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February 1st, 2012, 06:05 AM   #1
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polynomials

Hi,

My problem

P=A +B*K +C*K^2 +D*K^3 +E*K^4 +F*K^5

how to express K in terms of the other variables

or simply

K= ????

please help me out
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February 1st, 2012, 06:56 AM   #2
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Re: polynomials

In general this is not possible. Look up Abel, Galois, or quintic equation. For certain specific values you might be able to get a formula, but in general it cannot be done.
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February 1st, 2012, 08:40 AM   #3
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Re: polynomials

Quote:
Originally Posted by kunju
P=A +B*K +C*K^2 +D*K^3 +E*K^4 +F*K^5
Usually shown this way:
A*K^5 + B*K^4 + C*K^3 + D*K^2 + E*K + F = 0

If you're lucky, all variables except F will equal 1, and F will equal -5
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February 1st, 2012, 10:01 AM   #4
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Re: polynomials

Quote:
Originally Posted by mrtwhs
In general this is not possible. Look up Abel, Galois, or quintic equation. For certain specific values you might be able to get a formula, but in general it cannot be done.
kunju, what makes you think that there is a formula for solutions to a fifth degree polynomial?
If you have ax^2 + bx + c = 0, then you can solve for x..
Likewise, if you have ax^3 + bx^2 + cx + d = 0...
But not for a fifth degree polynomial..
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February 1st, 2012, 10:30 AM   #5
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Re: polynomials

Or any higher degree (than four) for that matter.
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February 1st, 2012, 01:38 PM   #6
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Re: polynomials

Well. . .

We do have a solution to the quintic equation. However, we say that there is no solution in the sense that there is not a solution involving only the elementary operations of addition, subtraction, multiplication, division, exponentiation, and root extraction. There does not exist a solution to the quintic with solely these operations.

However, there are far more behemous operations involving advanced analytical functions. We can express the solution in terms of these functions.

This is a lot like how, when learning a new language, you can conceive of something complex to express. However, you can't really express it until you learn the more advanced aspects of the language.
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