anonymous
  • anonymous
Guys what is partial fraction decomposition?
Mathematics
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schrodinger
  • schrodinger
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zzr0ck3r
  • zzr0ck3r
\[\frac{1}{x(x+1)}=\frac{a}{x}+\frac{b}{x+1}\]for some a,b
zzr0ck3r
  • zzr0ck3r
this is an example of partial fraction decomposition of 1/(x(x+1))
anonymous
  • anonymous
so what is the aim of a partial fraction decomposition?

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terenzreignz
  • terenzreignz
Useful in particular, when integrating. It's usually easier to integrate when the denominator-polynomial is only a linear one. It just results in a logarithm.
terenzreignz
  • terenzreignz
...unless your context isn't calculus D: If so, then... sorry ^_^
zzr0ck3r
  • zzr0ck3r
lets figure out what a,b are. multiply everything by x(x+1) \[1=a(x+1)+bx=ax+a+bx=x(a+b)+a\]here is the "trick we have x(a+b)+a = 1 since we have no x's on the RHS we know that a+b=0 and since the rhs has only numbers we know that a=1 so a=1, and b =-1 so \[\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\]
zzr0ck3r
  • zzr0ck3r
As @terenzreignz said, I would rather integrate the RHS and not the LHS
terenzreignz
  • terenzreignz
There are always exceptions, though :) Like... \[\Large \int \frac{1}{1-x^2}dx\]
terenzreignz
  • terenzreignz
But... yeah, that's the long and short of it... sometimes it's easier to deal with sums of simpler fractions than just one 'big' fraction (big here meaning a rather complicated polynomial for a denominator)
anonymous
  • anonymous
thank you guys ^.^

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