How to prove periodicity of $\sin(x)$ or $\cos(x)$ starting from the Taylor series expansion?

Solution 1:

A rough sketch for (a) could be

1. The power series converge everywhere.
2. $\sin(x)^2+\cos(x)^2=1$. (As noted below, it is simpler to do this after step 3.)
3. $\frac{d}{dx}\sin(x)=\cos(x)$ and $\frac{d}{dx}\cos(x)=-\sin(x)$.
4. The differential equation $\frac{d^2y}{dx^2} = -y$ determines $y$ for all $x$ if you know $y$ and $\frac{dy}{dx}$ at any one particular $x$.
5. There is a positive $\theta$ such that $\cos(\theta)=0$ and $\sin(\theta)=1$.
6. $\cos(\theta+x)=-\sin(x)$ for this particular $\theta$.
7. cos and sin both have period $4\theta$.

For part (b), you have to determine the period numerically in general. Of course the answer is $2\pi$, but proving this depends on what your definition of $\pi$ is. A popular definition is that $\pi$ is simply twice the smallest positive $\theta$ such that $\cos(\theta)=0$, in which case period $2\pi$ is just a tautology.

Solution 2:

Some ideas can be found in Rudin's Real and Complex Analysis, first chapter about the exponential function.

The exponential is defined to be

$$e^t=1+t+\frac{t^2}{2!}+...$$

Using the Taylor representation for $\sin, \cos$ it is easy to see that $e^{it}=\cos t+i\sin t$, so to prove that $\cos, \sin$ are periodic it is enough to prove that $t\mapsto e^{it}$ is periodic, and for this is enough to prove that there exists $t_0>0$ with $e^{it_0}=1$. Such a $t_0$ is $2\pi$ (since $\sin 2\pi=0$ and $\cos 2\pi=1$) and we are done.

If we don't know anything about the existence of $\pi$ and the values of $\sin,\cos$ in $2\pi$, then we can proceed as follows:

$\cos 0=1$; this is obvious from the series. $\cos 2< 1-\frac{4}{2}+\frac{16}{24}=-\frac{1}{3}$ (the inequality is easy to prove). From the series we can see that $\cos$ is a continuous function and therefore there exists a smallest $t_0>0$ with $\cos t_0=0$. Define $\pi=2t_0$. $|e^{it}|=1$ for every $t$ since $e^{-it}=\overline{e^{it}}$ and $e^{a+b}=e^a\cdot e^b$.

Then $\sin t_0 \in \{-1,1\}$ and since $\sin 't =\cos t>0$ on $(0,t_0)$ we deduce that $\sin t_0=1$. Therefore $e^{i \cdot \pi/2}=i$, and this means $e^{2\pi i}=1$.

Solution 3:

GH Hardy sketches a proof as follows (Hardy, Pure Mathematics, Section 224)

1. Use the sequences, which are absolutely convergent for all real $x$ to demonstrate the addition formulae (e.g. $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$)

   He then says "The property of periodicity is a little more troublesome."

2. Prove from the series for cosine that it changes sign just once in the interval (0,2) [I think this is really the key insight in the proof]

3. Call the zero $\pi/2$ and show that $\sin(\pi/2)=1$, $\cos(\pi)=-1$, $\sin(\pi)=0$

4. Use the addition formulae to establish periodicity.

Hardy cites a full proof in Whittaker and Watson's Modern Analysis, Appendix A.

Solution 4:

Start from the end of Beni Bogosel's first paragraph: We must find a positive $t$ such that $e^{it}=1$. Go through the usual proof that $$ \sum_{k=0}^{\infty} \frac{z^k}{k!} = \lim_{n\to\infty} \left(1+ \frac{z}{n}\right)^n$$

So $e^{it} = \displaystyle\lim_{n\to\infty} \left( 1+\frac{it}{n} \right)^n$. Since $1\leq \displaystyle\left(1+\frac{t^2}{n^2}\right)^n \leq \left(e^{t^2}\right)^{1/n}\to 1$, $|e^{it}|=1$.

Define $\rm si(x), co(x), ta(x) $ to be the usual trigonometric functions defined by points on the unit circle. Draw a picture to convince yourself that $\rm si(x) \leq x \leq ta(x) $ for small positive $x$, and $\rm ta(x) \leq x \leq si(x)$ for small negative $x$. Argue via squeeze theorem that $\rm ta(x) \sim x $ as $x\to 0$.

Now note $\arg(e^{it}) = \displaystyle\lim_{n\to\infty} \rm n\cdot ta^{-1}(t/n) = t$ . Thus, $e^{it}$ is the point on the unit circle whose argument is $t$, so has period equal to the circumference of the unit circle, $2\pi$.

Note that we don't have to depart from our geometric definition of $\pi$, instead it arises naturally. As freebies, we get that $\sin x = \rm si(x)$ and $\cos x = \rm co(x)$, i.e these power series definitions agree with the historical definition with triangles.