Prove a linear mapping of a polynomial function is a map

1. Dec 3, 2015

sa1988

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

Thanks!

2. Dec 3, 2015

Staff: Mentor

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.

3. Dec 3, 2015

sa1988

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

Such a little thing - thanks for pointing it out!

4. Dec 3, 2015

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.

5. Dec 3, 2015

Ray Vickson

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

6. Dec 3, 2015

Staff: Mentor

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