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Thread: newton's method?

  1. #1
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    newton's method?

    Am i approximate 2 ^ (1/5) correctly with newton's method? (assuming x0 = 1, find x1,x2)


    since 2 ^ (1/5) = x

    i get the function f(x) = x^5-2
    f'(x) = 5x^2

    therefore, x0 = 1


    x1 = 1- f(1)/f'(1) = = -2/5

    but when i find x2 = -2/5 - f(2/5)/f'(2/5) = -18.67

    i get a large negative number for x2 but x1 is like -2/5. Is this correct? The difference is so large :/
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  2. #2
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    Re: newton's method?

    Hi lc99

    Please note that your f'(x) is incorrect
    and also, substitutions a negative value in f(x) and f'(x) should include its negative sign....

    My advice in newton's method is write the whole equation first:

    {x_n} = {x_{n - 1}} - \frac{{f({x_{n - 1}})}}{{f'({x_{n - 1}})}}
    thus, in this case,
    {x_n} = {x_{n - 1}} - \frac{{{x_{n - 1}}^5 - 2}}{{5{x_{n - 1}}^4}}

    then you can substitute your value in this equation....

    Please try again
    Thanks from topsquark
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  3. #3
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    Re: newton's method?

    Opps!, i mistyped it, i had it written down as 5x^4 .
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  4. #4
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    Re: newton's method?

    when i found x1 = 6/5, i get really big numbers when i plug it back in to find x2. The numbers get very large. Am i suppose to be able to do this without a calculator cause this problem is part a midterm exam review. On the real exam, we don't get calculators
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  5. #5
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    newton's method?

    Hi lc99
    if you not able to use calculator, then simplify the expression is the best way to solve it....
    for example in this case,
    {x_n} = {x_{n - 1}} - \frac{{{x_{n - 1}}^5 - 2}}{{5{x_{n - 1}}^4}}
    {x_n} = {x_{n - 1}} - \left( \frac{{{x_{n - 1}}^5}}{{5{x_{n - 1}}^4}} - \frac{2}{{5{x_{n - 1}}^4}} \right)
    {x_n} = {x_{n - 1}} - \left( \frac{{{x_{n - 1}}}}{5} - \frac{2}{{5{x_{n - 1}}^4}} \right)
    {x_n} = \frac{4}{5}{x_{n - 1}} + \frac{2}{{5{x_{n - 1}}^4}}
    {x_2} = \frac{4}{5}\left( {\frac{6}{5}} \right) + \frac{2}{{5{{\left( {\frac{6}{5}} \right)}^4}}}

    then u can evaluate it by simplify it even more simple to get the final answer in fraction form!
    Thanks from lc99
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  6. #6
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    newton's method?

    In addition, the number in the both numerator and denomenator are getting larger because the answer is getting more precise to the {2}^{\frac{1}{5}} and this is not a rational number
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  7. #7
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    Re: newton's method?

    Hopefully, the ones on my exam will have cleaner numbers!!
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