Question

@sami-21 can i use \[\mathcal L \lbrace t^n f(t) \rbrace = (-1)^n \frac{d^n}{ds^n} \left[f(s)\right]\]in \[\mathcal L \lbrace t^{1/2} \rbrace\]

Answer 1

nooooopes ! it is something different

Answer 2

i know for sure i cant do \[\large \frac{n!}{s^{n+1}}\]

Answer 3

right?

Answer 4

yes you cannot use that as well !!

Answer 5

here gamma function values is required

Answer 6

uhh no we didnt have that

Answer 7

but fine...teach it to me

Answer 8

for such powers of t
you need
\[\Large L(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}\]

Answer 9

@mukushla am i right ?

Answer 10

what do these mean o.O

Answer 11

thats right

Answer 12

ok using n=1/2 we have in the above formula
\[\Large L(t^{1/2})=\frac{\Gamma(1/2+1)}{s^{1/2+1}}=\frac{\Gamma(3/2)}{s^{3/2}}\]

Answer 13

where
\[\Large \Gamma(3/2)=\frac{\sqrt{\pi}}{2}\]

Answer 14

wanna explain what that inverted L is and how to do it in latex?

Answer 15

WOAHHHH`

Answer 16

it is not possible for me to explain here in detail gamma function .but i can tell you that much so that you will easily be able to complete your Laplace course.
for this you have to remember the following
\[\Large \Gamma(1/2)=\sqrt{\pi}\]
just remember the following formula
\[\Large \Gamma(n+1)=n \Gamma(n)\]
let me explain below.

Answer 17

how to do that in latex? \(\gamma\)?

Answer 18

\[\Gamma\]

Answer 19

ahh found it

Answer 20

so if you need gamma(1/2+1) here n=1/2 using the above formula nd known value of gamma(1/2)=sqrt(pi}
\[\Large \Gamma(3/2)=\Gamma(1/2+1)=1/2\Gamma(1/2)=1/2\sqrt{\pi}\]
similarly for Gamma(3/2+1) here n=3/2 using above value of gamma(3/2 ) we have
\[\Large \Gamma(5/2)=\frac{3}{2}*\frac{\sqrt{\pi}}{2}\]
can you now tell me
\[\Large \Gamma(\frac{7}{2})\] ??

Answer 21

@lgbasallote

Answer 22

this gamma thing...when is it used again??

Answer 23

for example
\[\Huge L(t^{-1/2})\]
you will need this gamma thing

Answer 24

\[\Gamma (n)=(n-1)!\]

Answer 25

so basially when the given is \[\large \mathcal L \lbrace t^n \rbrace\]
yes?

Answer 26

@UnkleRhaukus yes you are right , but it will not be helpful if you have n=1/2 ,3/2,5/2...

Answer 27

you could use:
\[L(t^{n-1/2})=((1*3*5*...*(2n-1))\sqrt \pi)/(2^ns^{n+1/2})\]

Answer 28

where n is natural number

Answer 29

let's stick to one methid to avoid confusing teh asker shall we.... :(

Answer 30

so just set n=1, to get laplace transform of t^1/2

Answer 31

then you'd have to use \[\Gamma(z+1)=z\Gamma(z)\]
&
\[\Gamma \left(\frac12 \right)=\sqrt{\pi}\]

Answer 32

so this gamma thing is used whenever the given is L{t^n}???

Answer 33

i do not know my screen automatically becomes blank :(

Answer 34

refresh?

Answer 35

ok
just to write that all again was vanished by blanking of screen:(
here it is
when n=0,1,2,3,4,5...
then
\[\Large \Gamma(n+1)=n!\]
and you r known formula for n=0,1,2,3.. is used
\[\Large L(t^n)=\frac{n!}{s^{n+1}}\]

Answer 36

This is a standard result in any laplace transform table \[ \mathcal L\{t^r\}=\frac{\Gamma(r+1)}{s^{r+1}}\qquad; ~r>-1\]

Answer 37

but if n is n0t 0,1,2,3,4.......then
you will have to use
\[\Large f(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}\]

Answer 38

when r is a natural number n; \(\Gamma(n+1)=n!\)

Answer 39

WOAH...what's happening here??!! o.O

Answer 40

i was still asking a question...

Answer 41

is this gamma thingy done when the given is L{t^n}??

Answer 42

if n is a natural number you might as well go straight to n factorial
but if is is half integer use the gamma function and \(\Gamma(1/2)=\sqrt\pi\)

Answer 43

okay....in sami's earlier solution i was able to understand as far as \[\frac 12 \sqrt \pi\]
after that im lost

Answer 44

what bit dont you get? @lgbasallote ?

Answer 45

similarly for Gamma(3/2+1) here n=3/2 using above value of gamma(3/2 ) we have

\[\Gamma (\frac 52) = \frac 32 \times \frac{\sqrt{\pi}}{2}\]
and so on...

Answer 46

look at sami's earlier post...

Answer 47

ing to that one

Answer 48

i dont get what he did after \[\frac 12 \Gamma \frac 12 \implies \frac{\sqrt{\pi}}{2}\]

Answer 49

oh wait....he was just giving examples....stupid me...

Answer 50

i thought it was related to the problem

Answer 51

so how do i use the fact that \[\Gamma (1/2) = \sqrt \pi\]
and \[\Gamma (n+1) = n\Gamma (n)\]
to solve \[\mathcal L \lbrace t^{1/2} \rbrace\]

Answer 52

Read this
http://en.wikipedia.org/wiki/Gamma_function

Answer 53

\[\Gamma\left(\frac{7}{2}\right)=\frac{5}{2}\Gamma\left(\frac{5}{2}\right)=\frac{5}{2}\cdot\frac{3}{2}\Gamma\left(\frac{3}{2}\right)=\frac{5}{2}\cdot\frac{3}{2}\cdot\frac12\Gamma\left(\frac{1}{2}\right)=\frac{5}{2}\cdot\frac{3}{2}\cdot\frac12\cdot \sqrt{\pi}\]

Answer 54

wait i think i get it now... sami said \[f(t^n) = \frac{\Gamma(n+1)}{s^{n+1}}\]
so \[\mathcal L \lbrace t^{1/2} \rbrace = \frac{\Gamma(\frac 12 + 1)}{s^{1/2 + 1}}\]
right?

Answer 55

is that right @sami-21 ? because if yes...then i think im getting it now

Answer 56

and , simplify ..

Answer 57

nice.. \[\frac{1}{2s} \left(\frac{\pi}{s}\right)^{1/2} \quad s> 0\]
just like what the book says

Answer 58

@lgbasallote it is correct !!

Answer 59

@lgbasallote yes you are a fast learner :)

Answer 60

lol on contraire...it took me an hour to get what you said =)))

Answer 61

wow..are you a mathematician sir? your solutions are somehow...mathematical...idk hoq to explain it

Answer 62

That is the theory in general:
\[
\Gamma(z)=\int_0^\infty e^{-t} t^{z-1} dt\\
\cal L( t^{n})=\int_0^\infty e^{-s t} t^{n} dt\\
u= s t,\, du = s dt, \, dt =\frac {du}{ s}, \, t =\frac u s\\
\cal L( t^{n})=\int_0^\infty e^{-u} (u/s)^{n}\frac {du}{s}=\\
\frac 1 {s^{3/2}}\int_0^\infty e^{-u} u^{n}du=\\
\frac 1 {s^{n+1}}\int_0^\infty e^{-u} u^{n}du=\frac {\Gamma(n+1)}{s^{n+1}}
\]

Answer 63

is this the proof?

Answer 64

That is the theory in general:
\[
\Gamma(z)=\int_0^\infty e^{-t} t^{z-1} dt\\
\cal L( t^{n})=\int_0^\infty e^{-s t} t^{n} dt\\
u= s t,\, du = s dt, \, dt =\frac {du}{ s}, \, t =\frac u s\\
\cal L( t^{n})=\int_0^\infty e^{-u} (u/s)^{n}\frac {du}{s}=\\
\frac 1 {s^{n+1}}\int_0^\infty e^{-u} u^{n}du=\\
\frac 1 {s^{n+1}}\int_0^\infty e^{-u} u^{n}du=\frac {\Gamma(n+1)}{s^{n+1}}
\]

Answer 65

That is the proof in general. In your solution, you can use this fact without proving it.

Answer 66

i just wonder though...how did \[\Gamma (n+1) \] become n! if n is an integer?

Answer 67

The \(\Gamma\) funtcion is a generalization of the factorial function.

Answer 68

Look at the definition of the \(\Gamma \) function. (First line in my post)

Answer 69

ohh...nevermind i found out.. like for example n = 3
\[\Gamma (3+1) = 3\Gamma (3) \implies 3 \Gamma (2+1) \implies 3(2) \Gamma (1)\]

right?

Answer 70

Yes. \(\Gamma(1)=1\)

Answer 71

nice. thanks. that makes sense

Answer 72

In general \( \Gamma(n+1) = n! \) if n is a postive integer.

Answer 73

Try to prove that for any n integer or not
\[
\Gamma(n+1) = n \Gamma(n)
\]

Answer 74

n cannot be less than 0 yes?

Answer 75

Yes.

Answer 76

okay..let me try to prove that...

Answer 77

Yes if n is a negative integer, but n can be
for example -1/3

Answer 78

ohh it can be negative fraction but not negative integer...makes sense

Answer 79

Since I have to go for my hike
\[
\Gamma(z)=\int_0^\infty e^{-t} t^{z-1} dt\\
\Gamma(n+1)=\int_0^\infty e^{-t} t^{n} dt\\
u=t^n,\, dv = e^{-t} dt\\
du= nt^{n-1} dt, \, v=-e^{-t}\\
\Gamma(n+1)=\int_0^\infty e^{-t} t^{n} dt=\\
\left[- t^n e^{-t} \right]_{t=0}^\infty+ n \int_0^\infty e^{-t} t^{n-1} dt=0+ n\Gamma(n)


\]

I used integration by parts above.

Answer 80

@eliassaab thanks .you really made it clear to him.
@lgbasallote hope you got it now ! :)
Best of Luck with Laplace :)

Answer 81

yw