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Math Help - Newton's Method

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

    Given the function:

    f(x)=5e^{-x}-29x

    When, x_{0}=\ln(6)

    x_{1}=?

    I know...

    x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Newton's Method

    You will have:

    x_1=\ln(6)-\frac{5e^{-\ln(6)}-29\ln(6)}{-5e^{-\ln(6)}-29}

    Now, use the two identities:

    -\ln(y)=\ln\left(\frac{1}{y} \right) and e^{\ln(y)}=y to simplify.
    Thanks from alane1994
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  3. #3
    Junior Member alane1994's Avatar
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    Re: Newton's Method

    Thanks, I think it was the ln's that threw me... I hate those things...
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Newton's Method

    I advise you to learn to love the natural log function, as you will get to know it well during your study of the calculus.
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  5. #5
    Junior Member alane1994's Avatar
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    Re: Newton's Method

    Could you give me an answer. I have plugged it in two different times and gotten.
    -80.169265 and 23.752784.
    I have to go to 6 decimal places.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Newton's Method

    I get:

    x_1=\ln(6)-\frac{\frac{5}{6}-29\ln(6)}{-\frac{5}{6}-29}\approx0.077982108079
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  7. #7
    Junior Member alane1994's Avatar
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    Re: Newton's Method

    When you put it into your calculator, do you put it in all at once? Or do you break it up into segments to put in your calculator. Because I plugged the same thing you have there and I got a different answer...
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Newton's Method

    I enter the entire expression. Exactly what are you entering?
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  9. #9
    Junior Member alane1994's Avatar
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    Re: Newton's Method

    I believe that it was just misplaced parentheses.
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