# How does one show sin(x) is bounded using the power series?

Define the real valued function $$\sin:\mathbb{R} \rightarrow \mathbb{R}, \qquad given ~~by \qquad \sin(x) := x-\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$$

How does one show $\sin(x)$ is bounded using this definition? Note that you are not allowed to use the power series of $\cos(x)$ and try to show $\sin^2(x) + \cos^2(x) =1$ and then prove they are bounded. I want a direct proof using the power series of $\sin(x)$.

Remark: I am looking for a proof that will allow me to modify/mimic the arguments if I am given some DIFFERENT power series that also happens to be bounded (but which doesn't have all the nice properties of sin(x) and cos(x) ). That is the motivation for the question. Take the power series of $\exp(-x^2)$ for example. Why is that bounded?

• Could you use $\sin''x = -\sin x$ and then say $\sin''x\sin'x = -\sin(x)\sin'(x)$, so $(\sin'x)^2+ (\sin x)^2$ is a constant? It's almost same as Jack's answer, just avoid mentioning the name of $\cos x$ Nov 15 '14 at 12:51
• I added tag differential equation to draw attention from people familiar about the subject, due to my following comment(posted as an answer because it's too large for here) Nov 15 '14 at 13:24
• It seems to me that avoiding somewhat ad hoc methods is difficult. After all, the power series make perfect sense even when $x$ is a complex number or even a square matrix. With such extensions neither of the series of interest remain bounded. I'm sure you know all that. But it does indicate that tools from real analysis will be needed. Nov 15 '14 at 13:47
• The alternating series test shows that the power series is convergent for any (finite) x, and so it is bounded in any finite interval. Add to this a proof that the power series is periodic and it is therefore bounded everywhere (see e.g books.google.co.za/… Nov 15 '14 at 15:29
• Just a thought, but if you change a single coefficient of the power series for $\sin x$, the resulting series will fail to be bounded (for example, suppose we increase the $n^\mbox{th}$ term by $\epsilon$: the resulting series converges to $\sin x - \epsilon x^{2n+1}$, which is unbounded for all choices of $\epsilon$. In fact, this same argument holds if we change any finite number of coefficients). Wouldn't this suggest that any technique for proving $\sin$ was bounded from the power series would be difficult to extend to arbitrary power series?
– user88319
Nov 15 '14 at 19:42

The coefficient of $x^n$ is $\frac{i^n+(-i)^n}{2n!}$. (We don't really need complex numbers here, but that's a convenient way of explcitly describing the coefficient)

To ban the forbidden tricks, let us simply not use derivatives at all!

We have $$\sin 1 = 1-\frac1{3!}+\frac1{5!}\mp\ldots>1-\frac1{3!}=\frac56$$ because the summands are decreasing in absolute value. And $$\sin 4 =4-\frac{4^3}{3!}+\frac{4^5}{5!}\mp\ldots<4-\frac{4^3}{3!}+\frac{4^5}{5!}-\frac{4^7}{7!}+\frac{4^9}{9!}= -\frac{268}{405}$$ because all but the first few summands are decreasing in absolute value. By continuity, there exists a number $\pi\in(1,4)$ with $\sin\pi=0$. Then (absolute convergence justifies sum swapping) \begin{align}\sin(x+\pi)&=\sum_{n=0}^\infty \frac{i^n+(-i)^n}{2n!}(x+\pi)^n\\&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{i^n+(-i)^n}{2n!}\frac{n!}{k!(n-k)!}x^k\pi^{n-k}\\ &=\sum_{k=0}^\infty \frac{x^k}{k!}\sum_{n=k}^\infty\frac{i^n+(-i)^n}{2(n-k)!}\pi^{n-k}\\ &=\sum_{k=0}^\infty \frac{x^k}{k!}\sum_{m=0}^\infty\frac{i^{m+k}+(-i)^{m+k}}{2m!}\pi^m\\ \end{align} If $k\equiv0\pmod 4$, the inner series is $\sin\pi=0$. If $k\equiv 2\pmod 4$, it is $-\sin\pi=0$. If $k\equiv 1\pmod 4$, it is $c$, and if $k\equiv 3\pmod 4$ it is $-c$, where $c:=\sum_{m=0}^\infty\frac{i^{m+1}+(-i)^{m+1}}{2m!}\pi^m$. We conclude that $$\sin(x+\pi)=c\sin x.$$ Directly from the series, we see that $\sin$ is odd, i.e. $\sin(-x)=-\sin x$. Hence $$\sin(x-\pi)=-\sin(-x+\pi)=-c \sin(-x)=c\sin x$$ and ultimatley $$\sin x=\sin(x+\pi-\pi)=c^2\sin x$$ for all $x$. Then from $\sin(x+2\pi)=c\sin(x+\pi)=c^2\sin x=\sin x$, we see that $\sin$ is a periodic continuos function, hence bounded.

(Admittedly, this cannot be expanded to $e^{-x^2}$ in any way)

With the power series, it is obvious that: $$\frac{d}{dx}\sin x = \cos x,\qquad \frac{d}{dx}\cos x=-\sin x,$$ hence from: $$\frac{d}{dx}(\sin^2(x)+\cos^2(x)) = 2\sin(x)\cos(x)-2\sin(x)\cos(x)=\color{red}{0}$$ it follows that $\sin^2 x+\cos^2 x$ is constant and equal to $\sin^2 0+\cos^2 0 = 1$.

So neither $\sin x$ or $\cos x$ can exceed $1$ in absolute value.

As an alternative approach, notice that the De Moivre's identity: $$\cos x + i\sin x = e^{ix}$$ can be proven through series identities, hence: $$\sin x = \Im\left(e^{ix}\right)$$ gives that $\sin x$ is a $2\pi$-periodic function. Since $\sin x$ is obviously bounded on $[0,2\pi]$ as a continuos function, $\sin x$ is bounded on the whole real line.

• Isn't this using precisely the type of approach the OP says they want to avoid?
– Did
Nov 15 '14 at 13:32
• Both of them?! If one is not allowed to use anything, he cannot do anything, IMHO. Nov 15 '14 at 14:30

Termwise differentiation shows that $\sin$ satisfies the second order differential equation $$\sin''(x)=-\sin (x)\qquad(x\in{\mathbb R})\ .$$ When $0< x\leq2$ the neglected terms of $$\sin'(x)=1-{x^2\over2}+{x^4\over24}-\ldots$$ are decreasing in absolute value. It follows that $$\sin'(0)=1,\qquad \sin'(2)<1-{4\over2}+{16\over 24}=-{1\over3}<0\ .$$ Therefore there exists a $\sigma\in\ ]0,2[\$ with $\sin'(\sigma)=0$.

Consider now the auxiliary function $$u(t):=\sin(\sigma+t)-\sin(\sigma-t)\ .$$ One has $u(0)=u'(0)=0$, and $$u''(t)=\sin''(\sigma+t)-\sin''(\sigma-t)=-u(t)\qquad(t\in{\mathbb R})\ ,$$ which implies $u(t)\equiv0$ by the uniqueness theorem for IVPs. This gives $$\sin(2\sigma)=u(\sigma)+\sin(0)=0\tag{1}$$ and $$\sin'(2\sigma)=u'(\sigma)-\sin'(0)=-1\ .\tag{2}$$ From $(1)$ and $(2)$ one concludes that the auxiliary function $$v(t):=\sin t+\sin(2\sigma+t)$$ is the solution of the IVP $$v(t)+v''(t)=0, \qquad v(0)=v'(0)=0\ ,$$ and therefore vanishes identically. This shows that $$\sin(t+2\sigma)=-\sin(t)\qquad(t\in{\mathbb R})\ ,$$ from which we deduce $\sin$ has period $4\sigma$, and that $$|\sin x|\leq M:=\max_{0\leq t\leq\sigma}|\sin t|\qquad(x\in{\mathbb R})\ .$$

I guess the "general" way should be via the differential equation that the power series satisfies.

For example, if $f(x) = \sin x$, we get $f'' + f = 0$ and then we get $(f^2 + f'^2)'(x) = 0$.

If $f(x) = e^{-x^2}$, we get $f'(x)+2xf(x) = 0$(from its power series presentation) and $f(0) = 1$. Then remark that $f'(x) = -2xf(x)$, draw a graph beginning with $x=0$ and extend it to both sides, you can easily see the graph is(in an interval containing 0 for the moment) decreasing for $x>0$ and increasing for $x<0$ and the graph can never cross the line $y = 0$, because for $x >0$, $f'(x)$ and $f(x)$ have different sign and for $x<0$, $f'(x)$ and $f(x)$ have the same sign.

I guess we can't get a trick which works for all the different ODE's. Maybe there are some more general methods in ODE literature for identifying bounded solution, such as criteria for detecting the solution's periodicity. I know almost nothing about that

• I bet the OP was asking to prove that $\sin x$ is bounded on the whole real line. It is no big deal to prove that a continuous function is bounded on a closed interval. Nov 15 '14 at 12:27
• How does $\;|x|<2\;$ reflects in that inequality in the first line? Perhaps you meant $\;n\ge 2\;$ or something of the like? Nov 15 '14 at 12:27
• @Timbuc |x|<2 is used in the second line.
– Did
Nov 15 '14 at 12:29
• I just can't see it, @did...and the way things are written here doesn't make it clear, either. Nov 15 '14 at 12:32
• @JackD'Aurizio That is right, I want to show that sin(x) is bounded on the whole of R. Nov 15 '14 at 12:33