estimating the derivatives of $\sin z$ at $z=0$

Question

(a) Use Cauchy inequality to obtain estimate for the derivatives of $\sin z$ at $z=0$ and
(b) determine how good these estimates are

No examples are given except the proof on Cauchy Inequality. How do I work this out?

If I take $|z|=1$ as my circle, would $\displaystyle \max \left \{ \sqrt{\sin^2 x + \sinh^2 \left( \sqrt{1-x^2} \right) }\right \} n!$ be an approximation for my n-th derivative of $\sin z$ inside circle $|z| = 1 $?

a constant $M$ such taht $|f(z)|\leq M$ – Dominic Michaelis Apr 08 '13 at 11:06 — Dominic Michaelis, Apr 08 '13 at 11:06
Note that, $M$ depends on $r$. – Mhenni Benghorbal Apr 08 '13 at 11:40 — Mhenni Benghorbal, Apr 08 '13 at 11:40

Dominic Michaelis · Accepted Answer · 2013-04-08T11:47:43.610

1

At first we look at a nice definition to estimate $M$. As we know $$\sin(z)=\frac{1}{2i} \left(\exp(iz)-\exp(-iz)\right).$$

So when $|z|\leq 1 $ we have $|\sin(z)|\leq 2 e$. In fact we should have something like $$|\sin(z)|\leq \frac{1}{2}\cdot \left( e-\frac{1}{e}\right),$$ which gives you the estimate that $$|\sin'(0)|\leq 2 e $$ If you take the radius of your circle smaller, the approximation will be better. I don't know if this is what your are looking for.

The idea is that when $f$ is analytic we can write $f$ as $$f(z)=\sum_{k=0}^\infty a_k z^k$$ As we know that $$\sin(z)=\sum_{k=0}^\infty (-1)^k \frac{z^{2k+1}}{(2k+1)!}=z-\frac{z^3}{3!}\pm \dots$$

With that one we would get the estimate $$|\sin'(0)|\leq \frac{(r+\epsilon) \cdot 0!}{r}\approx 1$$ which gives a nice estimate for the first derivative.

edited Apr 08 '13 at 11:47

answered Apr 08 '13 at 11:30

Dominic Michaelis

19,935

probably!! if we take $r \to 0$ wouldn't our estimate grow? this $r^n$ at denominator is bothering me!! – Mula Ko Saag Apr 08 '13 at 11:33
If we take the same $M$ our estimate would grow (in fact it would grow freaking fast). But with a lower radius we can bound our $M$ stricter and with that we should be able to get a better approximation (but for that we need a better $M$). – Dominic Michaelis Apr 08 '13 at 11:37
looks like we need $|\sin z|$ as $z \to (0,0)$ – Mula Ko Saag Apr 08 '13 at 11:39
@MulaKoSaag added some details – Dominic Michaelis Apr 08 '13 at 11:48
I think $1!$ instead of $0!$, it seems okay for first derivative. any idea with higher derivatives? – Mula Ko Saag Apr 08 '13 at 11:55
oh right it is $1!$ sorry for that. For higher derivatives you get the same approximation but as you said the denominator will give no nice estimation. But well that is why it is called an approximation – Dominic Michaelis Apr 08 '13 at 11:59
ok ok .. thank you very much!! – Mula Ko Saag Apr 08 '13 at 12:00

score 1 · Answer 2 · answered Apr 08 '13 at 13:05

For example, by Cauchy:

$$f'(z_0)=\frac{1}{2\pi i}\oint\limits_C\frac{f(z)}{(z-z_0)^2}dz$$

witn $\,C\,$ any simple, closed path containing in its interior the point $\,z_0\,$ and wherein $\,f(z)\,$ is analytic (say, $\,C=$ the unit circle for $\,z_0=0\,\;,\;\;f(z)=\sin z\;$ in our case), and since

$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}\implies|\sin z|\le\frac{1}{2}(e^{-Im(z)}+e^{Im(z)})\le\frac{1}{2}(e+e^{-1})$$

so again by Cauchy:

$$|\sin'(0)|=\left|\frac{1}{2\pi i}\oint\limits_C\frac{\sin z}{z^2}dz\right|\le\frac{1}{2\pi}\oint\limits_C\left|\frac{\sin z}{z^2}\right|\,dz\le\frac{1}{4\pi}\oint\limits_C\frac{e+e^{-1}}{1}dz=\frac{e+e^{-1}}{2}\sim 1.86$$

estimating the derivatives of $\sin z$ at $z=0$

2 Answers2