Question

One question about the domain of a Laplace of a Particular function.

Answer 1

I have started to read something about laplaces on my own:)

Answer 2

\(\large\displaystyle {\cal L}\{\sinh(\alpha t)\}(s)=\lim_{N\to\infty}\int\limits_{0}^{N} e^{-st}\sinh(\alpha t)dt\)

\(\large\displaystyle {\cal L}\{\sinh(\alpha t)\}(s)=\frac{1}{2}\lim_{N\to\infty}\int\limits_{0}^{N} (e^{-st}e^{\alpha t}-e^{-st}e^{-\alpha t})dt\)

\(\large \displaystyle {\cal L}\{\sinh(\alpha t)\}(s)=\frac{1}{2}\lim_{N\to\infty}\int\limits_{0}^{N} (e^{(\alpha-s)t}-e^{-(\alpha+s) t})dt\)

Answer 3

\(\large \displaystyle {\cal L}\{\sinh(\alpha t)\}(s)=\frac{1}{2}\lim_{N\to\infty} \left[\frac{e^{(\alpha-s)t}}{\alpha-s}+\frac{e^{-(\alpha+s)t}}{\alpha+s}\right]_{t=0}^{t=N} \)

I will paste my question refering to here ****

Answer 4

\(\large \displaystyle =\frac{1}{2}\lim_{N\to\infty} \left[\frac{e^{(\alpha-s)t}}{\alpha-s}+\frac{e^{-(\alpha+s)t}}{\alpha+s}\right]_{t=0}^{t=N} \)

\(\large \displaystyle =-\frac{1}{2} \left[\frac{1}{\alpha-s}+\frac{1}{\alpha+s}\right] \)

\(\large \displaystyle =\frac{1}{2} \left[\frac{1}{s-\alpha}-\frac{1}{s+\alpha}\right] \)

\(\large \displaystyle =\frac{1}{2} \left[\frac{(s+\alpha)-(s-\alpha)}{s^2-\alpha^2}\right] \)

\(\large \displaystyle {\cal L}\{\sinh(\alpha t\}(s)= \frac{\alpha}{s^2-\alpha^2} \)

Just to finish the transform

Answer 5

Now, the question ...

Answer 6

For the limit to converge, all coefficients of t must be negative ((and not zero, because of the denominators that resulted after integration)), and that means that

\(\large \displaystyle \alpha-s<0 \) ---> \(\large \displaystyle \alpha<s \)
\(\large \displaystyle -(\alpha+s) <0 \) ---> \(\large \displaystyle s>-\alpha \)

And so it would be that:
\(\large \displaystyle s>| \alpha| \)

Answer 7

And everywhere I look I keep seeing just \(\large \displaystyle s> \alpha \).

Answer 8

(by the way, same question comes up for \(\large \displaystyle \cosh(\alpha t)\).... just is for sinh )

Answer 9

Ahh I am going to guess that \(\alpha\) is always just assumed to be greater than 0 since it being negative ultimately results in the same answer, like you have there \(|\alpha| <s\) .

For instance, if you picked an alpha less than 0 you'd be able to pick \(\alpha = - \beta\) and make it positive,

\[\sinh(\alpha t) = \sinh(-\beta t) = -\sinh (\beta t)\]

Probably not a very satisfying answer, sorry haha but at least you seem to be doing a pretty thorough job of self studying which is impressive. :D

Answer 10

sinh(X)= sin(ix)?

Answer 11

Well, I thought that hyperbolic easier.

Answer 12

Generically, you are just having e^(at+b)

Answer 13

it is im tryna determine the hyperbolix equation but i wanna do it myself thats how u get sinh from sin right

Answer 14

It might not work the imaginary way.

Answer 15

Hey stop trying to hijack idku 's question lol

Answer 16

lol sry :D

Answer 17

I won't derive Euler's Formula through Taylor series for e^x, it's pretty obvious without me.

\(\large\displaystyle {\cal L}\{\sinh(\alpha t)\}(s)=\lim_{N\to\infty}\int\limits_{0}^{N} e^{-st}i\sin(\alpha t)dt\)

the i is on the outside actually.

Answer 18

yes, I have seen that previously, while doing Dirichlet's test.

Answer 19

(for series conv)

Answer 20

But, then you need integration by parts twice to solve for this integral. Not the best way:)

Answer 21

In any case, Kainui, I guess, yes, \(\alpha>0\) is the presumption.
Otherwise, I don't see a lot of sense....

Answer 22

There's actually a third way of solving this integral without going to Euler's identity or doing integration by parts twice. You can invert a 2x2 matrix haha :P

Answer 23

Matrix? I haven't really done anything in Linear Algebra. I read about transformation, linear dependence, of course Gaussian elimination, "does S span R^n" and all introductory stuff...

Answer 24

Can you show me?

Answer 25

(what do I have to know to view this? I might not be eligible knowledge-wise)

Answer 26

Well if you can invert a 2x2 matrix, that's about all you need to know, it's sorta a strange trick if you're not used to it but once you know it it gets comfortable fast and you can use it to solve some differential equations as well.

Answer 27

Oh, take the inverse of.
Yes, I do.

Answer 28

I can take the inverse of any square-matrix!

Answer 29

Differentiation is a linear operator and the derivatives of \(e^{ax}\cos (bx)\) and \(e^{ax}\sin(bx)\) is closed, it's a 2D space if you wish to think of it that way, so you can use these as your basis vectors for your space, so the column vector:

\[ [ 3,-7]^\top \]
represents the function:

\[3 e^{ax} \cos(bx) -7 e^{ax}\sin(bx)\]

So here's where the fun comes in.

Answer 30

sorry to interrupt, but

we got s > a and s >-a

then how did that become s > |a| ????????????


for s > |a|, we must have s>a and s< -a, right?


since a is assumed to be positive,
s> a will eventually cover s> -a too.

Answer 31

If you know how the basis vectors transform, you can find the derivative matrix, since that's the main power of linear operators. Knowing how they change the basis vectors, you can change any linear combination with the transformation.

Once you know that, then you invert that derivative matrix, and that's your integral matrix, since differentiation is the inverse of integration. So multiply this inverted 2x2 integration matrix by the vector representing the function you wanna integrate and that's it... Hahaha maybe easier said than done?

Answer 32

Hartnn.
s>a
s>-a

If you just say s>a, then you will get a diverging Laplace for all a<0.
If you just say s<a, then you will get a diverging Laplace for all a>0.

So, from those two equations, the domain of s becomes s>|a|.

And, that was before we concluded that assumingly s>a.
(And I think they should write s>|a|, without that assumption, but what do I know?)

Answer 33

let me fix it.

Answer 34

assuming a>0.

Answer 35

I originally thought what @hartnn was thinking too and then at the end of writing my original response I had to delete it cause I saw what @idku was saying

Answer 36

what even? :O

where did s<a come from??? why would we say that?
if we had s<-a and s>a , only then s > |a| holds true!

go to basics, draw a number line if you need to!

s>a , s>-a implies s should be > a . thats what i know for sure...

I STILL DON'T SEE IT!! omg

Answer 37

Look at the limit, where I said "paste the question here" ....
The coefficients of t, all, must be negative.

Answer 38

(otherwise your laplace won't exist)

Answer 39

a-s < 0 --> a <s , s>a
-(a+s) < 0 --> a+s > 0 --> s > -a

s>a AND s>-a ---> s >a
>.<

Answer 40

Yes, that is assuming that a>0, but my question was before we made this assumption.

Answer 41

what will change in what i replied last if a was negative?

still s>a would hold good

Answer 42

a is negative

a-s < 0 --> a <s , s>a
-(a+s) < 0 --> a+s > 0 --> s > -a

s>a AND s>-a ---> s >a

Answer 43

-(a+s)

say, for example,
s=7
a=-8
(notice s>a)

-(-8+7)=1>0
limit diverges.

Answer 44

sorry for so much trouble, i just don't see it :P

Answer 45

lim N-->∞, e^(-(a+s)N)
converges only when -(a+s)<0.

lim N-->∞, e^(a-s)N)
converges only when a-s<0.

Answer 46

So, for both limits to converge, you need s>|a|, if the assumption a>0 isn't made.

Answer 47

@hartnn if you just say \(\alpha <s\) then this doesn't imply \(-\alpha < s\) since \(\alpha\) could be negative.

Answer 48

Specifically (lol) this could be true: \(\alpha < - \alpha\)

Answer 49

alright so for a negative,
s>a AND s>-a ---> s >a doesn't hold true...

Answer 50

So that's why I basically just said in my response that they probably just assume \(\alpha>0\) That's cause now YOU are saying \(|\alpha| < s\) because of your AND condition.

Answer 51

Why do I want to avoid this assumption a>0?

Well, sinh(at) is a generic hyperbolic sine functions
(pretty much, except that it lacks a generic coefficient/1st-deg-poly in front)

why do I want to neglect all cases when a<0?
Just because some "professors" are lazy to put |..| around a
(so that s>|a|, just as it should for limits to converge, for all a.)

Answer 52

This just went backwards hahaha

Answer 53

oh, 0-deg-poly (not 1st)

Answer 54

In any case, we have got off track:)

Answer 55

We kind of resolved the question already.

Answer 56

i need to referesh

Answer 57

you're right,
s > |a|

where did you see s >a only?

wiki says s>|a|
https://en.wikipedia.org/wiki/Laplace_transform

Answer 58

Oh, I have looked in the printed sources.

Answer 59

Another proof why they suck:) xD

Answer 60

So, Kainui, you want to set a transformation:

\(\large \displaystyle T(z_1, z_2) = (z_1e^{ax}\cos(bt)~,~z_1e^{ax}\sin(bt))\) ?

Answer 61

(using z for variables)

Answer 62

its always good to cross-verify with reliable sources like wiki (though sometimes wiki is also not that reliable)

Answer 63

Yeah, some of my instructors told me that they purposely posted fallacious info in wiki, and waited how long it took them to figure out and fix it. I will tell you, it wss a long long long while:)

Answer 64

Wait, kainui, so the matrix would be?

\(\large \displaystyle e^{ax}\cos(bx)~~~~~~~~~~~~~~~~~~~~e^{ax}\sin(bx)\)

\(\large \displaystyle \frac{d}{dx}e^{ax}\cos(bx)~~~~~~~~~~~~~\frac{d}{dx}e^{ax}\sin(bx)\)

Answer 65

The only thing I have done with matrix in calc/DE application so far, is Cramer's Rule for Variation of Parameters v_1', and v_2'.

Answer 66

Yeah so we have some derivative matrix with arbitrarily filled entries to solve for:

\[D=\begin{pmatrix} w & x \\ y & z \end{pmatrix}\]

We know the derivative of: \[D (e^{ax} \cos(bx)) = ae^{ax} \cos(bx)-be^{ax} \sin(bx)\]

So we can write this in terms of our vectors as:

\[\begin{pmatrix} w & x \\ y & z \end{pmatrix} \begin{pmatrix}1 \\ 0 \end{pmatrix}=\begin{pmatrix}a \\ -b \end{pmatrix} \]

So that's the first column of our matrix solved for. Try the next column or ask questions if you're lost.

Answer 67

Yeah you're on the exact right path I already had this typed up halfway and didn't see you had started so I kinda reexplained some stuff you're already figuring out haha

Answer 68

No, I just guessed that, because that is a similar set up for det(W) in Var of Par.

Answer 69

\(\large \displaystyle e^{ax}\cos(bx)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~e^{ax}\sin(bx)\)

\(\large \displaystyle ae^{ax}\cos(bx)-be^{ax}\sin(bx)~~~~~~~ae^{ax}\sin(bx)+be^{ax}\cos(bx)\)

Answer 70

Like that?

Answer 71

Yeah, like that. This is our linear transformation of the derivative from our basis onto itself again.

Answer 72

And what do I do to this matrix now?

Answer 73

I am just showing a way to solve for the entries of this matrix. Once you have it, you invert it to get the 'integration' matrix since you can invert a linear operator like this. Do you have the derivative matrix yet?

Answer 74

I'll just post it again, this is the steps I'm using to find the first column of our derivative matrix.
---

Yeah so we have some derivative matrix with arbitrarily filled entries to solve for:

\[D=\begin{pmatrix} w & x \\ y & z \end{pmatrix}\]

We know the derivative of: \[D (e^{ax} \cos(bx)) = ae^{ax} \cos(bx)-be^{ax} \sin(bx)\]

So we can write this in terms of our vectors as:

\[\begin{pmatrix} w & x \\ y & z \end{pmatrix} \begin{pmatrix}1 \\ 0 \end{pmatrix}=\begin{pmatrix}a \\ -b \end{pmatrix} \]

So that's the first column of our matrix solved for. Try the next column or ask questions if you're lost.

Answer 75

Alright, I will look at this.
I think I need to study more transformations before doing this particular example.
Also, its 3am here, kind of tired.

Kainui, thank you \(\displaystyle \lim_{x\to\infty} !^x\)

Answer 76

Have a good morning!

Answer 77

gtg

Answer 78

Thanks you too lol

Answer 79

:)