Question

i tried wolframming this but it didnt give me anything :/ \[\mathcal L \{ \cos (4t) u(t - \pi)\}\]
i got something like \[-\frac{s}{s^2 + 16}\]or something like that

Answer 1

\[\mathcal L\{f(t-c)u(t-c)\}=e^{-cs}F(s)\]but c isn't the same in both...

Answer 2

i think this is more of the \[\mathcal L \{ f(t) u(t-c)\} = e^{-cs} \mathcal L \{f(t+c) \}\]

Answer 3

yes it is of that form.

Answer 4

I guess, in which case \(c=\pi\) and \[f(t+\pi)=\cos(4(t+\pi))=\cos(4t+4\pi)=\cos(4t)\]

Answer 5

that's funny, it goes back where it started...

Answer 6

yes that was my main concern.. \[\cos[4(t+ \pi)]\]
i would like to know if this is equal to \[-\cos 4t\]

Answer 7

no, it's just \[\cos(4t)\]because\[\cos(4t+4\pi)=\cos(4t+2(2\pi))\]and adding any multiple f \(2\pi\) to the angle will take us back where we started

Answer 8

any *whole number* multiple of...

Answer 9

so I think the final answer is just\[{se^{-cs}\over16+s^2}\]

Answer 10

I could be wrong...

Answer 11

well there can be only one right...

Answer 12

turing test answer is correct.

Answer 13

replace c with pi in my answer :/\[{se^{-\pi s}\over16+s^2}\]

Answer 14

my teacher really loves teaching us wrong stuffs huh

Answer 15

a more rigorous proof that adding a whole number multiple of 2pi leaves the argument the same

Answer 16

yeah he is really loves that :P

Answer 17

well he still gives the grades...

Answer 18

usually when n=even
\[\Large \cos(n \pi)=1\]
when n is odd
\[\Large \cos(n \pi)=-1\]