Tag Info

New answers tagged

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

I shall describe an expansion for this integral $\mathcal{I}(a,b,c)$ in powers of $c^{-2}$ . To do so I will make a few changes of parameters first. Observe that the substitution $x=bk$ yields $$ \mathcal{I}(a,b,c) =\int_0^\infty dx \,e^{-a x^2/b^2} J_0(x)\frac{x^3}{b^{4} c^{2}+x^4}.$$ Defining $t=b^2/4a$, $\epsilon=1/b^4 c^2,$ and $I(\alpha,\epsilon) = b^4 ...


0

Thanks to Semiclassical, I was able to solve the integral, and it is indeed my original guess involving Bessel functions!


2

To compute the second integral, let $r=\sqrt{\alpha^2+\beta^2}$ and observe that we may write $(\alpha,\beta)=(r \cos\phi,r\sin\phi)$ for some particular $\phi$. The sum-to-products formula then gives $$\alpha \cos\theta+\beta\sin\theta = r \cos\phi\cos\theta+r\sin \phi \sin\theta = r\cos(\theta-\phi).$$ So the integral becomes $$\int_0^{2\pi}\,d\theta ...


1

This cannot be valid for $c,\alpha>0$ and all positive $x$, because for $x\rightarrow 0$ the LHS approaches $c$ while the RHS approaches $0$ for $\nu > 0!$ You have $$\frac{(\tfrac{1}{2}x)^{\nu}}{\Gamma(\nu+1)} \le I_{\nu}(x)$$ for all $x,\nu > 0$ and asymptotic expansions for large $x:$ $$I_{\nu}(x) \sim \frac{e^x}{\sqrt{2\pi x}}\left(1 + ...


1

A related technique. Here is an approach. Let's make the change of variables $\log_2(1-x)=-u$ which results in the integral $$ I = \int_0^1 \frac{{d}x}{1-\lfloor \log_2(1-x)\rfloor} = \int_0^{\infty} \frac{2^{-u}{d}u}{1-\lfloor -u\rfloor} =\ln(2)\sum_{k=0}^{\infty}\int_0^{\infty}\frac{2^{-u}{d}u}{1-(-k-1)} $$ $$ = ...


0

It is very easy. From the recent paper by Karbach et al.: http://arxiv.org/pdf/1407.0748v1.pdf substitute Eq(37) in Eq(13), replace $-iz$ to $z$, take the limit $z \to \infty$ and you will get the answer.


1

[Note: I don't have the "privilege" of adding this as a comment...] If you're new to Neural Networks, I recommend these course notes and Andrew Ng's Machine Learning course over at Coursera (or, for that matter, the lecture notes from his class at Stanford). Also, it may be helpful if you post the article to which you are referring...


7

Start by noting that $\lfloor\log_2(1-x)\rfloor = -(n+1)$ for all $x \in (1-2^{-n},1-2^{-(n+1)})$. Therefore, $\displaystyle\int_0^1 \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor} = \sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor}$ $= \displaystyle\sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} ...


2

This is the same as the integral with $1-x$ replaced by $x$, and this answer really solves the $x$ version of the integral. In the interval $(2^{-(n+1)},2^{-n})$ this function takes the value ${1}{1+n+1} = \frac{1}{n+2}$. So this integral is: $$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \frac{1}{n+2}$$ The general sum: $$\sum_{n=0}^\infty \frac {x^{n+1}}{n+2}$$ ...


3

Let us make the change of variables $t=\sinh \frac{x}{4}$. Since $$t^2+t^4=\frac{\sinh^2\frac{x}{2}}{4}=\frac{\cosh x-1}{8}$$ and $dt=\frac14 \cosh\frac{x}{4}dx$, the initial integral can be rewritten as \begin{align} I(n)&=\frac{e^{\frac18}}{4^{n+1}}\int_0^{\infty}\cosh\frac{x}{4}\sinh^{2n}\frac{x}{2}e^{-\frac18\cosh x}dx=\\ ...


2

Not a full answer, but a partial one: Notice that the powers of two in the denominator are increasing in periodic repetitions of 3, 2, 1. That is, in this case the exponents on the 2 are 2, 5, 7, 8, 11, 13, 14, 17, 19, 20.... This is sequence A047268 on the OEIS website, defined as the sequence of numbers congruent to {1, 3, 5} mod(6). I fiddled around ...


1

I would think that the parametric differentiation technique is probably easier to carry through: $$\frac{\partial}{\partial r}K_{\nu}(r)= -(1/2)(K_{\nu-1}(r) + K_{\nu+1}(r))......(1)$$ EDIT: From the answer by O.L., we know that: $$K_{\nu+1}(r)-K_{\nu-1}(r)=\frac{2\nu}{r}K_{\nu}(r)......(2)$$ Thus we can get rid of $K_{\nu-1}(r)$ and obtain: ...


0

Hint: $-\dfrac{\partial S}{\partial t}=ax^2+bx\dfrac{\partial S}{\partial x}+c\left(\dfrac{\partial S}{\partial x}\right)^2+\dfrac{1}{x-\alpha}+\dfrac{1}{\beta-x}$ $-\dfrac{\partial^2S}{\partial x\partial t}=2ax+bx\dfrac{\partial^2S}{\partial x^2}+b\dfrac{\partial S}{\partial x}+2c\dfrac{\partial S}{\partial x}\dfrac{\partial^2S}{\partial ...


1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


0

A function not depending on $t$ is a particular case of a function depending on $t$. In other words, if you need to find a $W(t,x)$ satisfying certain conditions and you manage to find $W_1(x)$ satisfying these conditions, then everything is ok. As for showing that the zero solution is (asymptotically) stable, just adopt the same proof as for the case of ...


1

Defining $I$ as the definite integral, $$I:= \int\limits_{0}^{1}\left[\frac{\zeta{(2)}-2\log^2{2}}{1-x}-\frac{1}{x(1-x)}\left(2\operatorname{Li}_2{\left(\frac{1-\sqrt{1-x}}{2}\right)}-\log^2{\left(\frac{1+\sqrt{1-x}}{2}\right)}\right)\right]\mathrm{d}x,$$ prove: $$I=\frac52\zeta{(3)}-2\zeta{(2)}\log{2}-\frac43\log^3{2}.$$ Substituting ...


1

From Kronecker delta:Alternative notation: Often, the notation $\delta_i$ is used. $$ \delta_{k} = \begin{cases} 0, & \mbox{if } k \ne 0 \\ 1, & \mbox{if } k=0 \end{cases} $$ Set $k=i-j-1$...


0

As pointed out in comments, you misread $\tan$ as $\tan^{-1}$. The answers are: $$ \begin{array}{c} z_x = \frac{1}{\tan\frac{x}{y}}\frac{1}{y}\sec^2\frac{x}{y}=\frac{\sec\frac{x}{y}\csc\frac{x}{y}}{y} \\ z_y = -\frac{1}{\tan\frac{x}{y}}\frac{x}{y^2}\sec^2\frac{x}{y}=\frac{-x \sec\frac{x}{y}\csc\frac{x}{y}}{y^2} \end{array} $$


2

A closed form for the integral is the Incomplete Beta function : $$\int{ \frac{1}{\sqrt{x+1}}{x^n}} dx = (-1)^{n+1}B_{-x}(n+1 , 1/2)+constant$$ The inverse function needs numerical calculus to be evaluated : http://www.dtic.mil/dtic/tr/fulltext/u2/a467901.pdf


0

\begin{align} \int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy} = \left\{ \begin{array}{l} \int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a < 0,\,\,\text{setting}\,\,a = 1 - r,r > 1 \\ \frac{1}{b}\left( {1 - d} \right)^b ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 0 \\ \int_d^1 {y^{r - 1} \left( ...


2

If we ignore all "coincidences", including associativity and commutativity where it applies, we can enumerate all valid exprsssions by arranging the $n$ numbers in one of $n!$ orders insertingof parentheses among them in $C(n-1)$$=\frac{(2n-2)!}{(n-1)!n!}$ ways inserting operators among the subexpressions in $4^{n-1}$ ways That gives us a total of ...


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} ...



Top 50 recent answers are included