# Linear operator on the set of polynomials

1. Jun 7, 2010

### clope023

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

Let L be the operator on P_3(x) defined by

L(p(x)) = xp'(x)+p"(x)

if p(x) = a_0(x)+a_1(x)+a_2(1+x^2) calculate L^n(p(x))

2. Relevant equations

stuck between 2 possible solutions

i) as powers of x decrease the derivatives of p(x) increase

ii) as derivatives of x decrease the derivatives of p(x) increase

3. The attempt at a solution

check with increasing powers

L^2(p(x)) = x^2p''(x)+xp"'(x)+p^(4)(x)

L^3(p(x)) = x^3p'''(x) + x^2p^(4)(x)+xp^(5)(x)

=> L^n(p(x)) = x^np^(n)(x)+x^(n-1)p^(n+1)(x)+x^(n-2)p^(n+2)(x)

when a superscript is applied to an x it is a power

when a superscript is applied to a p it is a derivative

I'm not quite sure which it should be

any help is appreciated, thank you

2. Jun 7, 2010

### CompuChip

What does
p(x) = a_0(x)+a_1(x)+a_2(1+x^2) ​
mean?

Are a_0, a_1 and a_2 numbers, multiplied by x? So basically,
$$p(x) = (a_0 + a_1) \cdot x + a_2 \cdot (1 + x^2) ?$$
Or are they functions of x and 1 + x2?

Note that, if p lies in P3, then p(k) = 0 for all k > 3.

3. Jun 7, 2010

### clope023

the a's are coefiicients multiplied by the x's, the first x is x to the power 0, the second is x to the power 1 and then third is (1+x^2)

so its a_0 by itelf + a_1 multiplied by x and a_2 multiplied by (1+x^2)

sorry for the confusion

4. Jun 7, 2010

### Staff: Mentor

Is there some reason p(x) is not a_0 + a_1*x + a_2 * x^2? That would seem more natural to me.

5. Jun 7, 2010

### CompuChip

OK, that makes it more clear.
Maybe it helps if you write out the first few lines explicitly.
For example,

$$L(p) = a_1 x + 2 a_2(1 + x^2)$$
$$L(L(p)) = \cdots ?$$

The pattern is quite hard to miss.

6. Jun 7, 2010

### vela

Staff Emeritus
I think the basis is a given in the problem because the matrix for L has a nice form in that basis.

7. Jun 8, 2010

### HallsofIvy

Staff Emeritus
p is a polynomial of degree 3 or less. The fourth derivative of such a polynomial is 0!