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RolyPoly

  • 2 years ago

Resolve into partial fraction \[\frac{x^2}{1-x^2}\]

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  1. RolyPoly
    • 2 years ago
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    \[\frac{x^2}{1-x^2} = \frac{x^2}{(1-x)(1+x)}\]How to continue??

  2. nitz
    • 2 years ago
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    \[x ^{2}/(1-x)(1+x)=A/(1-x)+B(1+x)\]

  3. nitz
    • 2 years ago
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    its B/(1+x)

  4. nitz
    • 2 years ago
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    ya?

  5. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    The degree of numerator must strictly be less than degree of denominator.

  7. hartnn
    • 2 years ago
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    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
    • 2 years ago
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    \(\large \frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)-1}{1-x^2} = -1-\frac{1}{1-x^2}\)

  9. hartnn
    • 2 years ago
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    or u can do that ^

  10. UnkleRhaukus
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    yes, sorry.

  13. hartnn
    • 2 years ago
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    but u got the point ?

  14. RolyPoly
    • 2 years ago
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    @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
    • 2 years ago
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    @hartnn Yes! When can I be as smart as you???!!!!!

  16. hartnn
    • 2 years ago
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    with just a little bit of practice :P

  17. UnkleRhaukus
    • 2 years ago
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    \[\frac{x^2}{1-x^2}=x^2\times\frac{1}{1-x^2} = x^2\times\frac{1}{(1-x)(1+x)}\] \[\frac{1}{(1-x)(1+x)}=\frac A{(1−x)}+\frac B{(1+x)}=\frac{A(1+x)+B(1-x)}{(1-x)(1+x)}\] \[1=A(1+x)+B(1-x)\] 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
    • 2 years ago
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    Excuse.. me.... When x =-1 1 = A(1+(-1)) + B(1-(-1)) 1 = 0 + 2B B = 1/2?

  19. UnkleRhaukus
    • 2 years ago
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    your right

  20. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    (x-1)=-(1-x)

  22. RolyPoly
    • 2 years ago
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    Such simple maths is killing me :\ Thanks!!!!!!!!!

  23. UnkleRhaukus
    • 2 years ago
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    im not sure about the -1

  24. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    A = B = 1/2 great work @RolyPoly

  26. hartnn
    • 2 years ago
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    1-x= -(x-1) again

  27. hartnn
    • 2 years ago
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    what u have done is correct.

  28. RolyPoly
    • 2 years ago
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    :'( Last but not least... \[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}-\frac{1}{2(x-1)}\]

  29. RolyPoly
    • 2 years ago
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    It needs tons of thousands of millions of practice!! :'( Thanks again!!!

  30. hartnn
    • 2 years ago
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    welcome ^_^

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