paradoxymoron said:
I believe your link is broken. However, I believe it was the one I referred to because I remember seeing it on the MIT OCW site.
Works for me <puzzle>.
I don't remember using a change of variables when taking the limit of a Reimann Sum as the length of the partitions approach zero in order to define the definite integral.
The reiman sums you started with were very simple - but the process holds.
I'm not sure I understand what you meant by using an integral as a "shortcut" method for a discrete sum. It's not like you can interchange sums and integrals freely, even if they converge. However, I might have misunderstood.
Just as a sum can approximate an integral, so an integral can approximate a sum.
The relation is exact in the appropriate limit.
See also:
http://johnmayhk.wordpress.com/2007/09/24/alpm-sum-an-infinite-series-by-definite-integrals/
... an example of using an integral to find the sum of an infinite series.
So is the Laplace Transform something that doesn't have a rigorous derivation? Did he just want to "transform" the discrete sum into an infinite one? And for some reason it just became useful in solving differential equations? What does it mean, though? What relationship does it have with the function it operates on?
The relationship of the transform to the function is in the definition - that's it. Like I said - compare with the Fourier transform. It's basically changing domains - but that needn't mean anything.
The Laplace transform is continuous into continuous - he was looking for a way to understand more difficult DEs.
An example to start you off...
If you have y'=f(x) but the f(x) is annoying, then writing ##f(x)=\sum a_nx^n## means that ##y=\sum a_nx^{n+1}/(n+1)## ... and then you can set out to look for the situations where that's an improvement.
Right away you can see that this depends on the way those ##a_n##'s work out ... which motivates treating it as a function.
So far, the only use I see the Laplace Transform has is to make things simpler; to turn a somewhat difficult differential equation into an algebraic equation. Does this mean that the Laplace Transform is just a way of turning the differential equation into an equation without meaning (but not useless) and then revert back using the inverse transform?
That is pretty much the only use I know for it.
Most of the maths you've learned - think: long division, multiplication by columns etc - is basically a way to turn a hard calculation into one that is somewhat removed but easier to do.
However, the Laplace transform is closely related to the Fourier transform's ability to change to a conjugate domain ... i.e. time into frequency, position into momentum, etc. Ultimately whether there is meaning or not depends on the context ... same with any maths.
Also see:
https://www.physicsforums.com/showthread.php?t=155709
http://math.stackexchange.com/questions/428408/physical-interpretation-of-laplace-transforms
We did not touch on Fourier Transforms in my intro to DE class. Should we have gone over it? I know a tiny scope about the general concept, but that's about it.
There is no special reason you'd meet Fourier transforms ahead of Laplace transforms. I met Fourier first in physics class and ran into Laplace in maths class studying DEs.