Prove the inequalities $1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$

Question

$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$

I think I should check the two sides. but how can I show that $1-\frac{x^2}{2}- \cos(x)\le 0$? And on the other side $0\le-\cos x+1-\frac{x^2}{2}+\frac{x^4}{24}$

Answer 1

This is a straight application of the Leibniz rule for the convergence of an alternating series.

The cosine series is alternating and for all alternating series you have that the partial sums are alternating upper and lower bounds for the value of the series.

If $a_n$ is decreasing to zero and $s_n=\sum_{k=0}^n(-1)^ka_k$ then $$ s_{2m+1}\le s_\infty\le s_{2n} $$ for all $m,n\in\Bbb N$.

Here $a_n=\frac{x^{2n+2}}{(2n+2)!}$, so $s_\infty=1-\cos x$ and this sequence is decreasing for $x^2<12$.

Answer 2

By Taylor-Lagrange, there is $\theta_x\in (0,1)$ s.t. $$\cos(x)=1-\frac{x^2}{2}\cos(\theta_x x).$$ Since $-\cos(\theta_x x)\geq -1$, we get $$\cos(x)\geq 1-\frac{x^2}{2}$$

In the same way, there is $\eta_x\in (0,1)$ s.t.

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}\cos(\eta_x x).$$

Since $\cos(\eta_x x)\leq 1$, we get $$\cos(x)\leq 1-\frac{x^2}{2}+\frac{x^4}{24}.$$

Answer 3

Using only elementary calculus:

Let $f(x) = \cos x - \left( 1 - \frac{x^2}{2} \right)$. Then $f(0) = 0$ and $f'(x) = -\sin x + x$. $f'(x) > 0$ for all $x > 0$ and $f'(x) < 0$ for all $x < 0$.

Thus

$$f(x) = \int_0^x f'(t) \ dt > 0 \quad\text{ for all } x > 0$$ and

$$f(x) = \int_0^x f'(t) \ dt = - \int_x^0 f'(t) \ dt > 0 \quad\text{ for all } x < 0$$

Hence $f(x) \geq 0$ for all $x$ and therefore $$1 - \frac{x^2}{2} \leq \cos x$$

Similarly, let $\displaystyle g(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cos x$. Then $g(0) = 0$, $g'(x) = -x + x^3/6 + \sin x$ and $g''(x) = -1 + x^2/2 + \cos x$.

By what we just showed above $g''(x) \geq 0$ for all $x$.

Thus $\displaystyle g'(x) = \int_0^x g''(t) \ dt > 0$ for all $x > 0$ and $g'(x) < 0$ for all $x < 0$. And hence (mirroring the analysis for $f$), $g(x) \geq g(0) = 0$ for all $x$.

Therefore

$$\cos x \leq 1 - \frac{x^2}{2} + \frac{x^4}{24}$$

Answer 4

You can use Taylor's theorem.

In general, under appropriate smoothness assumptions we have $ f(x) = \sum_{k=0}^n {1 \over k!} f^{(k)}(a) (x-a)^k+ {1 \over (k+1)!} f^{(k+1)}(\xi) (x-a)^{k+1}$, for some $\xi \in (a,x)$ (suitably interpreted to account for sign).

We let $a=0$ and note that $\cos^{(2k+1)}(0) = 0$ and $\cos^{(2k)}(0) = (-1)^k$ which gives $\cos x = \sum_{j=0}^n {1 \over (2j)!} (-1)^j x^{2j} + {1 \over (2(n+1))!} (-1)^{n+1} \xi^{2(n+1)}$, for some $\xi \in (-x,x)$.

If $n$ is odd ($n=1$ in the question), we see that the last term satisfies $0 \le {1 \over (2(n+1))!} (-1)^{n+1} \xi^{2(n+1)} \le {1 \over (2(n+1))!} x^{2(n+1)}$ from which we get the desired result.

Answer 5

HINT: Use the Taylor series expansion of cosine, along with the error theorem

Answer 6

Let me assume that you define $\cos(x)$ to be a function satisfying $\cos' = -\sin$ and $\sin' = \cos$. Then you can analyze the function $$f(x) = \cos(x) - 1 + \frac{x^2}{2}.$$ Taking derivative yields $$f'(x) = -\sin(x) + x.$$ The minimal/maximal of $f$ occurs at the place where $f'(x) = 0$. Let $\alpha$ be solution to $\sin(x) = x$. $$f(\alpha) \geq 0.$$ This is easily done by knowing that $\cos^2(x) + \sin^2(x) = 1$.

Answer 7

You can use the following corollary of the Mean Value theorem:

Let $I$ be an interval of $\mathbf R$, and $a\in I$. Suppose two differentiable functions $f$ and $g$ are defined on $I$ and satisfy

1. $f(a)\le g(a)$,
2. for all $x>a$ in $I$, $\;f'(x)\le g'(x)$.

Then for all $x>a$ in $I$, $\;f(x)\le g(x)$.