Recent content by Axiomer

I think pure math is the better choice. Most math courses that you will teach at the high school or community college level will be very hand wavy, but as a matter of principle any teacher should know the theory behind the material at a deep level. Should someone who doesn't know how limits are...

Is there a chance of getting a re-examination? Check your university's policies, I know that my school will generally let you retake a final exam if you do well but fail because of the final.

This is also where I learned measure theory from (2nd edition). I found this text great for both learning and as a reference. I haven't used any other measure theory textbooks, but I didn't feel the need to with this book handy. There is a nice chapter on probability theory, and a proof of the...

Hey all. I am an undergraduate math major in my third year who is planning to do a reading course in fluid mechanics next term. I have taken courses in rigorous advanced calculus, real and complex analysis, topology, and intro ODE and PDE. I would be taking a differential geometry course, which...

The UofT honours math program seems to be quite a bit more intensive than other Canadian schools, thoroughly covering Spivak's Calculus on Manifolds in second year, and doing galois theory as part of their 3rd year algebra course. The University of Waterloo seems to have a very solid math...

Oops. My bad!

f and |f| are continuous functions on a compact interval, hence are Riemann integrable. The Lebesgue integral is the Riemann integral in this case. By the fundamental theorem of calculus, |f(b) - f(a)| = |\int\limits_{[a,b]}f'(x)dm(x)| \leq \int\limits_{[a,b]}|f'(x)|dm(x)

Analytic (or holomorphic) means that that the function is complex differentiable on it's (open) domain. I would be interested in learning more about the motivation behind the complex line integral. To me, the complex line integral was presented as this dry definition from which all these...

Yes. Let ##A\subset ℝ^n##, and let ##x\in A## be an isolated point of A. By definition of isolated point, there is some ##r>0## s.t. ##B(x,r)\cap A = \{x\}##. Given any ##ε>0##, ##x\in B(x,ε)\Rightarrow B(x,ε)\cap A≠∅##. Moreover, ##B(x,r)\cap A = \{x\}\Rightarrow x+(\frac{1}{2}min\{ε,r\}...

That is how I first saw these concepts defined, and are probably the most practical definitions when dealing with R^n. Isolated points are indeed boundary points when working with R^n. A quick way to see that isolated points need not be boundary points is to take any non-empty set, X, and...

I suppose one might start by noticing that ##(b-a)## factors ##b^n-a^n##, since ##a## is clearly a root of the polynomial ##x^n-a^n##. So ##b^n-a^n=(b-a)p(a,b)## for some polynomial ##p(a,b)##. Then, to determine ##p(a,b)##, one could start doing long division to see the pattern. Alternatively...

##(b-a)(b^{n-1}+b^{n-2}a+...+a^{n-1}) = b(b^{n-1}+b^{n-2}a+...+a^{n-1})-a(b^{n-1}+b^{n-2}a+...+a^{n-1}) = (b^{n}+b^{n-1}a+...+ba^{n-1}) - (b^{n-1}a+b^{n-2}a^2+...+a^{n}) = b^n-a^n##

You're on the right track. Square both sides of ##x_1z_2+x_2z_1<2y_1y_2## to get ##x_1^2z_2^2+2(x_1z_1)(x_2z_2)+x_2^2z_1^2<4y_1^2y_2^2## (*) By the property of C, ##x_1z_1≥y_1^2## and ##x_2z_2≥y_2^2##. Applying this to (*) and rearranging gives ##x_1^2z_2^2+x_2^2z_1^2<2y_1^2y_2^2##, but...

For any measure space (X,\mathcal{S},μ), and any measurable function g:\rightarrow [-∞,∞], ∫|g|dμ=0\implies g=0 a.e. proof: Define A=\{x\in X: g(x)≠0\}. For all naturals n, define A_n=\{x\in X: |g(x)|>\frac{1}{n}\}. \frac{1}{n}μ(A_n)=∫\frac{1}{n}x_{A_n}dμ≤∫|g|dμ=0, so μ(A_n)=0 for all n...

That's a good point. Since the op didn't specify otherwise, I assumed we were talking about functions to the extended real line.