1
$\begingroup$

The task is to:

Prove that $x=\sin x+\frac{1}{4}$ has got a unique solution on $[\pi/4,\pi/2]$. Show that the iteration $x_0\in[\pi/4,\pi/2], x_{n+1}=\sin x_n+\frac{1}{4}$ is convergent to that solution.

I tried to show the first part using the graphs of functions $\sin x$ and $x-1/4$. But I am not sure if this is the best way (is it formal...). The next step is a bit harder to me. Should I somehow use the Picard-Lindelof theorem?

$\endgroup$
2
$\begingroup$

Let $$ f(x)=\sin x+\frac14. $$ If $\pi/4\le x\le \pi/2$ then $$ \frac\pi4<\frac{\sqrt2}{2}+\frac14\le f(x)\le1+\frac14<\frac\pi2. $$ Thus $f([\pi/4,\pi/2])\subset[\pi/4,\pi/2]$. Moreover $$ |f'(x)|=|\cos x|\le\cos\frac\pi4=\frac{\sqrt2}{2}<1,\quad x\in[\pi/4,\pi/2]. $$ The contraction principle,or Banach fixed point theorem, gives the result.

$\endgroup$

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.