Equality of a function and Taylor Series

Question

Does the following function have a Taylor series of the form given below:

$$\frac{1}{(1+(\eta z)^n)^p} = \sum_{i=0}^{\infty}\left(\frac{(-1)^i\Gamma(p+i)\eta^{in}}{\Gamma(p)\Gamma(i+1)}\right)z^{in}?$$

If yes, are there any further simplification(s) or an easier and proper form?

CODE:

clc; clear all; close all; format long

n=4.2; p=0.75; eta= 10.5; z=0.05; numTaylorTerms=10;

Approx=0;

for i=0:1:numTaylorTerms

 Approx=Approx+((-1)^(i)*gamma(p+i)/(gamma(p)*gamma(i+1)))*(eta*z)^(i*n);

end

Approx

Exact=1/(1+(eta*z)^n)^p

Answer 1

Just go with Newtons binomial series,

$$(1+x^n)^{-p}=\sum_{i=0}^\infty \binom{-p}{i} x^{in}=\sum_{i=0}^\infty\binom{p+i-1}{p-1}(-1)^ix^{in}$$

it converges to the value on the left side for $|x|<1$.

---

For the case $|x|>1$ use the same series in $x^{-1}$, i.e., after the transformation

$$(1+x^n)^{-p}=x^{-np}(1+x^{-n})^{-p}=x^{-np}\sum_{i=0}^\infty \binom{-p}{i} x^{-in}$$

---

One can use the binomial identities (see also the other answer)

$$\tbinom{-p}{i}=\frac{-p(-p-1)...(-p-i+1)}{i!}=(-1)^i\tbinom{p+i-1}{i}=(-1)^i\tbinom{p+i-1}{p-1}$$

to get binomials of positive numbers.

Answer 2

It looks too complicated and unfamiliar but it is exactly the opposite: It is simply the binomial expnasion for negative exponent: $$ (1+x)^{-m}=\sum_{i=0}^\infty \binom{-m}{i}x^{i}, $$ which is valid for $|x|<1$.

In particular, $$ \binom{-m}{i}=\frac{(-m)(-m-1)\cdots(-m-i+1)}{i!}=(-1)^i\frac{m(m+1)\cdots(m+i-1)}{i!} =(-1)^i\frac{\Gamma(m+i)}{\Gamma(i+1)\Gamma(m)}, $$ and hence $$ (1+x)^{-m}=\sum_{i=0}^\infty (-1)^i\frac{\Gamma(m+i)}{\Gamma(i+1)\Gamma(m)}x^{i}, $$ Set $m=p$ and $x=(\eta z)^n$ and obtain: $$ \frac{1}{\big(1+(\eta z)^n\big)^p}=\sum_{i=0}^\infty (-1)^i\frac{\Gamma(p+i)}{\Gamma(i+1)\Gamma(p)}(\eta z)^{ni}. $$