I'm studying for my Qualifying exams and this was one of the questions in the question bank under real analysis section. I'm currently stuck on this question. I think the answer is 2 but don't have a rigorous proof.
Find $\limsup\limits_{x\rightarrow\infty}\ (\sin(x)+\sin(\pi x))$.
My attempt: I tried to look at the sequence $x_n=\frac{1}{2}+2n$ but not sure how to calculate further and find the $\limsup$
On
You're off to a good start. If $x = 2n+\frac12$ then $\sin \pi x=1$. How big can we make $\sin x$ though? We'd like to say $x\approx 2m\pi+\frac\pi2$ so that $\sin x\approx 1$ and then conclude that the $\limsup$ is $2$. Then we'd have $$\pi\approx\frac{2n+1/2}{2m+1/2}=\frac{4n+1}{4m+1}$$ So, that's the plan of attack. Show that there is an infinite sequence of fractions of this form converging to $\pi$ and then conclude that there is a sequence $(x_n)$ such that $$\lim_{n\to\infty}\sin(x_n)+\sin(\pi x_n)=2$$
Can you carry out the plan?
EDIT
The above is incorrect. See the correction by Thomas Andrews in the comments.
On
Not an answer, but too,long for comment.
$2$ is certainly an upper bound, and is likely the supremes. What you want is:
$$x\approx \frac{\pi}{2}(4n+1)\\ \pi x\approx \frac{\pi}{2}(4m +1)$$
If $x=\frac 12(4m+1)$ then you need:
$$\pi\approx \frac{4m+1}{4n+1}\tag 1$$
The question is, how good an approximation can we get in this form? If $(4m+1)/(4n+1)$ is in the continued fraction for $\pi,$ then:$$-\frac1{2(4n+1)}<\frac12(4n+1)\pi-\frac12 (4m+1)<\frac1{2(4n+1)}$$ In which case $$\sin ((4m+1)/2)>1-\frac1{4(4n+1 )^2}$$ So if there are infinitely many such continued fractions, we can get closer and closer to $2.$
But I can’t think of any way to ensure the continued fractions will take this form infinitely often. Given the apparent randomness of the coefficients for $\pi,$ we might expect a convergent of this form in one out of ever $12$ convergents, but proving that will be hard.
The first example is:$$\frac{208341}{66317}$$ and $x=\frac{208341}{2},$ and, according to my calculator, $1-\sin x <\frac1{10^{11}}.$
The next is $$\frac{165707065}{52746197}.$$
But proving there are infinitely many will be hard.
So we might need less stringent conditions on $m,n.$ We need a sequence $(m_i,n_i)$ such that:
$$\pi(4m_i+1)-(4n_i+1)\to 0.$$
On
Here is the alternate answer I offered a while ago. The beginning is identical, but from $(3)$ on is an approach not using continued fractions.
Your original idea of looking at $x=\frac12+2n$ is in the right direction, because that gives $\sin(\pi x)=1$. The remaining piece is to seek $x\approx\left(\frac12+m\right)\pi$ so that $\sin(x)\approx1$.
What We Would Like
If $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$, then
$$
\begin{align}
1-\sin\left(2n+\tfrac12\right)
&=\left|\,\sin\left(\left(2m+\tfrac12\right)\pi\right)-\sin\left(2n+\tfrac12\right)\tag{1a}\,\right|\\[6pt]
&\le\left|\,\left(2m+\tfrac12\right)\pi-\left(2n+\tfrac12\right)\,\right|\tag{1b}\\[6pt]
&\le\epsilon\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $1-\sin(x)\ge0$ and $\sin\left(\left(2m+\tfrac12\right)\pi\right)=1$
$\text{(1b)}$: $|\sin(x)-\sin(y)|=2\left|\,\cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)\,\right|\le|x-y|$
$\text{(1c)}$: assumption about $m$ and $n$
Since $\sin\left(\left(2n+\tfrac12\right)\pi\right)=1$, we have $$ \overbrace{\sin\left(2n+\tfrac12\right)}^{\ge1-\epsilon}+\overbrace{\sin\left(\left(2n+\tfrac12\right)\pi\right)}^{=1}\ge2-\epsilon\tag2 $$ Thus, given an $\epsilon\gt0$, we would like to find $m,n\in\mathbb{Z}$ so that $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$.
Pigeonhole Approximations
Suppose $\pi$ is a positive irrational number. Choose an $N$ and break $[0,1)$ into $N$ equal sub-intervals $$ \left\{I_k=\left[\frac kN,\frac{k+1}N\right):0\le k\lt N\right\}\tag3 $$ Using $\{x\}=x-\lfloor x\rfloor$, consider the $N+1$ real numbers $$ \{\{j\pi\}:0\le j\le N\}\tag4 $$ which live in $[0,1)$. The Pigeonhole Principle says that there must be one of the $N$ sub-intervals, $I_k$, that contains at least two of the $N+1$ real numbers, $\{j_1\pi\}\lt\{j_2\pi\}$. Note that this does not imply that $j_1\lt j_2$, only that $$ 0\lt\{(j_2-j_1)\pi\}\lt\frac1N\tag5 $$ $(5)$ says that there are integers $p,q\in\mathbb{Z}$, with $|q|\le N$, so that $$ 0\lt p+q\pi\lt\frac1N\tag6 $$ The set $$ \begin{align} M &=\left\{j(p+q\pi):1\le j\lt\frac1{p+q\pi}\right\}\tag{7a}\\ &=\left\{\{jq\pi\}:1\le j\lt\frac1{\{q\pi\}}\right\}\tag{7b} \end{align} $$ has at least one element in each $I_k$ since the distance between elements of $M$ is $p+q\pi\lt\frac1N$. Furthermore, $M$ covers all of the $I_k$ because $j\gt\frac1{p+q\pi}\implies j(p+q\pi)\gt1$.
Since $N$ can be as large as we wish, we have shown that $\{\mathbb{Z}\pi\}$ is dense in $[0,1]$.
Getting What We Want
We can rewrite the condition $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$ from above as $$ \begin{align} \frac\epsilon2 &\ge\left|\,m\pi-n+\frac{\pi-1}4\right|\tag{8a}\\ &=\left|\{m\pi\}-\frac{5-\pi}4\right|\tag{8b} \end{align} $$ We can find an $m$ to satisfy $(8)$ because $\{\mathbb{Z}\pi\}$ is dense in $[0,1]$.
Your original idea of looking at $x=\frac12+2n$ is in the right direction, because that gives $\sin(\pi x)=1$. The remaining piece is to seek $x\approx\left(\frac12+2m\right)\pi$ so that $\sin(x)\approx1$.
What We Would Like
If $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$, then $$ \begin{align} 1-\sin\left(2n+\tfrac12\right) &=\left|\,\sin\left(\left(2m+\tfrac12\right)\pi\right)-\sin\left(2n+\tfrac12\right)\tag{1a}\,\right|\\[6pt] &\le\left|\,\left(2m+\tfrac12\right)\pi-\left(2n+\tfrac12\right)\,\right|\tag{1b}\\[6pt] &\le\epsilon\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: $1-\sin(x)\ge0$ and $\sin\left(\left(2m+\tfrac12\right)\pi\right)=1$
$\text{(1b)}$: $|\sin(x)-\sin(y)|=2\left|\,\cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)\,\right|\le|x-y|$
$\text{(1c)}$: assumption about $m$ and $n$
Since $\sin\left(\left(2n+\tfrac12\right)\pi\right)=1$, we have $$ \overbrace{\sin\left(2n+\tfrac12\right)}^{\ge1-\epsilon}+\overbrace{\sin\left(\left(2n+\tfrac12\right)\pi\right)}^{=1}\ge2-\epsilon\tag2 $$ Thus, given an $\epsilon\gt0$, we would like to find $m,n\in\mathbb{Z}$ so that $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$.
Results from Continued Fraction Approximations
Suppose that $\frac{p_{n-1}}{q_{n-1}}$ and $\frac{p_n}{q_n}$ are two consecutive continued fraction convergents for $\pi$. Then $$ \begin{align} \left|\,\frac{p_{n-1}}{q_{n-1}}-\pi\,\right|+\left|\,\pi-\frac{p_n}{q_n}\,\right| &=\left|\,\frac{p_{n-1}}{q_{n-1}}-\frac{p_n}{q_n}\,\right|\tag{3a}\\ &=\frac1{q_{n-1}q_n}\tag{3b} \end{align} $$ Explanation:
$\text{(3a)}$: $\pi$ is between any two consecutive convergents
$\text{(3b)}$: a property of continued fraction convergents: $p_{n-1}q_n-p_nq_{n-1}=(-1)^{n-1}$
For any $a,b\ge0$, since $p_{n-1}-q_{n-1}\pi$ and $p_n-q_n\pi$ have different signs, $(3)$ implies $$ \begin{align} |(ap_{n-1}+bp_n)-(aq_{n-1}+bq_n)\pi| &=|a(p_{n-1}-q_{n-1}\pi)+b(p_n-q_n\pi)|\tag{4a}\\[3pt] &\le\frac{\max(a,b)}{q_n}\tag{4b} \end{align} $$ Consider the matrix whose columns are consecutive convergents of $\pi$. By $(3)$, the determinant is $\pm1$.
Suppose $$ \begin{align} \begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}^{-1}\begin{bmatrix}1\\1\end{bmatrix} &=(-1)^{n-1}\begin{bmatrix}q_n&-p_n\\-q_{n-1}&p_{n-1}\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}\tag{5a}\\[3pt] &=\begin{bmatrix}(-1)^n(p_n-q_n)\\(-1)^{n-1}(p_{n-1}-q_{n-1})\end{bmatrix}\tag{5b}\\[3pt] &\equiv\begin{bmatrix}a\\b\end{bmatrix}\pmod4\tag{5c} \end{align} $$ where $0\le a,b\lt4$. Then $$ \begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\equiv\begin{bmatrix}1\\1\end{bmatrix}\pmod4\tag6 $$ Thus, we have from $(5)$ $$ ap_{n-1}+bp_n\equiv aq_{n-1}+bq_n\equiv1\pmod4\tag7 $$ and from $(4)$ $$ |(ap_{n-1}+bp_n)-(aq_{n-1}+bq_n)\pi|\le\frac{\max(a,b)}{q_n}\tag8 $$
Convergents and Examples
Here are some examples of generating $n$ and $m$ that satisfy $(1)$ for arbitrarily small $\epsilon\gt0$.
The convergents for $\pi$ start out $$ \frac31,\frac{22}7,\frac{333}{106},\frac{355}{113},\frac{103993}{33102},\frac{104348}{33215},\dots $$ Example $\bf{1}$
Using the first two convergents as columns: $$ \begin{bmatrix}3&22\\1&7\end{bmatrix}^{-1}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}15\\-2\end{bmatrix}\equiv\begin{bmatrix}3\\2\end{bmatrix}\pmod4 $$ and $$ \begin{bmatrix}3&22\\1&7\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=\begin{bmatrix}53\\17\end{bmatrix}\equiv\begin{bmatrix}1\\1\end{bmatrix}\pmod4 $$ and $$ |53-17\pi|=0.407075 $$ which is less than $\frac37$ as given in $(4)$. $$ \bbox[5px,border:2px solid #C0A000]{\sin\left(\frac{53}2\right)+\sin\left(\frac{53}2\pi\right)=1.9793576431} $$ Example $\bf{2}$
Using the third and fourth convergents as columns: $$ \begin{bmatrix}333&355\\106&113\end{bmatrix}^{-1}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}242\\-7\end{bmatrix}\equiv\begin{bmatrix}2\\1\end{bmatrix}\pmod4 $$ and $$ \begin{bmatrix}333&355\\106&113\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}1021\\325\end{bmatrix}\equiv\begin{bmatrix}1\\1\end{bmatrix}\pmod4 $$ and $$ |1021-325\pi|=0.0176124 $$ which is less than $\frac2{113}$ as given in $(4)$. $$ \bbox[5px,border:2px solid #C0A000]{\sin\left(\frac{1021}2\right)+\sin\left(\frac{1021}2\pi\right)=1.9999612255979} $$ Example $\bf{3}$
Using the fourth and fifth convergents as columns: $$ \begin{bmatrix}355&103993\\113&33102\end{bmatrix}^{-1}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}-70891\\242\end{bmatrix}\equiv\begin{bmatrix}1\\2\end{bmatrix}\pmod4 $$ and $$ \begin{bmatrix}355&103993\\113&33102\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}208341\\66317\end{bmatrix}\equiv\begin{bmatrix}1\\1\end{bmatrix}\pmod4 $$ and $$ |208341-66317\pi|=0.000008114318 $$ which is less than $\frac2{33102}$ as given in $(4)$. $$ \bbox[5px,border:2px solid #C0A000]{\sin\left(\frac{208341}2\right)+\sin\left(\frac{208341}2\pi\right)=1.99999999999176973} $$