Long division of the series

1. Apr 3, 2012

DEMJR

Not sure when this problem in my book says to calculate by long division the series 1/(1+x) = 1 - x + x^2 - x^3 + ..., and then integrating termwise between 0 and x.

I am really rusty on these types of problems and need help understanding how to even begin T.T. Thanks for the help.

2. Apr 3, 2012

Robert1986

I don't get the integrating between 0 and x part. But have you ever done long division of polynomials? That's all this is, one of the polynomials is $p(x) = 1 + x$ and the other one is $q(x) = 1$. Try working with some other polynomials first, then see if you can pick up the pattern doing it this way.

3. Apr 3, 2012

DEMJR

Not sure if I correctly implemented your response but here is what I tried:

1/(1+x) = 1 + 1/x

-x/(1+x) = -x - 1

x^2/(1+x) = x^2 + x

-x^3/(1+x) = -x^3 - x^2

However, this seems to be incorrect because everything cancels except the 1/x

1/(x+1) = 1 + 1/x - x - 1 + x^2 + x -x^3 - x^3 + ....

I must have misinterpreted your response and this is really starting to get to me. Shouldnt everything cancel to where 1/(x+1) = 1 + 1/x ?

4. Apr 3, 2012

Robert1986

No, you can't do this. You can't break up the denomnator like that. As a simple, example, take x = 0.

I suggest googling "long division of polynomials" and look at some examples, then try it with your problem.

5. Apr 3, 2012

HallsofIvy

1+ x)1- x+ x^2- x^3+ ...

Obviously 1 divides into 1 1 time so we have then subtract
1- x+ x^2- x^3+ ...
1+ x
__________________________
-2x+ x^2- x^3+....

And 1 divides into -2x -2x times. Multiplying and subtract
-2x+ x^2- x^3+ ...
-2x- 2x^2
_________________
3x^2- x^3+...

Now 1 divides into that 3x^2 times so multiplying and subtracting
3x^2- x^3+ x^4- x^5
3x^2+3x^3
__________________
-3x^3+ x^4- x^5

So far we have 1- 2x+ 3x^2. See the pattern?

6. Apr 3, 2012

rcgldr

long hand division example:
Code (Text):

1      - x + x^2 - x^3 + ...
-------------------------------
1 + x | 1
1  + x
------
- x
- x  - x^2
----------
x^2
x^2 + x^3
----------
- x^3
- x^3  - x^4
------------
x^4
...

7. Apr 3, 2012

DEMJR

Thanks. I see the pattern and realize my mistake in my previous post (so silly of me). I did it for the next few terms and got it to be 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5

How can I use the pattern to integrate termwise between 0 and x?

I forgot to mention in the original post that we are interested in -1 < x <= 1.

8. Apr 4, 2012

rcgldr

I think you're supposed to integrate each term, but since this is a sum, there's no reason these couldn't all be combined into one integral:

$$\int_0^x 1 dx - \int_0^x x dx + \int_0^x x^2 dx - \int_0^x x^3 dx \ + \ ...$$

Last edited: Apr 4, 2012