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

Question

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.

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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.

Answer 1

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).

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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.

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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 $u\neq v$.

Answer 6

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 $m=n$.

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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.

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