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anonymous

  • one year ago

Need Help! Use Newton's method to find all roots of the equation (x-2)^2=ln(x). Must be correct to 6 decimal places.

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    is this an on line class or do you really have to use newton's method?

  3. anonymous
    • one year ago
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    I really have to use newtons method...

  4. misty1212
    • one year ago
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    ok then you have to start with something like \[f(x)=\ln(x)-(x-2)^2\] and look for the zero of that one

  5. misty1212
    • one year ago
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    better still \[f(x)=x^2-4x+4-\ln(x)\]

  6. misty1212
    • one year ago
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    then you make a guess you got a good guess in mind?

  7. misty1212
    • one year ago
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    woah lot of algebra method you can use that too, but it is sometimes easier not to

  8. misty1212
    • one year ago
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    lets make a guess i guess 3

  9. misty1212
    • one year ago
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    \[f(3)=3^2-4\times 3+4-\ln(3)=-0.098612\]

  10. misty1212
    • one year ago
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    pretty close to zero already

  11. misty1212
    • one year ago
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    you gotta start somewhere right?

  12. misty1212
    • one year ago
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    figure it is easier to start with an integer

  13. DanJS
    • one year ago
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    for reference if needed... http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx

  14. misty1212
    • one year ago
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    now for the second approxmition it is \[3-\frac{f(3)}{f'(3)}\]

  15. misty1212
    • one year ago
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    \[f'(x)=2x-4-\frac{1}{x}\] so \[f'(3)=2\times 3-4-\frac{1}{3}=\frac{5}{3}\]

  16. misty1212
    • one year ago
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    so next guess is \[3+\frac{-0.098612}{\frac{5}{3}}\] at this point you really need a calculator

  17. misty1212
    • one year ago
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    this gets very very messy quick but you can do it with wolfram or a spreadsheet

  18. misty1212
    • one year ago
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    i get \[x_2=2.9408328\]

  19. misty1212
    • one year ago
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    now you have to repeat it so it is really going to be messy

  20. anonymous
    • one year ago
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    Yea, didnt get it...

  21. misty1212
    • one year ago
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    what did you not get?

  22. anonymous
    • one year ago
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    why are you using 3? whats so special about it

  23. misty1212
    • one year ago
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    oh , you have to make an educated guess first you can guess, graph, use wolfram (what i did) you have to do something to guess a number close to the zero of the function

  24. misty1212
    • one year ago
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    here is a picture of the function http://www.wolframalpha.com/input/?i=x%5E2-4x%2B4-ln%28x%29 click on "real values plot" so as not to get too confused you see it has two zeros

  25. misty1212
    • one year ago
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    one is very near 3, so i picked 3 as \(x_1\)

  26. mathmate
    • one year ago
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    ouch! I think it should be: \(\large 3\color{red}{-}\frac{-0.098612}{\frac{5}{3}}\)

  27. misty1212
    • one year ago
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    yeah you are right, it should be a plus

  28. misty1212
    • one year ago
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    so \[x_2=3.0591672\]

  29. misty1212
    • one year ago
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    you really want to make life easy on yourself here is what you can do for newton's method you are going to have to use a calculator anyways, so here is what we get if we make a first guess of 3 http://www.wolframalpha.com/input/?i=x-%28%28x-2%29%5E2-ln%28x%29%29%2F%282x-4-1%2Fx%29%2C+x%3D3

  30. misty1212
    • one year ago
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    click on the numeric answer, then rewrite with it substituted where i wrote x = 3

  31. misty1212
    • one year ago
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    i got this http://www.wolframalpha.com/input/?i=x-%28%28x-2%29%5E2-ln%28x%29%29%2F%282x-4-1%2Fx%29%2C+x%3D3.05917

  32. mathmate
    • one year ago
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    To resume how we could solve a problem using Newton's method. 1. define the equation in the form of f(x)=0, because Newton's method is for finding zeroes of an equation. Here (x-2)^2=ln(x) becomes f(x)=(x-2)^2-ln(x)=0 2. Graph the function and visually search for zeroes, which will be the potential initial values of the solutions. Also, visually check the graph to be convinced that all possible roots are counted. for the above f(x), we see that the graph is basically a quadratic, slightly modulated by the function ln(x), and which also restricted the domain to x>0. If we examine the graph between 1 and 4, we see that the roots are located near 1.5- and 3+. These will therefore be our initial approximations. Note: if a root of higher multiplicity is suspected, a different procedure from the following is required. 3. set up the iterative equation, x(n+1)=x(n) - f(x(n))/f'(x(n)) Here we have f(x)=(x-2)^2-ln(x) f'(x)=2(x-2)-1/x so the iterative equation is (x is used to stand for x(n) for simplicity) x(n+1)=x(n)-f(x)/f'(x) =x(n)- ((x-2)^2-ln(x))/)2(x-2)-1/x) =(xln(x)+x^3-5x)/(2x62-4x-1) 4. Apply the iteration solution a number of times to attain the required accuracy. Keep at least 2 decimals beyond the required accuracy for all calculations. Newton's method typically doubles the number of correct digits at each iteration, except for special circumstances such as multiple roots. Use calculator, Excel, Wolfram or any other tool to help you calculate accurately. Here the results are: A. for initial approximation of 3: 3 3.059167373200866 3.0571060546916 3.057103549998436 3.057103549994738 B. for initial approximation of 1.5 : 1.5 1.406720935135101 1.412369957251193 1.412391171725005 1.412391172023884 Clearly, the roots are 3.057104 and 1.412391 each to 6 decimal places.

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