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Help finding Constants for Taylor Series

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    The Taylor expansion of ln(1+x) has terms which decay as 1/n.
    Show, that by choosing an appropriate constant 'c', the Taylor series of
    (1+cx)ln(1+x)
    can be made to decay as 1/n2
    (assuming expansion about x=0)

    2. Relevant equations

    f(x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex] f(n)(0) [itex]\frac{x^{n}}{n!}[/itex]

    3. The attempt at a solution

    I used Maple to differentiate this function and find values at x=0 for several derivative:

    f(0)(0) = 0
    f(1)(0) = 1
    f(2)(0) = 2c-1
    f(3)(0) = -3c+2
    f(4)(0) = 8c-6
    f(5)(0) = -30c+24

    f(x)=[itex]\frac{(0)x^{0}}{0!}[/itex]+[itex]\frac{(1)x^{1}}{1!}[/itex]+[itex]\frac{(2c-1)x^{2}}{2!}[/itex]+[itex]\frac{(-3c+2)x^{3}}{3!}[/itex]+[itex]\frac{(8c-6)x^{4}}{4!}[/itex]+[itex]\frac{(-30c+24)x^{5}}{5!}[/itex] ....

    This is where I'm stuck... In order to get the terms decaying as 1/n2, I get different values of c for each term...

    c0=1
    c1=1
    c2=[itex]\frac{3}{4}[/itex]
    c3=[itex]\frac{8}{9}[/itex]
    c4=[itex]\frac{15}{16}[/itex]
    c5=[itex]\frac{24}{25}[/itex]

    And I need one constant c that will do it all. Any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 21, 2012 #2

    Simon Bridge

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    Did you construct the Taylor series for ln(1+x)?
    In what way do the coefficients decay as 1/n?
     
  4. Sep 21, 2012 #3

    jbunniii

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    Try simplifying the coefficients in your Taylor series to see if you can find a trend. Also, you can avoid the laborious task of calculating derivatives if you approach this another way. Hint: try substituting the Taylor series for ln(1+x) into the expression (1+cx)ln(1+x).

    P.S. There's no point trying to set each coefficient equal to 1/n^2. You won't get the coefficients to EQUAL 1/n^2, but you should be able to get them to decay at the same rate as 1/n^2.
     
  5. Sep 22, 2012 #4
    I did the differentiation method to more easily see the trend myself.
    I know ln(1+x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex](-1)[itex]^{n+1}[/itex][itex]\frac{x^{n}}{n}[/itex]

    Also, the taylor expansion of a polynomial is just that polynomial
    So, (1+cx)ln(1+x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=1}[/itex](-1)[itex]^{n+1}[/itex][itex]\frac{(1+cx)x^{n}}{n}[/itex]

    Rearranging this, I can get a general expression for each coefficient:
    an=(-1)n+1([itex]\frac{c}{n-1}[/itex]-[itex]\frac{1}{n}[/itex])

    But I'm stuck and don't know how to go about choosing this c. I'd imagine I'm going to need an equation which somehow relates an to an+1 and then solve for c, but I don't know what to do.

    I need to figure this out and really understand it, because I also have to do the same thing to make the function
    (1+ax+bx2)ln(1+x) decay as 1/n3
     
  6. Sep 22, 2012 #5

    jbunniii

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    OK, this looks promising. Let's rearrange it a bit:
    [tex]a_n = (-1)^{n+1}\left( \frac{nc - (n-1)}{n(n-1)} \right) =
    (-1)^{n+1}\left(\frac{n(c-1) + 1}{n^2 - n}\right)[/tex]
    This should give you a pretty good idea what [itex]c[/itex] should be.
     
    Last edited: Sep 22, 2012
  7. Sep 22, 2012 #6
    So, now I'm trying to solve this expression, substituting your equation into

    [itex]\frac{1/n^2}{1/(n+1)^2}[/itex]=an/an+1

    But this will still give different C's for different n's. I don't understand how I can find one exact c that will work for all n's. I still don't understand what to do with the equations I have.
     
  8. Sep 22, 2012 #7

    jbunniii

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    You won't be able to achieve this for every [itex]n[/itex], but you don't need to. "Decays as [itex]1/n^2[/itex]" means that this is true asymptotically (in the limit).

    Suppose I take [itex]c = 1[/itex]. Then
    [tex]|a_{n}| = \frac{1}{n^2 - n}[/tex]
    Clearly if [itex]n[/itex] is large, then the [itex]n^2[/itex] in the denominator is the dominant term, so asymptotically, [itex]|a_n|[/itex] decays like [itex]\frac{1}{n^2}[/itex].

    Equivalently,
    [tex]|a_{n}| = \frac{1}{n(n-1)}[/tex]
    For large [itex]n[/itex], the distinction between [itex]n[/itex] and [itex]n-1[/itex] is negligible, so [itex]\frac{1}{n(n-1)}[/itex] is almost the same as [itex]\frac{1}{n^2}[/itex].

    Now see if you can make this precise. Hint: try to quantify the relative error between [itex]\frac{1}{n(n-1)}[/itex] and [itex]\frac{1}{n^2}[/itex].
     
    Last edited: Sep 22, 2012
  9. Sep 22, 2012 #8
    Okay, that's exactly what I needed.

    When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.

    Thanks!
     
  10. Sep 22, 2012 #9

    jbunniii

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    Right, terminology like "such and such decays as so and so" pretty much universally means "in the limit". I edited my above comment with a hint regarding how to quantify this.
     
  11. Sep 22, 2012 #10

    Simon Bridge

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    Wasn't there a clue in the expansion for ln(1+x) to that effect? Does the expansion decay exactly as 1/n or just sort-of (in the limit) like that?
     
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