# Complex Taylor Expansion

#### Sonolum

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!

#### Redbelly98

Staff Emeritus
Homework Helper
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.

#### Sonolum

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?

#### Sonolum

Taylor Expansion: f(z) = sum (n=0 to infinity) (z-z0)n{f(n)(z0)}/{n!}
So is it blandly: f(z) = sum (n=0 to infinity) zn{f(n)(0)}/{n!}?

#### Sonolum

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??

#### Redbelly98

Staff Emeritus
Homework Helper
Redo f''''(z), it's not quite right.

#### Sonolum

f''''(0) = -3/(1+0)^4 = -3
Should be:
f''''(0) = -6/(1+0)^4 = -6

So I'l 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*(-6)/24 + ...
f(z) = z - (1/2)z2 + (1/3)z3 - (1/4)z4 + ....
And that resolved the problem! Excellent, thank you for finding my error!! ^_^

#### Redbelly98

Staff Emeritus
Homework Helper
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