Question

Laplace transform of f(t)=t^n

inductive proof?

Answer 1

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?

Answer 2

setup the integral and see if you can get a recurrence relation

Answer 3

\[F(s)=\int\limits_{0}^{\infty}t ^{n}e ^{-st}dt\]

Answer 4

\[\large \mathcal{L}\{t^n\} ~~=~~ \int\limits_0^{\infty} t^ne^{-st}\,dt\]

try by parts

Answer 5

so for \[\int\limits_{}^{}fdg=fg-\int\limits_{}^{}gdf\]

Answer 6

let f=t^n
dg=e^(-st)

Answer 7

looks good, keep going..

Answer 8

i'll be using equation mode so bare with me : )

Answer 9

\[F(s)=t ^{n}\frac{ -e ^{-st} }{ s }-\int\limits_{}^{}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\]

Answer 10

you would have to do by parts again right

Answer 11

nope, we're done

Answer 12

you just keep going on forever

Answer 13

\[\begin{align}\mathcal{L}\{t^n\} ~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}-\int\limits_{0}^{\infty}\frac{ -e ^{-st} }{ s }nt ^{n-1}dt\\~\\
~&=~t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty}+\dfrac{n}{s}\int\limits_{0}^{\infty} t^{n-1}e ^{-st}dt\\~\\
&=0+\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}
\end{align}\]

Answer 14

so we get the recurrence relation :
\[\large \mathcal{L}\{t^{n}\}~~=~~\dfrac{n}{s}\mathcal{L}\{t^{n-1}\}\]

Answer 15

isn't the first part (0-(-1))?

Answer 16

oh wait, the t out the front

Answer 17

makes it 0

Answer 18

yep, got it.

Answer 19

are you saying
\[t ^{n}\frac{ -e ^{-st} }{ s }\Bigg|_{0}^{\infty} = 0-(-1)\]

?

Answer 20

my mistake

Answer 21

so how is this a recurrence relationship?
Its been a while since ive done this

Answer 22

so the integral is actually the laplace of t^(n-1) right?

Answer 23

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}\]

Answer 24

could you simplify that more by saying that the laplace transform of t^0 is just the step change of the process; 1/s?

Answer 25

\[\frac{ n! }{ s ^{n+1} }\]

Answer 26

I think so, but im not so good with control theory...

Answer 27

yeah that would make sense because that is the final laplace transform of t^n

Answer 28

so im confused why you have: 2/s * 1/s *LT(t^0)
why is there a 1/s factor

Answer 29

if psble send me the meeting id, il explain it quick

Answer 30

can you only run this via pc? i have a mac

Answer 31

oh wait then

Answer 32

.exe is pc only i think

Answer 33

try this
http://download.teamviewer.com/download/TeamViewer.dmg

Answer 34

what do i do?

Answer 35

553 005 551

Answer 36

is the id

Answer 37

message me the password

Answer 38

thanks for that! i'll keep working on it.

cheers for the interactive help

Answer 39

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/

Answer 40

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!

Answer 41

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