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Riemann Sum Proof

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that:
    lim n->inf1/n*Ʃn-1k=0ekx/n

    =

    (ex-1)/x

    x>0

    2. Relevant equations

    That was all the information provided. Any progress i have made is below. I didn't want to add any of that to this section because this is all speculation on my part so far.

    3. The attempt at a solution
    I've been at this for awhile now, i feel as though i am getting close. I think i have all the "pieces" but i cant seem to put them together to prove the above statement.
    I know that the integral
    01etxdt is important because it integrates to
    (ex-1)/x
    but i'm not sure how to connect the summation to the integral to the answer
    I also have a feeling that the Theorem
    is relevant. But i'm not positive.

    Any help would be extremely appreciated!
     
  2. jcsd
  3. Dec 13, 2011 #2

    Dick

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    The thing your really need to notice to evaluate that sum is that e^(kx/n) is a geometric series. It's r^k where r=e^(x/n). There is a formula to sum a geometric series. Then, sure, your theorem tells you the limit as n->infinity of the sum is the integral. So now after you've simplified the sum try to find the limit.
     
    Last edited: Dec 13, 2011
  4. Dec 13, 2011 #3
    If i recall correctly that formula is
    I tried that but i hit a roadblock. I came up with 1-exn/n/ 1-ex/n. You can cancel the n's in the numerator's exponent of course. But that is where i hit my dead end. I couldn't manipulate it from there in any meaningful way, even using the 1/n factor in front of the summation.
    The reason i abandoned that course is because my professor hinted that i should look in the chapter of my book regarding Riemann Sums. But i've yet to dig up anything useful besides the Theorem i quoted in the OP
     
  5. Dec 13, 2011 #4

    Dick

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    That's a good start! So you've got a 1-e^x part. Now look at the denominator, lim n*(1-e^(x/n)) as n->infinity. That's an infinity*0 form, right? That suggests you might want to write it as (1-e^(x/n))/(1/n). That's a 0/0 form and you can apply l'Hopital's rule.
     
    Last edited: Dec 13, 2011
  6. Dec 13, 2011 #5
    But wouldn't taking the derivative of 1/n give us zero, because n is some constant approaching inf? So the derivative of (1-e^(x/n)) would be (xe^(x/n))/n and the derivative of 1/n would be 0, giving me ((xe^(x/n))/n )/0?
     
  7. Dec 13, 2011 #6

    Dick

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    Well, no. n isn't constant. It's a variable approaching infinity. x is the constant while you are taking the limit. You want to take d/dn of the numerator and denominator.
     
  8. Dec 13, 2011 #7
    So i am are applying L'Hôpital's rule to the fraction in the denominator of our function i.e applying it to (1-e^(x/n))/(1/n)
    So i get -xe^(x/n)/(n^2) on top and -1/n^2 on the bottom which simplifies to:
    xe^(x/n)?
     
  9. Dec 13, 2011 #8

    Dick

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    You missed a minus sign. But otherwise ok! Now take limit n->infinity of x*e^(x/n).
     
  10. Dec 13, 2011 #9
    as n approaches infinity x/n approaches zero, which means e^(x/n) approaches one which leaves me with x on the bottom!
    Incredible, i never would have seen that. Thanks so much Dick! You are a lifesaver!
     
  11. Dec 13, 2011 #10

    Dick

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    Yeah, it does work out nicely and you are welcome. But so far you've got (1-e^x)/x, and you want to get (e^x-1)/x. Better find that lost minus sign before you call it a done deal.
     
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