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Multiplication of Series

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the first three terms using multiplication of series:
    [tex]e^{x}*ln(1-\frac{x}{5})[/tex]

    2. Relevant equations
    Knowledge of Taylor Series

    3. The attempt at a solution
    Please see the attached pdf (when it gets cleared).

    Now, I know that polynomial multiplication is to be used with series... So, this is probably an error in the actual multiplication. Since I've found the series for ln(1-x/5) using the power series definition as well as the maclaurin series definition.

    I've been staring at this for the past hour and I really can't see where the mistake is.

    The answer that I came up with is:
    [tex]-\frac{x}{5}-\frac{11x^{2}}{50}-\frac{3x^{3}}{25}[/tex]

    Sincerely,

    NastyAccident
     

    Attached Files:

  2. jcsd
  3. Nov 5, 2009 #2
    So, the product of the sum of two power series A and B, with coefficients a_n, b_n in X is given by Sum from 0 to infinity of c_n x, where c_n is defined as follows (or really c_n is as follows just by multiplying it out)

    c_n = sum from 0 to n of (a_p)*(b_n-p). Why would that be true? (I'm sure you recognize this is as the definition of the product of two polynomials. Why does this definition still hold when we have an infinite series (That is, why does it still make sense.)

    Using this you should be able to easily compute these terms.
     
  4. Nov 6, 2009 #3
    So, in essence:
    [tex][\sum^{\infty}_{n=0}\left(a_{n}*x^{n}\right)][\sum^{\infty}_{p=0}\left(b_{p}*x^{p}\right)][/tex]

    [tex]\sum^{\infty}_{n=0}\left(\sum^{\infty}_{p=0}b_{p}*x^{p}\right)a_{n}*x^{n}[/tex]

    [tex]\sum^{\infty}_{n=0}\left(\sum^{\infty}_{p=0}a_{n}b_{p}*x^{p+n}\right)[/tex]

    [tex]\sum^{\infty}_{q=0}\left(\sum^{\infty}_{n+p=q}a_{n}b_{p}\right)*x^{q}[/tex]

    Thus, for my series:

    [tex]\sum^{\infty}_{q=0}\left(\sum^{\infty}_{n+p=q}\frac{1}{5^{n+1}}\frac{1}{p!}\right)*x^{q+1}[/tex]

    So, I should have the following terms:

    -x/5-x^2/(5*1*(1!))-x^2/(5^2*2*(0!))-x^3/(5^2*2*(2!))-x^3/(5^3*3*(1!))

    -x/5 - 11x^2/50 - 19x^3/1500

    Sincerely,

    NastyAccident.
     
  5. Nov 6, 2009 #4
    The formula is correct, however, I just forgot to have n = 0 while p = 2, r = 2

    So, in reality it is:
    -x/5-x^2/(5*1*(1!))-x^2/(5^2*2*(0!))-x^3/(5^2*2*(1!))-x^3/(5^3*3*(0!))-x^3/(5^1*1*(2!))

    Which simplifies to -x/5-11x^2/50-46x^3/375

    Thanks!

    Sincerely,

    NastyAccident
     
  6. Jun 23, 2010 #5
    The answer I get is:
    (-1/5)x -(11/50)x^2 - (46/375)x^3 -(29/625)x^4

    please let me know if it still matters for you

    best regards
     
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