Sine at Integer Values

Last Thursday, I was introducing to my students the notions of limit superior and limit inferior. In particular, I was emphasizing that a sequence (x_{n})_{n=1}^{\infty} of reals may not converge, but \limsup_{n\rightarrow\infty}x_{n} and \liminf_{n\rightarrow\infty}x_{n} always exist in the extended real number system. After giving some fairly basic examples of such sequences, one student asked me what the \limsup is of the sequence defined by

\displaystyle x_{n}:=\sin(n),\indent\forall n\in\mathbb{Z}^{\geq0}

I was a bit taken aback by the question because, aside from \sin(0), I did not (and do not) know off the top of my head what the values \sin(n) are. Nevertheless, I thought it was a great question–I love getting asked questions that require me to think a bit before answering–and asked the student if I could get back to him.

Well, it turns out that another one of my students also thought it was a great question and was eager to find out the answer himself. He did some investigating on Math.StackExchange, which led to the answer that

\displaystyle\limsup_{n\rightarrow\infty}\sin(n)=1

More interestingly, the set \left\{\sin(n):n\in\mathbb{Z}^{\geq 0}\right\} is dense in [-1,1]. Proving this result is the goal of today’s post.

Lemma 1. If \alpha\in\mathbb{R} is irrational, then the set

\displaystyle\mathbb{Z}^{\geq 0}-\alpha\cdot\mathbb{Z}^{\geq 0}=\left\{n-\alpha m: n,m\in\mathbb{Z}\right\}

is dense in \mathbb{R}.

Proof. Since every real number x can be written as the difference

x=\begin{cases}{y+2\pi l} & {x\geq 0}\\ {y-n} & {x<0}\end{cases}

where y\in[0,2\pi] and l\in\mathbb{Z}^{\geq 0}, we may assume that x\in[0,2\pi]. Fix \theta\in [0,2\pi], and let \varepsilon>0 be given. Replacing \varepsilon by a smaller \varepsilon'>0, if necessary, we may assume that \varepsilon<1. Choose N\in\mathbb{N} sufficiently large so that \frac{2\pi}{N}<\epsilon. Partition the interval [0,2\pi] into N intervals of length \frac{2\pi}{N} given by

\displaystyle\left(\frac{2\pi j}{N},\frac{2\pi(j+1)}{N}\right],\indent\forall j=0,\cdots,N-1

By the pigeonhole principle, there exist two distinct natural numbers n_{1},n_{2} such that

\displaystyle n_{1}-2m_{1}\pi,n_{2}-2m_{2}\pi\in\left(\frac{2\pi j}{N},\frac{2\pi(j+1)}{N}\right]

for some j\in\left\{0,\cdots,N-1\right\} and natural numbers m_{1},m_{2}\in\mathbb{N}. Since \pi is irrational, n_{1}-2\pi m_{1}\neq n_{2}-2\pi m_{2}. Without loss of generality, we may assume n_{1}>n_{2} so that

\displaystyle n_{1}-2\pi m_{1}>n_{2}-2\pi m_{2}\Longrightarrow \delta:=(n_{1}-n_{2})-2\pi(m_{1}-m_{2})>0

Observe that m_{1}>m_{2}, since \pi is irrational. Since the disjoint half-open intervals

\displaystyle\left(k\delta,(k+1)\delta\right],\indent\forall k=0,\cdots,M-1

contain [0,2\pi], for M\in\mathbb{N} satisfying M\delta>2\pi, we see that there exists k\in\left\{0,\cdots,M\right\} such that

\displaystyle x\in\left(k\delta,(k+1)\delta\right]

Setting n=k(n_{1}-n_{2}) and m=k(m_{1}-m_{2}), we conclude that \left|x-(n-2\pi m)\right|<\varepsilon. \Box

Lemma 2. \pi is irrational.

The following proof is due to British mathematician Mary Cartwright.

Proof. For each n\in\mathbb{Z}^{\geq 0}, define

\displaystyle I_{n}(x):=\int_{-1}^{1}(1-u^{2})^{n}\cos(xu)du,\indent\forall x\in\mathbb{R}

Observe that

\displaystyle I_{0}(x)=\int_{-1}^{1}\cos(xu)du=x^{-1}\sin(xu)\mid_{u=-1}^{u=1}=x^{-1}[\sin(x)-\sin(-x)]=2x^{-1}\sin(x)

and

\begin{array}{lcl}I_{1}(x)=\displaystyle\int_{-1}^{1}(1-u^{2})\cos(xu)du&=&\displaystyle x^{-1}(1-u^{2})\sin(xu)\mid_{u=-1}^{u=1}+x^{-1}\displaystyle\int_{-1}^{1}2u\sin(xu)du\\[.9 em]&=&-2x^{-2}u\cos(xu)\mid_{u=-1}^{u=1}\displaystyle+2x^{-2}\int_{-1}^{1}\cos(xu)du\\[.9 em]&=&-\displaystyle4x^{-2}\cos(x)+4x^{-3}\sin(x)\end{array}

We can integrate by parts twice to obtain a recurrence relation for I_{n}, with n\geq2. Indeed,

\begin{array}{lcl}I_{n}(x)&=&x^{-1}(1-u^{2})^{n}\sin(xu)\mid_{u=-1}^{1}+x^{-1}\int_{-1}^{1}2nu(1-u^{2})^{n-1}\sin(xu)du\\[.7 em]&=&-\displaystyle x^{-2}2nu(1-u^{2})^{n-1}\cos(xu)\mid_{u=-1}^{1}+x^{-2}\int_{-1}^{1}2n\left[(1-u^{2})^{n-1}-2(n-1)u^{2}(1-u^{2})^{n-2}\right]\cos(xu)du\\[.7 em]&=&\displaystyle(2n+4n(n-1))x^{-2}\int_{-1}^{1}(1-u^{2})^{n-1}\cos(xu)du-4n(n-1)x^{-2}\int_{-1}^{1}(1-u^{2})^{n-2}\cos(xu)du\\[.7 em]&=&\displaystyle2n(2n-1)x^{-2}I_{n-1}(x)-4n(n-1)x^{-2}I_{n-2}\end{array}

For each n\in\mathbb{N}, set J_{n}(x):=x^{2n+1}I_{n}(x), so that the above recurrence relation can be restated as

\displaystyle J_{n}(x)=2n(2n-1)J_{n-1}(x)-4n(n-1)x^{2}J_{n-2}(x),\indent\forall n\geq2

and J_{0}(x)=2\sin(x), J_{1}(x)=-4x\cos(x)+4\sin(x).

By induction, it follows that for each n\in\mathbb{Z}^{\geq0},

\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x)=n!\left[P_{n}(x)\sin(x)+Q_{n}(x)\cos(x)\right],

where P_{n} and Q_{n} are polynomials with integer coefficients and degree \leq 2n. Suppose that \pi is rational so that \frac{\pi}{2}=\frac{b}{a}, where a,b\in\mathbb{N}. Since \deg(P_{n})\leq 2n, we see that the RHS of the equation

\displaystyle\frac{b^{2n+1}}{n!}I_{n}\left(\frac{\pi}{2}\right)=P_{n}\left(\frac{\pi}{2}\right)a^{2n+1}

is an integer. Since the integrand in the definition of I_{n}(\frac{\pi}{2}) is nonnegative and bounded by 1, we see that 0< I_{n}(\frac{\pi}{2})<2 for all n\in\mathbb{Z}^{\geq0}. As n\rightarrow\infty, the expression

\displaystyle\frac{b^{2n+1}}{n!}I_{n}\left(\frac{\pi}{2}\right)\rightarrow0,

So for sufficiently large n, there exists an integer P_{n}(\frac{\pi}{2})a^{2n+1}\in(0,1), which is a contradiction. \Box

Now that we know that \pi is irrational, and therefore 2\pi is irrational, we can complete the proof that countable set \left\{\sin(n):n\in\mathbb{Z}^{\geq 0}\right\} is dense in [-1,1]. Fix x\in[-1,1]. Since \sin() is a continuous function, by the intermediate value theorem, there exists \theta\in[0,2\pi] such that x=\sin(\theta). By lemma 2, there exist integers n,m\in\mathbb{Z}^{\geq 0} such that \left|\theta-(n-2\pi m)\right|<\delta, for \delta>0 given. Since \sin() is continuous, for \varepsilon>0 given, there exists \delta>0 such

\displaystyle\left|\theta-\theta'\right|<\delta\Longrightarrow\left|\sin(\theta)-\sin(\theta')\right|<\varepsilon,

which implies that

\displaystyle\left|x-\sin(n-2\pi m)\right|=\left|\sin(\theta)-\sin(n-2\pi m)\right|=\left|\sin(\theta)-\sin(n)\right|<\varepsilon,

where we use the 2\pi periodicity of \sin().

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2 Responses to Sine at Integer Values

  1. Pingback: Sine at Prime Values | Math by Matt

  2. Pingback: Minkowski Sums in Euclidean Space | Math by Matt

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