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Newton's Method General Formula

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    The statement [itex]\sqrt[4]{a}=x[/itex] means that [itex]x^{n}=a[/itex].

    Using this, we can approximate the radical [itex]\sqrt[n]{a}[/itex] by approximating the

    solution to the equation [itex]x^{n}-a=0[/itex].

    Consider the function [itex]f(x)=x^{n}-a[/itex].

    We can use Newton's Method to approximate where f(x)=0 and thus approximate the radical

    [itex]\sqrt[n]{a}[/itex].


    a) Use Newton's Method with the function [itex]f(x)=x^{n}-a[/itex]

    to obtain a general formula approximating [itex]\sqrt[4]{a}[/itex].


    b) Enter [itex]\sqrt[4]{100}[/itex] using your calculator and give the approximation to the

    accuracy found by your calculator.


    c) Use the formula found in (a) and make a table of values to approximate

    [itex]\sqrt[4]{100}[/itex] to the same accuracy as your calculator. Use 3

    as your initial guess.


    d) How many iterations are required to obtain this same accuracy?


    e)The fundamental theorem of algebra guarantees 4 solutions to x^4-100=0.

    you just found one. Are there more real solutions? Use your tools of calculus to sustain

    your answer.


    2. Relevant equations

    [itex]x_{2}=x_{1}-\frac{f(x_{1}}{f'(x_{1}}[/itex]

    3. The attempt at a solution

    a)[itex]f(x)=x^{n}-a[/itex]

    Am I supposed to assume a is constant here? If so, then:

    [itex]f '(x)=nx^{n-1}[/itex]

    [itex]x_{2}=x_{1}-\frac{x^{n}-a}{nx^{n-1}}[/itex]


    b) Easy enough, plugged it in to the calc to get 3.16227766.


    c) I want to make sure I have part a) right before making a chart.


    d) This will be simple after part c)


    e) I can tell I will be stuck on this final part. Any tips to help me get started?
     
  2. jcsd
  3. Jun 17, 2011 #2

    I like Serena

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    Hi crybllrd! :smile:

    You have (a) right!
    Just fill in n=4 and a=100 and you're good to go!

    Cheers! :wink:
     
  4. Jun 17, 2011 #3
    Thanks for the reply.

    I'm working on part c). She has a set up a chart like this:

    2aiej68.jpg


    I guess I'm confused as to where I use n=4 and n=iteration
     
  5. Jun 17, 2011 #4

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    Yes. It's a pity n is used for 2 different things.

    Let's just say that the n in your formula needs to be replaced by 4, while all the occurrences of n in your table should have been named k or i.
     
  6. Jun 17, 2011 #5
    Oh ok, not sure why my teacher would do that to us :P

    However, my numbers are not converging.

    My n=1 is 3.175925926

    n=2 is 2.986439461

    n=3 is 3.191986488

    n=4 is 2.970703369

    I stopped there because they should be converging to 3.16227766.

    Here's what I am using:

    f(x)=x^4-100

    f '(x)=4x^3

    so, [itex]x-\frac{x^{4}-100}{4x^{3}}[/itex], where x is the previous approximation (x1=3)

    Am I missing something?
     
  7. Jun 17, 2011 #6

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    Not sure what you did, because starting from n=2 I get a different sequence, which is converging.

    To make it explicit:

    [itex]x_{k+1} = x_k-\frac{{x_k}^{4}-100}{4{x_k}^{3}}[/itex]

    and

    [itex]3.1759-\frac{3.1759^{4}-100}{4 \cdot 3.1759^{3}} = 3.162[/itex]
     
  8. Jun 17, 2011 #7
    Not sure what I did either, but I redid it and got it in 3 iterations.

    For the final part, I should have 4 solutions to x^4-100=0,

    3.16227766 being one of them, and -3.16227766 would be another.

    How would I go about getting the others? I'm assuming they are not real solutions.
     
  9. Jun 17, 2011 #8

    I like Serena

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    Quite right.

    The other 2 roots are:
    3.16227766 i
    and
    -3.16227766 i

    where i is the imaginary constant, or sqrt(-1).

    What they would have intended (I think), is that you can find the negative root from the fact that looking at the graph of y = x^4 - 100, you'll see 2 real solutions. ;)
     
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