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Thread: numerical analisys question..

  1. #1
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    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
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  2. #2
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    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
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by transgalactic View Post
    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
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