# 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