# Tag Info

2

$f(1+a)=a^2\sin(n\pi(1+a))=a^2\sin(n\pi+n\pi a)=a^2(\sin n\pi\cos n\pi a+\cos n\pi\sin n\pi a)$ $=a^2\cos n\pi\sin n\pi a$, but $f(1-a)=a^2\sin(n\pi(1-a))=a^2\sin(n\pi-n\pi a)=a^2(\sin n\pi\cos n\pi a-\cos n\pi\sin n\pi a)$ $=-a^2\cos n\pi\sin n\pi a$; so the function does not appear to be even about 1. For example, if $n=1$, $f(3/2)=-1/4$ while ...

0

Reading this problem again, here is a simple construction to map the monomial $x^q$ to $q^{r-1}x^{q-1}$. Consider the formal derivative operator $D(x^q) =q x^{q-1}$; then $(Dx)(x^q)=(q+1)x^q$. Consequently $$(Dx)^r D(x^q) = (Dx)^r(qx^{q-1})=(Dx)^{r-1}q^2 x^{q-1}=\cdots = q^r x^{r-1}.$$ (Equivalently we may write this as $D(xD)^{r-1}$.)

1

It seems to me that the operator you seek would be: $$f(x^p)=\frac{1}{x}\int_0^x t^p dt$$ this has the desired properties. Note that this operator is actually independant of $p$, and can be written more generally, for some $g\in\Bbb R[x]$ $$f(g)=\frac{1}{x}\int_0^x g(t)dt$$

1

EDIT: I'm quite sure that I misunderstood the OP's intention. However, the comments to this answer may still be of interest and so have made this a wiki answer. No such function exists. We want $f(x^q)=x^q/(q+1)$ for all $q$; consider in particular $q=6$. Then we require \begin{align} f(x^{6}) &=f((x^2)^3))=f((x^3)^2))\\ ...

2

(This is more a comment than answer, but I couldn't get MathJax to properly show it in comments) Here is a nice identity (equation (21) of this paper with $x=-1/7$): $$_2F_1 \left(a,a+\frac{1}{2};\frac{4a+5}{6};-\frac{1}{7}\right)=\left(\frac{7}{4}\right)^a {_2}F_1 \left(\frac{a}{3},\frac{a+1}{3};\frac{4a+5}{6};-27\right)$$ It's an example of a cubic ...

18

Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector $$\vec{y}(z)=\left(\begin{array}{c} y_1 \\ y_2 \end{array}\right)=\left(\begin{array}{c} _2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z) \end{array}\right).\tag{1}$$ This is a single-valued ...

3

The integral converges only when $a$ is an odd multiple of $\pi/2$, and does not seem to vanish even for those $a$. For large $\left|z\right|$ it is known that $$J_0(z) = \sqrt{\frac2{\pi\left|z\right|}} \bigl(\cos (\left|z\right|-\frac\pi4) + O(1/\left|z\right|) \bigr).$$ Therefore we have for large $x$ (either $x>0$ and $x<0$): $$... 16$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$After first multiplying and dividing the integrand by 2, substitute x=\frac{t}{2}: ... 5 From the Jacobi-Anger expansion:$$e^{i z\cos\theta} = J_0(z) + 2\sum_{n=1}^{+\infty}i^n J_n(z) \cos(n\theta)$$we have, by considering the imaginary part:$$\sin(z\cos\theta) = 2\sum_{m=0}^{+\infty}(-1)^m J_{2m+1}(z)\cos((2m+1)\theta)\tag{1}$$and we can remove the cosine-dependent term by exploiting the identities: ... 1 There is probably a typo in : \sin(1)=2\sum_{k=1}^\infty J_{2k+1}(1) because \sin(1)=0.841471... and 2\sum_{k=1}^\infty J_{2k+1}(1)=0.0396292... An exact similar relationship is :$$\sin(1)=2\sum_{k=\infty0}^\infty (-1)^kJ_{2k+1}(1)$$More general results can be found in : http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/23/01/ From ... 1 Ah, I guess the problem is that f blows up on the imaginary axis, so z^{v-1/2}f(z) does not go to 0 uniformly as z goes to infinity. 2 Probably not what you want, but consider this: Let h : \mathbb R[X] \to \mathbb R[X] be defined by$$h(P)= X P'(X) \,.$$Define now h_1=h and recursively$$h_r=h_{r-1} \circ h \,.$$Then, h, h_r are independent of q and satisfy$$h_r(X^q)=q^rX^q$$Take f_r(P(X)) = \frac{h_r(P(X))}{X}. Note that h is a linear function and h(\mathbb P_n) ... 1 If your question is on the generalisation of the following result by Polya (J fur die reine angwt math, 1921): " Let F\in \mathbb{Z}[[x]] with non zero radius of convergence and suppose that xF^{\prime}(x) is a rational function, then F is a rational function" obtained by replacing the word "rational" by "algebraic", then I think that this is proved ... 4 I would not call this "part of mathematics" beautiful nor aesthetic, although sometimes it can be a pleasant waste of time. Most of the computable integrals, sums and products can be found in the books like Gradshteyn-Ryzhik or Prudnikov-Brychkov-Marychev. Programs like Mathematica or Maple, as well as theoretical physicists, solve this kind of problems ... 4 Continuing from O.L.'s answer, the following is an evaluation of$$\int_{0}^{\infty} \frac{\sin 2x}{x} \text{Ci}(x) \ dx .$$First notice that by making the substitution  \displaystyle u = \frac{t}{x},$$ \text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos t}{t} \ dt = - \int_{1}^{\infty} \frac{\cos xu}{u} \ du.$$Therefore,$$ \int_{0}^{\infty} \frac{\sin ...

3

I bet that the answers to the first and third question (about convergence and zeroes) are affirmative, since the function $$f(z) = \left(1-\frac{x^2}{4}+\frac{x^4}{64}\right)\mathbb{1}_{[0,1]}(x)+\sqrt{\frac{2}{\pi x}}\cos(x-\pi/4)\mathbb{1}_{[1,+\infty)}(x),$$ by following Abramowitz and Stegun, is a very good approximation for $J_0(x)$, but I do not think ...

1

I was in the end able to derive the correct expression as... $$\frac{(1/n)_k}{(1+1/n)_k}=\frac{\frac1n(\frac1n+1)(\frac1n+2)\cdots(\frac1n+k-1)}{(1+\frac1n)(1+\frac1n+1)(1+\frac1n+2)\cdots(\frac1n+1+k-2)(\frac1n+1+k-1)} =\frac{1}{nk+1}$$

2

Deriving from integral approach is easier: \begin{align} \int(1+x^n)^{-\frac{1}{m}}~dx &=\int_0^x(1+t^n)^{-\frac{1}{m}}~dt+C\\ &=\int_0^{x^n}(1+t)^{-\frac{1}{m}}~d(t^\frac{1}{n})+C\\ &=\dfrac{1}{n}\int_0^{x^n}t^{\frac{1}{n}-1}(1+t)^{-\frac{1}{m}}~dt+C\\ &=\dfrac{1}{n}\int_0^1(x^nt)^{\frac{1}{n}-1}(1+x^nt)^{-\frac{1}{m}}~d(x^nt)+C\\ ...

1

Okay, I think I've got it. The solution to your integral is likely $$f(z)=\sqrt{\pi}e^{z^2}(\operatorname{sgn}(\operatorname{Re}(z)) - \operatorname{erf}(z) ).$$ To see this, let first $\operatorname{Im}(z)=:v> 0$ and denote $z=u+iv$. First, note that $$\int_{-\infty}^\infty \frac{e^{-t^2}dt}{t+z}= e^{-z^2} \int_{-\infty+iv}^{\infty+iv} ... 2 For the integral$$I:=\int_0^z {\frac {\exp {(w\cdot \frac{a^2-1}{2a^2})}}{\sqrt {(z-w) w}} dw} $$by completing the square in the denumenator and simplify we get$$\sqrt{\frac{1}{4} z-\frac{1}{4} z+zw-w^2}=\frac{1}{2} z\sqrt {1-(1-\frac{2w}{z})^2}$$Now using the substitution$$\sin\theta = 1-\frac {2w}{z} $$Therefore  I  reduces to$$ I=\int_{\frac ...

1

Some notes to consider: Let $\beta \rightarrow ia$ to obtain the form \begin{align} F(a, \gamma) = \int y \cos(a y^{2}) \ J_{0}(\gamma y^{2}) \ dy \end{align} or, more generally, \begin{align}\tag{1} F(a, \gamma) = \int y e^{ia y^{2}} \ J_{0}(\gamma y^{2}) \ dy. \end{align} Let $t = y^{2}$ for which (1) becomes \begin{align}\tag{2} F(a, \gamma) = ...

2

This identity follows from the following distribution relation on the Wikipedia page: $$\displaystyle\sum_{m=0}^{n-1}\zeta\left(z,a+\frac{m}{n}\right)=n^z\zeta\left(z,na\right).\tag{1}$$ It suffices to set therein $a=\frac{q}{2}$, $n=2$ and differentiate once with respect to $z$.

0

I'll take the easy way out and point you to the book where a solution to your problem is given along with its proof: Khalil, Nonlinear Systems. The short answer to your question is yes. Your Lyapunov function does not have to have an explicit $t$ dependence. The general result goes as follows. If you can find a function $V(t,x)$ that is lower and upper ...

1

I am guessing that you would like to know $\lim_{s \to 0+} f(s) = \lim_{s \to 0+} \int_0^\infty a(t) e^{-st}dt$. Let $(s_n)$ be any sequence of positive numbers converging to $0$ and set $f_n(t) = a(t)e^{-s_nt} \to a(t)$ as $n \to \infty$. Moreover, $|f_n(t)| \leq |a(t)|$ so if $a(t)$ is Lebesgue integrable, then the limit equals $\int_0^\infty a(t) dt$ ...

4


1

We have $$\tag+\frac{1-e^X}X = K \iff 1-KX = e^X \iff (1-KX)\cdot e^{-X} = 1$$ We want to have something of the form $Ye^Y$ to apply $W$, hence we write $$e^{-X} = e^{-\frac{KX}K} = \frac{e^{\frac 1K - \frac{KX}K}}{e^{1/K}} = \frac{e^{\frac{1-KX}K}}{e^{1/K}}$$ So, we divide $(+)$ by $K$ to get $$\frac{1-KX}{K} \cdot e^{\frac{1-KX}K} = \frac{e^{1/K}}K ... 1 The solution given by user1337 in the comments is correct. To derive it rewrite your equation$$ \frac{1-e^{X}}{X} = K$$to$$ \frac{1-KX}{K}e^{\frac{1-XK}{K}} = \frac{e^{\frac{1}{K}}}{K}$$This is now on the form W(Z) e^{W(Z)} = Z with W(Z) = \frac{1-XK}{K} and Z = \frac{e^{\frac{1}{K}}}{K}. 2 A reference: Chaudhry, M.A. et al. Asymptotics and closed form of a generalized incomplete gamma function. 12 Generalized Laguerre polynomials have the generating function$$\sum_{n=0}^{\infty}\xi^n L_n^{(t)}(x)=(1-\xi)^{-t-1}e^{-\frac{\xi x}{1-\xi}}. Multiplying this identity by $\xi^{t-1}$ and integrating w.r.t. $\xi$ from $0$ to $1$, one finds \begin{align}\sum_{n=0}^{\infty}\frac{L_n^{(t)}(x)}{t+n}&=\int_0^1\left(\frac{\xi}{1-\xi}\right)^{t-1} ...

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