Let $\large f(z)=\sqrt{cosz}$ with the branch of the square root chosen so that $f(0)=1$.
Consider the power series expansion of $f(z)$ in powers of $z$.
Part 1) Compute the first three non-zero terms of the power series explicitly.
Part 2) What is the radius of convergence of this series?
My work:
I think that I am possibly not doing the expansion correctly.
First off, write
$$f(z) = \sqrt{cosz} = \large e^{\large \frac{1}{2}log(cos(z))}$$
To fulfill $f(0) = 1$, I think the principal branch, having $-\pi \ Arg(z) \le \pi$ works just fine.
Then my series expansion is (using the above equation, and using the exponential function)
$$f(z) = \sqrt{cosz} = e^{\large \frac{1}{2}log(cos(z))}$$
$$= \large \sum_{n=0}^{\infty} \frac{(\large \frac{1}{2}log(cos(z))^n}{n!}$$
$$= 1+ \frac {1}{2}log(cos(z)) + \frac {1}{8}log^2(cos(z)) + ... $$
Have I computed the first three terms correctly? Or, have I not said much at all with this expansion, and that I have possibly missed the point?
Thanks,
You have expanded in $\log\cos z,$ whereas the assignment is probably to expand in $z.$
If you know the series expansion of $\cos,$ then the series of its square root can be computed term by term from
$$\left(\sum_{n=0}^\infty a_nz^n\right)^2=\sum_{n=0}^\infty\left(\sum_{i=0}^na_ia_{n-i}\right)z^n$$
where the sum between brackets on the left hand side is the required expansion.
In degree zero this gives $a_0=1$ (we discard $a_0=-1$ because it represents the other branch of the square root). The next equation is $a_0a_1+a_1a_0=0,$ from which you deduce $a_1=0,$ and so on.
The radius of convergence is determined by the maximum disk around the origin that does not contain any singularities. The closest singularities to the origin occur at $z=\pm\frac\pi2.$