Power series and Laplace transform

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SUMMARY

The discussion centers on the relationship between discrete power series and continuous Laplace transforms. It establishes that while both approaches can converge to similar functions under certain conditions, they are not equivalent. Specifically, the discrete power series converges to 1/(1-x) for -10. The assertion that substituting s=-ln(x) yields equivalent results is proven incorrect, as the processes of summation and integration cannot be interchanged without proper justification.

PREREQUISITES
  • Understanding of discrete power series and their convergence properties
  • Familiarity with the Laplace transform and its applications
  • Knowledge of the Dirac delta function and its role in continuous functions
  • Basic calculus, particularly integration and summation techniques
NEXT STEPS
  • Study the properties of the Dirac delta function and its applications in transforms
  • Learn about the conditions under which series and integrals can be interchanged
  • Explore advanced topics in Laplace transforms, including inverse transforms
  • Investigate convergence criteria for power series in different contexts
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Mathematicians, engineers, and students studying advanced calculus, particularly those interested in the applications of transforms in solving differential equations and analyzing systems.

ricard.py
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As it can be read here, http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series
the Laplace transform is a continuous analog of a power series in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by exp(-s).

Therefore, computing a discrete power series or a continuous laplace transform should converge to the same function, is it right?

Let's apply it for the simplest case: a(x)=1

  • For the discrete power series it converges to 1/1-x (provided that -1<x<1)

  • For the continuous power series it converges to 1/s (provided that s>0)
Now, this two should be equivalent right? If you substitute s=-ln(x) you get
-1/ln(x), which is not the same as 1/1-x.

What I am doing wrong?
 
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ricard.py said:
As it can be read here, http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series
the Laplace transform is a continuous analog of a power series in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by exp(-s).

Therefore, computing a discrete power series or a continuous laplace transform should converge to the same function, is it right?

Let's apply it for the simplest case: a(x)=1

  • For the discrete power series it converges to 1/1-x (provided that -1<x<1)

  • For the continuous power series it converges to 1/s (provided that s>0)
Now, this two should be equivalent right? If you substitute s=-ln(x) you get
-1/ln(x), which is not the same as 1/1-x.

What I am doing wrong?

"Replace the discrete parameter n by the continuous parameter t" is not that straightforward.

Doing that literally would require that
<br /> \int_0^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty f(n)e^{-ns}<br /> which, as your example shows, is in general false.
What is true is that if you define f(t) = \sum_{n=0}^\infty a_n\delta(t - n) where \delta is the Dirac delta distribution then
<br /> \int_{0^{-}}^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty a_n e^{-ns}.<br /> Observe that here f(t) = 0 for all non-integer t and that f(t) is not technically defined for integer t; it is not the case that f(n) = a_n.
 
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In general, your statement "computing a discrete power series or a continuous laplace transform should converge to the same function" is incorrect. Interchanging two limit process does NOT always give the same result.
 

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