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Homework Help: Use Differentiation to Find a Power Series Representation for:

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    a.) f(x) =1/ ( (1+x)^2 )
    what is the radius of convergence?

    b.) Use part a.) to find a power series for

    f(x)=1/ ( (1+x)^3)

    c.) Use part b.) to find a power series for
    f(x) =x^2 /( (1+x)^3)

    2. Relevant equations
    I want to check my work.
    I used properties of functions defined by power series.

    3. The attempt at a solution
    ## \dfrac {d} {dx}\left[ \left( -1\right) \left( 1+x\right) ^{-1}\right] =\sum _{n=0}\left( -1\right) \left( -1\right) ^{n}x^{n}##

    after differentiating:
    1/ ( (1+x)^2) =SUM[ -n x^(n-1) (-1)^n ]

    ##\dfrac {d} {dx}\left[ \left( 1+x\right) ^{-2}\right] =\dfrac {d} {dx}\left[ \sum _{n=0}-nx^{n-1}\left( -1\right) ^{n}\right]##

    2/( (1+x)^3)= SUM [ n(n-1) x^(n-2) (-1)^n ], and multiplying both sides by (1/2) yields the desired fuction.

    Only multiply both sides by x^2 and the desired function is found.

    Is this correct?

    Thank you.
  2. jcsd
  3. May 14, 2012 #2
    For part a, the resulting series after differentiation is correct, assuming your lower bound of summation is n=1. Otherwise, the Taylor series is SUM[(-1)^n (n+1) x^n), and the lower bound is n=0. To find radius of convergence, I would suggest Ratio test.
    In part b, your process is correct, but the series that you use from part a is incorrect (see above). Also, both sides should be multiplied by -1/2.
    In part c, the process is correct. It may be easier to check your work with series using http://wolframalpha.com. Make sure you take note of the bounds of integration, since having the wrong ones changes your sum.
  4. May 14, 2012 #3
    Why do we need to change the indexing? Does it have to do with :
    f(x)=a0+a1(x-c)+ a2(x-c)^2 + a3(x-c)^3+...=SUM{n=0 to infinity [an (x-c)^n ]}
  5. May 14, 2012 #4


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    The above line should read:

    [itex]\displaystyle \dfrac {d} {dx}\left[ \left( -1\right) \left( 1+x\right) ^{-1}\right] =\dfrac {d} {dx}\left[ \ \sum _{n=0}^{\infty}\left( -1\right) \left( -1\right) ^{n}x^{n}\ \right][/itex]​
    Which gives
    [itex]\displaystyle \left( 1+x\right) ^{-2} =\sum _{n=1}^{\infty}-nx^{n-1}\left( -1\right) ^{n}[/itex]​

    You didn't answer: "what is the radius of convergence?"
  6. May 14, 2012 #5
    You are right.
    ROC=1 , since |-x|<1,
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