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Math Help - Series question

  1. #1
    Junior Member
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    Berkeley, California
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    Lightbulb Series question

    I am having trouble solving a problem involving a series.

    The problem is the following:

    x \displaystyle\sum_{n=1}^{\infty}\ na_n x^{n-1} + \displaystyle\sum_{k=0}^{\infty}\ a_k x^k

    The problem wants me to rewrite the given expression as a sum who generic term involves x^n

    For some reason I can't get the same answer that is in the back of the book. I especially don't know what to do when I have the variable "n" with the variable "k". I've tried substituting (n+1) in the place of "n" but that still does work. That method worked in similar problems, so I feel I'm missing out on a subsequent step.

    Can anyone please explain thi s to me?

    Thank you!
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  2. #2
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    Anchorage, AK
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    Two Hints

    First, what happens when you use the distributive property on the product x\cdot\sum_{n=1}^{\infty}na_nx^{n-1}?

    Second, what is the value of na_nx^n when n=0?

    --Kevin C.
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  3. #3
    Member
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    didnot understand

    Quote Originally Posted by TheBerkeleyBoss View Post
    I am having trouble solving a problem involving a series.

    The problem is the following:

    x \displaystyle\sum_{n=1}^{\infty}\ na_n x^{n-1} + \displaystyle\sum_{k=0}^{\infty}\ a_k x^k

    The problem wants me to rewrite the given expression as a sum who generic term involves x^n

    For some reason I can't get the same answer that is in the back of the book. I especially don't know what to do when I have the variable "n" with the variable "k". I've tried substituting (n+1) in the place of "n" but that still does work. That method worked in similar problems, so I feel I'm missing out on a subsequent step.

    Can anyone please explain thi s to me?

    Thank you!
    i didnot get you. first what is meant by generic term? it would be helpful if you could cite a simple example of a similar problem.
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  4. #4
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    Berkeley, California
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    If I factor the "x" in using the distributive property, I get the following result:

    \displaystyle\sum_{n=0}^{\infty}\ na_n x^{n} with the index of summation set equal to zero, correct? If n=0 in the above variables, then the whole thing is equal to zero?

    I'm still confused. I know the answer is supposed to be (n+1)a_n x^{n}. I can get everything to look right by using n = (n+1) except for the a_n part. I figured it would be a_{n+1} but it is not. I'm guessing the a_k x^k has something to do with it.
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  5. #5
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    Anchorage, AK
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    My first hint was that x\cdot\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=1}^{\infty}x\cdot{n}a_nx^{n-1}=\sum_{n=1}^{\infty}na_nx^n.

    The second was to note that
    \sum_{n=0}^{\infty}na_nx^n=0a_0+1a_1x+2a_2x^2+\dot  s
    =0+1a_1x+2a_2x^2+\dots
    =1a_1x+2a_2x^2+\dots
    \sum_{n=0}^{\infty}na_nx^n=\sum_{n=1}^{\infty}na_n  x^n
    Since the n=0 term (not the entire sum, just that one term) is zero, we can add it to the series without changing the value of the series.
    Thus, we go from
    x\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{k=0}^{\infty}a_kx^k
    to
    \sum_{n=1}^{\infty}na_nx^n+\sum_{k=0}^{\infty}a_kx  ^k
    to
    \sum_{n=0}^{\infty}na_nx^n+\sum_{k=0}^{\infty}a_kx  ^k
    Now, just rename the variable "k" to "n"
    \sum_{n=0}^{\infty}na_nx^n+\sum_{n=0}^{\infty}a_nx  ^n,
    and combine the sums, grouping terms of the same n
    \sum_{n=0}^{\infty}\left(na_nx^n+a_nx^n\right),
    =\sum_{n=0}^{\infty}(n+1)a_nx^n.
    Does that make sense?

    --Kevin C.
    Last edited by TwistedOne151; March 22nd 2010 at 12:28 PM. Reason: typo: mising latex backslash
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