Suppose f is a analytic function such that $f(0) = 1 = f'(0)$ and $/f^{(k)}(0) \in \mathbb{R} \; \forall k$

Show that $v(r,\theta) = Im(f(re^{i\theta})) = r \sin(\theta)(1+O(r))$ for small $r$.

I need to show this, but I don't believe it's true, here's why:

I know that $f(re^{i\theta}) = \sum a_kr^k(\cos(k\theta)+i\sin(k\theta)) \implies v(r,\theta) = \sum a_kr^k \sin(k\theta)$

where $a_k = \frac{f^{(k)}(0)}{k!}$.

Now, $v(r,\theta) = 1+r\sin(\theta)+\sum_{k\geq 2}a_kr^ksin(k\theta) = r\sin(\theta)\left(1+\frac{1}{r\sin(\theta)}+\sum_{k\geq 2}a_kr^{k-2}\frac{sin(k\theta)}{sin(\theta)}\right)$

Now, $\frac{1}{r\sin(\theta)}$ is not limited by $Cr$ for any constant $C$ for small r, so, how can $\frac{1}{r\sin(\theta)}+\sum_{k\geq 2}a_kr^{k-2}\frac{sin(k\theta)}{sin(\theta)} = O(r)$ ?

I'm sorry, I'm not used to the big Oh notation, regardless I don't know how that statement could be true.

  • $\begingroup$ You seem to be correct, this would be the case only if $f(0)=0$. $\endgroup$ – DinosaurEgg Oct 2 '18 at 20:52
  • $\begingroup$ What does $/f^{(k)}(0)$ mean? $\endgroup$ – zhw. Oct 4 '18 at 15:48
  • $\begingroup$ @zhw. It means the k-th derivative of f on 0. I've found out today that the I was asked to prove a false statement because of a typo of my teacher. He meant to use a lower case 'o', not a upper case one. $\endgroup$ – Matheus barros castro Oct 16 '18 at 20:31

You've made a mistake: The constant term in the expansion of $v(re^{i\theta})$ isn't $1,$ it's $0.$ Thus

$$\text {Im } f(re^{i\theta}) = r\sin \theta + a_2r^2\sin (2\theta) + \cdots.$$

To finish, you may find the inequality $|\sin (n\theta)|\le n|\sin \theta|$ useful (you can prove this by induction).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.