Prove a linear mapping of a polynomial function is a map

[sa1988]

Homework Statement

The bane of all physicists... 'Proof' questions...

So we have the mapping,

Δ : P₃→P₃
Δ[P(x)] = (x²-1) d²P/dx² + x dP/dx

And I need to prove that this is a linear mapping

Homework Equations

Linear maps must satisfy:
Δ[P(x+y)] = Δ[P(x)] + Δ[P(y)]
and
Δ[P(αx)] - αΔ[P(x)]

The Attempt at a Solution

I'm not sure what to do. I've tried working through the actual mapping, performing the differential operations on the polynomial:
ax³+bx²+cx+d
and on:
a(x+y)³+b(x+y)²+c(x+y)+d
then expanded the brackets to see if I could separate the x and y terms to show that it's possible to pull them apart and demonstrate equality with Δ[P(x)] + Δ[P(y)]
But it doesn't work. My best bet is that I've done something wrong regarding the part where I need to do
d²P/d(x+y)². I've never really had to work in that way before. Maybe I've gone wrong completely.

Any advice?

Thanks!

 

[Mark44]

Homework Statement

So we have the mapping,

Δ : P₃→P₃
Δ[P(x)] = (x²-1) d²P/dx² + x dP/dx

And I need to prove that this is a linear mapping

Homework Equations

Linear maps must satisfy:
Δ[P(x+y)] = Δ[P(x)] + Δ[P(y)]

The above isn't right.
It should be Δ[P(x) + P(y)] = Δ[P(x)] + Δ[P(y)]
IOW, the Δ map carries a sum of functions to the sum of the transformations of the functions.

and
Δ[P(αx)] - αΔ[P(x)]

The Attempt at a Solution

I'm not sure what to do. I've tried working through the actual mapping, performing the differential operations on the polynomial:
ax³+bx²+cx+d
and on:
a(x+y)³+b(x+y)²+c(x+y)+d
then expanded the brackets to see if I could separate the x and y terms to show that it's possible to pull them apart and demonstrate equality with Δ[P(x)] + Δ[P(y)]
But it doesn't work. My best bet is that I've done something wrong regarding the part where I need to do
dP²/d(x+y)². I've never really had to work in that way before. Maybe I've gone wrong completely.

Any advice?

Thanks!

 

[sa1988]

The above isn't right.
It should be Δ[P(x) + P(y)] = Δ[P(x)] + Δ[P(y)]
IOW, the Δ map carries a sum of functions to the sum of the transformations of the functions.

Bingo, you've got it. I remember now.

Such a little thing - thanks for pointing it out!

 

[RUber]

Taking the differential on $P(x) = Ax^3 + Bx^2 + Cx + D$ should give:
$\Delta P(x) = (x^2-1)(6Ax+2B)+x(3Ax^2+2Bx+C) = 9Ax^3 +4Bx^2+(C-6A)x-2B$
As Mark pointed out, you want to compare two polynomials $P_1(x) = ax^3 + bx^2 + cx + d$ and $P_2(x) = hx^3 + ix^2 + jx + k$, not polynomials of different variables.
This should make sense, because a polynomial of greater than first order is known to not be a linear function.

 

[Ray Vickson]

The above isn't right.
It should be Δ[P(x) + P(y)] = Δ[P(x)] + Δ[P(y)]
IOW, the Δ map carries a sum of functions to the sum of the transformations of the functions.

Shouldn't that be $\Delta [P_1(x) + P_2(x)] = \Delta P_1(x) + \Delta P_2(x) \: \forall x$?

 

[Mark44]

Shouldn't that be $\Delta [P_1(x) + P_2(x)] = \Delta P_1(x) + \Delta P_2(x) \: \forall x$?

Now that you mention it, yes...
Thanks for the correction.