Paul's Homework 2

Comments:

GREAT job (except 1.5's). Let me know tonight if you have any questions. I am going to sleep and dream of korean cooking.

p.50

1.5a. Bah, hate nmeumonics. I did not calculate it, but by proposition 5.1 when you take the dot product of the vector with x it should be 0 (likewise when you dot it with y). The "physics" reason this is important is we want a direction orthogonal to two directions (it is also a way to "complete" a space in R^3). The dotproduct with y and x cross y is strictly positive, so something went wrong (thought it did work with x).

1.5b. Same problem. Its ok. Most math majors can't add anymore (reminds me of a very funny lecture where the prof just waited until we calculated the integral via normal integration methods, and even the top international math olympiads forgot).

1.6a. Good reasoning! (guessing you drew the triangle in R^3// awesome that you used the correct formula for cosine) I wonder if you can use the norm of the cross product divided by 2 in this case. But then again, that would only work for certain triangles, those formed from bisecting a parallelogram). I really meant to give you 12, but you would definitely get that since it is far easier than this question.

14a. Good! Thank you for working them out seperately. Just remember to finish the punchline at the very end: "Thus x cross y = ....". Actually, you did do it for the second part. Good show (this was a midterm problem for the non honor sequence).

14b. Whoa, this is a cool fact that I never knew. Now you know to appreciate associativity : ).

p. 71

7a. BAH. Need to put page numbers on top. Good but here is a more concise/legible/rigorous arguement: let x be a vector in U union V. Then by definition of union, x is in U, or x is in V.

If x is in U, then you can draw a ball around x such that the ball is in U. This is because U is open. Since U is contained in U union V, this ball is in U union V.

If x is in V, then you can draw a ball around x such that the ball is in V. This is because V is open. Since V is contained in U union V, this ball is in U union V.

In both cases, you can draw a ball around x contained in U union V. Since you could have picked ANY vector x in U union V, for any point in U union V, you can put a ball around it so it is contained in U union V. Then by definition, U union V is open.

Proof for intersection is far simpler (take the smaller of two balls): try to write the proof in this form. Also, for this proof, you assumed that you can use the same delta for balls. Not always true. Some sets are much much smaller than others.

7b. Spot on (see, 5 seconds). Think about if you were intersecting/unioning an infinite number of open sets instead of two and if it still works. Maybe you can use the same arguement as above for one of these cases (cough cough).

p.106 3+4. BAH! More computation. At least you correctly used chain rule. Very important for engineering and life. And universe. 1a,b,c. Same comment. I think I am going to ask if Otis/David Philipson still has the Shifrin solutions from grading H series. Or I can stop being lazy and do it myself. But the methodology is right!

2. Same thing: I know they give you the methodology in the previous example. But this is INCREDIBLY important if you are a physicist and engineer. You must know how to calculate these things.