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amberjordan2006

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  • 11 months ago
  • 11 months ago

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

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

    • 11 months ago
  3. zepdrix
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    Can you? :D Heh

    • 11 months ago
  4. zepdrix
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    Ummmmmmmmmmmm

    • 11 months ago
  5. zepdrix
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    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.

    • 11 months ago
  6. zepdrix
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    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.

    • 11 months ago
  7. zepdrix
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    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

    • 11 months ago
  8. zepdrix
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    Hmm I dunno :c

    • 11 months ago
  9. zepdrix
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    Yah the site gets a lot of visitors.. slows down sometimes :(

    • 11 months ago
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