## RolyPoly 3 years ago Resolve into partial fraction $\frac{x^2}{1-x^2}$

1. RolyPoly

$\frac{x^2}{1-x^2} = \frac{x^2}{(1-x)(1+x)}$How to continue??

2. nitz

$x ^{2}/(1-x)(1+x)=A/(1-x)+B(1+x)$

3. nitz

its B/(1+x)

4. nitz

ya?

5. RolyPoly

But that doesn't seems to work.. Here's what I mean: A(1+x) + B(1-x) = x^2 (A-B)x +(A+B) = x^2 => A-B =0 , A+B =0 Not reasonable.

6. hartnn

The degree of numerator must strictly be less than degree of denominator.

7. hartnn

so, one thing u can do is, resolve 1/(1-x^2) into partial fractions and then multiply x^2 to both terms.

8. hartnn

$$\large \frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)-1}{1-x^2} = -1-\frac{1}{1-x^2}$$

9. hartnn

or u can do that ^

10. UnkleRhaukus

A(1+x) + B(1-x) = x^2 for x= -1 2 B = x^2 B = x^2/2 for x= 0 A + B = x^2 for x= 1 2A = x^2 A = x^2/2

11. RolyPoly

Hmmm... -1 -1 = -2?! Shouldn't it be $\frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)+1}{1-x^2} = -1+\frac{1}{1-x^2}$

12. hartnn

yes, sorry.

13. hartnn

but u got the point ?

14. RolyPoly

@UnkleRhaukus A(1+x) + B(1-x) = x^2 for x= -1 2 B = (-1)^2 B = 1/2 for x= 0 A + B = 0 for x= 1 2A = 1^2 A = 1/2 And it doesn't work here :S

15. RolyPoly

@hartnn Yes! When can I be as smart as you???!!!!!

16. hartnn

with just a little bit of practice :P

17. UnkleRhaukus

$\frac{x^2}{1-x^2}=x^2\times\frac{1}{1-x^2} = x^2\times\frac{1}{(1-x)(1+x)}$ $\frac{1}{(1-x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1-x)}{(1-x)(1+x)}$ $1=A(1+x)+B(1-x)$ for $$x=1$$ $$1=2A$$ $$A=1/2$$ for $$x=-1$$ $$1=-2B$$ $$B=-1/2$$ $\frac{1}{(1-x)(1+x)}=\frac{1/2}{(1−x)}-\frac {1/2}{(1+x)}$ $\frac{x^2}{(1-x)(1+x)}=\frac{x^2}{2(1−x)}-\frac {x^2}{2(1+x)}$

18. RolyPoly

Excuse.. me.... When x =-1 1 = A(1+(-1)) + B(1-(-1)) 1 = 0 + 2B B = 1/2?

19. UnkleRhaukus

20. RolyPoly

http://www.wolframalpha.com/input/?i=Resolve+into+partial+fraction%3A+x%5E2+%2F+%281-x%5E2%29 I don't know how to get that -1/(2(1-x))

21. UnkleRhaukus

(x-1)=-(1-x)

22. RolyPoly

Such simple maths is killing me :\ Thanks!!!!!!!!!

23. UnkleRhaukus

im not sure about the -1

24. RolyPoly

Here it goes~ $\frac{x^2}{1-x^2}$$=\frac{-(-x^2)}{1-x^2}$$=\frac{-(1-x^2)+1}{1-x^2}$$=-1+\frac{1}{1-x^2}$ $\frac{1}{1-x^2}=\frac{A}{1+x}+\frac{B}{1-x}$A(1-x)+B(1+x)=1 A = B = 1/2 So, $\frac{1}{1-x^2}=\frac{1}{2(1+x)}+\frac{1}{2(1-x)}$ Thus, $\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}+\frac{1}{2(1-x)}$

25. UnkleRhaukus

A = B = 1/2 great work @RolyPoly

26. hartnn

1-x= -(x-1) again

27. hartnn

what u have done is correct.

28. RolyPoly

:'( Last but not least... $\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}-\frac{1}{2(x-1)}$

29. RolyPoly

It needs tons of thousands of millions of practice!! :'( Thanks again!!!

30. hartnn

welcome ^_^