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I want to show $$ (z \frac{d}{dz} + a_j){}_p F_q(a_1,...,a_j,...a_p;b_1,...,b_q;z) = a_j {}_p F_q(a_1,...,a_j+1,...a_p;b_1,...,b_q;z), $$ as is found here.

My problem is, that I'm stuck at this step: $$(z \frac{d}{dz} + a_j){}_p F_q(a_1,...,a_p;b_1,...,b_q;z) =\\ z \frac{a_1...a_p}{b_1...b_q}F_q(a_1+1,...,a_p+1;b_1+1,...,b_q+1;z) + a_j{}_p F_q(a_1,...,a_p;b_1,...,b_q;z)$$

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For simplicity we will show this for $a_1$ and denote $\mathbf a=\{a_2,\dots,a_p\}$, $(\mathbf a)_k=\prod_{l=2}^p(a_l)_k$, $\mathbf b=\{a_1,\dots,a_q\}$, and $(\mathbf b)_k=\prod_{l=1}^q(b_l)_k$. Then, $$ \begin{align} (z \tfrac{d}{dz} + a_1){_pF_q}(a_1,\mathbf a;\mathbf b;z) &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(z \tfrac{d}{dz}z^k + a_1z^k)\\ &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(zkz^{k-1}+ a_1z^k)\\ &=\sum_{k=0}^\infty\frac{(a_1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}(k+ a_1)z^k. \end{align} $$ But $$ (k+ a_1)(a_1)_k =a_1\frac{(a_1+k)\Gamma(a_1+k)}{a_1\Gamma(a_1)} =a_1\frac{\Gamma(a_1+1+k)}{\Gamma(a_1+1)} =a_1(a_1+1)_k, $$ and so $$ (z \tfrac{d}{dz} + a_1){_pF_q}(a_1,\mathbf a;\mathbf b;z) =a_1\sum_{k=0}^\infty\frac{(a_1+1)_k(\mathbf a)_k}{(\mathbf b)_k\,k!}z^k=a_1{_pF_q}(a_1+1,\mathbf a;\mathbf b;z), $$ which is the desired result.

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  • $\begingroup$ thank you so much! It really helped me. $\endgroup$
    – user12345
    Jun 16 at 19:02

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