Partial Fractions?

  • Thread starter hqjb
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  • #1
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Homework Statement


In one of the workings of a question I couldn't solve(from solution sheet) there was one step I couldn't understand.

Homework Equations



[itex]\frac{1}{x^2+x+1} = \frac{1-x}{1-x^3}[/itex]

The Attempt at a Solution



Tried partial fractions(unfactorable denominator) and could'nt get it, finally got it through trial and error but is there a method where you apply to all such expressions to simplify it?
 

Answers and Replies

  • #2
NascentOxygen
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You're wanting to factorize (1-x³). You can see that x=1 solves it, so take out (x-1) as a factor.

How to do this? Do you know the "long division" method for dividing (1-x³) by (x-1)? This will give you the other factor.

:: EDIT:: I assumed you were working from the RH expression, wanting to find the LHE. Maybe you weren't. Which do you regard as the "simplest" expression here? :smile:
 
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  • #3
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You're wanting to factorize (1-x³). You can see that x=1 solves it, so take out (x-1) as a factor.

How to do this? Do you know the "long division" method for dividing (1-x³) by (x-1)? This will give you the other factor.

:: EDIT:: I assumed you were working from the RH expression, wanting to find the LHE. Maybe you weren't. Which do you regard as the "simplest" expression here? :smile:

Haha sorry if I didnt put it into context properly. I wanted to obtain the RHE from the LHE, it wasnt a proving question but rather the solution manual skipped steps from LH-> RH and I'm not sure how he did it and if I can replicate it on a similar question without hindsight.

Edit : It's for Mclaurin's series for I needed it in 1\(1-x) form.
 
  • #4
Mentallic
Homework Helper
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If you want to go from the LHS to the RHS of [tex]\frac{1}{x^2+x+1}\equiv \frac{1-x}{1-x^3}[/tex] then what you need is for the numerators to be equivalent, but at the same time you can't change the LHS expression.

So what you can do is use [tex]\frac{1}{x^2+x+1}\cdot \frac{1-x}{1-x}[/tex] to get the numerator the way you want it, and simultaneously keeping the expression equal.
 
  • #5
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Alright thanks!
 

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