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<math>F(e^{ln(x)}) = F(x)=\int_{0}^{\infty} f(t)e^{ln(x)t} \ dt</math> | <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 | + | 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> | <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. | 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. | ||
Revision as of 09:06, 3 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 analogue of the discrete power 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 are elements 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 discretely defined function $ a $ to continuously defined 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 likewise expect $ f(t) $ to be defined at all those points.
(3) Define variable $ s $ in terms of $ x $
By setting $ x^{t} $ to the more easily integrable $ e^{ln(x)t} $, and realizing that $ e^{ln(x)} $ only depends on $ x $, we obtain:
$ F(e^{ln(x)}) = F(x)=\int_{0}^{\infty} f(t)e^{ln(x)t} \ dt $
Finally, noting that the integral is only guaranteed to converge if the exponential is to a negative power (which implies that $ ln(x) $ must be $ < 0 $), we arbitrarily set $ s = -ln(x) $ or $ -s = ln(x) $, which leaves us with:
$ F(e^{-s}) = F(s)=\int_{0}^{\infty} f(t)e^{-st} \ dt \ | \ \forall s > 0 $
The integral is not necessarily defined if $ s=0 $ (eg, if $ f(t)=t $). Also, I'm not sure how to deal with the case when $ s $ is undefined (ie $ x<0 $), but Prof. Mattuck just asserts that $ 0 < x < 1 $ to avoid this case.
