Maths..help..pls...
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(meaning my answer)
30 September 2004, 08:56 PM
#7
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The plot looks correct but 0 isn't a solution to the equation y=0
As x->0 y->inf but strictly y is undefined at 0.
Best use a numerical approximation (e.g. Newton-Raphson) to find the roots.
As x->0 y->inf but strictly y is undefined at 0.
Best use a numerical approximation (e.g. Newton-Raphson) to find the roots.
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30 September 2004, 08:59 PM
#8
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Ooops sorry the roots at +/- 1 are obvious
1 to the power of anything is 1, so you get (1-1) or (-1+1)
Newton-Raphson is overkill!!!
As x gets large x^-12 will be small and -x^6 will dominate so y-> -inf
1 to the power of anything is 1, so you get (1-1) or (-1+1)
Newton-Raphson is overkill!!!
As x gets large x^-12 will be small and -x^6 will dominate so y-> -inf
30 September 2004, 09:03 PM
#9
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kinda guessed that..... 
but if there are 3 roots as ajm suggests u might have been right..
I dont need to solve it... just need to be able to recognise what order it is and hence how many roots you would expect.. been a while since i did this stuff and cant remember.
but if there are 3 roots as ajm suggests u might have been right..I dont need to solve it... just need to be able to recognise what order it is and hence how many roots you would expect.. been a while since i did this stuff and cant remember.
30 September 2004, 09:09 PM
#10
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Conventional polynomials of the type
Ax^n + Bx^(n-1) + Cx^(n-2) + ... + K = 0
are of order n and have n roots (some may be imaginary)
The equation you've shown doesn't conform to this type as it includes negative powers of x and cannot be analysed in the same way!
Ax^n + Bx^(n-1) + Cx^(n-2) + ... + K = 0
are of order n and have n roots (some may be imaginary)
The equation you've shown doesn't conform to this type as it includes negative powers of x and cannot be analysed in the same way!
30 September 2004, 09:15 PM
#11
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For example
y = x^12 - x^6
This equation is a 12th order polynomial, has 12 roots, 6 of them zero, 2 real (+/-1) and 4 imaginary (+/- 0.5 +/- 0.866i)
Are you sure you don't mean this? The equation you put up cannot be analysed in the same way.
y = x^12 - x^6
This equation is a 12th order polynomial, has 12 roots, 6 of them zero, 2 real (+/-1) and 4 imaginary (+/- 0.5 +/- 0.866i)
Are you sure you don't mean this? The equation you put up cannot be analysed in the same way.
Last edited by Sprint Chief; 30 September 2004 at 09:18 PM.
30 September 2004, 09:27 PM
#12
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yes it was x^-12
probably there 2 confuse me.... calcluation program comes up with 18 roots though..
what about y = x^6 +1/x
probably there 2 confuse me.... calcluation program comes up with 18 roots though..
what about y = x^6 +1/x
Last edited by krazy; 30 September 2004 at 09:32 PM.
30 September 2004, 09:45 PM
#14
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These polynomials don't conform to the conventional definition and need to be solved in a different way - you may have trouble getting an analytical answer here! You can use techniques such as deconvolution or partial-fraction expansion but this is an ill-posed problem that can produce unstable results (i.e. minute changes in input parameters result in large swings in output)
Sorry can't help much more here...
Sorry can't help much more here...
