Homework Help: 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

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