Question

is \[\huge \frac{- (-1)^n}{n! (s+a)^2} = \frac{1}{(s+a)^{n+1}}\]
if so..how can i show it?

Answer 1

@sami-21 please abandon my very difficult question and help me here instead :DDD hehe

Answer 2

try it with some arbitrary values

Answer 3

Put \(n=1\), \(s=1\) and \(a=0\) and check identity.

Answer 4

maybe i should state my solution and how i got this?

Answer 5

basically..i was trying to prove \[\mathcal L \{ \frac{t^n e^{-at}}{n!} \} = \frac{1}{(s+a)^{n+1}}\]using the formula \[\mathcal L \{t^n f(t) \} = (-1)^n \frac{d^n}{ds^n} (\mathcal L \{f(t) \} )\]

Answer 6

cute symbols got into cute question now :D

Answer 7

ahh laplace transforms

Answer 8

yeahh

Answer 9

maybe i did a wrong step? or maybe i shouldnt have used that formula?

Answer 10

do you know what laplace is?

Answer 11

laplace transform is

\[\int f(t)e^{-st}dt=F(s)\]

Answer 12

although.. i do know \[\mathcal L \{e^{-at} t^n \} = \frac{n!}{(s+a)^{n+1}}\]
but i still want to prove it some other way...

Answer 13

and yes im familiar of teh general form (if that's what it's called)

Answer 14

so to prove i think you'd have to do
\[\int \frac{e^{-st}t^ne^{-at}}{n!}\]dt

Answer 15

\[\int \frac{e^{(-s-a)t}t^n}{n!}dt\]

Answer 16

so i cant use \[\large \mathcal L \{ t f(t) \} = (-1)^n \frac{d^n}{ds^n} (F(s) )?\]
i can only use the "general equation"?

Answer 17

that should be L{t^n f(t)}

Answer 18

wouldn't that be usng another proof to prove another proof? proofception?

Answer 19

well im not really asked to prove it...im practicing my skills by proving the table in my book

Answer 20

best way would to use the definition of laplace's transform. however the reason why the book just gives you those is because the proof my be a huge problem... i'd check my book for the proof but it's outside=/

Answer 21

all of these transforms on the table derive from the laplace transform definition

Answer 22

well im just practicing my use and mastery of that formula i wrote...so i was asking if it was applicable

Answer 23

it's possible

Answer 24

so is my result right? how can i change it to the other form?

Answer 25

is it too hard? should i move on and come back to it later on?

Answer 26

so let me see you divided by n! and it was put into the laplace... i don't think that's right

Answer 27

i'm not 100% sure but ithink it should be the laplace of the top over n!

Answer 28

which would make sense because

Answer 29

if you have n! as a denominator it'd cancel with the top to get one

Answer 30

yes exactly