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Given a function $F(z)$, the function $F(z-m)$ is called the Taylor shift of $F(z)$.
If $F(z)$ is a (general) hypergeometric function, is $F(z-m)$ some simple transformation of a hypergeometric function?
Typical transforms of $F(z)$ are $A(z) F(B(z))$ where $A(z), B(z)$ are polynomials.

More precisely, suppose $F(z)={}_pF_q((a);(b);z)$ where $(a),(b)$ are the parameters, is $F(z-m)$ is some simple transformation of another hypergeometric function. But the Taylor shift is already in this form: choose $A(z)=1, B(z)=x-m$. Call this the trivial transformation. My question is whether non-trivial transformations always exist.

For example, $F(z)={}_1F_0(a;;z)$, I can show the non-trivial transformation, $$ F(z-m) = (1+m)^{-a} F\big(\frac{z}{1+m}\big) $$

But does the Taylor shift of $F(z)={}_0F_1(;b;z)$ have a nontrivial transformation?

Taylor shift is very important for efficient algorithms. Also, any pointers or references would be appreciated. This is my first question in mathoverflow. Thanks to the editor who fixed my latex (removing indents).

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  • $\begingroup$ You get the LaTeX to process by not indenting. $\endgroup$ Jan 22, 2023 at 15:12
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    $\begingroup$ $F(z-m)$ is evidently equal to the "simple transform" of $F$, namely $A(z)F(B(z))$, where $A=1$ and $B(z)=z-m$ are polynomials. So what are you asking about? $\endgroup$ Jan 22, 2023 at 17:06
  • $\begingroup$ Perhaps the question can be phrased as “is there always a non trivial transformation” equivalent to the Taylor shift $\endgroup$ Jan 22, 2023 at 17:59
  • $\begingroup$ Thanks for the latex Hint about not indenting. (I was copying and pasting my question). $\endgroup$
    – chee
    Jan 23, 2023 at 1:12
  • $\begingroup$ Alexandre, Thanks for pointing out that original question was self-answering in a trivial way. It is not clear what I wanted in the final form. $\endgroup$
    – chee
    Jan 23, 2023 at 1:18

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