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Math Help - Numerical method,finding root in iteration method.

  1. #1
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    Numerical method,finding root in iteration method.

    hello i have problem with solution. please clear my doubt.

    my doubt is how did the writer get x0 = 3.8
    how did he get x0 value.
    Attached Thumbnails Attached Thumbnails Numerical method,finding root in iteration method.-codecogseqn.gif  
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  2. #2
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    Quote Originally Posted by avengerevenge View Post
    hello i have problem with solution. please clear my doubt.

    my doubt is how did the writer get x0 = 3.8
    how did he get x0 value.
    Hi avengerevenge,

    Maybe from graphing

    f(x)=2x-log_{10}x-7=0

    Or..

    10^{2x-log_{10}x}=10^7

    \frac{10^{2x}}{10^{log_{10}x}}>10^7

    \frac{10^{2x}}{x}>10^7

    10^{2x}>x10^7

    x=1,\ 10^2<10^7

    x=2,\ 10^4<(2)10^7

    x=3,\ 10^6<(3)10^7

    x=4,\ 10^8>(4)10^7

    3<x<4
    Attached Thumbnails Attached Thumbnails Numerical method,finding root in iteration method.-root-iteration.jpg  
    Last edited by Archie Meade; March 6th 2010 at 04:05 AM.
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  3. #3
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    Your attachment is hard to read but it says:
    "Find real root of 2x- log_{10}(x)= 7 by iteration method

    The answer is x= \frac{1}{2}(log_{10}(x)+ 7)

    x_0= 3.8
    x_1= \frac{1}{2}(log_{10}(x)+ 7)= 3.7893
    x_2= \frac{1}{2}(log_{10}(x)+ 7)= 3.7893

    The answer is 3.7893."

    How did he get x_0= 3.8? Pretty much an "educated" guess. Under the right conditions, if x_0 is any number reasonably close to the answer to begin with, this sequence will converge to the solution to the equation. I suspect that this author already knew the answer and chose 3.8 because it was already correct to the first decimal place.

    If I were coming to this equation without knowing the answer I might argue "log(x) is always smaller than x so what happens if I ignore it? The equation becomes 2x= 7 and x= 7/2= 3.5. I think I will try x_0= 3.5." Then I would get:

    x_0= 3.5

    x_1= \frac{1}{2}(log_{10}(3.5)+ 7)= 3.7720

    x_2= \frac{1}{2}(log_{10}(3.7720)+ 7)= 3.7883

    x_3= \frac{1}{2}(log_{10}(3.7883)+ 7)= 3.7892

    x_4= \frac{1}{2}(log_{10}(3.7892)+ 7)= 3.7893

    x_5= \frac{1}{2}(log_{10}(3.7893)+ 7)= 3.7893

    Since those last two iterations are the same to the four decimal places I am keeping, I can stop here, confident that my answer is correct at least to three decimal places. It took a couple more iterations because my first value was not as close to the correct answer as 3.8 but I still get the same answer.

    Notice that if I had started with x_0= 1, my first step would have been
    x_1= \frac{1}{2}(log_{10}(1)+ 7)= 7/2= 3.5

    And then the iteration would be the same as above.


    If I had started with something really bad, say x_0= 100, then I would have

    x_1= \frac{1}{2}(log_{10}(100)+ 7)= 27/2= 13.5
    x_2= \frac{1}{2}(log_{10}(13.5)+ 7)= 4.0652
    x_3= \frac{1}{2}(log_{10}(4.0652)+ 7)= 3.8045
    x_4= \frac{1}{2}(log_{10}(3.8045)+ 7)= 3.7901

    and I am heading right back to 3.7893.
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    Hi avengerevenge,

    Maybe from graphing

    f(x)=2x-log_{10}x-7=0

    Or..

    10^{2x}-x=10^7
    Hopefully, not! 10^{2x- log_{10}x}= \frac{10^{2x}}{x}, not 10^{2x}- x.

    10^{2x}>10^7

    2x>7\ \Rightarrow\ x>3.5
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