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Math Help - Help needed with mathematical induction problem

  1. #1
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    Help needed with mathematical induction problem

    The question is:

    Prove (by mathematical induction) that:

    \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:

    \int_0^\infty{x^{k+1}}{e^{-ax}}{dx}. And I used integration by parts to get the expression \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:

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

    But I don't think that's correct. Any help appreciated! Many thanks.
    CP
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  2. #2
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    Re: Help needed with mathematical induction problem

    It appears that you did the integration by parts incorrectly.

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

    So, \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

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

    \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.
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  3. #3
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    Re: Help needed with mathematical induction problem

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

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

    \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:

    \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.
    Last edited by CrispyPlanet; October 19th 2013 at 04:35 PM.
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    Re: Help needed with mathematical induction problem

    Quote Originally Posted by CrispyPlanet View Post
    Hi SlipEternal, thanks for your response. I did the same as you for the LHS. I got:

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

    \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:

    \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 k=10 for instance, your expression means \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.

    \dfrac{(k+1)k(k-1)(k-2)...1}{a^{k+2}} = \dfrac{(k+1)!}{a^{k+2}} = S(k+1)
    Thanks from CrispyPlanet
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  5. #5
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    Re: Help needed with mathematical induction problem

    Quote Originally Posted by CrispyPlanet View Post
    The question is:

    Prove (by mathematical induction) that:

    \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 \int_0^\infty xe^{-ax}dx= 1/a^2. Did you do the integral on the left "by parts"?

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

    \int_0^\infty{x^{k+1}}{e^{-ax}}{dx}. And I used integration by parts to get the expression \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:

    \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 \frac{n(n-1)(n-2)\cdot\cdot\cdot (2)(1)}{a^{n+1}}.
    Replacing "n" by "k+ 1" that becomes \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.
    Thanks from CrispyPlanet
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    Re: Help needed with mathematical induction problem

    Quote Originally Posted by SlipEternal View Post
    Do you understand what the ... means? If k=10 for instance, your expression means \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.

    \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.
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  7. #7
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    Re: Help needed with mathematical induction problem

    Quote Originally Posted by HallsofIvy View Post
    You can't just say "S(1) is true". If n= 1 then S(1) is \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 \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
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