Solve Laplace Transform & MacLaurin Series Problems

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Homework Help Overview

The discussion revolves around finding the Laplace transform of the function t * sin(t) and the Maclaurin series for the function f(t) = sin(t)/t, with f(0) = 1. Participants explore various methods and properties related to these transforms and series expansions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using tables of transforms and properties rather than definitions for the Laplace transform. There are attempts to express sin(t) in terms of complex exponentials and to derive the transform for t * sin(t). Questions arise about the validity of using complex numbers in this context. Additionally, there are inquiries about the Maclaurin series for f(t) and the convergence of the series.

Discussion Status

Some participants have provided guidance on using properties of Laplace transforms and the derivative of a transform. There is acknowledgment of confusion regarding the divergence of series and the need to find appropriate points for series expansion. Multiple interpretations of the series and transform properties are being explored.

Contextual Notes

Participants note potential gaps in their understanding of series convergence and the application of complex numbers in Laplace transforms. There is also mention of possible missed lectures that may have contributed to misunderstandings of the properties discussed.

SqueeSpleen
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I have two questions:

I had to find the Laplace transform of:
[itex]t \cdot sin(t)[/itex]
Not by definition, using a table of transforms and the properties.
I did:
[itex]sin(t) = i \cdot sinh(it) = i \cdot \frac{e^{t}}{2}-i \cdot \frac{e^{-t}}{2}[/itex]
Then
[itex]t \cdot sin(t) = it \cdot \frac{e^{t}}{2}-it \cdot \frac{e^{-t}}{2}[/itex]
And
[itex]t^{n}e^{-at}=\frac{n!}{(s+a)^{n+1}}[/itex]
So
[itex]\frac{i}{2} t^{1}e^{-it}=\frac{i}{2} \frac{1}{(s+i)^{2}}=\frac{i}{2} \frac{1}{s^{2}+2si-1}[/itex] and [itex]\frac{i}{2} t^{1}e^{it}=\frac{i}{2} \frac{1}{(s-i)^{2}}=\frac{i}{2} \frac{1}{s^{2}-2si-1}[/itex]
[itex]\frac{i}{2} \frac{s^{2}-2si-1-(s^{2}+2si-1)}{(s^2+1)^2}=\frac{2s}{(s^2+1)^2}[/itex]
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:
[itex]f(t)=\frac{sin(t)}{t}[/itex] if [itex]t \neq 0[/itex]
[itex]f(t)=1[/itex] if [itex]t = 0[/itex]
Find the Mac Laurin serie of the function and check that [itex]L\left\{f(t)\right\}=arctan(\frac{1}{s})[/itex] s >1
I find the following serie:
[itex]\sum_{n=0}^{\infty} (-1)^n (\frac{1}{s})^{2n+1}\frac{1}{2n+1}[/itex]
But it diverges for s>1, so it's useless to my purpose.
And I had problems trying to compute a Taylor Serie of [itex]arctan(\frac{1}{s})[/itex], I don't know what point to use.

Sorry if I made some gramatical mistakes, I don't speak English very well.
 
Last edited:
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SqueeSpleen said:
I have two questions:

I had to find the Laplace transform of:
[itex]t \cdot sen(t)[/itex]
Not by definition, using a table of transforms and the properties.
I did:
[itex]sen(t) = i \cdot senh(it) = i \cdot \frac{e^{t}}{2}-i \cdot \frac{e^{-t}}{2}[/itex]
Then
[itex]t \cdot sen(t) = it \cdot \frac{e^{t}}{2}-it \cdot \frac{e^{-t}}{2}[/itex]
And
[itex]t^{n}e^{-at}=\frac{n!}{(s+a)^{n+1}}[/itex]
So
[itex]\frac{i}{2} t^{1}e^{-it}=\frac{i}{2} \frac{1}{(s+i)^{2}}=\frac{i}{2} \frac{1}{s^{2}+2si-1}[/itex] and [itex]\frac{i}{2} t^{1}e^{it}=\frac{i}{2} \frac{1}{(s-i)^{2}}=\frac{i}{2} \frac{1}{s^{2}-2si-1}[/itex]
[itex]\frac{i}{2} \frac{s^{2}-2si-1-(s^{2}+2si-1)}{(s^2+1)^2}=\frac{2s}{(s^2+1)^2}[/itex]
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:
[itex]f(t)=\frac{sen(t)}{t}[/itex] if [itex]t \neq 0[/itex]
[itex]f(t)=1[/itex] if [itex]t = 0[/itex]
Find the Mac Laurin serie of the function and check that [itex]L\left\{f(t)\right\}=arctan(\frac{1}{s})[/itex] s >1
I find the following serie:
[itex]\sum_{n=0}^{\infty} (-1)^n (\frac{1}{s})^{2n+1}\frac{1}{2n+1}[/itex]
But it diverges for s>1, so it's useless to my purpose.
And I had problems trying to compute a Taylor Serie of [itex]arctan(\frac{1}{s})[/itex], 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?
 
Yes, I forget to translate them when I originally created the thread.
 
SqueeSpleen said:
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?
 
Last edited:
SqueeSpleen said:
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
[tex]\sin(t)=i⋅\sinh(it)=i⋅\frac{e^t}{2}−i⋅\frac{e^{−t}}{2} \; \leftarrow \text{ wrong!}[/tex]
It should be
[tex]\sin(t) = \frac{1}{2i} \left( e^{it} - e^{-it} \right)<br /> = - \frac{i}{2} e^{it} + \frac{i}{2} e^{-it}[/tex]
 
SqueeSpleen said:
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).
 
SqueeSpleen said:
And I had problems trying to compute a Taylor Serie of [itex]arctan(\frac{1}{s})[/itex], 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.
 
vela said:
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.

vela said:
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.
[itex]t^n f(t) \leftrightarrow (-1)^n F^{(n)}(s)[/itex]
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
 
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