1. ## numerical analisys question..

A.
show that $\displaystyle \frac{x}{2}=\sin x$ has a root on [1,2]

use the two steps of the crossing method that contains the root.
B.
find the root of the equation using newton referson methos.

solution of the book:
f(0)=0
f(pi/3)>0
so it rises and the is no root in [0,pi/3] interval

then they say that in [pi/3,pi] f(x) going down because f'(x)<0

so there is a solution in this interval

but how does f'(x)<0 shows that there is a soltion??

in order to see that there is a solution we need to have two point one positive one negative and that f(x) is going down

why they check from 0 to pi
and not from [1,2] like asked in the question

2. after doing two steps in the cutting technique
i see that the solution lies [1.75,2]
which number to put in the referson formula to find the solution

3. Originally Posted by transgalactic
A.
show that $\displaystyle \frac{x}{2}=\sin x$ has a root on [1,2]

use the two steps of the crossing method that contains the root.
B.
find the root of the equation using newton referson methos.

solution of the book:
f(0)=0
f(pi/3)>0
so it rises and the is no root in [0,pi/3] interval

then they say that in [pi/3,pi] f(x) going down because f'(x)<0

so there is a solution in this interval

but how does f'(x)<0 shows that there is a soltion??

in order to see that there is a solution we need to have two point one positive one negative and that f(x) is going down

why they check from 0 to pi
and not from [1,2] like asked in the question
My guess is that you are looking at the wrong solution in the book.

Put

$\displaystyle f(x)=\sin(x)-\frac{x}{2}.$

Then $\displaystyle f(1)\approx 0.34$ and $\displaystyle f(2)\approx-0.09$ and as $\displaystyle f(x)$ is continuous and changes sign on $\displaystyle [0,1]$ it has a root in $\displaystyle [0,1]$

Now $\displaystyle f(1.5) \approx 0.25$ , so we replace the bottom limit to get a new interval $\displaystyle [1.5,2]$ for our root.

Next step the new trial point is $\displaystyle 1.75$ and $\displaystyle f(1.75)\approx 0.11$, so again we replace the bottom limit to get the new interval $\displaystyle [1.75,2]$.

CB