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I need to simplify the expression written below to get the $ x $ term in its simplest form: $$ \ E=\left(\sum_{k=1}^b \sum_{n=0}^\infty Z_n(a,k) x^\frac{n+k}{2} \right)^t ,\ $$ where $$ \ Z_n(a,k)=\frac{\Gamma[a-k]}{{(1+k-a)_n}n!} \left(\frac{ab}{gb+\omega}\right)^{k+n} . \ $$ $ (.)_n $ denotes the Pochhammer symbol, and $ a,g,\omega $ are positive constants. $ b $ and $ t $ are natural numbers. The infinite summation (power series) can be solved by using the relations given in this Link. However, a finite summation ( from $ k=1 $ to $ b $ ) is involved in the above expression. I thought of using multinomial theorem, but I got confused.

I will greatly appreciate any hint or help. Thanks.

Abdulgani
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1 Answers1

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The expression $ E $ can be rewritten as: $$ E=\left(\sum_{n=0}^\infty Z_n(a,1)x^\frac{n+1}{2}+\dots+\sum_{n=0}^\infty Z_n(a,b)x^\frac{n+b}{2} \right)^t \tag{1} $$ Now, using Multinomial theorem, $ (1) $ can be simplified as: $$ E= \sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \left(\sum_{n=0}^\infty Z_n(a,1)x^\frac{n+1}{2} \right)^{r_1} \dots \left( \sum_{n=0}^\infty Z_n(a,b)x^\frac{n+b}{2} \right)^{r_b} \tag{2} $$

From $(0.314)$ in Gradshteyn and Ryzhik book, $ (2)$ can be written as: $$ E=\sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \left(\sum_{n=0}^\infty Z_n^\left({r_1}\right)(a,1)x^\frac{n+r_1}{2} \right) \dots \left( \sum_{n=0}^\infty Z_n^\left({r_b}\right)(a,b)x^\frac{n+br_b}{2} \right) \tag{3}$$

where $z^\left(r\right) $ denotes $z$ is convolved $ (r-1) $ times with itself. Finally, $ (3) $ is simplifed with the help of $ (0.316) $ in Gradshteyn and Rydzhik book as:

$$ E=\sum_{r_1+ r_2+\dots+r_b =t} \dbinom{t}{r_1, r_2 ,\dots , r_b} \sum_{n=0}^\infty C_n \left(Z_n^\left({r_1}\right)(a,1), \dots , Z_n^\left({r_b}\right)(a,b), \right)x^\frac{n+r_1+\dots+{br_b}}{2} \tag{4} $$

$ C_n(Z_1,Z_2) $ indicates the convolution of $Z_1$ and $Z_2$, and can be calculated as explained in this Link.

Abdulgani
  • 53