math problem

[Claudio Irrgang]

Hello,

I had a question:

Suppose:

K2 = a + b + c
K1 = a b + b c + c a
K0 = a b c

I want to acheive "2a - b - c", "2b - a - c", "2c - a - b" using any
combinations of K2, K1, K0 using any operations (+, -, *, ÷, roots, logs,
etc.) under real numbers.

1) How do I know if such combinations of K2, K1, K0 will get me to "2a - b -
c"?
2) If such a combination exists, how would I figure it out?

Thanks,

Claudio

[Greg Neill]

"Claudio Irrgang" <lay...@com.cvicad> wrote in message
news:kmxub.36041$j1d....@news04.bloor.is.net.cable.rogers.com...

You have three equations in three unknowns.
Solve for a, b, and c in terms of K0, K1, and K2.

Note: this will be a very messy result, with multiple
cases depending upon the values that the variables
take on.

After you have each of a, b, and c as functions of K0, K1, and K2,
you can write any combination of a, b, and c that you want.

[Norbert Marrek]

"Claudio Irrgang" <lay...@com.cvicad> wrote in message
news:kmxub.36041$j1d....@news04.bloor.is.net.cable.rogers.com...

Symmetric functions can always be expressed in terms of
the elementary symmetric polynoms, e.g. a^2 + b^2 + c^2
is a combination of K0, K1 and K2.

As 2a -b -c is not symmetric, I think, you cannot find
a combination in K0, K1 K2.

Bye,
Norbert Marrek

[Ken Pledger]

In article
<kmxub.36041$j1d....@news04.bloor.is.net.cable.rogers.com>,
"Claudio Irrgang" <lay...@com.cvicad> wrote:

> ....

> K2 = a + b + c
> K1 = a b + b c + c a
> K0 = a b c
>

That is, a, b, and c are the three roots of the cubic equation

x^3 - (K2)(x^2) + (K1)x - K0 = 0.

To see this, multiply out (x - a)(x - b)(x - c) and compare the
coefficients.

> I want to acheive "2a - b - c", "2b - a - c", "2c - a - b" using any
> combinations of K2, K1, K0 using any operations (+, -, *, ÷, roots, logs,
> etc.) under real numbers.

Since 2a - b - c = 3a - K2, and you know K2, finding 2a - b - c
is essentially the same problem as finding a separately. Similarly
finding your other two expressions amounts to find b and c. So
you're really after the solution of a general cubic equation, which you
can find at various web sites such as:

http://www.sosmath.com/algebra/factor/fac11/fac11.html

http://mathworld.wolfram.com/C/CubicEquation.html

Ken Pledger.