Revision as of 16:15, 30 September 2012 by Namh (Talk | contribs)

Practice material for Exam 1 collaboration space

You can easily talk about math here, like this:

$ e^{i\theta} = \cos \theta + i \sin \theta. $

Is this the Cauchy Integral Formula?

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a} \ dz $


This isn't directly related to the practice exam, but is concerning a fact discussed in class. In one of the first lessons an important fact was provided. Namely, Suppose u is continuously a differentiable function on a connected open set $ \Omega $ and that $ \nabla u \equiv 0 $ Then u must be constant on $ \omega $.

How/Why is

$ 0 = \int\nabla u\ ds $

If we have a domain, $ \Omega $ and $ \gamma $ a curve in $ \Omega $, where $ A, B $ are end points of $ \gamma $, from vector calculus, we have

$ u(B) - u(A) = \int_\gamma \nabla u \ ds $.

but the integral is zero since $ \nabla u $ is zero. Hence $ u(A) = u(B) $ for any arbitrary point $ A $ and $ B $ and $ u $ is constant in $ \Omega $.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang