Apply the Mean Value Theorem to show that:

$$|\cos(\frac{x_1}{2}) - \cos(\frac{x_2}{2})| \le \frac12|x_1 - x_2|,\ \text{ for any $x_1$ and $x_2$ in $[0, 2\pi]$}.$$

What I have done is this: use $f'(c) = [f(b) - f(a)]/(b-a)$. Here $2\pi$ is $b$ and 0 is $a$. I then treated both $x_1$ and $x_2$ as $c$ like this: $f'(x_1) = [f(2\pi) - f(0)]/2\pi. I found the function values at $2\pi$ and at $0$.

I got $[(\cos\pi - \cos\pi)]/(2\pi-0)$ which equals $0$. I then took the derivative with respect to $x_1$. So, $f'(x1) = -(1/2\sin(x_1/2) - \cos(x_2/2) = 0$.

I did the same thing again, this time using $x_2$ and $f'(x_2)$ and I got $0$ again.

So, by applying the MVT to $x_1$ and $x_2$ I get $0 \le (1/2)|x_1 - x_2|$.

I really do not think this process is correct.

  • $\begingroup$ Write your version of the MVT as (by taking absolute values): $$|f(x_1)-f(x_2)|=|f'(c)|\cdot |x_1-x_2|$$ Let $f(x)=\cos(x/2)$ and $c\in [0,2\pi]$. Now, what is the maximum value of $|f'(c)|$? $\endgroup$ – projectilemotion Sep 5 '19 at 21:48
  • $\begingroup$ Welcome to MSE. Please use MathJax to format your questions. $\endgroup$ – saulspatz Sep 5 '19 at 21:52
  • $\begingroup$ To find the maximum value of |f ' (c)|, using f(x) = cos(x/2): [f(2π) - f(0)] / 2π. I get | -1/2π | which is 1/2π for the maximum value? $\endgroup$ – Tim Sep 5 '19 at 22:07
  • $\begingroup$ Nope. What do you get if you differentiate $f(x)$? $\endgroup$ – projectilemotion Sep 5 '19 at 22:11
  • $\begingroup$ f ' (x) = -1/2 sin (x/2) $\endgroup$ – Tim Sep 5 '19 at 22:17

Claim: If $0 \leq x_1 \lt x_2 \leq 2\pi$, then $$\left \vert \cos \frac{x_1}{2}- \cos \frac{ x_2}{2} \right \vert \leq \frac 12 \vert x_1 - x_2 \vert.$$

Proof: Define $f(x)= \cos \frac x2$. Then $f'(x)= - \frac 12 \sin \frac x2.$ Thus, for any $x_1, x_2 \in [0, 2\pi]$, the Mean Value Theorem (applied with $a=x_1, b= x_2$) tells us $\exists c$ with $x_1 \lt c \lt x_2$ such that

$$\left \vert \frac {f(x_1)-f(x_2)}{x_1-x_2} \right \vert \leq \vert f'(c) \vert.$$

Substituting, we have

$$\left \vert \frac{\cos \frac {x_1}{2} - \cos \frac{x_2}{2}}{x_1 - x_2} \right \vert \leq \frac 12 \vert \sin \frac c2 \vert \leq \frac 12.$$

The last inequality holds because $\vert \sin \frac c2 \vert \leq 1$ for any (real) value of $c$. Now just multiply through by $\vert x_1-x_2 \vert$ and the proof is complete.


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