How do you write (x-1)(x-2)(x-3)...(1) as a gamma function?
Assume x is a natural number.
My guess is that it may be (x-1).
I'm looking for an answer in the same format as my guess.
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Is $x$ a real number? – garondal Oct 03 '22 at 21:23
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3The Gamma function is a bit shifted with respect to the factorial. $\Gamma(x)=(x-1)!$ – plop Oct 03 '22 at 21:24
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Why would it be shifted? – Rashid Abramov Oct 03 '22 at 21:28
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1The problem is that your question is not well-defined. Let for example $x=5.2$. Then $(5.2-1)(5.2-2)(5.2-3)(5.2-4)(5.2-5)$ and you will not get a $1$ as last factor. – garondal Oct 03 '22 at 21:29
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How should it be defined for the last factor to be 1? – Rashid Abramov Oct 03 '22 at 21:30
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2For "How should it be defined for the last factor to be $1$?" you would need $x$ to be a natural number. – Robert Lee Oct 03 '22 at 21:32
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"Simple" is a subjective term. While you might not find this post helpful, others do. – Rashid Abramov Oct 16 '22 at 23:05
1 Answers
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Everyone is reading FrustratedSnake's question incorrectly. Since $x$ is a natural number (read the OP completely), the expression in the OP is just $\Gamma (x)=(x-1)!$.
Kamal Saleh
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(Note that the OP did not originally say $x$ is a natural number---consult the edit history.) – Jakob Streipel Oct 03 '22 at 23:10
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@prets Yes of course that's why the other answer was considering reals in general. But I saw this question later so yeah :) – Kamal Saleh Oct 03 '22 at 23:54