Partial Fractions

  • Thread starter Gear300
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  • #1
Gear300
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For a rational function, (x^2+1)/(x^2-1) = (x^2+1)/[(x+1)(x-1)], if we were to split it into partial fractions so that (x^2+1)/(x^2-1) = A/(x+1) + B/(x-1) = [A(x-1) + B(x+1)]/(x^2-1)...solving for A and B get us A = -1 and B = 1. This would mean that (x^2+1)/(x^2-1) = 2/(x^2-1)...which doesn't seem right; x^2+1 is 2 greater than x^2-1, so (x^2+1)/(x^2-1) can be rewritten as 1 + 2/(x^2-1), which is 1 greater than the function I got earlier. Why is there a discrepancy?
 

Answers and Replies

  • #2
eok20
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The problem is that you're assuming that (x^2 + 1)/(x^2 - 1) can be written in the form A/(x+1) + B/(x-1) which is simply not the case. Indeed, A/(x+1) + B/(x-1) = [A(x-1) + B(x+1)]/(x^2-1) as you say, but A(x-1) + B(x+1) is linear whereas the numerator in the original function is quadratic. Did this help?
 
  • #3
Defennder
Homework Helper
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In general, when you have an 'improper' function in fraction form, where the highest power of the variable in the numerator is greater or equal to that of the denominator, you should do polynomial long division, until the highest power of the numerator is lesser than that of the denominator. So in this case,

[tex]\frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1}[/tex]
 
  • #4
Gear300
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thanks for the posts...I see...so since it can not be written in that form, I should do the long division instead...but why doesn't the method I used work out...I noticed that the function in the numerator is always 2 greater than the denominator, so they never touch...would that have something do with it?
 
  • #5
Defennder
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I can't decipher what you're saying here. For one thing, note that you cannot even split the fraction into partial fractions if the greatest power in the numerator is greater or equal to that of the denominator. You must ensure that it is less than that of the denominator.

Secondly, note that you're values of A and B are wrong; try substituting them into the partial fractions you decomposed the original fraction into and see if you can arrive at the function.
 

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