Difference between revisions of "Origin of the Laplace Transform" - Rhea

@@ Line 1: / Line 1: @@
-'''Origin of [[Laplace transform|Laplace Transform]]''' [https://kiwi.ecn.purdue.edu/rhea/index.php/User:Green26 (alec green)]
-----
-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 analogue of the discrete power series.
-Below I've merely summarized his explanation.
-----
-(1) '''Power series = discrete summation'''
-We start with this power series:
-<math>A(x) = \sum_{n=0}^{\infty} a(n)x^{n} \ \mid  \ a(n) \in \R \ \ \ \forall \ n \in \N</math>
-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 are elements of' 'the set of natural numbers'.
-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>.
-(2) '''Integral = continuous summation'''
-Now we'll make the following conversions:
-*from discretely defined function <math>a</math> to continuously defined function <math>f</math>
-*from discrete dependent variable <math>n</math> to continuous dependent variable <math>t</math>
-to arrive at:
-<math>F(x)=\int_{0}^{\infty} f(t)x^{t} \ dt  \ \mid  \ f(t) \in \R \ \ \ \forall \ t \in (0,\infty)</math>
-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 likewise expect <math>f(t)</math> to be defined at all those points.
-(3) '''Define variable <math>s</math> in terms of <math>x</math>'''
-By setting <math>x^{t}</math> to the more easily integrable <math>e^{ln(x)t}</math>, and realizing that <math>e^{ln(x)}</math> only depends on <math>x</math>, we obtain:
-<math>F(e^{ln(x)}) = F(x)=\int_{0}^{\infty} f(t)e^{ln(x)t} \ dt</math>
-Finally, noting that the integral is only guaranteed to converge if the exponential is to a negative power (which implies that <math>ln(x)</math> must be <math>< 0</math>), we arbitrarily set <math>s = -ln(x)</math> or <math>-s = ln(x)</math>, which leaves us with:
-<math>F(e^{-s}) = F(s)=\int_{0}^{\infty} f(t)e^{-st} \ dt \ | \ \forall s > 0</math>
-The integral is not necessarily defined if <math>s=0</math> (eg, if <math>f(t)=t</math>).  Also, I'm not sure how to deal with the case when <math>s</math> is undefined (ie <math>x<0</math>), but Prof. Mattuck just asserts that <math>0 < x < 1</math> to avoid this case.

Difference between revisions of "Origin of the Laplace Transform" - Rhea

Latest revision as of 06:40, 5 November 2012

Alumni Liaison