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Problem 51 from (GRE) math subject test: practice test book

  1. Oct 3, 2009 #1

    Problem 51 from GRE math test practice book or test form code: GR0568
    The problem is as follows (ps. I'm sorry, I do not have LaTeX or something)

    If |_ x _| denotes the greatest integer not exceeding x, then 'the integral from zero to infinity' of |_ x _| * e ^ (-x) dx = ?

    The answer is 1/(e - 1)

    I thought that the way to solve the problem would be:
    1. the limit of |_ x _| and x, as x -> infinity, is x.
    2. integrate x * e ^ (-x)
    3. plug in the limits, to get the answer
    But this method gives me 1

    Please help, and sorry for the problem description (not using LaTeX/etc)
  2. jcsd
  3. Oct 3, 2009 #2


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    Over each interval [n,n-1) your function is going to be strictly less than the function xe-x, so you wouldn't expect the integrals to be the same.

    Try integrating over each such interval, then summing up all the integrals
  4. Oct 4, 2009 #3
    You meant [n,n+1) right? Also, I'm curious. Are they really expecting you to be able to know or be able to calculate what the infinite series you end up with sums up to?
  5. Oct 4, 2009 #4
    The hint to write it as a sum of integrals was already given, and I think it kind of defeats the purpose and is bad form when you flat out do the problem. Especially when the original poster has not responded.
  6. Oct 4, 2009 #5
    thanks everyone...

    @ jbunniii: i must be like crazy dense because i still can't figure how
    [itex] \sum_{n=0}^{\infty} n e^{-n} = e / (e - 1)^(2) [/itex]
    which would make your method lead to the answer...
    @ n!kofeyn: i did respond, but deleted it after seeing jbunniii's post (thats why he/she responded again)...i didn't want to keep asking questions when everyone seemed to get it....i thought it was some error on my part...thnx again
    Last edited: Oct 4, 2009
  7. Oct 4, 2009 #6


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    Actually he had responded, saying that he thought the answer was a geometric series, which wasn't quite right. Unfortunately he deleted that post after I replied to it, which is bad form IMHO.

    I've deleted my offending posts.
  8. Oct 4, 2009 #7
    ps...i really dont get it, and I'm not looking for easy handouts...it just so happens that for some reason, i can't do this particular problem, and would like to know how to, prior my test...
    i deleted the post because i was changing it to respond to a new post, i think this is my first time posting (cant recall), but my apologies, next time, I'll leave all my posts/comments
    Last edited: Oct 4, 2009
  9. Oct 4, 2009 #8
    Haha. I apologize then. You're right. It becomes quite confusing when posts are deleted as it is difficult to see the context of responses.
  10. Oct 4, 2009 #9
    looking over, i finally get it:
    i need to change the exponent....Taylor series....(someone should have hinted at that lol)
    thnx i appreciate all your help guys
  11. Aug 25, 2010 #10
    hey guys

    the solution is something like this

    |x|= 0 when 0<=x<=1
    |x|=1 when 1<=x<=2
    and so on

    So break the integral and we have

    F(0 to 1) 0*e^-x + F(1 to 2) 1*e^-x+ F(2 to 3)............................................

    i am using F as the symbol for the integral.

    now F e^-x = -e^-x
    so the integral becomes-

    -{(e^-2 - e^-1) +2(e^-3 - e^-2)+ 3(e^-4 - e^-3)+ 4(............................}

    = -{-e^-1 -e^-2- e^-3- e^-4+...........................}

    =e^-1 +e^-2+ e^-3+ e^-4+...........................}

    this is a geometric progression with 1/e as the multiplier and

    for an infinite GP, sum= a/(1-r)
    we have: i/e/(1-1/e)========>1/(1-e)

    Hope that solves the problem

    this is NOT a taylor series, just a GP

    good one guys
  12. Apr 24, 2012 #11
    the solution is right but there are a little mistake i think...... there would be

    |x|= 0 when 0<=x<1
    |x|=1 when 1<=x<2

    instead of
    |x|= 0 when 0<=x<=1
    |x|=1 when 1<=x<=2
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