BMAD 63 Report post Posted March 3, 2014 If P(x) and Q(x) have 'reversed' coefficients, for example: P(x) = x^{5}+3x^{4}+9x^{3}+11x^{2}+6x+2 Q(x) = 2x^{5}+6x^{4}+11x^{3}+9x^{2}+3x+1 What can you say about the roots of P(x) and Q(x)? 1 1 Quote Share this post Link to post Share on other sites
0 LVan Toren 2 Report post Posted March 21, 2014 (edited) 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 Quote Share this post Link to post Share on other sites
0 superprismatic 11 Report post Posted March 3, 2014 The roots of P are the reciprocals of the roots of Q. Quote Share this post Link to post Share on other sites
0 superprismatic 11 Report post Posted March 24, 2014 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? Quote Share this post Link to post Share on other sites
0 BMAD 63 Report post Posted March 24, 2014 good point! Quote Share this post Link to post Share on other sites
0 LVan Toren 2 Report post Posted March 25, 2014 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)= a_{0} x^{n} +a_{1} x^{n-1}+...+a_{n-2} x^{2}+a_{n-1} x + a_{n} will have a zero root if a_{n} = 0, but then Q(x)= 0 x^{n} +a_{n-1} x^{n-1}+...+a_{2} x^{2}+a_{1} x + a_{0} 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. Quote Share this post Link to post Share on other sites
If P(x) and Q(x) have 'reversed' coefficients,
for example:
P(x) = x^{5}+3x^{4}+9x^{3}+11x^{2}+6x+2
Q(x) = 2x^{5}+6x^{4}+11x^{3}+9x^{2}+3x+1
What can you say about the roots of P(x) and Q(x)?
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