Question

lgba Problem

Answer 1

\[\mathcal L \{t^nf(t) \} = (-1)^n\frac{d^n}{ds^n}[\mathcal L \{f(t)\}]\]

Answer 2

@lgbasallote

in your proof you decided to take n=1 derivative .. the only way to prove it this way would to be prove that both the left side and right side are true for all \[n>1\] and then test a point

Answer 3

..ok..im not following but maybe if i see the next step i will

Answer 4

\[\mathcal L \{t^ne^{-at} \} = (-1)^n\frac{d^n}{ds^n}[\mathcal L\{e^{-at}\}]\]

Answer 5

on the side you did this you have

\[\frac{(-1)^n(-1)}{n!(s+a)^2}\]

however this is only when n=1 .... it alternates and chance from each derivative

Answer 6

OHHHH yeah...i only did first derivative

Answer 7

so how should it have been?

Answer 8

well for one if you use this way you'd have to prove both of all n for n>1 by using the definition of laplace transform. Once you prove that both are that , you can test any n >1 and prove that they're =

Answer 9

maybe the only way to prove it is using the fact that \[\large \mathcal L \{e^{at} f(t)\} = f(s-a)\]
huh?

Answer 10

or beter yet you could use the fact that e^{-as} can be shown as a series

Answer 11

..uhmmm im totally illiterate in series...

Answer 12

if you can figure out the pattern of which n derivatives goes

Answer 13

well you have to find out as you take n derivatives, a pattern of e^{-at}

Answer 14

if you can find the pattern of that, you can say that \[\frac{d}{ds^n}[e^{-at}]=\sum_{n=1}^{\infty}\]

Answer 15

if you look as you take derivatives lets see what you get

Answer 16

...wow.....

Answer 17

o wait nvm not e^{-at} it'd be htis

\[\frac{d}{ds^n}[\mathcal L \{e^{-at} \}]\] now figure out the patern of this