Taylor series for a complex function

[jimmycricket]

Homework Statement

Find the 5 jet of the following function at z=0:

$f(z) = \frac{sinhz}{1+exp(z^3)}$

Homework Equations

$\frac{1}{1-z}=\sum_{n=0}^\infty z^n$ where $z=-exp(z^3)$

The Attempt at a Solution

I have tried to multiply the series for sinhz by the series for $\frac{1}{1-(-exp(z^3))}$ but to no avail since I end up with an infinite string of z s to the power of 3. I've also tried to simply substitute the series for sinhz and exp(z^3) directly into the fraction but cannot separate the fraction to give separate terms. Any suggestions for another method?

 

[LCKurtz]

Homework Statement

Find the 5 jet of the following function at z=0:

What is a 5 jet?

 

[jimmycricket]

Sorry I should have said the 5 jet is the expansion up to and including powers of z to the fifth power and no further.

 

[Dick]

Sorry I should have said the 5 jet is the expansion up to and including powers of z to the fifth power and no further.

You can't apply the series to 1/1+exp(z^3). exp(z^3) isn't small near z=0. What does the taylor expansion of exp(z^3) look like?

 

[jimmycricket]

$f(z)=exp(z^3)$ = $\sum_{n=0}^\infty\frac{z^{3n}}{n!}$

 

[jimmycricket]

$\sum{0}^infty\z^n$

 

[jimmycricket]

edit: only just learning to use latex forget that last post

 

[Dick]

$f(z)=exp(z^3)$ = $\sum_{n=0}^\infty\frac{z^{3n}}{n!}$

Good. Now just write the terms of order <= 5 and substitute that into the denominator.

 

[jimmycricket]

$\sum_{n=0}^\infty\frac{z^{3n}}{n!}$ = $1+z^3+\frac{z^6}{2}+\frac{z^9}{6}+\frac{z^{12}}{24}$

 

[Dick]

$\sum_{n=0}^\infty\frac{z^{3n}}{n!}$ = $1+z^3+\frac{z^6}{2}+\frac{z^9}{6}+\frac{z^{12}}{24}$

Too many terms. z^6 has power greater than 5 already.

 

[jimmycricket]

$1+z^3$

 

[jimmycricket]

Do I now expand $*\sum(-1)^n(1+z^3)^n$

 

[Dick]

Do I now expand $*\sum(-1)^n(1+z^3)^n$

You can't again. (1+z^3) isn't a small number either. You need to substitute (1+z^3) for e^(3z) into 1+e^(3z) and THEN think about expanding.

 

[jimmycricket]

I end up with $\frac{1}{2}*\frac{1}{1-(-\frac{z^3}{2})}$ . Now I presume I expand $\frac{1}{2}\sum (\frac{-z^3}{2})^n$ to get $\frac{1}{2}\sum(-1)^n\frac{z^{3^n}}{2^n}$ = $\frac{1}{2}(1-\frac{z^3}{2})$. Now multiplying by the series for sinhz we get $(z+\frac{z^3}{6}+\frac{z^5}{120})(\frac{1}{2}-\frac{z^3}{4})$ = $\frac{z}{2}+\frac{z^3}{12}-\frac{z^4}{4}+\frac{z^5}{240}$

 

[Dick]

I end up with $\frac{1}{2}*\frac{1}{1-(-\frac{z^3}{2})}$ . Now I presume I expand $\frac{1}{2}\sum (\frac{-z^3}{2})^n$ to get $\frac{1}{2}\sum(-1)^n\frac{z^{3^n}}{2^n}$ = $\frac{1}{2}(1-\frac{z^3}{2})$. Now multiplying by the series for sinhz we get $(z+\frac{z^3}{6}+\frac{z^5}{120})(\frac{1}{2}-\frac{z^3}{4})$ = $\frac{z}{2}+\frac{z^3}{12}-\frac{z^4}{4}+\frac{z^5}{240}$

And that's right. Good job! Your series formula for 1/(1-z) only converges if |z|<1. You can't apply it to things that aren't small.

 

[jimmycricket]

Thank you.