+1})=-n-\frac{1}{2}-\frac{1}{3 n}+\frac{7}{12
n^2}+O\left(\frac{1}{n^3}\right)$$ Combining the above $$\log(u_{n+1})-\log(u_{n})=\log \left(\frac{u_{n+1}}{u_n}\right)=-1+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Taylor again $$\frac{u_{n+1}}{u_n}=\exp\left({\log \left(\frac{u_{n+1}}{u_n}\right) }\right)=\frac{1}{e}+\frac{1}{3 e n^2}+O\left(\frac{1}{n^3}\right)$$
Edit
Making the problem more general with $$v_n= \left(\frac{n}{n+a}\right)^{(n+b)^2}$$ and using the same steps, we should find $$\frac{v_{n+1}}{v_n}=e^{-a}+\frac{a e^{-a} \left(a^2-3 a b+3 b^2\right)}{3
n^2}+O\left(\frac{1}{n^3}\right)$$
| {
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Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated
| Here is how I would solve it, First, I would eliminate x.
\begin{eqnarray*}
x &=& 3 - y - z \\
f(x,y,z) &=& (3 - y - z)^3 + y^3 + z^3
\end{eqnarray*}
then I would apply the second derivative test. More information about the second derivative test can be found at the following URL:
http://faculty.csuci.edu/brian.sittinger/2nd_DerivTest.pdf
First, we find the partial derivatives:
\begin{eqnarray*}
f_y &=& -3(3 - y - z)^2 + 3y^2
f_z &=& -3(3 - y - z)^2 + 3z^2
f_yz &=& 6(3 - y - z) + 6y \\
f_yy &=& -6(3 - y - z) + 6y \\
f_zz &=& -6(3 - y - z) + 6z
\end{eqnarray*}
Now, we find the critical points:
\begin{eqnarray*}
-3(3 - y - z)^2 + 3y^2 &=& 0 \\
-3(3 - y - z)^2 + 3z^2 &=& 0 \\
3y^2 - 3z^2 &=& 0 \\
y^2 &=& z^2 \\
\end{eqnarray*}
Therefore, $y = z = 0$ is a critical point. However, it is a minimum not a maximum. We also have an infinity number of critical points, so I am not
sure how to proceed.
Bob
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Area between $y^2=2ax$ and $x^2=2ay$ inside $x^2+y^2\le3a^2$ I need to find the area between $y^2=2ax$ and $x^2=2ay$ inside the circle $x^2+y^2\le3a^2$. I know it's an integral but I can't seem to find the right one.
| The entire figure scales proportionally to $a$, so we may set $a=1$ and multiply the area thus obtained by $a^2$ at the end.
The desired area when $a=1$ is twice the area of the blue region $B$ above plus the area of the circular sector $C$ bounded by the black lines $y=\sqrt2x$ and $x=\sqrt2y$. $B$ is bounded by $y=\frac x{\sqrt2}$ from above and $y=\frac{x^2}2$ from below, so
$$B=\int_0^{\sqrt2}\left(\frac1{\sqrt2}x-\frac12x^2\right)\,dx$$
$$=\left[\frac1{2\sqrt2}x^2-\frac16x^3\right]_0^{\sqrt2}=\frac{\sqrt2}6$$
The angle $\theta$ between the two black lines satisfies
$$\tan\theta=\frac{\sqrt2-1/\sqrt2}{1+\sqrt2\cdot1/\sqrt2}=\frac{\sqrt2}4$$
which implies $\cos\theta=\frac{2\sqrt2}3$ and (since the sector radius is $\sqrt3$)
$$C=\frac{r^2\theta}2=\frac32\cos^{-1}\frac{2\sqrt2}3$$
Finally, for a given $a$ the total area of the region in question is
$$(2B+C)a^2=\left(\frac{\sqrt2}3+\frac32\cos^{-1}\frac{2\sqrt2}3\right)a^2$$
$$=0.981159\dots×a^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
| Even shorter:
You know $\sin x = \cos^2 x$. Then
$\cos^8 x + 2 \cos^6 x + \cos^4 x = \sin^4 x + 2 \sin^3 x + \sin^2 x = (\sin^2 x +\sin x)^2 = 1^2 = 1$.
| {
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"question_score": "2",
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Prove if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$ My solution is:
Multiply $x > 3$ with $x$, yielding $x^{2} > 9$
Multiply $y < 2$ with $2$, yielding $2y < 4$
Thus, based on the above $2$ yielded inequalities, we can prove that if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$.
Is this a correct proofing steps?
| For a roundabout way to prove it, which is overkill in this case, but may prove useful in other cases, note that the blue terms are positive since $x -3\gt 0$ and $2-y \gt 0\,$, therefore:
$$
x^2-2y =\left((x-3)+3\right)^2 - 2\left(-(2-y)+2\right) = \color{blue}{(x-3)^2} + 6\color{blue}{(x-3)} + \color{red}{9} +2\color{blue}{(2-y)} - \color{red}{4} \gt \color{red}{5}
$$
| {
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Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\begin{cases}
x^2-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\
=\begin{cases}
\frac{1}{y}-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\=
\begin{cases}
\frac{1-y^3}{y} = 0 \\
-\\ \end{cases}
\\=
\begin{cases}
y^3 = 1 \\
-\\ \end{cases}
\\=
\begin{cases}
y = 1 \\
x =1\\ \end{cases} $$
And so $\sqrt{2i} = 1+i$, but my book states the solution is $\sqrt{2i} = 1+i$ and $\sqrt{2i} = -1-i$.
What did I forget?
| $2i=1+2i+i^2=(1+i)^2$. The roots are $\pm(1+i)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the domain of $\arcsin\left(\frac{3-x}{\sqrt{9-x^2}}\right)$
Find the domain of
$$\arcsin\left(\dfrac{3-x}{\sqrt{9-x^2}}\right)$$
This question is given in my book. The answer is as follows:
For the function to be defined, $-1\le3-x\le1$ and $9-x^2>0$. Solve these inequalities and take their common region.
$$2\le x\le4\tag1$$
$$-3 < x < 3\tag2$$
The answer is $2\le x<3$. My question is why have they not taken $-1\le\dfrac{3-x}{\sqrt{9-x^2}}\le 1$, in which case the answer would have been different. Please help.
| For $x\ne3,\dfrac{3-x}{\sqrt{9-x^2}}=\sqrt{\dfrac{3-x}{3+x}}$
Now we need $$-1\le\sqrt{\dfrac{3-x}{3+x}}\le1\iff0\le\dfrac{3-x}{3+x}\le1$$
If $\dfrac{3-x}{3+x}=0\implies x=3\ \ \ \ (1)$
$\dfrac{3-x}{3+x}>0\iff(3-x)(3+x)>0\iff(x-3)(x+3)<0\iff-3<x<3\ \ \ \ (2)$
$\dfrac{3-x}{3+x}\le1\iff0\ge\dfrac{3-x}{3+x}-1=\dfrac{3-x-(3+x)}{3+x}=-\dfrac{2x}{3+x}$
By $(2),x+3>0,$ so we need $-2x\le0\iff x\ge0\ \ \ \ (3)$
By $(1),(2),(3):0\le x<3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Completing the square of $x^2 - mx = 1$ is not giving me the right answer. This is my attempt
$$
\begin{align}
x^2 - mx &= 1 \\
x^2 - mx - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - \frac{4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2 - 4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) &= \frac{m^2 - 4}{4} \\
\left(x - \frac{m}{2}\right)^2 &= \frac{m^2 - 4}{4} \\
\sqrt{\left(x - \frac{m}{2}\right)^2} &= \sqrt{\frac{m^2 - 4}{4}} \\
x - \frac{m}{2} &= \pm \frac{\sqrt{m^2 - 4}}{\sqrt{4}} \\
x &= \frac{m}{2} \pm \frac{\sqrt{m^2 - 4}}{2} \\[20pt]
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 - 4}}{2} \\
x_1 &= \frac{m - \sqrt{m^2 - 4}}{2} \\[16pt]
x_2 &= \frac{m}{2} + \frac{\sqrt{m^2 - 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 - 4}}{2} \\
\end{align}
$$
However, the correct answer according to the text is:
$$
\begin{align}
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 + 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 + 4}}{2} \\
\end{align}
$$
Why $\sqrt{m^2 + 4}$ instead of $\sqrt{m^2 - 4}$ ???
| When you combined $$-\frac{m^2}{4} - \frac{4}{4}$$ into one fraction, you wrote it as $$-\frac{m^2-4}{4}.$$ You should have gotten $$-\frac{m^2+4}{4}$$ to make both terms appropriately negative.
| {
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Given $a+b+c+d=4$ To find Minimum value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$ Given $a+b+c+d=4$ where $a,b,c,d \in \mathbb{R^{+}}$
find Minimum value of $$S=\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$$
I have no clue to start...any hint?
| By AM-GM $$\sum_{cyc}\frac{a}{b^3+4}=\frac{a+b+c+d}{4}-\sum_{cyc}\left(\frac{a}{4}-\frac{a}{b^3+4}\right)=$$
$$=1-\sum_{cyc}\frac{b^3a}{4\left(\frac{b^3}{2}+\frac{b^3}{2}+4\right)}\geq1-\frac{b^3a}{4\cdot3\sqrt[3]{\left(\frac{b^3}{2}\right)^2\cdot4}}=$$
$$=1-\sum_{cyc}\frac{b^3a}{12b^2}=1-\frac{1}{12}(ab+bc+cd+da)=$$
$$=1-\frac{1}{12}(a+c)(b+d)\geq1-\frac{\left(\frac{a+b+c+d}{2}\right)^2}{12}=1-\frac{1}{3}=\frac{2}{3}.$$
For $(a,b,c,d)\rightarrow(2,2,0,0)$ we see that $S\rightarrow\frac{2}{3}$,which says that $\frac{2}{3}$ is infimum
and the minimum does not exist.
Done!
| {
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Indentify the plane defined by $Re(z\overline{z})=\frac{lm(\overline{z})}{2}$ I tried:
$$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 = \frac{-y}{2} \Leftrightarrow \\
-2x^2 = y$$
But my book states:
Circunference of center $(0,-\frac{1}{4})$ and radius $\frac{1}{4}$.
What went wrong?
| $$Re(z\overline{z})=\frac{lm(\overline{z})}{2} \Leftrightarrow \\
Re((x+yi)(x-yi))=\frac{lm(x-yi)}{2} \Leftrightarrow \\
Re(x^2+y^2)= \frac{lm(x-yi)}{2} \Leftrightarrow \\
x^2 + y^2 = \frac{-y}{2} \Leftrightarrow \\
x^2 + (y+\frac{1}{4})^2 = \frac{1}{16}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:
$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
Where $a\ge b$
Examples:
Where $F(1,1)= \sqrt{2}$, $F(2,1)=\sqrt{2-\sqrt{3}}$, $F(3,1)={1\over \sqrt{3}}(2-\sqrt{2})$, $...$
How do we evaluate the closed form for $(1)?$
| You may like this method. Let
$$I(b)=\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right) $$
and then
\begin{eqnarray}
I'(b)&=&\int_{0}^{\pi/2}\bigg(\frac1{a+b\sin^2(x)}+\frac1{a-b\sin^2(x)}\bigg) \mathrm dx\\
&=&\frac{\pi}{2\sqrt a}(\frac1{\sqrt{a+b}}+\frac{1}{\sqrt{a-b}}).
\end{eqnarray}
So
$$ I(b)=\frac{\pi}{2\sqrt{a}}\int_0^b(\frac1{\sqrt{a+t}}+\frac{1}{\sqrt{a-t}})dt=\pi\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt a}$$
and hence
$$ F(a,b)=\frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt a}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving System of Congruences Solve
$x^{} \equiv 5 \pmod {7}$
$x^{} \equiv 5 \pmod {9}$
$x^{7} \equiv 5 \pmod {32}$
This is what I have done so far:
$x=7k+5$
Substitute in x in the second congruence statement
$7k+5 \equiv 5 \pmod {9}$
Solving this results in
$k \equiv 0 \pmod {9}$
then
$k=9j$
$x=63j+5$
I'm lost from here on. I don't know how to deal with the $x^7$ congruence statement.
| By Euler's totient theorem, for all odd integer $x$:
$$x^{16} \equiv 1 \pmod{32}$$
Trivially, $x$ is odd, for if $x$ is even, then so would be $x^7$.
Then,
$$\begin{array}{rcll}
x^7 \equiv 5 & \pmod{32} \\
(x^7)^7 \equiv 13 & \pmod{32} \\
x^{49} \equiv 13 & \pmod{32} \\
x^{16\times3+1} \equiv 13 & \pmod{32} \\
(x^{16})^3 \cdot x \equiv 13 & \pmod{32} \\
(1)^3 \cdot x \equiv 13 & \pmod{32} \\
x \equiv 13 & \pmod{32} \\
\end{array}$$
You should be able to continue now.
| {
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Binomial expansion of negative exponents. Let's say I have to expand $(1+x)^{-1}$ using binomial expansion.
Using the theorem, I get:
$$(1+x)^{-1} = 1-x+x^2-x^3+x^4-x^5+x^6+....+{\infty}$$
Substituting $x$ for $1$, I get:
$$\frac{1}{2}= 1-1+1-1+1-1+1+....+{\infty}$$
A similar result arises with higher power of the exponent
For $(1+x)^{-2}$ we get:
$$(1+x)^{-2} = 1-2x+3x^2-4x^3+5x^4-6x^5+7x^6+....+{\infty}$$
Substituting $x$ for $1$, I get:
$$\frac{1}{4}= 1-2+3-4+5-6+7+....+{\infty}$$
How does this makes sense? Help please!
| The equality $$\sum_{n=0}^\infty x^n = \frac{1}{x+1}$$
is only true if $|x<1|$, so you cannot substitute $x=1$ into it and expect the result to work.
| {
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How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
How many solutions does the equation $x^2-y^2=5^{29}$ have, given that $x$ and $y$ are positive integers?
Someone told me that it has $(29+1)/2=15$ solutions. How come? Any other method to solve this?
| Note $x^2-y^2=(x+y)(x-y)$ so both of these must be factors of $5^{29}$, i.e. powers of $5$.
Let $x+y=5^a$ for some $0 \leq a \leq 29$. Then by the factorisation above, $x-y=5^{29-a}$ and this always has a solution in rational numbers, namely $x=\frac{5^a+5^{29-a}}{2}$, $y=\frac{5^a-5^{29-a}}{2}$.
We do require these to be integers, but this is always true as $5^k$ is always odd and odd+odd=even.
The only other restriction is that they are positive and whilst $x$ necessarily is, $y$ will only be positive when $29-a \leq a$, i.e. $a>14$. Hence any $15 \leq a \leq 29$ works and we have $15$ solutions.
Note that in general factorisations, we could have had $x+y=-5^a$, $x-y=-5^{29-a}$, but this is not allowed as both $x$ and $y$ are positive.
| {
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Inequality Question-Maximum Problem:
Let $a, b, c, d, e, f$ be nonnegative real numbers such that $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6$ and $ab + cd + ef = 3$. What is the maximum value of $a+b+c+d+e+f$?
How would I do this? Would we need to use Cauchy-Schwarz or any of those types of inequalities?
Edit: My question is different from the possible duplicate because the answers on that question are based on Lagrangian multipliers and mine is based on Cauchy-Schwarz
| Use Cauchy-Schwarz on the vectors
$$x = \begin{pmatrix}a+b \\
c+d \\
e+f\end{pmatrix}, \quad y = \begin{pmatrix}1 \\
1 \\
1\\ \end{pmatrix}$$
Then $$x\cdot x = (a+b)^2 + (c+d)^2 + (e+f)^2 \\= a^2 + b^2 + c^2 + d^2 + e^2 + f^2+ 2ab+2cd+2ef \\=6+2\times3 \\=12$$
Also, $y\cdot y = 3$, and $x \cdot y = a+b+c+d+e+f$, the quantity you're looking to maximise. Cauchy-Schwarz says $(x \cdot y)^2 \leq (x\cdot x)(y \cdot y)$, so
$$ (a+b+c+d+e+f)^2 \leq 12 \times 3 \\a+b+c+d+e+f \leq 6$$
Also, observe that $a=b=c=d=e=f=1$ achieves this value.
| {
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Sum of all possible values of $\gcd(a-1,a^2+a+1)$
Find the sum of all possible values of
$$\gcd(a-1,a^2+a+1)$$
where $a$ is a positive integer.
| HINT:
Let $a-1=b\iff a=?$
$$a^2+a+1=(b+1)^2+b+1+1=b^2+3b+3$$
$$(a-1,a^2+a+1)=(b,b^2+3b+3)=(b,3)$$
| {
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Finding the probability generating function of $P(X=n)$ You toss a fair coin repeatedly until heads appears three times.
Suppose the third head appears on the $X$-th toss. Find the
probability distribution of $X$, that is find a formula for $F_{X}(n) =
P(X = n)$. Hence find a formula for the generating function
$P_{X}(s) = E(s^
X)$ of $X$. Using the generating function, or
otherwise, find $E(X), E(X^2
)$ and $Var(X)$.
I want to find the Probability Generating Function
I find that the probability $X = n$ is :
$$P(X=n) = \frac{1}{2^n}\binom{n-1}{2}$$
Following that :
$$P_{X}(s) = \sum_{n=3}^\infty \frac{1}{2^n}\binom{n-1}{2}s^n =\frac{s^3}{2^3} \sum_{m = 0}^\infty \binom{m+2}{2}\frac{s^m}{2^m}$$
Where : $\,\,\,\,\,\,\,$ $ n = m+3$
So I would believe that this is a geometric progression where $a = 1\,\,\,r = \frac{s}{2}$
So I got $$= \frac{s^3}{2^3}(\frac{1}{1-\frac{s}{2}})$$
However the solution gives :
$$= \frac{s^3}{2^3}(\frac{1}{(1-\frac{s}{2})^3})$$
Can someone explain why I am wrong and the general approach to this solution.
| The problem is that the series is not geometric, but we could try to turn it into a geometric:
$$P_{X}(s) = \sum_{n=3}^\infty \frac{1}{2^n}\binom{n-1}{2}s^n =\frac{s^3}{2^3} \sum_{m = 0}^\infty \binom{m+2}{2}\frac{s^m}{2^m}$$
$$= a\sum_{m = 0}^\infty (m+2)(m+1)\frac{s^m}{2^m},\text{ where }a=\frac{s^3}{2^4}$$
Let $S_n=\sum_{m = 0}^n (m+2)(m+1)\frac{s^m}{2^m} $
$$X_n=S_{n}-\frac{s}{2}S_{n}=2-(n+2)(n+1)\frac{s ^{n+1}}{2^{n+1}}+2\sum_{m = 1}^{n} (m+1)\frac{s^{m}}{2^{m}} $$
$$Y_n=X_{n}-\frac{s}{2}X_n=(1-\frac{s}{2})^2S_n$$
$$=(1-\frac{s}{2})\bigg(2-(n+2)(n+1)\frac{s ^{n+1}}{2^{n+1}}\bigg)$$$$+2s-2(n+1)\frac{s^{n+1}}{2^{n+1}}+2\sum_{m=2}^{n}\frac{s^m }{2^m}$$
Now you get a geometric one, and you could solve it, and then solve back to $S_n$
Or, since now you only want the $n\rightarrow \infty$, take the limit on the above one, and you will get it. You could also see clearly why you have extra $(1-\frac{s}{2})^2$ in your denominator.
So to summarize the series $S_n$ you get is NOT geometric series, but the series of $(1-\frac{s}{2})^2S_n$ is a geometric series, and you could apply the geometric summation only to $(1-\frac{s}{2})^2S_n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding Singular Value of an SVD Say you're given the following SVD:
$B=
\begin{bmatrix}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & d \\
1 & e\\
\end{bmatrix}
\begin{bmatrix}
9 & 0 & 0 \\
0 & f & g \\
\end{bmatrix}
\begin{bmatrix}
-2/3 & -1/3 & a \\
1/3 & 2/3 & b \\
-2/3 & 2/3 & c \\
\end{bmatrix}^T
$
How does one find the values for $f$ and $g$. I've found the values for $a,b,c,d,e$ simply because they are orthogonal matrices which rotate. So you can find the corresponding perpendicular vectors and use the determinant to find out which, when filled in, make them rotating matrices.
However, I'm having trouble figuring out the values for $f, g$. I suppose $g = 0$ because it's not in the diagonal of $\Sigma$, thus it has to be $0$?
Also I've gotten as tip to use $BB^T$ to calculate the eigenvalues and singular values, however I'm still having trouble understand how to approach this.
| This is a nice problem to probe understanding of the singular value decomposition. My upvote goes to @Roland for his succinct answer. The following elaboration is to help other readers who may need help with other parts.
Problem
Input matrix:
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\end{array}
\right]
\in \mathbb{C}^{2\times 3}_{2}
$$
Complete the singular value decomposition
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\end{array}
\right]
=
% U
\left[
\begin{array}{rrr}
0 & d \\
1 & d \\
\end{array}
\right]
% S
\left[
\begin{array}{ccc}
9 & 0 & 0 \\
0 & f & g \\
\end{array}
\right]
% V
\frac{1}{3}
\left[
\begin{array}{rrr}
-2 & 1 & 3a \\
1 & -2 & 3b \\
2 & -2 & 3c
\end{array}
\right]^{*}
%
$$
Observations: the matrix has full row rank, and a column rank defect of 1. This is an underdetermined system.
Tools
The Fundamental Theorem of Linear Algebra
A matrix $\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$ induces for fundamental subspaces:
$$
\begin{align}
%
\mathbf{C}^{n} &=
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
\mathbf{C}^{m} &=
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$
Because the matrix $\mathbf{A}$ has full row rank, we have a trivial null space:
$$
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} = \mathbf{0}
$$
For this matrix the subspace decomposition takes the form
$$
\begin{align}
%
\mathbf{C}^{3} &=
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
\mathbf{C}^{2} &=
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)}
%
\end{align}
$$
Singular value decomposition
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V*
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right] \\
%
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cccc|cc}
\sigma_{1} & 0 & \dots & & & \dots & 0 \\
0 & \sigma_{2} \\
\vdots && \ddots \\
& & & \sigma_{\rho} \\\hline
& & & & 0 & \\
\vdots &&&&&\ddots \\
0 & & & & & & 0 \\
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right] \\
%
& =
% U
\left[ \begin{array}{cccccccc}
\color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{v_{1}^{*}} \\
\vdots \\
\color{blue}{v_{\rho}^{*}} \\
\color{red}{v_{\rho+1}^{*}} \\
\vdots \\
\color{red}{v_{n}^{*}}
\end{array} \right]
%
\end{align}
$$
The $\rho$ singular values are ordered and satisfy
$$
\sigma_{1} \ge \sigma_{2} \ge \dots \sigma_{1} > 0
$$
The column vectors are orthonormal basis vectors:
$$
\begin{align}
% R A
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} &=
\text{span} \left\{
\color{blue}{u_{1}}, \dots , \color{blue}{u_{\rho}}
\right\} \\
% R A*
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
\color{blue}{v_{1}}, \dots , \color{blue}{v_{\rho}}
\right\} \\
% N A*
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
\color{red}{u_{\rho+1}}, \dots , \color{red}{u_{m}}
\right\} \\
% N A
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} &=
\text{span} \left\{
\color{red}{v_{\rho+1}}, \dots , \color{red}{v_{n}}
\right\} \\
%
\end{align}
$$
Resolve constants
The constants are listed in alphabetical order. But the fastest way to solve the problem, after a careful first reading, is $(a,b,c)$, then $(f,g)$, then $(d,e)$.
a, b, c
Just from dimensional considerations we know
$$
\begin{align}
% R A*
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
\color{blue}{v_{1}}, \color{blue}{v_{2}}
\right\} \\
% N A
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} &=
\text{span} \left\{
\color{red}{v_{3}}
\right\}
\end{align}
$$
The missing vector in the matrix $\mathbf{V}$ is the lone null space vector. The quickest way to find the solution is to use the cross product of the first two column vectors.
$$
\begin{align}
%
\color{blue}{v_{1}} \times
\color{blue}{v_{2}}
&=
\color{red}{v_{3}} \\
%
\frac{1}{3}
\left[
\begin{array}{rrr}
-2 \\ 1 \\ -2
\end{array}
\right]
\times
\frac{1}{3}
\left[
\begin{array}{rrr}
-1 \\ 2 \\ -2
\end{array}
\right]
&=
\frac{1}{9}
\left[
\begin{array}{rrr}
-6 \\ -6 \\ 3
\end{array}
\right]
%
\end{align}
$$
To be in a domain matrix, this vector must be normalized.
The constants are
$$
\boxed{
(a, b, c) = \frac{1}{3} \left( -2, -2, 1 \right)
}
$$
The matrix for the row space is
$$
\mathbf{V} =
\frac{1}{3}
\left[
\begin{array}{rrr}
\color{blue}{-2} & \color{blue}{1} & \color{red}{-2} \\
\color{blue}{1} & \color{blue}{-2} & \color{red}{-2} \\
\color{blue}{-2} & \color{blue}{-2} & \color{red}{1} \\
\end{array}
\right]
$$
d e
Options abound for the second range space vector for
$$
\mathbf{U} =
\color{blue}{\mathbf{U}_{\mathcal{R}}} =
\left[ \begin{array}{cc}
\color{blue}{u_{1}} & \color{blue}{u_{2}}
\end{array} \right]
$$
Method 1: Since $\color{blue}{u_{1}}$ and $\color{blue}{u_{2}}$ are orthogonal we can write
$$
\color{blue}{u_{2}}
=
\color{blue}{\left[
\begin{array}{rrr}
1 \\ 0
\end{array}
\right]}
$$
That is,
$$
\boxed{
(d, e) = (1, 0)
}
$$
Method 2: Resolve the eigensystem for
$$
\mathbf{W} = \mathbf{A}\mathbf{A}^{*} =
\left[
\begin{array}{cc}
9 & 0 \\
0 & 81 \\
\end{array}
\right]
$$
The first part of this task would be to find the eigenvalues. But they are sitting on the diagonal:
$$
\lambda \left( \mathbf{W} \right) = \left\{ 9, 81 \right\}
$$
The subtlety is to recall that the singular values are ordered:
$$
\sigma = \sqrt{\left\{ 81, 9 \right\}} = \left\{ 9, 3 \right\}
$$
This means that the eigenvector we seek, $\color{blue}{u_{2}}$, is associated with the smallest eigenvalue $\lambda = 9$.
Method 3: This method also involves an out of alphabetical order solution. Given the SVD,
$$
\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
\qquad \Rightarrow \qquad
\mathbf{U}_{k} = \sigma^{-1}_{k} \mathbf{A} \mathbf{V}_{k}, \ k = 1, \ \rho
$$
That is
$$
\color{blue}{u_{2}} = \frac{1}{\sigma_{2}} \mathbf{A} \color{blue}{v_{2}} =
\frac{1}{3}
\left[
\begin{array}{rrr}
1 & -2 & -2 \\
-6 & 3 & -6 \\
\end{array}
\right]
\frac{1}{3}
\left[
\begin{array}{r}
1 \\ -2 \\ -2 \\
\end{array}
\right]
=
\color{blue}{\left[
\begin{array}{rrr}
1 \\ 0
\end{array}
\right]}
$$
f g
By construction the $\Sigma$ matrix has the form
$$
\left[
\begin{array}{ccc}
9 & 0 & 0 \\
0 & \sigma_{2} & 0 \\
\end{array}
\right]
$$
Immediately,
$$
\boxed{g=0}
$$
If you are solving for the constants out of alphabetical order as suggested earlier, then you would look for the second singular value by finding the eigenvalues of the product matrix
$$
\mathbf{W} = \mathbf{A}\mathbf{A}^{*} =
\left[
\begin{array}{cc}
9 & 0 \\
0 & 81 \\
\end{array}
\right]
$$
The eigenvalues are the diagonal elements:
$$
\lambda \left( \mathbf{W} \right) = \left\{ 9, 81 \right\}
$$
Remember the singular values are ordered:
$$
\sigma = \sqrt{\left\{ 81, 9 \right\}} = \left\{ 9, 3 \right\}
$$
Therefore
$$
\boxed{f=3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum of $xy(1-\frac{z^2}{(x+y)^2})+yz(1-\frac{x^2}{(y+z)^2})+zx(1-\frac{y^2}{(z+x)^2})$ Let $x,y,z>0$ and $x^2+y^2+z^2=1$. What is the maximum value of
$$xy\left(1-\frac{z^2}{(x+y)^2}\right)+yz\left(1-\frac{x^2}{(y+z)^2}\right)+zx\left(1-\frac{y^2}{(z+x)^2}\right)?$$
We can get an upper bound of $1$ if we throw away the minus terms and use the fact that $xy+yz+zx\leq x^2+y^2+z^2$. However, when $x=y=z$ the value is $3/4$, while if some variable approaches zero the value is at most $1/2$.
| For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3}{4}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}xy\left(1-\frac{z^2}{(x+y)^2}\right)\leq\frac{3}{4}(x^2+y^2+z^2)$$ or
$$\sum_{cyc}\left(\frac{1}{4}z^2-\frac{z^2xy}{(x+y)^2}\right)\leq\sum_{cyc}(x^2-xy)$$ or
$$\sum_{cyc}\frac{z^2(x-y)^2}{(x+y)^2}\leq2\sum_{cyc}(x-y)^2$$ or
$$\sum_{cyc}(x-y)^2\left(2-\frac{z^2}{(x+y)^2}\right)\geq0.$$
Let $x\geq y\geq z$.
Hence,
$$y^2\sum_{cyc}(x-y)^2\left(2-\frac{z^2}{(x+y)^2}\right)\geq y^2\sum_{cyc}(x-y)^2\left(1-\frac{z^2}{(x+y)^2}\right)=$$
$$=(x+y+z)y^2\sum_{cyc}\frac{(x-y)^2(x+y-z)}{(x+y)^2}\geq$$
$$\geq(x+y+z)\left(\frac{y^2(x-z)^2(x+z-y)}{(x+z)^2}+\frac{y^2(y-z)^2(y+z-x)}{(y+z)^2}\right)\geq$$
$$\geq(x+y+z)\left(\frac{x^2(y-z)^2(x-y)}{(x+z)^2}+\frac{y^2(y-z)^2(y-x)}{(y+z)^2}\right)=$$
$$=\frac{(x+y+z)(y-z)^2(x-y)(x^2(y+z)^2-y^2(x+z)^2)}{(x+z)^2(y+z)^2}=$$
$$=\frac{(x+y+z)z(y-z)^2(x-y)^2(2xy+xz+yz)}{(x+z)^2(y+z)^2}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Q: Convergence of improper double integral Consider the improper double integral
$$I=\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+4y^2)^{\alpha}}\ dx\ dy$$
I want to determine the range of $\alpha$ , such that the integral above is convergent. Using transformation to polar coordinates I get $
I=\iint_{r>0}\frac{rsin(r^2)}{r^{2\alpha}(1+3sin^2\theta)^{\alpha}}\ dr\ d\theta
$ (I am not sure, if $r$ - determinant of the Jacobian is applied here). So it is $
I=\iint_{r>0}\frac{sin(r^2)}{r^{2\alpha-1}(1+3sin^2\theta)^{\alpha}}\ dr\ d\theta
$. For $\lim{r\to 0} \ \alpha>2 \ (2\alpha-3<1) $ and for $\lim{r\to \infty} \ \alpha<1 \ (2\alpha-1>1)$. But I don't know what to do with $\frac{1}{(1+3sin^2\theta)^{\alpha}}$ and how to
substantiate this range of $\alpha$. Any help would be appreciated.
Edit: $$\frac{1}{4^{\alpha}}{\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}\ dx\ dy} \leq \iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}
{(x^2+4y^2)^{\alpha}}\ dx\ dy \leq {\iint_{x^2+y^2>0}\frac{sin(x^2+y^2)}{(x^2+y^2)^{\alpha}}\ dx\ dy}$$ Can I do it like that or should I add module?
| It is more practical to "confine the mess to the numerator" then switch to polar coordinates:
$$\begin{eqnarray*} I &=& \frac{1}{2}\iint_{\mathbb{R}^2\setminus\{(0,0)\}}\frac{\sin\left(x^2+\frac{1}{4}y^2\right)}{(x^2+y^2)^{\alpha}}\,dx\,dy \\[0.2cm]&=&\frac{1}{2}\,\text{Im}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{e^{i\rho^2\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)}}{\rho^{2\alpha-1}}\,d\theta\,d\rho\\[0.2cm]&=&\frac{1}{4}\,\text{Im}\int_{0}^{+\infty}\int_{0}^{2\pi}\frac{e^{iz\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)}}{z^{\alpha-1}}\,d\theta\,dz\tag{1}\end{eqnarray*}$$
and $\cos^2\theta+\frac{1}{4}\sin^2\theta$ is bounded between $\frac{1}{4}$ and $1$ for every $\theta$. By Dirichlet's test the integral $\int_{0}^{+\infty}\frac{e^{ikz}}{z^{\alpha-1}}\,dz$ is convergent for any $\alpha\in(1,2)$, and with such assumption:
$$ I = \frac{\Gamma(2-\alpha)}{4}\,\sin\left(\frac{\pi \alpha}{2}\right)\int_{0}^{2\pi}\frac{d\theta}{\left(\cos^2\theta+\frac{1}{4}\sin^2\theta\right)^{2-\alpha}} \tag{2}$$
where the last integral is finite and can be computed through the residue theorem. It is interesting to point out that the limit of the previous expression as $\alpha\to 2^-$ simply equals $\frac{\pi^2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$?
I need to find $\theta$ bounds when calculating the volume enclosed between sphere $x^2+y^2+z^2=4a^2$ and cylinder $x^2+(y-a)^2=a^2$. The final answer must be that volume=$\frac{48\pi-64}{9}$.
Here's an illustration:
Let $$V=\int\int\int_Bf(\theta,r,z)dV$$
Because of the symmetry of volume relative of the $y$ axis we can calculate the volume of the upper half then multiply it by 2.
Then:
$$
0\le z\le \sqrt{4a^2-x^2-y^2} \le \sqrt{4a^2-r^2}
$$
Because our the given surfaces don't have their center at the origin we need to adjust the $r$. We can retrieve it from the cylinder equation, $r=2a\sin\theta$. So:
$$
0\le r\le 2a\sin\theta
$$
Finally, the projection of the surface onto the $xy$ plane is a circle. So I think that $0\le \theta \le 2\pi$.
But I don't get the correct answer with this angle range.
If we evaluate the triple integral we'll get to this expression in the end:
$$
E=-\frac{8a^3}{3}(\frac{1}{3}\cos^2\theta \sin\theta+\frac{2}{3}\sin\theta-\theta)
$$
For example:
$$
\int_0^{2\pi}E=\frac{16\pi}{3}a^3
$$
which is not the answer I should've got.
However if I choose the following bounds for $\theta$:
$$
\int_0^{\frac{\pi}{2}}E=\frac{12\pi-16}{9}a^3
$$
The last bounds give the volume of the quarter of the original volume so if we multiply it by $4$ we get the desired answer:
$$
\frac{12\pi-16}{9}a^3\cdot 4=\frac{48\pi-64}{9}
$$
Please explain why my bounds are not good.
| Step 2 is incorrect because both limits vanish $E$ at $0, 2\pi$ . Steps 3 and 4 are correct evaluating between limits $ \int _{\pi/2}^0 $ and multiplying by $4.$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of the following series. My professor asked me to prove that the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is convergent and find the sum.
But I ended up proving it divergent. Here's my work :
Notice that $\sqrt{1+\frac{1}{n}}<\frac{4}{\sqrt{2}}-1$ for all positive integer $n$. This implies $\sqrt{2n}\times\sqrt{n+1}<4n-\sqrt{2}\times n$ for all positive integer $n$, i.e., $\sqrt{2n}\times(\sqrt{n}+\sqrt{n+1})<4n$ for all positive integer $n$.
Finally, we get $\frac{1}{4n}<\frac{1}{\sqrt{2n}(\sqrt{n}+\sqrt{n+1})}=\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ for all positive integer $n$.
Since $\sum_{n=1}^{\infty}\frac{1}{4n}$ is divergent, it follows that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is divergent by Comparison test.
Can someone tell me what's wrong with my proof ? Is the series really DIVERGENT ?
| Yes the series diverges,
as $2n + 2 \geq n $ and $\sqrt{2n} \leq 2\sqrt{n}$,so $\sqrt{2n} + \sqrt{2n+2} \leq 2\sqrt{n} + \sqrt{n} = 3 \sqrt{n}$,
So,$\frac{1}{(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3\sqrt{n}}$,
$\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3n}$ or $\frac{1}{3n} \leq \frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ ,so by the comparision test the series diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this limit of an integral Find $$\lim_{n\to\infty}n\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)$$
it is obvious
$$\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}})dx=B\left(\dfrac{n+1}{2n},\dfrac{2n+1}{4n}\right)$$
| $\lim_{n\to\infty}n^2\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)
$
I'll proceed very naively.
Since,
for large $n$,
$z^{1/n}
=e^{\ln z/n}
=1+\frac{\ln z}{n}+\frac{\ln^2 z}{2n^2}+O(\frac1{n^3})
=1+\frac{\ln z}{n}+O(\frac1{n^2})
$,
$\begin{array}\\
(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}
&=(1+\frac{\ln \sin x}{n}+O(\frac1{n^2}))(1+\frac{\ln \cos x}{2n}+O(\frac1{n^2}))\\
&=1+\frac{2\ln (\sin x)+\ln(\cos x))}{2n}+O(\frac1{n^2})\\
&=1+\frac{\ln (\sin^2 x\cos x))}{2n}+O(\frac1{n^2})\\
\end{array}
$
so
$\begin{array}\\
\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\frac{\ln (\sin^2 x\cos x)}{2n}+O(\frac1{n^2}))dx\\
&=\frac{\pi}{2}+\frac1{2n} \int_{0}^{\frac{\pi}{2}}\ln (\sin^2 x\cos x)dx +O(\frac1{n^2})\\
&=\frac{\pi}{2}+\frac1{2n} (-\frac32 \pi\ln 2) +O(\frac1{n^2})
\qquad\text{(according to Wolfy)}\\
&=\frac{\pi}{2} -\frac{3\pi \ln 2}{4n}+O(\frac1{n^2})\\
\text{so}\\
n(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx
-\frac{\pi}{2})
&= -\frac{3\pi \ln 2}{4}+O(\frac1{n})\\
\end{array}
$
Therefore
the limit in the question
is $\infty$
(unless I have made a mistake,
in which case
more terms may need to be taken
in the expansion).
| {
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"url": "https://math.stackexchange.com/questions/2306526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area between $r=4\sin(\theta)$ and $r=2$ I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.
I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$
Is this correct? Or did I find the area for the following region
| You are just intersecting two circles with the same radius, going through the center of each other.
The area of a circle sector with radius $R=2$ and amplitude $60^\circ$ is $\frac{1}{6}\pi R^2=\frac{2\pi}{3}$, while the area of an equilateral triangle with side length $2$ is given by $\sqrt{3}$, hence the area of the circle segment by the difference of these objects is $\frac{2\pi}{3}-\sqrt{3}$.
These results are enough to solve your question without integrals:
$$\color{red}{\mathcal{A}}=2\sqrt{3}+4\left(\frac{2\pi}{3}-\sqrt{3}\right)=\color{red}{\frac{8\pi}{3}-2\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Modulus Function issue.
I'm asked to solve $$|x+1|>x^2-5$$
My attempt,
For my basic inequality skill, this is a very easy question.
Since $$|x|=\left\{\begin{matrix}
x, x \geq0 & \\
-x,x<0&
\end{matrix}\right.$$
Then for $x<-1$
$x^2+x-4<0$
$\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$
Then for $x\geq-1$
$x+1 >x^2-5$
$x^2-x-6 <0$
$-2 < x<3$
So combine the ranges,
I got $\frac{-1-\sqrt{17}}{2}<x<3$
So, a senior told me that basically my solution is correct, but instead of $x< -1$, I should write $x \leq -1$. Why? According to the definition of modulus function which shows that I'm correct. But why he said that actually the = sign can be for both. So, he asked me don't be bothered too much by this trivial issue. I really don't understand why he said so. Can anyone explain it for me? Thanks in advance.
| It's $x+1>x^2-5$ or $x+1<-x^2+5$ without any additional cases.
because if $x^2-5<0$ then the inequality is obviously true.
We get $-2<x<3$ or $\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$, which gives your answer:
$$\frac{-1-\sqrt{17}}{2}<x<3$$
I think your solution is right, but your way is bad.
For example, solve the following inequality:
$$|x^3+x-1|>2-x$$
By your way we need to solve two inequalities: $x^3+x-1\geq0$ and $x^3+x-1<0$.
I don't say that it's hard (sometimes it's just impossible!), I say that it's not necessary.
Indeed, $|x|>a\Leftrightarrow x>a$ or $x<-a$ for all $a\in\mathbb R$
because for $a<0$ the inequality $|x|>a$ is obvious.
Thus, $|x^3+x-1|>2-x$ gives $x^3+x-1>2-x$ or $x^3+x-1<-2+x$,
which is $x^3+2x-3>0$ or $x^3+1<0$, which is
$x^3-x^2+x^2-x+3x-3>0$ or $(x+1)(x^2-x+1)<0$, which is
$(x-1)(x^2+x+3)>0$ or $x<-1$, which gives the answer:
$$(-\infty,-1)\cup(1,+\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307887",
"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}{4}} &=
\frac{\cos(2x)}{\frac{4x-\pi}{4}} \\&=
\frac{4\cos(2x)}{4x-\pi} = \,\,???\end{align}$$
What do I do next?
| Let $u = x-\pi/4$ then you want $$\lim_{u\to 0} \frac{\cos(2(u + \pi/4))}{\sin u}=-\lim_{u \to 0} \frac{\sin 2u}{\sin u } = -\lim_{u \to 0} 2\cos u=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is there any easy way to calculate this?
Please someone help.
| Rewrite the sum $\sum_{i=1}^{\infty }\frac {1}{(2i)(2i+1)(2i+2)}$ as
$$ \sum_{i=1}^{\infty }\frac {(2i+1)-2i}{(2i)(2i+1)(2i+2)} = \sum_{i=1}^{\infty }\frac {1}{(2i)(2i+2)} -\sum_{i=1}^{\infty }\frac {1}{(2i+1)(2i+2)} $$
Or using partial fractions
$$ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) - \sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right) $$
The left sum telescopes to $1$, so we get that $ \frac{1}{4} \sum_{i=1}^{\infty} \left(\frac{1}{i} - \frac{1}{i+1}\right) = \frac14$
For the right sum
$$\sum_{i=1}^{\infty} \left(\frac{1}{2i+1} - \frac{1}{2i+2}\right) = \frac13 - \frac14 + \frac15-\frac16 + \dots $$
We use series expansion for $\ln(1+x)$
$$ \ln(1+x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}x^k}{k} = x-\frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4} + \dots $$
Plug in $x=1$, to get that
$$ \ln 2 = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \dots $$
Or
$$\frac13 - \frac14 + \frac15-\frac16 + \dots = \ln 2 - \frac12 $$
and that's our right sum so the final sum is equal to
$\frac14 - \ln2 + \frac12 = \frac34-\ln2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?
| The discriminant of the quadratic form $u^2 + uv + 4 v^2$ is $-15.$
$$ (-15|11) = (-4|11) = (-1|11) (4|11) = (-1|11)=-1, $$
the last part because $11 \equiv 3 \pmod 4.$
Therefore, if
$$ u^2 + uv + 4 v^2 \equiv 0 \pmod {11}, $$ it follows that both $u,v$ are divisible by $11.$ General theorem with proof at Prime divisors of $k^2+(k+1)^2$
So, with integer $x,$ $x^2 + x + 4$ is the same as $x^2 + xy + 4 y^2$ restricted to $y=1;$ that is, this $y$ is not divisible by $11.$ So $x^2 + x + 4$ cannot be divisible by $11$
I like this way of writing things, in quadratic forms this behavior is called "anisotropic." My experience is that existence of a solution that, however, violates some condition, is easier for the student to deal with than "infinite descent" arguments. All is the same, just a matter of emphasis.
Sigh. Same thing: $x^2 + 1$ is not divisible by any prime $q \equiv 3 \pmod 4.$ The discriminant of $u^2 + v^2$ is $-4.$ For such $q,$ if $q | (u^2 + v^2)$ then both $q|u$ and $q | v.$ However, it is not possible to have $q | 1,$ so $q$ cannot divide $x^2 + 1$ either.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$f(\frac{1}{n})=a_n$ for which sequence holomorphic? Let $D=\{z\in\mathbb{C}\colon |z|<1\}$. Is there a holomorphic function $f\colon D\rightarrow\mathbb{C}$ with $f(\frac{1}{n})=a_n$ $(n=2,3,4,...)$ where $a_n$ is one of the sequences ?
a) $a_n=0,\frac{1}{2},0,\frac{1}{4},0,\frac{1}{6},...$
b) $a_n=\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{6},\frac{1}{6},...$
c) $a_n=\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7},...$
Has anyone an idea on how to tackle this question ?
| If our function is holomorphic the derivatives at zero will all exist, and and any sequence of points approaching zero should give us the same derivative. So what happens for a)?
$\displaystyle\lim_{n \to \infty}\frac{f(\frac{1}{2(n+1)})-f(\frac{1}{2n})}{\frac{1}{2(n+1)}-\frac{1}{2n}}=\displaystyle\lim_{n \to \infty}\frac{\frac{1}{2(n+1)}-\frac{1}{2n}}{\frac{1}{2(n+1)}-\frac{1}{2n}}=1$, but $\displaystyle\lim_{n \to \infty}\frac{f(\frac{1}{2(n+1)+1})-f(\frac{1}{2n+1})}{\frac{1}{2(n+1)+1}-\frac{1}{2n+1}}=\displaystyle\lim_{n \to \infty}\displaystyle\frac{0}{\frac{1}{2(n+1)+1}-\frac{1}{2n+1}}=0$ which is a contradiction so no such function exists.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse of the matrix $$U(a,b)=\left(\begin{matrix}a&b&b&b\\b&a&b&b\\b&b&a&b\\b&b&b&a\end{matrix}\right)$$
Is there an easy way to find the inverse of $U(1,2)$ , a trick to solve this problem easy?(its part of an exam with answers)
| Note that
$$U(a,b)U(c,d) = U(ac+3bd, bc+ad+2bd)$$
This can be seen because, if we let $A$ be the matrix each of whose entries are $1$'s, since $U(a,b)=(a-b)I+bA$, and since $A^2=4A$
$$((a-b)I+bA)((c-d)I+dA)$$
$$ = (a-b)(c-d)I+(b(c-d)+d(a-b)+4bd)A$$
$$= ((ac+3bd)-(bc+ad+2bd))I+(bc+ad+2bd)A$$
So, if $U(a,b)^{-1}$ exists and is equal to $U(c,d)$ for some $c,d$, then we would have $ac+3bd=1$ and $bc+ad+2bd=0$. The second equation gives us
$$bc = -(a+2b)d$$
The determinate of $U(a,b)$ is actually $(a-b)^3(a+3b)$, so suppose $U(a,b)\neq 0$, so that neither of these factors are zero. Plugging the above into the first equation,
$$-a(a+2b)d+3b^2d=b$$
$$\implies d = \frac{b}{-a^2-2ab+3b^2} = -\frac{b}{(a-b)(a+3b)}$$
Finally,
$$c = \frac{a+2b}{(a-b)(a+3b)}$$
Therefore, assuming $a\neq b$ and $a\neq -3b$,
$$U(a,b)^{-1}=U\left(\frac{a+2b}{(a-b)(a+3b)} , -\frac{b}{(a-b)(a+3b)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
} |
equation of common tangent touching circle and the parabola I am trying to find the equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the x-axis is :
Equation of tangent on parabola: $y=mx+a/m$ here a = 1;
so $y = mx + 1/m$
centre of the circle (3,0) and r = 3. Now
distance of point from line is, $3 = |3m+1/m|/(1+m^2)^(1/2)$
I am having difficulty in finding the value of m. Please help me!!!
| \begin{align}
\frac{\displaystyle \left|3m+\frac{1}{m}\right|}{\sqrt{m^2+1}}&=3\\
\left(3m+\frac{1}{m}\right)^2&=9(m^2+1)\\
9m^2+6+\frac{1}{m^2}&=9m^2+9\\
\frac{1}{m^2}&=3\\
m&=\frac{\pm1}{\sqrt{3}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\
&= \cos^{-1}(0)=90° \\
\alpha_2 &= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\
&= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\
\alpha_3 &= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\
&= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5°
\end {aligned}$$
As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts?
| It's enough to show the following:
$$(2,1,-5)\cdot(4,7,3)=0$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Numbers of the form $8k^2-1$ in a sequence defined by $a_0=-1$, $a_1=1$ and $a_{n+2}=6a_{n+1}-a_n$.
Suppose a sequence $\{a_n\}_{n\in\mathbb{N}}$ is defined by $a_0=-1$, $a_1=1$ and $$a_{n+2}=6a_{n+1}-a_n$$ for all $n\in\mathbb{N}$. Find all $n$ such that $a_n$ is of the form $8k^2-1$ for some $k\in\mathbb{N}$.
The problem comes from Bulgaria National Olympiad 2003, Problem 3:
Given the sequence $\{y_n\}_{n\in\mathbb{N}_+}$ defined by $y_1=y_2=1$ and
$$y_{n+2}=(4k-5)y_{n+1}-y_n+4-2k,\qquad n\in\mathbb{N}_+.$$
Find all $k\in\mathbb{Z}$ such that $y_n$ is a perfect square for all $n\in\mathbb{N}_+$.
I attempted to solve this problem by using the fact that $y_2=2k-2$ and $y_3=8k^2-20k+3$ are perfect squares. Let $2k-2=(2u)^2$ and $8k^2-20k+3=v^2$, and we have negative Pell's equation $$(8u^2-1)^2-2v^2=-1.$$ The general solution to $x^2-2y^2=-1$ is given by $x_n+y_n\sqrt{2}=(1+\sqrt{2})^{2n+1}$, thus leading to my original question.
Obviously, $a_0=-1$ and $a_2=7$ is of the form $8k^2-1$. How do I prove that these are the only solutions?
| Work in progress ...
Proposition 1. $a_{2k} \equiv -1 \pmod{8}$
Proof. Leaving the theoretical material aside, characteristic polynomial of the recurrent sequence is
$$x^2-6x+1=0$$
with the solutions $x_1=3-2\sqrt{2},x_2=3+2\sqrt{2}$, thus
$$a_n=A(3-2\sqrt{2})^n+B(3+2\sqrt{2})^n$$
or, given the initial conditions
$$a_n=\left(-\frac{1}{2}-\frac{1}{\sqrt{2}}\right)(3-2\sqrt{2})^n+\left(-\frac{1}{2}+\frac{1}{\sqrt{2}}\right)(3+2\sqrt{2})^n$$
One thing to mention is that
$$(3-2\sqrt{2})^n=C-D\sqrt{2}\\
(3+2\sqrt{2})^n=C+D\sqrt{2}$$
both $C,D \in \mathbb{Z}$ and
$$a_n=2D-C$$
But
$$C=3^{n}+\binom{n}{2}3^{n-2}(2\sqrt{2})^2+\binom{n}{4}3^{n-4}(2\sqrt{2})^4+...=3^{n}+8Q$$
$$D=\binom{n}{1}3^{n-1}2+\binom{n}{3}3^{n-3}2^3\sqrt{2}^2+\binom{n}{5}3^{n-5}2^5\sqrt{2}^4+...=n3^{n-1}2+8R$$
Thus
$$a_n+1=2D-C+1=n3^{n-1}4-3^{n}+1+16R-8Q$$
Or if
$$3^{n-1}(4n-3)+1\equiv 0 \pmod{8} \Rightarrow a_n \equiv -1 \pmod{8}$$
which is true only for even $n=2k$, since
$$4n-3=8k-3 \equiv -3 \pmod{8}\Rightarrow 3^{n-1}(4n-3)+1 \equiv -3^{2k}+1 \equiv 0 \pmod{8} \tag*{$\blacksquare$}$$
Note for odd $n=2k+1 \Rightarrow 4n-3=8k+1 \equiv 1 \pmod{8} \Rightarrow$ $3^{n-1}(4n-3)+1 \equiv 3^{2k}+1 \not\equiv 0 \pmod{8}$, since $9 \equiv 1 \pmod{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ and the straight lines $OP,OQ$ make angles $\alpha,\beta$ with the $x$-axis where $O$ is the origin then $\frac{\tan\alpha}{\tan\beta}=$
$(A)-1$
$(B)-2$
$(C)2$
$(D)\sqrt2$
Let $P$ be $(x_1,y_1)$ and $Q$ be $(x_2,y_2)$.
Equation of tangent is $\frac{y-y_1}{x-x_1}=\frac{3x_1^2}{2y_1}$
It passes through $(x_2,y_2)$,so $\frac{y_2-y_1}{x_2-x_1}=\frac{3x_1^2}{2y_1}$
I need to find out $\frac{\tan\alpha}{\tan\beta}=\frac{\frac{y_1}{x_1}}{\frac{y_2}{x_2}}=\frac{y_1x_2}{x_1y_2}$
I am stuck here.
| Simplify the expression a bit.
Notice that tangent of an angle is just the slope it makes.
So, we need to find the ratio of the slopes.
Let $P$ be $(1,1)$.
We have the derivative of $y=\displaystyle \sqrt{x^3} = \frac32\sqrt{x}$ .
Therefore, the tangent line is $(y-1)=\frac{3}{2}(x-1)$.
Because $y^2=x^3$ is concave upward for $y>0$, we know that the line will intersect the function again at a Q such that $y<0$.
Solving for Q, we have $\displaystyle -x^{\frac{3}{2}}-1=\frac{3}{2}x-\frac{3}{2}$, or $x^{\frac32}+\frac{3}{2}x-\frac{1}{2}=0$.
We simplify as $\frac{1}{2}(\sqrt{x}+1)^2(2\sqrt{x}-1)=0$.
Because we have $\sqrt{x} \neq 1$, we must have $2\sqrt{x}=1$,and therefore $x=\frac{1}{4}$, and $Q=\displaystyle \left(\frac14, -\frac18\right)$.
We have the slope of $OP$ is $1$ and the slope of $OQ$ is $\displaystyle -\frac12$.
Therefore, finally, $\displaystyle \frac{\tan \alpha}{\tan \beta}=-2$, or (B)
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating the Fourier series to find the Fourier series of $\frac{1}{2}x^2$
The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series
$$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2}\sin\left(\frac{2\pi x}{l} \right) + \frac{1}{3}\sin\left(\frac{3\pi x}{l} \right) - \ldots \right)$$
Integrate the series term-by-term to find the Fourier series for $\frac{1}{2}x^2$, up tp a constant of integration (which is then the $\frac{1}{2}A_0$ term in the cosine series). Find the $A_0$ using the standard formula to completely determine the series.
Attempted solution - We have
\begin{align*}
&\sum_{m=1}^{\infty}\int_{0}^{l}\frac{2l}{\pi}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l}\right)dx\\
&= \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\int_{0}^{l}\sin\left(\frac{m\pi x}{l}\right)dx\\
&= \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\left[-\frac{l}{\pi m}\left(\cos\left(\frac{m\pi x}{l}\right)\Big|_0^l\right)\right]\\
&= -\frac{2 l^2}{\pi^2}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}\left(\cos\left(\frac{m \pi l}{l}\right) - 1\right)\\
\end{align*}
I am not sure where to go from here, any suggestions are greatly appreciated.
| You're trying to integrate $x$ to $x^2/2$, implying you want to use the indefinite operator $\int dx$, not a definite integration on a period. This gives you a series of cosines, viz. $$\frac{x^2}{2}=A-\frac{2l^2}{\pi^2}\sum_{m\ge 1}\frac{\left( -1\right)^{m+1}}{m^2}\cos\frac{m\pi x}{L}$$ for some constant $A$ obtainable by setting $x=0$, viz. $$A=\frac{2l^2}{\pi^2}\sum_{m\ge 1}\frac{\left( -1\right)^{m+1}}{m^2}=\frac{2l^2}{\pi^2}\eta\left( 2\right)=\frac{2l^2}{\pi^2}\frac{\pi^2}{12}=\frac{l^2}{6}.$$
| {
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"answer_id": 0
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Find the limit $\lim\limits_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\,dx$ $$\lim_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\, dx$$
Some help please. I don't have any idea. Thank you!
| Using the formula for the tangent of a sum, we get
$$
\arctan(\alpha\tan(x))=x+\arctan\left(\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right)
$$
Since $\left|\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right|\le\frac{|\alpha-1|}{2\sqrt{\alpha}}$, arctan never has to go through a singularity, so this is a nice, smooth function.
$$
\require{cancel}
\begin{align}
\int_0^n\frac1{1+n^2\cos^2(x)}\,\mathrm{d}x
&=\int_0^n\frac{\sec^2(x)}{1+n^2+\tan^2(x)}\,\mathrm{d}x\\
&=\cancel{\left.\frac1{\sqrt{n^2+1}}\arctan\left(\frac{\tan(x)}{\sqrt{n^2+1}}\right)\right]_0^n}\\
&=\left.\frac1{\sqrt{n^2+1}}\left(x-\arctan\left(\frac{\left(\sqrt{n^2+1}-1\right)\tan(x)}{\sqrt{n^2+1}+\tan^2(x)}\right)\right)\right]_0^n\\
&=\frac1{\sqrt{n^2+1}}\left(n-\arctan\left(\frac{\left(\sqrt{n^2+1}-1\right)\tan(n)}{\sqrt{n^2+1}+\tan^2(n)}\right)\right)
\end{align}
$$
Since $\arctan$ is between $-\frac\pi2$ and $\frac\pi2$, the limit as $n\to\infty$ is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Complex integral Let $C$ denote the unit circle centered at the origin in Complex Plane
What is the value of
$$ \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,$$ where the integral is taken anti-clockwise along $C$?
*
*0
*1
*2
*3
What I have answered is 0 because it seems like $f(z)$ is analytic at 0 hence by Cauchy's Theorem.
| We want the integral
$$\frac{1}{2\pi i}\int_C |1+z+z^2|dz=\frac{1}{2\pi}\int_0^{2\pi}|1+z+z^2|z~d\theta$$
since $z=e^{i\theta}$.
Now consider the absolute value portion,
$$
\begin{align}
|1+z+z^2|
&=\sqrt{(1+z+z^2)(1+z+z^2)^*}\\
&=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})}\\
&=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})\frac{z^2}{z^2}}\\
&=\sqrt{\left(\frac{1+z+z^2}{z} \right)^2}\\
&=\sqrt{\left(\frac{1}{z}+1+z \right)^2}\\
&=\sqrt{(1+2\cos\theta)^2}\\
&=|1+2\cos\theta|
\end{align}
$$
We can return to solve the integral
$$
\begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2|dz
&=\frac{1}{2\pi}\int_0^{2\pi}|1+2\cos\theta|(\cos\theta+i\sin\theta)~d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi}|1+2\cos\theta|\cos\theta~d\theta\\
&=\frac{1}{3}+\frac{\sqrt{3}}{2\pi}\approx0.60900
\end{align}$$
This is in agreement with our own numerical solution as well as that of others noted previously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding values of $t$ I have this equation -
$3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0 $
I'm struggling on how to find the values of $t$
First, I power both sides by $2$ to make the power be a whole number .
I get
$3t - \frac{2}{5} t^{-3} = 0 $
From here I'm stunned and stuck . Can I get a hint ! Thanks !
| You can't square.
$(a + b)^2 \ne a^2 + b^2$ but instead equals $a^2 + 2ab + b^2$ so that won't make things simpler.
Also squaring both both sides of an equation will add extraneous solutions. For example, the problem $x + 1 = 2$ has one solution: $x = 1$. But if I chose to square both sides $(x+1)^2 = 4^2$ then $x^2 + 2x + 1 = 4$ and $x^2 + 2x - 3 = 0$ so $(x+3)(x-1) = 0$ so $x = 1$ or $x = -3$. So somehow we now have two solutions. Where did $x = -3$ come from and is it a valid solution?
Notice when we say $a = b$ there is only one possibility: $a$ and $b$ are the same thing. But if we square both sides to get $a^2 = b^2$ there are two possibilities: $a$ and $b$ are the same thing, or $a$ and $b$ are negatives of each other.
So $x + 1 = 2$ means $x + 1 $ and $2$ are the same thing. But $(x+1)^2 = 2^2$ means either $x + 1$ and $2$ are the same thing, OR $x+1$ and $-2$ are the same thing. If $x+1 = -2$ we get ..... $x =-3$... which was the new answer that came out of nowhere. It is not a valid answer because $x + 1 \ne -2$. We call that an extraneous solution and this often (usually) occurs if we square both sides.
Squaring both sides isn't wrong but we must test for extraneous solutions.
$3t^{\frac 12} - \frac 25 t^{-\frac 32} = 0$
$(3t^{\frac 12} - \frac 25 t^{-\frac 32})^2 = 0$
$9t - 2*3t^{\frac 12}*\frac 25 t^{-\frac 32} + \frac {4}{25}t^{-3} = 0$
$9t - \frac {12}{5t} + \frac {4}{25t^3} = 0$
And ... that's really not any easier... and it wil have extraneous solutions so ... let's not do it.
If getting rid of the fractional powers are *really * a concern (why are they?) then to get rid of them simply do substitution.
Let $r = t^{\frac 12}$.
Then
$3t^{\frac 12} - \frac 25 t^{-\frac 32} = 0 \implies$
$3r - \frac 25 r^{-3} = 0\implies$
$(3r - \frac 25 r^{-3})5r^3 = 0 *5r^3 \implies$
$15r^4 - 2 = 0\implies$
$r^4 = \frac 2{15}\implies$
$r = \pm \sqrt[4]{\frac 2{15}}\implies$
$t^{\frac 12} = \pm \sqrt[4]{\frac 2{15}}$. But $t^{\frac 12} \ge 0$ so
$t^{\frac 12} = \sqrt[4]{\frac 2{15}}\implies$
$t = \sqrt{\frac 2{15}}$.
But there really isn't any reason to be afraid of the half powers.
Simply put everything over a common denominator:
$3t^{\frac 12} - \frac 25t^{-\frac 32} = 0 \implies$
$\frac {15t^2 - 2}{5t^{\frac 32}} = 0\implies$
$15t^2 - 2 = 0\implies$
$t = \pm \sqrt {\frac {2}{15}}$.
But as $t^{1/2}$ implies $t$ is non-negative.
$t = \sqrt{\frac {2}{15}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get:
$$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$
$$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$
$$\sum_{r=-2}^{0}{r^3} = -9 = e$$
$$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$
$$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$
After some algebra we now know $a= -\frac12, b = -1, c = 1, d = \frac32, e = -9$:
$$\therefore \sum^{n}_{r=-2}{r^3}= -\frac12n^4 -n^3 + n^2 +\frac32n -9$$
We define $\sum^{n}_{r=0}{r^3}$:
$$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=-\frac12n^4 -n^3 + n^2 +\frac32n -9 -(-9)$$
$$-\frac12n^4 -n^3 + n^2 +\frac32n = \frac14(-2n^4-4n^3+4n^2 + 6n)$$
$$\frac14n^2\biggl(-2n^2-4n+4+\frac6n\biggl) = \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
$$ \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
My problem:
I got $d = \frac32$ and as far as i can see since it need to end up with $d = 0$ as the smallest term that can be left is a $n^2$ term as $\frac14n^2(n+1)^2$ is multiplied by $n^2$ meaning $e$ has to cancel out (which it does) but i don't know how to get rid of $d$, if you could explain it to me it would be much appreciated
| $$\begin{align}
&n=0: &e&=-9\tag{1}\\\\
&n=1: &a+b+c+d&=1\tag{2}\\
&n=-1: &a-b+c-d&=0\tag{3}\\\\
\frac 12[(2)+(3)]: &&a+c&=\frac 12\tag{4}\\
\frac 12[(2)-(3)]: &&b+d&=\frac 12\tag{5}\\\\
&n=2: &16a+8b+4c+2d&=9\tag{6}\\
&n=-2: &16a-8b+4c-2d&=1\tag{7}\\\\
\frac 12 [(6)+(7)]: &&16a+4c&=5
\quad\Rightarrow a=\frac 14, c=\frac 14
\;\;\ \text{using }(4)\\
\frac 14[(6)-(7)]: &&4b+d&=2
\quad\Rightarrow b=\frac 12, d=0
\quad\text{using }(5)\\\\
&&\sum_{r=0}^n r^3&=\left(\sum_{r=-2}^n r^3\right) -9\\
&& &=an^4+bn^3+cn^2+dn+e-(-9)\\
&& &=\frac 14 n^2+\frac 12 n^3+\frac 14 n^2\\
&& &=\frac 14n^2(n+1)^2\\
&& &=\left(\frac 12 n(n+1)\right)^2\end{align}$$
See also this question I posted.
If we can show that $n^2$ and $(n+1)^2$ are factors of the sum of odd powers $p(>1)$ of the first $n$ integers, then for $p=3$ (i.e. sum of cubes) we would just have to determine the constant ($\frac 14$, by putting $n=1$) and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Can a solved Sudoku game have an invalid region if all rows and columns are valid? Given a $9 \times 9$ solved Sudoku game with $3 \times 3$ regions, is it possible that one (or more) of the regions are invalid if all rows and columns are valid (i.e. have a unique sequence of $1-9$)?
| Yes, it can happen that all $3 \times 3$ regions are invalid:
\begin{array}{|ccc|ccc|ccc|} \hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 \\ \hline
4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 \\
5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 \\
6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 \\ \hline
7 & 8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 \\
8 & 9 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
9 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Irreducible polynomials in GF(p) Given is the polynomial $\varphi(X)=X^4+1$.
Now there are two tasks:
(1) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \equiv1$ (mod 4).
(2) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \equiv3$ (mod 4).
My ideas until now:
(1) I try to find $a,b \in \mathbb F_p [X]$, such that $X^4+1 = (X^2+a)(X^2+b)$. Not to use the middleterm $+cX$ was a hint. Now $(X^2+a)(X^2+b)=X^4+aX^2+bX^2+ab=X^4+(a+b)X^2+ab$. This should equal $X^4+1$, so I know, that $a+b \equiv 0$ mod p and $ab \equiv 1$ mod p. This means, that $a$ and $b$ must be units in $\mathbb F_p$, more exactly, $a$ is the multiplicative inverse element of $b$ in $\mathbb F_p$. $a+b \equiv 0$ reasons, that $a$ is the additive inverse to $b$.
But know I don't know, how I can go on with conclusion. I hope, somebody can help me.
| The two congruences give you
$$b^2 \equiv -1 \pmod{p}$$
This is the same as asking when is $-1$ a quadratic residues mod $p$? If you are familiar with Legendre symbol, then perhaps you might have seen
$$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=\begin{cases}1 & \text{ if } p \equiv 1 \pmod{4}\\-1 & \text{ if } p \equiv 3 \pmod{4}\end{cases}$$
So if $p \equiv 1 \pmod{4}$, then we have $a,b$ such that $a+b \equiv 0$ and $ab \equiv 1$. The existence of such $a,b$ implies that $a^2 \equiv -1 \pmod{p}$. Therefore,
$$x^4+1 =(x^2+a)(x^2-a)$$
part(2)
Claim: When $p \equiv 3 \pmod{4}$, then one of $2$ or $-2$ will be a quadratic residue. For this consider $2^{\frac{p+1}{4}}$. It can be shown that this will be the square root of either $2$ or $-2$.
If $2$ happens to be a quadratic residue mod $p$, i.e. if there exists an $a$ such that $a^2 \equiv 2 \pmod{p}$, then we can factor as follows:
$$x^4+1 = x^4+1+a^2x^2-a^2x^2=(x^2+1)^2-a^2x^2=(x^2+1+ax)(x^2+1-ax).$$
Now consider the case when $2$ is not a quadratic residue, then try to see if you can prove that $-2$ will definitely be a quadratic residue. Then you can use the idea above to complete the factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different numbers. But I can't find a way to solve.
| Let $a$ and $b$ be the sides of the rectangle. We know the perimeter of a rectangle is $ 2(a+b) $, and the area is$(ab)$.
Since these two expressions have to be equal, let us solve the equation:
$$2(a+b) = ab$$
$$2a = ab - 2b$$
$$2a = b(a-2)$$
$$b = \frac{2a}{a-2}$$
Since $b$ is an integer, $\frac{2a}{a-2}$ must be one as well. This means that $a$ is even, because otherwise when $a$ is odd, $a$ is not divisible by $a-2$, excluding the case $a=3$. This means that $b = \frac{2*3}{3-2} = 6$, which is a valid solution.
We can approach this by doing casework: checking all even numbers from $a=4$ (when $a=2$, $a-2 = 0$).
When $a=4$, $b=\frac{2*4}{4-2}=4$, but this solution does not have distinct integers.
When $a=6$, $b=\frac{2*6}{6-2}=3$. Therefore, this is the same solution mentioned earlier.
Therefore, one solution to the problem is when $a=6,b=3$. Now check if the values of $a$ and $b$ satisfy the original expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the derivative of the inverse function, $Dg(0,1)$. Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation
\begin{equation} f(x,y)=(x^2-y^2,2xy)\end{equation}
Parts (a) and (b) of the problem asked me to show that $f$ is one-to-one on the set $A$ consisting of all $(x,y)$ with $x>0$, and to find set $B=f(A)$, which was easy enough to do. Part (c) asks
If $g$ is the inverse function, find $Dg(0,1)$.
What I thought to do was to find the matrix $Df$ and find the inverse of it, but I am finding that very challenging to do (here I thought I had my linear algebra skills in check!) Because what I get is that
$Df=
\begin{bmatrix}
2x &2y\\
-2y & 2x
\end{bmatrix}$
And when I try to find the inverse, i get to the step
$
\begin{bmatrix}
x & y & 1/2 & 0\\
0 & \frac{x^2+y^2}{x} & \frac{y}{2x} & 1/2
\end{bmatrix}$
And i feel like maybe this is not the correct method to finding the inverse...?
Thank you!
| First, fix the differential ($f(x,y)=(u(x,y),v(x,y))\;;u=x^2-y^2\;;v=2xy$):
$$Df=
\begin{pmatrix}
\partial u/\partial x &\partial u/\partial y\\
\partial v/\partial x & \partial v/\partial y
\end{pmatrix}=
\begin{pmatrix}
2x &-2y\\
2y & 2x
\end{pmatrix}$$
Now, with the invaluable aid of the commentators, the inverse is
$$Dg(f(x,y))=\dfrac{1}{\vert Df\vert}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=\dfrac{1}{4(x^2+y^2)}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=$$
$$=\dfrac{1}{2(x^2+y^2)}\begin{pmatrix}x &y\\ -y & x\end{pmatrix}$$
We have to find the values for wich $f(x,y)=(0,1)$, so is, $x^2-y^2=0$ and $2xy=1\implies (x,y)=(1/\sqrt{2},1/\sqrt{2})$ or $(x,y)=(-1/\sqrt{2},-1/\sqrt{2})$. But $x\gt0$, so we drop the solution with the negatives. Then,
$$Dg(0,1)=Dg(f(1/\sqrt{2},1/\sqrt{2}))=\dfrac{1}{2}\begin{pmatrix}\sqrt{2}/2 &\sqrt{2}/2\\-\sqrt{2}/2 &\sqrt{2}/2\end{pmatrix}=$$
$$Dg(0,1)=\dfrac{1}{4}\begin{pmatrix}\sqrt{2} &\sqrt{2}\\-\sqrt{2} & \sqrt{2}\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
| Please see the attached photo for a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Try this notation for less clutter:
Let $$S_n=z^n+\frac 1{z^n}$$
It can be easily shown that
$$S_n^2=S_{2n}+2$$
Hence
$$S_4=S_2^2-2=47
\qquad \Rightarrow S_2=7\\
S_2=S_1^2-2=7\qquad \Rightarrow S_1=3$$
Also, $$S_aS_b=S_{a+b}S_{a-b}$$
Putting $a=3, b=1$,
$$S_3S_1=S_4+S_2\\
S_3=\frac {S_4+S_2}{S_1}=\frac {47+7}3=\color{red}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.
My attempt:
| We'll prove that for all triangle
$$r\leq\sqrt{\frac{A}{3\sqrt3}}.$$
Indeed, let $AB=c$, $AC=b$ and $BC=a$.
Hence, $A=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$, which gives $r=\frac{2A}{a+b+c}$.
Thus, we need to prove that
$$\frac{2A}{a+b+c}\leq\sqrt{\frac{A}{3\sqrt3}}$$ or
$$12\sqrt3A\leq(a+b+c)^2.$$
But by Heron formula we have
$$A=\sqrt{p(p-a)(p-b)(p-c)},$$
where $p=\frac{a+b+c}{2}$.
Thus, we need to prove that
$$(a+b+c)^2\geq12\sqrt3\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a+c-b}{2}\cdot\frac{b+c-a}{2}}$$ or
$$(a+b+c)^3\geq27(a+b-c)(a+c-b)(b+c-a)$$ or
$$\frac{(a+b-c)+(a+c-b)+(b+c-a)}{3}\geq\sqrt[3]{(a+b-c)(a+c-b)(b+c-a)},$$
which is AM-GM.
The equality occurs for $$a+b-c=a+c-b=b+c-a$$ or
$$a=b=c,$$
which says that $r$ gets a maximal value, when our triangle is an equilateral triangle,
which says that $\measuredangle BAC=60^{\circ}$.
Done!
If the following is obvious for you, then I am ready to delete it.
The equality occurring in our AM-GM just for $a+b-c=a+c-b=b+c-a$ we can understand by the following way.
Let $a+b-c=x^3$, $a+c-b=y^3$ and $b+c-a=z^3$.
Hence, $x$, $y$ and $z$ are positives and the equality case gives
$$\frac{x^3+y^3+z^3}{3}=xyz$$ or
$$x^3+y^3+z^3-3xyz=0$$ or
$$x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0$$ or
$$(x+y)^3+z^3-3xy(x+y+z)=0$$ or
$$(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y+z)$$ or
$$(x+y)^2-(x+y)z+z^2-3xy=0$$ or
$$x^2+y^2+z^2-xy-xz-yz=0$$ or
$$2x^2+2y^2+2z^2-2xy-2xz-2yz=0$$ or
$$(x-y)^2+(x-z)^2+(y-z)^2=0$$ or
$$x=y=z,$$
which gives $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{2+2\sin(A)\cos(A)}{\sin(A)(1+\cos(A))}
\end{align}
What should be done from here?
| $$\frac{\sin^2A+(1+\cos(A))^2}{\sin(A)(1+\cos(A))} = \frac{\sin^2A+1+2\cos(A) + \cos^2(A)}{\sin(A)(1+\cos(A))}$$
Now, usse $$\sin^2(A) + \cos^2(A) = 1.$$
Take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Remainder and long division I have been thinking about this. Lets say we take 7 divided by 3, we know the remainder is 1. However, if we let x=7 and x-4=3, and we take x/(x-4), after performing long division the remainder is 4. Why is it not 1?
| Notice:
$7=4*3+(-5) $
$7=3*3+(-2) $
$7=2*3+1$
$7=1*3+4$
$7=0*3+7$
$7=-1*3+10$
Etc. So which one qualifies as "the" remainder? Why is it $1$? Whis it not $4$, $10$ or $-2$?
Well, because the remainder is defined to be at least $0$ but strictly less than the divisor.
The 1 is "the" remainder. The rest are congruences. And there are an infinite number of them.
So $\frac x {x-4}= 1 + \frac 4 {x-4} $. Or in other words $x=1*(x-4)+4$.
But is $4$ "the" remainder? Or is $4$ just a congruence?
The depends on whether $4 < x-4$ or not. If $x=7$ then $4> x-4$ and $4$ is not "the" remainder. It is a congruence.
In particular this is $7=1*3+4$.
... which is $7=1*3+3+1$ which is $7=(1+1)*3+1$ which is $7=2*3+1$
Note: we also have: $x = 2*(x-4)+(4-(x-4))=2*(x-4)+(8-x) $.
If we use that expression, then the remainder is $8-x$ which is $1$ for $x=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
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