2
$\begingroup$

Prove that the equation $\cos x - x -1/2 = 0$ has a unique real solution.

My solution: I was starting with a function $F(x) = \cos x - 1/2$ and the interval $[0,\pi/4]$ and trying to show that the fixed point theorem is applicable here. If I can find this solution to a few decimal places then I have shown that there is unique real solution. However, I am having some trouble applying the Fixed Point Theorem to $F(x)$. Any help would be greatly appreciated.

$\endgroup$
  • $\begingroup$ There are quite a few fixed point theorems, which one are you using? There is one where, if you can show that $F(x)$ is a contraction map, that you will indeed have a fixed point. Don't forget about the power of the Intermediate Value Theorem here as well. $\endgroup$ – DaveNine Sep 24 '15 at 15:06
  • $\begingroup$ Where did you get that exercise? $\endgroup$ – Nash Nov 16 '16 at 16:18
3
$\begingroup$

An other way

Let $f(x)=\cos(x)-\frac{1}{2}$. $$|f(x)-f(y)|=|\cos(x)-\cos(y)|.$$ Using the mean value theorem, we can show that $$|\cos(x)-\cos(y)|\leq \frac{\sqrt 2}{2}|x-y|$$

and thus $f$ is a contraction mapping. By Banach fixe point theorem, $f$ has a unique fix point, and thus, there is a unique $x$ such that $f(x)-x=0$ what prove the claim.

ADDED :

By mean value theorem, there is a $c=c_{x,y}\in]0,\pi/4[$ such that $$\cos(y)-\cos(x)=-\sin(c)(y-x)\implies -|\sin(c)||y-x|\leq \cos(y)-\cos(x)\leq |\sin(c)||y-x|\implies |\cos(x)-\cos(y)|\leq |\sin(c)||y-x|\leq \sin(\pi/4)|y-x|=\frac{\sqrt 2}{2}|y-x|.$$

$\endgroup$
  • $\begingroup$ Can you edit to show how you got <= √2/2 |x−y| $\endgroup$ – YesItsMario Sep 24 '15 at 15:29
  • $\begingroup$ Ok, it's done :-) $\endgroup$ – idm Sep 24 '15 at 15:37
2
$\begingroup$

Let $f(x)=\cos x-x-1/2$. $f'(x)\le0$ implies $f$ is monotonic. And the zeros of $f'$ are isolated, so $f$ is strictly monotonic. Note that $f(0)>0$ and $f(1000)<0$.

$\endgroup$
  • $\begingroup$ This proves there are some zeroes, not that there is a unique zero. $\endgroup$ – Did Sep 25 '15 at 17:46
  • $\begingroup$ @Did Thanks. I have filled that gap. $\endgroup$ – Eclipse Sun Sep 26 '15 at 0:39
1
$\begingroup$

$F(x) = \cos x - \frac{1}{2}$ then $F'(x) =-\sin x \Rightarrow \vert -\sin x \vert \leq \frac{\sqrt{2}}{2}<1 $ and $F'(x) =-\sin x<0$ in $[0,\frac{ \pi}{4}]$ thus max happen in $0$ and min in $\frac{ \pi}{4}$ where $f(0)=\frac{1}{2}$ and $f(\frac{ pi}{4})=\frac{\sqrt{2}}{2}-\frac{1}{2}$ thus $F$ maps $[0,\frac{ \pi}{4}]$ to $[0,\frac{ \pi}{4}]$. Now apply the Fixed Point Theorem to F(x). finished

$\endgroup$
0
$\begingroup$

Another approach to this problem can be:

Let $f(x)=\cos x$ and $g(x)=x+\frac{1}{2}$

The first and the most important thing we note here is that these two functions are continuous

Now, at $x=0$ $$f(0)=1>\frac{1}{2}=g(0)$$ At $x=π/2$ $$f(π/2)=0<\frac{π+1}{2} =g(π/2)$$

Thus, there has to exist a real point P where these two points meet, i.e. at some real $x=u$, $$f(u)=g(u)$$ $$\implies \cos u=\frac{1}{2}+u$$ $$\implies \cos u-u-\frac{1}{2}=0$$ And thus we have proved the existence of a real solution for this equation. $$Q.E.D.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.