Expert: Scotto - 10/29/2011

Question
QUESTION: Hello Sir,
        for a set of n  quadratic polynomials qi(x)=ai(x^2) +bi(x) +c,s.t n>3,is it possible that not exist
a Q(x) from the set s.t
Q(x)*a=Q0(x)*a0  +  Q1(x)*a1   +  .... +  Q(n-1)(x)*a(n-1);
(a linear combo of the rest of the quadratics)?
Key Point:its in the case :
n>3,
its fairly easy for n<=3.
Thanks.

ANSWER: Let me rewrite the equation as Q(x)*d = Q0(x)*d0  +  Q1(x)*d1   +  .... +  Q(n-1)(x)*d(n-1).
and assume that Q(x) = ax^2 + bx + c.

This generates the equations a0*d0 + a1*d1 + ... + an*dn = a,
b0*d0 + b1*d1 + ... + bn*dn = b, and
c0*d0 + c1*d1 + ... + cn*dn = c.

This would generate 3 equations with 3 unkwons, and that can be solved.

It would seem like if we had more equations, any thing beyond the first three constants could be taken as 0, resulting in our 3 equations with 3 unknowns from the first three equations.


---------- FOLLOW-UP ----------

QUESTION: How about this proof i wrote?
proof try:
axiom(0):all operations over the real number domain;
axiom(1):exp(k*x)=a^(k*x);

exp(kx) is linearly independent of all other exp(ki*x) s.t ki!=k;
!= means not equal to.
where exp(kx) is the function in general( not the function  evaluated at x);

axiom(2):two different exponential functions can share at most one point.
axiom(3):two different polynomials of degree x can share at most x-1 point.exists point not shared.
claim:
for pi(var)=ai(var)+bi;
ai=exp(ki*xo);
bi=exp(ki*x1);
A system of equations
equ0,equ1,...,equi..,equ(n-1);
equi=> pi(x0)var0 +  pi(x1)var1 +...+pi(n-1)var(n-1)=vali;
for equ' where equ' could be any  equi 0<=i<n;
equ'!=a0*equ0+ a1*equ1+....+a(n-1)*equ(n-1);
for any non trivial ai(i.e system is linearly independent);
Proof:
Now exp(kx) is the function evaluated at xj;
let exp(knew*xj)=a0*exp(k0*xj)+a1*exp(kj*xj)+....+a(n-1)*exp(k(n-1)*xj);
for some exp(k'xi) which could be any exp(ki*xi);
0<=i<n;
if exp(k'xj)=exp(knew*xj) then
for any other xl s.t xl!=xj;
exp(k'xl)!=exp(knew*xl); by axiom(2);

therefore since the
p'(var) will  differ from the linear combination of the rest of the pi(var) in at least one coefficient,it is not equal to it.

therefore for p'(var) which could be any pi(var):
p'(var)!=a0*p(var0)+a1*p(var1)+.....+a(n-1)*p(var(n-1));

The points the pi(x) are evaluated at ; xi , are s.t xi!=xj; if i!=j;(the points are different allowing for use of axiom(3)),
and n>3,and the pi are different:

equ'!=a0*equ0+ a1*equ1+....+a(n-1)*equ(n-1);
Q.E.D.Thanks.

Answer
From what I read, this is very similar to what I wrote, but more to the point.
You're saying for a big factorial, we need several power of integers close to each other,
and that doesn't occur outside of 8 = 2³ and 9 = 3² with the next number, 10, being 2*5.