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Homework Help: Another power series problem

  1. Nov 30, 2004 #1
    This one involves differentiation and integration of a power series.

    I need to find the first derivative, second derivative, and integral of this.

    [tex]f(x)=\sum(\frac{x}{2})^n[/tex]


    if [tex]\frac{d}{dx}f(x)=\sum na_n(x-c)^{n-1}[/tex]

    I guess I'm having problems figuring out what [tex]a_n[/tex] is.

    Shouldn't it be 2^n and give

    [tex]\sum\frac{n}{2^n}(x)^n[/tex] I know this is wrong, how is the book getting
    [tex]\sum(\frac{n}{2})(\frac{x^n}{2})[/tex]

    I think if I can just figure out how to find [tex]a_n[/tex] I'll be all set, I hope
     
    Last edited: Nov 30, 2004
  2. jcsd
  3. Nov 30, 2004 #2
    Use the theorems

    [tex]\int \left( \sum f \right) = \sum \left( \int f \right)[/tex]

    and

    [tex] D \left( \sum f \right) = \sum \left( D f \right)[/tex]

    --J
     
  4. Nov 30, 2004 #3
    I totally screwed up the first function, it's correct now.

    The bottom line for me is, if this is my problem
    [tex]f(x)=\sum(\frac{x}{2})^n[/tex]
    what is [tex]a_n[/tex] and how do we know?

    Thanks a lot for any help, this stuff is really driving me crazy. The book doesn't have nearly enough examples and the solutions manual just plain sucks.
     
  5. Nov 30, 2004 #4
    You're going to need to start including the indices for the summations.

    Regardless, I don't know why you're so worried about determining an [itex]a_n[/itex]. It's not necessary and not particularly convenient, either. Just use the theorem I posted above, i.e.

    [tex]\frac{d}{dx} \left( \sum_{n=0} \frac{x^n}{2^n} \right)= \sum_{n=0}\frac{d}{dx} \left( \frac{x^n}{2^n} \right) [/tex]

    --J
     
  6. Nov 30, 2004 #5

    James R

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    What you have there is a general fact for the derivative of any power series of the form:

    [tex]f(x) = \sum\limits_0^N a_n (x - c)^n[/tex]

    Because a power series can be differentiated term-by-term, the derivative is:

    [tex]f'(x) = \sum\limits_0^N n a_n (x - c)^{n-1}[/tex]

    In your example, if you want to apply the formula mindlessly

    [tex]a_n = \frac{1}{2^n}, \qquad c = 0[/tex]
     
  7. Nov 30, 2004 #6
    Just a note, the f' series could just as well start at n = 1, since the n = 0 term contributes nothing. Your book seems to use this fact.

    --J
     
  8. Nov 30, 2004 #7
    Thanks, I think I see where I've been making most of my mistakes with power series in general, and why they have been giving me fits. In some situations I've been thinking of n as a variable and trying to treat it as such.
     
  9. Nov 30, 2004 #8

    HallsofIvy

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    To answer your original question, in
    [tex]f(x)=\sum(\frac{x}{2})^n= \sum\frac{1}{2^n}x^n[/tex]

    [tex] a_n= \frac{1}{2^n}[/tex]
     
  10. Nov 30, 2004 #9
    one more thing is still bothering me though.

    When I take the derivative of
    [tex]\sum_{n=0}\frac{d}{dx} \left( \frac{x^n}{2^n} \right) [/tex]

    I come up with
    [tex]\sum_{n=1}\left( \frac{n}{2^n} \right)\left( \frac{x}{2} \right) ^{n-1} [/tex]

    The book has
    [tex]\sum_{n=1}\left( \frac{n}{2} \right)\left( \frac{x}{2} \right) ^{n-1} [/tex]

    I can see that they would be equivalent at n=1, but the terms after that are not going to match. I think I'm still missing something important here.
     
  11. Nov 30, 2004 #10

    James R

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    [tex]\frac{d}{dx} \left( \frac{x^n}{2^n} \right) = \frac{n}{2^n}x^{n-1} = \frac{n}{2}\frac{x^{n-1}}{2^{n-1}} = \left(\frac{n}{2}\right)\left(\frac{x}{2}\right)^{n-1}[/tex]
     
  12. Nov 30, 2004 #11
    Thanks James, makes sense now. I see my mistake now.
     
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