sHello to everyone.What I am trying (and I can not find) is an expression in series the following trio with rational exponent: $$(1+x+y)^n$$ where n is a rational number, $$\mid x\mid<1,\mid y\mid<1$$
As it is known from the binomial theorem: $$(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k=_2F_1(-n,1,1,x)$$ What I'm looking for is a similar expression for the trinomial illustrated above. Remembering some notion on multinomial theorem i proceeds this: $$(1+x+y)^n=\sum_{k=0}^\infty\sum_{s=0}^k\binom{n}{k}\binom{k}{s} x^{k-s} y^{s}$$ Do you think this result is correct?If it were correct, how can I rewrite it in the form of hypergeometric function of several variables as shown before for the binomial theorem? Thanks very much for your help and for your time.
With me always check!
Of course, you could just write :)
$_{2}F_{1}\left(-n,1;1;\left(x+y\right)\right)$
or $_{1}F_{0}\left(-n;-;\left(x+y\right)\right)$
But there is a named two variable form in $x^{n}y^{s}$
To facilitate the conversion we rephrase in terms of rising Pochhammer functions; the ones used in hypergeometric function definition:
$\left(\begin{array}{c} \alpha\\ k \end{array}\right)$ $=\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{k!} $
$\left(1+z\right)^{\alpha}={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{k!}z^{k}$
$\left(1+\left(x+y\right)\right)^{\alpha}={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{k!}\left(x+y\right)^{k}$
$I=\left(1+\left(x+y\right)\right)^{\alpha}={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{k!}\left(1+\frac{y}{x}\right)^{k}\left(x\right)^{k} $
$={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{k!}\left(x\right)^{k}{\displaystyle \sum_{s=0}^{k}}\left(\begin{array}{c} k\\ s \end{array}\right)\left(\frac{y}{x}\right)^{s} $ $={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{1}\left(x\right)^{k}{\displaystyle \sum_{s=0}^{k}}\frac{1}{s!\left(k-s\right)!}\left(\frac{y}{x}\right)^{s} $
$={\displaystyle \sum_{k=0}^{\infty}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{1}{\displaystyle \sum_{s=0}^{k}}\frac{1}{s!\left(k-s\right)!}\left(x^{k-s}\right)\left(y\right)^{s} $
We reformat it to
Bateman's Special functions 5.7.1 http://apps.nrbook.com/bateman/Vol1.pdf
or
Wolfram's Horn's list http://mathworld.wolfram.com/HornFunction.html
$I={\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{s=0}^{k}}\frac{\left(-1\right)^{k}\left(-\alpha\right)_{k}}{(k-s)!s!}\left(x\right)^{k-s}\left(y\right)^{s}$
Let $n=k-s$
$={\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{s=0}^{k}}\frac{\left(-1\right)^{n+s}\left(-\alpha\right)_{\left(n+s\right)}}{(n)!s!}\left(x\right)^{n}\left(y\right)^{s}$
$={\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{s=0}^{k}}\frac{\left(-\alpha\right)_{\left(n+s\right)}}{(n)!s!}\left(-x\right)^{n}\left(-y\right)^{s}$
$I=F_{2}\left(\alpha,1,1,1,1,-x,-y\right) $