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5

This is just an outline of an answer. Call the integral to be calculated $I(a)$. First write $$I(a) = 2\int_0^{\pi/2} \sqrt{\cos^2 k + a^2 \sin^2 k} \, dk$$ by symmetry. The difficulty here is that you can't just apply Taylor's formula for $a \to 0$, because values of $k$ near $\pi/2$ make a significant contribution to the integral and $\cos k$ is small ...


5

Hint. Your integral may be reduced to a generalized incomplete gamma function that has been previously studied. Step 1. Integrating with respect to $y$ gives $$ \begin{align} \int_0^\infty \exp\left(-\frac{y^2}{z^2}-\frac{x^2}{y^2}\right)dy&=\exp{\left(-\frac{2 x}{ z}\right)}\int_0^\infty \exp{-\left(\frac{y}{z}-\frac{x}{y}\right)^2}dy\\\\ ...


4

We have: $$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$ while on the other hand, if we set: $$ A(n,m)\triangleq\int_{0}^{1} x^n ...


3

$$e ^{\ln x } = e^{\frac{1}{x}} \Rightarrow \frac{1}{x} e^{\frac{1}{x}} = 1\Rightarrow x = \frac{1}{W(1)} $$ Where $W(1)$ is the Omega constant.


3

First thing we should note is: $$\lim_{n\to\infty}\frac{U_n(x)^2}{U_{n-1}(x)^2+U_n(x)^2}=\lim_{n\to\infty}\frac{1}{\left(\dfrac{U_{n-1}}{U_n}\right)^2+1}$$ So, we look at $\displaystyle\lim_{n\to\infty}\frac{U_{n-1}}{U_n}$. Wikipedia gives the formula: $$U_n=\frac{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}{2\sqrt{x^2-1}}$$. Thus, ...


2

Whittaker and Watson's A Course of Modern Analysis is a standard source for these types of problems.


2

Hint: $$B(n,a) = \int_{0}^{1} t^{n-1}(1-t)^{a-1} dt = \frac{\Gamma(n)\Gamma(a)}{\Gamma(n + a)}$$ Where $B$ is the Beta Function and $\mathcal {Re} (a) > 0$.


2

$$\begin{align*} \ln x &= \frac1x\\ x &= e^{1/x}\\ 1 &= \frac1x e^{1/x}\\ W(1) &= \frac1x\\ x &= \frac1{W(1)} \end{align*}$$ Where $W$ is the inverse function of $we^w$. Alternatively, $$\begin{align*} x\ln x &= 1\\ (\ln x)e^{\ln x} &= 1\\ \ln x &= W(1)\\ x &= e^{W(1)} \end{align*}$$ and WolframAlpha confirms $e^{W(1)} ...


2

I can't comment because apperently just guys with super high level can help you. About this approach with the mean valued integral theorem: I've tried it.It won't work simply because $\xi$ depends on $n$ if it was a constant for any $n$ I would agree with you, this limit is 1. But the sequence $a_n$ you created is not necessarly monotonic. Because $\xi$ is a ...


2

By the duplication formula, $$\Gamma\left(\frac{2}{50}\right) = C\cdot \Gamma\left(\frac{1}{50}\right)\Gamma\left(\frac{13}{25}\right)$$ hence: $$\Gamma\left(\frac{1}{50}\right) = \frac{\Gamma\left(\frac{1}{25}\right)}{C\cdot \Gamma\left(\frac{13}{25}\right)}.$$


2

All you need is continuity at $0$. If for some $c$, $f(c) = 0$, then it is easy to show that $f(x) = 0$ for all $x$. So suppose $f(x) \neq 0$ for any $x$. Now $f(x) = (f(x/2))^2 \gt 0$. Thus it makes sense to talk about $g(x) = \log f(x)$. This satisfies $g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation and continuity at $0$ implies $g(x) ...


2

As user mickep already pointed out in his comment, the integral can be rewritten as $\dfrac{\sqrt\pi}2~e~I''(1)$ where $I(a)=\displaystyle\int_1^{+\infty}\exp\bigg(\!\!-az-\frac{2x}z~\bigg)~dz$. Now, if the lower integration limit would have been $0$ instead of $1$, then we'd have $J(a)=2~\sqrt{\dfrac{2x}a}~K_1\Big(2\sqrt{2ax}\Big)$ which after being ...


2

Start from the fact that, for each $i$, the density of the full random vector $(X_j)_j$ can be factored as $$P(X_i=x_i,\hat X_i=\hat x_i)=h(\hat x_i)\cdot\mathrm e^{a_i(x_i,\hat x_i)x_i},\qquad$$ for some positive function $h$ whose exact value is irrelevant, with $$\hat X_i=(X_j)_{j\ne i},\qquad\hat x_i=(x_j)_{j\ne i},\qquad a_i(x_i,\hat ...


1

From @Haderlump Let's define $x_i$ on $x$ being $b$ as $x^b$. Since $$ P(x_j,j \ne i) = P(x^0) + P(x^1) $$ We can do $$ P(x_i = 1 | x_j,j \ne i) = \frac{P(x^1)}{P(x^0) + P(x^1)} $$ It's better to see $P(x^1)$ as divider, and cleaner to get rid of the +1 in $1/1 + e^{-a_i}$, so we flip the division and subtract one, $$ 1 / P(x_i = 1 | x_j,j \ne i) = ...


1

Yes indeed: Rewriting the $\sin(z^2)$ Term by the help of Eulers formula, we get $$ I(y,x,t)=\frac{1}{2i}\int_0^{\beta(x,y,t)}(e^{i z^2 }-e^{-i z^2 })e^{-\alpha(x,y)z^2}\mathrm{d}z $$ with $\alpha(x,y)=\frac{4xy}{y^2-x^2}$ and $\beta(x,y,t)=\frac{y^2-x^2}{4t}$. By a simple rescaling and by using the definition of the Error function ...


1

Outline: Use integration by parts to calculate $\int_0^\infty x^3 e^{-\alpha x^2}\,dx$. We know this is $\frac{1}{2A}$, so we can calculate $\alpha$. Now use a couple of integrations by parts to calculate $\int_0^\infty x^4 e^{-\alpha x^2}\,dx$. Not elegant, but it does the job.


1

It is not clear what you mean by that it should describe a circle, can you elaborate that? If you want a rotational symmetry you can use use the radius $r=\sqrt{x^2+y^2}$ as an argument: $F(x,y) := f(\sqrt{x^2+y^2})$ but in this of course only uses the part of $f$ with a positive argument, that means that $f(x,0) = f(|x|)$. EDIT: $f(x) = ...


1

The function you give is the second version of my sigmoid function, which works better than the one shown in the video. To get the effect shown, I put three NTSF in series. The first takes values from -1 to 1 converted to the range 0 to 1. The output is converted back to the range -1 to 1, and the second NTSF applied. This output is then converted to the ...


1

$% Predefined Typography \newcommand{\paren} [1]{\left({#1}\right)} \newcommand{\bparen}[1]{\bigg({#1}\bigg)} \newcommand{\brace} [1]{\left\{{#1}\right\}} \newcommand{\bbrace}[1]{\bigg\{{#1}\bigg\}} \newcommand{\floor} [1]{\left\lfloor{#1}\right\rfloor} \newcommand{\bfloor}[1]{\bigg\lfloor{#1}\bigg\rfloor} \newcommand{\mag} ...


1

I intend to develop some of N3buchadnezzar's ideas and correct some changes of sign which he has lost. The result will generalize the calculations that 17762 and Arbias Hashani have started. From the definition of the $\Gamma$ function and a change of variable I get that \begin{equation*} \Gamma(\alpha) = \int_{0}^{\infty}x^{\alpha -1}e^{-x}\, dx = 2 ...


1

$$ \Gamma(x+1)=x\Gamma(x). $$ So \begin{align} \Gamma(5+a) & = (4+a)\Gamma(4+a) \\[8pt] & = (4+a)(3+a)\Gamma(3+a) \\[8pt] & = (4+a)(3+a)(2+a)\Gamma(2+a) \\[8pt] & = (4+a)(3+a)(2+a)(1+a)\Gamma(1+a) \\[8pt] & = (4+a)(3+a)(2+a)(1+a)a\Gamma(a) \end{align} and then cancel a factor from the numerator and denominator. And similarly for $n\ne4$. ...


1

From $$\psi(1-z)=-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k} $$ and $\psi(2-z)=\psi(1-z)+\frac{1}{1-z}$ it follows that: $$\psi(2-z) = \frac{1}{1-z}-\gamma-\sum_{k\geq 1}\zeta(k+1) z^{k}=(1-\gamma)+\sum_{k\geq 1}(1-\zeta(k+1))\,z^k\tag{1}$$ now replace $z$ with $z+2$ to get the stated identity.


1

You could have $f(x,y) = a(y) x + b(y)$ for any continuous functions $a(y)$ and $b(y)$ on $\mathbb R$. No reason for them to be constant. EDIT: With the new assumptions, you could have $f(x,y) = a x y + b x + c y + d$.


1

$\cos(n x) = T_n(\cos(x))$, where $T_n$ is the $n$'th Chebyshev polynomial. Taking the derivative, $ n \sin(n x) = T_n'(\cos(x)) \sin(x)$. Thus $$ \dfrac{\sin(nx)}{\sin(x)} = \dfrac{T_n'(\cos(x))}{n} $$ For odd $n$, $T_n$ is an odd function, so $T_n'$ is even, and thus $T_n'(\cos(x)) = S_n(\cos^2(x)) = S_n(1-\sin^2(x))$ where $S_n$ is a polynomial.


1

HINT: Compute the radius of convergence of the series. Can we consider limit in infinity for this series?


1

The sigmoid squashing function is the same as the sigmoid function. The term sigmoid squashing function is favored in the neural net community. The logistic function is the classical squashing function. Other sigmoid functions include: arctangent, the hyperbolic tangent, the Gudermannian function, and the error function. It is called the squashing ...


1

RHS: lim$_{h\rightarrow 0}\dfrac{f(h)-f(0)}{h}=\dfrac{\sin h-h}{h^2}$ On applying L'Hospital's Rule twice we get limit 0 Similarly LHS: $\dfrac{f(0)-f(-h)}{h}=\dfrac{h-\sin h}{h^2}(\dfrac{0}{0})$ On applying L'Hospital's Rule twice we get limit 0 Thus differentiable at 0


1

The continued fraction representation $x=[a_0;a_1,a_2,\ldots]$ is unique when $x$ is irrational. If $x$ is rational, there are exactly two continued fraction representations: $x=[a_0;a_1,\cdots,a_n]=[a_0,a_1,\cdots,(a_{n}-1),1]$, so it's easy to see the question-mark function is indeed independent of continued fraction representation.


1

A partial answer for now. Let: $$B_{m,l}(x)\triangleq\frac{d^m}{dx^m}P_l\frac{d^{m+1}}{dx^{m+1}}P_{l+1}-\frac{d^m}{dx^m}P_{l+1}\frac{d^{m+1}}{dx^{m+1}}P_{l}.$$ We have: $$B_{0,l}(x) = P_l P_{l+1}'-P_{l+1}P_l'$$ so: $$(1-x^2) B_{0,l}(x) = P_l ((1-x^2)P_{l+1}') - P_{l+1}((1-x^2)P_{l}') $$ and since the Legendre differential equation gives: ...


1

Standard trick $$ x\rightarrow \alpha x\\ y\rightarrow \beta y. $$ It will turn out you will not need to transform $y$. In any case $$ \bar{x}^2y'' + \bar{x}y' + \left(4\alpha^2\bar{x}^2 - \frac{9}{25}\right)y = 0 $$ let $\alpha = \frac{1}{2}$ we obtain $$ x^2y'' + xy' + \left(x^2 - \frac{9}{25}\right)y = 0 $$ which is bessel if we have the $x\rightarrow ...



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