What is the Taylor expansion of ln(1+z)?

[Sonolum]

Homework Statement

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

Homework 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}

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]

Do you realize that f⁽ⁿ⁾(z0) means the n^th 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]

Oh, wait... No chain rule, neh?

Thanks for this link: https://www.physicsforums.com/showthread.php?t=111961, it's the same thing! ^_^

 

[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]

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]

You're welcome