# Tag Info

## New answers tagged special-functions

0

Notes: The paper referenced is unclear about the parameters $p$ and how $z$ is defined. It would seem that somehow $p \sim \Gamma(0)$. This would work if another parameter is defined, say $\alpha$, such that $p = \alpha \, \Gamma(0) \to \beta$ where $\beta$ is a finite quantity. It is of interest to note that \begin{align} I &= \int_{0}^{\infty} ...

0

The statement, an excerpt from Hilbert Curve does seem a bit vague, but this is how I interpret it. Say you have a space filling curve $\gamma$ in the unit square with a length $d$ that ends at $(x_1, y_1)$. If you follow that curve further to a length $d + \epsilon$ you will end up at a nearby point $(x_2, y_2)$ because it is continuous. On the other ...

1

To find the degree of this polynomial was Problem 11403 in the Dec. 2008 issue of the American Mathematical Monthly. The function was there called $f_n(x)$ and its degree is indeed $\lfloor n/2\rfloor.$ In your notation we have $E[(X_n-n)^k]=f_k(n).$ Here is the first paragraph of the published solution (March 2011): The degree of $f_n$ is ...

1

Try looking at Vilenkin's introductory 1968 book on Representation Theory and Special Functions. Then look at his 3 volumes on the same subject with a more detailed treatment.

0

Write $x=\lfloor x\rfloor+\{x\}$. Then we have to solve $$3\lfloor x\rfloor+2\{x\}=1\ .$$ There is no solution with $\lfloor x\rfloor<0$ or $\lfloor x\rfloor\geq1$. When $\lfloor x\rfloor=0$ we need $2\{x\}=1$, which implies $x={1\over2}$. It is easy to check that this is indeed a solution.

1

Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n$,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases}$$ The given equation implies $\;1=\bigl\lfloor\lfloor x\rfloor+ ... 1$x-1<\lfloor x\rfloor \leq x$, so$3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies$x\in [1/3, 2/3)$, so$\lfloor x\rfloor=0$, then$2x=1$,$x=1/2$. 1 Rewrite the given equation as $$\lfloor x \rfloor = 1-2x$$ Note that$1-2x$must be an integer. So$x=\frac{k+1}{2}$. Now simplification is much easier by considering two cases (1)$k$is even (2)$k$is odd. OR alternatively you can do as follows for$x > 1/2$the right side is negative but left side is non-negative so no equality. Likewise for$x ...

3

Note that the function $f(x) =\lfloor x \rfloor + 2x$ is strictly increasing. It is always between $3x-1$ and $3x$. Solving $3x-1=1$ and $3x=1$ gives that $x$ is between $\frac13$ and $\frac23$, because $f$ is strictly increasing. So $\lfloor x \rfloor = 0$, since $\frac13 < x < \frac23$. So we solve $2x=1$ so $x=\frac12$.

1

The op $D^{-1}D_{h,x}$ and its inverse $DD_{h,x}^{-1}$ are classic operators associated with the Bernoulli numbers (and polynomials) and the Euler-Maclaurin formula.

1

I think it is a correct answer: $$e^{hD}f = \sum \frac{h^n}{n!}D^n f = f(x+h)$$ so $$D_{h,x}f = \frac{f(x+h)-f(x)}{h}$$ Given $$g(x) = \frac{f(x+h)-f(x)}{h}$$ we may reconstruct $f(x)$ as $$f(x) = h\sum_{n=1} g(x - nh)$$ Now $g(x - nh) = e^{-nhD}g$, that is $$f = h\sum e^{-nhD}g = h\sum (e^{-hD})^n g= h(\frac{1}{1 - e^{-hD}} - 1)g = \frac{h}{e^{hD} - 1}g$$ ...

0

$D^{-1}_{h,x}(f) = e^{-h\frac{d}{dx}}(hf+1)$

1

If $\mathcal{I} = D^{-1}$ then wouldn't $$D^{-1}_{h,x}[f] = \mathcal{I} \left[ \frac{\ln(fh+1)}{h} \right],$$ because then $$D^{-1}_{h,x}\left[D_{h,x}[f]\right] = f?$$

3

Let's make a substitution $\sin(\theta)=y$ $$I(x,z)=\int_{0}^{\sin(z)}\frac{J^2_{1}(x y)}{y^2\sqrt{1-y^2}}dy$$ Because $z\approx0.1$ we may that assume $\sin(z)\approx z$. Furthermore we perform another subsitution $\frac{y}{z}=q$ yielding $$I(x,z)=z\int_{0}^{1}\frac{J^2_{1}(x z q)}{(zq)^2\sqrt{1-(zq)^2}}dq$$ As a first step to simplify the problem we ...

0

By making a few reductions the series can be seen in the form \begin{align} S = \sum_{n=1}^{\infty} \sin\left(\frac{\pi n}{2}\right) \, \frac{\zeta(n+1)}{(2\pi)^{n+1}} \, \frac{\Gamma(z)}{\Gamma(z-n)} \, \left[\psi^{0}(z-n)-\psi^{0}(z)\right] = \sum_{n=0}^{\infty} \frac{\zeta(2n+2) \, (1-z)_{2n+1}}{(2\pi)^{2n+2}} \, \left(\sum_{k=1}^{2n+1} ...

0

I don't know if a local argument is enough in this case, anyway one may reason as follows. First of all you need to prove that if $\{\vec{x}_1,\vec{x}_2,\ldots,\vec{x}_N\}$ are equally spaced on $2\mathbb{S}^1$, then your minimum points $y_\min$ can take one of $N$ equivalent places: for every couple $\vec{x}_a$ and $\vec{x}_b$ of consecutive points, there ...

0

Here we provide the answer for a symmetric (ie unskewed) L\'{e}vy stable distribution with mean zero. Denote the PDF and the CDF of the stable distribution by $\rho_\mu(\xi)$ and by $\Phi_\mu(\xi)$ respectively.Then we have: \begin{eqnarray} &&\tilde{\mathcal I}_\mu(A) := \int\limits_{-1/A}^\infty \log(1+A \xi) \rho_\mu(\xi) d\xi \nonumber\\ ...

6

As an alternative, we may apply the substitution $x\mapsto\sinh^2{x}$. \begin{align} \int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x &=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\ &=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\ ...

3

We can write $$\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\left(1+ta^{-n}\right)\right]_{t=1},$$ which implies that $$\sum_{n=0}^{\infty}\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\prod_{n=0}^{\infty}\left(1+ta^{-n}\right)\right]_{t=1}= \left[\frac{\partial}{\partial t}\,\ln\left(-t;a^{-1}\right)_{\infty}\right]_{t=1},$$ where ...

1

Your function is called the Generalized Exponential Integral $E_p(x)$ and is described in http://dlmf.nist.gov/8.19: $$E_p(x) = x^{p-1}\int_x^\infty \frac{e^{-t}}{t^p}\: d t \; = \int_1^\infty \frac{e^{-xt}}{t^p}\: d t\; ,$$ Without a specific reference to the actual open source code, I guess that the discrete recurrence formula based on $E_1$ or your ...

6

By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align} The first ...

2

By expanding $\frac{1}{1+a^n}$ as a geometric series, $$\sum_{n\geq 1}\frac{1}{1+a^n}=\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{a^{mn}}=\sum_{l\geq 1}\frac{1}{a^l}\sum_{d\mid l}(-1)^{\frac{l}{d}+1}=\sum_{l\geq 1}\frac{g(l)}{a^l}$$ where $g(l)$ is an arithmetic function counting the difference between the number of odd divisors and the number of even ...

5

Let: $$f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx.$$ We have, by integration by parts: $$f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$ hence: $$\left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$ $f(0)=2\sqrt{2}-1$ and $\lim_{s\to +\infty}f(s)=0$. We have: $$f(s)=\sum_{n\geq 0}\frac{(-1)^n ... 3 I tried 2 ways to find a closed form, though unsuccessful up to this point. 1st trial. Let I denote the integral, and write$$ I = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-(2n+1)x})}{x (1 + e^{-x})^{2}} \, dx. $$In order to evaluate the integral inside the summation, we introduce new functions I(s) and ... -2 Those are all idempotent functions. More examples are here. 1 Assuming that you are talking only about \Bbb R\to\Bbb R functions, for a such function, this equation holds:$$f(x)=x\;\forall x\in f(\Bbb R)$$So, for any nonempty A\subset \Bbb R define f(x)=x for x\in A, and f(x) to be any y\in A otherwise. Any such function will hold f\circ f=f. The sets A correspondig to your examples are, ... 1 Hint. You have an Euler-type integral, you may then rewrite your initial integral in terms of the Appel hypergeometric function$$ F_1(a,b_1,b_2,c;x,y)=\dfrac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1\frac{v^{a-1}(1-v)^{c-a-1}}{(1-xv)^{-b_1}(1-yv)^{-b_2}}~dv $$where \Re c>\Re a>0 with the change of variable$$ v=\frac{u}{1-e^{-\lambda x}}. $$0 The Gudermannian function can be defined as:$$ \text{gd}(x)= \int_0^x\frac{dt}{\cosh(t)}= 2\arctan(\tanh(\frac{1}{2}x)) $$Furthermore, notice that:$$\text{gd}'(x)=\frac{1}{\cosh(x)}$$with this in mind, the integral can be rewritten as:$$\int_{0} ^{+\infty}\frac{\text{gd}(2x)(\text{gd'}(x))^2}{x}dx$$From here on out it gets messy. Trying to compute ... 1 y=\phantom{}_2 F_1(a,b;c;z) is the regular solution of the ODE:$$z(1-z)y''+\left[c-(a+b+1)z\right]y'-ab y=0 $$hence y=\phantom{}_2 F_1(-n,n;1/2;z) is the regular solution of the ODE:$$z(1-z)y''+\left(\frac{1}{2}-z\right)y'+ n^2 y=0 $$and y=\phantom{}_2 F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2}\right) is the regular solution of the ODE: ... 2 The elements in a must be non-negative. (Take a=-1 for a start of a counterexample) If a\ge 0 then$$ a^Tb_i = \sum_{j=1}^d a_j b_{i,j} \le \sum_{j=1}^d a_j \max_i b_{i,j}= a^T (\max b_i). $$Now take the maximum on the left-hand side. 1 We have:$$\sum_{k=1}^{n-1}\frac{1}{k^2(n-k)^2}=\frac{1}{n^2}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}$$so:$$ \text{Li}_2(x)^2 = \sum_{n\geq 2}\left(\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}\right) x^n\tag{1}$$and since:$$ \int_{0}^{1}\frac{x^n \log x}{1-x^2}\,dx = -\sum_{m\geq ...

1

This is one posibility: $$f(x)=\frac{e^{ax}-1}{e^{ax}+e^{as/2}}.$$ It is clear that $0\le f(x)<1$, $f(0)=0$ and $\lim_{x\to\infty}f(x)=1$. Moreover $$f'(x)=\frac{b \bigl(e^{\frac{b s}{2}}+1\bigr) e^{b x}}{\bigl(e^{\frac{b s}{2}}+e^{b x}\bigr)^2}>0\quad\text{and}\quad f''(s/2)=0.$$ Finally $$2<b<4\quad\text{and}\quad ... 0 Firstly I just wanted to clarify what you meant by "twice absolutely continuous", my assumption is that means "twice absolutely continuously differentiable". If that is the case, then the fact that the question is posed on \Bbb R is important, of course f\in L^2(\Bbb R) is a strong assumption. Assume Hf=-f''+x^2f\in L^2(\Bbb R), this tells us there ... 0 The power series given by Robert Israel converges when the absolute value of x is less than 1/e, as n! grows about as fast as the nth power of n/e (Stirling's Formula). This does give a solution, if you replace x by x-c in the series. 1 The function you are looking for (for c=0) is \dfrac{-W(-x)}{1 + W(-x)} where W is the Lambert W function. 0 I barely understand this, but that won't stop me from giving my understanding of the problem. I'll assume that you mean for the derivative of this theoretical function at the single value c to be equal to the above. In general you want.$${{d^n f} \over {d t^n}}=n^{n-1}$$Your solution will be a function, but without knowing initial conditions, the ... 2 Assuming that we know the Taylor series of \text{Ai}^2(x) and \text{Bi}^2(x) in a neighbourhood of the origin, the problem boils down to evaluate a weigthed sum of values of the Beta function. Since the Airy functions are the fundamental solutions of:$$ y'' = x y $$if y=\text{Ai} or y=\text{Bi} we have:$$ \frac{d}{dx}y^2 = 2yy',\quad ...

1

Let us compute a simpler integral first. We replace the square root in the denominator by a first power. We have : \begin{eqnarray} I=\int\limits_0^1 \frac{1}{\Gamma(x)} dx = \lim_{n\rightarrow \infty} \int\limits_0^1 \frac{x^{(n+1)}}{n! n^x} d x = \lim_{n\rightarrow \infty} \frac{(-1)^{n+1}}{n!} \int\limits_0^1 \frac{(-x)_{(n+1)}}{ n^x} d x =\\ ...

0

With the definition of the Dirichlet's $\eta$ as in your post $\sum_{k=1}^\infty \eta(k+1)$ the whole expression is divergent because the $\eta(k+1)$ converge to $1$. But if you rewrite your formula $$\sum_{k=1}^\infty (\eta(k+1)-1)$$ this shall converge because $\eta(1+k)-1$ converges quickly to zero when $k$ increases. Finally we get for this ...

0

You should look at that source of all wisdom: Kato's Perturbation theory of linear operators, which, in the first chapter, discusses the perturbation of eigenvalues of a matrix (before plunging into infinite dimensions in subsequent sections). Apply that discussion to the companion matrix of your polynomial. Of course, for quartics, there is an explicit ...

0

Hint: use the formula: $$\cos(ix)\cos(jx) = \frac{1}{2}\left(\cos((i+j)x)+\cos((i-j)x)\right)$$ and consider that: $$\sum_{k=0}^{n}\cos(kx) = \text{Re} \sum_{k=0}^{n}\left(e^{ix}\right)^k.$$

1

I do not think that closed form solution of the positive root of equation $$f(a,x)=x^a -(1 - e^{-x})=0$$ could be obtained. It seems to me that, beside the trivial $x=0$, the existence of a positive solution requires $a >1$. In such a case $f(a,0)=0$, $f(a,1)=\frac 1e$. The derivative of the function cancels for $$x_*=(a-1) W\left(\frac 1{(a-1) ... 1 Hint: ~\displaystyle\int_0^\infty\exp\bigg(-ax-\frac bx\bigg)~\frac{dx}x~=~2K_0\Big(2\sqrt{ab}\Big),~ for positive values of a and b. See Bessel function for more information. -1 Check regularity of the singular points: 2t^3y''+(5t^2-t)y'+(t^2-t+1)y=0 y''+\left(\dfrac{5}{2t}-\dfrac{1}{2t^2}\right)y'+\left(\dfrac{1}{2t}-\dfrac{1}{2t^2}+\dfrac{1}{2t^3}\right)y=0 \therefore the position(s) of finite singular point(s) in this question is t=0 only \lim\limits_{t\to ... 0 there is the "approximate functional equation" https://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_formula which is described in the Tichmarsh and on the french wikipedia of zeta-Riemann function but not on the english one. someone should add a link to that Riemann Siegel formula to the wiki. 0 Here we provide an answer for the Tsallis' case only. Denoting the unknown integral by \tilde{\mathcal I}_q we clearly have(see the answer to the other question for justification): \tilde{\mathcal I}_q(A) = \frac{1}{\Gamma(\frac{1}{q-1}-\frac{1}{2})} \int\limits_0^\infty \frac{d s}{s} s^{\frac{1}{q-1}-\frac{1}{2}} e^{-s} \tilde{\mathcal ... 2 I'd try to use:$$\displaystyle \int\limits_{0<x+y<1} x^{u-1}y^{u-1}(1-x-y)^{v-1} \,dx\,dy= \frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}$$for reals u,v \in \mathbb{R}. Hence,$$\displaystyle \begin{align}\sum\limits_{n,k=1}^{\infty} \frac{\Gamma(k+u)^2\Gamma(n+v)}{\Gamma(2k+2u+n+v)} &= \int\limits_{0<x+y<1} ...

0

When I look at $$\lambda = -\sum_{c=1}^{n-1} \sum_{k=c}^n {k \choose c} \frac{(-1)^k}{k!} f^{k-c}U(-c,k-2c+1,-f)\phi(n,k)$$ I see that $1 \le c \le n-1$ and $c \le k \le n$. (Though I am suprised that $c$ goes up to $n-1$, not $n$.) Therefore $1 \le k \le n$ and $1 \le c \le k$ and $c \le n-1$. The sums could then be rearranged as $$\lambda = ... 1$$\sum_{c=1}^{n-1}\,\sum_{k=c}^{n}=\sum_{k=1}^{n}\,\sum_{c=1}^{\min(k,n-1)} The first sum is in a triangular grid, less the point $k=n$, $c=n$. When summing first on $k$, $k$ starts at the variable index $c$ and ends at $n$. The outer sum in this case extends over the permissible values for $c$. Now, when summing first on $c$, $c$ starts at $1$ and ...

Top 50 recent answers are included