MHB How is the Gamma Function Related to Factorials?

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The Gamma function, defined as \(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\), relates to factorials through the property \(\Gamma(r+1)=(r+1)\Gamma(r)\). The initial problem involves proving that \(\Gamma(0)=1\), which is straightforward, and then tackling the recursive relationship in part b, where integration by parts is suggested. This leads to the conclusion that \(\Gamma(n)=n!\) for natural numbers \(n\), which can be proven using mathematical induction. The discussion also highlights the importance of understanding the definitions of the Gamma and factorial functions to avoid confusion.
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I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.
 
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skatenerd said:
I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.
Hint for b.: use integration by parts on $\Gamma(r+1)$.

Hint for c.: induction.
 
b) $\displaystyle \Gamma(r+1)=\int_0^{\infty}x^{r+1}e^{-x}\,dx$

Using IBP, we may define:

$\displaystyle u=x^{r+1}\,\therefore\,du=(r+1)x^r\,dx$

$\displaystyle dv=e^{-x}\,\therefore\,v=-e^{-x}$

and we have...?

edit: pipped at the post! (Tmi)
 
Last edited:
Thanks for the responses, still a little confused though.
I tried working out the integration by parts for myself and ended up with this:

\(\int_{0}^{\infty}x^{r+1}e^{-x}=-e^{-x}x^{r+1}+(r+1)\int_{0}^{\infty}x^re^{-x}\)

It seems like I'm on the right track but I guess I am overlooking something.
 
What you have is actually:

$\displaystyle \int_{0}^{\infty}x^{r+1}e^{-x}=\left[-e^{-x}x^{r+1} \right]_0^{\infty}+(r+1)\int_{0}^{\infty}x^re^{-x}\,dx$
 
Woops! Guess I've never thought to solve that part out with its bounds without solving the whole integral yet.
Anyways, solving that out, I got zero, and that in turn proves what I needed to prove. Thanks a bunch!
One last question, how do you make those longer integral symbols in Latex? They look a lot nicer than these little ones... \(\int f(x) dx\)
 
I use the tags

Code:
$\displaystyle insert LaTeX code here$
 
Sorry, I just realized I still have no idea how to start part c. Can I just plug in n into \(\Gamma(r)\) ? I still just can't really see any way of relating that integral \(\Gamma(n)\) to n!.
 
skatenerd said:
I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.

In order to avoid misunderstanding the 'Gamma Function' is usually defined as...

$\displaystyle \Gamma (r) = \int_{0}^{\infty} x^{r-1}\ e^{- x}\ dx$ (1)

... and the 'Factorial Function' as...

$\displaystyle r! = \int_{0}^{\infty} x^{r}\ e^{- x}\ dx$ (2)

The properties of course are very similar, but it is important don't have confusion. Your example implies the Factorial Function... Kind regards $\chi$ $\sigma$
 
  • #10
You have stated that:

$\displaystyle n\in\mathbb{N}$

but have (real) after it. I am assuming we are to let n be a natural number instead.

I would begin the proof by induction by demonstrating the validity of the base case:

$\displaystyle \Gamma(1)=1!$

Using the result from part b) we may state:

$\displaystyle \Gamma(0+1)=(0+1)\Gamma(0)$

Using the result from part a) we now have:

$\displaystyle \Gamma(1)=1=1!$

So, the base case is true. Now, state the induction hypotheses $\displaystyle P_k$:

$\displaystyle \Gamma(k)=k!$

From part b) we know $\displaystyle \Gamma(k)=\frac{\Gamma(k+1)}{k+1}$

Now, substitute to finish the proof by induction.
 
  • #11
Thanks for all that you guys. And yes I don't know why I put real I did in fact mean natural. I feel like this teacher assumes we all took a class on proofs, but I haven't learned any of these things yet! Guess I'm going to have to take that class soon.
 
  • #12
Induction is sometimes taught in Precalculus, but I suppose it may be optional and up to the discretion of the instructor.

It is a very useful method, and I recommend if you have spare time to give it a look. (Handshake)
 

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