How might I figure out the $n$th term, and the general Maclaurin/Taylor series, for $\sin^3 x$?

Question

I'm trying to work out the solution to finding the Maclaurin series of $\sin^3x$. While I am used to finding terms from exponential transformations, I've mainly had experience doing it with forms related to $e^x$ and $\frac{1}{1-x}$. I've gotten the first terms from this work:

$$\frac{d}{dx}(\sin^3x)(0) = 3\sin^2x\cos x = 3\sin{0}\cos{0} = 0$$ $$\frac{d^2}{dx^2}(\sin^3x)(0) = 3(\sin^2x\cos{x} -\sin^3x) = 0$$ $$\frac{d^3}{dx^3}(\sin^3x)(0) = 3\left(2\cos \left(2x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(2x\right)-3\sin ^2\left(x\right)\cos \left(x\right)\right) = 6$$ $$\frac{d^4}{dx^4}(\sin^3x)(0)= 3\left(-8\sin \left(2x\right)\cos \left(x\right)-4\cos \left(2x\right)\sin \left(x\right)+3\sin ^3\left(x\right)\right) = 0$$ $$\frac{d^5}{dx^5}(\sin^3x)(0)= 3\left(-8\left(2\cos \left(2x\right)\cos \left(x\right)-\sin \left(x\right)\sin \left(2x\right)\right)-4\left(-5\cos \left(x\right)+6\cos ^3\left(x\right)\right)+9\sin ^2\left(x\right)\cos \left(x\right)\right) = -60$$ $$\frac{d^7}{dx^7}(\sin^3x)(0)=3\left(98\cos \left(2x\right)\cos \left(x\right)-49\sin \left(x\right)\sin \left(2x\right)-324\cos \left(x\right)+408\cos ^3\left(x\right)-27\sin ^2\left(x\right)\cos \left(x\right)\right) = 546$$

Essentially I'm getting these terms after working out the equation for the first 7 derivatives. $$\frac{6}{3!}x^3-\frac{60}{5!}x^5+\frac{546}{7!}x^7+ \dots$$

I know the best way to write the series would be as $\displaystyle\left(\sum_{n=0}^\infty \frac{(-1)^nx^{1+2n}}{(1+2n)!}\right)^3$, but I'm trying to figure out a way to get a nth term as well as a simpler summation within the first degree. Is it posible, and if so, how would I go along with that? Thank you so much!

Answer 1

I made a tiny mistake in my comment, here is the correct expression:

$$\sin (x) = \frac{e^{ix}- e^{-ix}}{2i}$$

so:

$\sin^3(x) = \frac{e^{3ix} -e^{-3ix} -3(e^{ix}-e^{-ix})}{-8i} \iff -8i\sin^3(x) = e^{3ix} -e^{-3ix} -3(e^{ix}-e^{-ix})$

we just need to remember

$$e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$$

and so

$e^{3ix} - e^{-3ix} = \sum_{n=0}^{+\infty} \frac{(3ix)^n - (-3ix)^n}{n!} = \sum_{k=0}^{+\infty} \frac{2(3ix)^{2k+1}}{(2k+1)!} = 2i \sum_{k=0}^{+\infty} \frac{(-1)^{k}(3x)^{2k+1}}{(2k+1)!}$

and

$e^{ix} - e^{-ix} = \sum_{n=0}^{+\infty} \frac{(ix)^n - (-ix)^n}{n!} = \sum_{n=0}^{+\infty} \frac{2(ix)^{2k+1}}{(2k+1)!} = 2i\sum_{n=0}^{+\infty} \frac{(-1)^k(x)^{2k+1}}{(2k+1)!}$

so

$-8i\sin ^3(x) = 2i[\sum_{k=0}^{+\infty} \frac{(-1)^{k}(3x)^{2k+1}}{(2k+1)!} -3\sum_{n=0}^{+\infty} \frac{(-1)^k(x)^{2k+1}}{(2k+1)!}] = 2i \sum_{k=0}^{+\infty} \frac{(-1)^{k}[(3x)^{2k+1}-3x^{2k+1}]}{(2k+1)!}$

$$\boxed{\sin^3(x) = -\frac34 \sum_{k=0}^{+\infty} \frac{(9^k-1)(-1)^{k}x^{2k+1}}{(2k+1)!}}$$

notice that $k = 0 \implies 9^k-1 = 0$ so the term on $x^1$ is zero but after it we move on $2$.

Answer 2

I looked at this a little more, and while it'd be nice to find one singular series, it's better if I split it in 2 at first glance (taking Anne's idea).

$$\sin^3 x = \frac{3\sin{x} - \sin{3x}}{4}$$

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

$$\land \sin{3x} = \sum_{n=0}^\infty \frac{(-1)^n 3^{1+2n} x^{1+2n}}{(1+2n)!}$$

$$\therefore \sin^3 x = \frac{1}{4} \left( 3\sum_{n=0}^\infty \frac{(-1)^n x^{1+2n}}{(1+2n)!} - \sum_{n=0}^\infty \frac{(-1)^n 3^{1+2n} x^{1+2n}}{(1+2n)!} \right)$$

Looking at the first four terms for each:

$$= \frac{1}{4}\left( (3(x) - 3x) + \left(3\frac{-1x^3}{3!} - \frac{-1 \cdot 3^3 x^3}{3!}\right) + \left(3\frac{1x^5}{5!} - \frac{1 \cdot 3^5 x^5}{5!}\right) + \left(3\frac{-1x^7}{7!} - \frac{-1 \cdot 3^7 x^7}{7!}\right) + \dots \right)$$ $$= \frac{3}{4}\left( (1(x) - x) + \left(\frac{-1x^3 - -1 \cdot 3^2 x^3}{3!} \right) + \left(\frac{1x^5 - 1 \cdot 3^4 x^5}{5!} \right) + \left(\frac{-1x^7 - -1 \cdot 3^6 x^7}{7!}\right) + \dots \right)$$ which can be made into $$-\frac{3}{4} \left(\sum_{n=0}^\infty \frac{(-1)^n x^{1+2n} (9^{n}-1)}{(1+2n)!} \right)$$

Answer 3

Very much inspired by hellofriends' answer. For $n>1$,

$$\frac{d^n}{dx^n}\sin(x) = \frac{d^{n-1}}{dx^{n-1}}\cos(x) = -\frac{d^{n-2}}{dx^{n-2}}\sin(x)$$ So we have that

$$\frac{d^{2k}\sin}{dx^{2k}}(0) = 0 \qquad \text{and}\qquad \frac{d^{2k+1}\sin}{dx^{2k+1}}(0) = (-1)^{k}$$ And $\sin(x) = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$ by Taylor's theorem.

Now we use the exponential definition of $\sin$ to get $$\sin(x)^3 = \frac{(e^{ix} - e^{-ix})^3}{(2i)^3} = \frac{e^{3ix} - e^{-3ix} - 3e^{ix} + 3e^{ix}}{-8i} = \frac{3\sin(x) - \sin(3x)}{4}$$

Now we can just change the argument of our $\sin(x)$ expansion to get

$$\sin(x)^3 = \frac{1}{4}\sum_{k=0}^\infty \frac{3x^{2k+1} - (3x)^{2k+1}}{(2k+1)!}(-1)^{k} = \frac{3}{4}\sum_{k=0}^\infty \frac{(-1)^{k}(1-9^{k})x^{2k+1}}{(2k+1)!}$$

Answer 4

Another approach to add to the pile:

Start with the identities

$$ \sin^3 x = \frac{3}{4}\sin x - \frac{1}{4}\sin 3x $$ $$ \sin^{(n)}(x) = \sin(x - n\pi/2) = \cos(x)\sin(n\pi/2) + \sin(x)\cos(n\pi/2) $$

Let $f(x) = \sin^3 x$, $g(x) = \sin x$, and $h(x) = \sin 3x$. We can easily calculate

$$ h^{(n)}(x) = 3^n g^{(n)} (x) $$

Furthermore,

$$ g^{(n)} (0) = \sin \frac{n\pi}{2} $$

and so

$$ f^{(n)} (0) = \frac{3}{4} (1-3^{n-1}) \sin \frac{n\pi}{2} $$

This is correct, but you may not be satisfied because of the $\sin$ bit. Let's try breaking $n$ into even and odd cases.

When $n = 2k$, $$\sin \frac{n\pi}{2} = \sin \pi k = 0 $$ When $n = 2k+1$, $$\sin \frac{n\pi}{2} = \sin \pi \left(k + \frac{1}{2}\right) = (-1)^k$$

Putting all together, we have

$$ f^{(2k)} (0) = 0 $$ $$ f^{(2k+1)}(0) = \frac{3}{4}(1-3^{2k})(-1)^k $$

Thus,

$$ \sin^3 x = \frac{3}{4}\sum\limits_{k=0}^\infty \frac{(-1)^k(1-9^k)}{(2k+1)!}x^{2k+1} $$

Answer 5

Using the Cauchy product,

$$\begin{align*} \sin(x) &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} \\[2ex] \sin^2(x) &= \sum_{m=0}^\infty \sum_{n=0}^m \frac{(-1)^n}{(2n+1)!} \cdot \frac{(-1)^{m-n}}{(2(m-n)+1)!} x^{2m+2} \\[1ex] &= \sum_{m=0}^\infty \frac{(-1)^m}{(2m+2)!} \sum_{n=0}^m \binom{2m+2}{2n+1} x^{2m+2}\\[1ex] &= \sum_{m=0}^\infty \frac{(-1)^m}{(2m+2)!} \cdot 2^{2m+1} x^{2m+2} \\[2ex] \sin^3(x) &= \sum_{m=0}^\infty \sum_{n=0}^m \frac{(-1)^n 2^{2n+1}}{(2n+2)!} \cdot \frac{(-1)^{m-n}}{(2(m-n)+1)!} x^{2m+3} \\[1ex] &= \sum_{m=0}^\infty \frac{(-1)^m}{(2m+3)!} \sum_{n=0}^m 2^{2n+1}\binom{2m+3}{2n+2} x^{2m+3} \\[1ex] &= \sum_{m=0}^\infty \frac{(-1)^m}{(2m+3)!} \cdot \frac34 \left(3^{2m+2}-1\right) x^{2m+3} \end{align*}$$

where the finite sums are computed using the method shown here.

Answer 6

Since $ \ \sin x \ $ is an odd function, $ \ u \ = \ \sin^3 x \ $ is as well, so its Maclaurin series has the form $ \ \sum_{k=0}^\infty \ a_{2k+1} \ x^{2k+1} \ \ . \ $ We establish that $ \ u' \ = \ 3·\sin^2 x·\cos x \ \ $ and $$ \ u'' \ = \ 6·\sin x · \cos^2 x \ - \ 3·\sin^3 x \ \ = \ \ 6·\sin x·(1 - \sin^2 x) \ - \ 3·\sin^3 x $$ $$ = \ \ 6 · \sin x \ - \ 9·\sin^3 x $$ [there is an error in your expression], so we may write $ \ u'' \ + \ 9u \ = \ 6·\sin x \ \ . \ $ The Maclaurin series for sine is the familiar $$ \ \sum_{k=0}^\infty \ \frac{(-1)^k \ · \ x^{2k \ + \ 1}}{(2k \ + \ 1)!} \ \ . \ $$

Differentiating the "template series" for $ \ u \ $ twice causes what would be the first term $ \ (k \ = \ 0) \ , $ $ 2k·(2k + 1)·a_{2k+1} \ x^{2k - 1} \ \ , \ $ to vanish, so we must perform an "index shift" to express the second derivative as $$ u'' \ \ = \ \ \sum_{k=0}^\infty \ (2k + 2)·(2k + 3)·a_{2k + 3} \ x^{2k + 1} \ \ = \ \ 6·a_3·x \ + \ 20·a_5·x^3 \ + \ 42·a_7·x^5 \ + \ \ldots \ \ ; $$ this also "shifts" the general term in the series for $ \ u \ $ to be $ \ a_{2k + 3} \ x^{2k + 3} \ \ . \ $ We then seek to find the coefficients which satisfy $$ \sum_{k=0}^\infty \ (2k + 2)·(2k + 3)·a_{2k + 3} \ x^{2k + 1} \ + \ 9 \sum_{k=0}^\infty \ a_{2k+3} \ x^{2k+3} \ \ = \ \ 6 \sum_{k=0}^\infty \ \frac{(-1)^k \ · \ x^{2k \ + \ 1}}{(2k \ + \ 1)!} \ \ . $$ The first term is exceptional, since only the series for $ \ u'' \ $ has a linear $ \ (x^1) \ $ term, so we have $ \ (2·0 + 2)·(2·0 + 3)·a_3·x \ = \ 6x \ $ to match the first term $ \ 6x \ $ of the series for $ \ 6·\sin x \ \ . \ $ Hence, $ \ a_3 \ = \ 1 \ $ and there is no linear term in the series for $ \ u \ \ (a_1 \ = \ 0 \ ) \ . $

By "matching" terms of like exponents, the next few equations of this sequence are $$ 20·a_5 \ + \ 9·a_3 \ \ = \ \ -1 \ \ \ , \ \ \ 42·a_7 \ + \ 9·a_5 \ \ = \ \ \frac{1}{20} \ \ \ , \ \ \ 72·a_9 \ + \ 9·a_7 \ \ = \ \ -\frac{1}{840} \ \ \ , \ \ \ldots $$ $$ \Rightarrow \ \ a_5 \ \ = \ \ \frac{-1 \ - \ (-9)}{20} \ \ = \ \ -\frac12 \ \ \ , \ \ \ a_7 \ \ = \ \ \frac{1/20 \ + \ 9/2}{42} \ \ = \ \ \frac{91/20 }{42} \ \ = \ \ \frac{13 }{120} \ \ \ , \ \ldots \ \ . $$

The general term for the series is then given by $$ (2k + 5)!·a_{2k+5} \ \ = \ \ (-1)^{k + 1}·6 \ - \ 9·a_{2k+3}·(2k + 3)! \ \ , \ \ a_3 \ = \ 1 \ \ , $$ which, with a bit of effort, can be changed from a recursion relation into $$ a_{2k + 5} \ \ = \ \ \frac{(-1)^{k + 1}·6·\left(\frac{9^{k+2} \ - \ 1}{9 \ - \ 1} \right)}{(2k \ + \ 5)!} \ \ , $$ the terms containing a geometric series.

We thus obtain the Maclaurin series $$ \sin^3 x \ \ = \ \ x^3 \ + \ \sum_{k=0}^\infty \ \frac{(-1)^{k + 1} · 3 · (9^{k+2} \ - \ 1)}{4 · (2k \ + \ 5)!} · x^{2k \ + \ 5} \ \ , $$ or, since $ \ \frac{(-1)^0 \ · \ 3 \ · \ (9^1 - 1)}{4 \ · \ 3!} \ · \ x^3 \ = \ x^3 \ $ is consistent with the general term, we can include it in the summation through an "index shift" with $$ \sin^3 x \ \ = \ \ \sum_{k=0}^\infty \ \frac{(-1)^{k} · 3 · (9^{ k + 1} \ - \ 1)}{4 · (2k \ + \ 3)!} · x^{2k \ + \ 3} $$ $$ = \ \ x^3 \ - \ \frac12·x^5 \ + \ \frac{13}{120}·x^7 \ - \ \frac{41}{3024}·x^9 \ + \ \frac{671}{604,800}·x^{11} \ \ldots \ \ . $$

(In the approach taken here, this could be seen as the series solution to an inhomogeneous second-order linear ODE.)