n-degree polynomial problems

• Jun 15th 2005, 06:21 AM
Mathman24
n-degree polynomial problems
Hi

I'm tasked with finding two polynomials p(x) and q(x) of degree three according to the following conditions:

p(0)= -1, p'(0)= 1, q(1)= 3, q'(1)= -2, q(0)= p(0)= -1, q'(0)= p'(0)= 1.

Okay, write p(x)= a0+ a1x+ a2x2+ a3x3 and
q(x)= b0+ b1x+ b2x2+ b3x3

p(0)= a0= -1 and p'(0)= a1= 1. Also, since q(0)= a(0), b0= -1 and since q'(0)= p'(0)= 1, b1= 1.

So far we have p(x)= -1+ x+ a2x2+ a3x3 and since that is all the information given about p(x), I don't see anyway of determining a2 or a3.

We have q(x)= -1+ x+ b2x2+ b3x3
so q(1)= -1+ 1+ b2+ b3= 3
and q'(x)= 1+ 2b22+ 3b2 so
q'(1)= 1+ 2b2+ 3b2= -2.

I'm told that the resulting polynomials are:

p(x) = (2 + s - 2t) x^3 + (3 + 2s - 3t) x^2 + s*x + t

q(x) = (-6 + s + 2t) x^3 + (9 - 2s - 3t) x^2 + s*x +t

where s,t belong to R.

My question is how do I get from my result to the final result ??

Hope there is somebody out there who can guide me to answer :-)

Sincerley

Fred
• Jun 17th 2005, 01:00 PM
hpe
The conditions determine q uniquely, since
Quote:

We have q(x)= -1+ x+ b2x2+ b3x3
so q(1)= -1+ 1+ b2+ b3= 3
and q'(x)= 1+ 2b22+ 3b2 so
q'(1)= 1+ 2b2+ 3b2= -2.
and therefore b2+b3=3, 2b2+3b3=-3, b2= 12, b3=-9. So $q(x) = -1+x+12x^2-9x^3$.

You only have two conditions for p(x), so $p(x) = -1+x+sx^2+tx^3$, where s and t are arbitrary real parameters. The solution formulae $p(x) = (2 + s - 2t) x^3 + (3 + 2s - 3t) x^2 + s\cdot x + t$ and $q(x) = (-6 + s + 2t) x^3 + (9 - 2s - 3t) x^2 + s\cdot x +t$ can't be correct, since for example this would mean p(0) = t and q(0) = t. Perhaps there is a typo somewhere.