2
$\begingroup$

$$\sin{2z^2}$$ $$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$

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?

$\endgroup$
0
4
$\begingroup$

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 $R=+\infty$

$\endgroup$
8
  • $\begingroup$ What are you comparing sin(2z^2) to via the first theorem? $\endgroup$
    – Jwan622
    Nov 23 '20 at 3:34
  • $\begingroup$ What's this theorem called? $\endgroup$
    – Jwan622
    Nov 23 '20 at 3:46
  • $\begingroup$ @Jwan622 Not sure what the theorem is called. I guess you can refer to it by "composition property of power series". Whenever you have a convergent power series $\sum_{n\geq 0} a_nx^n$ with some radius of convergence $R$ and interval of convergence $I$, then when you plug in a function $h(x)\in I$ for $x$ you get another convergent power series with same radius and interval of convergence: $\sum_{n\geq 0} a_n (h(x))^n$. No comparison needed. $\endgroup$ Nov 23 '20 at 3:55
  • $\begingroup$ And you're comparing this power series with $\sin{x}$ right? $\endgroup$
    – Jwan622
    Nov 23 '20 at 4:18
  • 1
    $\begingroup$ and we know previously that the power series for sin(x) has a radius of conv = $\infty$ correct? $\endgroup$
    – Jwan622
    Nov 23 '20 at 4:38
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I didn't expect that! Thanks for teh answer $\endgroup$
    – Jwan622
    Nov 23 '20 at 5:33
  • $\begingroup$ You just are taking the limit in the last line right? This isn't Cauchy Hadamard right? You're just taking the limit of n as n goes to infty and if the limit is 0, that means the radius of convergence is infinite for any z right? $\endgroup$
    – Jwan622
    Nov 23 '20 at 5:37
  • $\begingroup$ I’m just taking the limit $\lim_{n\to\infty}[4z^2/(2n+2)(2n+3)]$. Nothing even as modern as nineteenth-century mathematics here. $\endgroup$
    – Lubin
    Nov 23 '20 at 13:24
  • $\begingroup$ I see and intuitively if this limit of these terms go to 0, then that means the sequence terms eventually goes to 0 and so it converges no matter what z we haev $\endgroup$
    – Jwan622
    Nov 24 '20 at 16:22

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.