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## New answers tagged hypergeometric-function

1

It's best to reduce the number of parameters (also I changed the notation $\delta v \to \delta$ to avoid confusion). Substitute: $$u= z t$$ $$I(z ; v, \delta) = z^2 \int^1_0 \text{d}t \, \sqrt{1 - t^2} \exp\left[-\frac{(v/z-t)^2}{2 \,\delta^2/z^2}\right]$$ Denoting: $$\alpha = v/z, \qquad \beta=\delta/z$$ Now we only need to find: $$J(\alpha, \beta)=\... 3 Using the information in OEIS sequence A008309 I found that$$ {}_2F_1(1-n, 1-x; 2; 2) = \sum_{k=1}^n\sum_{j=k}^n x^{k-1}2^{j-1}{n-1 \choose j-1}S_1(j,k)/j! $$where S_1(.,.) is the Stirling number of the first kind. If you use the definition of the Hypergeometric function then you get$$ {}_2F_1(1-n, 1-x; 2; 2) = \sum_{k=0}^n \frac{2^k(1-n)_k (1-x)_k}{...

7

Before attacking the integral, I mention something about cubic theta function. The whole solution heavily exploits tools from modular forms. The "footnote" contains more information. The three cubic theta functions are defined by \begin{aligned} a(q) &= \sum_{m,n} q^{m^2+mn+n^2}\\ b(q) &= \sum_{m,n} \zeta_3^{m-n} q^{m^2+mn+n^2}\\ c(q) &= \sum_{... 5 Not an answer, but an extended comment for now. This hypergeometric function is a special case, and some complicated quadratic and cubic transformations apply to it. See this like for reference: https://dlmf.nist.gov/15.8. The formulas 15.8.25 and 15.8.26 both apply here. However, the most interesting one is so called Ramanujan’s Cubic Transformation (15.... 0 Note that(3ax^6+6x^5-9ax^4-4x^3+9ax^2+6x-3a)/x^3=3a(x-1/x)^3+6(x-1/x)^2+8,$$hence solvable. 2 First, multiply by q to simplify the denominator:$$qS_q(x)=\sum_{k=0}^\infty\frac{x^{2k+1}}{\Gamma\left[\frac{2k+q+1}q\right]}$$If all you want is hypergeometric functions, then just look at the ratio between terms. Let k=qm+n and a_{m+1}/a_m:$$\frac{x^{2qm+2n+2q+1}}{\Gamma\left[\frac{2qm+2n+3q+1}q\right]}\div\frac{\Gamma\left[\frac{2qm+2n+q+1}q\...

0

Unfortunately, I don't understand the sense to make a sum of series out of one series. At least these sub-series should have something particular, not only another structure. Perhaps it's clearer to build a closed form using the incomplete gamma function. $x!:=\Gamma(1+x) ~ ; ~~z>0$ $\displaystyle E_z(x):=\sum\limits_{k=0}^\infty\frac{x^k}{(kz)!}\hspace{... 2 This is not an answer, but some information in attempt to clear the air. Using the first definition for the limit function, we have: $$f(z)=(1-z)^2F_2(\alpha;1,1;\alpha,\alpha;z,z)$$ Using the integral definition for this Appell function we obtain: $$f(z)=(1-z)^2(\alpha-1)^2 \int_0^1 \int_0^1 \frac{(1-u)^{\alpha-2}(1-v)^{\alpha-2} du dv}{(1-z (u+v))^\... 0 You are right, the series is a little messed up, but there's no reason to try guessing the proper form. We can always use the general way to get the hypergeometric form of a series. Consider:$$f(z)=z \Gamma(3)\left[\frac{1}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^2}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{... 2 The last formula is valid as the limit for$m$approaching an integral value. For your values of the parameters, the function becomes a finite sum: $$U(n, n + m + 1, z) = \frac {m! \hspace {1px} z^{-n}} {(n - 1)!} \sum_{k = 0}^m \frac {(n + k - 1)!} {(m - k)!} \frac {z^{-k}} {k!}.$$ There are some contrived ways to write a power function as a G-function; ... 1 The integrand is$\Gamma^2(-\alpha - s) g(s)$, where$g$is regular at$-\alpha \not \in \mathbb Z\$. Then -\operatorname*{Res}_{s = -\alpha} \Gamma^2(-\alpha - s) g(s) = -\operatorname*{Res}_{s = -\alpha} \frac {\pi^2 g(s)} {\Gamma^2(1 + \alpha + s)} \, \csc^2 \pi (\alpha + s) = \\ -\frac d {ds} \frac {g(s)} {\Gamma^2(1 + \alpha + s)} \bigg\rvert_{s = -\... 1 I have some idea that might help you. In the book by Abramowitz and Stengun, there is a definition of the Hypergeometric function in terms of Gamma functions (15.1.1): \begin{align*} _2F_1(a,b;c;z)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}\sum_{n=0}^\infty\frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)}\frac{z^n}{n!}. \end{align*} For the Gamma function, we have the ... 2 This answer is meant to connect the ones given by @Harry Peter and @Start wearing purple, clarifying a few questions emerged in the comments. The integral of interest can be evaluated in the way pointed out by @Harry Peter, without forgetting to set some conditions on the parameters. First of all, for \left|z\right|<1 we can use the power series ... 0 "Are there more infinite product identities like this one for hypergeometric functions?" Well there is the obvious\frac{\sin x}{x}=\,_0F_1\left(;\tfrac32;-\tfrac{x^2}{4}\right)=\prod_{n\ge1}\left(1-\frac{x^2}{\pi^2n^2}\right)$$and the corresponding product for the cosine. 0 Too long for comments. Considering$$f=\, _2F_1\left(\frac{1}{2}-\frac{x}{2},1-\frac{x}{2};2;1\right) \qquad \text{and} \qquad g=\frac{2^{x+1}\, \Gamma \left(x+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (x+2)}$$I was naively thinking that the series expansions of their logarithms could be of interest. The problem is that, for \log(f), I have only be able ... 1 To recap my findings from @martycohen's guidance, I got to this result for the inverse Laplace transform I need:$$ \mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!}\frac{\gamma(k+1,at)}{\Gamma(k+1)}.$$The book "An Introduction to the Classical Functions of Mathematical Physics" by Temme (1996) ... 4 Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].  I'll blindly try to reverse the order of summation and see what happens. \begin{array}\\ S(u, v) &=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\sum_{k=l}^\infty \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\... 2 tl;dr: Wolfram is giving the general solution that works even when n isn't an integer. Long version: Let's look at the definition of _2F_1:$$ _2F_1(a,b;c;x) = \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c+k)}\frac{x^k}{\Gamma(k+1)} Plugging in a = 1, b = n+1, c = n+2 gives \begin{align} -\frac{x^{n+1}}{n+... 0 I remember observing something like this myself. I believe the key will be to expand out your fraction in a similar way as this: \frac{1}{x-1} = -\frac{1}{1-x} = -\left(1+x+x^2 \cdots\right)$$As then the deriviative will be an infinite of fractions which correspond to the geometric series. For the case above we have:$$ \left(\frac{1}{1-x}\right)' = 1+...

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