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Daniele Tampieri
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Given a function $F(z)$, the function $F(z-m)$ is called the Taylor shift of $F(z)$. If
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).

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).

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).

edited the title
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chee
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Is Can the Taylor shift of a hypergeometric function alsoalways be expressed as a simplenontrivial transformation of another hypergeometric function?

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 aanother hypergeometric function. One trivial answer is 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).

Is the Taylor shift of a hypergeometric function also a simple transformation of another hypergeometric function?

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 a hypergeometric function. One trivial answer is 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 $F(z)={}_0F_1(;b;z)$ 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).

Can the Taylor shift of a hypergeometric function always be expressed as a nontrivial transformation of another hypergeometric function?

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).

Rephrase to make the question non-trivial.
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chee
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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 a hypergeometric function. One trivial answer is 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 thatthe non-trivial transformation, $$ F(z-m) = (1+m)^{-a} F\big(\frac{z}{1+m}\big) $$ Taylor

But does $F(z)={}_0F_1(;b;z)$ 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. Not sure how Thanks to getthe editor who fixed my latex to be processed(removing indents).

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 a hypergeometric function.

For example, $F(z)={}_1F_0(a;;z)$, I can show that $$ F(z-m) = (1+m)^{-a} F\big(\frac{z}{1+m}\big) $$ Taylor shift is very important for efficient algorithms. Also, any pointers or references would be appreciated. This is my first question in mathoverflow. Not sure how to get latex to be processed.

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 a hypergeometric function. One trivial answer is 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 $F(z)={}_0F_1(;b;z)$ 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).

edited body
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Robert Israel
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deleted 11 characters in body
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don't know how to get mathJax to work.
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chee
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chee
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