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I have to prove some result. And for that, I need to prove this new problem. To prove, $c_{1}\Gamma(z+b_{1})+c_{2}\Gamma(z+b_{2})+\ldots+c_{n}\Gamma(z+b_{n})=0$ has at most $(n-1)$ real positive solutions, where $c_{i}$ are arbitrary non zero real numbers and $b_{i}$ are arbitrary positive numbers, s.t. $b_{i+1}>b_{i}>0$ for all $i$ and $\Gamma(z)$ is gamma function.

Gerry Myerson
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VSP
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    But $\Gamma(x)-\Gamma(x+1000)=0$ has at least $1000$ real solutions... – Pietro Majer Jul 19 '18 at 13:21
  • I have deleted a few tags that seemed to me not to be justified by anything in the statement of the question. If OP can add to the body of the question some indication of the relevance of those tags, then OP is encouraged to edit those tags back in. – Gerry Myerson Jul 19 '18 at 23:53
  • Also, I wonder whether there is anything special about $\Gamma$ that would make the allegation true, other than its rapid growth. That is, maybe the allegation is true with $\Gamma$ replaced by any function that grows rapidly enough. – Gerry Myerson Jul 19 '18 at 23:58
  • @PietroMajer , in your equation $b_{1}=0$, i want all $b_{i} >0$ – VSP Jul 20 '18 at 02:34
  • Thanks to all for replying, @PietroMajer, I think you are right, we will have many real solutions but I am interested in positive solutions only, I have edited my question now. – VSP Jul 20 '18 at 02:44
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    I'm still not happy with a few things here. The title: why does it say "polynomial", when there is no polynomial anywhere? The complex-variables tag: if $z$, $c_i$, and $b_i$ are all real, what does complex analysis have to do with it? The linear-algebra tag: in what way is this a question about Linear Algebra? – Gerry Myerson Jul 20 '18 at 04:15
  • @Gerry Myerson, Ya thats right..thanks...my fault. Let me edit it. – VSP Jul 20 '18 at 09:49

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With the new constraints, this follows from the proof of MO question 306366: Prove that the matrix $[\Gamma(\lambda_{i}+\mu_{j})]$ is nonsingular which you posted earlier today and is already answered. If there were $n$ solutions $z_1<\cdots<z_n$ then $\det[\Gamma(z_i+b_j)]$ would vanish; but that determinant is always positive.

Noam D. Elkies
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  • Thanks for replying, u r right, but I just want an alternate proof of it. Any alternative way. – VSP Jul 20 '18 at 09:53
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    If you want something more than what's in your question, prince, you should edit your question so it says what you really want. – Gerry Myerson Jul 20 '18 at 10:52