# Computation of $\left.\dfrac{\text d^{100}}{\text dz^{100}}\cos(z^2)\right|_{z=0}$

I was solving the following contour integral, $$\oint_{|z|=r,\forall r>0}\dfrac{\cos(z^2)+\sin z}{z^{101}}\text d z=\dfrac{2\pi i}{100!}f^{(100)}(0),$$ where $$f(z)=\cos(z^2)+\sin z$$.

Sine's derivative is direct by induction: $$\left.\dfrac{\text d^{100}}{\text dz^{100}}\sin(z)\right|_{z=0}=\left[\sin(z)\right|_{z=0}=0.$$

But I'm not too sure about $$\cos(z^2)$$.

• That equality for the contour integral doesn't look right. Besides, it would be much easier to simply analyze the Taylor series for both functions. Dec 4, 2023 at 18:45
• Forgot the factorial, sorry. Dec 4, 2023 at 18:46
• Why don't you apply the MacLaurin series for $\cos z$ plug in $z^2$ and read off the coefficient at the power $z^{100}$ ? This coefficient multiplied by $100!$ will give you the derivative you are looking for. Dec 4, 2023 at 19:23

As suggested in the comments, you can solve it via Taylor series too (obvious but didn't think of that): $$\cos z^2=\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n)!}z^{4n}.$$

Thus, $$\dfrac{f^{(4n)}(0)}{(4n)!}=\dfrac{(-1)^n}{(2n)!}\iff f^{(4n)}(0)=(-1)^n\dfrac{(4n)!}{(2n)!}.$$

Finally, evaluating at $$n=25$$ in order to find the $$100^{\text{th}}$$ derivative at $$z=0$$, we get that $$f^{(100)}(0)=-\dfrac{100!}{50!},$$

meaning that the integral is equal to $$-\dfrac{2\pi i}{50!}.$$

• The same can be done with sine. Dec 4, 2023 at 19:35

Well, you can use Faa di Bruno:

$${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,1!^{m_{1}}\,m_{2}!\,2!^{m_{2}}\,\cdots \,m_{n}!\,n!^{m_{n}}}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left(g^{(j)}(x)\right)^{m_{j}}},$$ where the sum is over $$m_1+2m_2+\cdots n\cdot m_n = n$$ and so consider $$f = cos(z)$$ and $$g = z^2$$.

Notice that $$g^{(3)}(z)=0$$ and so the sum really goes for $$m_1+2m_2 = 100$$, now if $$m_1\neq 0$$ then $$(2z)^{m_1}=0$$, hence the only term that survives is $$m_2 = 50$$ and so

$$\frac{d^{100}}{dz^{100}}cos(z^2)=\frac{100!}{50!2^{50}}cos^{50}(z^2)2^{50}=-\frac{100!}{50!}cos(0)=-\frac{100!}{50!}$$