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## Main Question or Discussion Point

Question about quadratics of quadratics

can any 4th degree polynomial be expressed as a quadratic of a quadratic function?

or in the more general case, can any polynomial of degree 2^n be expressed as n-many quadratic functions of quadratic functions?

and given a polynomial of degree 2^n is there a way to find the coefficients of the quadratics?

i tried finding a way to reduce the 4th degree poly into a quadratic of a quadratic. but the main problem is when trying to find the coefficients that i have 5 equations in 6 unknowns, so the system to find the coefficients isnt determined. why and what does this mean?

r=(R4)X^4+(R3)X^3+(R2)X^2+(R1)X+R0

p[q[X]]=P2(Q2*X^2+Q1*X+Q0)^2+P1(Q2*X^2+Q1*X+Q0)+P0

in order for r[x] to equal p[q[x]]:

R4=(P2)(Q2)

R3=2(P2)(Q2)(Q1)

R2=(P2)(2(Q2)(Q0)+(Q1)^2)+(P1)(Q2)

R1=2(P2)(Q0)(Q1)+(P1)(Q1)

R0=(P2)(Q0)^2+(P1)(Q0)+(P0)

im wondering if i could just set one of the coefficients of either quadratic equal to zero. but even then the system is unsolveable by matrices, and im not sure what terms are allowed to be zero.

can any 4th degree polynomial be expressed as a quadratic of a quadratic function?

or in the more general case, can any polynomial of degree 2^n be expressed as n-many quadratic functions of quadratic functions?

and given a polynomial of degree 2^n is there a way to find the coefficients of the quadratics?

i tried finding a way to reduce the 4th degree poly into a quadratic of a quadratic. but the main problem is when trying to find the coefficients that i have 5 equations in 6 unknowns, so the system to find the coefficients isnt determined. why and what does this mean?

r=(R4)X^4+(R3)X^3+(R2)X^2+(R1)X+R0

p[q[X]]=P2(Q2*X^2+Q1*X+Q0)^2+P1(Q2*X^2+Q1*X+Q0)+P0

in order for r[x] to equal p[q[x]]:

R4=(P2)(Q2)

R3=2(P2)(Q2)(Q1)

R2=(P2)(2(Q2)(Q0)+(Q1)^2)+(P1)(Q2)

R1=2(P2)(Q0)(Q1)+(P1)(Q1)

R0=(P2)(Q0)^2+(P1)(Q0)+(P0)

im wondering if i could just set one of the coefficients of either quadratic equal to zero. but even then the system is unsolveable by matrices, and im not sure what terms are allowed to be zero.