Can Induction Prove Gamma Function Convergence for p≥0?

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Homework Help Overview

The discussion revolves around the convergence of the gamma function, specifically for values of p greater than or equal to 0. The original poster is exploring whether mathematical induction can be applied to prove this convergence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining the use of induction to establish the convergence of the gamma function. There are attempts to clarify the assumptions regarding the values of p for which the convergence holds, particularly questioning the range of p and the implications of the induction steps.

Discussion Status

The conversation is ongoing, with participants providing insights into the induction process and its application to the gamma function. Some guidance has been offered regarding the need to demonstrate convergence for values of p in the interval [0, 1) to extend the proof to all real numbers p ≥ 0.

Contextual Notes

There is a noted concern about the initial assumption of convergence only for natural numbers and the need to address convergence for real numbers in between. The participants are also considering the implications of their assumptions on the overall proof.

mekkomhada
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I just learned induction in another thread and I'm curious if it can be used to prove that the gamma function converges for [itex]p\geq0[/itex]. I'm not sure if it can be used in this way. Is this wrong?

Gamma Function is defined as:
[tex]\Gamma(p+1)=\int_0^\infty e^{-x}x^p \,dx[/tex] We're trying to show that this converges for [itex]p\geq0[/itex]

Smallest case, p=0:
[tex]\Gamma(1)=1[/tex] converges

Assume the following converges:
[tex]\Gamma(p)=\int_0^\infty e^{-x}x^{p-1} \,dx[/tex]

Using integration by parts we find:
[tex]\Gamma(p+1)=p\Gamma(p)[/tex]

So since
[tex]\Gamma(p)[/tex] converges
then
[tex]\Gamma(p+1)=p\Gamma(p)[/tex] must also converge
 
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mekkomhada said:
Assume the following converges:
[tex]\Gamma(p)=\int_0^\infty e^{-x}x^{p-1} \,dx[/tex]

For which values of p are you assuming it holds? Let's assume it holds for all [itex]1\leq p \leq p'[/itex] for some real number [itex]p'\geq 1[/itex].
Using integration by parts we find:
[tex]\Gamma(p+1)=p\Gamma(p)[/tex]

So since
[tex]\Gamma(p)[/tex] converges
for [itex]p\in [1,p'][/itex]
then
[tex]\Gamma(p+1)=p\Gamma(p)[/tex] must also converge
for [itex]p\in[1,p'][/itex].

Together with the first step of the induction process, you've shown that [itex]\Gamma(p+1)[/itex] converges for p=0,1,2,3,4,..., but not for the real numbers in between.
If you can show that [itex]\Gamma(p+1)[/itex] converges for [itex]0\leq p <1[/itex] your induction shows it to be true for all real numbers [itex]p\geq 1[/itex]. Can you see why?
 
Last edited:
For which values of p are you assuming it holds? Let's assume it holds for all [itex]1\leq p \leq p'[/itex] for some real number [itex]p' \geq 1[/itex].

Ah right, that was my assumption that it was true for some real number greater than 1. I should have been more explicit.

To prove that [itex]\Gamma(p+1)[/itex] converges for [itex]0\leq p <1[/itex] couldn't I just say that [itex]\Gamma(1+1) \geq \Gamma(0+1)[/itex] and since [itex]\Gamma(1+1)[/itex] converges then the gamma function must converge for [itex]0 \leq p \leq 1[/itex]?
 
looks like u only showed that it converges for all natural numbers
 

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