Find power series if you know its laplace transformation

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The discussion focuses on determining a power series from its Laplace transformation, specifically the transformation given as -s^{-1}e^{-s^{-1}}. The initial attempt involved using the Laplace transformation of the Heaviside function but faced challenges in manipulating the exponent. The solution progressed by expanding the exponential as a series and applying the inverse Laplace transformation, leading to a series representation. The final result for the power series is expressed as the sum of terms involving factorials, specifically summing (-1)^{k+1}/(k!)^2. The conversation concludes with a correction on the series expansion, emphasizing the importance of accurate factorial representation.
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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