Find power series if you know its laplace transformation

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Homework Help Overview

The discussion revolves around determining a power series given its Laplace transformation, specifically the transformation of the form ##-s^{-1}e^{-s^{-1}}##. Participants are exploring the relationship between Laplace transforms and power series, particularly in the context of inverse transformations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenges in manipulating the Laplace transformation and consider expanding the exponential function as a series. There are attempts to invert the transformation term by term and questions about correctly including terms in the resulting series.

Discussion Status

The discussion is active, with participants providing guidance on the correct expansion of the exponential function and engaging in back-and-forth clarification. Some participants have made progress in deriving terms of the series, while others are questioning the accuracy of their expansions.

Contextual Notes

There are indications of confusion regarding the proper form of the series expansion and the treatment of certain terms, such as the inclusion of constants in the series. Participants are also navigating the constraints of homework rules that may limit the sharing of complete solutions.

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Homework Statement


a) Determine power series ##\sum _{n=0}^{\infty }a_nt^n## if you know that its laplace transformation is ##-s^{-1}e^{-s^{-1}}##
b) Determine function ##g## that this power series will be equal to ##J_0(g(t))##

Homework Equations


The Attempt at a Solution



Hmmm, I am having some troubles with this laplace transformation in part a).

Well, I know that Laplace transformation of Heaviside function ##H_C(t)## is ##\frac{1}{s}e^{-Cs}##

Knowing this I get almost the same as the problem says: ##-H_1## ---> ##-\frac{1}{s}e^{-s}##. But I have absolutely NO idea what to do to get ##\frac{1}{s}## in the exponent function. If I just power both sides of the equation everything else collapses...

So how can I deal with this?
 
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Expand the exponential as a series then invert the transform term by term.
 
Hmmmm...

##-\frac{1}{s}e^{-1/s}=-\frac{1}{s}(1-\frac{1}{s}+\frac{1}{2}(\frac{1}{s})^2-\frac{1}{3}(\frac{1}{s})^3+\frac{1}{4}(\frac{1}{s})^4- ...)=-\frac{1}{s}+(\frac{1}{s})^2-\frac{1}{2}(\frac{1}{s})^3+\frac{1}{3}(\frac{1}{s})^4-\frac{1}{4}(\frac{1}{s})^5+ ...##

Using inverse Laplace transformation gives me:

##-1+t-\frac{1}{4}t^2+\frac{1}{18}t^3-\frac{1}{96}t^4+\frac{1}{600}t^5-\frac{1}{4320}t^6+...##

BUT I can't fine a way to include that -1 into series:

##-1+\sum _{k=1}^{\infty }\frac{(-1)^{k+1}}{(k+1)!-k!}t^k##
 
You didn't expand the exponential function correctly. It should be n! in the denominator, not n.
 
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uh, jup, you are right.

The result is ##\sum _{k=0}^{\infty }\frac{(-1)^{k+1}}{(k!)^2}t^k##.

Thank you, vela!
 

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