| Posted on Saturday, 31 May, 2003 - 08:27 pm: |
Questions in review: 1> What is the path of integration for integrals in the real plane, i.e. integrals like òab f(x) dx? Is it the real axis? If yes then is the path unique and if so why cannot I define a different path? 2> The geometric interpretation of òab f(x) dx is area under the curve (as is logically implied by the summing of areas of infinite number of rectangles). What is the geometrical interpretation of òz1z2 f(z)dz? Is it the same? If yes then how (since the path from z1 to z2 is arbitrary choice) and if no then why not? 3> Why is it that the value of òz1z2 f(z)dz is independent of path if f(z) is analytic? (I am quite aware of the mathematical proof but I would like it if it were interpreted in geometrical sense.)
love arun
P.S-> i know that these questions seem totally hidious but then me ole' brain successfully disfunctions most of the times.
| Posted on Saturday, 31 May, 2003 - 11:16 pm: |
1. I would have thought that for real integrals, either you must accept that f(x) is only defined on the real line (and so your path of integration must stay on the real line, since this is the only place where the function exists), or you invent a new analytic function g(z) such that whenever Im(z)=0, g(z)=f(z). Then you can choose your path of integration freely, and by the uniqueness property you mentioned in (3), the answer you get is the same as it would be if you went along the real line.
2. I can't offer a geometric interpretation of the integral, but I tend to think of it in a physical way, as the potential difference between the start and finish points in a field.
3. This then gives a nice way of thinking of the uniqueness of the value of the integral; Take a conservative field (e.g. gravitational or electrostatic, N.B. NOT magnetic or electromagnetic). It can be represented as a scalar at each point in space defined with an arbitrary additive constant, i.e. it has a scalar potential. (The magnetic field has a vector potential and the E-M field a four-vector potential.) This potential is obtained by the path integral of the field itself (a vector), and the uniqueness of the integral (i.e. having a unique scalar potential at every point) means that you can't do work by going around a closed loop. This isn't quite the same as your problem, since the physics analogy works for integrating vectors, but the idea is simiar to complex numbers - real and imaginary parts are like orthogonal components of a vector.
I hope these points, even if not all directly related to your questions, help to put the abstract results into context.
Alex.
| Posted on Saturday, 31 May, 2003 - 11:19 pm: |
I'm not sure if there is a geometrical interpretation of complex integrals in the same way as there is for real ones. Do you know about Cauchy's formula for complex functions with singularities? If so, there is something like an interpretation of a circuit integral (starting point the same as the ending point). Basically, the integral round the circuit counts the flow in or out of the area bounded by the circuit where the singularities can be sources or sinks. Cauchy's formula says that if a function at a point z0 is of the form g(z)+a-1/(z-z0)+a-2/(z-z0)2+ ¼ where g(z) is analytic at z0 then the integral around a small circle around z0 is 2pi a-1. a-1 is called the residue of the pole at z0. So, for example, the integral of 1/z around the circle of radius 1 is 2pi. (This is related to logarithms. For real numbers, log(z)=ò1z 1/z dz - the same is true for complex numbers. The only problem is, because 1/z has a singularity at 0 there are multiple paths from 1 to any z, depending on how many times you go around 0. The values of log(z) you get all differ by a multiple of 2pi, these are the different branches of the log function. So, if a closed path (like a circle or rectangle) bounds an area A, and if all the singularities in that area have real singularities (i.e. a-1 is real) then interpreting singularities with positive residues as, e.g., liquid flowing in a rate a1 and singularities with negative residues as liquid flowing out at rate -a1 then the complex integral around the boundary of A is the rate of fluid flow out of the area A. If the residues are complex, you have to imagine complex rates of fluid flow. This isn't quite as hard as it seems, just imagine the real component is the amount of water and the imaginary component is the amount of red dye. If you interpret the complex integral like this your 3rd question has a sort of answer, which is that the integral is independent of the path you take because if youhave two paths, say P1 and P2 with the same starting and ending points. Create a third path Q by joining the end of P1 to the end of P2, and you get a circuitbounding an area A. If there are no singularities within this area, then there are nosources and sinks in the fluid flow, so thenet flow of liquid in or out is 0. So the integral along Q is 0. But the integral along Q is the integral along P1 minus the integral along P2 (because in Q you go along P2 backwards, changing the sign). Since the difference is zero, the two integrals are the same. You can make some of the above stuff work in n dimensions and the equivalent of Cauchy's theorem is called Stokes' theorem. I'm not sure I understand your first question.
| Posted on Sunday, 01 June, 2003 - 07:48 pm: |
Alexander,
i am not sure if i understand your answer to my first question that well.(Unless you are trying to switch from real to complex and then come back to real plane)
As for the second and third questions,i think i see the point you are trying to make.
love arun
| Posted on Sunday, 01 June, 2003 - 07:57 pm: |
Dan,
its hard to know that there is no geometrical interpretation for complex integrals.On the other hand,the circuit interpretation makes a lot of sense too.Infact it makes for easier understanding.
As for my first question,
i meant why is the path for normal integration like integral from a to be of f(x)dx always along real axis.Why cannot i define paths like these
Why cannot i integrate f(x) along paths shown above?
Why is it possible for me to define different paths from z1 to z2? and why the same is not possible in real plane?
love arun
| Posted on Sunday, 01 June, 2003 - 08:04 pm: |
Arun, what do you mean by 'real plane'? The real numbers are 1-dimensional.
David
| Posted on Sunday, 01 June, 2003 - 08:14 pm: |
(*stares,blinks,clears his eyes and reads again*
"The real numbers are 1-dimensional.")
OK!!My idiocracy has reached its limits these days.
Thanks David.That sentence was an eye-opener.
(oo)
love arun
P.S-> is this normal with students doing complex analysis or is it just me??