## A cool thing about polynomials and squared roots

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

Ashbash
Posts: 184
Joined: Thu Oct 11, 2007 5:46 am UTC

### A cool thing about polynomials and squared roots

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.

Woxor
Posts: 506
Joined: Mon May 07, 2007 11:28 pm UTC

### Re: A cool thing about polynomials and squared roots

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
Posts: 184
Joined: Thu Oct 11, 2007 5:46 am UTC

### Re: A cool thing about polynomials and squared roots

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

GBog
Posts: 114
Joined: Fri Jun 01, 2007 4:57 pm UTC

### Re: A cool thing about polynomials and squared roots

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

CFT
Posts: 66
Joined: Mon Apr 14, 2008 11:33 am UTC

### Re: A cool thing about polynomials and squared roots

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

thornahawk
Posts: 64
Joined: Wed Oct 29, 2008 3:56 am UTC
Location: somewhere out in the SEA
Contact:

### Re: A cool thing about polynomials and squared roots

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.

E-mail!

t0rajir0u
Posts: 1178
Joined: Wed Apr 16, 2008 12:52 am UTC
Location: Cambridge, MA
Contact:

### Re: A cool thing about polynomials and squared roots

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.

Return to “Mathematics”

### Who is online

Users browsing this forum: No registered users and 8 guests