How do you compute the Laplace Transform of [[x]] - I think that it has something to do with the greatest integer function.
L { [[x]] } = ???
*****Thanks for all of your help!!!! I really appreciate it!!!*****
L { [[x]] } = ???
*****Thanks for all of your help!!!! I really appreciate it!!!*****
-
Just use the definitions and write down the first few integrals:
1
∫ 0 e^(-st) dt +
0
2
∫ 1 e^(-st) dt +
1
3
∫ 2 e^(-st) dt +
2
4
∫ 3 e^(-st) dt + ...
3
You should get
0 + (1/s)[e^(-s) - e^(-2s) + 2e^(-2s) - 2e^(-3s) + 3e^(-3s) - 3e^(-4s) +- ...]
which simplifies to
(1/s)[e^(-s) + e^(-2s) + e^(-3s) + e^(-4s) + ,,,]
The series in the square brackets is an infinite geometric sum with a = r = e^(-s) , which converges to e^(-s)/(1 - e^(-s)) for e^(-s) < 1; that is, for s > 0. Multiply the numerator and denominator by e^s to get the simpler form 1/(e^s - 1). Thus,
L { [[x]] } = 1/(s(e^s - 1))
1
∫ 0 e^(-st) dt +
0
2
∫ 1 e^(-st) dt +
1
3
∫ 2 e^(-st) dt +
2
4
∫ 3 e^(-st) dt + ...
3
You should get
0 + (1/s)[e^(-s) - e^(-2s) + 2e^(-2s) - 2e^(-3s) + 3e^(-3s) - 3e^(-4s) +- ...]
which simplifies to
(1/s)[e^(-s) + e^(-2s) + e^(-3s) + e^(-4s) + ,,,]
The series in the square brackets is an infinite geometric sum with a = r = e^(-s) , which converges to e^(-s)/(1 - e^(-s)) for e^(-s) < 1; that is, for s > 0. Multiply the numerator and denominator by e^s to get the simpler form 1/(e^s - 1). Thus,
L { [[x]] } = 1/(s(e^s - 1))
-
.
-
Ans
1/s^2
1/s^2