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satellite73 Group Title

partial fraction madness, continuation

  • one year ago
  • one year ago

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

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

    • one year ago
  3. satellite73 Group Title
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    @mahmit2012 method with derivatives

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

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

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

    • one year ago
  7. alienbrain Group Title
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    they have something ugly...

    • one year 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

    • one year 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

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

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

    • one year ago
  12. alienbrain Group Title
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    what's your method for B

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

    • one year 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

    • one year ago
  15. satellite73 Group Title
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    but is sure is snapp!

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

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

    • one year ago
  18. alienbrain Group Title
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    a slower... sane approach

    • one year 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

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

    • one year ago
  21. alienbrain Group Title
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    but that's not right i guess

    • one year ago
  22. satellite73 Group Title
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    wowee zowee

    • one year ago
  23. satellite73 Group Title
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    how cool is that? i love it

    • one year 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

    • one year ago
  25. sauravshakya Group Title
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    |dw:1344137128472:dw|

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

    • one year 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 \(B=-1\)

    • one year ago
  28. mahmit2012 Group Title
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    |dw:1344137486006:dw|

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

    • one year 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

    • one year ago
  31. satellite73 Group Title
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    yeah i am sure

    • one year ago
  32. mahmit2012 Group Title
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    |dw:1344137568037:dw|

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

    • one year ago
  34. mahmit2012 Group Title
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    |dw:1344137675922:dw|

    • one year 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

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

    • one year ago
  37. alienbrain Group Title
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    back, open study lagged me out

    • one year 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

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

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

    • one year ago
  41. satellite73 Group Title
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    ok lets go back back back

    • one year ago
  42. satellite73 Group Title
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    we know \(A=6\)

    • one year 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 \(Bx(x+1)\)

    • one year 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 \(Bx^2\)

    • one year 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 \(B=-1\)

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

    • one year 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 :)

    • one year 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

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

    • one year ago
  50. satellite73 Group Title
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    well, if they are linear

    • one year 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

    • one year ago
  52. alienbrain Group Title
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    ya

    • one year 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

    • one year ago
  54. alienbrain Group Title
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    eh.. integral of 6/x ?

    • one year ago
  55. mathslover Group Title
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    @vishweshshrimali5

    • one year ago
  56. alienbrain Group Title
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    oh nm

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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