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Math Help - Application of Newton's Method

  1. #1
    Junior Member alane1994's Avatar
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    Application of Newton's Method

    Hello!
    I have a homework problem that is involving Newton's Method.
    You have two equations that make up a graph. I had to graph them.

    y=\frac{1}{x} and y=64-25x^2

    From there I had to approximate where the functions intercept(x value). On the online homework program that my class uses, I approximated the values of -1.6,0.01, and 1.6 for where they intercept.

    Then it asks the graphs intersect when x is approximately_________.
    There are three answers, I got them all wrong.
    The answers were -1.607756 , 0.015626 , 1.592130.
    What I want to know is how do you find those values? Please help, this confuses me greatly.
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  2. #2
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    Re: Application of Newton's Method

    Quote Originally Posted by alane1994 View Post
    Hello!
    I have a homework problem that is involving Newton's Method.
    You have two equations that make up a graph. I had to graph them.

    y=\frac{1}{x} and y=64-25x^2

    From there I had to approximate where the functions intercept(x value). On the online homework program that my class uses, I approximated the values of -1.6,0.01, and 1.6 for where they intercept.

    Then it asks the graphs intersect when x is approximately_________.
    There are three answers, I got them all wrong.
    The answers were -1.607756 , 0.015626 , 1.592130.
    What I want to know is how do you find those values? Please help, this confuses me greatly.
    Newtons method is used to approximate real zeros of a function.

    We can define a new function by subtracting the two equations gives to get

    f(x)=25x^2+\frac{1}{x}-64=\frac{25x^3-64x+1}{x}

    We want to find the zero's of this function.

    Newton's method is given by

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

    So f'(x)=50x-\frac{1}{x^2}=\frac{50x^3-1}{x^2}

    This gives

    x_{n+1}=x_n-\frac{\frac{25x_{n}^3-64x_{n}+1}{x_{n}}}{\frac{50x_{n}^3-1}{x_{n}^2}}=\frac{25x_{n}^4-64x^2_{n}+x_{n}}{50x_{n}^3-1}

    This will zero in on each root. Each time you do the process again you will get closer to exact answer.

    If you use your seed of 1.6 you will get the accuracy you want after 2 interations.

    Let x_1=1.6 then plug this into the above to get x_2 and repeat...

    Here is a spread sheet that will allow you to play around with different seed values.

    Newton.xls
    Thanks from alane1994
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Application of Newton's Method

    I would set:

    \frac{1}{x}=64-25x^2 and then define:

    f(x)=25x^2-64+\frac{1}{x}=0 and we find:

    f'(x)=50x-\frac{1}{x^2}

    and so we will have:

    x_{n+1}=x_n-\frac{25x_n^2-64+\frac{1}{x_n}}{50x_n-\frac{1}{x_n^2}}

    Looking at a graph of f(x) we see the smallest root is near x=-1.6 so we set:

    x_0=-1.6

    What I do on my TI-89 Titanium is the following:

    -1.6 [ENTER]

    result: -1.6

    ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

    result: -1.60777453839

    Now, just keep hitting [ENTER] to get successive approximations:

    -1.60775601298

    -1.60775601287

    -1.60775601287

    The limit of accuracy of the calculator has been reached.

    Now, the second root appears to be just above zero, so we'll use x_0=0.01

    0.01 [ENTER]

    result: 0.01

    ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

    result: 0.013600430022

    0.015363748612

    0.015622071623

    0.015626489293

    0.015626490543

    0.015626490543

    Now, the third root seem to be near x_0=1.6

    1.6 [ENTER]

    result: 1.6

    ans(1)-(25ans(1)^2-64+1/ans(1))/(50ans(1)-1/ans(1)^2) [ENTER]

    result: 1.59214916585

    1.59212952245

    1.59212952233

    1.59212952233

    So, we know the 3 roots are approximately:

    x=-1.60775601287,\,0.015626490543,\,1.59212952233

    Round as required.
    Thanks from alane1994
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  4. #4
    Junior Member alane1994's Avatar
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    Re: Application of Newton's Method

    Thanks alot to both of you. So basically, you just do really monotonous work over and over until you get two like values. Then you have your root? Also, another question? When does calculus become second nature? You guys seem to just know how to do it right off! I love math and always will, but calculus is just really difficult for me to pick up for some reason...
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Application of Newton's Method

    It just takes time and practice. You'll get it...very few people just pick it right up!
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  6. #6
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    Re: Application of Newton's Method

    Quote Originally Posted by alane1994 View Post
    Thanks alot to both of you. So basically, you just do really monotonous work over and over until you get two like values. Then you have your root? Also, another question? When does calculus become second nature? You guys seem to just know how to do it right off! I love math and always will, but calculus is just really difficult for me to pick up for some reason...
    Exactly how I feel about algebra hahaha. Those guys to you, is like you to me haha
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