Proof for Inverse Roots Polynomial

If you don't have time or don't feel like readings lot (or what ever reason you may have) there is an abridged version at the bottom that is much easier to read.

In this note, I will prove that for all polynomials f(x)=anxn+an1xn1+...+a0f(x)=a_n x^n+a_{n-1}x^{n-1}+...+a_0, with an and a00a_n ~ and ~ a_0 \neq 0 and x=0 is not a root, that after being divided by GCD of the coefficients (gg) or multiplied by a factor of kk such that the new coefficients are both integral and co-prime, the function with the smallest possible integral inverseinverse roots has the form f(x)=p(a0xn+a1xn1...+an)f'(x)=p(a_0x^n+a_1x^{n-1}...+a_n) where p is a constant and p=1p=1 if and only if f(x)f(x) has all group wise co-prime integer coefficients.

(In the short, I'm proving that the coefficients are reversed if the roots are inverted (AKA: r1r_1 becomes 1r1\dfrac{1}{r_1})


By Vieta's, we have an1an=r1+r2+r3+rn-\dfrac{a_{n-1}}{a_n}=r_1+r_2+r_3 \dots+ r_n for f(x)f(x)

Now, assume that we didn't know that f(x)f'(x) had the reverse coefficients of f(x)f(x)

Thus f(x)=bnxn+bn1xn1...+b0f'(x)=b_nx^n+b_{n-1}x^{n-1}...+b_0

However, we know that the roots of f(x)f'(x) are 1ri\dfrac{1}{r_i} where 0in0\leq i \leq n.

For f(x)f'(x), we have bn1bn=1r1+1r2+1r3+1rn-\dfrac{b_{n-1}}{b_n}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} \dots+ \dfrac{1}{r_n}

If we multiply both the top and bottom of every fraction on the R.H.S. by every root other than the one in its denominator, we will have (a1an)(a0an)a1a0-\dfrac{\left(\dfrac{a_1}{a_n}\right)}{\left(\dfrac{a_0}{a_n}\right)}\Rightarrow -\dfrac{a_1}{a_0} (This is too ugly to show in "proper" math format here. So I'll show a simpler and proper version below)

a1a0=bn1bn\therefore -\dfrac{a_1}{a_0}=-\dfrac{b_{n-1}}{b_n}

It can be easily observed that bn=p(a0)b_n=p(a_0) where p is a constant and bn1=p(a1)b_{n-1}=p(a_1).

This process can be repeated to show that ai+1ai=bn(i+1)bn(i)\dfrac{a_{i+1}}{a_{i}}=\dfrac{b_{n-(i+1)}}{b_{n-(i)}} where 0i<n0\leq i < n (NOTE: this is the one time where i<ni<n rather than ini \leq n because there is no b1b_{-1} nor an+1a_{n+1}

We can see that the ratio's of the aia_i is equal to the ratio's of the bib_i in reverse order. Going back to our other equation a1a0=bn1bn\dfrac{a_1}{a_0}=\dfrac{b_{n-1}}{b_n}. Assuming that f(x)f(x) has all group wisegroup ~ wise co-prime coefficients, a1a0\dfrac{a_1}{a_0} could be reducible by a factor of h. However, this leads to two contradictions: 1. every other ratio would have to be reduced by h or f(x)f(x) would be changed. 2. If every other coefficient was reduced then some would become non-integers if they were all group wise co-prime (if they weren't then this would contradict a condition earlier).

Also since we're looking for an f(x)f'(x) with the smallest possible co-prime integer coefficients, bn1bn\dfrac{b_{n-1}}{b_n} should be irreducible. These statements imply that bn1=a1b_{n-1}=a_1 and bn=a0b_n=a_0.

Once again, we can use this logic to prove that ai=bnia_i=b_{n-i} where 0in0\leq i \leq n.

Thus we have proved that for integral coefficients of f(x)f(x), f(x)=a0xn+a1xn1...+anf'(x)=a_0x^n+a_1x^{n-1}...+a_n

If f(x)f(x) doesn't have integer coefficients, then we can multiply by a factor of pp such that the coefficients become integral, thus our final result is f(x)=p(a0xn+a1xn1...+an)f'(x)=p(a_0x^n+a_1x^{n-1}...+a_n).


Proper Version for omitted part

bn1bn=1r1+1r2+1r3+1rn=i=1n((i=1n(ri))ri)1(i=1n(ri))-\dfrac{b_{n-1}}{b_n}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} \dots+ \dfrac{1}{r_n}=\displaystyle \sum_{i=1}^{n} \left(\dfrac{\left( \displaystyle \sum_{i=1}^{n} (r_i) \right)}{r_i}\right)\cdot \dfrac{1}{\left( \displaystyle \sum_{i=1}^{n} (r_i) \right)}

i=1n((a0an)ri)1(a0an)=i=1n(1ana0ri)ana0\Rightarrow \displaystyle \sum_{i=1}^{n} \left(\dfrac{\left(\dfrac{a_0}{a_n}\right)}{r_i}\right)\cdot\dfrac{1}{\left(\dfrac{a_0}{a_n}\right)}= \displaystyle \sum_{i=1}^{n} \left(\dfrac{1}{a_n}\cdot \dfrac{a_0}{r_i}\right)\cdot\dfrac{a_n}{a_0}

(1an(a1))ana0=a1a0\Rightarrow \left(\dfrac{1}{a_n}\cdot (a_1)\right)\cdot\dfrac{a_n}{a_0}=\dfrac{a_1}{a_0}


Pretty version for those like myself who hate looking at ugly summations:

Polynomial ax3+bx2+cx+dax^3+bx^2+cx+d has roots p,q,r.

ba=p+q+r-\dfrac{b}{a}=p+q+r

ca=pq+qr+rp\dfrac{c}{a}=pq+qr+rp

da=pqr-\dfrac{d}{a}=pqr

Polynomial hx3+gx2+jx+khx^3+gx^2+jx+k has roots 1p,1q,1r\dfrac{1}{p}, \dfrac{1}{q}, \dfrac{1}{r}

-\dfrac{g}{h}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=\dfrac{(pq)}{(pq)r}+\dfrac{(rp)}{(rp)q}+\dfrac{(pq)}{(pq)r}=\dfrac{pq+qr+rp}{pqr}=\Rightarrow -\dfrac{\left(\dfrac{c}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \pink{-\dfrac{c}{d}}}

\dfrac{j}{h}=\dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{rp}=\dfrac{(r)}{(r)pq}+\dfrac{p}{(p)qr}+\dfrac{(q)}{(q)rp}=\dfrac{p+q+r}{pqr}=\dfrac{\left(\dfrac{b}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \green{\dfrac{b}{d}}}

-\dfrac{k}{h}=\dfrac{1}{pqr}=\dfrac{1}{\left(-\dfrac{d}{a}\right)}=\boxed{\color \red{-\dfrac{a}{d}}}

Assuming a monic polynomial (we can divide by h because dividing by a constant won't change the roots)

hx3+gx2+jx+khx^3+gx^2+jx+k has equivalent roots to x^3+\dfrac{g}{h}x^2+\dfrac{j}{h}x+\dfrac{k}{h}=x^3+\color \pink{\dfrac{c}{d}}x^2+\color \green{\dfrac{b}{d}}x+\color \red{\dfrac{a}{d}}

Multiplying through to rationalize the denominators by d.

dx3+cx2+bx+a\boxed{dx^3+cx^2+bx+a}

Note by Trevor Arashiro
4 years, 11 months ago

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Why not tranform the equation by the substitution x=1yx=\dfrac{1}{y} Its a 1 line proof after that.

Aneesh Kundu - 4 years, 11 months ago

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That transformation is certainly a much quicker approach. Can you add more details about what you did?

For a problem along the same lines, check out Squaring the roots of a polynomial.

Calvin Lin Staff - 4 years, 11 months ago

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I'm not sure exactly what you mean, but by your substitution, since we're solving for roots, isn't x=x=\infty

Trevor Arashiro - 4 years, 11 months ago

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Why x=x=\infty?

Aren't we just inverting the roots?

Aneesh Kundu - 4 years, 11 months ago

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@Aneesh Kundu Because you substitute \x=1y\x=\dfrac{1}{y}. And since we're finding roots y=0

Trevor Arashiro - 4 years, 11 months ago

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@Trevor Arashiro NOOOOOOO!!!

I just used yy to take a different variable. I did not mean the yy in y=f(x)y=f(x)

Aneesh Kundu - 4 years, 11 months ago

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@Aneesh Kundu Ahh, I see. I was confused there.

Trevor Arashiro - 4 years, 11 months ago

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Note: You should state that x=0 x = 0 is not a solution, or that equivalently a00 a_0 \neq 0 . This would also justify allowing you to divide by a0 a_0 .

Calvin Lin Staff - 4 years, 11 months ago

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Ahh, thanks, thats a good point.

Trevor Arashiro - 4 years, 11 months ago

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Thanks bruh

Loexx Manncch - 4 years, 11 months ago

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Chee Bra.

Trevor Arashiro - 4 years, 11 months ago

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