Question

Find the Laplace Transform of the given function. Assume that term-by-term integration of the infinite series is permissible.

Answer 1

\[\huge f(t)=1+\sum_{k=1}^{\infty}(-1)^ku_k(t)\]

Answer 2

@ganeshie8

Answer 3

Should I begin by expanding the summation and then just find the Laplace Transform of the first few terms?

Answer 4

.

Answer 5

Hmm. I think I can rewrite it as
\[\mathcal{L}[f(t)]=\frac{1}{s}+\sum_{k=1}^{\infty}(-1)^k~\mathcal{L}[u_k(t)]\]But what to do with that, no clue..

Answer 6

\[\mathcal{L}[f(t)]=\frac{1}{s}+\sum_{k=1}^{\infty}(-1)^k~\frac{e^{-cs}}{s}\]

Answer 7

Maybe?

Answer 8

\[\mathcal{L}(f(t-a)u(t-a)) = e^{-as}F(s)\]

Answer 9

in your case \(f(t-a) = 1\)

Answer 10

must be a typo, you mean

\[\mathcal{L}[f(t)]=\frac{1}{s}+\sum_{k=1}^{\infty}(-1)^k~\frac{e^{-\color{red}{k}s}}{s}\]

Answer 11

Ah yes, yes X) oops..

Answer 12

looks good to me, lets check with wolfram

Answer 13

How do I rid the sum?

Answer 14

http://www.wolframalpha.com/input/?i=+laplace+transform+%5Csum%5Climits_%7Bk%3D1%7D%5E%7B3%7D+%28-1%29%5Ek*%5Ctheta%28t-k%29

look at alternate forms

Answer 15

Hmm, but during the test I cannot check Wolfram =/ I managed to find the same question elsewhere, and they have the same steps I do until about halfway and then they start doing something I don't understand.

Answer 16

@ganeshie8

Answer 17

hey

Answer 18

wolfram is just for double checking your answer so that you feel confident

Answer 19

it is just a geometric series :

\[\sum\limits_{k=0}^{\infty} \clubsuit^k = \dfrac{1}{1-\clubsuit}\]

for \(|\clubsuit|\lt 1\)

Answer 20

ohh wow haha, thanks!