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RolyPoly Group Title

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

  • one year ago
  • one year ago

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

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

    • one year ago
  3. nitz Group Title
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    its B/(1+x)

    • one year ago
  4. nitz Group Title
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    ya?

    • one year ago
  5. RolyPoly Group Title
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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.

    • one year ago
  6. hartnn Group Title
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    The degree of numerator must strictly be less than degree of denominator.

    • one year ago
  7. hartnn Group Title
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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.

    • one year ago
  8. hartnn Group Title
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    \(\large \frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)-1}{1-x^2} = -1-\frac{1}{1-x^2}\)

    • one year ago
  9. hartnn Group Title
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    or u can do that ^

    • one year ago
  10. UnkleRhaukus Group Title
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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

    • one year ago
  11. RolyPoly Group Title
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    Hmmm... -1 -1 = -2?! Shouldn't it be \[\frac{-(-x^2)}{(1-x^2) } = \frac{-(1-x^2)+1}{1-x^2} = -1+\frac{1}{1-x^2}\]

    • one year ago
  12. hartnn Group Title
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    yes, sorry.

    • one year ago
  13. hartnn Group Title
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    but u got the point ?

    • one year ago
  14. RolyPoly Group Title
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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

    • one year ago
  15. RolyPoly Group Title
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    @hartnn Yes! When can I be as smart as you???!!!!!

    • one year ago
  16. hartnn Group Title
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    with just a little bit of practice :P

    • one year ago
  17. UnkleRhaukus Group Title
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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 \(x=-1\) \(1=-2B\) \(B=-1/2\) \[\frac{1}{(1-x)(1+x)}=\frac{1/2}{(1−x)}-\frac {1/2}{(1+x)}\] \[\frac{x^2}{(1-x)(1+x)}=\frac{x^2}{2(1−x)}-\frac {x^2}{2(1+x)}\]

    • one year ago
  18. RolyPoly Group Title
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    Excuse.. me.... When x =-1 1 = A(1+(-1)) + B(1-(-1)) 1 = 0 + 2B B = 1/2?

    • one year ago
  19. UnkleRhaukus Group Title
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    your right

    • one year ago
  20. RolyPoly Group Title
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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))

    • one year ago
  21. UnkleRhaukus Group Title
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    (x-1)=-(1-x)

    • one year ago
  22. RolyPoly Group Title
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    Such simple maths is killing me :\ Thanks!!!!!!!!!

    • one year ago
  23. UnkleRhaukus Group Title
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    im not sure about the -1

    • one year ago
  24. RolyPoly Group Title
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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, \[\frac{x^2}{1-x^2}=-1+\frac{1}{2(1+x)}+\frac{1}{2(1-x)}\]

    • one year ago
  25. UnkleRhaukus Group Title
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    A = B = 1/2 great work @RolyPoly

    • one year ago
  26. hartnn Group Title
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    1-x= -(x-1) again

    • one year ago
  27. hartnn Group Title
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    what u have done is correct.

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

    • one year ago
  29. RolyPoly Group Title
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    It needs tons of thousands of millions of practice!! :'( Thanks again!!!

    • one year ago
  30. hartnn Group Title
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    welcome ^_^

    • one year ago
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