Question

Determine all pairs of polynomials f and g with real coefficients such that \(x^2g(x)=f(g(x))\)

Answer 1

Let g(x) be a0+a1x+a2x^2+a3x^3+...+anx^n.
Let f(x) be b0+b1x+b2x^2+b3x^3+...+bnx^n.

a0x^2+a1x^3+a2x^4+...+anx^(n+2) = f(a0+a1x+a2x^2+...+anx^n)
a0x^2+a1x^3+a2x^4+...+anx^(n+2) = b0 + b1(a0+a1x+a2x^2+...+anx^n) + b2(a0+a1x+a2x^2+...+anx^n)^2 + b3(a0+a1x+a2x^2+...+anx^n)^3 + ... + bn(a0+a1x+a2x^2+...+anx^n)^n

Nah just kidding, wrong approach here.

Answer 2

ok ;)

Answer 3

Maybe using calculus?
x^2g(x) = f(g(x))
2xg(x)+x^2g'(x) = f'(g(x))g'(x)
2xg(x) = f'(g(x))g'(x) - x^2g'(x)
2xg(x) = g'(x)[f'(g(x))-x^2]

Nah this is not going the right way...

Answer 4

How do I separate f and g?! :O

Answer 5

jus thinking about making a new function \(h(x)=(fog)(x)=f(g(x))\)

Answer 6

What's the difference...

Answer 7

f(g(x)) = h(x)
g(x) = f^-1(h(x))

mother of G-d...

Answer 8

x^2f^-1(h(x))=h(x)

Answer 9

Still a chain function here...

Answer 10

@ganeshie8

Answer 11

no clue... still thinking..

Answer 12

first pair :
g = x
f = x^3

seem to satisfy

Answer 13

maybe \(f(x)=x^{n-2},g(x)=x^n\)

Answer 14

Let f's order be m and g's order be n.
O(x^2g(x)) = O(x^2)+O(g(x)) = n+2
O(f(g(x)) = O(f(x))O(g(x)) = mn

x^2g(x)=f(g(x))
O[x^2g(x)]=O[f(g(x))]
n+2=mn
m=(n+2)/n

n+2=mn
mn-n=2
n(m-1)=2
n=2/(m-1)

Answer 15

Note that m and n must be integers.

Answer 16

looks great !
so m = 2, 3

Answer 17

Thanks @ganeshie8 for making me come to the answer.

Answer 18

And @Jonask too.

Answer 19

After some trial-and-errors we can say that n can be 1 or 2

Answer 20

ok looks very good so far,the order's are working out properly

Answer 21

And as ganeshie said m can be 2 or 3

Answer 22

A million thanks to @ganeshie8 and a billion of them to @Jonask for asking this question :)

Answer 23

When m=2, n=2;
When m=3, n=1.

Answer 24

how to conclude # of pairs of functions...
there can be many functions wid those degree... or im not seeing something obvious uhmmm

Answer 25

When m=2 and n=2:
Let f(x) and g(x) be ax^2+bx+c and dx^2+ex+f respectively.
x^2g(x)=f(g(x))
x^2(dx^2+ex+f)=a(dx^2+ex+f)^2+b(dx^2+ex+f)+c
blah blah blah

Answer 26

m,n has to be integers ,so that makes only two pairs

Answer 27

When m=3 and n=1:
Let f(x) and g(x) be ax^3+bx^2+cx+d and ex+f respectively.
x^2g(x)=f(g(x))
x^2(ex+f)=a(ex+f)^3+b(ex+f)^2+c(ex+f)+d
blah blah blah

Answer 28

These two pairs are actually solvable, but I'm looking for an easier way to find out all the pairs.

Answer 29

a trillion thanks to @kc_kennylau for coming up with the "orders"

Answer 30

6 variables thats not gonna be smooth

Answer 31

but this could help \(0=f(g(0)),g(1)=f(g(1)\dots\)

Answer 32

And g(-1)=f(g(-1))

Answer 33

so basically we can get all variables with this conditions,i am looking for a computerised solution

Answer 34

@ganeshie8 any idea?