Solve Laplace Transform & MacLaurin Series Problems

[SqueeSpleen]

I have two questions:

I had to find the Laplace transform of:
t \cdot sin(t)
Not by definition, using a table of transforms and the properties.
I did:
sin(t) = i \cdot sinh(it) = i \cdot \frac{e^{t}}{2}-i \cdot \frac{e^{-t}}{2}
Then
t \cdot sin(t) = it \cdot \frac{e^{t}}{2}-it \cdot \frac{e^{-t}}{2}
And
t^{n}e^{-at}=\frac{n!}{(s+a)^{n+1}}
So
\frac{i}{2} t^{1}e^{-it}=\frac{i}{2} \frac{1}{(s+i)^{2}}=\frac{i}{2} \frac{1}{s^{2}+2si-1} and \frac{i}{2} t^{1}e^{it}=\frac{i}{2} \frac{1}{(s-i)^{2}}=\frac{i}{2} \frac{1}{s^{2}-2si-1}
\frac{i}{2} \frac{s^{2}-2si-1-(s^{2}+2si-1)}{(s^2+1)^2}=\frac{2s}{(s^2+1)^2}
I want to know other way of calculate this without using the definition, because I don't know if I'm meant to use complex numbers.

The other problem is:
f(t)=\frac{sin(t)}{t} if t \neq 0
f(t)=1 if t = 0
Find the Mac Laurin serie of the function and check that L\left\{f(t)\right\}=arctan(\frac{1}{s}) s >1
I find the following serie:
\sum_{n=0}^{\infty} (-1)^n (\frac{1}{s})^{2n+1}\frac{1}{2n+1}
But it diverges for s>1, so it's useless to my purpose.
And I had problems trying to compute a Taylor Serie of arctan(\frac{1}{s}), I don't know what point to use.

Sorry if I made some gramatical mistakes, I don't speak English very well.

 

[Ray Vickson]

I have two questions:

I had to find the Laplace transform of:
t \cdot sen(t)
Not by definition, using a table of transforms and the properties.
I did:
sen(t) = i \cdot senh(it) = i \cdot \frac{e^{t}}{2}-i \cdot \frac{e^{-t}}{2}
Then
t \cdot sen(t) = it \cdot \frac{e^{t}}{2}-it \cdot \frac{e^{-t}}{2}
And
t^{n}e^{-at}=\frac{n!}{(s+a)^{n+1}}
So
\frac{i}{2} t^{1}e^{-it}=\frac{i}{2} \frac{1}{(s+i)^{2}}=\frac{i}{2} \frac{1}{s^{2}+2si-1} and \frac{i}{2} t^{1}e^{it}=\frac{i}{2} \frac{1}{(s-i)^{2}}=\frac{i}{2} \frac{1}{s^{2}-2si-1}
\frac{i}{2} \frac{s^{2}-2si-1-(s^{2}+2si-1)}{(s^2+1)^2}=\frac{2s}{(s^2+1)^2}
I want to know other way of calculate this without using the definition, because I don't know if I'm meant to use complex numbers.

The other problem is:
f(t)=\frac{sen(t)}{t} if t \neq 0
f(t)=1 if t = 0
Find the Mac Laurin serie of the function and check that L\left\{f(t)\right\}=arctan(\frac{1}{s}) s >1
I find the following serie:
\sum_{n=0}^{\infty} (-1)^n (\frac{1}{s})^{2n+1}\frac{1}{2n+1}
But it diverges for s>1, so it's useless to my purpose.
And I had problems trying to compute a Taylor Serie of arctan(\frac{1}{s}), I don't know what point to use.

Sorry if I made some gramatical mistakes, I don't speak English very well.

What is the function $sen(t)$? I have never heard of it. Do you mean $\sin(t)$? I have also never heard of $senh(t)$, but I do know about $\sinh(t)$. Is that what you mean?

 

[SqueeSpleen]

Yes, I forget to translate them when I originally created the thread.

 

[SqueeSpleen]

But it diverges for s>1, so it's useless to my purpose.

It isn't, I don't know why I confused the divergence of t with the diverge of s, it has nothing to do with it.
So, I guess all I have to do is to find a good point to calculate the serie.
It's a good idea to calculate the McLaurin serie of arctan(x) then use it to arctan(1/x) changing x by 1/x and restricting the x values?

 

[Ray Vickson]

It isn't, I don't know why I confused the divergence of t with the diverge of s, it has nothing to do with it.
So, I guess all I have to do is to find a good point to calculate the serie.
It's a good idea to calculate the McLaurin serie of arctan(x) then use it to arctan(1/x) changing x by 1/x and restricting the x values?

Your very first step is incorrect: you wrote
\sin(t)=i⋅\sinh(it)=i⋅\frac{e^t}{2}−i⋅\frac{e^{−t}}{2} \; \leftarrow \text{ wrong!}
It should be
\sin(t) = \frac{1}{2i} \left( e^{it} - e^{-it} \right)<br /> = - \frac{i}{2} e^{it} + \frac{i}{2} e^{-it}

 

[vela]

I want to know other way of calculate this without using the definition, because I don't know if I'm meant to use complex numbers.

Your table should list a property for finding the Laplace transform of t f(t) in terms of F(s), the transform of f(t).

 

[vela]

And I had problems trying to compute a Taylor Serie of arctan(\frac{1}{s}), I don't know what point to use.

Use the fact that the derivative of arctan x is $\frac{1}{1+x^2}$. Expand the latter as a series and then integrate it term by term.

 

[SqueeSpleen]

Use the fact that the derivative of arctan x is $\frac{1}{1+x^2}$. Expand the latter as a series and then integrate it term by term.

Thanks.

Your table should list a property for finding the Laplace transform of t f(t) in terms of F(s), the transform of f(t).

I guess it's it.
t^n f(t) \leftrightarrow (-1)^n F^{(n)}(s)
It was named "derivative of a transform", so I didn't noticed it the first time because I didn't pay attention to this propertie (perhaps we didn't see it in the lectures, perhaps we saw it was the day I was sick).
Thank you again, it clarified a lot and it was a lot shorter than my previous approach xD