2
$\begingroup$

Do any of you know a simple form for the confluent hypergeometric function with positive integers that involve simple functions?

What I actually need to compute is $_1F_1(n,n + m,z)$.

I know for $m=0$, $_1F_1(n,n,z)=e^z$ and for $m=1$, the hypergeometric function is related to the incomplete gamma.

Are there some results for $m\ge 2$?

Many thanks in advance for my first question.

P

$\endgroup$
1
$\begingroup$

many thanks for help. Having a look at some other properties Gradshteyn and Ryzhik, see 3.383. Case $1^{11}$, pag 347, I think I found the result I was looking for. Here it is:

$_1F_1(n,n+m,z)=\frac{z^{1-n-m}}{B(m,n)}\int_0^zt^{n-1}(z-t)^{m-1}e^tdt$.

Developing the binomial the integral becomes:

$\int_0^zt^{n-1}(z-t)^{m-1}e^tdt=\sum_{\ell=0}^{m-1}z^\ell(-1)^{m-1-\ell}\beta(m-1,\ell)\int_0^zt^{n+m-\ell-2}e^tdt$

where $\beta(m-1,\ell)$ denotes the binomial coefficient.

The last integral can be solved by parts and is known. One can also write it in terms of incomplete gamma $\gamma(a,x)$ and finally would get a finite sum:

$_1F_1(n,n+m,z)=\frac{(-1)^nz^{1-n-m}}{B(m,n)}\sum_{\ell=0}^nz^{\ell}\beta(m-1,\ell)\gamma(n+m-\ell-1, -z)$.

Please note that $\gamma(n+m-\ell-1, -z)$ is real because $n+m-\ell-1$ is an integer.

I checked numerically and it works. Just a remark, I tried using the results in Wikipedia https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions but I do not get the correct answer. Something seems wrong in the results that are mentioned.

Thanks once more. P

$\endgroup$
0
$\begingroup$

For $m = 2$, it involves both $e^z$ and the incomplete gamma function. Let,

$$_1F_1(n,n+2,z) =\,_1F_1(v-1,v+1,z)$$

for symmetry. Then,

$$_1F_1(v-1,v+1,z) = v\,e^z - n(-v)^{-v}(v + z - 1)\big(\Gamma(v) - \Gamma(v, -z)\big)$$

For $m>2$, you can find more identities using WA.

$\endgroup$
0
$\begingroup$

As an example of using the contiguous relations from http://dlmf.nist.gov/13.3#i

We convert:

$_{1}F_{1}(n,n,z)=e^{z}$

$_{1}F_{1}(n,n+1,z)=n\cdot z^{-n}\cdot\gamma(n,-z)$

To

$_{1}F_{1}(3,2,z)$

--

We can take http://dlmf.nist.gov/13.3#E2

$_{1}F_{1}(3,2,z)=\frac{-b\cdot\left(1-b-z\right)\cdot_{1}F_{1}\left(3,3,z\right)-z\cdot\left(b-a\right)\cdot_{1}F_{1}(3,4,z)}{b\cdot\left(b-1\right)}$

$=\frac{-3\cdot\left(-2-z\right)\cdot_{1}F_{1}\left(3,3,z\right)-z\cdot0\cdot_{1}F_{1}(3,4,z)}{2\cdot3}$

$_{1}F_{1}(3,2,z)=\frac{\left(2+z\right)}{2}e^{z}$

Sort of a chess game with an infinite board and new rules. We can always start from $_{1}F_{1}(n,n,z)=e^{z}$ and $_{1}F_{1}(n,n+1,z)=n\cdot z^{-n}\cdot\gamma(n,-z)$ Lets consider: n=1,m=4 Then using 13.3.2

$2\cdot _{1}F_{1}(1,1,z)+2\cdot (-1-z)\cdot _{1}F_{1}(1,2,z)=-z\cdot (1)\cdot _{1}F_{1}(1,3,z)$

$6\cdot _{1}F_{1}(1,2,z)+3\cdot (-2-z)\cdot _{1}F_{1}(1,3,z)=-z\cdot (2)\cdot_{1}F_{1}(1,4,z)$

Notice that in the straight forward application we have singularities that occur when b<2 or =a . These can result in the calculation being 0 times the answer we want. The cases b<=0 is easy to toss out since it's not allowed in the series expression. b=1 has to be handled by something like 13.3.4. b=a is also of this nature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.