Question

Calculate linear transformation D^nS^n and S^nD^n (n= 1,2,3......) where D is differentiation and S is integration transformations.
Please, help

Answer 1

@Kainui

Answer 2

Isn't this just the identity matrix?

Answer 3

how?

Answer 4

Well integrate x^2 you get

\[\frac{ 1 }{ 3 } x^3 + C_1\]
integrate again we get:

\[\frac{ 1 }{ 12 } x^4 + C_1x+C_2\]

Now take the derivative twice and you'll get back x^2.

So sure this is a specific case, but it should sort of show how you could really pick anything and it would be just itself again.

Answer 5

if P(x) is a polynomial degree k, then, we must consider k>n, k <n , and k = n
and D^n S^n \(\neq\) S^n D^n

Answer 6

because in the case k>n, S^n first will end at a polynomial degree (k-n) + a long tail of constants, then, take derivative back to get the original one

Answer 7

but if derivative first, at the n times, and k <n the polynomial = 0 , how to take the next step?
for example k =2 and n =10, let say x^2 and S^10 D^10 (x^2) at D^7 the value of it disappears

Answer 8

Since the operator has the integral BEFORE the derivative operator you will never be losing terms. If you differentiate first, then you'd run into problems like you describe.

Because the power of D is the same as S, it will always bring us back since each one exactly undoes the other amount.

Unless I'm misunderstanding the problem...

Answer 9

OH I was only answering the first part D^nS^n and not the other part.

Alright one second lol.

Answer 10

You'll get back a polynomial of the same degree, but depending on the power of the operator that will really decide the amount of terms get changed/added and what you get back will just be a general formula for a polynomial that you need k initial/boundary conditions to get back the exact thing you started with.

Answer 11

OK, that means I have to argue all cases which can happen, right? justify my answer in each case, right? even I get 0, it's a polynomial still, right?

Answer 12

hey, help me one more, ok?

Answer 13

If A and B are linear transformations such that AB - BA commutes with A, then \(A^kB-BA^k=kA^{k-1}(AB-BA)\)
I stuck at k in the front from the right hand side. why is it there?

Answer 14

because of that k, everything goes wrong to me. :(

Answer 15

are you sleeping? I am sleeping, too. hehehe

Answer 16

Since we know it commutes, you can play around with it a little...

A(AB-BA)=(AB-BA)A
distribute the A's\[A^2B-ABA=ABA-BA^2\]Oh hey add them together and the ABA part cancels, that's interesting... :)\[2A(BA-BA)=(A^2B-BA^2)\]

Answer 17

That was a quick nap i took btw lol.