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satellite73 Group Title

partial fraction madness, continuation

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    method for finding \(b\) in \[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}\]

    • 2 years ago
  2. mathslover Group Title
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    in terms of a and c and x ?

    • 2 years ago
  3. satellite73 Group Title
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    @mahmit2012 method with derivatives

    • 2 years ago
  4. Libniz Group Title
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    what's wrong with ordinary method of finding partial differential?

    • 2 years ago
  5. satellite73 Group Title
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    @mahmit2012 had a snap method, but i don't understand it

    • 2 years ago
  6. mathslover Group Title
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    just a question: do wolfram has solution?

    • 2 years ago
  7. alienbrain Group Title
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    they have something ugly...

    • 2 years ago
  8. satellite73 Group Title
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    we get \(a=\frac{6}{1}=6\), and \(c=\frac{-9}{-1}=9\) but \[b=5+0-\frac{6}{(-1)^2}=-1\] step i don't get

    • 2 years ago
  9. alienbrain Group Title
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    http://www.wolframalpha.com/input/?i=integrate+%285x%5E2%2B20x%2B6%29+%2F+%28x%5E3%2B2x%5E2%2Bx%29

    • 2 years ago
  10. alienbrain Group Title
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    oh the original question was an integral, btw.

    • 2 years ago
  11. satellite73 Group Title
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    something to do with a derivative but i don't see it

    • 2 years ago
  12. alienbrain Group Title
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    what's your method for B

    • 2 years ago
  13. Libniz Group Title
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    is there link to where it was done in snappier way?

    • 2 years ago
  14. satellite73 Group Title
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    looks like the method was, take \[5x^2+20x+6\] divide by \(x\) get \[5x+20+\frac{x}{6}\] take the derivative get \[5-\frac{6}{x^2}\] and then evaluate at \(x=-1\) why this works is going to bug me

    • 2 years ago
  15. satellite73 Group Title
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    but is sure is snapp!

    • 2 years ago
  16. sauravshakya Group Title
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    I think we can solve this making three equations....

    • 2 years ago
  17. alienbrain Group Title
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    i meant did you have your own way, satellite?

    • 2 years ago
  18. alienbrain Group Title
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    a slower... sane approach

    • 2 years ago
  19. alienbrain Group Title
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    i plugged in A and C.. expanded it all out... plugged in 1 for x and got B=-3

    • 2 years ago
  20. satellite73 Group Title
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    ooooooooooooooh!!!! multiply both sides by \((x+1)^2\)!!!!!

    • 2 years ago
  21. alienbrain Group Title
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    but that's not right i guess

    • 2 years ago
  22. satellite73 Group Title
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    wowee zowee

    • 2 years ago
  23. satellite73 Group Title
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    how cool is that? i love it

    • 2 years ago
  24. satellite73 Group Title
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    @alienbrain yes i wrote my method before solved \(6+B=5\) but it required some visualization

    • 2 years ago
  25. sauravshakya Group Title
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    |dw:1344137128472:dw|

    • 2 years ago
  26. satellite73 Group Title
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    it is already solved, i was trying to understand @mahmit2012 approach

    • 2 years ago
  27. satellite73 Group Title
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    now i get it, maybe i can remember is @alienbrain do not multiply out, you will make an algebra mistake, it is a pain we know \(6+B=5\) and so \(B=-1\)

    • 2 years ago
  28. mahmit2012 Group Title
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    |dw:1344137486006:dw|

    • 2 years ago
  29. satellite73 Group Title
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    @mahmit2012 my less sophisticated method was to visualize the coefficient of \(x^2\) as \(6+B\)

    • 2 years ago
  30. sauravshakya Group Title
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    @satellite73 are u sure your answer is correct.... because I got a=15 b=-10 and c=1

    • 2 years ago
  31. satellite73 Group Title
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    yeah i am sure

    • 2 years ago
  32. mahmit2012 Group Title
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    |dw:1344137568037:dw|

    • 2 years ago
  33. satellite73 Group Title
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    yes i see thanks i didn't understand that you multiplied on both sides

    • 2 years ago
  34. mahmit2012 Group Title
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    |dw:1344137675922:dw|

    • 2 years ago
  35. satellite73 Group Title
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    i think i would not want to use this method if i had to take the derivative repeatedly

    • 2 years ago
  36. mahmit2012 Group Title
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    this is the shortest method to get cofficiant in partial fraction problem >

    • 2 years ago
  37. alienbrain Group Title
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    back, open study lagged me out

    • 2 years ago
  38. satellite73 Group Title
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    yeah it was getting laggy hope we have worn this topic out sufficiently we have the answer and a plethora of methods

    • 2 years ago
  39. alienbrain Group Title
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    so 5=6 + B are you just combining like terms there

    • 2 years ago
  40. alienbrain Group Title
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    good thing you can't throw a math book at me:)

    • 2 years ago
  41. satellite73 Group Title
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    ok lets go back back back

    • 2 years ago
  42. satellite73 Group Title
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    we know \(A=6\)

    • 2 years ago
  43. satellite73 Group Title
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    first term on the left is \(A(x+1)^2=6(x+1)^2\) and we also have a term that looks like \(Bx(x+1)\)

    • 2 years ago
  44. satellite73 Group Title
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    whatever all the other mess is when you multiply out, the only way to get \(x^2\) is from the first term, which will give you \(6x^2\) and from the second term which will give you \(Bx^2\)

    • 2 years ago
  45. satellite73 Group Title
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    on the other hand we know we have to end up with \(5x^2\) that tells you \(6+B=5\) and so \(B=-1\)

    • 2 years ago
  46. alienbrain Group Title
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    ok, i got it... but the method seems to vary for each letter

    • 2 years ago
  47. alienbrain Group Title
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    i was hoping to avoid critical thinking and just be able to do it by a process :)

    • 2 years ago
  48. satellite73 Group Title
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    ok you can use one method if you like you can multiply this mess out and solve a 3 by 3 system, but i try to avoid that if at all possible because it is a ton of work try to look for values of \(x\) that will make everything zero obviously that will not work all the time when you have repeated factors but i find it much easier to use that method as far as i can once you have some of the constants, the others will be easier to find, even if you have a system of equations

    • 2 years ago
  49. satellite73 Group Title
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    if you have no repeated factors, the it always works

    • 2 years ago
  50. satellite73 Group Title
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    well, if they are linear

    • 2 years ago
  51. satellite73 Group Title
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    it is also very easy to make an algebra mistake when multiplying out and combining like terms, much few mistakes plugging in numbers and evaluating in any case if you were going to use "one method" you would have ended up with the equation \(A+B=5\) but think of how much easier if you know \(A=6\) already and we got \(A=6\) with almost no work

    • 2 years ago
  52. alienbrain Group Title
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    ya

    • 2 years ago
  53. satellite73 Group Title
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    ok enough partial fractions for one night i wish @eliassaab was here to show another method it has to do with taking the limit as \(x\to \infty\) which for a rational function you can pretty much eyeball, and then determining what a constant is from that

    • 2 years ago
  54. alienbrain Group Title
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    eh.. integral of 6/x ?

    • 2 years ago
  55. mathslover Group Title
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    @vishweshshrimali5

    • 2 years ago
  56. alienbrain Group Title
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    oh nm

    • 2 years ago
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