@Riker Keep loose, stretch, and take a short walk.... Next time you jog six miles, before you stop, walk a quarter mile or so afterwards, and stretch, to cool down. Helps to flush the lactic acid from one's muscles.
I did, SBA: I more or less believe it is true, because you can look at the Taylor series and see that the coefficients decrease (faster than polynomial, right?), and so it's probably easy to get a step function that does it. Then, mollify the step function to get something $C^\infty$ and if you play your cards right, you can probably get it to differ from the step function on a very small set.
The trick is to get the relevant step function, but, since you're a fan of big numbers, you probably could get it to work with the Busy Beaver function? :P
Mostly because I don't think about it very much. And I know that it's very hard to be an interesting teacher (and effective), and so I feel like, if I'm not putting a lot of effort in, then I probably am not doing that.
I suppose in some sense I haven't been given a lot of room to take risks; just because of the boring details about how my assignments so far have run.
And maybe I'm just experiencing a kind of impatience.
I am hoping to have a more traditional class next year, where I might have some more flexibility.
and the other two were also "for scientists and engineers", the third being Calc II and most recently the linear algebra again
Calc II is a very frustrating course to teach, as a mathematician. The students basically have to do proofs but they are given no guidance about how to do proofs.
Business calculus is frustrating for different, more selfish reasons :P
Yeah now I understand how frustrating must be teaching a course where there are a lot of proofs but not being properly a course, let's say of real analysis
I think that 'handing out worksheets' is something of a local maximum of effectiveness. If you want to improve a little, you have to really adjust the strategy a lot.
haha so I'm reviewing my notes from the Joint Meetings and the next post I'm supposed to write is "Pitching Coxeter Groups to a Curious Undergraduate". It strikes me more as a talk about how I'm supposed to write a sequence on Coxeter groups, rather than a talk I should actually write up itself >.<
His suggested program is (1) cube group and symmetry groups of a simplex, (2) classification theorem (3) group presentations and (4) infinite coxeter groups
The laziness is really in writing (1). Which has the unfortunate property of being the first step...
The thing is, if you feel good about group presentations, it's very easy to just dive straight into Coxeter stuff. But he argued that most students aren't, so you have to get to it somehow
@SBM Logic lesson: Let the domain of x, y be the set of all people. Let $L(x, y)$ mean "x loves y". How do the following statements differ?: $\forall x, \exists y (L(x, y))$ vs. $\exists y, \forall x(L(x, y))$
$\sup (\mathbb N^2)$ would be $\sup\{ (x, y) \mid x, y\in \mathbb N\}$ so unless you have an ordering relation which compares any two ordered pairs (a, b), (c, d) and defining when one is greater than the other, vice versa, is undefined. $\sup \mathbb N$ does not exist, so nor does its square.
But unless you can provide a partial order on ordered pairs of such integers (finite or infinite) under an ordering relation on the natural numbers that defines when $(a, b)\leq(c, d),\;\; a, b, c, d \in \mathbb N,$ and that is reflexive, antisymmetric and transitive, $$((a, b) \leq (a, b),$$ $$((a, b) \leq (c, d)) \rightarrow (c, d)\leq (a, b),$$ $$((a, b) \leq (c, d), \land (c, d) \leq (e, f)) \rightarrow (a, b)\leq (e, f)$$ there can be no sup $\mathbb N\times N = \mathbb N^2$.
@SBM See the article in Wikipedia, on "partially ordered set". Also, the second of the three properties I listed above should be "antisymmetric": If $(a, b)\leq (b, a),$ and $(b, a)\leq (a, b),$ then $(a, b) = (b, a)$.
@JoErNanO I was simply playing on Erics comment: "a right natural combinatorialist" ... no political angle intended. E.g., If we favor right cosets over left cosets, or vice versa, this isn't a political bias, but a group theory bias.
Well, maybe @Mith champions $2$ (quintessential factor of every even number) because of 2's inferiority complex, since it is the only even prime), because prime numbers are dominated by odds with only one exception, by the odds!!
But in any case, the point is that a group by definition only has one operation (it gets a second one from the inverses, but that one is not associative)
@SBM Best to start with reviewing material on binary relations which are well defined. And also, only an associative binary relation on a set which is closed under that binary relation, can be group operations.
Key ideas in defining a group is the identity element (for example, the real numbers under addition have 0 as it's identity), because every real number, added to $0$, remains unchanged. We also need that each element in the group has an inverse (in this case, with $0$ as the identity of the reals under addition, that means inverse of some element $x \in \mathbb R$ is $-x$, because $x+(-x) = 0$.
@EricStucky That's one reason "minus" is a problematic group operator, unless treated as the negation of a number. Since minus is not an associative binary operator.
Well, division is a problem on many sets, for example, the set of integers. There does not exist closure of division on integers. For example let $a=2, b=3$. The problem becomes that $2\div 3 = \frac 23 \notin \mathbb Z$.
@shredalert Are you referring to my example statement $\forall x, \exists y(L(x, y))$, and its translation? If so, you are being naughty! Although it is true that $\forall x(L(x, \text{amWhy}))$
Maybe I've lost you; What about a person that has a problem so s/he decides to solve it with a bash script. As a result, s/he has two problems (in addition to the problems s/he invariably has). :P
@JoErNanO This would all make more sense if you were to phonetically sound out loud my user name. a....m....Why
What does this rebus, drawn with pretzels, say? (This is what happens when I get bored with pretzels on the table... ;P)
Note: The 'scheduled series' mentioned is this. It still needs to be solved! Finally solved by @randal'thor! :D
Overall picture:
Individual pictures:
Note: That bla...
@Mithrandir My mom played along with me and my brother and sister, (way back when) by "illustrating" our last name (as it might appear in a cave, using stick figures and such): the sketch of what was clearly meant to represent a young woman; (sounds like) the sketch of a farm animal (I'll leave this as a puzzle), and a stick figure wearing "skis" on a slope.