How to Integrate Termwise Between 0 and x?

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The discussion revolves around integrating the series 1/(1+x) = 1 - x + x^2 - x^3 + ... termwise between 0 and x using long division of polynomials. Participants clarify that the integration involves treating each term separately, leading to the expression of the integral as a sum of individual integrals. The correct approach is to integrate each term from 0 to x, resulting in a series of integrals that can be combined. The user realizes their earlier misunderstanding of the polynomial long division process and begins to see the correct pattern. Ultimately, the focus is on applying the integration to the derived series correctly.
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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.
 
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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.
 
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:

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