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  1. #1
    dan
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    help with Newton

    Is there some one who would get a kick out of explaining Newtonís method to me? I've read some different things about it but I didnít understand the calculus Is it possible to explain it without using calculus?
    dan
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  2. #2
    Eater of Worlds
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    What seems to be the trouble?. Newton's Method is one of the easiest things in calculus.

    It's just a matter of iterations.

    Let's give an example:

    Use Newton to find the 'real' solutions of x^{3}-x-1=0

    Let f(x)=x^{3}-x-1\;\ and\;\ f'(x)=3x^{2}-1

    Graph the function, x^{3}-x-1

    You can see that y=0 when x is between 1 and 2.

    Make something in that region your initial guess.

    Try 1.5

    x_{2}=1.5{-}\frac{(1.5)^{3}-1.5-1}{3(1.5)^{2}-1}=1.34782609

    Now use the result you get from this and sub back into the equation:

    What is wrong with this website that I keep getting these errors saying the image is too big?. I've never seen this on another site. The code is fine as far as I can tell..

    x_{3}=1.34782609- \frac{(1.34782609)^{3}-(1.34782609)-1}{3(1.34782609)^{2}-1}=1.32520040

    Continue in this fashion until you arrive at approximations which are so close they are virtually unchanged.

    We end up with:

    x_{1}=1.5\;\ x^{2}=1.34782609\;\ x^{3} =1.32520040\;\ x^{4}= 1.32471817\;\ x^{5}=1.32471796\;\ x^{6}=1.32471796

    See, the last two are the same.

    No need to continue. . The solution is \approx{1.32471796}
    Last edited by galactus; July 15th 2006 at 02:30 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by galactus

    What is wrong with this website that I keep getting these errors saying the image is too big?. I've never seen this on another site. The code is fine as far as I can tell..
    The TeX system seems to not like long (not that long) strings of TeX.
    Placing pairs of [ /math] and [ math] (without extra spaces) s seems to
    fix the problems usually (as I have done in your post).

    RonL
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  4. #4
    Eater of Worlds
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    Thank you Cap'N. I will keep that in mind. I have not run into that problem before, regardless of the length of the string. Thanks for the fix up.
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  5. #5
    dan
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    Unhappy help with newton

    Ok,
    What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book so... I don't even know what f'(x) means and how you get there from f(x)
    Dan
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  6. #6
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    Quote Originally Posted by dan
    Ok,
    What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book so... I don't even know what f'(x) means and how you get there from f(x)
    Dan
    I would be more strict and tell you to study "Pre-calculus" however they call the course.
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  7. #7
    Eater of Worlds
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    Quote Originally Posted by dan
    Ok,
    What seems to be the problem is...I know practically no calculus and my dad wont let me take a course until I finish my college algebra book so... I don't even know what f'(x) means and how you get there from f(x)
    Dan
    Poor Dan. I didn't realize that. Math is done in steps; You need to finish your algebra or pre-caculus and then go to Calc I. BTW, f'(x) is the derivative of a function f(x). What brought up Newton anyway?. You need to learn differentiation before you can use Newton's Method. Do a Google search.
    Good luck.
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  8. #8
    Super Member malaygoel's Avatar
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    Someone Please help.
    How will you explain the working of Newton's Method?I can't figure out why it works!!

    Keep Smiling
    Malay
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  9. #9
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    Sorry, dan!

    I don't even know what f'(x) means . . .

    You're saying, "Can someone explain a B\flat\text{ minor} chord without using Music Theory?
    . . You see, I don't read music."

    Answer: No.
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  10. #10
    Eater of Worlds
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    Most any calc book will explain how it works. It's not that complicated.

    Off the top of my head.

    The solutions of f(x)=0 are the values of x where the graph crosses the x-axis.

    Suppose that x=c is some solution we are looking for. Even if we can't find c exactly, it is usually possible to approximate it by graphing and using the Intermediate Value Theorem.

    If we let, say, x_{1} be our initial approximation, then we can improve by moving along the tsngent line to y=f(x) at x_{1} until we meet it at a point x_{2}.

    Repeat.

    One thing we have to do is derive some sort of formula so we can use ol' Newton.

    Start with point-slope form of a line:

    y-f(x_{1})=f'(x_{1})(x-x_{1})

    If f'(x_{1})\neq{0}, then this line is not parallel to the x-axis and crosses it at some point (x_{2},0).

    Sub this in our point-slope form:

    -f(x)=f'(x_{1})(x_{2}-x_{1})

    Solve for x_{2}

    x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{2})}

    We keep going until we see that the approximation is

    x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}, n=1,2,3,......
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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