Question

If L1, L2 are linear differential operators, then
L1L2=L2L1, true or false.
They said that it's false but I don't get. Please give me a counterexample to see it's false.

Answer 1

Hey @Loser66 are these distances?

Answer 2

oh , it lacks of the condition that "If L1, L2 are linear differential operators, then L!L2=L2L1"

Answer 3

ok

$$
L1 = ax + bx\\
L2 = cx + dx\\
L1L2=a(cx+dx)+d(cx+dx)\\
$$
See if this equals
$$
L2L1=c(ax+bx)+d(ax+bx)
$$

Answer 4

@ybarrap so, you put all L2 as x in L1L2? I mean L1(L2) = L1 (x)?

Answer 5

I don't get the logic in your interpretation, L1L2 =......

Answer 6

For L1L2, first you apply the "rule" L2 to x: say, y=ax+bx then you apply the "rule" L1 to the results of L2: cy+dy.

Sort of like a composition \(L1L2=(L1\circ L2)x\)

Answer 7

$$
(L1\circ L2)x=L1(L2(x))
$$

Answer 8

One more question, why don't you choose L1= ax' +bx since it is linear differential? why ax + bx (it's just in C^0)?

Answer 9

If you take L1 and L2 as differential operators, then, for example, \(L1(x^2)=2x\) and \(L2(L1(x^2))=2x\). But now instead of \(x^2\), use any function of \(x\). So \(L1 = (af(x)+bg(x))'\) is one application of the differential operator. The simple format I used above was just to simplify the basic idea. Your next step would be to apply L2 to these results \(L2L1 = (c(af(x)+bg(x))'+d(af(x)+bg(x))')'\).

Answer 10

got you, thanks a lot. I know how to apply.

Answer 11

np