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1

Since $\mathrm{erf}$ is analytic/holomorphic over the real line (in particular, you have a power series representation that converges for all reals), you have the deep result of (complex) analysis that there is a unique analytic continuation of $\mathrm{erf}$ to the entire complex plane. Thus, In what ways can I extend the error function to accept ...


2

Exact expression for the solution is: $$\gamma=\frac{\pi/4}{e^{\pi/4}-1}=0.6581842...$$ EDIT The two branches are related in a trascendental way. Their difference can be used to solve other trascendental equations. If $y(x)=\frac{x}{1-e^x}$ then : $x=W_0(ye^y)-W_{-1}(ye^y)=y-W_{-1}(ye^y)$ for $-1< x <0 $ $x=W_{-1}(ye^y)-W_{0}(ye^y)=y-W_{0}(ye^y)$ ...


4

In terms of Lambert function, there are two roots which are $$x_1=-\frac{1}{2} W\left(-\frac{1}{3}\right)\approx 0.309$$ $$x_2=-\frac{1}{2} W_{-1}\left(-\frac{1}{3}\right)\approx 0.756$$ If you want to compute accurately these roots, you could solve $$f(x)=6x-e^{2x}=0$$ using Newton method. Even if we already know the results, you can notice that $f(x)$ ...


4

The Lambert W function has infinitely many branches. For $-1/e < x < 0$, both the "$-1$" branch $W_{-1}$ and the "$0$" branch $W_0$ are real; both are $-1$ at $-1/e$, but $W_{-1}(x)$ decreases to $-\infty$ as $x$ increases to $0$ while $W_0(x)$ increases to $0$. Here's a plot, with $W_0$ in red and $W_{-1}$ in blue. For numerical approximations, ...


0

$\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{a^nb^{n+m}}{(m+n)^2\Gamma(m+n)}{_2F_2}\left(m+n,m+n;m+n+1,m+n+1;-b\right)$ $=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^nb^{n+m}\Gamma(m+n)(m+n)_k(m+n)_k(-b)^k}{\Gamma(m+n+1)\Gamma(m+n+1)n!(m+n+1)_k(m+n+1)_kk!}$ ...


2

$$\int_a^b{f(x)dx}=\int_a^{b}{f(x)[H(x-a)-H(x-b)]dx}=\int_0^{\infty}{f(x)[H(x-a)-H(x-b)]dx}$$


1

Your question is incomplete, one need to read page 46 of Sneddon's book to figure out what the problem really want. To summarize, what the book want is start from an equation of the form $$z = f(u)\quad\text{ where }\quad u = \frac{xy}{z}$$ derive a PDE for $z$ which doesn't involve the function $f(u)$ explicitly. The tool you need is chain rule for ...


0

I try this today. $p= \dfrac{\partial z}{\partial x}$ and $q= \dfrac{\partial z}{\partial y} $ So $ p= \dfrac{\partial f}{\partial x }\left( \dfrac{yz-pxy}{z^2} \right) $ and $ q= \dfrac{\partial f}{\partial y }\left( \dfrac{xz-pxy}{z^2} \right) $ Is that right? After that, I try many minupulations... All attempts, I needed divided by ...


2

Your problem can be rephrased as $$\arg\max_y\{\sum_{r_1\in \Omega_1}p(r_1,s_2)u(y,r_1,s_2)\}\ge \arg\max_x\{\sum_{r_1\in \Omega_1}p(r_1,t_2)u(x,r_1,t_2)\}\text{ if }s_2>t_2.$$ Note that if $s_2>t_2$ and $x>y$, we will have $$u(x,r_1,s_2)-u(y,r_1,s_2)>u(x,r_1,t_2)-u(y,r_1,t_2).$$ Denote the left part to be $a_{r_1}$ and right part be $b_{r_1}$, ...


0

It is better to refer the exact definition given in your course, because it is not a standard notion. One of possibilities is that if $\vec x(t)$ is the solution of your differential equation, then you will have $$\frac{d}{dt}V(\vec x(t))\le 0$$ for Lyapunov function and $$\frac{d}{dt}V(\vec x(t))< 0 \text{ whenever }x\ne 0$$ for strong Lyapunov ...


1

If you are dealing with this numerically, use expm1(n*log1p(x)) (if those functions are available in your math library) to get high precision results. expm1(x)=exp(x)-1 and log1p(x) = ln(1+x) both for small x without cancellation.


1

Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt. \end{align} Setting $x=1$ leads to \begin{align} \Gamma'(1) = \int_{0}^{\infty} e^{-t} \, \ln(t) \, ...


2

The Weierstrass product for the $\Gamma$ function gives: $$\Gamma(z+1)=e^{-\gamma z}\cdot\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}$$ hence by considering $\frac{d}{dz}\log(\cdot)$ of both terms we get: $$ \psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{2}$$ and by evaluating the ...


1

If you want to use the Gamma function the substitution is $a x^2 = t$, so “$dx = \frac{1}{2\sqrt{a}}t^{-1/2}dt$". Then the integral appears as, $$\frac{1}{2\sqrt{a}} \int_0^\infty dt\,t^{-1/2}e^{-t} = \frac{1}{2\sqrt{a}}\Gamma(1/2)\ .$$ That's all.


3

The trick (at least one of them) is to write: $$I = \int_0^\infty {e^{-ax^2}} dx$$ and since the variable doesn't matter, $$I = \int_0^\infty {e^{-ay^2}} dy$$ So that $$I^2=\int_0^\infty {e^{-ax^2}} dx\int_0^\infty {e^{-ay^2}} dy=\int_0^\infty\int_0^\infty {e^{-a(x^2+y^2)}}dx dy$$. We then switch to circular polar coordinates so that $dxdy=r drd\theta$ and ...


0

It is important to know in what sense you want the best approximations, and whether you are looking only for linear combinations of the functions or not. As an example, you can take a linear combination $\tilde{f} = \gamma g + \delta h$ of the 2 approximations $g$ and $h$, choose a set of grid points, use the values of the numerical solution and $\tilde{f}$ ...


2

Yes, you can do it directly using integration by parts to get to $(x+y)B(x+1,y)=x B(x,y)$. With this, you then actually want to go the other way: write the formula as $$ B(x,y) = \frac{x-1}{x+y-1} B(x-1,y). $$ Now you iterate/use induction to get to the formula $$ B(m,n) = \frac{(m-1)(m-2) \dotsm (m-(m-1))}{(m+n-1)(m+n-2)\dotsm (m+n-(m-1))} B(0,n) = ...


1

$$\gamma\simeq0.658184273600902103771558680728257154392531172950\ldots$$


0

If you allow the floor function, $\lfloor x\rfloor$, you could use $$\left\lfloor \frac{1}{(A-B)^2+1}\right\rfloor$$


2

No It is not In terms of elementary functions, But we can write in Infinite Series form. Using $$\displaystyle \sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..........$$ So $$\displaystyle \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........$$ So $$\displaystyle \int \frac{\sin x}{x}dx = \int ...


1

For $\epsilon\leq 1/e$, $$T \geq e^{-W_{-1}(-\epsilon)}=-\epsilon^{-1}W_{-1}(-\epsilon)$$ where $W_{-1}$ is the second branch of the Lambert $W$ function. The Lambert $W$ functions are branches of the inverse of $f(z)=ze^z$. It cannot be expressed in terms of elementary functions, so this is the best you can do. Answer courtesy Wolfram Alpha.


0

Legendre polynomials are orthogonal.here So it may work to calculate that integral by using integrating by parts twices.


0

Playing around I found an answer to my own question. For the $k=3$ case, you can construct a solution from the third roots of unity. $$f_3(n)=\frac{1}{3}\bigg(1 + \big(e^{2i\pi(1/3)}\big)^n + \big(e^{2i\pi(2/3)}\big)^n \bigg) $$ The second and third term, since they are third roots of unity, equal unity if $3|n$, which satisfies the first part of the ...


1

It might be easier to start with one of the other integral representations, but here's one way. We'll begin with some manipulations to strip away the lower order terms: $$ \begin{align} &\int_0^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\ &\qquad = \left(\int_0^1 + \int_1^\infty\right) t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\ &\qquad = \int_0^1 t^{a-1} ...


1

Hint: Expand $\ln(1-u)$ into its Mercator series, and then reverse the order of summation and integration.


0

I have found this quick paper that gave an identity I was not aware of, viz. \begin{align} \Gamma\left(\alpha\right)&=\frac{\Gamma\left(\alpha+1\right)}{\alpha},\tag{1} \end{align} so it seems that if $\alpha=-3.1$, then \begin{align} \Gamma\left(-3.1\right)&=\frac{\Gamma\left(-2.1\right)}{-3.1}\\ &=\frac{\Gamma\left(-1.1\right)}{6.51}\\ ...


1

Plugging $(3)$ into $(2)$, we get $$ \frac{\Gamma(a-n)}{\Gamma(a)}\stackrel?=\frac{(-1)^n\Gamma(1-a)}{\Gamma(1-a+n)}$$ or equivalently $$\tag 4 \Gamma(a-n)\Gamma(1-a+n)\stackrel?=(-1)^n\Gamma(1-a)\Gamma(a)$$ From Euler, we know $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}$ for all $x\in\mathbb C-\mathbb Z$, hence $(4)$ reduces to the valid (outside $\mathbb ...


2

$$f(n,k)=1+ H(n-k),$$ where $H$ is the Heavyside function: $$H(x)=\begin{cases}1,&x\ge0,\\0,&x<0.\end{cases}$$


1

If $f$ is general, then $X$ may not be compact. Pick an arbitrary, strictly convex function $g$ with $g(0.5) = 0$, and define $f(x) = g(x)$ for $x \neq 0.5$ and $f(0.5) = 1$. You have $\tilde f = g$ and $X = [0,1]\setminus \{0.5\}$ which is not compact. If $f$ is continuous, then the result is true, because if $\tilde f(x)<f(x)$ then this happens on a ...


6

$$f(x) = \sum_{k=0}^{\infty} \frac{k x^{k+1}}{(k+1)^2} $$ $$f'(x) = \sum_{k=0}^{\infty} \frac{k x^{k}}{k+1} = \sum_{k=0}^{\infty} x^k - \sum_{k=0}^{\infty} \frac{x^k}{k+1} = \frac1{1-x} + \frac{\log{(1-x)}}{x}$$ The sum is then $$f \left ( \frac12 \right ) = \int_0^{1/2} dx \left [\frac1{1-x} + \frac{\log{(1-x)}}{x} \right ] = \log{2} - ...


1

All you need is a normal number. They should be easy to find, because almost all reals are normal. Unfortunately, it is hard to prove any given number is normal unless it is constructed to be so and then it probably won't look "random". The Champernowne constant $0.123456789101112131415...$ fits this. It is known to be normal in base $10$, so if you break ...


1

$L^2~=~\displaystyle\lim_{p\to\infty}~\dfrac{\sqrt[P]{\Gamma\big(p+a,~b\big)}}p~=~\dfrac1e~,\qquad\forall~a,b\in\mathbb C.~$ This follows from the fact that $|b|<p$ and $~\displaystyle\lim_{p\to\infty}~\frac{\gamma\big(p,~p\big)}{\Gamma\big(p,~p\big)}~=~1,~$ in conjunction with Stirling's approximation.


0

There are other exact values of erf(x) besides 0 and plus or minus infinity this graph there are horizontal asymptotes at y=1 and y=-1 because The Z-score is a probability that certain data fall within a certain interval. For example a Z-score of 1 is represented by erf(1) = 0.682 which means that plus or minus 1 standard deviation from the mean will ...


5

Gauss multiplication formula works, indeed. If we take $z=\frac{1}{n}$ in: $$\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right) = (2\pi)^{\frac{n-1}{2}} \,n^{\frac{1}{2}-nz}\,\Gamma(nz)\tag{1}$$ we get: $$ \prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right) = (2\pi)^{\frac{n-1}{2}} \,n^{-\frac{1}{2}}=\frac{(\sqrt{2\pi})^n}{\sqrt{2\pi n}}.\tag{2} $$


0

This link http://www.dtic.mil/dtic/tr/fulltext/u2/a252517.pdf is a PhD-thesis using H-function in the study of statistical distributions. There are other papers with applications in this area, such as The Distribution of Products, Quotients and Powers of Independent H-Function Variates by Bradley D. Carter and Melvin D. Springer SIAM J. Appl. Math., ...


1

Fox–Wright function or even the Fox H-function are very useful on dealing the infinte series expressions whose its coefficient involves for example the gamma functions of linear expressions in the index $n$. For example: Definite Integral of $e^{ax+bx^c}$ Definite integration of a high order exponential function mixed with rational function Can this ...


0

Fractional linear transforms have a very useful property called 3-transitivity. If $(z_1, z_2, z_3)$ and $(w_1, w_2, w_3)$ are two sets of distinct points in $\mathbb{C} \cup \infty$, then there is a fractional linear transform between them (i.e. a transform such that $z_i \mapsto w_i$). Furthermore, the transform is unique ; though this means that in ...


2

A hyperbolic trigonometry approach. Set $$ C(x)=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\quad\text{and}\quad S(x)=\sum_{k=1}^\infty\frac{x^{2k-1}}{(2k-1)!} $$ It suffices to show that $C(x)>S(x)$, for every $x\in\mathbb R$. First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that $$ ...


1

Using termwise differentiation one finds that $\exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):\equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $x\mapsto e^x$, which is positive when $x=0$, is positive on ...


0

The series expansion is $$ e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $$ For $x\ge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive. For $x<0$ notice that: $$ \frac1{e^x}=e^{-x} $$ So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.


0

For $\lambda=1$ we have $$ f(x,y) = \begin{cases} 1 &\mbox{if } \quad0<x<1, y=1, \\ 1+y & \mbox{if } \quad\text{otherwise}\end{cases} $$ and $$ L(3/2;f)=(0,1)\times\{1\}\bigcup [0,1]\times[0,1/2]. $$ Since $L(3/2;f)$ is not convex, $f$ is not quasiconvex on $C$ for $\lambda=1$.


2

You just need to prove the reflection formula: $$ \psi(1-z)-\psi(z)=\pi\cot(\pi z)\tag{1}$$ then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $\Gamma$ function: $$\Gamma(t+1) = e^{-\gamma t}\prod_{n=1}^{+\infty}\left(1+\frac{t}{n}\right)^{-1}e^{\frac{t}{n}}\tag{2}$$ leading to: $$ ...


1

The original $S(t)$ has flexes at $t=\log(2\pm\sqrt 3)$. Now define $$ Z(t)=S(kt) $$ for some stretch scalar $k\in\mathbb R^+$, and convince yourself that $Z'''(t)=0$ iff $S'''(kt)\cdot k^3=0$ iff $S'''(kt)=0$. Hence $Z(t)$ has flexes at $$ t=\frac{\log(2\pm\sqrt 3)}{k}=\pm\frac{1}{A} $$ and for $A,k>0$ this leads to $$ k=A\log(2+\sqrt 3) $$ which makes ...



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