reversed coefficients

[BMAD]

If P(x) and Q(x) have 'reversed' coefficients,

for example:

P(x) = x⁵+3x⁴+9x³+11x²+6x+2

Q(x) = 2x⁵+6x⁴+11x³+9x²+3x+1

What can you say about the roots of P(x) and Q(x)?

[LVan Toren]

If P(x)= a0 xn +a1 xn-1+...+an-2 x2+an-1 x + an

then Q(x)= an xn + an-1 xn-1 + ... + a2 x2 + a1 x1 + a0 with reversed coefficients

Let x=1/y

then P(x)= a0 1/yn +a1 1/yn-1+...+an-2 1/y2+an-1 1/y + an

P(x)= 1/yn ( a0 +a1 y1+...+an-2 yn-2+an-1 yn-1 + an yn)

P(x)= 1/yn Q(y)

If x1 is a root for P(x), then P(x1) = 0

but y1=1/x1 will also give Q(y1) = 0

For every root of P there is a reciprocal of Q

Edited March 24, 2014 by bonanova
spoiler

[superprismatic]

The roots of P are the reciprocals of the roots of Q.

[superprismatic]

If P(x)= a0 xn +a1 xn-1+...+an-2 x2+an-1 x + an

then Q(x)= an xn + an-1 xn-1 + ... + a2 x2 + a1 x1 + a0 with reversed coefficients

Let x=1/y

then P(x)= a0 1/yn +a1 1/yn-1+...+an-2 1/y2+an-1 1/y + an

P(x)= 1/yn ( a0 +a1 y1+...+an-2 yn-2+an-1 yn-1 + an yn)

P(x)= 1/yn Q(y)

If x1 is a root for P(x), then P(x1) = 0

but y1=1/x1 will also give Q(y1) = 0

For every root of P there is a reciprocal of Q

One possibility you hadn't addressed: What about roots which are 0, and possibly with multiplicity?

[BMAD]

good point!

[LVan Toren]

The roots of P are the reciprocals of the roots of Q.

A root of P that is zero has no reciprocal, so Q will have a root less.

P(x)= a₀ xⁿ +a₁ x^n-1+...+a_n-2 x²+a_n-1 x + a_n will have a zero root if a_n = 0, but then

Q(x)= 0 xⁿ +a_n-1 x^n-1+...+a₂ x²+a₁ x + a₀ is of a degree less and will have a root less (and possibly more than one if also a_n-1=0 etc).

On the other hand one can introduce zero roots in Q by adding zero coëfficiënts in front of P.