anonymous
  • anonymous
Laplace transform of f(t)=t^n inductive proof?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
So 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
  • ganeshie8
setup the integral and see if you can get a recurrence relation
anonymous
  • anonymous
\[F(s)=\int\limits_{0}^{\infty}t ^{n}e ^{-st}dt\]

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ganeshie8
  • ganeshie8
\[\large \mathcal{L}\{t^n\} ~~=~~ \int\limits_0^{\infty} t^ne^{-st}\,dt\] try by parts
anonymous
  • anonymous
so for \[\int\limits_{}^{}fdg=fg-\int\limits_{}^{}gdf\]
anonymous
  • anonymous
let f=t^n dg=e^(-st)
ganeshie8
  • ganeshie8
looks good, keep going..
anonymous
  • anonymous
i'll be using equation mode so bare with me : )
anonymous
  • anonymous
\[F(s)=t ^{n}\frac{ -e ^{-st} }{ s }-\int\limits_{}^{}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\]
anonymous
  • anonymous
you would have to do by parts again right
ganeshie8
  • ganeshie8
nope, we're done
anonymous
  • anonymous
you just keep going on forever
ganeshie8
  • ganeshie8
\[\begin{align}\mathcal{L}\{t^n\} ~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}-\int\limits_{0}^{\infty}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\\~\\ ~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}+\dfrac{n}{s}\int\limits_{0}^{\infty} t^{n-1}e ^{-st}dt\\~\\ &=0+\dfrac{n}{s}\mathcal{L}\{t^{n-1}\} \end{align}\]
ganeshie8
  • ganeshie8
so we get the recurrence relation : \[\large \mathcal{L}\{t^{n}\}~~=~~\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}\]
anonymous
  • anonymous
isn't the first part (0-(-1))?
anonymous
  • anonymous
oh wait, the t out the front
anonymous
  • anonymous
makes it 0
anonymous
  • anonymous
yep, got it.
ganeshie8
  • ganeshie8
are you saying \[t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty} = 0-(-1)\] ?
anonymous
  • anonymous
my mistake
anonymous
  • anonymous
so how is this a recurrence relationship? Its been a while since ive done this
anonymous
  • anonymous
so the integral is actually the laplace of t^(n-1) right?
ganeshie8
  • ganeshie8
so we get the recurrence relation : \[\begin{align}\mathcal{L}\{t^{n}\}~~&=~~\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}\mathcal{L}\{t^{n-2}\}\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}*\dfrac{n-2}{s}\mathcal{L}\{t^{n-3}\}\\~\\ &=\cdots\\~\\ &=~~\dfrac{n}{s}*\dfrac{n-1}{s}*\dfrac{n-2}{s}\cdots*\dfrac{2}{s}*\dfrac{1}{s}*\mathcal{L}\{t^{0}\}\\~\\ &=~~\dfrac{n!}{s^n}*\mathcal{L}\{t^{0}\}\\~\\ \end{align}\]
anonymous
  • anonymous
could you simplify that more by saying that the laplace transform of t^0 is just the step change of the process; 1/s?
anonymous
  • anonymous
\[\frac{ n! }{ s ^{n+1} }\]
ganeshie8
  • ganeshie8
I think so, but im not so good with control theory...
anonymous
  • anonymous
yeah that would make sense because that is the final laplace transform of t^n
anonymous
  • anonymous
so im confused why you have: 2/s * 1/s *LT(t^0) why is there a 1/s factor
ganeshie8
  • ganeshie8
if psble send me the meeting id, il explain it quick
anonymous
  • anonymous
can you only run this via pc? i have a mac
ganeshie8
  • ganeshie8
oh wait then
anonymous
  • anonymous
.exe is pc only i think
ganeshie8
  • ganeshie8
try this http://download.teamviewer.com/download/TeamViewer.dmg
anonymous
  • anonymous
what do i do?
anonymous
  • anonymous
553 005 551
anonymous
  • anonymous
is the id
ganeshie8
  • ganeshie8
message me the password
anonymous
  • anonymous
thanks for that! i'll keep working on it. cheers for the interactive help
ganeshie8
  • ganeshie8
np :) check this video when free http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/
anonymous
  • anonymous
i'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
  • IrishBoy123
you 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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