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Expansion of a Taylor Series

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Expand f(x) = x/(x+1) in a taylor series about a=10.

    2. Relevant equations

    f(x) = Ʃ (f^n(a)*(x-a)^n / n!

    3. The attempt at a solution

    I'm having a hard time arriving at the correct answer..I think I'm definitely getting lost somewhere along the way. Here's what I've got so far:


    I started by computing the derivatives.
    f'(x) = 1/(x+1)^2

    f''(x) = (-2(x+1))/(x+1)^4

    Then evaluating each at 10:

    f'(10) = 1/121

    f''(10) = -22/14641

    and f(10) = 10/11

    Then, using the above equation,

    10/11 + 1/121 * (x-10) + (-22/11^4 * (x-10)^2)/2 + ....

    This doesn't really take me in the right direction at all, though. I know x/1-x is near the form 1/1-x which I need for a power series expansion. Should I be trying to represent it as such?

    Hope this is clear! I'm quite confused!
     
  2. jcsd
  3. Apr 19, 2012 #2

    micromass

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    Do you know the series expansion of

    [tex]\frac{1}{1-y}[/tex]

    ??

    Try to write your function in that form.
     
  4. Apr 19, 2012 #3
    Ok, I had a feeling that was the direction I should be heading in. So, writing it in that form, I get

    1/1 - (-1/x)

    and so my series becomes from n=0 to infinity, (-1/x)^n

    Am I on the right track at least?
     
  5. Apr 19, 2012 #4

    micromass

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    Maybe I'll do an instructive example:

    [tex]\frac{1}{x+10}=-\frac{1}{-10-x}=-\frac{1}{-18-(x-8)}=\frac{1}{18}\frac{1}{1-(x-8)/(-18)}=\frac{1}{18}\sum \frac{1}{(-18)^n}(x-8)^n[/tex]

    I hope I didn't make any typos. But that's basically it.
     
  6. Apr 19, 2012 #5
    Ok, I follow all of that except where you factor out 1/18. Shouldn't the 1 in the denominator, 1-(x-8)/(-18) be negative?

    And then your series emerges from the fact that 1/1-x = summation x^n, where -(x-8)/-18 is your x, correct?

    So, that's kinda what I tried...here it is step by step...

    x/1+x = 1/(1/x+1) = 1/(1- (-1/x)

    And my x for the series is (-1/x)
     
  7. Apr 19, 2012 #6

    micromass

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    OK, I might have made some mistakes in my post, but you get the point I see.

    But what you did is NOT allowed at all. You did [itex]\frac{1}{x+1}=\frac{1}{x}+1[/itex] which is very, very, very wrong!!

    What you should do is forget the x in the numerator for a second and try to make something out of

    [tex]\frac{1}{1+x}[/tex]

    We'll worry about the numerator later.
     
  8. Apr 19, 2012 #7
    Hm, I'm confused. I just factored an x out of the denominator, and canceled it with the numerator, but that's not ok?

    Ok, but ignoring the numerator for now...it would just be

    1/1-(-x), and so for my series, x = -x ?


    Thanks for the help by the way! :) I really appreciate it.
     
  9. Apr 19, 2012 #8

    micromass

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    Your second step is not ok.

    Remember that you would like to expand the series around a=10.
    What you're doing now is fine, except that your expansion is around a=0 now.
     
  10. Apr 19, 2012 #9
    Ahh! That's right!

    So in that case,

    Ʃ(-x-10)n
     
  11. Apr 19, 2012 #10

    micromass

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    No, that's not correct.
     
  12. Apr 19, 2012 #11
    Ah, ok, in that case, I'm lost. For taylor series, the derivative is involved, but I don't see what I'd be taking the derivative of? And I know that Taylor series is summation of f'(a)(x-a)/n!, but I'm not sure how to get this into those terms.

    Edit// I take that back...I don't think the derivative is involved at all, and this is just a straight-forward power-series. Although, I'm still not sure how to represent it in terms of a. Does my a affect the original function immediately, or do I incorporate it as part of the series later?
     
    Last edited: Apr 19, 2012
  13. Apr 19, 2012 #12
    I've been able to make sense of my teacher's solution up to a point...

    He starts with,

    ((x-10)+10)/((x-10) + 11)

    Now, I can see how in doing that, he hasn't changed the original problem at all, and he has incorporated the a=10 right away. But...I don't really see how exactly the a is being incorporated, other than arbitrarily being thrown into the function? It could just as easily have been (x-5 + 5)/(x-5+6) for that matter, right?

    From there, he breaks the above up into a sum of two fractions, and then sets (x-10) = u. And from there he proceeds to set up a series, but I don't even understand the aforementioned step, so I can't quite figure out what's going on with the series.
     
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