Question

Calculate the linear transformation \(D^nS^n~~and~~S^nD^n,\) n= 1,2,3,.....on arbitrary element of P.
D : differentiation; S: integration transformations

Answer 1

I know it is = P; just not know how to put in logic.
D S =1 or DS =0?
Please, help

Answer 2

Is \(P\) the vector space of polynomials?

Answer 3

yes

Answer 4

Well, my strategy would be to simply take an arbitrary polynomial of degree \(k\), and apply the two linear transformations on it. So for the transformation \(D^nS^n\), integrate the polynomial \(n\) times, and then take the derivative \(n\) times. Since everything is a polynomial, this shouldn't be too bad.

Answer 5

linear transformation D^n S^n = (DS)^n , is it right?

Answer 6

For that particular product of transformation I think that's true, but it's not true in general. For example, \(S^nD^n\neq(SD)^n\).

Answer 7

I don't think so.
Let say D*D*D*D*..........DS*S*S*S.........S= DS at the end

Answer 8

And you are right if the transformations do not cancel out like this D and S

Answer 9

So?? in this particular case, what should I conclude?

Answer 10

I still think what you should do is just take an arbitrary polynomial, and apply the linear transformations. So let \[p(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_1x+a_0\]and apply \(D^nS^n\) and \(S^nD^n\). One note for \(S^nD^n\) though, you need to consider the case when \(k\ge n\) and when \(k<n\) separately.

Answer 11

Ok, let do S^nD^n first, I have to take differentiate of p(x) n times. if n>k, at the time n = k_i I get a constant, then n = k_i +1 , I get 0. the case is done.

Answer 12

Hold on. If \(n>k\), then \(D^n(p(x))=0\). But \(S^n(0)=g(x)\) where \(g(x)\) is a polynomial of degree \(n-1\).

Answer 13

Let make the simplest case
S^3D^3(x+1) = S^3 D^2(2)=S^3D(0) = S^3 (0) no more polynomial to take integral, right?
this 0 is a number 0, not polynomial degree 0

Answer 14

*S^3D^2(1)

Answer 15

In this particular vector space, 0 is the zero vector, which is the zero polynomial.

Answer 16

I 'm not with you then. p(x) = \(\sum_k a_kx^k\) and \(a_k \) is a scalar. so, if the n>k, the term n=k is a number on D function

Answer 17

Even if you do just consider 0 as a number, then \(\int0\,dx=a\) where \(a\in\mathbb{R}\) is a constant

Answer 18

and then , at the end up, how can we compute the degree of polynomial?
If n is >>k, like p(x) = x+1, after twice, we get 0 for D, and start S. after twice we get x+ c ( another scalar, not 1 anymore)
3rd round = x^2 /2 + cx + b
4th round = x^3/6 +cx^2/2 + bx + d......
can it go on that way?

Answer 19

What you should eventually see, is that if \(n>k\), is that \(S^nD^n(p(x))=g(x)\) where \(g(x)\) is an arbitrary polynomial of degree \(n-1\).

Answer 20

I got you :) I will try that way. thank you