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

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: • I would try to prove it's decreasing in $m$ May 11, 2022 at 9:56
• @Exodd, That might be one proof strategy, but I believe that we need another representation of $S_m(x)$ that allows certain monotonicity in $m$ to be discovered. The current form of $S_m(x)$ seems not adequate for that purpose. May 11, 2022 at 13:17

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 $$S_{m+1}(x) = S_{m}(x) - \frac{2(2m)! }{2^{4m}(m!)^2(2m+3)} [2\cos(x)]^{2m+2}$$

• That is indeed a crucial hint! Motivated from this, I can now recognize $S_m(x)$ as simply $$S_m(x) = T_{2m}(\cos x),$$ where $T_{2m}$ is the degree $2m$ Taylor polynomial of $\sqrt{1-x^2}$. The monotonicity then follows from the fact that all the coefficients of the Maclaurin series of $\sqrt{1-x^2}$, except for the constant term, are negative. I will update my posting accordingly. May 12, 2022 at 9:46

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. $$\square$$

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

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

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 $$\frac {a_m}{b_m} \sim \frac 1{2m^2}$$

• Thank you! I haven't thought about investigating the extra terms of the Taylor series $S_m(x)-\sin(x)$, so it's nice to see that they also have nice patterns. This and other observations together seem strongly hinting a combinatorial origin that I am currently not aware of. May 12, 2022 at 9:02
• @SangchulLee. You are more than welcome ! For once, I bring something to you ! This is a big event. I shall cotinue. Cheers -:) May 12, 2022 at 9:07
• Motivated from Exodd's answer, I realized that $S_m(x)$ is simply $$S_m(x) = T_{2m}(\cos x),$$ where $T_{2m}$ is the degree $2m$ Taylor polynomial of $\sqrt{1-x^2}$. So there is not much secret in the inequality I found, but at least It was a fun question! May 12, 2022 at 9:48
• @SangchulLee. It was a very interesting problem. Thanks for having posted it. Cheers :-) May 12, 2022 at 12:03