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Old 2017-04-25, 04:16   #2
Romulan Interpreter
LaurV's Avatar
Jun 2011

927510 Posts

Going to differentiation may be a bit tricky and not exactly "elementary" (as we learn polynomials in school much earlier than differentiation). One simpler way is to express the two in zero, getting \(c=t\), then express them in 1 and -1 (or 2, etc) and get \(b=s\), \(a=r\) etc.

We have: \( f(0)=g(0),\ \ \text{i.e.}\ \ a\cdot 0^2+b\cdot 0+c=r\cdot 0^2+s\cdot 0+t\) from which \(s=t\), then express them in 1 and eliminate t=c from the both sides:\( f(1)=g(1),\ \ \text{i.e.}\ \ a\cdot 1^2+b\cdot 1+c=r\cdot 0^2+s\cdot 0+t\) or \(a+b=r+s\). Doing the same for -1, we get \(a-b=r-s\) and now we add them together or subtract them together, and get a=r and b=s.

This also proves that it is enough for two polynomials of degree n to be the same in n+1 values (as they have n+1 coefficients), to be the same in all their domain. (i.e. if two polynomials of degree n in R have the same values in n+1 points, they are the same in all R, following a similar procedure, and induction).

Again, congratulations for your effort and for your patience to write another excellent and educative post!

Last fiddled with by LaurV on 2017-04-25 at 04:32 Reason: fixing \tex thingies
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