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

Tags:
1. Oct 3, 2009

martizzle

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

2. Oct 3, 2009

Office_Shredder

Staff Emeritus
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

3. Oct 4, 2009

n!kofeyn

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?

4. Oct 4, 2009

n!kofeyn

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.

5. Oct 4, 2009

martizzle

thanks everyone...

@ jbunniii: i must be like crazy dense because i still can't figure how
$\sum_{n=0}^{\infty} n e^{-n} = e / (e - 1)^(2)$
@ 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
6. Oct 4, 2009

jbunniii

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.

7. Oct 4, 2009

martizzle

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
8. Oct 4, 2009

n!kofeyn

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.

9. Oct 4, 2009

martizzle

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

10. Aug 25, 2010

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

11. Apr 24, 2012

tanay

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