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Homework Help: Just couldnt figure this out = (2)

  1. Oct 31, 2008 #1
    I just don't understand why it goes like this...why 'therefore'?

    i know (x^2)f''(x)+xf'(x)+((x^2)-1)f(x)) = x(xf''(x)+f'(x)+xf(x)) - f(x), but why it goes to zero when taking (d^n/dx^n)?

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  2. jcsd
  3. Nov 1, 2008 #2


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    "It is given that xf"+ f'+ xf= 0"

    and then it concluded
    [tex]\frac{d^n}{dx^n}(x^2f"+ xf '+ (x^2-1)f)= 0[/tex]
    for any positive integer n and for x in (-r, r).

    There is clearly something missing. Where did that "r" come from? Was there something more before this?
  4. Nov 2, 2008 #3
    Sorry, i shouldve made things clearer.
    Here is everything before this. r is the radius of convergence.

    Attached Files:

  5. Nov 2, 2008 #4
    Hmmm..it's actually on a solution sheet. I stuck at that point when i tried to go through it again before midterm...

    the question is here, just in case. Thank you soooo much!

    Attached Files:

  6. Nov 2, 2008 #5
    can anyone tell me...

    If xf"+f'+xf = 0

    Is d^n/dx^n (xf"+f'+xf) = 0 ??

    cos the solution says d^n/dx^n [x(xf"+f'+f)-f] = 0...that puzzles me. --'
  7. Nov 3, 2008 #6


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    Yes, since

    [tex]\frac{d^n}{dx^n} (xf''+f'+xf)=\frac{d^n}{dx^n}(0)=0[/tex]

    ; that is, since the quantity your taking the nth derivative of is by supposition always zero...

    That puzzles me too!....

    [tex]\frac{d^n}{dx^n} [x(xf''+f'+xf)-f]=\frac{d^n}{dx^n}[(0)-f]=\frac{-d^nf}{dx^n} \neq 0[/tex]

    ... perhaps that line in your solutions manual was a typo, and it was supposed to be an[itex]x^2f(x)[/itex] instead of an [itex](x^2-1)f(x)[/itex]....are you sure you are looking at the solution to the given problem and nota different problem by accident?
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