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I would like to prove $$\cosθ\leq-1＋\frac{1}{2} (θ-π)^2$$ using Taylor exapansion or Taylor theorem.

I can prove this enequality by differentiating $f(θ)＝\cosθ＋1-\frac{1}{2} (θ-π)^2$, but I would like to prove this by using Taylor expansion.

I know $\cosθ＝-1＋\frac{1}{2} (θ-π)^2＋O((θ-π)^3)$ using big $O$, but this kind of estimation may not be suitable for inequality.

Robert Z
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  • If you use approximated Taylor expansion near $\theta=\pi$, you will not get anywhere, as Taylor approximation is just useful locally. To do that instead, you should maybe write $\cos(\theta)$ as an infinite series of its Taylor expansion, subtract the terms on the right hand side, and prove by hand that what remains is negative – Lorenzo Pompili Jul 01 '21 at 17:13

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Since $\cos(\theta)=-\cos(\theta-\pi)$, you want to prove that you always have$$-\cos(\theta-\pi)\leqslant-1+\frac{(\theta-\pi)^2}2,$$which is equivalent to$$(\forall\alpha\in\Bbb R):\cos(\alpha)\geqslant1-\frac{\alpha^2}2.$$If $|\alpha|\geqslant\sqrt2$, this is clear, since then $1-\frac{\alpha^2}2\leqslant-1$. So, suppose that $|\alpha|\leqslant\sqrt2$. Then\begin{align}\cos(\alpha)-\left(1-\frac{\alpha^2}2\right)&=\frac{\alpha^4}{4!}-\frac{\alpha^6}{6!}+\frac{\alpha^8}{8!}-\frac{\alpha^{10}}{10!}+\cdots\\&=\frac{\alpha^4}{4!}\overbrace{\left(1-\frac{\alpha^2}{5\times6}\right)}^{\phantom0\geqslant0}+\overbrace{\frac{\alpha^8}{8!}\left(1-\frac{\alpha^2}{7\times8}\right)}^{\phantom0\geqslant0}+\cdots\\&\geqslant0.\end{align}

José Carlos Santos
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  • Why can we calculate the infinite sum by ()ing like yours? How to justify it? –  Jul 02 '21 at 04:06
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    Given any convergent series $a_0+a_1+a_2+a_3+\cdots$ with sum $s$, you also have $(a_0+a_1)+(a_2+a_3)+\cdots=S$. That's because if $S_n=a_0+a_1+\cdots+a_n$, then by definition, $S=\lim_{n\to\infty}S_n$ and so$$(a_0+a_1)+(a_2+a_3)+\cdots=\lim_{n\to\infty}S_{2n-1}=S.$$ – José Carlos Santos Jul 02 '21 at 06:30