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I was requested to find the Maclaurin series of $\sin(5x^2)$. I attempted to find the derivatives of this function in hopes of finding a pattern. However, the derivatives become more and more complicated, without any pattern emerging —not to my eyes, at least. For example, the first derivatives are

  • $f^{1}(x)=10x\cos(5x^2)$
  • $f^{2}(x)=10(-10x^2\sin(5x^2) + \cos(5x^2))$
  • $f^{(3)}(x)= 10(-100x^3\cos(5x^2)-30x\sin(5x^2))$

The third derivative is already ugly, let alone the fourth. When evaluated at $0$, they give $f^{1}(0)=0, f^{2}(0)=10, f^{3}(0)=0, f^{4}(0)=0,...$ and I didn't go beyond that because derivatives got too nasty already.

Just to clarify, it is not laziness what stops me when facing an "ugly" expression. If it must be computed, fair enough. But I have a strong feeling I'm making this harder than it should be. Maybe there is a method to find the Maclaurin expansion for this series that I'm unaware of. My hunch tells me I should simply use the Maclaurin expansion of $\sin x$:

$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}+...$$

However, I does not seem justifiable to simply let $u=\sin 5x^2$ and $\sin u=u-\frac{u^3}{3!}+\frac{u^5}{5!}+...$ because the coefficients of this expansion are the $n$th derivatives of $\sin$ with respect to $x$, not $u$.

In short, I am quite lost and I've runned out of ideas regarding this expansion. Any help is much appreciated.

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3 Answers 3

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Good observation that the coefficients in the expansion of $\sin x$ depends on the derivatives of $\sin x$, whereas the coefficients of the series for $\sin 5x^2$ should in general depends on the derivatives of $\sin 5x^2$. In short, the reason why it works is the following.

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

This series is valid for any fixed real $x$ (the radius of convergence is infinite). So if I choose a fixed $x$, I should expect that the value of $\sin x$ will be correctly given by the series on the right. Moreover, suppose I decide I want to compute $\sin 5x^2$. Since $5x^2$ is also some real number, and the above formula is true for all real numbers, then it should be true that

$$\sin (5x^2) = (5x^2) - \frac{(5x^2)^3}{3!} + \frac{(5x^2)^5}{5!} - \frac{(5x^2)^7}{7!} + \cdots$$

and indeed, it is true. Why this is not a cheat: if you expand and simplify, you will find that the correct series for $\sin 5x^2$ is given by

$$5x^2 - \frac{125x^6}{3!} + \frac{3125x^{10}}{5!} - \frac{78125x^{14}}{7!} + \cdots$$

This implies something about the derivatives of $f(x) = \sin 5x^2$. In particular, $f(0) = 0$ and $f'(0) = 0$, but $f''(0) = 10$ (as $10/2! = 5$). Then the derivatives of order 3 through 5 all evaluate to zero at $x = 0$, while $f^{(6)}(0) = -125 \cdot 6!/3!$, and so on. This isn't a cheat because we don't know the value of these derivatives at any other point, just at zero. It's not like we've been given a formula for $f'(x)$, $f''(x)$, and so on, but the series above implies these particular values for those unknown functions, at one particular point.

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  • $\begingroup$ I mark your answer as correct because it specifically adresses my particular doubt regarding the use of substitution, instead of simply stating that substitution is the way to go. A very clear answer all around. Thank you. $\endgroup$
    – lafinur
    Commented Oct 4, 2022 at 23:10
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    $\begingroup$ So, to complete the picture, should the expansion of $\sin x$ not be valid for all $x \in \mathbb{R}$, such substitution trick would not be valid but for some restricted set of $x$? In this case, al $x$ such that $5x^2$ belong to the set of $\mathbb{R}$ for which the expansion holds? $\endgroup$
    – lafinur
    Commented Oct 4, 2022 at 23:12
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    $\begingroup$ You are right; $\sin$ has an infinite radius of convergence, but if a function $f(x)$ had a radius of convergence $|x| < r$, then you would expect the substitution $f(5x^2)$ to only work when $|5x^2| < r$, or equivalently $|x| < \sqrt{r/5}$ $\endgroup$
    – Rob
    Commented Oct 4, 2022 at 23:16
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Take as given that $$ \sin(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{(2n+1)!} = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \dotsb. $$ Let $t = 5x^2$. Then \begin{align} \sin(5x^2) &= \sin(t) \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{(5x^2)^{2n+1}}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{5^{2n+1}x^{4n+2}}{(2n+1)!} \\ &= 5x^2 - \frac{125 x^6}{3!} + \frac{3125 x^{10}}{5!} - \frac{5^7 x^{14}}{7!} + \dotsb. \end{align}

More generally, suppose that $f$ is a function which is analytic at zero (i.e. it has a Maclaurin series expansion—note that this can be extended to a function which is analytic at any point via a translation), and let $r > 0$ denote the radius of convergence of the series $$ f(t) = \sum_{n=0}^{\infty} a_n t^n. $$ For any $t = u(x)$, it will follow that $$ f(t) = f(u(x)) = \sum_{n=0}^{\infty} a_n u(x)^n, $$ so long as $|u(x)| < r$. There really isn't a lot to prove here: by assumption, the value of $f(t)$ is given by the power series for any $|t| < r$. So fix some $x$ such that $|u(x)| < r$, and the claimed result follows.

In the case of $\sin(t)$, the radius of convergence of the Maclaurin series is infinite (the Maclaurin series converges for all $x \in \mathbb{R}$ (or $\mathbb{C}$, for that matter)). Hence $$ \sum_{n=0}^{N} (-1)^n \frac{(5x^2)^{2n+1}}{(2n+1)!} \to \sin(5x^2)$$ for all $x \in \mathbb{R}$.

For an example of a function which does not have a Maclaurin series which is always convergent, consider $t \mapsto \log(t+1)$. The Maclaurin series is given by $$ \log(t+1) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{t^n}{n},$$ and converges absolutely on the interval $(-1,1)$ (the radius of convergence is $1$). From this, it follows that (for example) $$ \log(5x^2) = \log((5x^2-1)+1) = \sum_{n=1}^{\infty} (-1)^n \frac{(5x^2-1)^n}{n}, $$ which will converge as long as $$ |5x^2 - 1| < 1 \iff 0 < |x| < \sqrt{\frac{2}{5}}. $$

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Hint: $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$. Substituting $5x^2$ for $x$ in the formula to get the desire series.

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