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

Solution 1:

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)

Solution 2:

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.

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

Solution 4:

(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