Paul's Phone Meetings

Date: August 3rd

Time: 8-9 PM, PDT Summary: Summary: Went over first homework assignment: matrix properties and norm properties. In particular, we reviewed 1.2 Question 8 which used infinite dot products, cosine, and an interesting number theory application.

Date: August 10

Time: 8-8:15 PM, PDT

Summary: Went over homework 2 (wow, never knew cross products were not associative). Especially covered opened and closed sets and closure under intersection and union. Also explained difference between arbitrary union/intersection and finite union/intersection. Recommended he uses arbitrary choice over induction since the infinite collection of open sets need not be countable.

Date: August 11

Time: 7-8 PM, PDT

Summary: Went over material that will be on the test, arranged the 24 hour test time, and reviewed last few weeks material.

Date: August 13

Time: 1-1:15 PM, PDT

Summary: Clarified question 6 on test: intersection of infinitely many OPEN sets.

Question about infinitely many union: Cannot argue question 6a via induction since methodology fails for 6b.

Question 8: Proof does not follow by polyomial properties since square root of x is NOT a polynomial.

Date: August 18

Time 7-8PM PDT

Summary: Went over epsilon-delta proofs (evil wizard). Will look at picture in article Epsilon delta in particular the delta arrows and epsilon arrows. Went over proofs of exercise 2.3.1 and 2.3.2.

Proof 2.3.1:

Suppose that the limit as x approaches a of f(x) were not unique (call them L, m). Then by definition of a limit, for some given epsilon, there is a corresponding delta_1 such that whenever we have ||x-a||  0. This determines delta_1 and delta_2. Choose delta smaller than delta 1 and delta 2. Then we know that |f(x)-m| < epsilon and |f(x)-L| < epsilon: adding and dividing by 2 gives us: ||L-f(x)||+||f(x)-m||< ||L-m||. Triangle inequality states that ||x-y|| <=||x-z||+||z-y|| (derive from p.12 by plugging in x-z for x and z-y for y). Plugging in L, f(x), and m yields ||L-f(x)||+||f(x)-m||>=||L-m||. But we cannot have a>=b and a < b simulatneously, a contradiction.

Proof 2.3.2: By p. 76  , we have to show it is continuous at every point a. Fix a point a. f(a)=||a|| is what I suspect is the limit and it obviously exists. Now I want to show that given any epsilon > 0 ||f(x)-||a|| ||< epsilon I can find a delta,which depends on epsilson such that whenever ||x-a||<delta we are guarenteed to have ||f(x)-||a|| ||< epsilon. By second part of (p.14) 1. 2.17, || f(x)-||a|| || < ||x-a||. Ok, guess delta is epsilon. Then ||x-a|| < delta =epsilon. Are we guarenteed ||f(x)-||a|| ||< epsilon? Absolutely since || f(x)-||a|| || < ||x-a|| < delta=epsilon, and cutting out the middle inequalities, || f(x)-||a|| || < epsilon. Thus we can always find a delta that guareentees the function will be within epsilon of the limit ||a||. Since a was an arbitrary point (it could have been any point) the function is continuous at every point and so we are done.

Date: September 21

Time 11AM-12 Taiwan Time

Summary: Went over last two homeworks: talked about curves and physics and went over normal vector to hyperplane in 4 dimensions problem http://www.youtube.com/watch?v=vJRMr5xx7sA and the nonsingular/singular matrix problem http://www.youtube.com/watch?v=8Y7fjcAtR2g. Discussed college applications, college plans, and multivariable calendar. Also talked about real analysis: Bolzano Weisterstrauss and Leon Simon Notes. Also asked about setup of extremun problems and p.207 number 2 and 4. Philip will present these two problems online.