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Math Help - Newtons Method

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    Newtons Method

    Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
    Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

    I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coe236 View Post
    Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
    Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

    I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?
    there are two solutions for x. one is around 2.5 (a little more), the other is between 0.2 and 0.5.

    first rewrite: e^x - 5x = 0

    thus we let f(x) = e^x - 5x

    now we guess a root. let's say we're going after the one around 2.5. so we guess 2.5 (that is, we let x_n = 2.5)

    then x_{n + 1} = 2.5 - \frac {f(2.5)}{f'(2.5)}

    you will get a value for x_{n + 1}

    then repeat the same process, using the last value you got as x_n, and keep doing it until you realize the values settling (you always get the same first three decimal places).

    do this again for the other root, guess 0.5


    below is the graph
    Attached Thumbnails Attached Thumbnails Newtons Method-newton.jpg  
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  3. #3
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    Quote Originally Posted by coe236 View Post
    Use Newtons method to find the two solutions of e^x=5*x to three decimal places.
    Newtons Method: X(n+1) = Xn - f(Xn)/f'(Xn)

    I just dont understand how you would set this up. Do you set both the approximations equal to each other to solve or do you solve them individually and keep checking?
    Hello,

    1. make a rough sketch: You'll notice that there are 2 points of intersection thus

    2. the equation e^x-5x = 0 have 2 solutions.

    3. consider the function f(x) = e^x-5x~\implies~f'(x) = e^x-5

    then you have:

    x_{n+1}=x_{n}-\frac{e^{x_n}-5x_n}{e^{x_n}-5}

    Now use your calculator:
    Code:
    start:             0                  2
    2. step:       0.25                3.092876765
    3. step:       0.25915...          2.70696...
       .
       .
    answer:        0.259171101..       2.542641358...
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  4. #4
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    ohh i understand, I kept solving for x and that messed me up. That sounds simple enough. Thanks guys
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  5. #5
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    Quote Originally Posted by coe236 View Post
    ohh i understand, I kept solving for x and that messed me up. That sounds simple enough. Thanks guys
    Hello,

    if (and only if) your calculator has a \boxed{ANS}-key then these kinds of problems are simpler than they appear.

    With your problem type:

    0 = [remark: Now the zero is stored in the ANS]
    ANS - ((e^ANS-5*ANS)/(e^ANS-5)) = [now you have the 2nd result]
    = [you only hit the \boxed{=}-key]
    =
    ...
    = until the results don't change

    You have finished the complete problem within seconds!
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  6. #6
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    hehe, yeah thats going to be useful method to use on exams. Thankyou =]
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