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Failure of L’Hospital’s rule to $\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)$? I rewrote the limit
$$\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right) =\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)}{\frac 1x},$$
noting that both the numerator and denominator tend to zero as $x\rightarrow \infty$ so that the conditions for L’Hospital’s rule are fulfilled. However, differentiating top and bottom then yields:
$$\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)}{\frac 1x}=\lim_{x\rightarrow\infty} \frac{\frac{1}{1+\left(\frac{x+1}{x+2}\right)^2}}{-\frac 1{x^2}}=\lim_{x\rightarrow\infty} \frac{-x^2(x+2)^2}{(x+2)^2+(x+1)^2}=\infty,$$
which is incorrect. What went wrong?
| Alternatively, with the trigonometric identity,
$$ \tan^{-1} (a) - \tan^{-1}( b) = \tan^{-1} \left( \dfrac{ a - b}{1 + ab} \right) $$
Set $a = \frac{x+1}{x+2} $ and $ b = 1 , $
$$ \tan^{-1} \left( \dfrac{x+1}{x+2} \right) - \dfrac \pi4 = \tan^{-1} \left( \dfrac1{-2x-3} \right) $$
So the limit is $$ L :=\lim_{x\to\infty} x \cdot \tan^{-1} \left( \dfrac1{-2x-3} \right) = \lim_{x\to\infty}-x \cdot \tan^{-1} \left( \dfrac1{2x+3} \right) $$
Can you finish it off from here?
| {
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"url": "https://math.stackexchange.com/questions/4283482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$ in two ways gives different answers
Evaluate
$$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$$
I started by rationalising followed by dividing numerator and denominator by $x$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow-\infty}\frac{{(\sqrt{4x^2-x}+2x)}{(\sqrt{4x^2-x}-2x)}}{(\sqrt{4x^2-x}-2x)}$
=$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$
=$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$
and we get $\frac{-1}{√4-2}$ i.e.-1/0 form.
The answer however is $\frac{1}{4}$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow\infty}{\sqrt{4x^2+x}-2x}$
=$\lim_{x\rightarrow\infty}\frac{{(\sqrt{4x^2+x}-2x)}{(\sqrt{4x^2+x}+2x)}}{(\sqrt{4x^2+x}+2x)}$
=$\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2+x}+2x}$
=$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$
=$\frac{1}{4}$
What is wrong in first method?
| Hint note that if $x < 0$ then $-2x > 0$, but $-2 < 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let ABCD be a square, E be midpoint of AB. Let F be on BC and G be on CD and EF is parallel to AG. Prove FG is tangent to the incircle of the square.
I tried using OC and finding the slope so that maybe OC would be perpendicular to FG. I know that EF and AC will have the same slope. I also know that I have to use the distance formula but I am really stuck on how to prove a line is tangent to an incircle.
| Line $EF$ has the parametric form $\pmatrix {-1 \\ 0}+\mu\pmatrix {1+x \\ 1}$.
Line $AG$ has the parametric form $\pmatrix {-1 \\ -1}+\lambda\pmatrix {1+x \\ 1}=\pmatrix {1 \\ y}$.
$\therefore\lambda=\dfrac{2}{1+x}$
$\therefore y=-1+\dfrac{2}{1+x}=\dfrac{1-x}{1+x}$
Line $FG$ has the parametric form $\pmatrix {x \\ 1}+\alpha\pmatrix {1-x \\ -1+\dfrac{1-x}{1+x}}=\pmatrix {x \\ 1}+\alpha\pmatrix {1-x \\\dfrac{-2x}{1+x}}$.
The length of line $FG$ is $\sqrt{(1-x)^2 + \left(\dfrac{2x}{1+x}\right)^2} = \dfrac{1+x^2}{1+x}$.
Expand $\triangle FCG$ so that the hypotenuse is $1$ (i.e. multiply all sides by $\dfrac{1+x}{1+x^2}$).
Place the corner $C$ at the origin of the circle, with the $y$ side along the $x$-axis. This touches the circle, and also $\triangle FCG$ at $90^\circ$ (because the rotation is $90^\circ$), and when
$x+a(1-x)=\dfrac{2x}{1+x^2}$
$a(1-x)=\dfrac{x-x^3}{1+x^2}$
$a=\dfrac{x(1+x)}{1+x^2}$
we have that
$1+\dfrac{x(1+x)}{1+x^2}\left(\dfrac{-2x}{1+x}\right)=\dfrac{1-x^2}{1+x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
| A better approach would be to treat the given expression as a quadratic in $x$ and, so we can re-write the expression as $ 4x^2 -2x(y+2) +3(y-1)$. Set it equal to zero, and solve for $x$ using the quadratic formula.
$x = \frac{ 2(y+2) \pm \sqrt{4(y+2)^2-48(y-1)}} {8}$
This gives us the solutions $x=\frac{3}{2} $ and $x=\frac{y-1}{2}$
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Unsure how to solve this quadratic formula problem and how to expand and substitute into the formula Find all values of a and b that will make $(ax+5)^2 = 16x^2 + bx + 25$ true for all $x$.
I believe that this is an equation that needs to be solved using the quadratic formula but am not sure how to expand and substitute into the formula. Any pointers/tips would be helpful.
| The quadratic formula is not needed here
\begin{align*}
(ax+5)^2 &= 16x^2 + bx + 25\\
\implies
b &= a^2 x- 16 x + 10 a \\
\implies \mathbf{(ax+5)^2} &= 16x^2 + \bigg( a^2 x- 16 x + 10 a \bigg)x + 25\\
&=\mathbf{(ax+5)^2}
\end{align*}
We can see from
$\quad (ax+5)^2=(ax+5)^2\quad $
how all values of $\space x\space$ require that $\quad b = a^2 x- 16 x + 10 a $.
On the other hand, expanding both sides and comparing coefficients,
$$a^2 x^2 + 10 a x + 25
=16x^2 + bx + 25\\
\implies\quad a=\pm4, \quad b=10a=\pm40$$
these results are easily verified in a spreadsheet. This is also confirmed by
\begin{align*} b &= a^2 x- 16 x + 10 a\\
&= 4^2 x- 16 x + 10\cdot 4\\
&=40
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show this sequence always is rational number let $\{a_{n}\}$ such $a_{1}=-8$,and such
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that
$$a_{n}\in Q,\forall n\in N^{+}$$
I try let $a_{2}=x$,and for $n=1$, then we have
$$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$
and for $n=2$ I get $a_{3}=-\dfrac{389017}{4913}$ and so on
The first observation
i want use this well known identity: $$a^3+b^3+c^3=3abc ~~~~~~~~~~~~~~~~~~if~~~~~ a+b+c=0$$
let $a=4\sqrt[3]{a_{n}},b=5\sqrt[3]{a_{n+1}},c=-3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$,so
$$64a_{n}+125a_{n+1}-189(a_{n}+1)(a_{n+1}+1)=-180\sqrt[3]{7a_{n}a_{n+1}(a_{n}+1)(a_{n+1}+1)}$$
ADD it by 2021,11.6.PM.18:05
The second observation
and Now I have found this interesting:
if $x,y,p,q,a,b\ge 0$, Hölder's inequality :
$$(x^3+y^3)(p^3+q^3)(a^3+b^3)\ge (xpa+yqb)^3$$
$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$
the condition it's
$$(a_{n}+1)(1+a_{n+1})(4^3+5^3)= (4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}})^3$$
or
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
but this sequence $a_{n}<0$ or $a_{n}>0$,so this Hölder inequality seem can't hold
| This isn't an answer, really more of a musing on the problem, a full expression for $a_{n+1}$ in terms of $a_n$, and a way forward. Assuming that this problem is well posed (that is, $a_n$ can always be calculated from $a_{n-1}$ without ambiguity) then to get this expression, define $a_n=b_n^3$. Then the formula relating the $n$th and $(n+1)$th members (after cubing both sides) becomes
$$0=(4 b_n + 5 b_{n+1})^3 - 27 (7 (b_n^3 + 1) (b_{n_+1}^3 + 1))$$
$$=-189 b_{n+1}^3 b_n^3-125 b_n^3+240 b_{n+1} b_n^2+300 b_{n+1}^2 b_n-64 b_{n+1}^3-189$$
Since we are assuming that this cubic in $b_{n+1}$ has one real solution (otherwise the problem would be ill-posed) we can solve for $b_{n+1}$ in terms of $b_n$:
$$b_{n+1}=3 \sqrt[3]{\frac{7}{2}} \left(\sqrt[3]{-\frac{125 b_n^6+61 b_n^3-64}{\left(189
b_n^3+64\right){}^2}-\frac{23625 b_n^9+27721 b_n^6+8192 b_n^3+4096}{\left(189
b_n^3+64\right){}^3}}+\sqrt[3]{\frac{125 b_n^6+61 b_n^3-64}{\left(189
b_n^3+64\right){}^2}-\frac{23625 b_n^9+27721 b_n^6+8192 b_n^3+4096}{\left(189
b_n^3+64\right){}^3}}\right)-\frac{100 b_n}{-189 b_n^3-64}$$
We then use the above expression to get an expression for $a_{n+1}$ which is
$$a_{n+1}=\frac{\left(100 \sqrt[3]{a_n}-45\ \sqrt[3]{49a_n^2 \left(a_n+1\right)}-48
\sqrt[3]{7\left(a_n+1\right){}^2}\right){}^3}{\left(189
a_n+64\right){}^3}$$
One method to go forward here is to assume that $a_n=r^3$, and then it is sufficient to show that
$$a_n+1=7q^3$$
(for some rational $q$). This is because if this is the case then $a_{n+1}$ can also be written as $s^3$ for some rational $s$. Of course, it is not enough to simply assume $a_n=r^3$ because there are many rational $r$ for which the above equation does not hold (for example $r=1$). It is only for some $r$ (for example $r=-2$) where the above expression evaluates correctly and showing that this will always be the case for $a_n$ seems to be the difficult thing.
| {
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"url": "https://math.stackexchange.com/questions/4298049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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proving $\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} = 2^n\cos^{n}(\frac{x}{2})\left(\cos(\frac{nx}{2})+i\sin(\frac{nx}{2})\right)$ How can I prove the following where i is the imaginary unit.
$$\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} = 2^n\cos^{n}(\frac{x}{2})\left(\cos(\frac{nx}{2})+i\sin(\frac{nx}{2})\right)$$
| $$\left[\left(\cos(x)+1\right)+i\sin(x)\right]^{n} \implies\left(2\cos^{2}(\frac{x}{2})+2i\sin(\frac{x}{2})\cos(\frac{x}{2})\right)^{n}\\ \implies2^n\cos^n(\frac{x}{2})\left(\cos(\frac{x}{2}) + i\sin(\frac{x}{2})\right)^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum likelihood estimator for beta distribution I'm trying to derive MLE for $\theta $ parameter in beta distribution $B(\theta, 2)$:
$$f(x, \theta) = \theta(\theta+1)x^{\theta-1}(1-x) {1}_{\{(0, 1)\}}(x)$$
My strategy will be a standard one - calculate derivative and look for zero of a function.
$$L(X, \theta) = \theta^n (\theta+1)^n (x_1x_2...x_n)^{\theta-1} (1-x_1)(1-x_2)...(1-x_n){1}_{(0,1)}(x_1){1}_{(0,1)}(x_2)...{1}_{(0,1)}(x_n)$$
$$\ln L(X, \theta) = \ln(\theta^n) + \ln(\theta+1)^n + \ln(x_1...x_n)^{\theta-1}+\ln((1-x_1)...(1-x_n)) + \ln({1}_{(0,1)}(x_1)...,{1}_{(0,1)}(x_n))$$
$$\ln L(X,\theta) = n\ln(\theta) + (\theta - 1)\ln(x1,...,x_n) + \ln((1-x_1)...(1-x_n)) + \ln({1}_{(0,1)}(x_1)...,{1}_{(0,1)}(x_n))$$
$$\frac{\partial \ln L(X, \theta)}{\partial \theta} = \frac{n}{\theta} + \frac{n}{\theta} + \ln(x_1,...x_n)$$
$$\frac{\partial \ln L(X, \theta)}{\partial \theta} = 0 \Leftrightarrow \frac{n}{\theta} + \frac{n}{\theta + 1} + \ln(x1 \cdot x2 \cdot ... \cdot x_n)= 0$$
$$(\theta + 1)n + \theta n + \theta(\theta + 1)\ln(x1 \cdot x2 \cdot ... \cdot x_n) = 0$$
$$\theta^2 \ln(x1 \cdot x2 \cdot ... \cdot x_n) + \theta (\ln(x1 \cdot x2 \cdot ... \cdot x_n) + 2n) + n =0$$
$$\theta^2 \cdot \frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n} + \theta (\frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n} + 2) + 1 = 0$$
For simplicity denote $a_n:= \frac{\ln(x1 \cdot x2 \cdot ... \cdot x_n)}{n}$
$$\theta^2a_n + \theta(a_n + 2) + 1 = 0$$
$$\Delta = (a_n + 2)^2 - 4a_n = a_n^2 + 4 \Leftrightarrow \sqrt{\Delta} = \sqrt{a_n^2 + 4}$$
We have two solutions:
$$\theta_1 = \frac{-a_n - 2 - \sqrt{a_n^2 + 4}}{2a_n}$$
$$\theta_2 = \frac{-a_n - 2 + \sqrt{a_n^2 + 4}}{2a_n}$$
I know that $a_n < 0$ since $X_1, X_2,...,X_n \in [0, 1] \Rightarrow \frac 1 n \sum_{i = 1}^n \ln(X_i) < 0$.
I wanted to show somehow that one of them is meaningless - for example by showing that $\theta_1$ or $\theta_2$ only take negative values but I don't think its true.
My question is: Do I have two maximum likelihood estimators or I should get rid of one?
| For $a_n < 0$, $$\frac{-a_n - 2 \pm \sqrt{a_n^2 + 4}}{2a_n} = -\frac{1}{2} - \frac{1}{a_n} \mp \frac{1}{2}\sqrt{1 + \frac{4}{a_n^2}}.$$ Notice the reversal in sign due to the fact that $a_n < 0$ implies $a_n = -\sqrt{a_n^2}$. Therefore, as $a_n \to -\infty$, we have $\theta_1 \to -\frac{1}{2} + \frac{1}{2} = 0$, but $\theta_2 \to -\frac{1}{2} - \frac{1}{2} = -1$. This would suggest $\theta_1$ is the correct root.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to evaluate $232^2-62^2\times14$ by factoring or using identities or...? The expression $232^2-62^2\times14$ can be calculated directly ($53824-3844
\times14=8$). But is it possible to evaluate it for example by factoring or using identities?
Here is what I have tried,
$$(58\times4)^2-62^2\times14=58^2\times16-62^2\times14=(60-2)^2(15+1)-(60+2)^2(15-1)$$
Or
$$58^2\times16-62^2\times14=29^2\times64-31^2\times56=(30-1)^2\times8^2-(30+1)^2(8\times7)$$
But I can't see an elegant way to get $8$ from either of the calculatins.
| The numbers involved are easy to factor, so take out the common factors to get
$$232^2-62^2\times14=(2^3\times29)^2-(2\times31)^2\times(2\times7)
=2^3\times(2^3\times29^2-7\times31^2).$$
The last factor looks similar to a difference of two squares; note that
\begin{eqnarray*}
2^3\times29^2-2^3\times31^2&=&2^3\times(29^2-31^2)\\
&=&2^3\times(29+31)\times(29-31)\\
&=&-2^4\times60\\
&=&-960,
\end{eqnarray*}
and so
$$2^3\times29^2-7\times31^2=-960+31^2=1.$$
It follows that
$$232^2-62^2\times14=2^3\times1=8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302120",
"timestamp": "2023-03-29T00:00:00",
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Can two circles intersect each other at right angles, such that one circle passes through the center of the other circle? More precisely, do there exist two intersecting circles such that the tangent lines at the intersection points make an angle of 90 degrees, and one of the circles passes through the center of the other? I think the answer is no, and that this can only be the case if the circle passing through the center of the other circle is actually a line. I'm not sure how to show this though. Can someone give a proof or a counterexample?
| Let
\begin{align}
\vec{r}_1(\alpha)&=R\left(\cos \alpha, \sin \alpha\right)\\
\vec{r}_2(\beta)&=\lambda R\left(1+\cos \beta, \sin \beta\right)
\end{align}
be equations of the given circles where $\lambda$ and $R$ are both positive. The parameters $\alpha$ and $\beta$ span from $0$ to $2\pi$.
Assume their intersections occur at $\alpha = a$ and $\beta = b$, we have
\begin{cases}
R\cos a = \lambda R \left(\cos b + 1\right)\\
R\sin a = \lambda R \sin b
\end{cases}
With some relatively easy algebra manipulations, we have the solution for the upper intersection as follows.
\begin{cases}
\cos a = \frac{1}{2\lambda}\\
\cos b = \frac{1-2\lambda^2}{2\lambda^2}\\
\sin a=\frac{\sqrt{4\lambda^2 -1}}{2\lambda}\\
\sin b =\frac{\sqrt{4\lambda^2 -1}}{2\lambda^2}
\end{cases}
Next, find the direction vector of the tangents as follows.
\begin{cases}
\vec{t}_1=\left.\frac{\mathrm{d}\vec{r}_1}{\mathrm{d}\alpha}\right|_{\alpha=a}=R\left(-\sin a, \cos a\right)\\
\vec{t}_2=\left.\frac{\mathrm{d}\vec{r}_2}{\mathrm{d}\beta}\right|_{\beta=b}=\lambda R\left(-\sin b, \cos b\right)
\end{cases}
Now find the angle between the tangents.
\begin{align}
\vec{t}_1 \cdot \vec{t}_2 &= |\vec{t}_1| |\vec{t}_2| \cos \theta \\
\lambda R^2 \left(\cos a \cos b + \sin a \sin b \right)
&= \lambda R^2 \cos \theta\\
\cos \theta
&=\cos a \cos b + \sin a \sin b\\
&= \frac{1}{2\lambda} \times \frac{1-2\lambda^2}{2\lambda^2} + \frac{\sqrt{4\lambda^2 -1}}{2\lambda}\times \frac{\sqrt{4\lambda^2 -1}}{2\lambda^2}\\
&= \frac{1-2\lambda^2+4\lambda^2-1}{4\lambda^3}\\
&=\frac{2\lambda^2}{4\lambda^3}\\
&=\frac{1}{2\lambda}
\end{align}
They can be perpendicular each other (or $\theta=90^\circ$) if one of the circles has an infinite radius ($\lambda \to \infty$).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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In how many distinguishably different ways can a pair of indistinguishable dice come up? The answer is $21$, but why doesn't $(6\times 6)/2$ work? It accounts for the overlap where $2-4$ and $4-2$ are the same cases.
| Suppose we have a pair of indistinguishable white dice. There are $\binom{6}{2}$ outcomes in which they show different numbers and $\binom{6}{1}$ outcomes in which they show the same number. Hence, there are a total of
$$\binom{6}{2} + \binom{6}{1} = 21$$
outcomes when two indistinguishable dice are thrown.
Suppose we now paint one of the dice red. If there are two different outcomes, there are two ways to do this since we can either paint the higher number or the lower number red. If the outcomes are the same, it does not matter which die is painted red since both dice will still show the same number. Thus, if we make the dice distinguishable, we obtain
$$2\binom{6}{2} + \binom{6}{1} = 36$$
possible outcomes as we should expect since there are six possible outcomes for each die.
Thus, the problem with your approach, as user525966 pointed out in the comments that it only makes sense to divide the number of outcomes on distinct dice by $2$ when the two outcomes are different. If one die is red and the other is white, there are $6 \cdot 5 = 30$ such outcomes. On the other hand, if the distinct dice show the same outcome, they still would show the same outcome if we painted both dice white. Thus, if we start with the $36 = 6^2$ possible outcomes for two distinct dice, we can obtain the number of possible outcomes for indistinguishable dice by dividing the number of outcomes in which the two dice display different outcomes by $2$ and then add the number of outcomes in which both dice display the same number.
$$\frac{6 \cdot 5}{2} + 6 = \binom{6}{2} + \binom{6}{1} = 21$$
| {
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To determine the integration of $ \int_{0}^{+\infty} \exp\!\Big(-\Big(\frac{ax^2+bx+c}{gx+h}\Big)\Big) dx$. What is the integration of the following function:
$$ \int_\nolimits{0}^{+\infty} \exp\!\bigg(-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) \bigg)dx.$$
What I have done is as follows:
Here, $\kappa=c-\Big(\frac{bg-ah}{g^2}\Big)h$.
\begin{align*}\implies & \int_{0}^{+\infty} \exp\!\bigg(-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg)\bigg) dx \\ =& \int_{0}^{+\infty}\exp\!\bigg(-\bigg(\frac{a}{g}x+\frac{b g-a h}{g^2}\bigg)\bigg)\exp\!\bigg(-\frac{\kappa}{gx+h}\bigg)dx.\end{align*}
I am finding it difficult to proceed further from here. Is this approach correct or is there any other intuitive way to solve this problem? Thanks in advance for the help.
| Defining the incomplete Bessel function as
$$
K_\nu(x,y) = \int_1^{\infty} t^{-\left(\nu+1 \right)}e^{-\left(xt + \frac{y}{t} \right)}\, \mathrm{d}t
$$
We get
\begin{align}
\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx+h}} \, \mathrm{d}x & \overset{\color{purple}{x = \frac{h}{g}(t - 1)}}{=}\frac{h}{g}e^{\frac{2ah}{g^2}-\frac{b}{g}} \int_{1}^{\infty}e^{-\left[\left( \frac{ah}{g^2}\right)t + \left(\frac{ah}{g^2}- \frac{b}{g} + \frac{c}{h} \right)\frac{1}{t} \right]}\, \mathrm{d}t\\
& =\boxed{\frac{h}{g}e^{\frac{2ah}{g^2}-\frac{b}{g}} K_{-1}\left(\frac{ah}{g^2},\frac{ah}{g^2}- \frac{b}{g} + \frac{c}{h}\right)}
\end{align}
For the special case of $h=0$ you can obtain a closed-form in terms of a Modified Bessel function.
Notice that
$$
\frac{ax^2+bx+c}{gx} = \frac{a}{g}x + \frac{c}{gx} + \frac{b}{g}
$$
Now, for the integral to converge we require that the $\frac{ax^2+bx+c}{gx} \to + \infty$ when $x \to \infty$. This condition is met when $\frac{a}{g}>0$, so we'll do the rest of the analysis assuming this condition holds true. Additionally, we don't want $\frac{ax^2+bx+c}{gx}\to -\infty$ when $x \to 0^+$ since this would also make the integral divergent. To avoid this we also require that $\color{Purple}{\frac{c}{g}>0}$. With this we see that
$$
\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx}} \, \mathrm{d}x = e^{-\frac{b}{
g}}\int_{0}^{\infty} e^{-\left(\frac{a}{g}x + \frac{c}{gx} \right) } \, \mathrm{d}x \overset{\color{Purple}{u = \frac{g}{c}x}}{=}\frac{c}{g}e^{-\frac{b}{g}} \int_{0}^{\color{Purple}{+\infty}} e^{-\left(\frac{\left(\color{green}{\frac{2}{|g|}\sqrt{ac}}\right)^2}{4}u + \frac{1}{u}\right)} \, \mathrm{d}u
$$
Now, from the Digital Library of Mathematical Functions we know that
$$
K_{\nu}(x) = \frac{1}{2}\left( \frac{1}{2}x\right)^{\nu}\int_{0}^{\infty} e^{-\left( \frac{x^2}{4t} + t\right)} \frac{1}{t^{\nu+1}} \, \mathrm{d}t, \qquad \forall x \in \mathbb{R}
$$
which means that
$$
\frac{4}{x} K_{1}(x)=\int_{0}^{\infty} e^{-\left( \frac{x^2}{4t} + t\right)} \frac{1}{t^{2}} \, \mathrm{d}t \overset{\color{blue}{u = \frac{1}{t}}}{=} \int_{0}^{\infty} e^{-\left(\frac{\color{green}{x}^2}{4} u + \frac{1}{u} \right)} \, \mathrm{d}u
$$
So combining everything, if $\frac{a}{g}, \frac{c}{g} >0$ then:
$$
\boxed{\int_{0}^{\infty} e^{-\frac{ax^2+bx+c}{gx}} \, \mathrm{d}x = 2\sqrt{\frac{c}{a}}e^{-\frac{b}{g}}K_{1}\left(\frac{2}{|g|}\sqrt{ac}\right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306458",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Write $x^k - y^k$ as the product of two factors. Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \cdots + xy^{k-2}}_{k-2} +y^{k-1})$
It's not exact what the middle terms should be, so I'm not sure how to write the expression for the second factor. I think I may need to use the parity of $k$ and some mention of mirroring, though mirroring sounds wrong.
| Here is a variation without using induction. We want to use the sigma symbol and start with an example for small $n=4$ to better see what's going on.
We have
\begin{align*}
x^4-y^4&=(x-y)(x^3+x^2y+xy^2+y^3)\\
&=(x-y)(x^3y^0+x^2y^1+x^1y^2+x^0y^3)\tag{1}\\
&=(x-y)\sum_{k=0}^3 x^{3-k}y^k
\end{align*}
Since we can write $x^3=x^3y^0$ and $y^3=x^0y^3$ each term in (1) has a representation $x^{3-k}y^k$, $0\leq k\leq 3$ which can be conveniently written using the sigma symbol $\sum$.
For the general case with integral $n>0$ we have
\begin{align*}
\color{blue}{x^n-y^n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots + xy^{n-2}+y^{n-1})\\
&=(x-y)(x^{n-1}y^0+x^{n-2}y^1+\cdots + x^1y^{n-2}+x^0y^{n-1})\\
&\,\,\color{blue}{=(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}\tag{2}
\end{align*}
We prove (2) algebraically. We obtain for integral $n>0$ by starting with the right-hand side of (2)
\begin{align*}
\color{blue}{(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}&=x\sum_{k=0}^{n-1} x^{n-1-k}y^k-y\sum_{k=0}^{n-1} x^{n-1-k}y^k\tag{3.1}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=0}^{n-1} x^{n-1-k}y^{k+1}\tag{3.2}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=1}^{n} x^{n-k}y^{k}\tag{3.3}\\
&\,\,\color{blue}{=x^n-y^n}
\end{align*}
and the claim follows.
Comment:
*
*In (3.1) we multiply out $(x-y)\sum \Box=x\sum \Box - y\sum\Box$.
*In (3.2) we simplify by multiplying with $x$ resp. $y$. Note, we have $x\sum x^{n-1-k}y^k=\sum x^{1+(n-1-k)}y^k=\sum x^{n-k}y^k$.
*In (3.3) we shift the index $k$ in the right-hand side by one. We so have lower bound $k=1$ and upper bound $k=n$. We compensate this shift by substituting $k\to k-1$ within the sum. We now have the convenient situation that we have a telescoping sum with alike terms in the left-hand sum and the right-hand sum for $k=1,\ldots,n-1$, so that everything cancels out besides $x^n-y^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2\leqq1 \; , 2x^2+y^2-2x=0$ i'm trying to find the volume of the solid which is the intersection of the solid sphere $x^2+y^2+z^2\leqq1 \; , 2x^2+y^2-2x=0$
My attempt:
I tried with cylindrical coordinates $$x=rcos(\phi)\;,y=rsin(\phi)\;,z=z$$
$$(rcos(\phi))^2+(rsin(\phi))^2+z^2=r^2+z^2 \leqq 1 \Rightarrow -\sqrt{(1-z^2)} \leqq r \leqq \sqrt{(1-z^2)} $$
I'm stuck here, because i don't know how to use $2x^2+y^2-2x=0$ to find the other limits of integration.
| Well, first of all, the second equation describes a shifted elliptical cylinder. From $2x^2+y^2-2x=0$, divide by $2$ and complete the square, yielding
$$
(x-1/2)^2+y^2/2=1/4,
$$
which writing in standard form yields
$$
\frac{(x-1/2)^2}{1/4}+\frac{y^2}{1/2}=1.
$$
Hence, the natural $(x,y)$ parameterization is
\begin{align}
x(r,\theta)&=\frac12+\frac{r}{4}\cos\theta \\
y(r,\theta)&=\frac{r}{2}\sin\theta
\end{align}
with $0\leq r \leq 1$ and $0\leq \theta \leq 2\pi$. Then calculate the Jacobian and produce a triple integral. Using the two equations, you should be able to solve for the $z$ bounds in terms of $x$ and $y$ (and therefore, in terms of $r$ and $\theta$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdot i.$
We now need to find the sum $S=x_1+x_2$ and the product $P=x_1\cdot x_2$ to form an equation $x^2-Sx+P=0$ which has the solutions $x_1$ and $x_2$.
\begin{align*}
sum = x_1+x_2&=6m \tag{1}\\
\iff(r+n\cdot i)+(r-n\cdot i)=2r&=6m \tag{2}\\
\implies r&=3m \tag{3}
\end{align*}
The $product = x_1\cdot x_2=(r+n\cdot i)(r-n\cdot i)=r^2+n^2 = (\cdots).\quad $
From here I don't know what relation to find between $\Delta$ and $x_1\cdot x_2.$
I tried: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$
| You already concluded from the negative discriminant that the zeroes of the quadratic polynomial must be a "conjugate pair" $ \ \alpha \ + \ \beta·i \ \ . $ The difference of the zeroes is
$$ x_1 \ - \ x_2 \ \ = \ \ \frac{\sqrt{\Delta}}{a} \ \ \Rightarrow \ \ 2·\beta·i \ \ = \ \ \frac{\sqrt{-36}}{a} \ \ = \ \ \frac{6i}{a} \ \ \Rightarrow \ \ \beta \ = \ \frac{3}{a} \ \ . $$
The sum of the zeroes is given as
$$ x_1 \ + \ x_2 \ \ = \ \ 2·\alpha \ \ = \ \ 6m \ \ \Rightarrow \ \ \alpha \ = \ 3m \ \ $$
(as you observed).
If we choose to construct the corresponding monic polynomial, we would set $ \ a = 1 \ \ , $ producing the conjugate pair of zeroes $ \ x_{1,2} \ = \ 3m \ \pm \ 3i \ \ . $ Multiplying out the factors $ \ (x \ - \ 3m \ - \ 3i) \ · \ (x \ - \ 3m \ + \ 3i) \ \ $ gives us the polynomial shown in user2661923's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What's the measure of the radius of the biggest circle below? For reference: In figure A is the tangency point::
LE=2(TE),
$\overset{\LARGE{\frown}}{AN}=60^o$
$\frac{(TE)^2}{R-r}= 10$
Calculate R.(Answer:80)
My progress
$\triangle AON: \angle AON = 60^o, AO=r, ON=r \implies\\
AN =r \therefore \triangle AON(equilateral)$
$\triangle ATL(right): (LN+r)^2=(3x)^2+(r+OT)^2 \implies\\
LN^2+2LNr+r^2 = 9x^2+r^2+2rOT+OT^2 \rightarrow\\
LN^2+2LNr=9x^2+2rOT+OT^2$
....???
I"freeze" here...
Can I say that A, O. T and O1 are collinear?
| First note that $A, O $ and $O_1$ are collinear and as $A, O$ and $T$ are collinear, all four points $A, O, T$ and $O_1$ must be collinear. As $\angle O_1AL = 60^\circ$ and $O_1A = O_1L = R$, $\triangle O_1AL$ is equilateral with side length $R$.
Now there are multiple ways to get to the answer. One approach would be,
$LT = \frac{R \sqrt 3}{2} \implies TE = \frac{R}{2 \sqrt 3}$
$OT = \frac{R}{2} - r$
$OT^2 + TE^2 = OE^2 \implies \left(\frac{R}{2} - r\right)^2 + \left(\frac{R}{2 \sqrt3}\right)^2 = r^2$
And we obtain $R = 3r$
Now we also know that $TE^2 = \frac{R^2}{12} = 10(R-r) = \frac{20 R}{3}$.
$\therefore R = 80$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interval $[0,2\pi]$ is equal to_____
My approach is as follow
$a = \sin x - 1;b = \cos x - 1;c = \sin x$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
$ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right)$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} \Rightarrow 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right) = 0$
$ \Rightarrow 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
How do we approach from here
| Starting with: $ 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
$$ ab(c + a + b) + bc(a+b+c) + ac\left( {a + c} \right) = 0$$
$$ b(c + a + b)(a+c) + ac\left( {a + c} \right) = 0$$
$$ (a+c)\bigl(b(c + a + b)+ac\bigr) = 0$$
$$ (a+c)(bc + ab + b^2+ac) = 0$$
$$ (a+c)\bigl(c(a+b) + b(a+b)\bigr) = 0$$
$$ (a + b)(a + c)(b + c) = 0$$
| {
"language": "en",
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"source": "stackexchange",
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Proving $\varepsilon$-$\delta$ for multivariables I have a pracitce question that I have been troubling proving: $\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$ using the $\varepsilon$-$\delta$ definition. But here is my attempt of the solution.
If $\displaystyle \lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$, then we say that $\forall \varepsilon > 0 \exists \delta > 0 \forall(x, y) \in \mathbb{R}^2$, if $\sqrt{(x - 2)^2 + (y - 1)^2} < \delta$, then we want to show that $\displaystyle \left|\dfrac{x^2 - 2xy}{x^2 - 4y^2} - \dfrac{1}{2}\right| < \varepsilon$.
First, I tried to apply some algebraic manipulation for the given expression:
\begin{equation}
\left|\dfrac{x^2 - 2xy}{x^2 - 4y^2} - \dfrac{1}{2}\right| = \left|\dfrac{x}{x + 2y} - \dfrac{1}{2}\right| = \left|\dfrac{2x - (x + 2y)}{2(x + 2y)}\right| = \left|\dfrac{x - 2y}{2(x + 2y)}\right| = \dfrac{|x - 2y|}{2|x + 2y|}
\end{equation}
I am a little confused on how to proceed from here, so I would like some assistance. Thanks
|
$\lim_{(x, y) \to (2, 1)} \dfrac{x^2 - 2xy}{x^2 - 4y^2} = \dfrac{1}{2}$
Actually, the above assertion is false, for the following reason. In any (small) neighborhood of radius $\delta$ around $(2,1)$, you will find $(x,y) \neq (2,1)$ such that $x - 2y = 0$. If you examine the original fraction that is being interrogated:
$\displaystyle \frac{x^2 - 2xy}{x^2 - 4y^2} = \frac{x(x - 2y)}{(x + 2y)(x - 2y)}$
The assertion that the above fraction equates to
$\displaystyle \frac{x}{x + 2y}$ is false, for any $(x,y) \neq (2,1)$ in the small neighborhood, where $x = 2y$.
Therefore, the only way to proceed is to add the stipulation that with
$\displaystyle f(x,y) = \frac{x^2 - 2xy}{x^2 - 4y^2}$,
you are only being asked to consider the limit as $(x,y) \to (2,1)$ for those specific values of $(x,y)$ such that $f(x,y)$ is well defined. For the present problem, this signifies the constraints of :
*
*$0 < |(x,y) - (2,1)| < \delta$
*$(x - 2y) \neq 0.$
Further, it is understood that although both the numerator and denominator above are going to $(0)$ as $(x,y) \to (2,1),~$ an $\varepsilon, \delta$ approach is required, rather than (for example) L'Hopital's Rule.
With the added stipulation on $f(x,y)$ discussed above, I agree with your algebraic manipulation and will use that as a starting point.
This means that if
$~\displaystyle \frac {|x - 2y|}{|2(x + 2y)|} < \varepsilon$
then $~\displaystyle \left| ~\frac{x^2 - 2xy}{x^2 - 4y^2} - \frac{1}{2} ~\right| < \varepsilon$.
$\underline{\textbf{Preliminary Results}}$
*
*In general, if you have a function $f(r)$, and a positive number $A$,
then the constraint $|f(r)| < A$ is equivalent to
$-A < f(r) < A.$
*The constraint that $0 < |(x,y) - (2,1)| < \delta$,
implies that all three of the following constraints are satisfied:
$|x - 2| < \delta$
$|y - 1| < \delta$
$(x,y) \neq (2,1)$.
*As indicated, the constraint to be satisfied is
$~\displaystyle \frac {|x - 2y|}{|2(x + 2y)|} < \varepsilon$.
*To make things very easy on myself, I will impose the artificial constraint that
$\delta \leq (1/10)$.
Focusing on #2 above, you have that
*
*$-\delta < (x - 2) < \delta \implies (2 - \delta) < x < (2 + \delta).$
*Similarly, $(1 - \delta) < y < (1 + \delta)$.
Examining the numerator in #3 above, you have that the following lower and upper bounds (not necessarily the tightest bounds possible) are established:
$(2 - \delta) - 2(1 + \delta) < (x - 2y).$
This simplifies to $-3\delta < (x - 2y)$.
$(2 + \delta) - 2(1 - \delta) > (x - 2y).$
This simplifies to $3\delta > (x - 2y)$.
Examining the denominator in #3 above, you have that the following lower and upper bounds (not necessarily the tightest bounds possible) are established:
$2[(2 - \delta) + 2(1 - \delta)] < 2(x + 2y)$.
This simplifies to $8 - 6\delta < 2(x + 2y)$.
$2[(2 + \delta) + 2(1 + \delta)] > 2(x + 2y)$.
This simplifies to $8 + 6\delta < 2(x + 2y)$.
Putting this all together, you have that :
*
*$-3\delta < (x - 2y) < 3\delta$
*$8 - 6\delta < 2(x + 2y) < 8 + 6\delta$.
However, because I have imposed the artificial constraint that $\delta \leq (1/10)$, I can immediately conclude that
$7 < 2(x + 2y) < 9.$
I can express a lower bound for the fraction in #3 above by using a lower bound for the numerator and (since the numerator is negative), a lower bound for the denominator.
Therefore, I know that
$\displaystyle \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)}.$
Similarly, I know that
$\displaystyle \frac{3\delta}{7} > \frac{x - 2y}{2(x + 2y)}.$
This implies that whenever constraints #2 and #4 above are both satisfied, that:
$$ \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)} < \frac{3\delta}{7}.\tag1 $$
It is desired that (1) above should imply that constraint #3 above is satisfied. Setting $\delta = \varepsilon$ would imply that
$$ -\varepsilon = - \delta < \frac{-3\delta}{7} < \frac{x - 2y}{2(x + 2y)} < \frac{3\delta}{7} < \delta = \varepsilon. $$
Therefore, a workable final specification is :
$\displaystyle \delta = \min\left[ ~\varepsilon, ~\frac{1}{10} ~\right].$
| {
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Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both sides of the equation and continue from there, I tried by solving the equation:
$(x^3+x)(ax^3+bx^2+cx+d) = 1$
But the solution I get is that $a=0,b=1,c=1,d=0$ which doesn't produce the inverse element.
What is the inverse element of $(1010)$ in this field and how can I arrive at that solution?
Any help is greatly appreciated.
| You can use the Extended Euclidean algorithm to find the inverse of $x^3+x$ modulo $x^4+ x^3+1$ in $GF(2)[x]$
Given $a$ and $b$ you will find $u$ and $v$ such that
$$au+bv=\gcd(a,b)$$
in our example $\gcd(a,b)=1$ because $a$ is an irreducible polynomial and $\deg(a)>\deg(b)$. So we have
$$bv\equiv 1\pmod(a)$$
and $v$ is the multiplicative inverse.
We have to do some polynomial long divisions to find the greatest common divisor. From these divisions we can construct the representation of the greatest common divisor.
(x^4+ x^3+1) : (x^3+x) = x+1
x^4 +^2
------------
x^3 + x^2 +1
x^3 + x
------------
x^2+x+1
so we have
$$ x^4+ x^3+1=(x^3+x) ( x+1)+ x^2+x+1$$
and further
$$x^2+x+1=(x^4+ x^3+1) + (x^3+x) ( x+1)$$
we continue this process and get
(x^3+x) : (x^2+x+1) = x + 1
x^3+x^2+x
----------
x^2
x^2+x+1
-------
x+1
$$ (x^3+x) = (x^2+x+1)(x+1) +(x+1) $$
$$x+1 = (x^3+x) + (x^2+x+1)(x+1)$$
x^2+x+1 : x+1 = x
x^2+x
-------
1
$$x^2+x+1 =( x+1 )x +1$$
and so, by substituting the previous results
$$1= (x^2+x+1) +( x+1 )x \\
= (x^2+x+1)+ ( (x^3+x) + (x^2+x+1)(x+1))x\\
=(x^2+x+1)(x^2+x+1)+(x^3+x)x\\
=((x^4+ x^3+1) + (x^3+x) ( x+1))(x^2+x+1)+(x^3+x)x\\
=(x^4+ x^3+1)(x^2+x+1) + (x^3+x)(( x+1))(x^2+x+1)+x)\\
=(x^4+ x^3+1)( x^2+x+1)+ (x^3+x)(x^3+x+1)$$
So
$$(x^3+x)(x^3+x+1)=1 \pmod{x^4+ x^3+1}$$
and the inverse of $1010$ in $GF(2^4)$ is $1011$.
Another way to calculate the inverse, at least for such small fields, to use the fact that
$$r\cdot r^{\text{ord} (r) -1}\equiv r^{\text{ord}( r)} \equiv 1$$
in a field. $\text{ord}(r)$ is a divisor of the order of the multiplicative group of the field. In our case this means
$$\text{ord}(x^3+x) \in \{3,5,15\}$$
and so
$$(x^3+x)^{-1} in \{(x^3+x)^2, (x^3+x)^4, (x^3+x)^{14}\}$$
So first we calculate $$(x^3+x)^2=x^6+x^2=x^3+x+1$$ and further
$$(x^3+x)^3=(x^3+x+1)(x^3+x)=1$$ and we have found the inverse $x^3+x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use Principle of Mathematical Induction to show that for all $n\:\in \mathbb{N}$, $a_n=2^{n+2}\cdot 5^{2n+1}+3^{n+2}\cdot 2^{2n+1}$ is divisible by 19 Here's what I did:
Base Case: $n = 1$, $2^3*5^3+3^3*2^3$ = $1216$ = $19(64)$
Induction Hypothesis (IH): $a_k=2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}$ for some $k\:\in \mathbb{N}$.
Inductive Step: $a_{k+1}=2^{\left(k+1\right)+2}\cdot \:5^{2\left(k+1\right)+1}+3^{\left(k+1\right)+2}\cdot \:2^{2\left(k+1\right)+1}$
$a_{k+1}=2^{\left(k+2\right)}\cdot 2\cdot \:5^{2k+1}\cdot 5^2+3^{k+2}\cdot 3\cdot \:2^{2k+1}\cdot 2^2$
$a_{k+1}=2^{k+2}\:\cdot 5^{2k+1}\cdot 50+3^{k+2}\cdot 2^{2k+1}\cdot 12$
How do I prove that the latter statement is divisible by 19?
(I've tried to set up the equation in a way where I know that $2^{k+2}\:\cdot 5^{2k+1}$ and $3^{k+2}\cdot 2^{2k+1}$ is divisible by 19. I'm not sure about the 50 and the 12.)
| You're almost there! Hint for the finish:
$$
2^{k+2}\cdot 5^{2k+1}\cdot 50+3^{k+2}\cdot 2^{2k+1}\cdot 12\\
= 12\left(2^{k+2}\cdot 5^{2k+1}+3^{k+2}\cdot 2^{2k+1}\right) + 38\left(2^{k+2}\cdot 5^{2k+1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding the sum of geometric progression Evaluate the sum:
$$\sum_{x=0}^\infty x(x-1) {2+x \choose x}(0.008)(0.8)^x $$
I was able to make this into:
$$0.004\sum_{x=0}^\infty x(x-1) (x+1)(x+2)(0.8)^x $$
Let $x=n-2$ then $n=x+2$:
$$0.004\sum_{n=2}^\infty n(n-1)(n-2)(n-3)(0.8)^{n-2} $$
$$0.004\sum_{n=4}^\infty (n)_3~(0.8)^{n-2} $$
Let $j=n-4$ then $n=j+4$:
$$0.004\sum_{j=0}^\infty (j+4)_3~(0.8)^{j+2} $$
$$0.00256\sum_{j=0}^\infty (j+4)_3~(0.8)^{j} $$
I thought I could just use the formula for geometric sum on this but I know the answer's $192 $ and right now my solution's pretty far off.
| $$(x-1) x \binom{x+2}{x}=\frac{1}{2} (x-1) x (x+1) (x+2)$$ Rewriting
$$\frac{1}{2}(x-1) x (x+1) (x+2)= \frac{1}{2}x (x - 1) (x - 2) (x - 3) + 4 x (x - 1) (x - 2) + 6 x (x - 1) $$
$$S=\sum_{x=0}^\infty x(x-1) {2+x \choose x}p^x=\frac{1}{2}p^4\sum_{x=0}^\infty x (x - 1) (x - 2) (x - 3)p^{x-4}+$$ $$4p^3\sum_{x=0}^\infty x (x - 1) (x - 2) p^{x-3}+6p^2\sum_{x=0}^\infty x (x - 1) p^{x-2}$$
$$S=\frac{1}{2}p^4\Bigg[\sum_{x=0}^\infty p^x\Bigg]''''+4p^3\Bigg[\sum_{x=0}^\infty p^x\Bigg]'''+6p^2\Bigg[\sum_{x=0}^\infty p^x\Bigg]''$$
Since
$$\sum_{x=0}^\infty p^x=\frac 1 {1-p}$$ just compute the second, third and fourth derivatives and get
$$S=\frac{12 p^2}{(1-p)^5}$$ If $p=0.8$ then $S=24000$ which, multiplied by $0.008$ gives $192$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
$$G_{3} = \sqrt[3]{a_{1}×a_{2}×a_{3}},$$
$$A_{3}\ge G_{3}.$$
However, I am not sure how to do this.
I have tried to substract one side from the other:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} - \frac{3}{a+b+c} > 0.$$
I have also tried to reverse both sides of inequality.
I would appreciate if you could just give me a hint.
| Since $a > 0$, $b > 0$, and $c > 0$, we have
$$ 0 < a +b < a+b + c, $$
and therefore upon dividing both sides of this inequality by $(a+b)(a+b+c) > 0$, we get
$$
0 < \frac{1}{a+b+c} < \frac{1}{a+b},
$$
which implies
$$
\frac{1}{a+b} > \frac{1}{a+b+c}. \tag{1}
$$
Similarly, we have the inequalities
$$
\frac{1}{b+c} > \frac{1}{a+b+c} \tag{2}
$$
and
$$
\frac{1}{c+a} > \frac{1}{a+b+c}. \tag{3}
$$
Using (1), (2), and (3), we obtain
\begin{align}
\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a } &> \frac{1}{a + b + c} + \frac{1}{a + b + c} + \frac{1}{a + b + c} \\
&= \frac{3}{a + b + c},
\end{align}
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the area of this function. I want to calculate the region bounded by $x^2-2xy+y^2+x+y=0$ and $x+y+2=0$.
I have drawn a sketch of this region in Geogebra:
I noticed I can use $\iint1dD$ to solve for the integral. I can apply the mapping: $(x^2-2xy+y^2)=(x-y)^2=u$, and $x+y=v$, so $x^2-2xy+y^2+x+y=u+v=0$ and $x+y+2=v+2=0$.
This is the new equation I have. Then I can evaluate $\int_{0}^{2}\int_{-x}^{0}1dA$ yes?
Or maybe I need to add the Jacobian as well...
| The answer is: $\frac{8\sqrt{2}}{3}$.
Explanation:
We want to calculate the area of the region $D$ bounded by $$x^{2}-2xy+y^{2}+x+y=0\quad \text{and}\quad x+y+2=0.$$
The area of the region $D$ is given by $$\color{red}{\boxed{A(D)=\iint_{D}{\rm d}A}}.$$
Setting the change of variable,
$$u\longmapsto u(x,y)=(x-y)^{2} \quad \text{and}\quad v\longmapsto v(x,y)=x+y$$
so the area of the new region $D^{*}$ given by the change of variable is $$\color{red}{\boxed{A(D^{*})=\iint_{D^{*}}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|{\rm d}A^{*}}}.$$
We have that, $$x^{2}-2xy+y^{2}+x+y=0 \iff u+v=0$$ and
$$ x+y+2=0 \iff v+2=0.$$
Then,
$$\frac{\partial (x,y)}{\partial (u,v)}=\det \begin{bmatrix} x_{u} & x_{v}\\ y_{u} & y_{v}\end{bmatrix}=\det \begin{bmatrix}-\frac{1}{2\sqrt{u}} & 1\\ \frac{1}{2\sqrt{u}} & 1 \end{bmatrix}=-\frac{2}{2\sqrt{u}}=-\frac{\sqrt{u}}{u}.$$
Hence,
$$A(D^{*})=\iint_{D^{*}}\left|-\frac{\sqrt{u}}{u} \right|{\rm d}A^{*}=\int_{0}^{2}\int_{-2}^{-u}\frac{\sqrt{u}}{u}{\rm d}v{\rm d}u=\int_{0}^{2}\frac{2-u}{\sqrt{u}}{\rm d}u=\frac{8\sqrt{2}}{3}\approx 3.7712.$$
Therefore,
$$\color{red}{\boxed{A(D)=A(D^{*})}}=\frac{8\sqrt{2}}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4331437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ Find the solution of ${\left( {{x^2} - 7x + 11} \right)^{\left( {{x^2} - 13x + 42} \right)}} = 1$
My approach is as follow $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$
Case 1: $f\left( x \right) = 1;x = 2,5$
Case 2: $f\left( x \right) \ne 0;g\left( x \right) = 0;x = 6,7\& f\left( 6 \right) \ne 0;f\left( 7 \right) \ne 0$
Case 3: $f\left( x \right) = - 1;g\left( x \right) =$ even ;$x = 3,4\& f\left( 3 \right) = 12;f\left( 4 \right) = 6$
I have one doubt in Case 3, I am analysing the case $x^x$, in such scenario $x>0$.
Also if $y=a^x$, then if $a=0,x \in R-0$ and $a>0$
$\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ is also a type of exponential function, they I presume that f(x) is not negative so why they have taken $f(x)=-1$ for case 3, please elaborate
| $$\left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )}=1\Leftrightarrow \left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )} =\left ( x^2-7x+11 \right )^0$$
$$\left ( x^2-7x+11-1 \right )\left ( x^2-13x+42 \right )=0\Leftrightarrow \left ( x^2-7x+10 \right )\left ( x^2-13x+42 \right )=0$$
$$x^2-7x+10=0\Rightarrow x=\left \{ 2,5 \right \}$$
$$x^2-13x+42=0\Rightarrow x=\left \{ 6,7 \right \}$$
You need to check the roots:
$$x^2-7x+11>0$$
All four roots fit!
| {
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"timestamp": "2023-03-29T00:00:00",
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Contour integral $\oint_{|z|=1}\frac{z^2\sin(1/z)}{z-2}dz$ How do I compute this integral?
$$\oint_{|z|=1}\frac{z^2\sin(1/z)}{z-2} \, dz$$
I tried substituting $1/z$ with $z$ and ended up with
$$\oint_{|z|=1}\frac{\sin(z)}{z^3(1-2z)} \, dz$$
At this point I thought of using the residue theorem and got $2i\pi(2-4\sin(\frac{1}{2}))$ but the correct answer should be $\frac{i\pi}{6}$. Can someone help me?
| Where are you getting $\pi i/6$? I am also getting a residue of
$$2-4\sin\left(\frac{1}{2}\right)$$
We can see this directly using power series centered at zero:
$$\frac{z^2\sin(1/z)}{z-2}=$$
$$z^2\cdot\left(\frac{1}{1!z}-\frac{1}{3!z^3}+\cdots\right) \cdot\left(\frac{-1/2}{1-z/2}\right)=$$
$$z^2\cdot\left(\frac{1}{1!z}-\frac{1}{3!z^3}+\cdots\right)\cdot \frac{-1}{2}\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\cdots\right)$$
We can now see term by term that the coefficient of $1/z$ will be:
$$\frac{-1}{2}\left(\frac{-1}{3!2^0z}+\frac{1}{5!2^2z}-\frac{1}{7!2^4z}+\cdots\right)$$
Which is clearly:
$$2-4\cdot\sin\left(\frac{1}{2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $1If $a$, $b$, $c$ are three positive real numbers such that:
$a+b>c$
$b+c>a$
$c+a>b$
$a+b+c=2$
$a$, $b$ and $c$ might be or might not be equal in value
Show that:
$$1<ab+bc+ca-abc<\frac{28}{27}$$
From the first half of the question, I realised that $a$, $b$ and $c$ are the sides of a triangle whose perimeter is $2$. However, I'm unable to solve this problem, even after using some standard inequalities.
Can someone help me with this problem?
| Proof:
Let
$$x = \frac{a + b - c}{2} > 0, \quad y = \frac{b + c - a}{2} > 0, \quad z = \frac{c + a - b}{2} > 0.$$
Correspondingly, $a = z + x, b = x + y, c = y + z$ (the so-called Ravi's substitution).
We have $a + b + c = 2(x + y + z)$. Thus, $x + y + z = 1$.
Also, we have
\begin{align*}
&ab + bc + ca - abc \\
=\,& (z + x)(x + y) + (x + y)(y + z) + (y + z)(z + x) - (z + x)(x + y)(y + z)\\
=\,& (z + x + y + z)(x + y) + (y + z)(z + x)(1 - x - y)\\
=\,& (1 + z)(1 - z) + [xy + z(x + y + z)]z \\
=\,& 1 - z^2 + (xy + z)z \\
=\,& 1 + xyz.
\end{align*}
First, $ab + bc + ca - abc = 1 + xyz > 1$.
Second, using AM-GM, we have $xyz \le (\frac{x + y + z}{3})^3 = \frac{1}{27}$ with equality if and only if $x = y = z = 1/3$.
Thus, we have $ab + bc + ca - abc \le \frac{28}{27}$ with equality if and only if $a = b = c = 2/3$.
We are done.
Remark: Actually, we can prove the following identity:
\begin{align*}
&ab + bc + ca - abc\\
\equiv\,& 1 + \frac{(a + b - c)(b + c - a)(c + a - b)}{8}\\
& + \frac{1}{8}(a + b + c - 2)(a^2 + b^2 + c^2 - 2ab - 2bc - 2ca + 2a + 2b + 2c + 4).
\end{align*}
| {
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how this simplification was done in the integral? I'm stuck at simplification
Here is the Integral ,
$$I=\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}} \ dx$$
To solve this ,
First let's Substitute, we got
$$x^2+2x+3=t^2$$
$$\implies {x}=\sqrt{(t^2-2)}-1$$
$$\implies dx=\frac{t}{\sqrt{t^2-2}}dt$$
Putting this back into the Integral followed by some simplification we have,
$$I=\int\sqrt{t^{2}-2}+\frac{1}{\sqrt{t^{2}-2}}-1 dt$$
Here, how they did it or how they "put back integral " and simplified it ? please, can anybody explain it step by step ?
| If you let $x=\sqrt{t^2-2}-1,$ then $\mathrm{d}x=\frac{t}{\sqrt{t^2-2}}\,\mathrm{d}t,$ $x^2+2x+3=t^2,$ and $x^2+x+1=x^2+2x+3-x-2=t^2-\left(\sqrt{t^2-2}-1\right)-2=t^2-1-\sqrt{t^2-2}.$ Therefore, $$\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}}\,\mathrm{d}x=\int\frac{t^2-1-\sqrt{t^2-2}}{\sqrt{t^2}}\frac{t}{\sqrt{t^2-2}}\,\mathrm{d}t=\int\frac{t^2-1-\sqrt{t^2-2}}{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t$$ $$=\int\frac{t^2-1}{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t-\int\sqrt{t^2}\,\mathrm{d}t=\int\sqrt{t^2-2}\sqrt{t^2}\,\mathrm{d}t+\int\frac1{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t-\int\sqrt{t^2}\,\mathrm{d}t=\int\sqrt{t^2-2}|t|\,\mathrm{d}t+\int\frac1{\sqrt{t^2-2}}|t|\,\mathrm{d}t-\int|t|\,\mathrm{d}t.$$
This is the correct way to do the substitution.
| {
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Identity for $\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d}$ I am looking for an identity for the following derivative:
$$\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d},$$
where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are column vectors.
My approach to expand the expression is as follows:
\begin{align*}
\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d}
&= [ ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} + \mathbf{c} \otimes\nabla(\mathbf{a} \cdot \mathbf{b}) ] \cdot \mathbf{d} \\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \} \cdot \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \}^\top \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \otimes \mathbf{c} \} \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + (\mathbf{c} \cdot \mathbf{d}) [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}].
\end{align*}
Here, $D \equiv \nabla^\top$, i.e., $\nabla$ stands for the gradient operator and $D$ denotes the Jacobian.
My question is, is this expansion correct? Moreover, in my understanding, the resulting expression is again a column vector. Is it right?
| I think I arrive at the same conclusion but in slightly different way.
Using differentials
$$
d\mathbf{f} = (d\mathbf{a})^T\mathbf{b} \mathbf{c}+
\mathbf{a}^T d\mathbf{b} \mathbf{c}+
\mathbf{a}^T\mathbf{b} d\mathbf{c}
$$
Now using
$d\mathbf{a}=\mathbf{J}_a d\mathbf{x}$... and rearranging terms,
you will find that the Jacobian of $\mathbf{f}$ which is a matrix, is
$$
\mathbf{J}_f =
\mathbf{c}
\left[
\mathbf{b}^T \mathbf{J}_a + \mathbf{a}^T \mathbf{J}_b
\right] +
\mathbf{a}^T\mathbf{b} \mathbf{J}_c
$$
The directional derivative is $\mathbf{J}_f \mathbf{d}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$
$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$
$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$
$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$
$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$
Substituting in the values for $x_1$ and $x_2$ we get:
$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$
$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$
$ = 0\le a^2-4a+4$
It seems as if I walked in circles through this process, can anyone help? Thanks in advance!
| you have that:
$$(x-x_1)(x-x_2)=0$$
since these are the roots, so:
$$x^2-(x_1+x_2)x+x_1x_2=0$$
so:
$$x_1+x_2=a\qquad x_1x_2=a-1$$
now we can say:
$$(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=(x_1^2+x_2^2)+2(a-1)=a^2$$
which finally gives us:
$$y=x_1^2+x_2^2=a^2-2a+2$$
now to find the minimum differentiate:
$$\frac{dy}{da}=2a-2$$
so for $y'=0$ we get $a=1$ giving $\min(x_1^2+x_2^2)=1$
In case you are not well versed on the use of derivatives, here is an alternative:
$$\begin{align}y=&a^2-2a+2\\
=&(a-1)^2+1\end{align}$$
from this we can see that the minimum of the function occurs where $(a-1)=0\Rightarrow a=1$
Also, here is a graph to visualise what is going on
| {
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Prove $\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2\left(x+y\right)\sqrt{xy}}$, with $x,\:y\in \mathbb{R}$ and $x\ne 0$ My thinking:
Starting with the basic fact $\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0$ which is trivial.
$\left(\sqrt{x}-\sqrt{y}\right)^4\ge 0\:$
$\rightarrow \:x^2-4\sqrt{yx}\sqrt{x}+6xy-4\sqrt{xy}\sqrt{y}+y^2\ge \:0\:$
$\rightarrow x^2+4\sqrt{yx}\sqrt{x}+6xy+4\sqrt{xy}\sqrt{y}+y^2\ge \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow $ $\left(\sqrt{x}+\sqrt{y}\right)^4\ge \: \:8\sqrt{yx}\sqrt{x}+8\sqrt{xy}\sqrt{y}$
$\rightarrow \frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge \sqrt{yx}\sqrt{x}+\sqrt{xy}\sqrt{y}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{8}\ge x\sqrt{xy}+y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^4}{4}\ge 2x\sqrt{xy}+2y\sqrt{xy}$
$\rightarrow \:\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2}\ge \sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$
$\rightarrow \:\left(\sqrt{x}+\sqrt{y}\right)^2\ge 2\sqrt{2x\sqrt{xy}+2y\sqrt{xy}}$ as needed. QED
I'm not sure if this is correct. If someone can provide some feedback that would be great!
| The given inequality is equivalent to the $t\ge0$ case of
$$(t+1)^4\ge8t(t^2+1)\tag1$$
(Let $t=\sqrt{y/x}$.) Taking the difference in (1),
$$(t+1)^4-8t^3-8t=(t-1)^4,$$
which proves (1) for arbitrary $t.$
| {
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How to solve the Diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$? How can one solve the following Diophantine equation in $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
| Since this is a Diophantine equation, we only seek for integer solutions. Notice that $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ implies $(y-x^2-x+3)(y+x^2+x-1)=-9$, we only have $6$ cases: $((y-x^2-x+3),(y+x^2+x-1))$ must be one of $(1,-9),(9,-1),(3,-3),(-1,9),(-9,1),(-3,3)$. However, after checking all these cases there are no solutions.
A simpler way to do this: modulo 4, we have
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6\equiv x^4 + 2x^3 +x^2-y^2-2y+2\equiv (x(x+1))^2-(y+1)^2+3\equiv 3-(y+1)^2$$
However, $3-(y+1)^2$ won't be zero modulo $4$ since the squares must be $0$ or $1$ modulo $4$, not $3$.
| {
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Rolling three dice and getting exactly two 4s - am I doing this right?
Toss $3$ dice. What's the probability that a "4" will come up exactly
twice?
Since we need two of the dice to come up 4, and the third can be any number except 4 we can have the following options:
[4, 4, $\neq$ 4], [4, $\neq$ 4, 4], or [$\neq$ 4, 4, 4]. Right?
P of getting a 4 is $1/6$ and P of not getting a 4 is $5/6$ (assuming we have a $6$-sided die and each event is independent).
Let's take the first option: [4, 4, $\neq$ 4]
$\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{5}{216}$ or $~0.02315$
And accounting for the other options is just multiplying by $3$ to get $0.06945$?
Did I do this correctly?
| You are very close, but in order to solve it you need to use the binomial distribution: $$\binom{\textrm{No. of attempts}}{\textrm{number of successes}} \cdot (\textrm{success chance})^{\textrm{No. of successes}} \cdot (1 - \textrm{success chance})^{\textrm{No. of failures}}.$$ In your case, it is
$$\binom{3}{2} \cdot \left(\frac{1}{6}\right)^{2} \cdot \left(\frac{5}{6}\right)^{1} = 3 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{15}{216} \approx 0.07=7\%$$
| {
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Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.
| You can do pairwise comparisons. For instance, to compare $\sqrt[4]4$ with $\sqrt[5]5$, you only need to compute $4^5$ and $5^4$. (And you can rule out $\sqrt[6]6$ easily by comparing it with $\sqrt[3]3$.)
| {
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Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| Specifically addressing your edit, $\sqrt{20 + 48i} = 2 \sqrt{5 + 12i}$.
Now if $\sqrt{5 + 12i} = a + bi$, then squaring both sides, $5 + 12i = (a^2 - b^2) + 2abi$, or that $5 = a^2 - b^2, 12 = 2ab$. Since $b = \frac{6}{a}$, you can form a quadratic with this information, which gives $a + bi = ±(3 + 2i)$.
| {
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Prove that $\frac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\frac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}$ Prove that $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=2\sqrt[4]{ax}, a>0, x>0,x\ne a.$$ My try $$\dfrac{a-x}{\sqrt{a}-\sqrt{x}}-\left(\dfrac{a+\sqrt[4]{ax^3}}{\sqrt{a}+\sqrt[4]{ax}}-\sqrt[4]{ax}\right)=\dfrac{\sqrt{a}^2-\sqrt{x}^2}{\sqrt{a}-\sqrt{x}}-\dfrac{a+\sqrt[4]{ax^3}-\sqrt[4]{a^3x}-\sqrt[4]{a^2x^2}}{\sqrt{a}+\sqrt[4]{ax}}$$ Can we do something more directly to arrive at the solution? Thank you!
| Just to make things easier to type let $c =\sqrt[4]a$ and $y=\sqrt[4]x$ so we are asked to prove
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)= 2cy$.
so lets do that
$\frac {c^4 - y^4}{c^2-y^2} -(\frac {c^4 + cy^3}{c^2 + cy} - cy)=$
$\frac {(c^2-y^2)(c^2+y^2)}{c^2-y^2} -\frac {c(c^3+ y^3)}{c(c+y)} + cy=$
$c^2+cy+y^2 -\frac {c^3+y^3}{c+y}=$
$c^2 + cy + y^2 - \frac {(c+y)(c^2 -cy+y^2)}{c+y}=$
$c^2 + cy + y^2 -(c^2 -cy +y^2) = 2cy$
That's it.
| {
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Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is not that brought me in a right way.
Do you have any idea how to solve this one?
| The two given equations imply that
\begin{align}
&(1-\tan x)^2=\frac{\sin^2y}{1-\sin^2y},\tag{1}\\
&\cos x=\frac{1}{\sqrt{2}\sin y}.\tag{2}
\end{align}
Suppose both $x$ lies inside the first quadrant. From $(2)$ we see that $y$ must lie inside the first or the second quadrant. Also,
\begin{align}
\tan x&=\frac{\sqrt{1-\cos^2x}}{\cos x}=\sqrt{2\sin^2y-1},\\
\sin^2y&=\frac{1+\tan^2x}{2}.\\
\end{align}
Substitute the expression for $\sin^2y$ into $(1)$, we obtain
\begin{aligned}
&(1-t)^2=\frac{(1+t^2)/2}{1-(1+t^2)/2},\\
&(1-t^2)(1-t)^2=(1+t^2),\\
&(1+t^2)-(1-t^2)(1-t)^2=0,\\
&t\left[t(t-1)^2+2\right]=0.
\end{aligned}
Since $x$ lies inside the first quadrant, $t=\tan x$ is nonnegative and the equation above has a unique root $t=0$. Hence $x=0,\ \sin^2 y=\frac{1+t^2}{2}=\frac12$ and $y=\frac{\pi}{4}$.
I guess the other cases (where $x$ or $y$ or both lie inside other quadrants) can be handled in a similar manner.
| {
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Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$ Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$
I actually managed to write a short proof, but I'd appreciate it if someone could check it for me and offer some feedback. I'm not very used to induction so I'm not sure the proof is very clear or correct. Hope that's appropriate for this forum!
My intuition was that starting with $n = 2 \rightarrow\frac{1}{1\times2} + \frac{1}{2\times1} = 1$ you can always add one more term by making $x_n = n + 1$ and $x_{n+1} = 1$, such as in
$\frac{1}{1\times2} + \frac{1}{2\times1} = \frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times1} = \frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + \frac{1}{4\times1}$ = ... = 1
So here's the proof:
Statement: for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$
Proof (by induction):
Base case:
For $n = 1$, we have that $\frac{1}{x_1} = 1$ for $x_1 = 1$
Inductive step:
If for some $n$:
$\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{nx_n} = 1$, for certain $x_1,x_2,...,x_n$
and we also have that $x_n = 1$
It follows that:
$\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n} = 1$
$\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+1} = 1$
$\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{n + 1 - n}{n(n+1)} + \frac{1}{n+1} = 1$
$\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n(n+1)} + \frac{1}{n+1} = 1$
Therefore, if $y_i = x_i$ for $1 \leq i \leq n-1$, $y_n = n + 1$ and $y_{n+1} = 1$, it follows that:
$\frac{1}{y_1} + \frac{1}{2y_2} + … + \frac{1}{ny_n} + \frac{1}{(n+1)y_{n+1}}= 1$
$$\tag*{$\blacksquare$}$$
Thanks in advance!
| I think the idea is correct, but you do not need induction.
You suggested that $\frac{1}{2*1}+\frac{1}{3*2}+...+ \frac{1}{n(n-1)}+\frac{1}{n}=1$. This is true: the sum is $\frac11-\frac12+\frac12-\frac13+....+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}=1$
| {
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Solve Summation of Catalan Convolution? Can someone help me solve this?
$$
\sum_{k=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{2 k+4} 4\left(\begin{array}{c}
2 k+3 \\
k
\end{array}\right)}{k+4}
$$
converges.
Wolfram alpha says it converges to $1$.
But actually, I have no idea, how to solve it.
| Using a version of the binomial power series
$$\sum_{k\geq 0}{2k+\alpha \choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^\alpha$$
With $\alpha=3$:
$$\sum_{k\geq 0}{2k+3\choose k}z^k=\frac 1 {\sqrt{1-4z}}\left(\frac {1-\sqrt{1-4z}}{2z}\right)^3$$
Consequently
$$\sum_{k\geq 0}{2k+3\choose k}\frac{z^{k+4}}{k+4}=\int_0^z \frac {t^3} {\sqrt{1-4t}}\left(\frac {1-\sqrt{1-4t}}{2t}\right)^3dt=\frac{\left(1-\sqrt{1-4z}\right)^4}{64}$$
Evaluating at $z=\frac 1 4$:
$$\sum_{k\geq 0}{2k+3\choose k}\frac{\left(\frac 1 2\right)^{2k+8}}{k+4}=\frac{1}{64}$$
Re-arranging the terms:
$$4\sum_{k\geq 0}{2k+3\choose k}\frac{\left(\frac 1 2\right)^{2k+4}}{k+4}=1$$
| {
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Can $(x^2+a^2)^{-\frac{3}{2}}$ be integrated without using trigonometric substitutions? I know that
$$
\int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx
$$
can be solved using a trigonometric substitution, but is there a trick to calculate this integral without using trigonometric substitutions? I think something in the same spirit with solving
$$
\int\frac{1}{(x^2+1)^2}\,dx
$$
by adding and subtracting $x^2$ to the numerator and then integrating by parts once one knows (and it is simple) how to integrate
$$
\int\frac{1}{(x^2+1)}\,dx
$$
Any suggestion or reference will be appreciated.
| A very simple approach would be:-
$$\int\frac{1}{(x^2+a^2)^{\frac{3}{2}}}\,dx=\int\frac{1}{(x^3)(1+\frac{a^2}{x^2})^{\frac{3}{2}}}\,dx$$
Substituting $(1+\frac{a^2}{x^2})$ as $t^2$,we get :
$$\int\frac{-1}{a^2t^2}dt=\frac{1}{a^2t} +c=\frac{x}{(a^2){(x^2+a^2)^{\frac{1}{2}}}} +c $$
| {
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Is there an elementary proof for $\int_{0}^{\infty} \frac{d x}{x^{n}+1}= \frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) $ When I first encounter the integral
$$\int_{0}^{\infty} \frac{d x}{x^{n}+1},$$
I am trying to resolve the integrand into partial fractions. Then I found it is very tedious and complicated and look for infinite series.
I first split the integral into 2 parts.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{x^{n}+1} =& \underbrace{\int_{0}^{1} \frac{d x}{x^{n}+1}}_{J} +\underbrace{\int_{1}^{\infty}\frac{d x}{x^{n}+1}}_{K}
\end{aligned}
$$
$$
\begin{aligned}
J &=\int_{0}^{1} \sum_{k=0}^{\infty}(-1)^{k} x^{n k} d x =\sum_{k=0}^{\infty}\left[\frac{(-1)^{k} x^{n k+1}}{n k+1}\right]_{0}^{1} =\sum_{k=0}^{\infty} \frac{(-1)^{k}}{n k+1}=\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
K&=\int_{1}^{\infty} \frac{1}{x^{n}+1} d x =\int_{1}^{\infty} \frac{x^{-n}}{1+x^{-n}} d x =\int_{1}^{\infty} x^{-n} \sum_{k=0}^{\infty}(-1)^{k}x^{-k n} d x=\sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{-(k+1) n+1} \end{aligned}$$
Rearranging and re-indexing yields
$$K=\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{-(k+1)+\frac{1}{n}}=\frac{1}{n} \sum_{k=-1}^{-\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}
$$
Grouping $J$ and $K$ yields $$\int_{0}^{\infty} \frac{d x}{x^{n}+1} =\frac{1}{n}\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}\right) $$
Using the theorem
$$
\pi \csc (\pi \alpha)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{k}}{k+\alpha}
$$
and putting $\alpha=\frac{1}{n}$ yields
$$\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$$
My question: Is there any elementary proof?
| Here is an elementary proof using partial fractions, avoiding complex analysis, infinite series, or special functions commonly employed to evaluate this integral
\begin{align}
& \int_{0}^{\infty} \frac{1}{1+x^{n}} dx\\
=&\int_{0}^{1} \frac{1}{1+x^{n}}dx+ \int_{1}^{\infty} \frac{1}{1+x^{n}}\overset{x\to 1/x}{dx }
= \int_{0}^{1} \frac{1+x^{n-2}}{1+x^{n}}dx \\
=& \int_0^1 \sum_{k=0}^{n-1} \frac{\frac2n \sin^2 \frac{(1+2k)\pi}n}{x^2-2x \cos \frac{(1+2k)\pi}n +1 }\ dx
=-\sum_{k=0}^{n-1} \frac{2\pi k}{n^2} \sin \frac{(1+2k)\pi}n\\
=& \ \frac d{da}\bigg(\sum_{k=0}^{n-1} \frac1n \cos \frac{(1+2k)\pi a}n \bigg)_{a=1}
=\frac d{da}\bigg(\frac{\sin 2\pi a}{2n\sin\frac{\pi a}n} \bigg)_{a=1}
=\frac{\pi}{n} \csc \frac{\pi}{n}
\end{align}
| {
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Why is $ 2 \cos(\frac{2π}{5}) + \left(2 \cos(\frac{2π}{5})\right)^2 = 1$? I was able to prove $(a)$, but I have trouble with $(b).$ Can somebody give me a hint?
Let $ \alpha := e^{
\frac{2πi}{5}} \in \mathbb{C}$, $K:=\mathbb{Q}(\alpha) $ and $f(X) := X^4 + X^3 + X^2 + X + 1$
(a) Prove that $f(\alpha) = 0$.
(b) Let $\overline{\alpha}$ be the complex conjugate of $\alpha$ and let $ \beta := \alpha + \overline{\alpha}$. Prove that $\beta + \beta^2 = 1$ and find the minimal polynomial of $\beta$ over $\mathbb{Q}$.
(c) Show that $\mathbb{Q}(\sqrt{5}) = \mathbb{Q}(\beta) ⊂ K$.
I was able to prove (a) because of
$$ 0 =\frac{\alpha^5−1}{\alpha−1}=\alpha^4+\alpha^3+\alpha^2+\alpha+1,$$
but now I am struggling with (b).
Well obviously: $\alpha = \text{cos} (\frac{2π}{5}) + i ~ \text{sin}(\frac{2π}{5}),$ so $\beta = 2 ~ \text{cos} (\frac{2π}{5})$
and therefore I have to show:
$$ 2 \text{cos} (\frac{2π}{5}) + \left(2 ~\text{cos} (\frac{2π}{5})\right)^2 = 1$$
But why is that? Or did I make a mistake?
My minimal polynomial over $\mathbb{Q}$ is then $ X^2 + X -1$.
| Note that $1,\alpha,\alpha^2,\alpha^3,\alpha^4$ are roots of $x^5-1=0$
Now we have $\alpha\bar{\alpha}=1\implies\alpha\bar{\alpha}=\alpha^5\implies\bar{\alpha}=\alpha^4$
Therefore $\beta=\alpha+\alpha^4$ and $\beta+\beta^2=\alpha+\alpha^4+\alpha^2+\alpha^8+2\alpha^5=2+\alpha+\alpha^2+\alpha^3+\alpha^4=2-1=\boxed{1}$
Hence proved
| {
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In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centroid of
$\triangle ABC$.
I am trying to do with Ceva's theorem: $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ and also with the result $\frac{AP}{PD}=\frac{AF}{FB}+\frac{AE}{EC}$ but having some difficulties. Please give any hint.
| This is not a trivial geometry problem... Let $x=AF/FB, y=BD/DC, z=CE/EA$, then we have $xyz=1$. Furthermore, we can calculate that the area of $AFB$ is $\frac{x}{(1+z+yz)(1+x)}$. Similarly we can calculate two others, so, we have
$$\frac{x}{(1+z+xz)(1+x)}=\frac{y}{(1+x+yx)(1+y)}=\frac{z}{(1+y+zy)(1+z)}$$
Notice that $(1+z+xz)=(xyz+z+xz)=z(xy+1+x)=z(xy+xyz+x)=zx(y+yz+1)$, so we have
$$\frac{x}{(1+x)}=\frac{yz}{(1+y)}=\frac{z^2x}{(1+z)}$$
So we have $x(1+z)=z^2x(1+x)$ and $x(1+y)=yz(1+x)$. Substitute $y=1/xz$, we have $x(1+1/xz)=z/(xz)(1+x)$, or, $x(1+xz)=z(1+x)$. This yields a linear equation of $z$, that is, $z=\frac{x}{1+x-x^2}$. Substitute it in, we have
$$(1+\frac{x}{1+x-x^2})=(\frac{x}{1+x-x^2})^2(1+x)$$
Or,
$$(1+2x-x^2)(1+x-x^2)=x^2(1+x)$$
Expanding, we have
$$x^4-4x^3-x^2+3x+1=0$$
Factorizing, we have
$$(x-1)(x^3-3x^2-4x-1)=0$$
One solution is $x=1$, the other positive solution is located $x>4$, where $z=\frac x{1+x-x^2}<0$ (which means $P$ is outside), contradiction. Therefore, the only valid solution is $x=y=z=1$.
| {
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How to compute $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}}$ Trying to prove $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$
I tried using ratio-root criteria with $a_n= \frac{1}{n^n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n}}$ and end up with $\frac{n^n}{(n+1)^{n+1}}\cdot 2(n+1)^2 \Pi_{k=1}^n\frac{\left(k^2\right)^{\frac{k}{n+1}} \left(1+\left(\frac{n+1}{k} \right)^2 \right)^{\frac{k}{n+1}}}{\left(k^2\right)^{\frac{k}{n}} \left(1+\left(\frac{n}{k} \right)^2 \right)^{\frac{k}{n}}}$ and can’t see how to simplify this to get the limit as the numerator and denominator seem to end up going to 1 and some kind of $e^{\Sigma \frac{1}{n}}$ but the $(n+1)^2$ seems to persist and send everything to infinity.
So I tried taking e^ natural log(I’ll denote them as log here, I don’t mind ln either) and got this sum in the exponent $-\log (n) + \Sigma_{k=1}^n \frac{k}{n^2}\log(n^2+k^2)$ comparison showed I could get a $\log 2$ and a $\log n$ But where is the $-1/2$ I need? My guess is that this approach won’t simplify further.
Am doing this for practice/review so would be interested in seeing solutions or hints as I already spent too much time on this. My hunch is that I’m either overlooking something or there’s some typo in the problem.
This is also equivalent to $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)^{\frac{1}{n^2}}(n^2+2^2)^{\frac{2}{n^2}} \cdots (n^2 + n^2)^{\frac{1}{n}} \right] = 2e^{-1/2}$
Try explaining this before calling this duplicates. I get something different:
$\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = n^{\frac{1}{n}}\left[(1 +(\frac{1}{n})^2)^{\frac{1}{n^2}}(1+(\frac{2}{n})^2)^{\frac{2}{n^2}} \cdots (1 + (\frac{n}{n})^2)^{\frac{1}{n}} \right]$.
How do you get $\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = (1 +(\frac{1}{n})^2)(1+(\frac{2}{n})^2)^{2} \cdots (1 + (\frac{n}{n})^2)^{n}$?
| This result for the logarithm that you've got can be calculated further. We have
\begin{align} & -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \log(k^2+n^2) = \\
&= -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \Big(2\log n + \log(1+\frac{k^2}{n^2})\Big) = \\
&= -\log(n) + \frac{2\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n} \sum_{k=1}^n\frac{k}{n}\log(1+\frac{k^2}{n^2}) = \\
&= \frac{\log n}{n} + \frac{1}{n} \sum_{k=1}^n \frac{k}{n}\log(1+\frac{k^2}{n^2})
\end{align}
In the limit
$$ \lim_{n\to\infty} (\dots) = 0 + \int_0^1 x\log(1+x^2) dx = \frac12(2\log 2 -1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376764",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 2,
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} |
Proving that $\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}$ when $n$ is very large This is an example from Mathematical Methods in the Physical Sciences, 3e, by Mary L. Boas.
My question is,
\begin{equation}
\frac{1}{n^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3}
\end{equation}
can also be written as,
\begin{equation}\frac{1}{(-n)^2} - \frac{1}{(n+1)^2} \approx \frac{2}{n^3} \end{equation}
so that $\Delta n = dn = -n-(n+1) = -2n-1$.
For $f(n) = 1/n^2$, $f'(n) = -2/n^3$, and
\begin{equation}
df = d(\frac{1}{n^2}) = f'(n)dn
\end{equation}
\begin{equation}
df = \frac{(2)(2n+1)}{n^3}
\end{equation}
Now, for very large $n$, $2n+1 \approx 2n$. Thus, \begin{equation}
df = \frac{4}{n^2}
\end{equation}
But, $4/n^2$ is not approximately equal to $2/n^3$ (required ans.) even if $n$ is very large.
So, please point out my mistake(s).
Thanks in advance (;
| You can also apply Lagrange's theorem to $f(x)=\frac{1}{x^2}$ in the interval $[n, n+1]$:
$$
f(n+1)- f(n) = f'(\xi_n)(n+1-n), \quad \xi_n \in (n, n+1),
$$
i.e.,
$$
\frac{1}{(n+1)^2} - \frac{1}{n^2} = -\frac{2}{\xi_n^3}, \quad \xi_n \in (n,n+1)
$$
or
$$
\frac{2}{(n+1)^3} \leq \frac{1}{n^2}-\frac{1}{(n+1)^2}\leq \frac{2}{n^3}.
$$
| {
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"answer_count": 4,
"answer_id": 1
} |
From homogeneous to non-homogeneous linear recurrence relation I'm trying to do the following exercise:
Find a non-homogeneous recurrence relation for the sequence whose general term is
$$a_n = \frac{1}{2}3^n - \frac{2}{5} 7^n$$
From this expression we can obtain the roots of the characteristic polynomial $P(x)$, which are $3$ and $7$, so $P(x) = x^2 - 10x + 21$ and $a_n = 10a_{n-1} - 21a_{n-2} \; \forall \; n \ge 2, \; a_0 = \frac {1}{10}, \; a_1 = -\frac{13}{10}$.
Now I don't know how to obtain a non-homogeneous recurrence relation given this homogeneous recurrence relation.
| Just start with your favorite term containing $a_n$ and $a_{n-1}$, say
$a_n - a_{n-1}$, calculate the difference, here
$$ a_n - a_{n-1} = \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac 25 7^{n-1} $$
giving you the inhomogeneous recurrence
$$ a_n = a_{n-1} + \frac 12 3^n - \frac 25 7^n - \frac 12 3^{n-1} + \frac 25 7^{n-1}, \quad n \ge 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380992",
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"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=} -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle 1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$
| If you like hypergeometric functions, there is an antiderivative
$$J_n=\int\frac{\log(x)}{\left(x^{2}+a\right)^{n}}\,d x=\frac x {a^n} \left( \,
_2F_1\left(\frac{1}{2},n;\frac{3}{2};-\frac{x^2}{a}\right) \log(x)-\,
_3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-\frac{x^2}{a}\right)
\right)$$ So
$$K_n=\int_0^\infty\frac{\log(x)}{\left(x^{2}+a\right)^{n}}\,d x=-\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}\,\frac 1{a^{n-\frac{1}{2}}}\left(\log \left(\frac{4}{a}\right)+H_{n-\frac{3}{2}}\right)$$ and if $a= 1$
$$I_n=\int_0^\infty\frac{\log(x)}{\left(x^{2}+1\right)^{n}}\,d x=-\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}\,\left(H_{n-\frac{3}{2}}+2 \log (2)\right)$$
For large values of $n$
$$I_n \sim -\frac{\sqrt{\pi }}{4}\,\frac{\log (n)+\gamma +2 \log (2)}{\sqrt{n}}\Bigg[1+\frac{3 (\log (n)+ \gamma +2 \log (2))-8}{8 (\log (n)+\gamma +2 \log (2))}\,\frac 1 n +\cdots\Bigg]$$ which is quite accurate ($0.22$% relative error for $n=5$ and $0.0047$% for $n=10$).
| {
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"answer_count": 8,
"answer_id": 3
} |
$\text { Show that } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}} $
Show that
$$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x$$ if $-1 \leq x \leq-\frac{1}{\sqrt{2}}.$
I tried solving this question a lot but I’m unable to. My answer comes different.
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\cos \theta$
$$
\begin{array}{l}
=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\
=\sin ^{-1}\left(2 \cos \theta \sqrt{\sin ^{2} \theta}\right) \quad\left(\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right)
\end{array}
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=-\sin ^{-1}(\sin 2\theta)
$$
(Using $\sin 2 x=2 \sin x \cos x$)
$=2 \theta \quad$ As $x=\cos \theta$
$=2 \times \cos ^{-1} x$
$=2 \cos ^{-1} x$
EDIT: If any confusion with Q writing
| hint
Let
$$f(x)=\arcsin(2x\sqrt{1-x^2})-2\arccos(x)$$
assuming $ f $ is differentiable at $ [-1,-\frac{1}{\sqrt{2}}] $,
prove that
$$f'(x)=0$$
So
$$f(x)=Cte=f(-1)=-2\pi$$
Other approach
Put $$x=\cos(\frac{\theta}{2})$$
| {
"language": "en",
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How can we determine the sum of a series where each term is a product of two integers using method of differences? I have been trying to find out the sum of a series up to the $n^{th}$ term but failed.$$S_n=1 \cdot 3+2 \cdot 4+3 \cdot 5+4 \cdot 6+ \ldots + n(n+2)$$
my work:
\begin{align*}
S_n & =\frac{n(n+2)(n+4)-(n-2)n(n+2)}{6}\\
& =\frac{n(n+2)(n+4)}{6}-\frac{n(n+2)(n-2)}{6}\\
& =V_n - V_0
\end{align*}
since $V_0=0$
$$S_n =\frac{n(n+2)(n+4)}{6}$$
but later I found that the actual sum is $$S_n=\frac{n(n+1)(2n+7)}{6}$$
Where did I make a mistake? I tried to find the sum in another way and that is by taking the sum of $n^2+2n$ and obtained the actual sum but couldn't find where I made a mistake in my first approach.
I am stuck with another problem too and that one is
$$S_n=1 \cdot 2+2 \cdot 5+3 \cdot 8+ \ldots + n(3n-1)$$
In this one I am completely clueless what to do since $n$ and $3n-1$ doesn't seem to be following a pattern between them.
| $S_n$ will be a cubic function of $n$, because there are $n$ terms of $2nd$ degree (quadratic). Then, we can write the formula for $S_n$ as $S_n = An^3 + Bn^2 + Cn + D$.
When $n=0$, $S_n = 0\cdot2 = 0 \Rightarrow D = 0$
When we plug in $n=1,2,3$, we get this system of equations:
\begin{align}
\tag{1}
A+B+C = 3
\newline
\tag{2}
8A+4B+2C = 11
\newline
\tag{3}
27A+9B+3C = 11
\end{align}
Solving this system gives $A=\frac{1}{3}, B = \frac{3}{2}, C = \frac{7}{6}$. Plugging these in to $S_n$, we get:
\begin{align}
S_n = \frac{2n^3+9n^2+7n}{6} = \frac{n(n+1)(2n+7)}{6}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the sum of natural numbers that have $5$ digits or less, and all of the digits are distinct? $1+2+3+\dots+7+8+9+10+12+13+\dots+96+97+98+102+103+104+\dots+985+986+987+1023+1024+1025+\dots+9874+9875+9876+10234+10235+10236+\dots+98763+98764+98765=$
The only thing I can do is to evaluate a (bad) upper bound by evaluating the sum of natural numbers from $1$ to $98765$, that is equal to $98765 \times (98765+1)/2=4,877,311,995$. So the desired answer is less than that.
Any help would be appreciated. Thanks.
| This is the approach of Empy2 (which I believe didn’t deserve any downvotes) spelled out.
Consider the numbers for each length separately.
First allow a leading zero. Then there are $\frac{10!}{(10-k)!}$ different numbers with $k$ digits, all distinct, and they sum to $10^k-1$ in pairs, so their sum is $\frac{10^k-1}2\cdot\frac{10!}{(10-k)!}$.
Now subtract the numbers with a leading zero. There are $\frac{9!}{(10-k)!}$ of these, their average digit is $5$, and the total value of their digits is $\frac{10^{k-1}-1}9$, so they sum to $5\left(10^{k-1}-1\right)\cdot\frac{8!}{(10-k)!}$.
Thus, the overall sum is
\begin{eqnarray}
&&\sum_{k=1}^5\left(
\frac{10^k-1}2\cdot\frac{10!}{(10-k)!}-5\left(10^{k-1}-1\right)\cdot\frac{8!}{(10-k)!}\right)\\
&=&
\sum_{k=0}^5\frac{10^k-1}2\cdot\frac{10!}{(10-k)!}-\sum_{k=0}^45\left(10^k-1\right)\frac{8!}{(9-k)!}
\\
&=&
\frac{10^5-1}2\frac{10!}{5!}+\sum_{k=0}^4\frac{8!\left(10^k-1\right)}{(10-k)!}\left(45-5(10-k)\right)
\\
&=&
\frac{10^5-1}2\frac{10!}{5!}+\sum_{k=0}^4\frac{8!\left(10^k-1\right)}{(10-k)!}(5k-5)
\\
&=&
1511984880+8479575
\\
&=&
1520464455\;.
\end{eqnarray}
The result is confirmed by this Java code.
| {
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"source": "stackexchange",
"question_score": "4",
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What am I doing wrong to finding the perpendicular distance between a point and a plane? From this post How to prove that a plane is a tangent plane to a sphere?, I went and attempted to show that $4x+3z+29=0$ is tangent to the sphere $(x+1)^2+(y-3)^2+z^2=25$. However I keep getting a different distance from the plane to the centre, the goal being to show that the distance between the plane and the centre is equal to the radius.
To find the distance of the plane from $(-1,3,0)$ I did the following:
The normal vector to the plane is $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$. Another random point on the plane is $(-2,0,-7)$. This means the vector from that random point to the centre is $\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}$. So from what I know, the perpendicular distance should then be the length of the projection of $\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}$ onto $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$. However when I do that I get:
$\left|\frac{\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}\cdot\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\sqrt{1^2+3^3+7^2}\sqrt{4^2+3^2}}\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}\right|$
=$\left|\frac{25}{\sqrt{1475}}\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}\right|$
$=\sqrt{\frac{625}{59}}$, which is clearly not the desired answer of $5$.
So what am I doing wrong in this calculation?
| The $\sqrt{1^2+3^2+7^2}$ should be $\sqrt{4^2+3^2}$.
The length of the projection of $\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}$ onto $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$ is given by
$$\left|\frac{\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}\cdot\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\color{red}{\sqrt{4^2+3^2}}}\frac{\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\sqrt{4^2+3^2}}\right|=5$$
where $\dfrac{\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}\cdot\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\color{red}{\sqrt{4^2+3^2}}}$ is the scalar projection of $\begin{bmatrix}
1 \\3\\7\\\end{bmatrix}$ onto $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$, and $\dfrac{\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}}{\sqrt{4^2+3^2}}$ is the unit vector in the direction of $\begin{bmatrix}
4 \\0\\3\\\end{bmatrix}$.
You might want to see here.
| {
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"source": "stackexchange",
"question_score": "2",
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Using diagonalisation of a symmetric on quadratic form Let $f (x, y) = 7x_2 + 4y_2 + 4xy$. Use the diagonalisation of a symmetric matrix to write this quadratic form in the form $λ_1u_2^2 +λ_2v_2^2$, with u and v linear combinations of x and y.
Here's what I have tried:
step 1. get the symmetric matrix
$$f (x, y) = 7x_2 + 4y_2 + 4xy=\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)$$
step 2. find the eigenvalues
$$det(A-\lambda I) \\
\left(\begin{matrix}7-\lambda&2\\2&4-\lambda\\\end{matrix}\right)=(7-\lambda)(4-\lambda)-4=(\lambda-3)(\lambda-8)$$
step 3. find the eigenvectors
case for $\lambda =3$
$$\left(\begin{matrix}7-3&2\\2&4-3\\\end{matrix}\right)=\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right)
$$
Row reduced
$$\left(\begin{matrix}4&2\\2&1\\\end{matrix}\right)=\left(\begin{matrix}1&\frac{1}{2}\\0&0\\\end{matrix}\right) \implies \left(\begin{matrix}x\\y\\\end{matrix}\right)=y_1\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right)$$
Doing the same for $\lambda =8$ (i'll skip the calculation), I get $\left(\begin{matrix}x\\y\\\end{matrix}\right)=y_2\left(\begin{matrix}2\\1\\\end{matrix}\right)$
Finding the orthogonal eigenvectors
$$u_1 = \frac{y_1}{||y_1||} = \frac{2}{\sqrt{5}}\left(\begin{matrix}-\frac{1}{2}\\1\\\end{matrix}\right) \\
u_2 = \frac{y_1}{||y_1||} = \frac{1}{\sqrt{5}}\left(\begin{matrix}2\\1\\\end{matrix}\right) \\
P=(u_1,u_2)=\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) \\
D = P^TAP = \left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right)\left(\begin{matrix}7&2\\2&4\\\end{matrix}\right)\left(\begin{matrix}-\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\\\end{matrix}\right) = \left(\begin{matrix}3&0\\0&8\\\end{matrix}\right) \\
x^TAx = \lambda^TD\lambda = \left(\begin{matrix}\lambda_1&\lambda_2\\\end{matrix}\right)\left(\begin{matrix}3&0\\0&8\\\end{matrix}\right)\left(\begin{matrix}\lambda_1\\\lambda_2\\\end{matrix}\right) = 3\lambda_1^2+8\lambda_2^2$$
However, my result does not look like it's in the same form as the question:
$λ_1u_2^2 +λ_2v_2^2 \ne 3\lambda_1^2+8\lambda_2^2$, how do I proceed from here?
| This is a longer comment rather than an full answer:
If you want to "diagonalize" a quadratic form, completing the square is simpler than your approach via the eigendecomposition. Note that
$$ f (x, y) = 7x^2 + 4y^2 + 4xy = 6 x^2 + (2 y +x)^2 = 6 u^2 + v^2$$
with
$$u = x, \qquad v = 2 y +x$$
is a diagonalization in terms of a linear combination of $x$ and $y$.
| {
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"question_score": "3",
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Inequality with real exponents I had the following problem on a past homework asignment:
For $x,p \in \mathbb{R}^+$
$$\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
So, I think one way to show this for $1\leq p$ would be with rules for real exponents for real exponents (since $x^{-1} >-1$):
$$(1+\frac{1}{x})^p \geq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
and since $(1+x)^p>0$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}>\frac{1}{(1+x)^{p}}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
For $0<p<1$ this argument doesn't hold since the inequality gets flipped:
$$(1+\frac{1}{x})^p \leq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p\leq 1+px^{-1}>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}\leq\frac{1}{(1+x)^p}+\frac{p}{x(1+x)^p}>\frac{1}{(1+x)^p}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{1}{x^p}-\frac{1}{(1+x)^p}\leq\frac{p}{x(1+x)^p}>\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}>\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
What would I need to do to complete this proof? I tried looking at $x^p(p+x+1)<(x+1)^{p+1}$ as well and got nowhere. Is this 'possible' to prove without the binomial series?
| Let $f(x)=\dfrac{1}{x^p}$, then $f'(x)=-\dfrac{p}{x^{p+1}}$. Your inequality becomes $$f'(x+1) \ge f(x+1)-f(x) = \dfrac{f(x+1)-f(x)}{(x+1)-x}$$
This follows easily from that $f$ is convex when $p>0$.
Edit: Since $f$ is convex, then for any $0<t<1$, $$f(x+t)=f((1-t)x+t(x+1))<(1-t)f(x)+tf(x+1) \implies \dfrac{f(x+1)-f(x+t)}{1-t}>f(x+1)-f(x)$$
Let $t \rightarrow 1^-$ and you get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\displaystyle \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$ I have a question which asks to find the value of:
$\displaystyle \tag*{} \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$
Where, $a,b >0$ and $n >2$
I tried to rationalize the denominator and arrived:
$\displaystyle \tag*{} \dfrac{1}{a-b}\int \limits _{0}^{\infty}\sqrt{x^n+a} - \sqrt{x^n+b} \ \ \mathrm dx $
Since individual integrals do not converge. But couldn't proceed any further. Any help would be greatly appreciated. Thank you!
| We'll evaluate the following auxiliary integral:
$$\int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x = \frac{\alpha^{\frac1n + \frac12} }{(n+2)\sqrt{\pi}}\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)\qquad \text{for} \quad \alpha >0, \, n>2$$
Proof: Taking the substitution $\sqrt{x^n +\alpha}- \sqrt{x^n} = \sqrt{\alpha t}$ gives $x = \alpha^{\frac1n}2^{-\frac{2}{n}}t^{-\frac{1}{n}}\left(1-t\right)^{\frac{2}{n}}$ which then gives $\mathrm{d}x =-\frac{1}{n}\alpha^{\frac1n}2^{-\frac{2}{n}}\left[ t^{-\frac1n -1}(1-t)^{\frac2n -1} + t^{-\frac1n}(1-t)^{\frac2n -1}\right] \mathrm{d}t $. Combining everything we get:
$\require{cancel}$
\begin{align*}
\int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[\int_{0}^{1} t^{\frac12-\frac1n -1}(1-t)^{\frac2n -1}\mathrm{d}t + \int_{0}^{1} t^{\frac12-\frac1n +1-1}(1-t)^{\frac2n -1}\mathrm{d}t\right]\\
& =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[B\left(\frac12-\frac1n, \frac2n\right) + B\left(\frac12-\frac1n+1, \frac2n\right)\right]\\
& =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[1 + \frac{\frac12 - \frac1n}{\frac12+\frac1n}\right]B\left(\frac12-\frac1n, \frac2n\right)\\
& =\frac{\alpha^{\frac1n + \frac12}\cancel{2^{-\frac{2}{n}}}}{\cancel{n}}\frac{\cancel{2}\cancel{n}}{n+2}\frac{\cancel{2^{\frac2n-1}}\cancel{\Gamma\left(\frac1n + \frac12\right)}\cancel{\Gamma\left(\frac12\right)}}{\cancel{\sqrt{\pi}}\cancel{\Gamma\left(\frac12 + \frac1n\right)}}B\left(\frac12-\frac1n, \frac1n\right)\\
& =\frac{\alpha^{\frac1n + \frac12} }{n+2}\frac{\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)}{\sqrt{\pi}}
\end{align*}
using that $B(x+1,y) = \frac{x}{x+y}B(x,y)$ and $B(x,2y) = \frac{2^{2y-1}\Gamma\left(y + \frac12\right)\Gamma\left(x+y\right)}{\sqrt{\pi} \,\Gamma\left(x+2y\right)}B(x,y)$. Note that the first result is a consequence of $\Gamma(z+1) = z\Gamma(z)$ and the second result is a consequence of Legendre's duplication formula.
With the previous result we can conclude the problem as follows:
\begin{align*}
I &=\frac{1}{a-b} \int_{0}^{\infty} \sqrt{x^n +a} -\sqrt{x^n +b} \, \mathrm{d}x\\
& = \frac{1}{a-b} \left[\int_{0}^{\infty} \sqrt{x^n +a}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x-\int_{0}^{\infty}\sqrt{x^n +b}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x\right]\\
& = \boxed{\frac{a^{\frac1n + \frac12 } - b^{\frac1n + \frac12 } }{(a-b)(n+2)\sqrt{\pi}} \Gamma \left(\frac12 - \frac1n\right) \Gamma \left( \frac1n\right)}
\end{align*}
| {
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"question_score": "8",
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Find the minimum of $a^4+2b^4+3c^4$ Suppose that $a, b,$ and $c$
are positive real numbers satisfying
$a+b+c=3$. Find the minimum of $$a^4+2b^4+3c^4$$
We know $f(x)=x^4$ is convex on the positive reals so by Jensen's Inequality, we have $$a^4+b^4+c^4\ge \frac{1}{27}$$
Hence $$a^4+2b^4+3c^4\ge \frac{1}{27}+b^4+2c^4$$
What now?
| In fact if you use Lagrange multipliers you get:
$\vec{\nabla}(a^4+2b^4+3c^4)=(4a^3,8b^3,12c^3)\ \ \propto\ \ \vec{\nabla}(a+b+c-3)=(1,1,1)$ which gives
$$a^3=2b^3=3c^3$$
same as Macavity $a^3:b^3:c^3=1:\frac 12:\frac 13$ ratio presentation.
Since $x\mapsto x^3\ $ bijective this is also $\ a=\sqrt[3]{2}\,b=\sqrt[3]{3}\,c\ $ and the sum becomes $$a+b+c=\Big(1+\frac 1{\sqrt[3]{2}}+\frac 1{\sqrt[3]{3}}\Big)\,a=3$$
which reveal the origin of this second LHS term you were asking about.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402274",
"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int \frac{dx}{x\sqrt{x^4-1}}$? $$\int \dfrac{dx}{x\sqrt{x^4-1}}$$
I need to solve this integration.
I solved and got $\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$, however the answer given in my textbook is $\dfrac12\sec^{-1}(x^2) + C$
How can I prove that both quantities are equal? Is there something wrong with my answer?
EDIT:
Here's my work:
$$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$
Let $x^4 - 1 = t^2$
$$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$
$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$
| I tell my students that inverse trig functions are angles. So if you write
$$\tan^{-1}\sqrt{x^4-1} = \theta,$$
then
$$\tan\theta = \sqrt{x^4-1}.$$
A right triangle that tells this story has $\theta$ as one angle, $\sqrt{x^4-1}$ as the opposite side and $1$ as the adjacent side. Using Pythagorean theorem we can work out the length $c$ of the hypotenuse:
$$(\sqrt{x^4-1})^2+1^2 = c^2$$
which shows that $c=x^2$.
So $\sec \theta = x^2/1$, that is $\sec^{-1}(x^2) = \theta = \tan^{-1}( \sqrt{x^4-1}).$
| {
"language": "en",
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"source": "stackexchange",
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Prove $8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab)$ Let $a\geq0, b\geq0,c\geq0$. Prove that:
$$8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab).$$
I tried to prove this inequality using only the inequality between the arithmetic mean and the geometric mean.
Does anyone succeed? Thanks!
| Without full expanding:
If $abc = 0$, clearly the inequality is true.
In the following, WLOG, assume that $a \ge b \ge c > 0$.
We split into two cases:
Case 1: $b^2 \ge ca$
Using Holder, we have
\begin{align*}
\mathrm{RHS} &\ge 27\left(\sqrt[3]{bc \cdot ca \cdot c^2} + \sqrt[3]{a^2\cdot b^2\cdot ab}\right)^3\\
&= 27\left(c\sqrt[3]{abc} + ab\right)^3.
\end{align*}
It suffices to prove that
$$3(c\sqrt[3]{abc} + ab) \ge 2\sqrt[3]{abc}\,(a + b + c)$$
or
$$3ab \ge \sqrt[3]{abc}\,(2a + 2b - c)$$
which is true. See Remark 1 at the end.
$\phantom{2}$
Case 2: $b^2 < ca$
Using Holder, we have
\begin{align*}
\mathrm{RHS} &\ge 27\left(\sqrt[3]{a^2 \cdot ca \cdot ab} + \sqrt[3]{bc\cdot b^2\cdot c^2}\right)^3\\
&= 27\left(a\sqrt[3]{abc} + bc\right)^3.
\end{align*}
It suffices to prove that
$$3(a\sqrt[3]{abc} + bc) \ge 2\sqrt[3]{abc}(a + b + c)$$
or
$$3bc \ge \sqrt[3]{abc}\,(2b + 2c - a)$$
which is true. See Remark 2 at the end.
We are done.
Remark 1:
Let $f(c) = \sqrt[3]{abc}\,(2a + 2b - c)$. We have
$f'(c) = \frac{2ab(a + b - 2c)}{3\sqrt[3]{a^2b^2c^2}} \ge 0$
for all $0 < c \le b^2/a$.
It suffices to prove that
$$3ab \ge \sqrt[3]{ab \cdot \frac{b^2}{a}}\,\left(2a + 2b - \frac{b^2}{a}\right)$$
or
$$a + b^2/a \ge 2b$$
which is true.
Remark 2:
Let $g(a) = \sqrt[3]{abc}\,(2b + 2c - a)$. We have
$g'(a) = - \frac{2bc(2a - b - c)}{3\sqrt[3]{a^2b^2c^2}} \le 0$
for all $a \ge b^2/c$.
It suffices to prove that
$$3bc \ge \sqrt[3]{\frac{b^2}{c} \cdot bc}\,\left(2b + 2c - \frac{b^2}{c}\right)$$
or
$$c + b^2/c \ge 2b$$
which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand... like just adding up the fractions until I get to $\frac{7}{8}$ and got $n=7$. Surely there must be a better way. Help appreciated here! Thanks, I really appreciate it.
| this is a telescopic series, this means that each term cancel part of other term, in your case you first need to apply partial fraction on the given expression:
$\sum_{k=1}^n \frac{1}{k(k+1)} =\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} $
this means that the first terms are:
$ k=1: 1-\frac{1}{2} $
$ k=2: \frac{1}{2}-\frac{1}{3} $
$ k=3: \frac{1}{3}-\frac{1}{4} $
$ ...$
$ k=n: \frac{1}{n}-\frac{1}{n+1} $
now you can see that when you sum all the term untill the n-th term you are getting:
$ 1-\frac{1}{n+1} $
now solve $ 1-\frac{1}{n+1} = \frac{7}{8} $ to get $ x=7 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Directional derivatives of $f(x,y) := \frac{2x^2y}{x^4 + y^2}$ in $(0,0)$ Given is the function $f: \mathbb{R^2} \to \mathbb{R}$ with
$ f(x,y) := \frac{2x^2y}{x^4 + y^2} \text{ for all } (x,y) \neq (0,0) \text{ and } 0 \text{ for all } (x,y) = (0,0)$
How can one prove that in all (bilateral) directional derivatives of $f$ exist in $(0,0)$?
I would have calculated the gradients
$$f_x = \frac{\partial}{\partial x} \frac{2x^2y}{x^4 + y^2} = \dfrac{4yx}{x^4+y^2}-\dfrac{8yx^5}{\left(x^4+y^2\right)^2}$$
and
$$f_y = \frac{\partial}{\partial y} \frac{2x^2y}{x^4 + y^2} = \dfrac{2x^2}{y^2+x^4}-\dfrac{4x^2y^2}{\left(y^2+x^4\right)^2}$$
and then I would have calculated $f_x(0,0)$ and $f_y(0,0)$, but that would lead to division by zero.
I also thought that we need a vector for directional derivatives.
What am I misunderstanding?
| According to the definition of directional derivative at the origin, if we let $v = (a,b)$, we get that:
\begin{align*}
D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\
& = \lim_{t\to 0}\frac{2t^{3}a^{2}b}{t^{3}(t^{2}a^{4} + b^{2})}\\\\
& = \lim_{t\to 0}\frac{2a^{2}b}{t^{2}a^{4} + b^{2}}\\\\
& = \frac{2a^{2}b}{b^{2}}\\\\
& = \frac{2a^{2}}{b}
\end{align*}
and we are done.
Hopefully this helps!
| {
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Effective method to solve $\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $ I want to know, is there an easy method to solve below equation $$\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$ I tried it by plotting and find the solution $x=0,1,-1$ then I tried to solve it by sustitution, multiplying by $(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x} \sqrt[3]{1-x} +\sqrt[3]{(1-x)^2})$ and so on ...
but I got stuck in solving it normally.
I am thankful if anyone can show me the clue.
Implicit: when I plot both sides , I feel if we name $$f(x)=\frac{(\sqrt[3]{1+x} -\sqrt[3]{1-x})}{(\sqrt[3]{1+x} +\sqrt[3]{1-x})}$$ then $$f^{-1}= \frac{x(x^2+3)}{3x^2+1} $$ so It is an effective way to solve $$f(x)=f^{-1}(x)=x$$
| $$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$
Use componendo-dividendo:
$$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x} + (\sqrt[3]{1+x} +\sqrt[3]{1-x})}{\sqrt[3]{1+x} -\sqrt[3]{1-x} - (\sqrt[3]{1+x} +\sqrt[3]{1-x})} = \frac{x(x^2+3) + 3x^2+1}{x(x^2+3) -(3x^2+1)} $$
$$ \frac{2\sqrt[3]{1+x}}{-2\sqrt[3]{1-x}} = \frac{x^3+3x^2+3x+1}{x^3-3x^2+3x-1}$$
$$ -\sqrt[3]{\frac{{1+x}}{{1-x}}} = \frac{(x+1)^3}{(x-1)^3} = -\left(\frac{1+x}{1-x}\right)^3$$
I think you can handle it from here
| {
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How to prove by induction $\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $ I'm trying to solve this problem about mathematical induction but every time I try to solve it I end up with an incorrect answer.
The problem is as follows:
Prove by induction : $$\frac 12 + \frac 14 + ... + \frac {1}{2^n} = 1- \frac {1}{2^n} $$
my work:
first I prove that it holds for n=1: $\frac 12 = 1- \frac {1}{2^1} = \frac 12$
Now assume it holds for n=k, prove for n=k+1 the sum is equal to $1 - \frac{1}{2^{k+1}}$
then $\frac 12 + \frac 14 +...+ \frac {1}{2^k} + \frac {1}{2^{k+1}} = 1-\frac {1}{2^k} + \frac {1}{2^{k+1}} = 1 - \frac {2^{k+1}+2^k}{2^{k+1}2^k} = 1- \frac {2(2^k)+2^k}{2(2^k)2^k} = 1 - \frac {3(2^k)}{2(2^k)2^k} = 1- \frac {3}{2(2^k)} = 1- \frac {3}{2^{k+1}} $
I would very much appreciate if at least someone can tell me where I'm wrong so I could try to solve it by myself. I think the mistake is at factoring $2^{k+1}+2^k$
| You went wrong when you combined the fractions. You should have $$1- \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{2^{k+1} - 2^k}{2^{k+1}2^k} = 1 - \frac{2 - 1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the number of $2\times2$ matrices whose sum of element is zero If the number of non skew symmetric matrices $A = [a_{ji}],$ where $a_{ji}\in\{-2,-1,0,1,2\} $ and the sum of elements is zero then find the number of such matrices.
I tried setting up a recursion using (-n....0....n) and adding -n-1 and n+1 but not able to build further
The answer given is 80 can anyone please help in how to approach such problems
| split into cases based on the number of zeros in your 4 elements, there can only be 0, 1, or 2.
The biggest case will be for 0 zeros, for which the elements would be from one of 3 sets,
{1, 1, -1, -1}, {2, 2, -2, -2} or {1, -1, 2, -2}
To find the number of matrices you just need the permutations of each of the sets
$$ n(0) = \frac{4!}{2!2!} + \frac{4!}{2!2!} + 4! \\ = 6 + 6 + 24 = 36$$
For 1 zero the sets have to be made up from
{1, 1, -2, 0} and {-1, -1, 2, 0}
$$n(1) = \frac{4!}{2!} + \frac{4!}{2!} = 12 + 12 = 24$$
for 2 zeros the sets have to be
{0, 0, 1, -1} or {0, 0, 2, -2}
$$n(2) = 12 + 12 = 24$$
out of these only 4 are skew-symmetric
$$
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 2 \\
-2 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & -2 \\
2 & 0 \\
\end{bmatrix}
$$
So total number of non-skew symmetric matrices is
$$ N_{tot}=36+24+24-4=80$$
| {
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Does the following definite integral exist? I encounter a problem in which I would need to deal with the folloing definite integral
$$I(t)=\int_{0}^{\infty} \mathrm dp \frac{p^2}{\omega^5} \sin^2\left(\frac{\omega t}{2}\right)$$
in which $$\omega=\sqrt{m^2+p^2}$$
where m is just some positive constant, t is also a parameter. But I would like to obtain the value of $I(t)$ when t is small. My approach to this problem is the following
$$I(t)=I(0)+I(0)'t+\frac{I(0)''}{2}t^2...$$
$$\frac{dI}{dt}|_{t=0}=0$$
However
$$I(0)''\propto \int dp \frac{p^2}{\omega^3}$$
which diverge like $\ln(p)$
Can anyone help me to find the expansion of I(t) in terms of t?
| With Mathematica I have:
$$\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{1}{2} \sqrt{m^2+p^2} t\right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\int_0^{\infty } \left(\frac{p^2}{2 \left(m^2+p^2\right)^{5/2}}-\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right) \, dp=\\\frac{1}{6 m^2}-\frac{\pi G_{1,3}^{2,0}\left(\frac{m
t}{2},\frac{1}{2}|
\begin{array}{c}
\frac{5}{2} \\
0,1,\frac{1}{2} \\
\end{array}
\right)}{8 m^2}$$
for: $t>0$
MMA code:
Integrate[(p^2 Sin[1/2 Sqrt[m^2 + p^2] t]^2)/(m^2 + p^2)^( 5/2), {p, 0, \[Infinity]}] == 1/(6 m^2) - (\[Pi] MeijerG[{{}, {5/2}}, {{0, 1}, {1/2}}, (m t)/2, 1/ 2])/(8 m^2)
Using MellinTransfrom:
$\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_t\left[\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right](s) \,
dp\right](t)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{1}{2} p^2 \left(m^2+p^2\right)^{-\frac{5}{2}-\frac{s}{2}} \cos \left(\frac{\pi s}{2}\right) \Gamma (s) \,
dp\right](t)=\\\mathcal{M}_s^{-1}\left[\frac{m^{-2-s} \sqrt{\pi } \cos \left(\frac{\pi s}{2}\right) \Gamma \left(1+\frac{s}{2}\right) \Gamma (s)}{8 \Gamma
\left(\frac{5+s}{2}\right)}\right](t)=\\\frac{\pi G_{1,3}^{2,0}\left(\frac{m t}{2},\frac{1}{2}|
\begin{array}{c}
\frac{5}{2} \\
0,1,\frac{1}{2} \\
\end{array}
\right)}{8 m^2}$
| {
"language": "en",
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MCQ about the product of $4$ consecutive odd numbers The product of four consecutive odd numbers must be …
(A) A multiple of 3, but not necessarily of 9.
(B) A multiple of 5 .
(C) A multiple of 7.
(D) A multiple of 9.
(E) A multiple of 3×5×7×9= 945.
Let $n\in \mathbb{Z}$, then
$(2n-1)(2n+1)(2n+3)(2n-3)$
$= (4n^2-1)(4n^2-9)$
$= 16n^4-40n^2+9$
$= 8n^2(2n^2-5)+9.$
For being a multiple of $9$, need $8n^2(2n^2-5)=9m$, for some suitable value of $m\in \mathbb{Z}$; or
$$2n^2-5=9m\cup n=3j, \text{for suitable} \,\,j\in \mathbb{Z}.$$
But, it gets more confusing to pursue further using my approach. Is my approach workable?
Want to add that if take case of showing not divisible by $5$, then : $x+9= 5k$, then $x= 5k-9$.
Hence, for $j, k\in \mathbb{Z}$, $$8n^2 = 5k-9\cup (2n^2-5)= 5j-9.$$
Both options are seemingly not possible, though not clear how to show theoretically the impossibility of both.
| Let $n$ be the first of the consecutive odd integers, with
$$f(n) = n(n+2)(n+4)(n+6)$$
If $n \equiv 0 \pmod{3}$, then $3 \mid f(n)$ (i.e., $f(n)$ is an integral multiple of $3$). If $n \equiv 1 \pmod{3}$, then $n + 2 \equiv 0 \pmod{3}$ so $3 \mid f(n)$. Finally, if $n \equiv 2 \pmod{3}$, then $n + 4 \equiv 0 \pmod{3}$ so $3 \mid f(n)$. Thus, in all cases, $f(n)$ is a multiple of $3$.
To show $f(n)$ is not necessarily a multiple of $9$, if $n \equiv 2 \pmod 9$ (e.g., $n = 11$), then $f(n) \equiv 2(4)(6)(8) \equiv 6 \pmod{9}$. Thus, option (A) is correct, plus options (D) and (E) are incorrect.
Note if $n \equiv 2 \pmod{5}$ (e.g., $n = 7$), then $f(n) \equiv 2(4)(6)(8) \equiv 4 \pmod{5}$, and if $n \equiv 2 \pmod{7}$ (e.g., $n = 9$), then $f(n) \equiv 2(4)(6)(8) \equiv 6 \pmod{7}$. Thus, these cases show that $f(n)$ is not necessarily a multiple of either $5$ or $7$, so both options (B) and (C) are not correct.
| {
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Combinatorial proof of $\sum_{k=0}^{n} k \binom{n+1}{k+1} n^{n-k} = n^{n+1}$ Show :
$$\sum_{k=0}^{n} k \binom{n+1}{k+1} n^{n-k} = n^{n+1}$$
for natural number $n$. I randomly discovered this identity, and managed to prove it using simple algebra. I tried a combinatorial proof of this, but it seems too difficult for me.
The RHS is basically distributing $n+1$ people to $n$ different groups where an empty group is possible, but I could not show that the LHS is the same. Picking $k+1$ people out of $n+1$ equals $\binom{n+1}{k+1}$, and distributing others($n-k$ people) is equal to $n^{n-k}$ ; and now I am stuck with that $k$. Also I have no idea what to do with $k+1$ people I just picked; if I distribute them to $n$ groups then it will be overlapped with other terms of the sum.
A proof using algebra is also welcome, just in case.
| $\underline{\text{Preliminary Results}}$
PR-1
$\displaystyle\sum_{i=0}^r \binom{r}{i}r^{r-i} = (1 + r)^r.$
This is directly from binomial expansion.
PR-2
$\displaystyle\sum_{k=0}^n \binom{n+1}{k+1}n^{n-k} = (1 + n)^{n+1} - n^{n+1}.$
Proof
First, re-index it to
$\displaystyle\sum_{k=1}^{n+1} \binom{n+1}{k}n^{n+1-k}.$
Then, re-express it as
$\displaystyle\left[\sum_{k=0}^{n+1} \binom{n+1}{k}n^{n+1-k}\right] - \binom{n+1}{0}n^{n+1}.$
Then, use PR-1 to convert this to
$\displaystyle (1+n)^{n+1} - n^{n+1}.$
PR-3
$\displaystyle k\binom{n+1}{k+1} = \left[(n+1) \times \binom{n}{k}\right] - \binom{n+1}{k+1}.$
Proof
$\displaystyle k\binom{n+1}{k+1} =
\frac{(n+1)!}{(n-k)!} \times \frac{k}{(k+1)!}$
$\displaystyle =~
\frac{(n+1)!}{(n-k)!} \times \left[\frac{k+1}{(k+1)!} - \frac{1}{(k+1)!}\right]$
$\displaystyle =~
\frac{(n+1)!}{(n-k)!} \times \left[\frac{1}{k!} - \frac{1}{(k+1)!}\right]$
$\displaystyle =~
\left[(n+1) \times \binom{n}{k}\right] - \binom{n+1}{k+1}.$
Using PR-3
$\displaystyle \sum_{k=0}^n k\binom{n+1}{k+1} n^{n-k} = S - T$
where $\displaystyle ~S = (n+1)\sum_{k=0}^n \binom{n}{k}n^{n-k}$
and $\displaystyle ~T = \sum_{k=0}^n \binom{n+1}{k+1} n^{n-k}.$
Using PR-1
$\displaystyle S = (n+1)\left[(1 + n)^n\right] = (n+1)^{n+1}.$
Using PR-2
$\displaystyle T = (1+n)^{n+1} - n^{n+1}.$
Thus, $\displaystyle S - T = n^{n+1}.$
| {
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If $P_n=\alpha^n+\beta^n\;, \alpha+\beta=1, \;\alpha \cdot \beta=-1,\;P_{n-1}=11,\; P_{n+1}=29$ Find $(P_n)^2,\;$ where $n\in \mathbb N$
If $P_n=\alpha^n+\beta^n\;, \alpha+\beta=1, \;\alpha \cdot \beta=-1,\;P_{n-1}=11,\; P_{n+1}=29$, $\alpha$ and $\beta$ are real numbers. Find $(P_n)^2,\;$ where $n\in \mathbb N$
My Approach:
Method $1$ :
$P_{n-1}\cdot P_{n+1}=11\cdot 29 \implies\; (\alpha^{n-1}+\beta^{n-1})\cdot (\alpha^{n+1}+\beta^{n+1})=\alpha^{2n}+\beta^{2n}+(\alpha \beta)^{n-1}\cdot(\alpha^2+\beta^2)$
$\implies \; 319=P_{2n}+3 (-1)^{n-1}$
$\implies\;P_{2n}=319-3(-1)^{n-1} \implies P_{2n}=319+3(-1)^n$
Now
$(P_{n})^2=(\alpha^n+\beta^n)^2=\alpha^{2n}+\beta^{2n}+2(\alpha \beta)^n\;\implies\;(P_{n})^2=P_{2n}+2(-1)^n$
$\implies \; (P_{n})^2=319+5(-1)^n$
$\implies (P_n)^2=324\;$ If $n$ is even and $(P_n)^2=314\;$ if $n$ is odd.
But given answer is $324$ only.
Method $2$:
Form a quadratic equation with sum of roots and product of roots is given
$\alpha^2-\alpha-1=0\quad$ and $\quad\beta^2-\beta-1=0$
$\implies \;\alpha^{n+1}=\alpha^n+\alpha^{n-1}\cdots(1)\quad$ and $\quad \beta^{n+1}=\beta^{n}+\beta^{n-1}\cdots (2)$
Now add both equation $(1)+(2)$
$\implies$ $P_{n+1}=P_{n}+P_{n-1} \; \implies \; P_{n}=18$
$\implies \; (P_n)^2=324$.
My doubt: What is wrong with my method $1$ ?
| Since $P_n = \alpha^n + \beta^n $, then $P_n$ satisfies the difference equation
$ (E - \alpha)(E - \beta) P_n = 0 $
where $E$ is the advance operator, i.e. $E\left(P_n\right) = P_{n+1}$
Hence, multiplying the operators, we get the difference equation
$P_{n+2} - (\alpha + \beta) P_{n+1} + \alpha \beta P_{n} = 0 $
Substituting the given values of $(\alpha + \beta)$ and $\alpha \beta$,
$P_{n+2} - P_{n+1} - P_{n} = 0 $
That is,
$P_{n+2} = P_{n+1} + P_n$
Now
$P_1 = \alpha + \beta = 1 $
$P_2 = \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2 \alpha \beta = 3 $
Therefore, the sequence $\{P_n\}$ is as follows:
$1, 3, 4, 7, 11, 18, 29, 47, ... $
We already see our terms $11$ and $29$ , hence the one between them is $ 18 $, whose square is $324$
| {
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"source": "stackexchange",
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Double integration in polar coordinates: $\int_{-\pi /4}^{\pi /4} \int_0^{a \sqrt{2 \cos (2 \theta )}} \sqrt{2a^2-r^2} r dr d\theta $ This is exercise 14 in Section 9.3 of Serge Lang's Calculus of Several Variables.
Question statement: The base of a solid is the area of one loop in Exercise 13(b)* and the top is bounded by the function (in terms of polar coordinates) $f^*(r,\theta)=\sqrt{2a^2-r^2}$. Find the volume.
*Note: Question 13(b): Find the area enclosed by $r^2 = 2a^2 \cos(2 \theta)$.
Solution attempt:
Let $V$ be the required volume. Then,
$$V=\displaystyle\int_{-\pi /4}^{\pi /4} \int_0^{a \sqrt{2 \cos (2 \theta )}} \sqrt{2a^2-r^2} r dr d\theta.$$
For the inner integral, substitute $u=2a^2-r^2$. Then $du = -2r dr$. Then the lower limit becomes $u=2a^2$, and the upper limit is $u=2a^2-r^2=2a^2(1-\cos(2\theta)=4a^2\sin^2\theta.$ Using this, I evaluate the inner integral as
$$\int_{4a^2 \sin^2 2 \theta}^{2a^2}\dfrac{\sqrt{u}}{2}du = \dfrac{2a^3}{3}(\sqrt{2}-4\sin^3 \theta).$$
Now, I see that $\sin^3 \theta$ is odd, and so that integral vanishes. This gives me $$V = \dfrac{\pi a^3 \sqrt{2}}{3},$$ as my answer.
However, this does not match the answer given in the book. The answer given is $$\dfrac{2\sqrt{2}\pi a^3}{3} - \dfrac{64}{9}a^3+\dfrac{40 \sqrt{2} a^3}{9}.$$ The book gives this as the integral $$2\displaystyle\int_{-\pi /4}^{\pi /4} \int_0^{a \sqrt{2 \cos (2 \theta )}} \sqrt{2a^2-r^2} r dr d\theta.$$
I am unsure where I went wrong in the evaluation of the integral, and why there is a factor of 2. Please, can someone help me identify where I went wrong? Thank you.
| You wrote
$$\int^{2a^2}_{4a^2\sin^22θ}
\frac{\sqrt u}{2}\,du=\color{red}{\frac{2a^3}{3}(\sqrt 2−4\sin^3θ)}
$$ in your arguments.
However, it should be $$\frac{2a^3}3(\sqrt 2−4\left|\sin^32θ\right|)
.$$
| {
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A problem with $\int_0^{\infty} \frac{x^a}{(1-x)^2} \ \mathrm dx$ I want to evaluate:
$\displaystyle \tag*{} I = \int_0^{\infty} \frac{x^a}{(1-x)^2} \ \mathrm dx$
Where $-1<a<1$. I tried to split the integral at $x=1$
$\displaystyle \tag*{}I= \int_0^1 \frac{x^a}{(1-x)^2} \ \mathrm dx + \int_1^\infty \frac{x^a}{(1-x)^2} \ \mathrm dx $
However, both the integrals diverge. Now I chose $\epsilon$ such that $\epsilon \ll 1$. So
$\displaystyle \tag*{} I= \int_0^{1-\epsilon} \frac{x^a}{(1-x)^2} \ \mathrm dx + \int_{1-\epsilon}^\infty \frac{x^a}{(1-x)^2} \ \mathrm dx$
Now, can anyone help me solve this? Is my steps correct at the first step? Any help would be appreciated.
| New Answer. Let $0 < \varepsilon < 1$, and consider the integral
\begin{align*}
I_{\varepsilon} = \int_{0}^{1-\varepsilon} \frac{x^a}{(x-1)^2} \, \mathrm{d}x + \int_{1+\varepsilon}^{\infty} \frac{x^a}{(x-1)^2} \, \mathrm{d}x.
\end{align*}
To analyze the behavior of this integral, we consider the counter-clockwise contour as below:
Then, as $ R\to\infty $,
\begin{align*}
0
&= \lim_{R\to\infty} \oint_{\text{contour}} \frac{z^a}{(z-1)^2} \, \mathrm{d}z \\
&= I_{\varepsilon}
+ \int_{\gamma_{\varepsilon}} \frac{z^a}{(z-1)^2} \, \mathrm{d}z
+ \int_{-\infty}^{0} \frac{z^a}{(z-1)^2} \, \mathrm{d}z.
\end{align*}
From this, we get
\begin{align*}
I_{\varepsilon}
&= - \int_{-\infty}^{0} \frac{z^a}{(z-1)^2} \, \mathrm{d}z - \int_{\gamma_{\varepsilon}} \frac{z^a}{(z-1)^2} \, \mathrm{d}z \\
&= - \int_{0}^{\infty} \frac{(-t)^a}{(1+t)^2} \, \mathrm{d}t + \int_{0}^{\pi} \frac{(1+\varepsilon e^{i\theta})^a}{(\varepsilon e^{i\theta})^2} \, i \varepsilon e^{i\theta} \mathrm{d}\theta \\
&= - e^{i\pi a} \int_{0}^{\infty} \frac{t^a}{(1+t)^2} \, \mathrm{d}t + \frac{i}{\varepsilon} \int_{0}^{\pi} (1+\varepsilon e^{i\theta})^a e^{-i\theta} \, \mathrm{d}\theta.
\end{align*}
Using the binomial series and the identity $\int_{0}^{\infty} \frac{t^a}{(1+t)^2} \, \mathrm{d}t = \frac{\pi a}{\sin(\pi a)}$ together, we get
\begin{align*}
I_{\varepsilon}
&= - e^{i\pi a} \frac{\pi a}{\sin(\pi a)} + \frac{i}{\varepsilon} \sum_{n=0}^{\infty} \binom{a}{n} \int_{0}^{\pi} (\varepsilon e^{i\theta})^n e^{-i\theta} \, \mathrm{d}\theta \\
&= - e^{i\pi a} \frac{\pi a}{\sin(\pi a)} + \frac{i}{\varepsilon} \biggl[ 2 \varepsilon + i \pi a \epsilon - \sum_{k=1}^{\infty} \binom{a}{2k} \frac{2}{2k-1} \varepsilon^{2k} \biggr] \\
&= \bbox[border:2px #E0E0FF solid; padding:5px;]{\frac{2}{\varepsilon} - \frac{\pi a}{\tan (\pi a)} - \sum_{k=1}^{\infty} \frac{2 a(a-1)\cdots(a-2k+1)}{(2k)!(2k-1)}\varepsilon^{2k-1}}
\end{align*}
Old Answer. We first extract the contribution from the double pole at $x = 1$:
\begin{align*}
I_{\varepsilon}
&= \biggl[ \int_{0}^{1-\varepsilon} \frac{x^a - 1}{(x-1)^2} \, \mathrm{d}x + \int_{0}^{1-\varepsilon} \frac{1}{(x-1)^2} \, \mathrm{d}x \biggr] + \biggl[ \int_{1+\varepsilon}^{\infty} \frac{x^a-1}{(x-1)^2} \, \mathrm{d}x - \int_{1+\varepsilon}^{\infty} \frac{1}{(x-1)^2} \, \mathrm{d}x \biggr] \\
&= \int_{0}^{1-\varepsilon} \frac{x^a - 1}{(x-1)^2} \, \mathrm{d}x + \int_{1+\varepsilon}^{\infty} \frac{x^a - 1}{(x-1)^2} \, \mathrm{d}x + \frac{2}{\varepsilon} - 1.
\end{align*}
Next, substituting $x \mapsto 1/x$ in the second integral,
\begin{align*}
I_{\varepsilon}
= \int_{0}^{1-\varepsilon} \frac{x^a - 1}{(x-1)^2} \, \mathrm{d}x + \int_{0}^{\frac{1}{1+\varepsilon}} \frac{x^{-a} - 1}{(x-1)^2} \, \mathrm{d}x + \frac{2}{\varepsilon} - 1.
\end{align*}
By noting that $\frac{1}{1+\varepsilon} = 1-\varepsilon + \mathcal{O}(\varepsilon^2)$ as $\varepsilon \to 0^+$ and $\frac{x^{-a} - 1}{(x-1)^2} = \mathcal{O}(|x-1|^{-1})$ as $x \to 1$, we easily find that
\begin{align*}
I_{\varepsilon}
&= \int_{0}^{1-\varepsilon} \frac{x^a - 1}{(x-1)^2} \, \mathrm{d}x + \int_{0}^{\color{blue}{1-\varepsilon}} \frac{x^{-a} - 1}{(x-1)^2} \, \mathrm{d}x + \frac{2}{\varepsilon} - 1 + \mathcal{O}(\varepsilon) \\
&= \int_{0}^{1-\varepsilon} \frac{x^a + x^{-a} - 2}{(x-1)^2} \, \mathrm{d}x + \frac{2}{\varepsilon} - 1 + \mathcal{O}(\varepsilon) \\
&= \int_{0}^{\color{blue}{1}} \frac{x^a + x^{-a} - 2}{(x-1)^2} \, \mathrm{d}x + \frac{2}{\varepsilon} - 1 + \mathcal{O}(\varepsilon),
\end{align*}
where the last step follows from $x^a + x^{-a} - 2 = \mathcal{O}(|x-1|^2)$ as $x \to 1$. The last integral can be computed explicitly, yielding
$$ \int_{0}^{1} \frac{x^a + x^{-a} - 2}{(x-1)^2} \, \mathrm{d}x = 1 - \frac{\pi a}{\tan (\pi a)}. $$
Combining altogether, we obtain the following asymptotic formula
$$ \bbox[border:2px #E0E0FF solid; padding:5px;]{I_{\varepsilon} = \frac{2}{\varepsilon} - \frac{\pi a}{\tan (\pi a)} + \mathcal{O}(\varepsilon)} $$
as $\varepsilon \to 0^+$.
| {
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Suppose that $a_n\neq0$ for all $n\in\mathbb{N}$ and $L=\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$. Prove that $a_{n}\to 0$. If $L<1$, how do I prove that $\lim_{n\rightarrow\infty}a_n=0$ ?
I tried setting $L$ as $1-\delta$, because
$$L<1\Longleftrightarrow L=1-\delta,\, \delta>0$$
So, by definition:
$$\forall\varepsilon>0, \exists N>0:n>N\Rightarrow\left|\left|\frac{a_{n+1}}{a_n}\right|-1+\delta\right|<\varepsilon$$
But I don't think that solving that inequality will help me at all, so I got stuck.
| Let $\epsilon > 0$. Let $N$ be large enough so that $\big| \frac {a_{n+1}} {a_n} \big| < 1 - \epsilon$ for all $n \ge N$. Then,
$$
\frac {1} {a_n} =
\frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}} \cdot
\frac {1} {a_{n+k}} \iff \\
\frac {a_{n+k}} {a_n} =
\frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}}
$$
So that,
$$
\Big| \frac {a_{n+k}} {a_n} \Big| =
\Big| \frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}} \Big| < \\
(1-\epsilon)^k \xrightarrow{k \rightarrow \infty} 0 \iff \\
a_{n+k} \xrightarrow{k \rightarrow \infty} 0
$$
I.e., $\lim_{n \rightarrow \infty} a_n = 0$.
| {
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proving or disproving that $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$
Prove or disprove the following: if N is a positive integer and $\epsilon > 0$ is such that $N < \frac{.01}{\epsilon}$, then $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$.
Clearly from the question, we must have $\epsilon <.01$. Using the binomial theorem, we need to prove $(1+\epsilon)^{N - 1} - 1 < 1.01 N\epsilon$. I tried to approximate this using Taylor's theorem, which states that $(1+\epsilon)^{N-1} = 1 + (N - 1)\epsilon + {N-1\choose 2} (1+x)\epsilon^2,$ where $x$ is between $0$ and $\epsilon$, but I'm not sure if this is good enough. Perhaps it might be useful to prove a stronger/weaker inequality?
Below is my attempt to use the mean-value theorem for the function $f(x) = (1+x)^{N-1}$ on the interval $[0,\epsilon]$. The derivative of this function is $(N-1)(1+x)^{N-2}$. By the mean value theorem, there exists some $c$ in $(0,\epsilon)$ so that $(N-1)(1+c)^{N-2} = \frac{(1+\epsilon)^{N-1} - 1}{\epsilon},$ so we need to show $(N-1)(1+c)^{N-2} \epsilon < 1.01 N\epsilon^2$. From above, $\epsilon < 0.01$, so $c < 0.01$. For $N = 1,2,$ the inequality is equivalent to $0 < 1.01\epsilon$ and $\epsilon < 2.02\epsilon$, both of which are obvious. So we may assume $N>2$ and hence $(1+c)^{N-2} < (1.01)^{N-2}.$ So $(N-1)(1+c)^{N-2}\epsilon < (N-1)(1.01)^{N-2}\epsilon.$ So we just need to show $(N-1)(1.01)^{N-2}\epsilon < 1.01N\epsilon\Leftrightarrow (1.01)^{N-3} < \frac{N}{N-1}.$ Let $f(x) = \ln x - \ln (x-1) - (x-3)\ln 1.01$. Clearly $f(3) > 0$. For $x\ge 3$, we have $f'(x) = \frac{1}{x(x-1)} - \ln 1.01.$
| Fact 1: Let $c, x$ be real numbers such that $0 < c < 1/100$ and $1 \le x \le \frac{1}{100c}$. Then
$$(1 + c)^{x - 1} - 1 < \frac{101}{100} x c.$$
(The proof is given at the end.)
According to Fact 1, the desired result follows.
Proof of Fact 1:
Denote $q = 101/100$.
Taking logarithm, it suffices to prove that
$$f(x) := \ln(1 + q x c) - (x - 1)\ln (1 + c) > 0.$$
Let
$$t := \frac{1/(100c) - x}{1/(100c) - 1} \in [0, 1].$$
We have $x = t\cdot 1 + (1 - t)\cdot 1/(100c)$.
Since $f(x)$ is concave (easy to prove), we have
$$f(x) = f(t\cdot 1 + (1 - t)\cdot 1/(100c))
\ge t f(1) + (1 - t)f(1/(100c)).$$
Also, we have $f(1) = \ln(1 + qc) > 0$ and
\begin{align*}
f(1/(100c)) &= \ln(1 + q/100) - (1/(100c) - 1)\ln(1 + c)\\
&\ge \ln(1 + q/100) - (1/(100c) - 1)c \tag{1}\\
&= \ln(1 + q/100) - 1/100 + c \\
&> \ln(1 + q/100) - 1/100 \\
&= \ln(1 + 101/10000) - 1/100\\
&> 0
\end{align*}
where we have used $\ln(1 + u) \le u$ for all $u \ge 0$ in (1).
Thus, $f(x) > 0$ for all $1 \le x \le 1/(100c)$.
We are done.
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If I define summation recursively, how do I prove formally that both of the following definitions are equivalent? First definition:
$\sum_{x=a}^{b} f(x)=f(b)+\sum_{x=a}^{b-1} f(x)$ if $a\leq b$, and equals $0$ otherwise
Second definition:
$\sum_{x=a}^{b} f(x)=f(a)+\sum_{x=a+1}^{b} f(x)$ if $a\leq b$, and equals $0$ otherwise
Intuitively, I have no doubt that both statements are always equal, but I am struggling to come up with a formal proof for it. I tried induction over $b$, but only got so far on working with the second definition:
$\sum_{x=a}^{b+1} f(x)=f(b+1)+\sum_{x=a}^{b}$
| First, we assume $a,b$ are integers (and so must be finite). If $a > b$, both sums yield $0$ (and thus agree), so assume $a \le b$.
A, intuitive direct approach is to apply the recursion. You have with the first definition:
$$
\begin{split}
\sum_{x=a}^b f(x)
&= f(a) + \sum_{x=a+1}^b f(x) \\
&= f(a) + f(a+1) + \sum_{x=a+2}^b f(x) \\
&= f(a) + f(a+1) + \ldots + f(b-1) + \sum_{x=b}^b f(x) \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b) + \sum_{x=b+1}^b f(x) \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b) + 0 \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b)
\end{split}
$$
The second definition yields
$$
\begin{split}
\sum_{x=a}^b f(x)
&= f(b) + \sum_{x=a}^{b-1} f(x) \\
&= f(b-1) + f(b) + \sum_{x=a}^{b-2} f(x) \\
&= f(a+1) + \ldots + f(b-1) + f(b) + \sum_{x=a}^a f(x) \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b) + \sum_{x=a-1}^a f(x) \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b) + 0 \\
&= f(a) + f(a+1) + \ldots + f(b-1) + f(b)
\end{split}
$$
so the sums agree.
A more formal approach is to use induction on the number of terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inverse of a function related area problem
My solution is based on the following diagram which I will illustrate in details
Now ${A_1} = \frac{5}{4};{A_3} = 1$
$g\left( {2x} \right) = 2f\left( x \right)$
$g\left( 2 \right) = 2f\left( 1 \right) = 2 \times 1 = 2$
$g\left( 4 \right) = 2f\left( 2 \right)$
$\int\limits_1^8 {xf'\left( x \right)dx} = \left. {xf\left( x \right)} \right|_1^8 - \int\limits_1^8 {\frac{d}{{dx}}x\left( {\int {f'\left( x \right)dx} } \right)dx} $
$\int\limits_1^8 {xf'\left( x \right)dx} = 8f\left( 8 \right) - f\left( 1 \right) - \int\limits_1^8 {f\left( x \right)dx} $
Not able to proceed from here
|
Lemma: If $f(x)$ is a continuous and increasing function and $a<b$, then:
$$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\, f(b)-a\, f(a). $$
To prove it, you just need to draw a picture or look at this Wikipedia page.
Now, following my comment, required integral = $ \displaystyle \int_{1}^8 x f'(x) dx= \int_{1}^8 g(x) dx=I\quad $ (say)
And, $\;f(n)=g(n)=n\;$ for $\; n=1,2,4,8.$
Using the above lemma on the given function $f$, we have
\begin{align}
&\int_{1}^{2} f(x) dx+\int_{f(1)}^{f(2)} g(x) dx=2f(2)-f(1)\\
\implies & \frac5{4} + \int_{1}^{2} g(x) dx = 3\\
\implies & \int_{1}^{2} g(x) dx=\frac7{4} \tag{1}
\end{align}
Now, $\;2f(x)=g(2x)\:$ and $\: \displaystyle\int_{1}^{2} f(x) dx=\frac5{4}$
\begin{align}
\implies & \frac{1}{2}\int_{1}^{2} g(2x) dx=\frac5{4}\\
\implies & \frac{1}{4}\int_{2x=2}^{2x=4} g(2x) d(2x)=\frac5{4}\\
\implies & \frac{1}{4}\int_{2}^{4} g(x) dx=\frac5{4}\\
\implies &\int_{2}^{4} g(x) dx=5 \tag{2}
\end{align}
Again using the lemma on $f$, we have
\begin{align} &\int_{2}^{4} f(x) dx+\int_{f(2)}^{f(4)} g(x) dx=4f(4)-2f(2)\\
\implies & \int_{2}^{4} f(x) dx + \int_{2}^{4} g(x) dx = 16-4\\
\implies & \int_{2}^{4} f(x) dx=12-5=7 \qquad \text{using (2)}
\end{align}
Again, $\;2f(x)=g(2x)\:$ and $\: \displaystyle \int_{2}^{4} f(x) dx=7$
\begin{align}
\implies & \frac{1}{2}\int_{2}^{4} g(2x) dx=7\\
\implies & \frac{1}{4}\int_{2x=4}^{2x=8} g(2x) d(2x)=7\\
\implies & \frac{1}{4}\int_{4}^{8} g(x) dx=7\\
\implies &\int_{4}^{8} g(x) dx=28 \tag{3}
\end{align}
From $(1), (2)$ and $(3)$
\begin{align} I&=\int_{1}^8 g(x) dx\\
&= \int_{1}^2 g(x) dx+\int_{2}^4 g(x) dx+\int_{4}^8 g(x) dx\\
&= \frac7{4} + 5 +28\\
&= \frac{139}4
\end{align}
Since, $\text{HCF}(139,4)=1\;$ required answer is $139+4=143$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maclaurin series of $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has all coefficients positive I had a similar problem posted here, and after experimentation with WA I noticed that the function $1-\left(\frac{\sin x}{x}\right) ^{\frac{2}{5}}$ has a Maclaurin expansion with all coefficients positive.
It seems to hold. A possible approach might using the product expansion for the function $\sin x$. Or maybe a differential equation.
Thank you for your interest!
$\bf{Added:}$ Denote the expression by $y(x)$. It's enough to show that $z\colon=y'$ has a positive expansion. We have
$$x\cdot z' = \left(\frac{x^2}{1 - x \cot x} + \frac{3}{5}( 1- x \cot x) - 2 \right)\cdot z$$
If the expression in the brackets has a positive expansion (it seems so) then we can show that $z$ has a positive expasion.
| Using the same approach as earlier
$$1-\left(\frac{\sin (x)}{x}\right)^a= \frac{a x^2}{6}\Bigg[1+\frac 1{60} \sum_{n=1}^\infty (-1)^n\frac{P_n(a)}{b_n }x^{2n}\Bigg]$$ where the first $b_n$ are
$$\{1,126,15120,997920,16345929600,\cdots\}$$ and the first $P_n(a)$ are
$$\left(
\begin{array}{cc}
n & P_n(a) \\
1 & 5 a-2 \\
2 & 35 a^2-42 a+16 \\
3 & (5 a-4) \left(35 a^2-56 a+36\right) \\
4 & 385 a^4-1540 a^3+2684 a^2-2288 a+768 \\
5 & 175175 a^5-1051050 a^4+2862860 a^3-4252248 a^2+3327584 a-1061376
\end{array}
\right)$$
and all coefficients are positive as long as $a \leq \frac 25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Simplify $\frac{1+\sin\alpha-2\sin^2\left(45^\circ-\frac{\alpha}{2}\right)}{4\cos\frac{\alpha}{2}}$ Simplify $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}$$
I am reading the solution of the authors and I really don't see how $$\dfrac{1+\sin\alpha-2\sin^2\left(45^\circ-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}=\dfrac{1+\sin\alpha-(1-\cos(90^\circ-\alpha))}{4\cos\dfrac{\alpha}{2}}$$ Which identity have they used?
| We start from the classical formula
$$\cos(a+b)=$$
$$\cos(a)\cos(b)-\sin(a)\sin(b)$$
which gives
$$\cos(2a)=\cos^2(a)-\sin^2(a)$$
$$=1-2\sin^2(a)$$
or
$$2\sin^2(a)=1-\cos(2a)$$
thus
$$2\sin^2(45-\frac{\alpha}{2})=$$
$$1-\cos(90-\alpha)=1-\sin(\alpha)$$
your expression becomes
$$\frac{1+\sin(\alpha)-(1-\sin(\alpha))}{4\cos(\frac{\alpha}{2})}=$$
$$\frac{2\sin(\alpha)}{4\cos(\frac{\alpha}{2})}=$$
$$\frac{2(2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2}))}{4\cos(\frac{\alpha}{2})}=$$
$$\sin(\frac{\alpha}{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{d^{2} y}{d x^{2}}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^{2} x=\cos ^{4} x$ Solve the differential equation $$\frac{d^{2} y}{d x^{2}}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^{2} x=\cos ^{4} x$$
My try:
I multiplied throughout with $\sec^2 x$, getting
$$\sec^{2} x \frac{d^{2} y}{d x}+\sec^{2} x \tan x \frac{d y}{d x}-3 \sec x \frac{d y}{d x}+2 y=\cos ^{2} x$$
Which can be written as:
$$\frac{d}{d x}\left(\sec ^{2} x \frac{d y}{d x}\right)-\tan x\left(\sec ^{2} x \frac{d y}{d x}\right)-3 \sec x \frac{d y}{d x}+2 y=\cos ^{2} x$$
Then I assumed $z=\sec^2 x\frac{dy}{dx}$, so we get
$$\frac{d z}{d x}-z \tan x-3 \sec x \frac{d y}{d x}+2 y=\cos ^{2} x$$
I am stuck now.
| The equation has trigonometric coefficients and singularities at the roots of the cosine. On the intervals between these roots, $\sin x$ is an equivalent parametrization to $x$, so try if $y(x)=u(\sin x)$ gives a more simple equation.
$$
y(x)=u(\sin x)\\
y'(x)=\cos x\,u'(\sin x)\\
y''(x)=\cos^2x\,u''(\sin x)-\sin x\, u'(\sin x)\\~\\
[\cos^2x\,u''(\sin x)-\sin x\, u'(\sin x)]+[(\sin x-3\cos^2x)\,u'(\sin x)]
+2\cos^2x\,u(\sin x)=\cos^4 x
$$
So indeed the equation simplifies. Naming the new parameter $s=\sin x$, the trigonometic functions can be completely replaced, giving
$$
u''(s)-3u'(s)+2u(s)=1-s^2
$$
This linear DE with constant coefficients has a general solution of the form
$$
u(s)=Ae^s+Be^{2s}+C+Ds+Es^2.
$$
Inserting gives the equation
$$
2C+(-3+2s)D+(2-6s+2s^2)E=1-s^2\\~\\
E=-\frac12\\
D=3E=-\frac32\\
2C-3D+2E=1\implies C=-\frac54
$$
Substituting back gives the result as
$$
y(x)=Ae^{\sin x}+Be^{2\sin x}-\frac54-\frac32\sin x-\frac12\sin^2x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$
My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
| What you have shown is that $|y| \leq \dfrac{2}{\sqrt{3}}$. While that does imply that $-\dfrac{2}{\sqrt{3}} \leq y \leq \dfrac{2}{\sqrt{3}}$, you cannot necessarily conclude that the range is $\left[-\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\right]$ since you obtained that result by squaring both sides of the equation, which is not a reversible step.
We are given the function
\begin{align*}
f(x) & = \frac{\sqrt{(x + 3)(x - 1)}}{x + 2}\\
& = \frac{\sqrt{x^2 + 2x - 3}}{x + 2}
\end{align*}
Observe that its domain of the function $f$ is $(-\infty, -3] \cup [1, \infty)$.
Differentiating gives
\begin{align*}
f'(x) & = \frac{\frac{1}{2\sqrt{x^2 + 2x - 3}}(2x + 2)(x + 2) - \sqrt{x^2 + 2x - 3}}{(x + 2)^2}\\
& = \frac{\frac{(x + 1)(x + 2)}{\sqrt{x^2 + 2x - 3}} - \sqrt{x^2 + 2x - 3}}{(x + 2)^2}\\
& = \frac{\frac{x^2 + 3x + 2 - (x^2 + 2x - 3)}{\sqrt{x^2 + 2x - 3}}}{(x + 2)^2}\\
& = \frac{x + 5}{(x + 2)^2\sqrt{x^2 + 2x - 3}}
\end{align*}
Setting the derivative equal to zero yields
\begin{align*}
f'(x) & = 0\\
\frac{x + 5}{(x + 2)^2\sqrt{x^2 + 2x - 3}} & = 0\\
x + 5 & = 0\\
x & = -5
\end{align*}
Thus, the function has a critical point at $x = -5$.
Observe that the denominator of the derivative function is always positive within the domain of the function. Therefore, the sign of the derivative depends only on the numerator. Hence, $f'(x) < 0$ if $x < -5$, $f'(-5) = 0$, and $f'(x) > 0$ if $-5 < x < -3$. Since the sign of the derivative changes from negative to positive at the critical point $x = -5$, the function has a relative minimum at $x = -5$ by the First Derivative Test.
Its relative minimum value is
\begin{align*}
f(-5) & = \frac{\sqrt{(-5)^2 + 2(-5) - 3}}{-5 + 2}\\
& = \frac{\sqrt{25 - 10 - 3}}{-3}\\
& = \frac{\sqrt{12}}{-3}\\
& = \frac{2\sqrt{3}}{-3}\\
& = -\frac{2}{\sqrt{3}}
\end{align*}
Also, note that $f'(x) > 0$ when $x > 1$. Therefore, $f(x)$ is strictly increasing on the interval $[1, \infty)$. Observe that $f(1) = 0$. Moreover,
\begin{align*}
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x - 3}}{x + 2}\\
& = \lim_{x \to \infty} \frac{|x|\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to \infty} \frac{x\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to \infty} \frac{\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{1 + \frac{2}{x}}\\
& = 1
\end{align*}
Since the function $f$ is continuous on the interval $[1, \infty)$, it assumes every value in the interval $[0, 1)$ by the Intermediate Value Theorem.
Observe that $f(-3) = 0$. Since $f$ assumes the value $-\dfrac{2}{\sqrt{3}}$ at $x = -5$ and is continuous on the interval $[-5, 0]$, it assumes every value in the interval $\left[-\frac{2}{\sqrt{3}}, 0\right]$ in the interval $[-5, 0]$. Moreover,
\begin{align*}
\lim_{x \to -\infty} f(x) & = \lim_{x \to -\infty} \frac{\sqrt{x^2 + 2x - 3}}{x + 2}\\
& = \lim_{x \to -\infty} \frac{|x|\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to -\infty} \frac{-x\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to -\infty} \frac{-\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{1 + \frac{2}{x}}\\
& = -1
\end{align*}
Since $f$ is continuous on the interval $(-\infty, -5]$, it assumes every value in the interval $\left[-\dfrac{2}{\sqrt{3}}, -1\right)$ in the interval $(-\infty, -5]$.
Therefore, the range of $f$ is
$$\left[-\frac{2}{\sqrt{3}}, -1\right) \cup \left[-\frac{2}{\sqrt{3}}, 0\right] \cup [0, 1) = \left[-\frac{2}{\sqrt{3}}, 1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Calculate the power series centered in $z_0=1$ I did this, I don't know if is correct
$$
\begin{gathered}
h(z)=\left(\frac{z}{z+1}\right)^{2} \\
u=z-1,z_0=1 \\
h(z)=\left(\frac{z}{z+1}\right)^{2}=z^{2} \frac{1}{(z+1)^{2}}=z^{2} \frac{1}{(1-(-z))^{2}} \rightarrow \text { geometric series } \\
\sum_{n=0}^{\infty} u^{n}=\frac{1}{1-u}, \quad|z|<1, \\
\\\frac{d}{d z} \frac{1}{1-u}=\frac{1}{(1-u)^{2}}
\\\left(\frac{z}{z+1}\right)^{2}=z^{2} \frac{d}{d u} \sum_{n=0}^{\infty} u^{n}=z^{2} \sum_{n=0}^{\infty} n u^{n-1}=z^{2} \sum_{n=0}^{\infty} n(z-1)^{n-1}=\sum_{n=0}^{\infty} n(z-1)^{n+1} \\
\left(\frac{z}{z+1}\right)^{2}=\sum_{n=0}^{\infty} n(z-1)^{n+1},|z-1|<1 \rightarrow|z|<2 \\
\\
\end{gathered}
$$
I need to use a geometric serie to Calculate the power series
I have a problem, the problem is in $z_0=1$ I don't know if I use it right
| Comment on your approach:
It is not very clear how each step performs, in addition the changes of variables are strange in the part that tries to adjust to the convergence radius. We can use only geometric series if we first make use of partial fractions and then study each series representation for each function obtained and we adjust the radius of convergence.
First use partial fraction
$$\left(\frac{z}{z+1}\right)^{2}=1-\frac{2}{z+1}+\frac{1}{(z+1)^{2}}$$
Now,
*
*The series expansion of a constant around every point is the same constant, in this case
$$\boxed{1=1,\quad \forall z\in \mathbb{C}}$$
*By Geometric Series we know that,
$$\sum_{n=0}^{+\infty}az^{n}=\frac{a}{z-1},\quad a\in \mathbb{C},\quad |z|<1$$
Now notice that
$$\frac{{\rm d}}{{\rm d}z}\left(\frac{-1}{z+1}\right)=\frac{1}{(1+z)^{2}}$$
and also,
\begin{align*}
\frac{-1}{z+1}&=\left(-\frac{1}{2}\right)\left(\frac{1}{1-\left(\frac{-z+1}{2}\right)}\right),\\ &=\left(-\frac{1}{2}\right)\sum_{n=0}^{+\infty}\left(\frac{-z+1}{2}\right)^{n},\\ &=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}(z-1)^{n},\quad |z-1|<2
\end{align*}
Then,
\begin{align*}
\frac{1}{(1+z)^{2}}&=\frac{{\rm d}}{{\rm d}z}\left(-\frac{1}{z+1}\right),\\&=\frac{{\rm d}}{{\rm d}z}\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}(z-1)^{n},\\&=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}n(z-1)^{n-1},\\
&=\sum_{n=1}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}n(z-1)^{n-1},\\
&\overset{k=n-1}{=}\sum_{k=0}^{+\infty}\left(-\frac{1}{2}\right)^{k+2}(k+1)(z-1)^{k},\\
&\overset{k=n}{=}\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n}
\end{align*}
where the differentiation is allowed because a power series can be differentiated term by term in any region that lies entirely inside its circle of convergence.
Therefore,
$$\boxed{\frac{1}{(1+z)^{2}}=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\quad |z-1|<2}$$
*
*Again by Geometric Series we have
\begin{align*}
-\frac{2}{z+1}&=2\left(-\frac{1}{2}\right)\left(\frac{1}{1-\left(\frac{-z+1}{2}\right)}\right),\\&=-\sum_{n=0}^{+\infty}\left(\frac{-z+1}{2}\right)^{n},\\ &=\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n},\quad |z-1|<2
\end{align*}
Therefore,
$$\boxed{-\frac{2}{z+1}=\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n},\quad |z-1|<2}$$
If $|z-1|<2$, we have
\begin{align*}
\left(\frac{z}{z+1}\right)^{2}&=1-\frac{2}{z+1}+\frac{1}{(z+1)^{2}},\\&=1+\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n}+\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\\&=1-1+\sum_{n=1}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n}+\frac{1}{4}+\sum_{n=1}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\\
&=\frac{1}{4}+\sum_{n=1}^{+\infty}\left[-\left(-\frac{1}{2}\right)^{n}+\left(-\frac{1}{2}\right)^{n+2}(n+1) \right](z-1)^{n},\\
&=\frac{1}{4}+\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{2^{n+2}}(n-3)(z-1)^{n}
\end{align*}
Therefore the series expansion at $z=1$ is given by,
$$\boxed{\left(\frac{z}{z+1}\right)^{2}=\frac{1}{4}+\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{2^{n+2}}(n-3)(z-1)^{n},\quad |z-1|<2}$$
Expanding we get
$$\left(\frac{z}{z+1}\right)^{2}=\frac{1}{4}+\frac{1}{2^{2}}(z-1)-\frac{1}{2^{4}}(z-1)^{2}+\frac{1}{2^{6}}(z-1)^{4}+\mathcal{O}(z-1)^{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality with interesting independent constants Let $b_1,\dots,b_{n-1}$ be integers satisfying $0 \le b_i \le n-i$ for each $i \in [n-1]$ such that $\sum_{i=1}^{n-1} b_i = \alpha \binom{n}{2}$ where $\alpha$ is constant strictly between $0$ and $1$. Prove that $\sum_{i=1}^{n-1} ib_i \ge c\alpha^2(n-1)\binom{n}{2}$ for some constant $c$ which is independent of $\alpha$ and $n$.
My approach is as follows: I basically tried to split the entire sum into $n-1$ terms like $b_k+\dots+b_{n-1}$ and bound it by using $b_i \le n-i$ as follows $b_k+\dots+b_{n-1}=b_1+\dots+b_{n-1}-(b_1+\dots+b_{k-1})\ge \alpha\binom{n}{2}-(n-1+\dots+n-k+1)=\alpha\binom{n}{2}-\frac{(k-1)(2n-k)}{2}$ (for $k \ge 2$). However, I couldn't finish this idea. Any help is appreciated! Thank you!
| Trivially the constant $c=0$ will satisfy the inequality for any $\alpha$ and $n$. For the following, I will pick $c = \frac 14$.
For the proof, I will relax the requirement of $\alpha$ to $0\le \alpha \le 1$. Hence every $b_1, \ldots, b_{n-1}$ may satisfy $0\le b_i \le n-i$ without further global constraints.
I will also rewrite the RHS of the inequality to be proven to
$$c\alpha^2 (n-1)\binom n2 = \frac{2c}{n}\alpha^2 \frac{n(n-1)}2\binom n2 = \frac{2c}{n}\left(\sum_{i=1}^{n-1} b_i\right)^2$$
For $n=1$, $0\ge 0$ is true.
For $n=2$, $0\le b_1 \le 1$ (which would not satisfy the stricter $\alpha$ requirement in the question), so
$$RHS = cb_1^2 = \frac{b_1}{4}b_1 \le b_1 = LHS$$
Induction hypothesis: assume that for some $n=k$, if there are $0\le b_i\le k-i$ for every $i=1,\ldots, k-1$, then
$$\sum_{i=1}^{k-1}ib_i \ge \frac{2c}{k}\left(\sum_{i=1}^{k-1}b_i\right)^2$$
For $n=k+1$, the $b_i$s satisfy $0\le b_i \le k+1-i$ for every $i=1,\ldots, k$. This means that, omitting $b_1$, the remaining $b_i$s satisfy $0\le b_{j+1}\le k-j$ for every $j = 1,\ldots, k$.
Let the $S$ be the sum of the remaining $b_i$s: $S = \sum_{i=2}^k b_i = b_2 + b_3 + \cdots + b_k$.
$$\begin{align*}
LHS &= \sum_{i=1}^k ib_i\\
&= \sum_{i=1}^k b_i + \sum_{i=2}^k (i-1)b_i\\
&= b_1 + S + \sum_{j=1}^{k-1} jb_{j+1}&&j=i-1\\
&\ge b_1 + S + \frac{2c}{k}\left(\sum_{j=1}^{k-1}b_{j+1}\right)^2 &&\text{induction hypothesis}\\
&= b_1 + S + \frac{2c}k S^2\\
RHS &= \frac{2c}{k+1} \left(\sum_{i=1}^kb_i\right)^2
= \frac{2c}{k+1} \left(b_1+S\right)^2
\end{align*}$$
Comparing the difference between both sides,
$$\begin{align*}
LHS - RHS &\ge \left(b_1+S+\frac{2c}k S^2\right) - \frac{2c}{k+1} \left(b_1+S\right)^2\\
&= b_1+S+\frac{2c}k S^2 - \frac{2c}{k+1}b_1^2 - \frac{4c}{k+1}b_1S - \frac{2c}{k+1}S^2\\
&= \left(1-2c\frac{b_1}{k+1}\right)b_1 + \left(1-4c\frac{b_1}{k+1}\right)S + 2c\left(\frac1k - \frac1{k+1}\right)S^2\\
&\ge 0\\
LHS &\ge RHS
\end{align*}$$
using $0\le b_1 \le k$, $0\le S$, and $0\le c \le \frac14$.
By induction, the inequality holds for all positive integer $n$ with appropriate $b_i$s, using constant $c=\frac 14$. The same inequality would also hold with additional constraint on $\alpha$.
| {
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all polynomials $P$ such that $P(P(x))=P(x^n)+P(x)-1$
Find all polynomials $P$ of degree $n$ such that $P(P(x))=P(x^n)+P(x)-1$.
I stumbled upon this problem while solving functional equations in polynomials. However, even after trying most of the methods I know, I'm still yet to solve it. Here's what I tried:
*
*Obviously, solutions exist, since the constant polynomial $P(x)=1$ satisfies the given equation.
*Considering the highest degree of both sides didn't give any results (they have the same degree).
*Taking the derivative of both sides just further complicated the equation.
*Substituting $x\mapsto P(x)$ didn't seem to help either.
*I tried to prove that $P$ is injective, but it just gave $P(x)=P(y) \Longrightarrow P(x^n)=P(y^n)$.
*Substituting $x\mapsto x^n$ and combining the result with the original equation yields $P(P(x^n))-P(P(x))=P(x^{n^2})-P(x)$. I tried to define some new polynomial from this equation to simplify it, but it didn't work.
*Trying for specific values of n, I found $P(x)=1$ and $P(x)=2x-1$ work when $n=1$, but no such polynomial of degree $2$ exists.
What should I do to solve this problem?
Also, I've noticed that the second-highest degree of RHS is $nm$, where $m$ is the second-highest degree of $P$. Is it possible to prove that LHS has a different second-highest degree?
| $\bullet\ $ If $n=0$, then $P=a$ for a constant $a$. Replacing in the equation, you get $a=a+a-1$, so $a=1$, so you get thet solution
$$P(X)=1$$
$\bullet\ $ If $n=1$, then $P(X)=aX+b$ for some constants $a\neq 0$ and $b$. Replacing in the equation, you get
$$a(aX+b)+b=aX+b+aX+b-1$$
so $a^2=2a$ and $ab+b=2b-1$. You get that $a=2$ and $b=-1$, so you get the solution
$$P(X)=2X-1$$
$\bullet\ $ If $n=2$, then $P(X)=aX^2+bX+c$ for some constants $a \neq 0$, $b$ and $c$. Replacing in the equation, you get that
$$a(aX^2+bX+c)^2 + b(aX^2+bX+c)+c=aX^4+bX^2+c + aX^2+bX+c - 1$$
i.e.
$$a^3X^4 + 2a^2bX^3+ (ab^2+2a^2c+ab)X^2 +(2abc+b^2)X + ac^2+bc+c$$ $$= aX^4+(a+b)X^2+ bX+2c - 1$$
Identifying the $X^4$ coefficient, you get $a^3=a$, so $a=\pm 1$.
Identifying the $X^3$ coefficient, you get then that $2b=0$, so $b=0$.
Identifying the $X^2$ coefficient, you get then that $2c=a$.
Identifying the constant coeffcient, you get that $ac^2 = c-1$, so $a(2c)^2=4c-4$, so $a^3 = 2a-4$, so $a=2a-4$, so $a=4$, contradiction.
So there is no solution for $n=2$.
$\bullet\ $ If $n\geq 3$ :
First, there is no solution of the form $P(X)=aX^n$. Indeed, for such a polynomial, you have $P(0)=0$, which contradicts the fact that $P(P(X))=P(X^n)+P(X)-1$ (just evaluate at $X=0$).
So if $P$ is solution, then you have $P(X)=aX^n + bX^m + Q(X)$, with $a,b \neq 0$, $m<n$ and $\deg(Q)<m$. For such a polynomial, one has
\begin{align*} &P(P(X)) \\
&=a(aX^n + bX^m + Q(X))^n + b(aX^n + bX^m + Q(X))^m + Q(aX^n + bX^m + Q(X))\\
& =a^{n+1}X^{n^2} + a^{n}bnX^{n(n-1)+m} + R(X)
\end{align*}
where $R$ has degree strictly less than $n(n-1)+m$. But one has also
$$P(X^n)+P(X)-1 = aX^{n^2} +S(X)$$
where $S$ has degree less than $\max(m,nm)$. But since $n \geq 3$, then $\max(m,nm) < n(n-1)+m$, so you got a contradiction.
Finally, the only solutions are
$$\boxed{P(X)=1 \quad \quad \text{and} \quad \quad P(X)=2X-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum possible sum of squares of two numbers with sum $k$? If the sum of two numbers is k. Find the minimum value of the sum of their squares.
This is my calculations so far.
a + b = k <-- google says that I should put x + y = k rather than a + b = k
a² + b² = y (but I don't know why should I do that)
(k - b)² + b² = y
k² - 2kb + 2b² = y <-- now I'm stuck here. I don't know which differentiation variable to take. If it is 'b' then why?
So here I am. Kindly show me how to solve the problem and where did I go wrong in my computations :P
| I focus on a calculus method given your tags on the question. We are given $a + b = k$. The sum of their squares is given by $a^2 + b^2$. Since $a + b = k$, we hat that $k - b = a$, hence
$$g(b) = a^2 + b^2 = (k-b)^2 + b^2 = k^2-2kb+b^2+b^2 = k^2 - 2kb + 2b^2\text{.}$$
$k$ is a fixed value that is known, so $b$ is the only unknown in the expression above. Differentiating the above expression with respect to $b$ and setting that equal to $0$, we obtain
$$g^{\prime}(b) = -2k + 4b = 0$$
or
$$4b = 2k \implies b = \dfrac{2}{4}k = \dfrac{k}{2} \text{.}$$
Since $b = \dfrac{k}{2}$ and $a + b = k$, it follows that $$a = k - b = k - \dfrac{k}{2} = \dfrac{k}{2}\text{.}$$
Hence from setting the derivative equal to $0$, we obtain $a = b = \dfrac{k}{2}$.
Next, we must demonstrate that this solution is a minimum. We have
$$g^{\prime\prime}(b) = 4 > 0$$
for any $b$, hence by the second derivative test, the solution $a = b = \dfrac{k}{2}$ must be a minimum.
Thus, the desired sum of squares is given by
$$a^2 + b^2 = \left(\dfrac{k}{2}\right)^2 + \left(\dfrac{k}{2}\right)^2 = 2 \cdot \dfrac{k^2}{4} = \dfrac{k^2}{2}\text{.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\iiint\frac{xyz\,dx\,dy\,dz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$ over $\{(x,\,y,\,z)\in[0,\,\infty)^3\mid x^2+y^2+z^2\le R^2\}$
Evaluate $$\iiint\frac{xyz\,dx\,dy\,dz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}$$ for $a>b>c>0$ where $\Omega:=\{(x,\,y,\,z)\in[0,\,\infty)^3\mid x^2+y^2+z^2\le R^2\}$
What do we do with the denominator here?
| Integrate in spherical coordinates
\begin{align}
&\iiint\frac{xyz}{\sqrt{a^2x^2+b^2y^2+c^2z^2}}\,dx\,dy\,dz\\
=& \int_0^{\pi/2} \int_0^{\pi/2}\int_0^R
\frac{r^4\sin \theta \cos\theta \sin^3\phi \cos\phi}
{\sqrt{ a^2\cos^2\theta\sin^2\phi+b^2\sin^2\theta\sin^2\phi +c^2\cos^2\phi}}dr d\theta d\phi\\
=& \int_0^{\pi/2} \int_0^{\pi/2}
\frac{R^5\sin^3\phi \cos\phi \ \sin \theta }
{5\sqrt{ a^2\sin^2\phi+ c^2\cos^2\phi +(b^2-a^2)\sin^2\phi\sin^2\theta}} d(\sin\theta)\ d\phi\\
=& \int_0^{\pi/2} \frac{R^5\left(\sqrt{c^2-(c^2-{b^2})\sin^2\phi}-\sqrt{c^2-(c^2-{a^2})\sin^2\phi}\right)\sin\phi}{5(b^2-a^2)} d(\sin\phi)\\
=&\ \frac{R^5(ab+bc+ca)}{15(a+b)(b+c)(c+a)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integer part equation with sum and radicals
For a fixed $n$, natural number greather or equal than $2$, solve the following equation for real $x$.
$$\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}$$
I have noticed that all but the last terms in the equations are integers, so should be the last one, so the modulus under the radical is an integer's perfect square. So, it's formula is:
$$x = \pm \sqrt{a^2 + 1}$$
Where $a$ is an integer. So I should proceed in finding $a$, but I could only prove that $a = 1$ is a solution.
| As you noted, we must have $|x^2-1|=a^2$ for some integer $a$.
Consider two cases . . .
Case $(1)$:$\;|x|\le 1$.
From $|x|\le 1$ we get $|x^2-1|=1-x^2$, hence
\begin{align*}
&
|x^2-1|=a^2
\\[4pt]
\implies\;\;&
1-x^2=a^2
\\[4pt]
\implies\;\;&
1-a^2=x^2
\\[4pt]
\implies\;\;&
1-a^2\ge 0
\\[4pt]
\implies\;\;&
a^2\le 1
\\[4pt]
\implies\;\;&
a^2=0\;\text{or}\;a^2=1
\\[4pt]
\implies\;\;&
x^2=1\;\text{or}\;x^2=0
\\[4pt]
\implies\;\;&
x\in\{\pm 1,0\}
\\[4pt]
\end{align*}
but then the equation
$$
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}
$$
fails since, for $x\in\{\pm 1,0\}$, the $\text{LHS}$ evaluates to $0$, while for $x=\pm 1$, the $\text{RHS}$ evaluates to $n-1$, and for $x=0$, the $\text{RHS}$ evaluates to $n$.
Thus for case $(1)$, there are no solutions.
Case $(2)$:$\;|x| > 1$.
From $|x| > 1$ we get $|x^2-1|=x^2-1 > 0$, hence $|a|\ge 1$.
Then we have
$$
|a|=\sqrt{x^2-1} < \sqrt{x^2+1}=\sqrt{a^2+2}\le\sqrt{a^2+2|a|} < \sqrt{a^2+2|a|+1}=|a|+1
$$
which implies $\lfloor \sqrt{x^2 + 1} \rfloor=|a|$.
But then for the equation
$$
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}
$$
the $\text{RHS}$ evaluates to $n|a|$ which is at least $n$, whereas for the $\text{LHS}$, noting that
$$
t-1 < \lfloor t \rfloor \le t
$$
holds for all real $t$, we get
\begin{align*}
&
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor
\\[4pt]
< \;&
n^2 x -\sum_{k = 1}^n \Bigl((2k -1)x -1\Bigr)
\\[4pt]
=\;&
n^2 x -(n^2x-n)
\\[4pt]
=\;&
n
\\[4pt]
\end{align*}
so the $\text{LHS}$ is less than the $\text{RHS}$, contradiction.
Thus for case $(2)$, there are no solutions.
Hence the given equation has no solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) .$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d x}{x^{6}+1} \\
&=\frac{1}{2} \oint_{\gamma} \frac{d z}{z^{6}+1} \\
&=\frac{1}{2} \cdot 2 \pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^{6}+1} , z_k \right)
\end{aligned}
$$
with its simple poles at $ z_{k}=e^{\frac{(2 k+i) \pi}{6} i}$, where $k=0,1,2$.
Now lets evaluate the residues of $\frac{1}{z^{6}+1} $ at $ z_{k}$.
$$
\operatorname{Res}\left( \frac{1}{z^{6}+1} , z_{k}\right)=\frac{1}{6 z _k^{5}}=\frac{z_{k}}{6 z_{k}^{6}}=-\frac{z_{k}}{6}
$$
Putting the residues back yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\pi i\left(-\frac{1}{6}\left(z_{0}+z_{1}+z_{2}\right)\right) \\
&=-\frac{\pi i}{6}\left(e^{\frac{\pi}{6}}+e^{\frac{\pi}{2} i}+e^{\frac{5 \pi}{6} i}\right) \\
&=-\frac{\pi i}{6}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i+i-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
Similarly, we can deal with the integral in general
$$I_{2n}=\int_{0}^{\infty} \frac{1}{x^{2n}+1} d x$$
with the same contour $\gamma$ and $n $ simple poles whose residues are
$$
\operatorname{Res}\left( \frac{1}{z^{2n}+1} , z_{k}\right)=\frac{1}{2n z _k^{2n-1}}=\frac{z_{k}}{2n z_{k}^{2n}}=-\frac{z_{k}}{2n}
$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{2 n}+1} d x&=\pi i \sum_{k=0}^{n-1} \operatorname{Res}\left(\frac{1}{z^{2 n}+1}, z_k\right) \\
&=\pi i \sum_{k=0}^{n-1}\left(-\frac{z _k}{2 n}\right)\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} z_{k}\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} e^{\frac{2 k+1}{2 n} \pi i}\\ &=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{1-e^{\frac{\pi i}{n}(n)}}{1-e^{\frac{\pi i}{\pi}}}\\
&=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{2}{1-e^{\frac{\pi}{n} i}}\\
&=-\frac{\pi i}{n} \frac{e^{\frac{\pi i}{2 n}}}{1-e^{\frac{\pi}{n} i}}\\&= -\frac{\pi i}{n} \cdot \frac{1}{e^{-\frac{\pi i}{2 n}}-e^{\frac{\pi i}{2 n}}}\\&= \frac{\pi}{2 n} \csc \frac{\pi}{2 n}
\end{aligned}
$$
However, I can’t use the same contour to deal with the odd one
$$\int_{0}^{\infty} \frac{1}{x^{2 n-1}+1} d x$$
as the simple pole $-1$ is on $\gamma_1$.
My Question: How can we evaluate the odd one?
| Let $\frac{1}{t}=1+x^{n} \Rightarrow \textrm{ then } x= \left(\frac{1}{t}-1\right)^{\frac{1}{n}}\textrm{ and } \displaystyle dx= \frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1}{n}-1}\left(-\frac{d t}{t^{2}}\right)$
$$
\begin{aligned}
I &=\int_{1}^{0} t\cdot\frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1}{n}-1}\left(-\frac{d t}{t^{2}}\right) \\
&=\frac{1}{n} \int_{0}^{1} t^{-\frac{1}{n}}(1-t)^{\frac{1}{n}-1} d t \\
&=\frac{1}{n} B\left(-\frac{1}{n}+1, \frac{1}{n}\right) \\
&=\frac{1}{n} \Gamma\left(-\frac{1}{n}+1\right) \Gamma\left(\frac{1}{n}\right) \\
&=\frac{\pi}{n} \csc \frac{\pi}{n}
\end{aligned}
$$
where the last result comes from the Reflection Theorem of Gamma Function: $$
\Gamma(x) \Gamma(1-x)=\pi \csc (\pi x) \quad \textrm{ for any }x\not \in Z.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Series representation of GCD(x, y) While playing around with WolframAlpha, I discovered that apparently the GCD of any integers x, y is equal to the following sum:
$$
x + y - xy + 2\sum_{k = 1}^{x-1} \lfloor \frac{ky}{x}\rfloor
$$
I've personally verified that this formula works.
My question is: where does this even come from??? I've searched everywhere and found absolutely nothing. Is there a proof somewhere? Or at least a sketch so I can start thinking about it?
| Let $x,y \in \mathbb{N}_0$, and let $d= \gcd(x,y)$. Let also $\displaystyle{S = \sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor}$.
Then one has
\begin{align*} S = \sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor & = \sum_{k = 1}^{x-1} \left\lfloor \frac{(x-k)y}{x}\right\rfloor\\
& = \sum_{k = 1}^{x-1} \left\lfloor y - \frac{ky}{x}\right\rfloor\\
& = \sum_{k = 1}^{x-1} y +\left\lfloor -\frac{ky}{x}\right\rfloor\\\end{align*}
so you get
$$ S=\sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor = y(x-1) + \sum_{k = 1}^{x-1} \left\lfloor -\frac{ky}{x}\right\rfloor \quad \quad \quad (*)$$
Now notice that for every $k$, one has
$$\left\lfloor -\frac{ky}{x}\right\rfloor =\left\lbrace\begin{array} $-1-\left\lfloor \dfrac{ky}{x}\right\rfloor & \text{if } x \not\mid ky\\ -\left\lfloor \dfrac{ky}{x}\right\rfloor & \text{if } x \mid ky\end{array}\right.$$
But for $k \in \lbrace 1, ..., x-1 \rbrace$, one has
$$ x \mid ky \quad \Longleftrightarrow \quad\frac{x}{d} \mid k\frac{y}{d} \quad\Longleftrightarrow \quad\frac{x}{d} \mid k\quad\Longleftrightarrow \quad k \in \left\lbrace \dfrac{x}{d}, \dfrac{2x}{d}, ..., \dfrac{(d-1)x}{d} \right\rbrace$$
So $\mathrm{Card} \lbrace k \in \lbrace 1, ..., x-1 \rbrace \text{ such that }x \mid ky \rbrace = d-1$, so one deduce that
$$\sum_{k = 1}^{x-1} \left\lfloor -\frac{ky}{x}\right\rfloor = \left(\sum_{k = 1}^{x-1} -1-\left\lfloor \frac{ky}{x}\right\rfloor\right) + d-1 = -x-S+d$$
Using $(*)$, you get finally that
$$S = y(x-1)-x-S+d = yx-y-x - S + d$$
so $$\boxed{\gcd(x,y)=x+y-xy+2\sum_{k = 1}^{x-1} \left\lfloor \frac{ky}{x}\right\rfloor}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using change of coordinates to find the exact value of an integral Use an appropriate change of coordinates to find the exact value of the integral
$$\int_{-\sqrt{3}}^{\sqrt{3}}\int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}}\int_{-3+x^2+y^2}^{3-x^2-y^2}x^2dzdydx$$
My work so far:
$-3+x^2+y^2\leq z\leq 3-x^2-y^2$
$-\sqrt{3-x^2}\leq y\leq \sqrt{3-x^2}$
$-\sqrt{3}\leq x \leq \sqrt{3}$
I can see that $y$ includes the entire circle, so that means
$0 \leq \theta \leq 2\pi$
Similarly, graphing the circle centered at the origin with radius $\sqrt{3}$, I can see that $0\leq r \leq \sqrt{3}$
I just need to change $z$ to polar now:
$3-x^2-y^2=3-r^2$
$-3+x^2+y^2=-3+r^2$
So the integral becomes:
$$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{-3+r^2}^{3-r^2} r(r\sin\theta)^2 dz dr d\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^3\sin^2\theta(3-r^2-(-3+r^2))drd\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^3\sin^2\theta(6-2r^2)drd\theta$$
$$=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} \sin^2\theta(6r^3-2r^5) drd\theta$$
$$=\int_{0}^{2\pi} \frac{9}{2}\sin^2\theta d\theta$$
$$=\frac{9\pi}{2}$$
Is my solution correct?
| Yes, I confirm that your calculation is correct.
By symmetries, one can either use the cylindrical coordinates for the triple integral or the polar coordinates after integrating with the variable $z$.
The iterated triple integral can be written as
$$
J=\iint_D (6-2x^2-2y^2)x^2\;dxdy
$$
where $D=\{(x,y)\mid x^2+y^2\le 3\}$. Applying the polar coordinates, one has
$$
\begin{align}
J&=\int_{0}^{2\pi}\int_0^{\sqrt3} (6-2r^2)r^3\cos^2(\theta) \;drd\theta\\
&=\left(\int_0^{\sqrt3} (6-2r^2)r^3\;dr\right)\cdot
\left(\int_{0}^{2\pi} \cos^2\theta\;d\theta\right)\\
&=\frac{9}{2}\cdot \int_0^{2\pi}\frac{1+\cos(2\theta)}{2}\;d\theta\\
&=\frac{9\pi}{2}
\end{align}
$$
| {
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Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfied.
Alternatively, if $\ell_1\leq\ell_2\leq\ell_3$ are the values of $a,b$ and $c$ ordered in ascending order, then the triangle exists iff $\ell_1+\ell_2\geq\ell_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in\mathbb R^2$ be the three vertices of the triangle, with $\|x\|=a,\,\|y\|=b$ and $\|x-y\|=c$. Then $c^2=\|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2=a^2+b^2-2\langle x,y\rangle$. Therefore $x^Ty=\langle x,y\rangle=\frac{1}{2}(a^2+b^2-c^2)$ and
$$\pmatrix{x^T\\ y^T}\pmatrix{x&y}=\frac{1}{2}\pmatrix{2a^2&a^2+b^2-c^2\\ a^2+b^2-c^2&2b^2}.\tag{1}$$
The RHS of $(1)$ must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain $x$ and $y$ and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of $(1)$ is positive semidefinite if and only if $(2a^2)(2b^2)-(a^2+b^2-c^2)^2\geq0$, by Sylvester's criterion. That is, the triangle exists if and only if
$$-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\geq0.\tag{2}$$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of $(2)$ is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both $(0)$ and $(2)$ are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why $(0)$ and $(2)$ are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?
| The equations in (0) can be equivalently expressed as
$$a + b - c \ge 0$$
$$a - b + c \ge 0$$
$$-a + b + c \ge 0$$
Since the product of non-negative factors must also be non-negative, we must have:
$$(a + b - c)(a - b + c)(-a + b + c) \ge 0$$
Of course, this product would also be positive if exactly two of the original three equations were false. To prove equivalence, we must show that this is impossible. Suppose that the first two were violated:
$$a + b - c < 0$$
$$a - b + c < 0$$
Adding up the two inequalities gives $2a < 0$, which contradicts the assumption that all of a triangle's sides are positive. By symmetry, this contradiction occurs regardless of which two inequalities are chosen to be false. Anyhow, multiplying everything out gives:
$$(a + b - c)(a - b + c)(-a + b + c) \ge 0$$
$$(a^2 - ab + ac + ab - b^2 + bc - ac + bc - c^2)(-a + b + c) \ge 0$$
$$(a^2 - b^2 + 2bc - c^2)(-a + b + c) \ge 0$$
$$-a^3 + a^2b + a^2c + ab^2 - b^3 - b^2c - 2abc + 2b^2c + 2bc^2 + ac^2 - bc^2 - c^3 \ge 0$$
$$-a^3 - b^3 - c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc \ge 0$$
So now, we have a form of the triangle inequality that's a cubic polynomial. It's a bit verbose, though, so let's see what happens if we include one more factor, $a + b + c \ge 0$.
$$(-a^3 - b^3 - c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc)(a + b + c) \ge 0$$
$$-a^4 - a^3b - a^3c - ab^3 - b^4 - b^3c - ac^3 - bc^3 - c^4 + a^3b + a^2b^2 + a^2bc + a^2b^2 + ab^3 + ab^2c + a^3c + a^2bc + a^2c^2 + a^2c^2 + abc^2 + ac^3 + ab^2c + b^3c + b^2c^2 + abc^2 + b^2c^2 + bc^3 - 2a^2bc - 2ab^2c - 2abc^2 \ge 0$$
$$-a^4 - b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2 \ge 0$$
Wow, that's a lot of intermediate terms that conveniently cancel out! Now we can just rearrange this a little.
$$2(a^2b^2 + a^2c^2 + b^2c^2) \ge a^4 + b^4 + c^4$$
$$a^4 + b^4 + c^4 \le 2(a^2b^2 + a^2c^2 + b^2c^2)$$
Q.E.D.
| {
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Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
This is related to the following problem:
Problem 2: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}.$$
Problem 2 is in "Problems From the Book", 2008, Ch. 2, which was proposed by Vasile Cartoaje.
See: Prove that $\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)} \le \frac{n(n-1)}{2(2n-1)^2}$
Background:
I proposed Problem 1 when I tried to find my 2nd proof for Problem 2.
It is not difficult to prove that
$$\frac{1}{(2n-1)^4} + \frac{16n^2(n-1)^2}{(2n-1)^4}\cdot \frac{x_ix_j}{1-x_i-x_j}
\ge \frac{x_ix_j}{(1-x_i)(1-x_j)}.$$
(Hint: Use $\frac{x_ix_j}{(1-x_i)(1-x_j)}= 1 - \frac{1}{1 + x_ix_j/(1-x_i-x_j)}$
and $\frac{1}{1+u} \ge \frac{1}{1+v} - \frac{1}{(1+v)^2}(u-v)$
for $u = x_ix_j/(1-x_i-x_j)$ and $v=\frac{1}{4n(n-1)}$. Or simply
$\mathrm{LHS} - \mathrm{RHS} = \frac{(4x_ix_jn^2 - 4x_ix_j n + x_i + x_j - 1)^2}{(2n-1)^4(1-x_i-x_j)(1-x_i)(1-x_j)}\ge 0$.)
To prove Problem 2, it suffices to prove that
$$\frac{1}{(2n-1)^4}\cdot \frac{n(n-1)}{2} + \frac{16n^2(n-1)^2}{(2n-1)^4}\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac{n(n-1)}{2(2n-1)^2} $$
or
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
For $n=2, 3, 4$, the inequality is true.
For $n=5, 6$, numerical evidence supports the statement.
Any comments and solutions are welcome and appreciated.
| An incomplete approach:
$\sum\limits_{1\le i < j \le n} A_i A_j=\frac{1}{2}[(\sum_{k=1}^{n} A_k )^2=\sum_{k=1}^{n} A_k^2]=$
Using integral representation, we can write
$\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} =\sum\limits_{1\le i < j \le n} \int_{0}^{1} x_i x_j t^{-x_i-x_j} dt=\int_{0}^{1} dt\sum\limits_{1\le i < j \le n} x_i t^{-x_i} ~x_j t^{-x_j}= \frac{1}{2} [(\sum_{k=1}^{n} \int_{0}^{1} x_k t^{-x_k})^2+\sum_{k=1}^{n} \int_{0}^{1} x_k^2 t^{-2x_k}]dt= \frac{1}{2} \left[\frac{x_k^2}{(1-x_k)^2}-\frac{x_k^2}{1-2x_k}\right]$
I wish to come back again......
| {
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Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$ While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ terms, that is, a single formula for the product $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)(x^{a^4}+1)\ldots(x^{a^n}+1)$$
After multiplying the first four terms I could see a pattern develop in the form $x^{a+a^2+a^3+a^4...a^n}+x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}+x^{a+a^2+a^3}+x^{a+a^2+a^4}+x^{a^2+a^3+a^4}+x^{a+a^3+a^4} ...+x^{a^{n-2}+a^{n-1}+a^n} + x^{a+a^2}
+x^{a+a^3}+x^{a+a^4}+x^{a^2+a^3}+x^{a^2+a^4}+x^{a^3+a^4}...x^{a^{n-1}+n}...+x^{a+a^2+a^3+....a^{n-2}+a^{n-1}+1}$
Now I can easily find the summation of powers in the first term but cant find a formula for the summation of $x^a+x^{a^2}+x^{a^3}+x^{a^4}...+x^{a^n}$ and also can't figure out how to account for the other terms.
| For the specific value of $a=1/2$ (as mentioned in the comments), one has
$$\prod_{j=1}^n \left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1},$$
which can be easily proven by induction on $n$: if this holds for $n$, then
$$\prod_{j=1}^{n+1}\left(x^{2^{-j}}+1\right)=\frac{x-1}{x^{2^{-n}}-1}\left(x^{2^{-(n+1)}}+1\right)=\frac{x-1}{x^{2^{-(n+1)}}-1},$$
so it holds for $n+1$.
| {
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Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\textrm{ and }$ $b+c$ are positive real numbers.
First of all, let’s ‘kill’ the term $\sin x$ and the square of $\cos x$ by identity and double angle formula.
$\displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x =& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b) \cos ^{2} x+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b)\left(\frac{1+\cos 2 x}{2}\right)+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[\frac{a-b}{2} \cos 2 x+\left(b+c+\frac{a-b}{2}\right)\right] d x \\\stackrel{2x\mapsto x}{=} & \frac{1}{2} \int_{0}^{\pi} \ln \left[\frac{a-b}{2} \cos x+\left(\frac{a+b+2 c}{2}\right)\right] d x\end{aligned}\tag*{} $
By my post, I found, by Feynman’s Integration Technique, that
$\displaystyle \int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right),\tag*{} $
where $\left|\frac{c}{b}\right|\geq 1$.
$\displaystyle I=\frac{\pi}{2} \ln \left[\frac{\frac{a+b+2 c}{2}+\sqrt{\left(\frac{a+b+2 c}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}}}{2}\right]\tag*{} $
Simplifying gives the result
$\displaystyle \begin{aligned}I&=\frac{\pi}{2} \ln \left[\frac{a+b+2 c+2 \sqrt{(a+c)(b+c)}]}{4}\right] \\&=\frac{\pi}{2} \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)^{2} \\&=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)\end{aligned}\tag*{} $
| We can write the initial integral in the form
$\displaystyle I=\int_0^\frac{\pi}{2}\ln(a\cos^2x+b\sin^2x+c)dx=\int_0^\frac{\pi}{2}\ln\big((a-b)\cos^2x+b+c\big)dx\tag*{}$
Using $1+\tan^2x=\frac{1}{\cos^2x}$
$\displaystyle I=\int_0^\frac{\pi}{2}\ln\Big((a-b)+(b+c)(1+\tan^2x)\Big)dx+\int_0^\frac{\pi}{2}\ln(\cos^2x)dx\tag*{}$
$\displaystyle =\int_0^\frac{\pi}{2}\ln\Big((a+c)+(b+c)\tan^2x\Big)dx+2\int_0^\frac{\pi}{2}\ln(\cos x)dx\tag*{}$
Using for the first integral the substitution $\displaystyle x=\arctan t\,$ and the well-known value of the second integral $\displaystyle \int_0^\frac{\pi}{2}\ln(\cos x)dx=-\frac{\pi}{2}\ln2$
$\displaystyle I=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln\Big((a+c)+(b+c)t^2\Big)}{1+t^2}dt-\pi\ln2\tag*{}$
Now we use $\displaystyle\,\ln(p+q\,t^2)=2\,\Re\ln(\sqrt p-i\sqrt q\, t)$ and close the contour in the upper-half of the complex plane, integrating along a big half-circle (radius $R$), counter-clockwise. It is easy to show that the integral along this half-circle $\to0$ at $R\to\infty$.
The function is single-valued in the upper half-plane (the only branch point of the logarithm is $z=-i$). Therefore, using the Cauchy's residue theorem
$\displaystyle I=\Re\,2\pi i\operatorname{Res}_{z=i}\frac{\ln\Big(\sqrt{a+c}-i\sqrt{b+c}\,z\Big)}{1+z^2}-\pi\ln 2\tag*{}$
$\displaystyle =\pi\ln\big(\sqrt{a+c}+\sqrt{b+c}\big)-\pi\ln2\tag*{}$
$\displaystyle \boxed{\,\,I=\pi\ln\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\,\,}\tag*{}$
| {
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$t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$. $t^{n}-1 = \prod^{n-1}_{k=0}(t-C^k)$ where $C=e^{\frac{2\pi i}{n}}$ in the ring $\mathbb{C}[t]$.
I wanted to make a proof by induction. I've checked it for $n = 3$:
\begin{equation}
\begin{split}
\prod^{3-1}_{k=0}(t-C^k) & = (t-1)(t-e^{\frac{2\pi i}{3}})(t-e^{\frac{4\pi i}{3}}) \\
& = (t-1)(t-\cos(\frac{2}{3}\pi)+i\sin(\frac{2}{3}\pi))(t-\cos(\frac{4}{3}\pi)+i\sin(\frac{4}{3}\pi)) \\
& = (t-1)(t+\frac{1}{2}+\frac{\sqrt{3}}{2}i)(t+\frac{1}{2}-\frac{\sqrt{3}}{2}i) \\
& = (t-1)(t^2+\frac{1}{2}t-\frac{\sqrt{3}}{2}it+\frac{1}{2}t+\frac{1}{4}-\frac{\sqrt{3}}{4}i+\frac{\sqrt{3}}{2}it+\frac{\sqrt{3}}{4}i+\frac{3}{4}) \\
& = (t-1)(t^2+t+1) = (t^3-1)
\end{split}
\end{equation}
This gave me an idea because
\begin{equation}
\prod^{n-1}_{k=0}(t-C^k) = (t-1)\prod^{n-1}_{k=1}(t-C^k)
\end{equation}
To show that
\begin{equation}
\prod^{n}_{k=1}(t-C^k) = \sum^{n}_{k=0}t^k
\end{equation}
But I can't seem to figure out how. What also makes this proof difficult is that if $n$ increases $C$ changes also. Can anybody help me?
| You are overcomplicating the question. Just note that the roots of $t^n-1$ are exactly $C^k$ with $k$ between $0$ and $n-1$. Just factorice the polynomial and get the result.
| {
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.
| As said in the comments, your answer looks good. If you wanted to generalize to the case $D\in\mathbb{R}$, which would be natural since your integral is invariant under $D\to -D$, your answer would then be $\frac{\pi}{2|D|^3}$.
| {
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Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$ How can I prove the following sequence converges?
$$\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$$
I tried everything.
Could not find any candidates for comparison test, and failed to find an upper bound.
Any hints will be appreciated.
| By Cauchy-Schwarz inequality:$$\begin{align}a_n&= \dfrac{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots + \dfrac{1}{\sqrt{n}}}{n^2}\\&\le \dfrac{\sqrt{n}\sqrt{1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}}}{n^2}\\&\le \dfrac{\sqrt{\ln(n+1)}}{n^{\frac{3}{2}}}\\&\le \dfrac{(n+1)^c}{n^{\frac{3}{2}}}\\&= \left(1+\dfrac{1}{n}\right)^c\cdot \dfrac{1}{n^{\frac{3}{2}-c}}\\&\le \dfrac{2^c}{n^{\frac{3}{2} - c}}\\&=b_n.\end{align}$$You can choose the constant $c > 0$ such that $\dfrac{3}{2} - c > 1$ or $0 < c < \dfrac{1}{2}$. Thus $\displaystyle \sum_{n=1}^\infty b_n$ converges by the comparing it to the $p$-series for $p > 1$. And by comparison test once again to the $b_n$-series, $\displaystyle \sum_{n=1}^\infty a_n$ converges.
| {
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Summing even and odd terms of binomial coefficients, with increasing powers of 2 and alternating signs For a complex number $z$, let $(1+z)^{15}=a_{0}+a_{1} z+\cdots+a_{15} z^{15}$. The value of
$$
\left(a_{0}-4 a_{2}+16 a_{4}+\cdots-2^{14} a_{14}\right)^{2}+\left(2 a_{1}-8 a_{3}+32 a_{5} \cdots-2^{15} a_{15}\right)^{2}
$$
While I was entirely stumped by this question, I found some similar questions on here, specifically A sum involving binomial coefficients, powers and alternating signs. This did not equip me to be able to solve the above question, though, since this problem drops odd terms and even terms, alternates signs, AND has a factor of increasing powers of $2$, not to mention squares the result and then sums them.
I don't see how the sum can even be expressed as a binomial to somehow use calculus, nor do I see any algebraic manipulations; I've only managed to embarrassingly restate the problem statement as
$$ \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$
| Let
$$F= \left(\sum_{k=0}^{7} 2^{2 k}({ }^{15} C_{2 k})(-1)^k\right)^{2}+\left(\sum_{k=0}^{7} 2^{2 k+1}\left({ }^{15} C_{2 k+1}\right)(-1)^k\right)^{2} $$
From binomial expansion we have
$S_1=\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k} x^{2k}=\frac{1}{2}[(1+ix)^{2n+1}+(1-ix)^{2n+1}]-(1+x^2)^{n+1/2} \cos (2n+1)t$,
$t=\tan^{-1}x.$
$S_2=\sum_{k=0}^{n} (-1)^k {2n+1 \choose 2k+1} x^{2k+1}=\frac{-i}{2}[(1+ix)^{2n+1}-(1-ix)^{2n+1}]=(1+x^2)^{n+1/2} \sin (2n+1)t.$
$F=S_1^2+S_2^2=(1+x^2)^{2n+1},\text{where} x=2, n=7$
So we get $F=5^{15}.$
| {
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} |
Integrate $\int\frac{3x}{x^5+x^4+1}dx$ I have an integral which I solved. But, I am not sure whether my answer is right or not. The integral is $$\int\frac{3x}{x^5+x^4+1}dx$$ My answer $$3\left(-\dfrac{\displaystyle\sum_{\left\{Z:\>Z^3-Z+1=0\right\}}\frac{\left(2Z^2-3Z-1\right)\ln\left(\left|x-Z\right|\right)}{3Z^2-1}}{7}\\+\dfrac{\ln\left(x^2+x+1\right)}{7}-\dfrac{4\arctan\left(\frac{2x+1}{\sqrt{3}}\right)}{7\sqrt{3}}\right)$$
| Maple agrees with your answer, except that $\ln(|x-Z|)$ is replaced by $\ln(x-Z)$. Maple does this because Maple considers $x$ to be a complex variable. Two out of three of your zeros $Z$ are complex numbers anyway, so $\ln(|x-Z|)$ would be wrong even if $x$ is a real variable.
A simpler example. Factor $(x^2+1) = (x+i)(x-i)$, then
apply partial fractions:
$$
\int \frac{dx}{x^2+1} =
\int\left(\frac{i}{2}\;\frac{1}{x+i}-
\frac{i}{2}\;\frac{1}{x-i}\right) dx ,
$$
which is questionably equal to
$$
A(x):=\frac{i}{2}\ln|x+i|-\frac{i}{2}\ln|x-i|+C .
$$
I claim this is wrong. Assuming $x$ is a real variable, compute the derivative
like this:
$$
\frac{d}{dx}\left(\frac{i}{2}\ln|x+i|\right)
=\frac{i}{4}\frac{d}{dx}\ln\big(|x+i|^2\big)
=\frac{i}{4}\frac{d}{dx}\ln\big(x^2+1\big)
=\frac{ix}{2(x^2+1)}
$$
and similarly
$$
\frac{d}{dx}\left(\frac{i}{2}\ln|x-i|\right)
=\frac{ix}{2(x^2+1)}
$$
so that
$$
\frac{d}{dx}\big(A(x)\big) = 0 .
$$
Thus $A(x)$ is not the antiderivative of $1/(x^2+1)$ .
The answer should be
$$
A_1(x) = \frac{i}{2}\ln(x+i)-\frac{i}{2}\ln(x-i)+C
$$
without the absolute values. Regardless of which branches of the logarithms you choose,
$$
\frac{d}{dx}\big(A_1(x)\big) = \frac{1}{x^2+1} .
$$
In fact, for the principal values of the logs, $A_1(x)$ is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Numbers of $2\times 2$ matrices $A$ with elements from the set $\{-1, 0, 1\}$ such that $A^2 =I$ where $I$ is an identity matrix of order $2$. Numbers of $2\times 2$ matrices $A$ with elements from the set $\{-1, 0, 1\}$ such that $A^2 =I$ where $I$ is an identity matrix of order $2$.
My Approach: Let $\;A=\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}\;$
$\implies A^2 =\begin{bmatrix}
a^2+bc & b(a+d) \\
c(a+d) & d^2+bc \\
\end{bmatrix}\; =\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix},\; $
$\implies$ $a^2+bc=1,\; b(a+d)=0,\; c(a+d)=0,\; d^2+bc=1$
Then i made $5$ cases:
Case $(1).$ $a+d=0\;, b\neq0,\;c\neq 0$
Here I obtained two matrices $a=0,\ d=0,\; b=1,\; c=1$ and $a=0,\; b=0,\; b=-1,\; c=-1$
Case $(2)$ $a+d=0,\;b=0,\;c\neq0$
Here I obtained $4$ matrices $a=1,\;d=-1,\;b=0,\;c=1\;$ and $a=1,\; d=-1,\; b=0, c=-1$ and $a=-1,\;d=1,\;b=0,\;c=1\;$ and $a=-1,\; d=1,\; b=0, c=-1$
Case $(3)$ $a+d=0,\ b\neq 0,\ c=0$
Here I obtained $4$ matrices $a=1,\;d=-1,\;b=1,\;c=0\;$ and $a=1,\; d=-1,\; b=-1, c=0$ and $a=-1,\;d=1,\;b=1,\;c=0\;$ and $a=-1,\; d=1,\; b=-1, c=0$
Case $(4)$ $a+d\neq 0,\;b=0,\;c=0$
Here i obtained two matrices $a=1,\;b=0,\;c=0,\;d=1$ and $a=-1,\;b=0,\;c=0,\;d=-1$
Case $(5)$ $a+d=0,\; b=0,\ c=0\;$
Here I obtained two matrices $a=1,\;d=-1, b=0, c=0\;$ and $a=-1,\; d=1,\; b=0,\; c=0$
I am obtaining $14$ Matrices in total but answer given is $16$.
What Case am I missing? and also is there any better method to solve these kind of problem?
I am also attaching solution provided by My institute I didn't understood there solution at all
| Some simplifications can be made. First, $\det(A)=ad-bc=\pm1$, so $ad=0$ and $bc=\pm1$ or $ad=\pm1$ and $bc=0$.
If $bc=0$, then since replacing $A$ with $A^t$ doesn't affect anything, we might as well assume $c=0$. If $ad=-1$ then all possibilities for $b$ work ($A$ has eigenvalues $\pm1$ so is conjugate to $\begin{pmatrix}1&\\&-1\end{pmatrix}$). Otherwise, you need $b=0$.
If $ad=0$, then by conjugating by $\begin{pmatrix}&1\\1\end{pmatrix}$, which swaps $a$ and $d$, we might as well assume $d=0$. Now,
$$A^2=\begin{pmatrix}a^2+bc&ab\\ac&bc\end{pmatrix}=I_2,$$
so $a=0$ and $bc=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{dx}{dt} = \frac{1}{y}$ and $\frac{dy}{dt} = \frac{1}{x}$ by integrable combinations method? I have a such system to solve using integrable combinations method:
$$
\frac{dx}{dt} = \frac{1}{y}
$$
$$
\frac{dy}{dt} = \frac{1}{x}
$$
And the right answer for it is:
$$
C_1 x^2 = 2t + C_2
$$
$$
y^2 = C_1 (2t + C_2)
$$
And I really didn't understand from what this answer goes, because then I divide these to equations I get for the first $x = C_1 y$ and for the second $y = C_2 x$, that's definitely wrong.
| \begin{align*}
\frac{dx}{dt} & =\frac{1}{y}\tag{1}\\
\frac{dy}{dt} & =\frac{1}{x}\tag{2}
\end{align*}
From (1) $dt=ydx$ and from (2) $dt=xdy$. Hence $ydx=xdy$ or $ydx-xdy=0$. But
$d\left( \frac{x}{y}\right) =\frac{ydx-x dy}{y^{2}}$. This shows that
$d\left( \frac{x}{y}\right) =0$ or $\frac{x}{y}=c$. Where $c$ is arbitrary
constant. Hence
$$
x=cy
$$
(2) now becomes
\begin{align*}
\frac{dy}{dt} & =\frac{1}{cy}\\
ydy & =\frac{1}{c}dt\\
\frac{1}{2}y^{2} & =\frac{1}{c}t+c_{2}\\
y^{2} & =\frac{2}{c}t+2c_{2}%
\end{align*}
Hence
$$
y=\pm\sqrt{\frac{2}{c}t+2c_{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all primes x such that f(x) is also a prime number. There is a function
\begin{align*}
f(x)=x^3 + x^2 +11x +2 \\
\end{align*}
Find all prime $x$ such that $f(x)$ is also a prime number.
I found that this is satisfied with an x value of 3 then the function is equal to 71, so both are primes, but I am unsure how to find other values or prove that there are no other existing solutions. I tried to use modular arithmetic, but I did not go so far.
| Mod 3: $\mod 3$ we know by Fermat's little theorem $x^3 \equiv x\pmod 2$ and so $f(x) = x^3 + x^2 + 11x + 2 \equiv x + x^2 -x -1 \equiv x^2 - 1\equiv (x+1)(x-1)\pmod 3$.
What does this tell us?
It tells us that if $x$ is divisible by $3$ then $f(x)\equiv -1 \pmod 3$. But if $x$ is not divisible by $3$ (i.e. if $x \equiv \pm 1 \pmod 3$) then $f(x)$ is divisible by $3$ (because $x\mp 1\equiv 0 \pmod 3$).
In other words $f(x)$ is divisible by $3$ if and only if $x$ is not divisible by three.
So for $x$ and $f(x)$ to both be prime we must have exactly one of them divisible by $3$. So either $x = 3$ (the only prime divisible by $3$) and $f(x) = 71$ which just happens to be prime. Or we have $f(x)=3$ and .... $x = ????$ well, if we assume $x\in \mathbb N$ then $x \ge 1$ and $f(x) \ge 1+1+11+2 > 3$ so that is not possible.
So $x = 3$ is the only case where both are prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$ I am trying to find a general form of
$$I_n = \int_{-\infty}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^n \ \mathrm{d}x$$
as well as methods of solving it (preferably elementary). This is easy enough to solve for individual $n$, which leads me to believe that there is likely a $\frac{\pi}{\sqrt{3}}$ term in the general expression, but I am having trouble finding a general solution. I have tried constructing a recurrence via IBP, but I always have an extra $\frac{(1 - x^2)^m}{(1 + x^2 + x^4)^n}$ term with $m \neq n$. I also tried creating a $2D$ recurrence, but that doesn't seem solvable, nor can I figure out how to incorporate the requirement of $m \leq 2n - 1$ (for the integral to be finite).
Additionally, I tried using a semicircular contour in the upper half-plane, but computing the residues at $z = e^{\frac{i\pi}{3}}, e^{\frac{2i\pi}{3}}$ seems very bashy.
I have tried using Mathematica, and that just outputs the input.
Any help would be great.
| A systemic procedure is to utilize the reduction formula
$$\int_0^\infty \frac{x^{2m}}{(1+x^2+x^4)^n}dx =I_{m,n}=-2I_{m-1,n}+ \frac{4n-2m-3}{2(n-1)}I_{m-1, n-1}$$
Then, given the initial values
$$I_{0,1}=I_{1,1}=\frac\pi{2\sqrt3}, \>\>\>I_{0,2}=\frac\pi{3\sqrt3}, \>\>\> I_{0,3}=\frac{13\pi}{48\sqrt3}, \>\>\>\cdots$$
we evaluate recursively
\begin{align}
&I_{1,2}=-2 I_{0,2}+\frac32 I_{0,1}=\frac\pi{12\sqrt3}\\
&I_{2,2}=-2 I_{1,2}+\frac12 I_{1,1}=\frac\pi{12\sqrt3}\\
&I_{1,3}=-2 I_{0,3}+\frac74 I_{0,2}=\frac\pi{24\sqrt3}\\
&I_{2,3}=-2 I_{1,3}+\frac54 I_{1,2}=\frac\pi{48\sqrt3}\\
&I_{3,3}=-2 I_{2,3}+\frac34 I_{2,2}=\frac\pi{48\sqrt3}\\
&\ \cdots
\end{align}
As a result
\begin{align}
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )dx=I_{0,1}- I_{1,1}=0\\
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^2dx=I_{0,2}- 2I_{1,2}+ I_{2,2} =\frac\pi{4\sqrt3}\\
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^3 dx= I_{0,3}- 3I_{1,3}+ 3I_{2,3} -I_{3,3}=\frac{3\pi}{16\sqrt3}\\
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^4 dx=\cdots=\frac{\pi}{6\sqrt3}\\
&\int_{0}^\infty \left (\frac{1 - x^2}{1 + x^2 + x^4}\right )^5 dx=\cdots\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$? This question was raised in comments of
Is the $(3,4,5)$ triangle the only rectangular triangle with this property?
and I was suggested to ask it as a separate one.
First some notation, let's write $(a,b,c)$ where $a<=b<c$ for a pythagorean triple, and let's write $(x,y,z)$ where $x<=y<z$ for the distances from the vertices to the center of the inner circle of the corresponding right triangle.
($XYZ$ should be lowercase but geogebra did not allow these labels)
The answer to above question proved (partially left to reader) the property:
$x * y = z$ if and only if $c - b = 1$
In comments was asked if a similar property would exist for $c - b = 2$
and it was confirmed (and proof was left to reader) that:
$x * y = 2 * z$ if and only if $c - b = 2$
Somewhat natural question then was raised (by me) if one can generalize for other values of $n >= 1$, that is:
for which $n$ (perhaps all) holds:
$x * y = n * z$ if and only if $c - b = n$
?
Thanks to @heropup for the suggestion (and the answers for $n=1,2$)
update
A simple computer programmed enumeration seems to confirm equivalence. At least for all $(a,b,c)$ with maximum $c <= 10000$. Note that it is not known to me (but perhaps it is known to others) if all $c - b$ cover all $n >= 1$.
So asking for all $n$ is a bit ambiguous since some $n$ might never occur.
An alternative, perhaps better, question rephrase is:
Prove equality
$$x * y = (c - b) * z$$
for all pythagorean triples.
| Let $(a,b,c)$ be a primitive Pythagorean triple with $a<b<c$. Then $\{a,b\}=\{2mn,m^2-n^2\}$ and $c=m^2+n^2$ for some coprime integers $m>n>0$ of different parity. If $2mn<m^2-n^2$, then $c-b=2n^2$ is twice a square. Moreover, if $m>3n$, then $2mn<(2/3)m^2<(8/9)m^2<m^2-n^2$, so for every $n$ there exists $(a,b,c)$ such that $c-b=2n^2$.
If $m^2-n^2<2mn$, then $c-b=(m-n)^2$, the square of an odd number. Let $k\ge7$ be odd, let $m=2k-2$, $n=k-2$; then $m,n$ are coprime of different parity, $m>n>0$, and $m-n=k$; also, $m^2-n^2=3k^2-4k$, $2mn=4k^2-12k+8$, and $m^2-n^2<2mn$ is equivalent to $k^2-8k+8>0$, which is $(k-4)^2>8$, which is true for $k\ge7$. So, for every odd $k\ge7$ there exists $(a,b,c)$ such that $c-b=k^2$. For $k=1,3,5$, respectively, we can take $(a,b,c)$ to be $(3,4,5)$, $(33,56,65)$, $(65,72,97)$ to get $c-b=k^2$.
Conclusion: for a primitive Pythagorean triple, $c-b$ can take on every odd square, and every twice-a-square, and only those values.
Given any positive integer $t$, the Pythagorean triple $(3t,4t,5t)$ has $c-b=t$, so, if non-primitive triples are allowed, then $c-b$ can take on any positive integer value.
Now, let's bring in $x,y,z$. Using the formulas at The distance from the incenter to an acute vertex of a right triangle, we have
$2x^2=(a+b-c)^2$
$2y^2=a^2+(c-b)^2$
$2z^2=b^2+(c-a)^2$
If $a=2mn<m^2-n^2=b$, $c=m^2+n^2$, then
$$
4(xy)^2=(2mn-2n^2)^2(4m^2n^2+4n^4)=(4n^2)(m-n)^2(4n^2)(m^2+n^2)=16n^4(m-n)^2(m^2+n^2)
$$
so $xy=2n^2(m-n)\sqrt{m^2+n^2}$, and
$$
2z^2=(m^2-n^2)^2+(m-n)^4=2m^4-4m^3n+4m^2n^2-4mn^3+2n^4=2(m-n)^2(m^2+n^2)
$$
so $z=(m-n)\sqrt{m^2+n^2}$. Thus, $xy=2n^2z=(c-b)z$, as requested.
If $a=m^2-n^2<2mn=b$, $c=m^2+n^2$, then
$$
4(xy)^2=(2mn-2n^2)^2((m^2-n^2)^2+(m-n)^4)=8n^2(m-n)^4(m^2+n^2)
$$
so $xy=\sqrt2n(m-n)^2\sqrt{m^2+n^2}$, and
$$
2z^2=4m^2n^2+4n^4=4n^2(m^2+n^2)
$$
so $z=\sqrt2n\sqrt{m^2+n^2}$. Thus, $xy=(m-n)^2z=(c-b)z$, again as requested.
If the triple is not primitive, the common factor cancels out in the end, and the conclusion remains.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A closed form for integral $\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$ $$\int_0^\infty \frac{\ln(1-e^{-\pi x})}{1+x^2} dx$$
let $u=e^{-\pi x}$, the integral goes to:
$$\pi \int_0^1 \frac{1}{u}\cdot\frac{\ln(1-u)}{\pi^2+\ln^2(u)}du$$
This looks a little like Gregory's coefficient, but not the same, because the integral is from $0$ to $1$, instead of from $0$ to $\infty$. How to do next?
| Similar to the evaluation of $\int_{0}^{\infty} \operatorname{Li}_{2}(e^{-\pi x}) \arctan(x) \, \mathrm dx$ here, we have $$ \begin{align} \int_{0}^{\infty} \frac{\ln(1-e^{-\pi x})}{1+x^{2}} \, \mathrm dx &= -\int_{0}^{\infty} \frac{1}{1+x^{2}} \sum_{n=1}^{\infty} \frac{e^{- \pi n x}}{n} \, \mathrm dx\\ &= -\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{e^{- \pi n x}}{1+x^{2}} \, \mathrm dx \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \left( \left(\frac{\pi}{2} - \operatorname{Si}(\pi n) \right) \cos(\pi n) + \operatorname{Ci}(\pi n ) \sin(\pi n ) \right) \\ &= \frac{\pi}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} \\ &= \frac{\pi}{2} \ln(2) + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n}, \end{align}$$
where $\operatorname{Si}(z)$ is the sine integral and $\operatorname{Ci}(z)$ is the cosine integral.
(Justification for interchanging the order of summation and integration comes from Tonelli's theorem.)
To show that $$\sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} = - \frac{\pi}{2}, $$ we can integrate the meromorphic function $$f(z) = \frac{\pi \csc(\pi z) \operatorname{Si}(\pi z)}{z}$$ around a rectangular contour with vertices at $z= \pm \left(N+ \frac{1}{2} \right) \pm i\left(N+ \frac{1}{2} \right),$ where $N$ is a positive integer.
As $|z| \to \infty$ in $ -\pi < \arg(z) < \pi$ , we have $$f(z) \sim \frac{\pi \csc(\pi z) \left(\frac{\pi}{2}- \frac{\cos(\pi z)}{\pi z} \right) }{z} = \frac{\pi^{2} \csc(\pi z)}{2z} - \frac{\cot(\pi z)}{z^{2}}.$$
(EDIT: As explained here, the asymptotic expansion of $\operatorname{Si}(z)$ for large complex $z$ can derived from the identity $$ \operatorname{Si}(z) = \frac{\pi}{2} - \cos(z) \int_{0}^{\infty} \frac{e^{-zt}}{1+t^{2}} \, \mathrm dt - \sin(z) \int_{0}^{\infty} \frac{te^{-zt}}{1+t^{2}} \, \mathrm dt \, , \quad \Re(z) >0$$ using Watson's lemma.)
Therefore, for reasons explained in the answers to this question, the integral vanishes on all sides of the rectangular contour as $N \to \infty$.
So we have $$ \begin{align} 0 &= 2 \pi i \sum_{n=-\infty}^{\infty}\operatorname{Res}[f(z), n] \\ &= 2 \pi i \left( \sum_{n=-\infty}^{-1} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} + \pi + \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} \right) \\ &= 2 \pi i \left(2 \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si}(\pi n)}{n} + \pi \right), \end{align}$$ and the result follows.
Alternatively, $$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \frac{\operatorname{Si} (\pi n)}{n} &= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \int_{0}^{\pi n} \frac{\sin t}{t} \, \mathrm dt \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \int_{0}^{\pi} \frac{\sin (n u)}{u} \, \mathrm du \\ &= \int_{0}^{\pi} \frac{1}{u} \sum_{n=1}^{\infty} (-1)^{n} \frac{\sin (n u)}{n} \, \mathrm du \\ &= - \int_{0}^{\pi} \frac{1}{u} \, \arctan \left(\tan\frac{ u}{2} \right) \, \mathrm du \\ &= - \int_{0}^{\pi} \frac{1}{u} \frac{u}{2} \, \mathrm du\\ &= - \frac{\pi}{2}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
How to find the sum of constants given the following systems of equations? Given that $q,r,s$, and $t$ are different constant values in the following systems of equations containing $a,b,c$, and $d$. Find the sum $q+r+s+t$.
$\frac{1}{qa+1} + \frac{1}{qb+1} + \frac{1}{qc+1}+ \frac{1}{qd+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} + \frac{1}{rc+1}+ \frac{1}{rd+1} = 1$
$\frac{1}{sa+1} + \frac{1}{sb+1} + \frac{1}{sc+1}+ \frac{1}{sd+1} = 1$
$\frac{1}{ta+1} + \frac{1}{tb+1} + \frac{1}{tc+1}+ \frac{1}{td+1} = 1$
I tried using a simpler version of the problem but I can't find the pattern.
I am thinking that there might be a pattern that I can get from finding the sum $q+r$ from this simpler version of the problem.
$\frac{1}{qa+1} + \frac{1}{qb+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} = 1$
I also tried adding the four equations that leads me to factor out the sum $q+r+s+t$ but it becomes more complicated. But I noticed that the sum of the each set of denominators per equation is equal to the product of all those denominators. However, I am stuck at looking for the pattern where I can use that fact.
Thanks in advance for your comments and suggestions on how to solve this particular algebra problem.
| Begin by noting that $q,r,s,t$ are the four roots of the equation $$\frac{1}{ax+1}+ \frac{1}{bx+1} + \frac{1}{cx+1} + \frac{1}{dx+1} =1$$
Simplifying, we get $$(bx+1)(cx+1)(dx+1)+(ax+1)(bx+1)(cx+1)+(cx+1)(dx+1)(ax+1)+(dx+1)(ax+1)(bx+1)=(ax+1)(bx+1)(cx+1)(dx+1).$$ Now, we know that for a general cubic equation $ux^4+vx^3+wx^2+m$, the sum of the roots is $-\dfrac vu$. So we do not need to simplify the whole equation, but just find the coefficients of $x^3$ and $x^4$.
Coefficient of $x^4=abcd$Coefficient of $x^3=0$
So $q+r+s+t=0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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