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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: frog0101 on October 11, 2018, 09:29:34 am

Title: Partial Fractions-When can you assume the coefficients will be independent of x
Post by: frog0101 on October 11, 2018, 09:29:34 am: Hi,
For partial fractions, when are you able to assume that the coefficients will all be independent of x even if there is a denominator of degree higher than 1?

Eg. In the Coroneos 100 Integrals challenge, the solutions that I have solve the following question:
$\int \frac{3x+2}{x(x+1)^{3}}dx=\int \frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}+\frac{D}{(x+1)^3} dx$

I thought they would need to have
$\frac{Ax+B}{(x+1)^2}+\frac{Cx^2+Dx+E}{(x+1)^3}$

with the other terms as well.

Thanks
Title: Re: Partial Fractions-When can you assume the coefficients will be independent of x
Post by: RuiAce on October 11, 2018, 09:42:37 am: That defeats the purpose. The partial fractions algorithm was designed so that by expanding the powers you would never need to walk into this issue, or else you'd be stuck finding coefficients forever.
$\frac{1}{\text{something}\times (ax+b)^n} = \text{something} + \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} +\dots + \frac{A_n}{(ax+b)^n}$
$\frac{1}{\text{something}\times (ax^2+bx+c)^n} = \text{something} + \frac{A_1x+A_2}{ax^2+bx+c} + \frac{A_3x+A_4}{(ax^2+bx+c)^2} + \dots + \frac{A_{2n-1}x+A_{2n}}{(ax^2+bx+c)^n}$
For that particular problem, the alternate would have been to do something like $ \int \frac{3x+2}{x(x+1)^3}dx = \int \frac{A}{x} + \frac{Bx^2+Cx+D}{(x+1)^3}\,dx $
Title: Re: Partial Fractions-When can you assume the coefficients will be independent of x
Post by: SEasternCry on October 11, 2018, 09:48:16 am: Quote from: frog0101 on October 11, 2018, 09:29:34 am
Hi,
For partial fractions, when are you able to assume that the coefficients will all be independent of x even if there is a denominator of degree higher than 1?

Eg. In the Coroneos 100 Integrals challenge, the solutions that I have solve the following question:
$\int \frac{3x+2}{x(x+1)^{3}}dx=\int \frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}+\frac{D}{(x+1)^3} dx$

I thought they would need to have
$\frac{Ax+B}{(x+1)^2}+\frac{Cx^2+Dx+E}{(x+1)^3}$

with the other terms as well.

Thanks

Sorry this is one of my first time answering. Someone teach me how to use tex? is it another program?
coefficients of x are always independent of x? I assume you mean whether there will be x on the numerator?
If that is so, then there's some good Eddie Woo videos about what partial fractions to use but I'll try to explain it here:
You can assume the numerator can be something like (ax+b) only if the adjacent denominator is an unfactorable polynomial that is fully simplified with the highest degree of x being greater than 1. [provided that there is two simplified fractions here] However, for more fractions here. There is no need for using the second line as you shown.^
EDIT: while making this post looks like someone answered lol.