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Trying to find all higher derivatives of $\cos(z)$, I go about using Cauchy's Inequality.

$$ \left|\ f^{(n)}(z_0) \right| \leq \frac{n!\ M_{R}}{R^{n}} $$

where $M_{R}$ is the maximum value of $\left|\ f^{(n)}(z_0) \right|$ and $R$ is the radius.

If i'm trying to determine the estimates, I believe i'm trying to solve for $M_{R}$.

For the sake of simplicity, let $ R = 1$, if then have $f(z) = \cos(z)$.

By definition, $$ \cos(z) = \frac{1}{2}(e^{iz}+e^{-iz}) $$

In my attempt of using is Cauchy's Inequality, I set it up as, $$ \left|\ f^{(n)}(z_0) \right| = \left|\ \frac{1}{2^{n}}(e^{inz}+e^{-inz}) \right| \leq n!\ M_{R} $$

When $z = 0$, everything reduces to $ M_{R} \geq \frac{2}{2^{n}n!} $.

I believe this is incorrect and i'm not sure how to go about this another way, there isn't a lot of examples involving Cauchy's Inequality as well. I just need to be steered in the right direction.

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  • $\begingroup$ $cos'=-\sin$, and $\sin'=\cos$. $\endgroup$ – David C. Ullrich Mar 2 '16 at 15:49
  • $\begingroup$ yes, that is right fundamentally. But i'm solving for ALL higher derivatives. $\endgroup$ – iron2man Mar 2 '16 at 16:04
  • $\begingroup$ Why does the answer of @DavidC.Ullrich not produce all higher derivatives by a simple induction argument? It seems you want something else. $\endgroup$ – Lee Mosher Mar 2 '16 at 16:16
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Use $$\cos^{(n)}(z)=\cos\left(z+{n\pi\over2}\right)\ .$$ On the other hand, from $$\cos(x+iy)=\cos x\cos(iy)-\sin x\sin(iy)=\cos x\cosh y+i \sin x \sinh y$$ and $\cosh^2 y=1+\sinh^2 y$ it follows that $$|\cos z|^2=\cos^2 x\cosh^2 y+\sin^2 x \sinh^2 y=\cos^2 x+\sinh^2 y\ .$$

Now put it all together.

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  • $\begingroup$ Awesome! This comes out to be very nice. Thank you! $\endgroup$ – iron2man Mar 2 '16 at 16:52

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