A cool thing about polynomials and squared roots

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Ashbash
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A cool thing about polynomials and squared roots

Postby Ashbash » Mon Apr 14, 2008 5:58 am UTC

No, that's not meant to be square roots. I really should have been studying, but for the past few hours I set out to find a generic formula for what you need to do to a polynomial if you square all the root values. Lemme show.

If the polynomial

anxn+an-1xn-1+an-2xn-2+.....+a0

has roots r0, r1, r2, ....., rn,

then the polynomial with roots (r0)2, (r1)2, (r2)2, ....., (rn)2, is such that every term can be worked out by this "formula"

(-1)n-k[(ak)2-2[(ak-1)(ak+1)-(ak-2)(ak+2)+(ak-3)(ak+3)-.....]]xk

where k is the value of the current term being operated on. In other words, a polynomial ax6+bx5+cx4+dx3+ex2+fx+g has roots r0, r1, r2, ....., r6. The polynomial with roots (r0)2, (r1)2, (r2)2, ....., (r6)2 is

a2x6-(b2-2ac)x5+(c2-2(bd-ae))x4+(d2-2(ce-bf+ag))x3+(e2-2(df-cg))x2+(f2-2eg)x+g2

Cool, huh? I wasted my life figuring that out on paper, I even had to learn the character for six in the greek alph. stigma. weird, it doesn't exist in the alphabet anymore, bet that confuses some people.

EDIT: fixerising. these sups and subs are retarded, you can barely see anything when you're editing.
Last edited by Ashbash on Mon Apr 14, 2008 7:05 am UTC, edited 1 time in total.

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Woxor
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Re: A cool thing about polynomials and squared roots

Postby Woxor » Mon Apr 14, 2008 6:13 am UTC

That's pretty cool, I'll have to try it out when I'm sober.
Spoiler:
I'm not too drunk to notice that you're missing an x^4 term, though. ;)

EDIT: lol, now you've got a plus sign in the superscript, after the x^4 term.
Last edited by Woxor on Mon Apr 14, 2008 7:41 am UTC, edited 1 time in total.

Ashbash
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Re: A cool thing about polynomials and squared roots

Postby Ashbash » Mon Apr 14, 2008 7:02 am UTC

ahh anus! i'll go fix it now...

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GBog
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Re: A cool thing about polynomials and squared roots

Postby GBog » Mon Apr 14, 2008 8:38 am UTC

A hint: We have [imath]\TeX[/imath] support now.

CFT
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Re: A cool thing about polynomials and squared roots

Postby CFT » Mon Apr 14, 2008 11:55 am UTC

If f(x) is your polynomial, then a solution might be:
f(sqrt(x))
There are 2 problems with this solution:
1. This isn't necessarily a polynomial.
2. Since sqrt(x) is multi-valued, the correct solution could be f(-sqrt(x))
These 2 problems can be solved with a single formula:
F(x) = f(sqrt(x)) * f(-sqrt(x))
F(x) is the polynomial you're looking for.
If f(r) = 0 and x = r2 then F(x) = 0 since either sqrt(x) = r or -sqrt(x) = r.
To prove it is a polynomial, write:
f(x) = E(x) + O(x)
Where E(x) contains the even degree elements, and O(x) contains the odd degree elements. Then:
f(sqrt(x)) * f(-sqrt(x)) = ( E(sqrt(x)) + O(sqrt(x)) ) * ( E(sqrt(x)) - O(sqrt(x)) ) = E(sqrt(x))2 - O(sqrt(x))2
E(sqrt(x)) is clearly a polynomial. Now define:
P(x) = O(x) / x
Then P(x) contains only elements of even degree, and:
O(sqrt(x))2 = x * P(sqrt(x)2
which is a polynomial.
So the solution is:
F(x) = E(sqrt(x))2 - x * P(sqrt(x))2

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thornahawk
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Re: A cool thing about polynomials and squared roots

Postby thornahawk » Thu Nov 13, 2008 7:20 am UTC

Odd... I seem to be seeing Graeffe somewhere...

~ Werner
John Dolan wrote:Cigarettes are insanely expensive and turn lots of poor people into cringing beggars.


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t0rajir0u
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Re: A cool thing about polynomials and squared roots

Postby t0rajir0u » Thu Nov 13, 2008 8:18 pm UTC

CFT wrote:F(x) = f(sqrt(x)) * f(-sqrt(x))
F(x) is the polynomial you're looking for.

The quickest way to convince yourself that this is the polynomial you want is to note that if f(x) is a product of terms of the form (x - r_k), then f(sqrt(x)) f(-sqrt(x)) is a product of terms of the form

(sqrt(x) - r_k)(- sqrt(x) - r_k) = (r_k^2 - x^2).

Of course, this might not convince you that if your original polynomial has, say, integer coefficients then this will still be true of your new polynomial, but that's what Vieta's and Newton's formulas (or Galois theory ;)) are for.


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