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Math Help - Newton's Method

  1. #1
    Member FalconPUNCH!'s Avatar
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    Newton's Method

    Use Newton's Method to approximate the given number correct to eight decimal places.

    ^{100}\sqrt{100}

    f(x) = x^{100} - 100

    f'(x) = 100x^{99}

    I know how to use Newton's Method but is there a faster way, a shortcut, to find it or am I going to have to start from x1 = 1?

    am I going to have to plug in numbers until I get two to match eight decimals?

    Second Problem:
    Use Newton's Method to approximate the indicated root of the equation correct to six decimal places.

    The positive root of 2cosx = x^4

    I know how to do Newton's Method but these two were giving me problems.
    Last edited by FalconPUNCH!; December 11th 2007 at 06:22 PM.
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  2. #2
    Member FalconPUNCH!'s Avatar
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    For the second one I get

    f(x) = 2cosx - x^4

    f'(x) = -2sinx - 4x^3

    I get X1 = 2 as my initial approximation but when I plug what I get into Newton's Formula I get some weird numbers and none of them are close to each other.

    For my second approximation I get 1.5022769 and I don't know but I feel that it's wrong and I don't think I should continue with the wrong answer.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    I just did the first one using Newton's Method and got what I was looking for. The second problem I posted seems to be a little tougher and I need help with that one.

    Edit: My only problem with the first one is I don't know what to use an initial approximation. It took me almost two pages to get the answer. Can anyone tell me a faster way to get an approximate number eight decimal places?
    Last edited by FalconPUNCH!; December 11th 2007 at 08:16 PM.
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  4. #4
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    Edit: My only problem with the first one is I don't know what to use an initial approximation. It took me almost two pages to get the answer. Can anyone tell me a faster way to get an approximate number eight decimal places?
    get a computer to do the calculating for you. I can't imagine why anyone would be learning Newton's method without some mathematical programming environment, eg. Haskell interpreter, R gui.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by badgerigar View Post
    get a computer to do the calculating for you. I can't imagine why anyone would be learning Newton's method without some mathematical programming environment, eg. Haskell interpreter, R gui.
    Yeah I can do that but I want to know how to find an initial approximation. Whatever I use is always far away from the actual root.
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  6. #6
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    Quote Originally Posted by FalconPUNCH! View Post
    I just did the first one using Newton's Method and got what I was looking for. The second problem I posted seems to be a little tougher and I need help with that one.

    Edit: My only problem with the first one is I don't know what to use an initial approximation. It took me almost two pages to get the answer. Can anyone tell me a faster way to get an approximate number eight decimal places?
    Hello,

    I would draw a rough sketch of the 2 graphs. For instance with #1:
    Draw the graph f(x)=2\cos(x)\ ,~-\pi < x < \pi anf p(x)=x^4. Use the estimated x-value of the intersection as an initial value.
    When I used x_0 = 1 I got the result with 8 decimals exact after 3 steps.

    to #2. If n is a large number then \sqrt[n]{n} \approx 1. If you use a value a little bit larger than 1 as an initial value it will be sufficient. But with your example the sequence of x-value converges very slowly. So when I used x_0 = 1.5 it took me more than 20 cycles to get a nearly exact value.
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  7. #7
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    I would draw a rough sketch of the 2 graphs. For instance with #1:
    Draw the graph f(x)=2\cos(x)\ ,~-\pi < x < \pi anf p(x)=x^4. Use the estimated x-value of the intersection as an initial value.
    When I used x_0 = 1 I got the result with 8 decimals exact after 3 steps.

    to #2. If n is a large number then \sqrt[n]{n} \approx 1. If you use a value a little bit larger than 1 as an initial value it will be sufficient. But with your example the sequence of x-value converges very slowly. So when I used x_0 = 1.5 it took me more than 20 cycles to get a nearly exact value.
    Thanks for helping me. Yeah for \sqrt[n]{n} it took me around 20 cycles. I'm going to probably get a number closer to one so I can fit it on half a page. Thanks for your assistance.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by badgerigar View Post
    I can't imagine why anyone would be learning Newton's method without some mathematical programming environment, eg. Haskell interpreter, R gui.
    I do it by hand daily. It puts hair on your chest.

    -Dan
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