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With the Taylor series representation of $\sin$ or $\cos$ as a starting point (and assuming no other knowledge about those functions), how can one:

a. prove they are periodic?

b. find the value of the period?

jprete
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Pierre Fermat
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  • The proof's somewhere in chapter 2 of Ahlfors's Complex Analysis, but I don't have my copy with me to check. – J. M. ain't a mathematician Sep 09 '11 at 13:59
  • Quite related... – J. M. ain't a mathematician Sep 09 '11 at 14:27
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    English aside: you don't know nothing else ... You mean we know nothing else or we don't know anything else . – GEdgar Sep 09 '11 at 16:31
  • @Pierre: GEdgar is correct that language purists will sneer at your double negative. But double negatives are a regular feature of many dialects of English, so don't be downhearted. GEdgar is wrong, however, in insisting on the first person plural ('we'). The second person ('you') is preferable here, because in fact we do know more about these functions. – TonyK Sep 09 '11 at 17:36
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    @Tony: in that case, "and without assuming anything else" might be a better choice of words, no? – J. M. ain't a mathematician Sep 09 '11 at 17:57
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    @Tony, linguistic peevery about double negatives is one thing -- but in mathematics we need to place greater demands on precision than ordinary language does. This is if for the pragmatic reason that we sometimes do have to utter a negated negation and have it understood as such. And, also in contrast to most nonmathematical conversation, we cannot rely on common sense to disambiguate, because we sometimes deliberately utter falsehoods for the purpose of proving things about these falsehoods. – hmakholm left over Monica Sep 09 '11 at 18:09
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    I think a proof of this is in Baby Rudin (i.e. Walter Rudin's Principles of Mathematical Analysis). Maybe in the "special functions" chapter. – Michael Hardy Sep 09 '11 at 18:32
  • @J.M.: Yes, your version is probably best. But OP's choice of pronoun was better than GEdgar's. – TonyK Sep 09 '11 at 18:34
  • This is done in the Prologue to Rudin Real and Complex Analysis. You can find a more detailed exposition of the same argument in the appendix on exp, sin and cos in Complex Made Simple. – David C. Ullrich Jan 05 '18 at 15:18

4 Answers4

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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.

hmakholm left over Monica
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    Proving 2. from the power series is a bit of work. – Mark Bennet Sep 09 '11 at 14:15
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    @Mark : I would suggest starting with proving 3 (straighforward), then differentiating $\cos^2+\sin^2$ to find that it is actually constant, and then evaluating at $0$. Another way (which I actually prefer) is to use the exponential to prove $|e^{it}| = 1$ for $t \in \mathbb{R}$, as Beni suggests. – Joel Cohen Sep 09 '11 at 14:28
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    @Joel: I like your suggestion, though, because it avoids going beyond the reals. – Mark Bennet Sep 09 '11 at 15:09
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    @Joel: I suggested the same thing for a similar problem. The periodicity follows from $\sin^2(x)+\cos^2(x)=1$ (insuring a closed path) and the flow equation $$\frac{\mathrm{d}}{\mathrm{d}x}\begin{pmatrix}\sin(x)\ \cos(x)\end{pmatrix}=\begin{pmatrix}0&1\-1&0\end{pmatrix}\begin{pmatrix}\sin(x)\ \cos(x)\end{pmatrix}$$ – robjohn Nov 09 '11 at 15:56
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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 $e^{2\pi i}=1$.

Beni Bogosel
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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.

Srivatsan
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Mark Bennet
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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.

Ragib Zaman
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