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Math Help - Approximate 5th root of 20

  1. #1
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    Approximate 5th root of 20

    f(x) = x^5 - 20

    f'(x) = 5x^4

    x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
    x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
    x2 = 1 - ( -19 / 5 )
    x2 = 24 / 5
    x2 = 4.08

    My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by TsAmE View Post
    f(x) = x^5 - 20

    f'(x) = 5x^4

    x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
    x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
    x2 = 1 - ( -19 / 5 )
    x2 = 24 / 5
    x2 = 4.08

    My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.
    You can use Taylor...
    x^1/5 ~ 1+(1/5)*(x-1) - (2/25)*(x-1)^2)+(6/125)*(x-1)^3+O((x-1)^4)
    You put x=20 and get the needed.
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  3. #3
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    I havent learned taylor, I want to do it using newtons method
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  4. #4
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    Your method is correct. You made a little slip. 24/5=4.8 rather than 4.08. Just keep going.

    On a decent calculator you can do this:

    1.5 =

    ANS - (ANS^5-20)/(5ANS^4) = = = = =

    It gives
    1.9901..
    1.8470..
    1.8213..
    and pretty soon you will have
    1.820564203
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  5. #5
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    Quote Originally Posted by TsAmE View Post
    f(x) = x^5 - 20

    f'(x) = 5x^4

    x2 = x1 - ( x1^5 - 20 ) / ( 5x1^4 )
    x2 = 1 - ( (1)^5 - 20 ) / ( 5(1)^4 )
    x2 = 1 - ( -19 / 5 )
    x2 = 24 / 5
    x2 = 4.08

    My answer is wrong and the correct answer is 1.82056420. I used an initial approximation of x1 = 1, as no approximation was given.
    Well, 24/5= 4.8, not 4.08 but what exactly was the problem? Surely not to do just one iteration? If you keep going, you get successive approximations that converge to 1.82...
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