Proving $\left|\sin x\right|\leq\frac{(2m+1)!}{2^{4m}(m!)^2}\left[\binom{2m}{m} - \sum_{k=1}^{m} \frac{2}{4k^2-1} \binom{2m}{m+k} \cos(2kx) \right]$

Question

Update. Based on @Exodd's answer, it turns out that the upper bound is equal to

$$ T_{2m}(\cos x) = \sum_{k=0}^{m} (-1)^k \binom{1/2}{k}\cos^{2k} x, $$

where $T_{2m}(x)$ is the degree $2m$ Taylor polynomial of $\sqrt{1-x^2}$. This completely settles down the questions asked below. (Check the community answer below.) Case closed!

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Old Question. While working to find a good upper bound of $\left|\sin x\right|$, I experimentally observed that

$$ \left|\sin x\right| \leq \bbox[color:navy]{ \frac{(2m+1)!}{2^{4m}(m!)^2} \biggl[ \binom{2m}{m} - \sum_{k=1}^{m} \frac{2}{4k^2-1} \binom{2m}{m+k} \cos(2kx) \biggr] } =: S_m(x) \tag{*} $$

holds for all $x \in \mathbb{R}$. However, I have no idea how to prove this, and a quick search on Approach Zero but it did not show anything. So, I am sharing my question with other users:

Question. Does the inequality $\text{(*)}$ really hold true? If so, how can we prove this?

Anyway, here are several observations:

1. For each $m$, $S_m(x)$ seems to be the unique trigonometric polynomial $\sum_{k=0}^{m} a_k \cos(2kx)$ satisfying $$ \sum_{k=0}^{m} a_k \cos(2kx) = \sin(x) + o\bigl(x-\tfrac{\pi}{2}\bigr)^{2m}. $$ This is actually how I conjectured the coefficients of $S_m(x)$. For each $m$, I determinted the values of $a_k$'s and tried to identify the patterns using OEIS. So, unfortunately I don't have a slightest idea as to how the coefficients of $S_m(x)$ arise.

2. By noting that $\frac{(2m+1)!}{2^{4m}(m!)^2}\binom{2m}{m+k} \to \frac{2}{\pi}$ as $m \to \infty$, we get $$ \lim_{m\to\infty} S_m(x) = \frac{2}{\pi} \biggl[ 1 - \sum_{k=1}^{\infty} \frac{2}{4k^2-1} \cos(2kx) \biggr]. $$ This is precisely the Fourier cosine series for $\left|\sin x\right|$. So, $\text{(*)}$ is consistent with the Fourier series for $\left|\sin x\right|$.

3. Surprisingly, $S_m(x)$ seems to be unimodal on $[0, \pi]$ (and hence on each $[k\pi, (k+1)\pi]$ for $k \in \mathbb{Z}$), as we can see from the figure below:

Answer 1

Writing down $S_{m-1}(x) - S_{m}(x) \ge 0$ you end up with the equivalent conditions $$\binom{2m}{m} + 2\sum_{k=1}^m \binom{2m}{m+k}\cos(2kx)\ge 0$$ or equivalently $$ \sum_{k=-m}^m \binom{2m}{m+k}\exp(2\text ikx)\ge 0 $$ but the LHS is just $[2\cos(x)]^{2m}$ that is always nonnegative.

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I think you can then characterize $S_m(x)$ recursively as $$ S_{m+1}(x) = S_{m}(x) - \frac{2(2m)! }{2^{4m}(m!)^2(2m+3)} [2\cos(x)]^{2m+2} $$

Answer 2

Based on @Exodd's answer, it turns out that $S_m(x)$ is equal to

$$ T_{2m}(\cos x) = \sum_{k=0}^{m} (-1)^k \binom{1/2}{k}\cos^{2k} x, $$

where $T_{2m}(x)$ is the degree $2m$ Taylor polynomial of $\sqrt{1-x^2}$. Indeed, noting that

$$ (-1)^k \binom{1/2}{k} = \frac{(-\frac{1}{2})(\frac{1}{2})\cdots(k-\frac{3}{2})}{k!} = \binom{2k}{k} \frac{1}{2^{2k}(1-2k)} < 0 \quad \text{for} \quad k = 1, 2, \ldots,$$

$T_{2m}(\cos x)$ is non-increasing in $m$ and converges to $\sqrt{1-\cos^2 x} = \left|\sin x\right|$ as $m \to \infty$. Moreover,

\begin{align*} T_{2m}(\cos x) &= \sum_{k=0}^{m} \binom{2k}{k} \frac{1}{2^{2k}(1-2k)} \, \cos^{2k}x \\ &= \sum_{k=0}^{m} \binom{2k}{k} \frac{1}{2^{4k}(1-2k)} \sum_{l=-k}^{k} \binom{2k}{k+l} e^{2lix} \\ &= \sum_{l=-m}^{m} A_{m,|l|} e^{2lix} \\ &= A_{m,0} + 2 \sum_{l=1}^{m} A_{m,l} \cos(2lx), \end{align*}

where $A_{m,l}$ is defined for $0 \leq l \leq m$ by

$$ A_{m,l} = \sum_{k=l}^{m} \binom{2k}{k} \frac{1}{2^{4k}(1-2k)} \binom{2k}{k+l}. $$

So it suffices to show that

Claim. $A_{m,l} = \tilde{A}_{m,l}$, where $\displaystyle \tilde{A}_{m,l} = \frac{(2m+1)!}{2^{4m}(m!)^2}\binom{2m}{m+l} \frac{1}{1-4l^2}$.

Indeed, it is easy to verify that $A_{m,m} = \binom{2m}{m} \frac{1}{2^{4m}(1-2m)} = \tilde{A}_{m,m}$. Moreover, for $0 \leq l < m$,

\begin{align*} \tilde{A}_{m,l} - \tilde{A}_{m-1,l} &= \binom{2m}{m}\binom{2m}{m+l} \frac{1}{2^{4m}(1-4l^2)} \biggl[ (2m+1) - \frac{2^4 m^2(m+l)(m-l)}{(2m)^2(2m-1)} \biggr] \\ &= \binom{2m}{m}\binom{2m}{m+l} \frac{1}{2^{4m}(1-2m)} \\ &= A_{m,l} - A_{m-1,l}. \end{align*}

Therefore the desired equality follows from the principle of mathematical induction. $\square$

Answer 3

$$S_m(x)=\frac{2\, \Gamma \left(m+\frac{1}{2}\right) \Gamma \left(m+\frac{3}{2}\right)}{\pi\, \Gamma(m+1)^2}-\frac{2 m (m+1) \Gamma \left(m+\frac{1}{2}\right) \Gamma \left(m+\frac{3}{2}\right)}{3 \pi \Gamma (m+2)^2}\,T_m(x)$$

$$T_m(x)=e^{-2 i x} \, _3F_2\left(\frac{1}{2},1,1-m;\frac{5}{2},m+2;-e^{-2 i x}\right)+$$ $$e^{2 i x} \, _3F_2\left(\frac{1}{2},1,1-m;\frac{5}{2},m+2;-e^{2 i x}\right)$$

Beside the fact that $S_m\left(\frac{\pi }{2}\right)=0$, for $0\leq x \leq \pi$ (explored by steps of $\frac \pi {180}$) and $1 \leq m \leq 10^4$, $\left(S_m(x)-\sin(x)\right)$ is a fast decreasing function which is always positive. For sure, the largest difference is at the boundaries.

$$S_m(0)=\frac{\Gamma \left(m+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (m+1)}$$

Expanded as series around $x=\frac \pi 2$ $$S_m(x)-\sin(x)=\frac {a_m}{b_m} \left(x-\frac{\pi }{2}\right)^{2 (m+1)}+O\left(\left(x-\frac{\pi }{2}\right)^{2(m+2)}\right)$$ where the $a_m$ are sequence $A098597$ and the $b_m$ are sequence $A046161$ in $OEIS$. All of them are positive and, as a very first approximation $$\frac {a_m}{b_m} \sim \frac 1{2m^2}$$