## amberjordan2006 Group Title . one year ago one year ago

1. zepdrix Group Title

Hmm interesting :O Ok let's get it plugged into the formula correctly, and see if we can figure this out. $\huge \mathscr{L}\left[e^{(t-1)^2}\right] =\int\limits_0^{\infty}e^{(t-1)^2}e^{-st}\;dt$

2. zepdrix Group Title

Hmm this one is a bit tricky. Let's first combine the exponentials using rules of exponents.$\huge \int\limits\limits_0^{\infty}e^{(t-1)^2-st}\;dt$

3. zepdrix Group Title

Can you? :D Heh

4. zepdrix Group Title

Ummmmmmmmmmmm

5. zepdrix Group Title

Hmmm I can't quite get a handle on this one... You can combine the -st into the equation and complete the square, I'm not really sure that gets us anywhere though.. hmmm @Mertsj @phi Maybe one of these smarty pants can help us.

6. zepdrix Group Title

As far as the 0 to infinity question goes... The Laplace Transform transforms us from one domain to another. Notice how the variable changes from t to s? There are many applications I'm sure, but one way we can think of it is going from the time domain to frequency. If you think of starting with time, we can't really integrate before time t=0. It doesn't have a nice interpretation, at least not in most cases. So we're just integrating over the entire domain of t, and it spits out a function of something else.

7. zepdrix Group Title

Let's look at the exponent a minute, maybe we can get somewhere with this.$\large (t-1)^2-st \qquad = \qquad t^2-2t+1-st$Which by combining the t terms, gives us,$\large t^2-(s+2)t+1$ From here.... I guess we could maybe complete the square and see if it takes us anywhere :\ It doesn't look like it will though.. hmmm

8. zepdrix Group Title

Hmm I dunno :c

9. zepdrix Group Title

Yah the site gets a lot of visitors.. slows down sometimes :(