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

Question

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 $\sin(x)$.

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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?

Answer 1

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.

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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.

Answer 2

(Old answer was deleted)

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

Answer 3

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 $$|\sin x|\leq M:=\max_{0\leq t\leq\sigma}|\sin t|\qquad(x\in{\mathbb R})\ .$$

Answer 4

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)