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$$\sin{2z^2}$$ $$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$
Let's replace z with $2z^2$ in the MacLaurin series:
$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$
$$= z^2 - \frac{2^3}{3!}z^6 + \frac{2^5}{5!}z^{10} -+ ... = z^2 - \frac{4}{3}z^6 + \frac{4}{15}z^{10} -+...$$
But what about the radius of convergence? How do I find this?
You can use the following theorem: If $h(x)$ is a function such that $h(x)\in I$ (where $I$ is defined below to be the interval of convergence of the Maclaurin series for $f(x)$ and if we have a Maclaurin series expansion for a function $f(x)$ (and which converges to $f(x)$), given by $f(x)=\sum_{n\geq 0}a_n x^n$ with radius of convergence $R$ and interval of convergence $I$ then the function $f(h(x))$ will have Maclaurin series expansion $$f(h(x))=\sum_{n\geq 0}a_n (h(x))^n$$ with same radius of convergence and same interval of convergence. Furthermore this Maclaurin expansion will converge to $f(h(x))$.
Hence, $$\displaystyle sin(2z^2)=\sum_{n\geq 0} (-1)^n\frac{2^{2n+1}}{(2n+1)!}z^{4n+2}$$ has radius of convergence $R=+\infty$ and interval of convergence $(-\infty, +\infty)$.
Alternatively if you forgot this theorem, you can redo all the work and directly use the ratio test to find the radius and interval of convergence of the composed Maclaurin series. Consider $$\displaystyle \lim_{n\to \infty} \Bigg |\frac{\frac{2^{2(n+1)+1}}{(2(n+1)+1)!}x^{4n+6}}{\frac{2^{2n+1}}{(2n+1)!}x^{4n+2}}\Bigg\vert =\lim_{n\to \infty} \frac{2^{2n+3}}{(2n+3)!}\frac{(2n+1)!}{2^{2n+1}}\bigg|\frac{x^{4n+6}}{x^{4n+2}}\bigg|$$ You're then left with: $$\lim_{n\to \infty} \frac{2^3}{(2n+2)(2n+3)}\vert x^4\vert$$
But the limit above is $0$ regardless of what $x$ can be and this shows that by the ratio test we have absolute convergence, hence convergence for all $x\in \mathbb{R}$ and the radius of convergence is $R=+\infty$
Isn’t it quite easy? Start with the standard Laurent series $$ \sin(z)=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\,, $$ make your substitution $z\to2z^2$, and get $$ \sin(2z^2)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}z^{4n+2}}{(2n+1)!}\,, $$ in which your $a_n$-term is $(-1)^n2^{2n+1}z^{4n+2}/(2n+1)!$. Now form $\mid a_{n+1}/a_n\mid$ to get $$ \left|\frac {a_{n+1}}{a_n}\right|=\frac{2^{2n+3}z^{4n+6}\big/(2n+3)!}{2^{2n+1}z^{4n+2}\big/(2n+1)!} =\frac{(2n+1)!}{(2n+3)!}\frac{2^{2n+3}z^{4n+6}}{2^{2n+1}z^{4n+2}}=\frac{4z^4}{(2n+2)(2n+3)}\,. $$ That rightmost thing clearly goes to zero, no matter what $z$ is, so your radius is infinite, as I’m sure you expected.
