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Newton Raphson Failure Help? A Level Maths watch

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    Hi the equation i am using for the failure is 1/x^3+4x^2-3x-8 with no brackets. The first iteration is around 4 with use of x0=1. but the equation on wolfram says 1 - 4x(x^5+2x^3-1)/8x^5-3x^4-3 which is equal to 4 when 1 is plugged in. This is the correct answer but i dont understand how they have got that equation? I know you need the equation on top then derivative on denominator so can someone help?
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    (Original post by liverpool2044)
    Hi the equation i am using for the failure is 1/x^3+4x^2-3x-8 with no brackets. The first iteration is around 4 with use of x0=1. but the equation on wolfram says 1 - 4x(x^5+2x^3-1)/8x^5-3x^4-3 which is equal to 4 when 1 is plugged in. This is the correct answer but i dont understand how they have got that equation? I know you need the equation on top then derivative on denominator so can someone help?
    It's still a bit ambiguous - is it x^-3 + 4x^2 - 3x - 8?
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    (Original post by SeanFM)
    It's still a bit ambiguous - is it x^-3 + 4x^2 - 3x - 8?
    X^3 + 4x^2 - 3x -8

    It doesnt work for a root between 0 and 1. It goes towards the root between 1 and 2.
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    (Original post by liverpool2044)
    X^3 + 4x^2 - 3x -8

    It doesnt work for a root between 0 and 1. It goes towards the root between 1 and 2.
    Without working it through I'm not sure, unless it comes from finding the derivative of 1/(x^3 + 4x^2 - 3x - 8).
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    (Original post by SeanFM)
    Without working it through I'm not sure, unless it comes from finding the derivative of 1/(x^3 + 4x^2 - 3x - 8).
    I got the derivative of that as (-3x^2-8x+3) / ( x^3 +4x^2 -3x -8 ) ^2
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    (Original post by liverpool2044)
    I got the derivative of that as (-3x^2-8x+3) / ( x^3 +4x^2 -3x -8 ) ^2
    Right.. so I would guess that Wolgram gets an answer that looks like what it does.. by multiplying everything out (including the ^2). Is that what you mean?
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    (Original post by SeanFM)
    Right.. so I would guess that Wolgram gets an answer that looks like what it does.. by multiplying everything out (including the ^2). Is that what you mean?
    Oh yeah possibly. I just dont really understand it because is that the derivative and original equation included or just the differentiation of it?
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    (Original post by liverpool2044)
    Oh yeah possibly. I just dont really understand it because is that the derivative and original equation included or just the differentiation of it?
    I'm not familiar with Wolfram or what you've actually put in, but if you've asked it to do the Newton-Raphson process then that's probably it (though it can get things wrong I imagine!)

    If they're both giving you the correct answer - though you've said it's 'about 4' when you've done it - then not to worry.
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    (Original post by SeanFM)
    I'm not familiar with Wolfram or what you've actually put in, but if you've asked it to do the Newton-Raphson process then that's probably it (though it can get things wrong I imagine!)

    If they're both giving you the correct answer - though you've said it's 'about 4' when you've done it - then not to worry.
    when i have manually done it though ive got an answer of 0.25 because i put it to the -1 and then chain rule which is different
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    (Original post by liverpool2044)
    when i have manually done it though ive got an answer of 0.25 because i put it to the -1 and then chain rule which is different
    I used the chain rule as well, but for the output I get 1 - (-1/8) = 9/8.

    Maybe you should double check your working, or if in doubt post it here
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    (Original post by SeanFM)
    I used the chain rule as well, but for the output I get 1 - (-1/8) = 9/8.

    Maybe you should double check your working, or if in doubt post it here
    im quite confused how its 4 then if you got different also, ill post the working tomorrow
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    (Original post by liverpool2044)
    im quite confused how its 4 then if you got different also, ill post the working tomorrow
    Okey dokey
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    Can we be 100% clear here? Are you trying to find solutions to x^3+4x^2-3x-8 = 0 or \dfrac{1}{x^3+4x^2-3x-8} = 0. (I really *hope* it's not the latter, since it has no solutions, but then again, if that is what you're trying to do, then well, "there's ya problem!").

    Or are you trying to solve something else entirely?
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    (Original post by DFranklin)
    Can we be 100% clear here? Are you trying to find solutions to x^3+4x^2-3x-8 = 0 or \dfrac{1}{x^3+4x^2-3x-8} = 0. (I really *hope* it's not the latter, since it has no solutions, but then again, if that is what you're trying to do, then well, "there's ya problem!".

    Or are you trying to solve something else entirely?
    it is the latter, i put the equation in to autograph and it says there are 3 roots present. I try newton raphson on the root between 0 and 1 but it converges to another root close to it.
    The equation f(x)=1/x3+4x2-3x-8 is shown on the graph andthe derivative of this is. The root I am going to find is the one in between[0,1]. X0 = 1 will be tried. The derivative of this is that -3x2-8x+3/ (x3+4x2-3x-8)2

    That's what i got up to?
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    Newton raphson is X0 - F(x) / F ' (x)
    Wolfram have done all of this into one substitution. The derivative they gave me for the 1 over part is -3x2-8x+3/ (x3+4x2-3x-8)2

    Can 4x(x^5+2x^3-1)/8x^5-3x^4-3 this be found with substitution because this is the right equation i need, just dont know how to get it
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    (Original post by liverpool2044)
    it is the latter, i put the equation in to autograph and it says there are 3 roots present.
    \dfrac{1}{x^3+4x^2-3x-8} = 0 has no roots.

    More generally, the only solutions for \dfrac{1}{f(x)} = 0 are where f(x) becomes infinite. Since here f(x) is a polynomial, it is never infinite and so there are no roots.

    There are 3 places where the fraction you have given changes sign, but this doesn't imply a root, in the same way that 1/x changes sign around the origin but it is not zero (or even definied) there. But Newton-Raphson won't help you find these places since the function is going to +/- infinity there rather than going to 0.
 
 
 
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