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I have to derive Kummer’s identity $$e^x\operatorname{M}\left(c-a,c;-x\right)=\operatorname{M}\left(a,c;x\right)$$ from Pfaff’s transformation $$\operatorname{F}\left(a,b,c;x\right)=\left(1-x\right)^{-b}\operatorname{F}\left(c-a,b,c;\frac{x}{x-1}\right)$$ Attempt: Since, Kummer's function $\operatorname{M}\left(a,c;x\right)$ is the special case of hypergeometric function $\operatorname{F}\left(a,b,c;x\right)$

So, Pfaff’s transformation will be $$\operatorname{M}\left(a,c;x\right)=\left(1-x\right)\operatorname{M}\left(c-a,c;\frac{x}{x-1}\right)$$ $$=\left(1-x\right)^{-1}\sum_{n=0}^{\infty}{\frac{(c-a)_n}{(c)_nn!}\left(\frac{x}{x-1}\right)^n}\\ =\sum_{n=0}^{\infty}{\frac{\left(1-x\right)^{-1-n}(c-a)_n}{(c)_nn!}\left(-x\right)^n}$$ how does $e^x$ comes in picture

Hisoka
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    The problem is that the Kummer function is not a special case of the hypergeometric function but rather $$ M(a,b,z) = \lim _{b \to + \infty } F(a,b;c;z/b). $$ Can you see it now? – Gary May 28 '21 at 18:01
  • @Gary its written in my text book – Hisoka May 28 '21 at 18:09
  • What is the statement in you book exactly? What relation does it give between the two functions? – Gary May 28 '21 at 18:13
  • @Gary $\operatorname{M}(a,c;x)={}_1\operatorname{F}_1(a,c;x)$, c is non negative – Hisoka May 28 '21 at 18:15
  • That is correct, but ${}_1 F_1$ is not the hypergeometric function you think it is. The Gauss hypergeometric function is ${}_2 F_1$, which is totally different. In general, we have the generalised hypergeometric function ${}_p F_q$. See https://dlmf.nist.gov/16.2 – Gary May 28 '21 at 18:17
  • i tried what u suggested i'm stuck at limit evaluation – Hisoka May 28 '21 at 18:20
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    Use the hypergeometric series + Tannery's theorem. Also $(1-x/b)^{-b} \to e^x$ as $b\to +\infty$. – Gary May 28 '21 at 18:21
  • $(1-x/b)^{-n}$ is left – Hisoka May 28 '21 at 18:31

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We have \begin{align*} M(a,c;x) & = \mathop {\lim }\limits_{b \to + \infty } F(a,b;c;x/b) \\ & = \mathop {\lim }\limits_{b \to + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} F\left( {c - a,b;c;\frac{{x/b}}{{x/b - 1}}} \right) \\ & = \mathop {\lim }\limits_{b \to + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} \sum\limits_{n = 0}^\infty {\frac{{(c - a)_n }}{{(c)_n n!}}\frac{{(b)_n }}{{b^n }}\frac{1}{{(1 - x/b)^n }}\left( { - x} \right)^n } \\ & = \mathop {\lim }\limits_{b \to + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} \sum\limits_{n = 0}^\infty {\frac{{(c - a)_n }}{{(c)_n n!}}\mathop {\lim }\limits_{b \to + \infty } \frac{{(b)_n }}{{b^n }}\mathop {\lim }\limits_{b \to + \infty } \frac{1}{{(1 - x/b)^n }}\left( { - x} \right)^n } \\ & = e^x \sum\limits_{n = 0}^\infty {\frac{{(c - a)_n }}{{(c)_n n!}} \cdot 1 \cdot 1 \cdot \left( { - x} \right)^n } \\ & = e^x M(c - a,c; - x). \end{align*} The change in the order of the summation and the limit operation can be justified by Tannery's theorem.

Gary
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