How to Integrate Termwise Between 0 and x?

  • Context: Undergrad 
  • Thread starter Thread starter DEMJR
  • Start date Start date
  • Tags Tags
    Division Series
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 10K views
DEMJR
Messages
14
Reaction score
0
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.
 
Physics news on Phys.org
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 [itex]p(x) = 1 + x[/itex] and the other one is [itex]q(x) = 1[/itex]. Try working with some other polynomials first, then see if you can pick up the pattern doing it this way.
 
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 ?
 
DEMJR said:
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 ?

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.
 
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?
 
long hand division example:
Code:
                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
                                        ...
 
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.
 
DEMJR said:
1/(1+x) = 1 - x + x^2 - x^3 + ...
DEMJR said:
How can I use the pattern to integrate termwise between 0 and x?
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:

[tex]\int_0^x 1 dx - \int_0^x x dx + \int_0^x x^2 dx - \int_0^x x^3 dx \ + \ ...[/tex]
 
Last edited: