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Complex Taylor Expansion

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Develop the Taylor expansion of ln(1+z).

    2. Relevant equations

    Taylor Expansion: f(z) = sum (n=0 to infinity) (z-z0)n{f(n)(z0)}/{n!}

    Cauchy Integral Formula: f(z) = (1/(2*pi*i)) <<Closed Integral>> {dz' f(z')} / {z'-z}

    3. The attempt at a solution

    I have NO idea how to start this problem. I know what a Taylor Series is, but I'm not sure how to apply that idea here...

    Do I just plug and chug into the Taylor Series expression with z0 = 0? If so, what am I doing with the f(n)(z0) stuff? We've done a bunch of stuff with residues in class, but I just can't see how all that is relating. There are several more problems in this section, and I haven't the slightest how to start them!! I'm hoping if I can get this one figured out, then I can extrapolate the method to the other problems (even though they're binomial and Laurent expansion).

    Can anybody help?! Thank you all so SO much in advance!
     
  2. jcsd
  3. Mar 1, 2009 #2

    Redbelly98

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    Do you realize that f(n)(z0) means the nth derivative of f(z), evaluated at z=z0?

    I.e.,
    f(1)(z0) means f'(z) at z=z0
    f(2)(z0) means f''(z) at z=z0
    etc.
     
  4. Mar 1, 2009 #3
    Yes, I realize that.

    in this case, f(z)=ln(1+z), f'(z) = (1+z)^(-1)*z', f''(z) = -(1+z)^(-2)*z' + (1+z)^-1*z'', by the chain rule, right?

    I understand the notation... But how is it that I "develop" the expansion?
     
  5. Mar 1, 2009 #4
    So is it blandly: f(z) = sum (n=0 to infinity) zn{f(n)(0)}/{n!}?
     
  6. Mar 1, 2009 #5
  7. Mar 1, 2009 #6
    Hang on, I'm still having a problem, can someone help? I'm getting:

    f(0) = ln(1) = 0
    f'(0) = 1/(1+0) = 1
    f''(0) = -1/(1+0)^2 = -1
    f'''(0) = 2/(1+0)^3 = 2
    f''''(0) = -3/(1+0)^4 = -3
    (and so on...)

    I can't quite figure out how to get it into the form sum(n=1 to infinity) [(-1)(n-1) ]*[(zn)/n], though, because I've got a (-1) and a (+1) for the first two terms...

    So I'll have:

    f(z) = z0f(0)(0) / 0! + z1f(1)(0) / 1! + z2f(2)(0) / 2! + z3f(3)(0) / 3! z4f(4)(0) / 4! + ....
    f(z)= 1*f(0) / 1 + z*f'(0)/1 + z2f''(0)/2 + z3f'''(0)/6 + z4f''''(0)/24 + ....
    f(z) = 1 * 0 / 1 + z * 1 / 1 + z2* (-1) / 2 + z3 * 2 / 6 + z4*(-3)/24 + ...
    f(z) = z - (1/2)z2 + (1/3)z3 - (1/8)z4 + ....

    I'm not seeing any way that I can get this into the correct form, so I must've messed up my differentiation??
     
  8. Mar 1, 2009 #7

    Redbelly98

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    Redo f''''(z), it's not quite right.
     
  9. Mar 1, 2009 #8
    Should be:
    f''''(0) = -6/(1+0)^4 = -6

    And that resolved the problem! Excellent, thank you for finding my error!! ^_^
     
  10. Mar 1, 2009 #9

    Redbelly98

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    You're welcome :smile:
     
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