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Origin of Laplace Transform [https://kiwi.ecn.purdue.edu/rhea/index.php/User:Green26 (alec green)]  
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'''Origin of [[Laplace transform|Laplace Transform]]''' [https://kiwi.ecn.purdue.edu/rhea/index.php/User:Green26 (alec green)]  
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In the first 15 minutes of this [http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/ MIT lecture], Arthur Mattuck delivers a clear illustration of what the Laplace transform really is: a continuous analogy of the discrete power series (specifically, the Maclaurin Series).
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Below I've merely summarized his explanation.
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----
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(1) '''Power series = discrete summation'''
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We start with this power series:
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<math>A(x) = \sum_{n=0}^{\infty} a(n)x^{n} \ \mid  \ a(n) \in \R \ \ \forall \ n \in \N</math>
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In case you're not familiar with all the above notation, here's the explicit translation starting starting after the summation term, where each quoted term corresponds to each symbol:  'such that' a(n) 'is an element of' 'the set of real numbers' 'for all' n 'which is an element of' 'the set of natural numbers'.
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Note that <math>a(n)</math> is a function here, and just defines the coefficient of each polynomial term in the power series, since a power series is <math>  = a(0) + a(1)x + a(2)x^{2} + ... + a(n)x^{n} + ... </math>.  However, because the power series is a discrete summation, <math>a(n)</math> is only guaranteed to be defined if <math>n</math> is a natural number (non-negative integer), as indicated above.  So for example, <math>a(23)</math> is defined, but not necessarily <math>a(-2)</math>, <math>a(.5)</math>, or <math>a(10.001)</math>.
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(2) '''Integral = continuous summation'''
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Now we'll make the following conversions:
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*from discrete function <math>a</math> to continuous function <math>f</math>
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*from discrete dependent variable <math>n</math> to continuous dependent variable <math>t</math>
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to arrive at:
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<math>F(x)=\int_{0}^{\infty} f(t)x^{t} \ dt  \ \mid  \ f(t) \in \R \ \ \forall \ t \in (0,\infty)</math>
  
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The only difference now is that we sum the contributions of <math>f(t)x^{t}</math> for all ''real numbers'' instead of all ''natural numbers'' from 0 to infiniti, and we can expect <math>f(t)</math> to be defined at all those points.
  
  
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(3) '''Define variable <math>s</math> in terms of <math>x</math>'''
  
[[ Laplace transform|Back to Laplace transform]]
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<math>F(s)=\int_{0}^{\infty} f(t)e^{-st} \ dt</math>

Revision as of 12:31, 2 November 2012

Origin of Laplace Transform (alec green)


In the first 15 minutes of this MIT lecture, Arthur Mattuck delivers a clear illustration of what the Laplace transform really is: a continuous analogy of the discrete power series (specifically, the Maclaurin Series).

Below I've merely summarized his explanation.


(1) Power series = discrete summation

We start with this power series:

$ A(x) = \sum_{n=0}^{\infty} a(n)x^{n} \ \mid \ a(n) \in \R \ \ \forall \ n \in \N $

In case you're not familiar with all the above notation, here's the explicit translation starting starting after the summation term, where each quoted term corresponds to each symbol: 'such that' a(n) 'is an element of' 'the set of real numbers' 'for all' n 'which is an element of' 'the set of natural numbers'.

Note that $ a(n) $ is a function here, and just defines the coefficient of each polynomial term in the power series, since a power series is $ = a(0) + a(1)x + a(2)x^{2} + ... + a(n)x^{n} + ... $. However, because the power series is a discrete summation, $ a(n) $ is only guaranteed to be defined if $ n $ is a natural number (non-negative integer), as indicated above. So for example, $ a(23) $ is defined, but not necessarily $ a(-2) $, $ a(.5) $, or $ a(10.001) $.


(2) Integral = continuous summation

Now we'll make the following conversions:

  • from discrete function $ a $ to continuous function $ f $
  • from discrete dependent variable $ n $ to continuous dependent variable $ t $

to arrive at:

$ F(x)=\int_{0}^{\infty} f(t)x^{t} \ dt \ \mid \ f(t) \in \R \ \ \forall \ t \in (0,\infty) $

The only difference now is that we sum the contributions of $ f(t)x^{t} $ for all real numbers instead of all natural numbers from 0 to infiniti, and we can expect $ f(t) $ to be defined at all those points.


(3) Define variable $ s $ in terms of $ x $

$ F(s)=\int_{0}^{\infty} f(t)e^{-st} \ dt $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva