If we take the differences then from the top differences we can derive the coefficients right to left.

f(x)=a+bx+cx^2:

x #1 f(x) #2 #3
1 a+b+c b+3c 2c [Top Differences]
2 a+2b+4c b+5c
3 a+3b+9c

Then #3/2=c , #2-3c=b , #1-b-c=a
But what are the formulas for this in general?
(Next I would/will write code to generate formulas.)
Are there simple formulas using the entire set of (all 6) differences?
Or take the differences of the top differences?

C-B

On Saturday, December 25, 2021 at 10:11:59 AM UTC-5, Charlie-Boo wrote:
> If we take the differences then from the top differences we can derive the coefficients right to left.
>
> f(x)=a+bx+cx^2:
>
> x #1 f(x) #2 #3
> 1 a+b+c b+3c 2c [Top Differences]
> 2 a+2b+4c b+5c
> 3 a+3b+9c
>
> Then #3/2=c , #2-3c=b , #1-b-c=a
> But what are the formulas for this in general?
> (Next I would/will write code to generate formulas.)
> Are there simple formulas using the entire set of (all 6) differences?
> Or take the differences of the top differences?
>
> C-B

algebraic NU( 1.0 );
algebraic polynomial( algebraic r[n], algebraic z)
{ if( n ) return ( r[n] + z ) * polynomial( r[n-1],z);
return NU;
}

This is in terms of roots, so pick roots and a z. Also this array presumes it is carrying n internally. To claim this is type algebraic might be flighty in some languages but probably C++ will do this. Anyway I didn't compile it and it obviously won't compile as is. Concept only. NU means Neutral Unity. The type 'algebraic' implies more than just the reals or the complex numbers; the nD version works fine and n can be zero. Really, I insist.