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Power Series Differentiation

  1. May 17, 2010 #1
    Consider the power series (n=1 to infinity) [tex]\Sigma[/tex] (x^n)/(n*3^n).
    (a) Find the radius of convergence for this series.
    (b) For which values of x does the series converge? (include the discussion
    of the end points).
    (c) If f(x) denotes the sum of the series, find f'(x) as explicitly as possible.

    I know how to do parts a and b and for the radius I got 3 and the interval looks like this:
    -3≤x<3
    But I don't understand the last part (c)
    I know that since (n=0 to infinity)f(x)= [tex]\Sigma[/tex] c_n(x-a)^n then (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1)
    I was looking thru various materials and first I was thinking about finding the representation for the power series but I was not sure how this would help me here since I don't have a function to begin with just the series. Then I was thinking about differentiating by terms but I didn't know how to incorporate the radius of convergence. Finally I did something that I pretty sure is wrong:
    (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1) = [tex]\Sigma[/tex] n*(1/(n*3^n))*(x^(n-1))=(1/(3^n))*(x^(n-1))=(x^n)/(x*3^n) for -3≤x<3
    But like I said it doesn't seem correct. Any suggestions?
     
  2. jcsd
  3. May 17, 2010 #2

    vela

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    You're close. You don't want to rewrite it the way you did, though. You got to here:

    [tex]f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{3^n}[/tex]

    Now pull out one power of 1/3 so the exponents on the x and the 3 will match, and then adjust the limits of the summation so the exponent is just n instead of n-1. Hopefully, you'll recognize the series and be able to write down the sum in explicit form.
     
  4. May 17, 2010 #3
    Thank you for the quick response.

    So we have that:
    f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n) from this it follows that
    f'(x)=(n=0 to infinity)∑ (x^n-1)/(3^n-1)
    f'(x)=(n=0 to infinity)∑ (x/3)^(n-1)
    f'(x)=(n=1 to infinity)∑ (x/3)^(n)

    I got confused with limits of the summation am I correct? When you said "Hopefully, you'll recognize the series and be able to write down the sum in explicit form" are you saying that I should use the interval of convergence i.e. -3≤x<3? I guess I'm not sure how to write it explicitly, I don't know how many terms I should write down (my guess is they have to be between -3 [including -3] and 3).
     
  5. May 17, 2010 #4

    vela

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    No, that's not correct at all. If you expand the first line, you get

    [tex]\sum_{n=1}^\infty \frac{x^{n-1}}{3^n} = \frac{1}{3}+\frac{x}{3^2}+\frac{x^2}{3^3}+\cdots[/tex]

    If you expand the second line, you get

    [tex]\sum_{n=0}^\infty \frac{x^{n-1}}{3^{n-1}} = \frac{3}{x}+1+\frac{x}{3}+\frac{x^2}{3^2}+\cdots[/tex]

    Those obviously aren't equal.

    Second, you don't use the interval of convergence for anything. If you do the calculation correctly, you should recognize the type of series it is and be able to write down the sum in closed form.
     
  6. May 17, 2010 #5
    Is this kind of manipulations permissible:
    f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n) from this it follows that
    f'(x)=(n=1 to infinity)∑ (x^n-1)/(3^n-1+1)
    f'(x)=(n=1 to infinity)∑ (1/3)*((x/3)^(n-1))
    f'(x)=(n=0 to infinity)∑ (1/3)*(x/3)^(n)

    If it is then I'm not sure if I see anything special here, in other words, I don't know how to proceed to find the closed expression for this.
     
    Last edited: May 17, 2010
  7. May 17, 2010 #6

    vela

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    That's good. The series is of the form

    [tex]\sum_{n=0}^\infty ar^n[/tex]

    It's a geometric series.
     
  8. May 18, 2010 #7
    I have the following DE to be solved through series
    y'+xy=1-x+x^2

    simply I can't find the right solution, which is :
    x-1+A[1-x^2/2+X^4/(4*2!)....
    Please help!
     
  9. May 18, 2010 #8

    vela

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    Edward, you should start a new thread for your problem as it has nothing to do with this thread. Also, the forum policy is that you need to make a reasonable attempt and show your work before you will get any help.
     
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