Help needed with mathematical induction problem

The question is:

Prove (by mathematical induction) that:

$\displaystyle \int_0^\infty{x^n}{e^{-ax}}{dx}=\frac{{n(n-1)(n-2)...1}}{a^{n+1}}}$

I've proved the equality for n=1 as my base step. For my induction step I said that S(1) is true. Suppose that S(n)=S(k). We need to prove that S(k+1) is true. So the LHS becomes:

$\displaystyle \int_0^\infty{x^{k+1}}{e^{-ax}}{dx}.$ And I used integration by parts to get the expression $\displaystyle \frac{k^4+2k^3+5k^2+2k...1}{a^{k+2}}$

However I'm not sure what the RHS expression becomes for S(k+1). I thought it might be:

$\displaystyle \frac{{k(k+1)(k-1)...1}}{a^{k+2}}$

But I don't think that's correct. Any help appreciated! Many thanks.

CP

Re: Help needed with mathematical induction problem

It appears that you did the integration by parts incorrectly.

Let $\displaystyle u = x^{k+1}, v = -\dfrac{e^{-ax}}{a}$. Then $\displaystyle u' = (k+1)x^kdx, v' = e^{-ax}dx$.

So, $\displaystyle \int_0^\infty x^{k+1}e^{-ax}dx = -\left. \dfrac{x^{k+1}}{ae^{ax}} \right]_0^\infty + \int_0^\infty \dfrac{k+1}{a}x^ke^{-ax}dx$

$\displaystyle \lim_{x\to \infty} \dfrac{x^{k+1}}{ae^{ax}} = 0$, so

$\displaystyle \int_0^\infty x^{k+1}e^{-ax}dx = \dfrac{k+1}{a}\int_0^\infty x^ke^{-ax}dx$

Apply the induction assumption to the RHS.

Re: Help needed with mathematical induction problem

Hi SlipEternal, thanks for your response. I did the same as you for the LHS. I got:

$\displaystyle \frac{k+1}{a}\int_0^\infty{x^k}{e^{-ax}}{dx}$. We know that:

$\displaystyle \int_0^\infty{x^k}{e^{-ax}}{dx}=\frac{k(k-1)(k-2)...1}{a^{k+1}}$. So for the LHS I end up with:

$\displaystyle \frac{k+1}{a}\frac{k(k-1)(k-2)...1}{a^{k+1}}$

I then multiplied this out to get what I stated in my first post. My problem is getting the RHS to equal this!

CP

EDIT: I just realised I may have made a mistake in multiplying the last expression out. I'll give it another go and make sure. But my RHS problem remains.

Re: Help needed with mathematical induction problem

Quote:

Originally Posted by

**CrispyPlanet** The question is:

Prove (by mathematical induction) that:

$\displaystyle \int_0^\infty{x^n}{e^{-ax}}{dx}=\frac{{n(n-1)(n-2)...1}}{a^{n+1}}}$

I've proved the equality for n=1 as my base step. For my induction step I said that S(1) is true.

You can't just **say** "S(1) is true". If n= 1 then S(1) is $\displaystyle \int_0^\infty xe^{-ax}dx= 1/a^2$. Did you do the integral on the left "by parts"?

Quote:

Suppose that S(n)=S(k). We need to prove that S(k+1) is true. So the LHS becomes:

$\displaystyle \int_0^\infty{x^{k+1}}{e^{-ax}}{dx}.$ And I used integration by parts to get the expression $\displaystyle \frac{k^4+2k^3+5k^2+2k...1}{a^{k+2}}$

However I'm not sure what the RHS expression becomes for S(k+1). I thought it might be:

$\displaystyle \frac{{k(k+1)(k-1)...1}}{a^{k+2}}$

But I don't think that's correct. Any help appreciated! Many thanks.

CP

What is shown "on the right" is $\displaystyle \frac{n(n-1)(n-2)\cdot\cdot\cdot (2)(1)}{a^{n+1}}$.

Replacing "n" by "k+ 1" that becomes $\displaystyle \frac{(k+1)(k+1-1)(k+1-2)\cdot\cdot\cdot (2)(1)}{a^{k+1+1}}= \frac{(k+1)(k)(k_1)\cdot\cdot\cdot(2)(1)}{a^{k+2}}$.

That **is** what you have- although for some reason you have the first two terms in the numerator reversed.

Re: Help needed with mathematical induction problem

Quote:

Originally Posted by

**CrispyPlanet** Hi SlipEternal, thanks for your response. I did the same as you for the LHS. I got:

$\displaystyle \frac{k+1}{a}\int_0^\infty{x^k}{e^{-ax}}{dx}$. We know that:

$\displaystyle \int_0^\infty{x^k}{e^{-ax}}{dx}=\frac{k(k-1)(k-2)...1}{a^{k+1}}$. So for the LHS I end up with:

$\displaystyle \frac{k+1}{a}\frac{k(k-1)(k-2)...1}{a^{k+1}}$

I then multiplied this out to get what I stated in my first post. My problem is getting the RHS to equal this!

CP

Do you understand what the ... means? If $\displaystyle k=10$ for instance, your expression means $\displaystyle \dfrac{11\cdot 10\cdot 9...1}{a^{12}} = \dfrac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{a^{12}}$. What you multiplied out is only the first four factors.

$\displaystyle \dfrac{(k+1)k(k-1)(k-2)...1}{a^{k+2}} = \dfrac{(k+1)!}{a^{k+2}} = S(k+1)$

Re: Help needed with mathematical induction problem

Quote:

Originally Posted by

**SlipEternal** Do you understand what the ... means? If $\displaystyle k=10$ for instance, your expression means $\displaystyle \dfrac{11\cdot 10\cdot 9...1}{a^{12}} = \dfrac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{a^{12}}$. What you multiplied out is only the first four factors.

$\displaystyle \dfrac{(k+1)k(k-1)(k-2)...1}{a^{k+2}} = \dfrac{(k+1)!}{a^{k+2}} = S(k+1)$

Thanks for that. I understand now.

Re: Help needed with mathematical induction problem

Quote:

Originally Posted by

**HallsofIvy** You can't just **say** "S(1) is true". If n= 1 then S(1) is $\displaystyle \int_0^\infty xe^{-ax}dx= 1/a^2$. Did you do the integral on the left "by parts"?

I should have said that I condensed my working a little in my first post. Yes, I did work out both sides and showed that they equal $\displaystyle \frac{1}{a^2}$ for my base step (using by parts).

I obviously don't understand factorials very well! But this has helped a lot. Thanks!

CP