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

inequalitytrigonometric series

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!

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:
    graph

Best Answer

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.

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

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