With an odd number of terms, the Taylor polynomial of cosine is greater than cosine, but with an even number of terms it's less than cosine

Question

If you use an odd number of terms, $$f_{4n}(x) = 1 - \frac{x^2}{2!} + ... + \frac{x^{4n}}{(4n)!},$$ this Taylor polynomial $f_{4n}(x) \geq \cos(x)$. With an even number of terms, $$f_{4n-2}(x) = 1 - \frac{x^2}{2!} + ... - \frac{x^{4n-2}}{(4n-2)!}$$ the Taylor polynomial $f_{4n-2}(x) \leq \cos(x).$ How do we prove this? The trouble is that the power series for $\cos$ is infinite, and there are infinitely-many negative terms and infinitely-many positive terms.

Answer 1

Suppose that you want to prove that$$(\forall x\in\Bbb R):\cos(x)\geqslant1-\frac{x^2}{2!}.$$This is the same thing as proving that$$(\forall x\in\Bbb R):\cos(x)-1+\frac{x^2}{2!}\geqslant0.$$Let $f(x)=\cos(x)-1+\frac{x^2}{2!}$. Then $f(0)=0$ and $f'(x)=-\sin(x)+x$. If it can be proved that $f'(x)\leqslant0$ when $x<0$ and that $f'(x)\geqslant0$ when $x>0$, we're done. But$$f''(x)=-\cos(x)+1\geqslant0.$$So, since $f'(0)=0$, and since $f'$ is increasing, we do indeed have that $f'(x)\leqslant0$ when $x<0$ and that $f'(x)\geqslant0$ when $x>0$. So, $(1)$ is indeed true.

Now, suppose that you want to prove that$$(\forall x\in\Bbb R):\cos(x)\leqslant1-\frac{x^2}{2!}+\frac{x^4}{4!}.$$Define$$g(x)=\cos(x)-1+\frac{x^2}{2!}-\frac{x^4}{4!}.$$ Then $g(0)=0$ and $g'(x)=-\sin(x)+x-\frac{x^3}{3!}$. If it can be proved that $g'(x)\geqslant0$ when $x<0$ and that $g'(x)\leqslant0$ when $x>0$, we're done. But $g''(x)=-\cos(x)+1-\frac{x^2}{2!}$, and we have already seen that then we always have $g''(x)\leqslant0$. So, since $g'(0)=0$, it is indeed true that $g'(x)\geqslant0$ when $x<0$ and that $g'(x)\leqslant0$ when $x>0$.

Can you take it from here?

Answer 2

Using $\sin' = \cos$ and $\cos' = -\sin$ get:

$$\sin(x) = \int_0^x \cos(t) dt$$

and

$$\cos(x) = 1-\int_0^x \sin(t) dt$$

From these, by induction, you can show that the successive sin and cos polynomials are alternatively above and below the corresponding functions.

Note: This does not originate with me. I saw this in Dorre's "100 Great Problems in Elementary Mathematics" which I highly recommend.

Answer 3

Alternate approach: use the Liebniz rule:

If $\langle a_n\rangle$ is a strictly decreasing sequence of positive terms, with limit $0$, then the alternating series
$\displaystyle \sum_{n=1}^\infty (-1)^{(n-1)}a_n$ converges.

Further, if $S$ denotes it sum and $s_n$ is its $n$-th partial sum then

$$0 < (-1)^n (S-s_n) < a_{n+1}.$$

This means that there is no reason to focus on the trigonometric association here. For any conditionally convergent alternating series, that is as described at the start of this answer, the partial sums will always oscillate from less than $S$ to greater than $S$ and back. Further, this oscillation will always get strictly closer and closer to $S$.

Of course, for the present problem, for any fixed value of $(x)$, you will have to wait a finite number of terms, before the absolute value of the terms start strictly decreasing.