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satellite73

  • 2 years ago

partial fraction madness, continuation

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  1. satellite73
    • 2 years ago
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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. mathslover
    • 2 years ago
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    in terms of a and c and x ?

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

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

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

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

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

  8. satellite73
    • 2 years ago
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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

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

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

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

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

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

  14. satellite73
    • 2 years ago
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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

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

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

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

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

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

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

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

  22. satellite73
    • 2 years ago
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    wowee zowee

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

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

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

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

  27. satellite73
    • 2 years ago
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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 \(B=-1\)

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

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

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

  31. satellite73
    • 2 years ago
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    yeah i am sure

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

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

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

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

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

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

  38. satellite73
    • 2 years ago
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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

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

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

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

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

  43. satellite73
    • 2 years ago
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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 \(Bx(x+1)\)

  44. satellite73
    • 2 years ago
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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 \(Bx^2\)

  45. satellite73
    • 2 years ago
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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 \(B=-1\)

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

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

  48. satellite73
    • 2 years ago
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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

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

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

  51. satellite73
    • 2 years ago
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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

  52. alienbrain
    • 2 years ago
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    ya

  53. satellite73
    • 2 years ago
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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

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

  55. mathslover
    • 2 years ago
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    @vishweshshrimali5

  56. alienbrain
    • 2 years ago
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    oh nm

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