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anonymous
 4 years ago
\[L^{1} (\frac{2s^3}{s^281})\]
The actual problem in the problem set is
\[L^{1} (\frac{2s^3}{s^481})\]
anonymous
 4 years ago
\[L^{1} (\frac{2s^3}{s^281})\] The actual problem in the problem set is \[L^{1} (\frac{2s^3}{s^481})\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[L^{1} (\frac{2s^3}{s^281})=L^{1} (2s+\frac{81}{s9}+\frac{81}{s+9})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't like Laplace /_\

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1i assume, thats correct, so whats the problem ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[L^{1}(2s)=trouble\] Btw.. I... suddenly... discovered that... I read the problem wrong :\

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But how to find \(L^{1}(2s)\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the correct question in the problem set: \[L^{1} (\frac{2s^3}{s^481})\]\[=L^{1} (\frac{2s^3}{s^481})\]\[=L^{1} (\frac{s}{s^2+9}+\frac{1}{2(s3)}+\frac{1}{2(s+3)})\]\[=cos(3t)+cosh(3t)\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, my computer restarted.... to get L^{1} (2s), u just replace 's' by 2s ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0L^{1} (s) also looks weird to me... Usually, it's a/s^n which I can do the inverse easily (easier, I mean)..

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ohh...i read that wrong .. u know whats L^{1} s ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1its \(\delta'(t)\) so, \(L^{1}(2s) = 2\delta'(t)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is \(\delta'(t)\)?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1u know what is \(\delta(t) ?\)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1Dirac delta function.. and ' means its derivative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I.. don't remember I've learnt such function...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1really ? http://en.wikipedia.org/wiki/Dirac_delta_function 0 everywhere except origin.....like an impulse.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I've only learnt int. from 0 to infty and from s to infty case, not the ve infty to +ve infty... Thanks for introducing a new friend to me though :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I want to get back my words. I don't know the name of this new friend, but he's just an old friend of mine :'( I'm sorry!!
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