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

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1. 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?

2. ganeshie8

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

3. anonymous

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

4. ganeshie8

$\large \mathcal{L}\{t^n\} ~~=~~ \int\limits_0^{\infty} t^ne^{-st}\,dt$ try by parts

5. anonymous

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

6. anonymous

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

7. ganeshie8

looks good, keep going..

8. anonymous

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

9. anonymous

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

10. anonymous

you would have to do by parts again right

11. ganeshie8

nope, we're done

12. anonymous

you just keep going on forever

13. 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}

14. ganeshie8

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

15. anonymous

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

16. anonymous

oh wait, the t out the front

17. anonymous

makes it 0

18. anonymous

yep, got it.

19. ganeshie8

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

20. anonymous

my mistake

21. anonymous

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

22. anonymous

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

23. 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}

24. 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?

25. anonymous

$\frac{ n! }{ s ^{n+1} }$

26. ganeshie8

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

27. anonymous

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

28. anonymous

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

29. ganeshie8

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

30. anonymous

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

31. ganeshie8

oh wait then

32. anonymous

.exe is pc only i think

33. ganeshie8
34. anonymous

what do i do?

35. anonymous

553 005 551

36. anonymous

is the id

37. ganeshie8

message me the password

38. anonymous

thanks for that! i'll keep working on it. cheers for the interactive help

39. ganeshie8
40. 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!

41. 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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