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This answer shows that $\cos(\kappa) = \sin(\kappa) = \frac{1}{\sqrt{2}}$ (but whilst $\cos$ and $\sin$ are yet to be defined, $c$ and $s$ are used).

The answer continues that if $0 < x < \kappa$, then $\frac{1}{\sqrt{2}} < x < 1$. Afterwards, it is claimed that $\frac{\kappa}{\sqrt{2}} < s(\kappa) < \kappa$ and $\frac{1}{\sqrt{2}} < \kappa < 1$.

Why do these bounds follow from a point of intersection?

johnsmith
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1 Answers1

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This does not directly answer the question but one can prove periodicity with another argument: for all $a\in{\mathbb R}$, one has \begin{equation} m:=\inf_{t\in[a, \infty)}|s(t)| = 0 \end{equation} Indeed, if $b>a$, the mean value theorem applied to $c(x)$ tells us that for a certain number $t\in (a,b)$, one has \begin{equation} m\le |s(t)| = |c'(t)| = \left|\frac{c(b)-c(a)}{b-a}\right|\le \frac{2}{b-a} \end{equation} which tends to $0$ when $b\to\infty$.

A consequence of this is that $s$ cannot increase monotonically on $(0,\infty)$ otherwise we would have $0<s(a)\le m$ for any $a>0$ and this is a contradiction. Let $\alpha$ be the largest number such that $s$ increases on $(0, \alpha)$. We have $c(\alpha) = s'(\alpha) = 0$ and $s(\alpha)> 0$ and this implies $s(\alpha) = 1$.

The differential equation $y''+y=0$ now implies that $c(x) = s(x + \alpha)$ because both hand sides of this equality satisfy the same Cauchy problem at $x=0$. As $c$ is even, we conclude that $s(\alpha - x) = s(\alpha + x)$ and it follows that $s(x) = -s(x + 2\alpha)$ by the same Cauchy problem argument. Hence $s(x + 4\alpha) = -(-s(x)) = s(x)$ and $s$ is periodic with period $4\alpha$.

Gribouillis
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  • I haven't taken a full course on the theory of differential equations, so I searched for an explanation of 'Cauchy problem'. This definition seems sufficient. Also, the answer I linked to in my question provides a way to evaluate $s(2x)$ (after the inequalities I asked about), which helps to verify $s(x) = -s(x + 2\alpha)$. – johnsmith Jan 22 '21 at 09:05
  • I use "Cauchy problem" as a synonym of "initial value problem". In our case it means that if you have the equation $y''+y=0$ and you give the two conditions $y(x_0)=\alpha$ and $y'(x_0)=\beta$, then there is a unique solution of the differential equation that satisfies these conditions. The proof here is very simple: if $y$ is a solution then $y'^2+ y^2$ is constant. In particular, if $\alpha=\beta=0$, the solution satisfies $y'^2+y^2=0$, hence the solution is $y=0$. – Gribouillis Jan 22 '21 at 09:09