Proving that the power series for the cosine function is greater than zero, for $x$ in $[0, \pi/2)$.

Question

I'm trying to prove the cosine power series $$\sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \;>\;0$$ for all $x \in [0, \pi/2)$. Here, $\pi$ is defined as the smallest positive real such that $\cos\frac{\pi}{2} = 0$.

My initial thoughts for an attempt at a solution:

Showing that the series achieves an absolute maximum and minimum at its endpoints, respectfully, and that both of these are positive. Thus by the intermediate value theorem, all points between must also be greater than zero.

I'm at a bit of a loss as to how to formally write this as a proof, however.

Answer 1

We can use geometry to show $\cos(x)$ is $C^{(\infty)}.$ Then we can use Lagrange Remainder Theorem to show the series equals the cosine function on all $\mathbb{R}$. Then we automatically get that the series is positive whenever $\cos(x)$ is positive.

Answer 2

A famous diagram-based argument shows any acute $x$ satisfies $0\le\sin x\le x$. Alternate use of $\cos x=1-\int_0^x\sin t dt,\,\sin x=\int_0^x\cos t dt$ allows us to prove by induction that $c_{2n+1}\le\cos x\le c_{2n}$ with $c_n:=\sum_{k=0}^n\frac{(-x^2)^k}{(2k)!}$, and a similar result for $\sin x$. Since the ratio test implies $\lim_{n\to\infty} c_n$ is finite for any acute $x$, by the squeeze theorem this limit is $\cos x\ge 0$.

Answer 3

Denote $f(x)= \cos x$. You have $f(0)=1$. As $f$ is continuous (converging power series are continuous), $f$ is strictly positive around $0$.

You can then conclude with $\pi$ definition.

Answer 4

$\cos x$, defined as its Maclaurin series, is an entire solution of the differential equation $f''(x)+f(x)=0$, $f(0)=1$, $f'(0)=1$. $\cos x$ is even and concave where positive, and we may denote with $\alpha$ the smallest positive zero of $\cos x$. The problem boils down to showing that $\alpha=\frac{\pi}{2}$. Now, by denoting through $\sin x$ the opposite of the derivative of $\cos x$, we have that $f''+f=0$ implies that $f^2+f'^2$ is constant, in particular equal to one (Pythagorean theorem, $\sin^2+\cos^2=1$). Since $\cos x$ is decreasing on $(0,\alpha)$ and the Pythagorean theorem plus the chain rule imply that the derivative of the inverse function of $\cos x$ is $-\frac{1}{\sqrt{1-x^2}}$, we have $$ \alpha = \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}.$$ Integration by parts relates $\alpha$ with the area of the unit circle: $$ \alpha = 2\int_{0}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{2} $$ and we are done: the series defining $\cos x$ is positive on $[0,\alpha)=\left[0,\frac{\pi}{2}\right)$.
As a side-effect, we have the hypergeometric identities $$ \frac{\pi}{2}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n},\qquad \frac{\pi}{2}=2-\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(n+1)(2n+3)}$$ provided by the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$ and $\sqrt{1-x^2}$.
The latter is faster-convergent, of course.