Proof of Taylor series for $\cos x$ is greater than $\cos x$ in $[-1,1]$ I would like to prove that Taylor series of $\cos x$ is greater or equal than $\cos x$ when $|x| \le 1$:
$$1-\frac12x^2+\frac1{24}x^4 \ge \cos x$$
I tried to think taking the as a function $g(x) = \cos x - \left(1-\frac12x^2+\frac1{24}x^4\right)$ and showing that is $> 0$, but I don't know how. Some help?
I also though at the Lagrange remainder. It should be: $$-\frac{\cos(c)}{6!}x^6$$ and it should be always negative for $x \in [-1,1]$. therefore this would prove that $\cos x$ is $\le$ than the Taylor expansion. is it right?
And returning to the previous approach, how could I prove it?
 A: Your function $g(x)$ is the remainder of the Taylor polynomial, for which there is are a few explicit formulas. In particular, I think you can show that the mean value form of the remainder is nonpositive.
With $f(x) = \cos(x)$ the remainder for the 4th degree Taylor polynomial is
$$\frac{f^{(5)}(\xi)}{5!}x^5 = \frac{-\sin \xi}{5!} x^5$$
for some $\xi$ between $0$ and $x$. This quantity is negative when $|x| \le 1$.
A: There's actually a part of the Leibniz Alternating Series Test that gives you this for free. When we have a positive sequence $a_n$ approaching $0$ monotonically, then
$$\sum_{n=0}^\infty (-1)^n a_n$$
converges to some finite $L$. Moreover (and this is the interesting bit),
$$\left|\sum_{k=0}^n (-1)^k a_k - L\right| \le a_{k+1}.$$
In other words, each partial sum is sufficiently close to the limit that adding the next term will be an over-correction. It means that each partial sum will alternate being lesser or greater than $L$. That is,
\begin{align*}
a_0 &\ge L \\
a_0 - a_1 &\le L \\
a_0 - a_1 + a_2 &\ge L \\
a_0 - a_1 + a_2 - a_3 & \le L,
\end{align*}
etc.
In our case, the Taylor series for $\cos$ is
$$\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}.$$
When $|x| \le 1$, note that $\frac{x^{2n}}{(2n)!}$ is a monotone decreasing sequence converging to $0$. Thus, we can say that the third partial sum,
$$1 - \frac{x^2}{2} + \frac{x^4}{24}$$
is greater than or equal to the limit $\cos(x)$.
A: Use Taylor's theorem, with, e.g. the Lagrange form of the remainder:
$$ f(x) - \sum_{k=0}^n \frac{(x-a)^k}{k!} f^{(k)}(a) = \frac{(x-a)^{n+1}}{(n+1)!} f^{(n+1)}(\xi) $$
for some $\xi \in (x,a)$. In particular, with $f=\cos$, $a=0$ and $n=5$,
$$ \cos{x} - \left( 1 + 0 - \frac{x^2}{2} + 0 + \frac{x^4}{24} + 0 \right) = \frac{x^6}{6!} (\cos^{(6)}{\xi}) = -\frac{x^6}{6!} \cos{\xi} \leq 0 , $$
since $\cos{\xi} > 0 $ on $(-\pi/2,\pi/2)$, so in fact the result holds on the whole of this interval.
A: Both sides are even functions, hence you only have to prove the inequality over $[0,1]$.
In order to do that, you may just integrate multiple times the convexity inequality $\sin(x)\leq x$, getting
$$ 1-\cos(x) \leq \frac{x^2}{2}, $$
$$ x-\sin(x) \leq \frac{x^3}{6}, $$
$$ \frac{x^2}{2}-1+\cos(x)\leq \frac{x^4}{24}$$
such that $\cos(x)\leq 1-\frac{x^2}{2}+\frac{x^4}{24}$ as wanted.
A: hint
Let $f(t)=\cos(t)$ and $0<x\le 1$.
$f$ is even, $C^4$ at $[0,x]$ and
$f^{(5)}$ exists at $(0,x)$, thus by
Taylor-Lagrange formula,
there exists $c\in (0,x)$ such that
$$f(x)=f(0)+xf'(0)+...\frac{x^4}{4!}f^{(4)}(0)+\frac{x^5}{120}f^{(5)}(c)$$
$$=1+0-\frac{x^2}{2}+0+\frac{x^4}{24}-\frac{x^5}{120}\sin(c)$$
But $$0<c<x\le 1$$
thus $\sin(c)>0$.
You can finish.
