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anonymous
 one year ago
Laplace transform of f(t)=t^n
inductive proof?
anonymous
 one year ago
Laplace transform of f(t)=t^n inductive proof?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we know the laplace transform of this, but could anyone explain how to use an inductive method. Would we suggest letting n=1, solving for that, n=2 solving for that, n=3 solving for that and then assume it for all cases?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2setup the integral and see if you can get a recurrence relation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[F(s)=\int\limits_{0}^{\infty}t ^{n}e ^{st}dt\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \mathcal{L}\{t^n\} ~~=~~ \int\limits_0^{\infty} t^ne^{st}\,dt\] try by parts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for \[\int\limits_{}^{}fdg=fg\int\limits_{}^{}gdf\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let f=t^n dg=e^(st)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks good, keep going..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'll be using equation mode so bare with me : )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[F(s)=t ^{n}\frac{ e ^{st} }{ s }\int\limits_{}^{}\frac{ e ^{st} }{ s }nt ^{n1}dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you would have to do by parts again right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you just keep going on forever

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\begin{align}\mathcal{L}\{t^n\} ~&=~t ^{n}\frac{ e ^{st} }{ s }\Bigg_{0}^{\infty}\int\limits_{0}^{\infty}\frac{ e ^{st} }{ s }nt ^{n1}dt\\~\\ ~&=~t ^{n}\frac{ e ^{st} }{ s }\Bigg_{0}^{\infty}+\dfrac{n}{s}\int\limits_{0}^{\infty} t^{n1}e ^{st}dt\\~\\ &=0+\dfrac{n}{s}\mathcal{L}\{t^{n1}\} \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so we get the recurrence relation : \[\large \mathcal{L}\{t^{n}\}~~=~~\dfrac{n}{s}\mathcal{L}\{t^{n1}\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isn't the first part (0(1))?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait, the t out the front

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2are you saying \[t ^{n}\frac{ e ^{st} }{ s }\Bigg_{0}^{\infty} = 0(1)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how is this a recurrence relationship? Its been a while since ive done this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the integral is actually the laplace of t^(n1) right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so we get the recurrence relation : \[\begin{align}\mathcal{L}\{t^{n}\}~~&=~~\dfrac{n}{s}\mathcal{L}\{t^{n1}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n1}{s}\mathcal{L}\{t^{n2}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n1}{s}*\dfrac{n2}{s}\mathcal{L}\{t^{n3}\}\\~\\ &=\cdots\\~\\ &=~~\dfrac{n}{s}*\dfrac{n1}{s}*\dfrac{n2}{s}\cdots*\dfrac{2}{s}*\dfrac{1}{s}*\mathcal{L}\{t^{0}\}\\~\\ &=~~\dfrac{n!}{s^n}*\mathcal{L}\{t^{0}\}\\~\\ \end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you simplify that more by saying that the laplace transform of t^0 is just the step change of the process; 1/s?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ n! }{ s ^{n+1} }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think so, but im not so good with control theory...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that would make sense because that is the final laplace transform of t^n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so im confused why you have: 2/s * 1/s *LT(t^0) why is there a 1/s factor

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2if psble send me the meeting id, il explain it quick

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you only run this via pc? i have a mac

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.exe is pc only i think

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try this http://download.teamviewer.com/download/TeamViewer.dmg

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2message me the password

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for that! i'll keep working on it. cheers for the interactive help

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2np :) check this video when free http://ocw.mit.edu/courses/mathematics/1803differentialequationsspring2010/videolectures/lecture19introductiontothelaplacetransform/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm doing a course in process control, and unfortunately we never came across laplace transforms in engineering math (my degree only uses 1st semester engineering math not second) but more learning the better!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you can also take an easy one \(\mathcal L \{1\} = \int_{0}^{\infty} e^{st} dt = \frac{1}{s}\) and diffeentiate it \(\large \frac{d}{ds} \int_{0}^{\infty} e^{st} dt = \int_{0}^{\infty} \frac{d}{ds}e^{st} dt = \frac{d}{ds} \frac{1}{s}\) over and over
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