# Is $\sin: \mathbb{N} \to \mathbb{R}$ injective?

I was trying to show that $\sin(x)$ is non-zero for integers $x$ other than zero and I thought that this result might emerge as a corollary if I managed to show that the result in question is true.

I think it's possible to demonstrate this by looking at the power series expansion of $\sin(x)$ and assuming that we don't know anything about the existence of $\pi$.

All of the answers below insist that the proposition '$\exists p,q \in \mathbb{Z}, \sin(p)=\sin(q)$' -where $\sin(x)$ is the power series representation-is undecidable without using the properties of $\pi$. If so this is a truly wonderful conjecture and I would like to be provided with a proof. Until then, I insist that methods for analyzing infinite series from analysis should suffice to show that the proposition is false.

Note: In a previous version of this post the question said "bijective" instead of "injective". Some of the answers below have answered the first version of this post.

• I think what you mean is: "is $\mathrm{sin} : \mathbb{N} \rightarrow \mathbb{R}$ injective?" – goblin Jun 5 '16 at 8:21
• The biconditional you have is not true. Take, for example, $a=0, b= \pi$. The question itself, though, seems quite interesting. – MathematicsStudent1122 Jun 6 '16 at 0:00
• If $\pi$ were rational $\sin$ would be periodic on ${\mathbb Z}$. Therefore you have to know something about $\pi$ in order to prove the claim. – Christian Blatter Jun 6 '16 at 9:24
• it does not make any sense to "not know about $\pi$", I am afraid. Are you supposed then not to know about the fact that the function $2\sqrt{1-x^2}$ is integrable in $[-1,1]$, then also? – Mariano Suárez-Álvarez Jun 25 '16 at 5:16
• For example, starting from tthe definition of $\sin$ as a power series, with a lot of work one can indeed show that it is a periodic function, and if $p$ is its period, to show that $p$ is irrational. The same thing can be done if one defines $\sin$ by inverting the primitive function of $(1-x^2)^{-1/2}$, as Jacobi would do. I suggest you look in a good textbook where trigonometric functions are defined from scratch and their properties obtained from the definition used (iirc, Spivak's book on calculus does this) – Mariano Suárez-Álvarez Jun 25 '16 at 5:31

## 6 Answers

Suppose that we define the function $\def\RR{\mathbb R}f:\mathbb R\to\mathbb R$ putting $$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$ for each $x\in\mathbb R$, after checking that the power series converges for all real $x$. It is easy to compute derivatives of functions defined by power series, and therefore it is trivial to show that $$f''(x)=-f(x) \tag{1}$$ for all $x\in\RR$. We have $$\frac{d}{dx}(f(x)^2+f'(x)^2) = 2f'(x)(f''(x)+f(x))=0$$ for all $x\in\RR$, so $f(x)^2+f'(x)^2$ is a constant function. Evaluating it at $0$, we see at once that $$f(x)^2+f'(x)^2=1 \tag{2}$$ for all $x\in\RR$.

Equation (1) tells us that $f$ and $f'$ are solutions to the linear differentia equation $$u''+u=0.$$ Their Wronskian is $$\begin{vmatrix}f&f'\\f'&f''\end{vmatrix}=-(f^2+f'^2)$$ is nowhere zero, according to (2), so that $f$ and $f'$ aare linearly independent. Since the differential equation is of order $2$, we see that $\{f,f'\}$ is a basis of its solutions.

Now fix $y\in\RR$ and consider the functions $g(x)=f(x+y)$ and $h(x)=f'(x)f(y)-f(x)f'(y)$. Using (1) we find at once that they are both solutions to out differential equation, and clearly $g(0)=h(0)$ and $g'(0)=h'(0)$, so the uniqueness theoreem of solutions to initial value problems tells us that $g$ and $h$ are the same function, that is, that $$f(x+y)=f'(x)f(y)-f(x)f'(y) \tag{3}$$ for all $x$ and, s ince $y$ was arbitrary, for all $y$. Compuing the derivative with respect to $y$ in this equation an using (1), we find that also $$f'(x+y)=f'(x)f'(y)+f'(x)f'(y) \tag{4}$$ for all $x$ and all $y$.

I claim that

there exists a $p>0$ such that $f'(p)=0$.

Once we have checked this, we may compute, using (3) and (4) that \begin{align} f(x+2p) &= f'(x+p)f(p)-f(x+p)\underbrace{f'(p)}_{=0}\\ &=(f'(x)\underbrace{f'(p)}_{=0}+f(x)f(p))f(p) \\ &= f(x)f(p)^2. \end{align} In view of (2) and out choice of $p$, we have $f(p)\in\{\pm1\}$, and therefore $f(p)^2=1$ and we have proved that $$f(x+2p)=f(x)$$ for all $x\in\RR$, that is, that $f$ is a periodic function and that $2p$ is one of its periods.

Let us prove our claim. To do so, suppose it is false, so that $f'(x)\neq0$ for all $x\in(0,+\infty)$. Since $f'(0)=1$ and $f'$ is continuous on $[0,+\infty)$, we see that in fact $f'(x)>0$ for all $x\in[0,\infty)$ and, therefore, that $f$ is strictly increasing on $[0,\infty)$. In view of (2), we have $f(x)\leq1$ for all $x\in[0,infty)$ so we know that the limit $\alpha=\lim_{x\to\infty}f(x)$ exists and is an element of $[0,1]$. As $f'(x)^2=1-f(x)^2$, $f'(x)^2$ converges to $1-\alpha^2$ as $x\to\infty$ and, since $f'(x)$ is positive on $[0,+\infty)$, we see that $\lim_{x\to\infty}f'(x)$ is the positive square root $\beta=\sqrt{1-\alpha^2}$.

Pick any $y\in\RR$. For all $x\in\RR$ we have that $$f(x+y)=f'(x)f(y)-f(x)f'(y).$$ Taking limits on both sides of this equality as $x\to\infty$ we find that $$\alpha=\beta f(y)-\alpha f'(y).$$ This holds for all $y$, so we may differentiate it with resprect to $y$, and we see that $$0=\beta f'(y)-\alpha f''(y)=\beta f'(y)+\alpha f(y).$$ This is absurd, as $f$ and $f'$ are linearly independent functions.

Let now $P$ be the set of positive real numbers $p$ such that $2p$ is a period of $f$. We have shown that $P$ is not empty. Let $\pi=\inf P$. If $\pi=0$, there exists a sequence $(p_i)_{i\geq1}$ of elements of $P$ converging to $0$, and then $$1=f'(0)=\lim_{i\to\infty}\frac{f(2p_i)-f(0)}{2p_i}=0.$$ Therefore $\pi$ is a positive number. Now we may use, for example, the argument of Bourbaki to show that $\pi$ is irrational. That works because our arguments above easily show that $f(x)>0$ if $x\in(0,\pi)$, which is one of the things needed, and we can do the required integration by parts using (1).

All this shows in fact that

if $f:\RR\to\RR$ is a twice differentiable function such that $f''=-f$ on $\RR$, $f(0)=0$ and $f'(0)=1$, then $f$ is periodic and its period is irrational.

What you want, follows form this.

Recall the following facts.

Fact 1. $\sin t=0$ if and only if $t=k\pi$ for some $k\in\Bbb Z$

Fact 2. $\cos t=0$ if and only if $t=k\pi+\frac{\pi}{2}$ for some $k\in\Bbb Z$

Also, recall the identity $$\sin u-\sin v=2\,\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)$$

Thus $\sin u=\sin v$ if and only if $u+v=k\pi+\frac{\pi}{2}$ or $u-v=k\pi$ for some $k\in\Bbb Z$. But the values \begin{align*} k\pi+\frac{\pi}{2} && k\pi \end{align*} are necessarily irrational for nonzero $k\in\Bbb Z$. Hence $\sin u\neq\sin v$ for $u,v\in\Bbb Z$ with $u\neq v$.

• @AidanRocke That's really a silly assumption to make: as Brian shows, it is very important to know about $\pi$ to do anything here. – Pedro Tamaroff Jun 6 '16 at 0:48
• @AidenRocke: The number $\pi$ is basically defined by its connection with $\sin$, and it's irrationality can be proved using the properties of $\sin$; why do you expect to use anything less? In principle, yes, we should be able to get lots of information out of the power series (and we can!), but your question is not just about $\sin$, it's about the relationship between $\sin$ and the positive integers. I find it entirely unsurprising that $\pi$ is intimately involved. – Will R Jun 6 '16 at 8:51
• @AidenRocke: I think you would probably just be rephrasing the same arguments. More to the point, I'm saying $\pi$ and $\sin$ are intimately connected, and a key ingredient in their relationship is the integers (since the integers are used to define periodicity). Perhaps you can come up with some general machinery that will apply to other series as well, but what makes you think they won't involve things like the least positive root of the function in question (which, for $\sin$, is $\pi$)? – Will R Jun 6 '16 at 11:25
• @AidenRocke Our knowledge of the irrationality of $\pi$ (which comes in large part thanks to the nice algebraic properties of $\sin$) far outweighs our ability to analyze very simple infinite series. We don't know whether $\sum_{n=1}^\infty 1/n^5$ is irrational, and it wasn't until 40 years ago that we discovered how to prove $\sum_{n=1}^\infty 1/n^3$ is irrational. Meanwhile series like $\sum_{n=1}^\infty 1/n^2$ and $\sum_{n=1}^\infty 1/n^4$ were known much earlier... precisely because they simplify to expressions involving $\pi$. – Erick Wong Jun 25 '16 at 3:22
• It's clear from geometry that $\sin$ has period $2\pi$. Then $\sin$ not being on injective on $\Bbb N$ would (more or less) immediately yield the rationality of $\pi$. Thus, injectivity is pretty much equivalent to $\pi$'s irrationality. – arctic tern Jun 25 '16 at 5:48

You have $\sin m=\sin n$ if and only if $m=n+2k\pi$ or $m=\pi-n+2k\pi$ (for some integer $k$.

In the first case, if $k\ne0$, you get $\pi=(m-n)/(2k)$ is rational. In the second case, $\pi=(m+n)/(2k+1)$.

Since $\pi$ is irrational, the only possibility is $k=0$ and $m=n$.

You can't prove injectivity without the knowledge that $\pi$ is irrational. Indeed, if $\pi=p/q$ with $p$ and $q$ positive integers, then $\sin(p)=\sin(p+2q\pi)=\sin(3p)$, so the function would not be injective.

Proofs of the irrationality of $\pi$ based on the power series expansion can be seen at Proof that $π$ is irrational

$\mathbb{N}$ is countable and $\mathbb{R}$ is uncountable, so there can never be a bijection

• Sorry. I didn't write the question properly the first time. – user93511 Jun 5 '16 at 8:44

It is not bijective because it is not surjective. There is not $x\in\mathbb N$ such that $\sin(x)=2$.

However, it is true that $\sin(x)=0$ for only one value of $x\in\mathbb N$. This is because

$$\forall x\in \mathbb R:\sin x = 0\iff x=k\pi$$

for some $k\in\mathbb N$, and $k\pi\in\mathbb N\iff k=0$. Indeed, you can see this because if there exists some $\mathbb N\ni k\neq0$ such that $k\pi\in\mathbb N$, then $k\pi=m$ for some $m\in\mathbb N$, and this means that $$\pi=\frac mk$$

for two integers $m,k$.

In other words, this means that $\pi$ is rational, something we know is not true.

• That doesn't imply that $x \to \sin x$ is injective. – anomaly Jun 25 '16 at 7:31
• @anomaly I never said it does... – 5xum Jun 25 '16 at 14:47
• @anomaly the question was fundamentally altered, as before, it was asking if the function is bijective. – 5xum Jun 25 '16 at 14:48
• ...then never mind. – anomaly Jun 25 '16 at 14:56

Bijective implies surjective and injective. $\sin: \Bbb{N} \to \Bbb{R}$ is not surjective since $\sin(x) = 1 \implies x = \pi/2 + 2\pi n \notin \Bbb{N}$.