Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only not deleted user 66131

Hypergeometric functions are the analytic functions defined by Taylor expansions of the shape $\sum_{n \geq 0} a_n x^n$, where $a_{n+1}/a_n$ is a rational function of $n$. This general family of functions encompasses many classical functions. The hypergeometric functions play an important role in many parts of mathematics.

4 votes

how to reduce the integral into hypergeometric function?

Letting $\beta=\frac{2\Pi}{1+\Pi^2}$, write the integrand as $$\sqrt{1+\Pi^2}\left(\sqrt{1+\beta \sqrt{1-y^{2}}}-\sqrt{1-\beta \sqrt{1-y^{2}}}\right).\tag1$$ Use the binomial expansion to express (1), …
T. Amdeberhan's user avatar
1 vote

Product of two hypergeometric functions

If you allow some relation between $m$ and $n$, then it is possible. Let $m+n=2\alpha+2$. Then \begin{align} &_2 F_1(\alpha, \alpha+1 ; m ; z)\, _2 F_1(\alpha, \alpha+1 ; n ; z) \\ =&\,\,_4F_3(\alpha …
T. Amdeberhan's user avatar
5 votes
Accepted

A contiguous ${}_3F_2(1)$ hypergeometric identity?

First, we convert the desired identity into equivalent forms: \begin{align} \sum_{i=0}^{\lfloor n/3\rfloor}(-1)^i \left\{ \prod_{j=1}^i{{n-3j\choose3}\over{n\choose3}-{n-3j\choose3}} -\frac{n}3\, \pro …
T. Amdeberhan's user avatar
9 votes
Accepted

Closed expression for hypergeometric sum

You may argue as GH from MO from your other post. the coefficient of $y^m$ in $(1-xy)^{-s}$ equals $\binom{s+m-1}{s-1}x^m$; the coefficient of $y^{\ell-m}$ in $(1-y)^{\ell-s-1}$ equals $\binom{s-m}{ …
T. Amdeberhan's user avatar
4 votes

speeding up Gosper and WZ algorithms

The group at RISC has been working aggressively towards developing improved algorithms for many computer algebra problems, including the Zeilberger's. So, it is a good place to ask. Meantime, I just …
T. Amdeberhan's user avatar