@TedShifrin Yes. Rosie is still a bit antsy about anything that bites. She jumps at mosquito bites much more than she used to, but on the whole, we are doing very well.
Doing fine, thanks, @robjohn. Still more tired than I'd like to be. I don't know if it's lasting COVID effects or due to blood pressure meds ... or just getting plum old. :D
I can relate to that :D Although my memory still seems to be better than that of plenty of people younger than I. ... But I've always stressed out over the DMV written test — all the arcane numbers ... how many feet from a fire hydrant, how many feet from a turn do I signal, etc. I just do these things with common sense ...
when he reviewed the first time he thought "i don't need to know the penalties for these various illegal things, because i'm not planning on doing any of them." they still test that, however.
i said he reminded me of calculus students who don't study because "none of this is going to be used in real life." you fail a test, you fail it in your real life.
shin: if you forget the rules of the road, just listen for honking and shouting and watch for hand gestures. people are very forthcoming with guidance if you need a refresher
ah, see, he did not have sufficient knowledge of the fact he must avoid running into things, for you see, a cellphone is a distraction, and can lead to running into things
learn this one simple rule, you won't believe how much it can keep you out of trouble. driving instructors hate me!
one problem though: according to the above set of equalities, I'm getting $f^{-1}(0)=\{(\bar x, \bar y): \|x\|^2\ge 0\}= R^{m+n}$, which can not be true as this would imply that $f\equiv 0$.
@TedShifrin apparently, the equality that I wrote earlier is not correct.
If V is a vector space and W is a subset of V that is a vector space, then W is a subspace of V.
I think this statement is true apparently but my book says false
Now, this is weird. I feel the book points it out because it is never mentioned in the question that W is a vector space which field. Due to this, ambiguity, I think the book says it's false
But with the context it becomes clear that W is a vector space over the field same as V is a vector space over.
The point of the exercise is to remind you that it's important to mention the field. Like $\Bbb{R}$ is a vector space over $\Bbb{Q}$ but not over $\Bbb{C}$
But the real thing is, equivalent and equal can sometimes differ. In this case, I think $R^2$ represents the set of all tuples of 2 real numbers and W is essentially a tuple of length 3 having the last element is zero. As the cardinalities of each tuple in V is different from the cardinality of each in W so no tuples in V are equal to tuples in W and vice versa. Wouldn't you agree with this, @DannyuNDos ?
But the real thing is, equivalent and equal can sometimes differ. In this case, I think $R^2$ represents the set of all tuples of 2 real numbers and W is essentially a tuple of length 3 having the last element is zero. As the cardinalities of each tuple in V is different from the cardinality of each in W so no tuples in V are equal to tuples in W and vice versa. Wouldn't you agree with this, @DannyuNDos ?
(This few lines are the reasoning)
@DannyuNDos Are these few lines of reasoning stands formal to you?
I have seen so many highly-upvoted questions about proving inequalities involving cycling over variable, for example this here + the related links. Can someone motivate the interest? Why are these things important/interesting?
So here's my question: I acknowledge that the direct limit of Euclidean spheres exists, and is contractible. But does the inverse limit exist? I think the answer is negative because spheres don't admit lower-dimensional spheres as retracts.
david: for whatever reason, such problems seem to be very popular in contests. outside of 'contest math' i do not know of any reason to care.
one desirable quality of a contest problem is that it be doable without advanced knowledge (particular, it ought to have at least one way of solving it that is both short and 'low tech'). another desirable quality is that a problem not be "too easy." so, random inequalities where you need to deploy a number of non-obvious but low tech 'tricks' fit the bill.
@DavidRaveh does it really need a reason/motivation? It could just be for fun like what I do
it’s like a puzzle and puzzles are fun
Also Leslie is right, I’ve encountered numerous such problems on contests. You won’t see them much outside of contests. For example you’ll never find problems like this in a standard high school mathematics textbook or even college entrance exam prep books
@DavidRaveh My guess is that there is a small cadre of folk on the site who really like these kinds of problems, and they have a kind of mutual admiration society.
The problems themselves are not, so far as I can tell, terribly interesting, nor do they lead to any deep results in analysis or anything. But I think that they come up on competitions, sometimes.
"1-parameter" means that there is a family of distributions, which are all the same, up to some parameter which changes the distribution in some way. For example, the normal distribution is a 2-parameter distribution (mean $\mu$ and variance $\sigma^2$); the exponential distribution is a 1-parameter family.
Also, doesn't that sentence describe the Poisson distribution?
(Though I suppose that it depends on how you present things... Do you take the parameter in the Poisson distribution to be the mean? or the reciprocal of the mean?)
I've never run into a distribution that "encodes" its mean and variance into one parameter I guess. Meaning that the mean and variance depend on the parameter! In most cases the mean is explicitly listed in the distribution function
Or it is almost an exponential distribution (an exponential distribution with parameter $\lambda$ has mean $1/\lambda$, and standard deviation $1/\lambda$, right?).
@geocalc33 Again, look at a Poisson or exponential distributions.
If you score better than 10 on a Putnam exam, for instance, you might be put a little closer to the top of the pile of applications at top-tier grad programs.
@Jakobian Oh, I think that these kinds of competitions are pointless. I don't like them---they feel antithetical to how I believe mathematics should be done (they are individualistic, rely on "tricks", and use time in a way that is unrealistic in the "real world"). I never participated in any of them.
My problem with mathematics competitions is that so many people seem to equate doing well in competition with being a good mathematician. I think that the correlation is overstated.
We hear about the people who do well in competition, then go on to do well in mathematics. We don't really ever hear about the people who do well in competition, and flame out as undergrads or graduates; nor do we really hear about the people who are fabulously successful and never participated in these kinds of competitions.
So the impression that one is left with is that these kinds of competitions are some kind of predictor of success. Seems like confirmation bias.
@XanderHenderson personally, I don't care about geometry problems at all either. They can't really be used anywhere outside of competition math, and usually the arguments for proving them are different from the kind of formal deduction in mathematics that I enjoy.
I was a competitive athlete 20 years ago. I'm in my 40s now. I have no expectation that I can keep up with a teenager, and have nothing to prove by trying.
Of course you are going to "win" any athletic competition against me. I don't have a 20 year old body any more. I don't know why I should care about that.
@Jakobian Cal Poly Pomona had a faculty vs students integration bee every year. I was only adjunct faculty, so not really eligible to compete. But they were fun to watch.
@冥王Hades Well, most people experience this thing called "empathy", wherein they put themselves into the shoes of another. If you are capable of that, you might understand how someone else feels when they lose, and feel bad for them.
@Joe it concerns theorem 3 in chapter 24. Does Spivak also write in the 4th edition that "$\{f_n\}$ is a sequence of functions which are differentiable on $[a,b]$"?
When I win someone in a math-related context in any sense, I feel good for about 10 seconds or maybe 20. But after that, I almost immediately felt: What's the point of all this?
@Joe Ok, I was a little confused by the closed interval $[a,b]$ rather than the open one, since he hasn't really talked about differentiability on closed intervals throughout the book
@sunny: It means that $f$ is differentiable on $(a,b)$ in the usual sense, $f$ is right differentiable at $a$, and $f$ is left differentiable at $b$. I believe Spivak first uses this convention when he introduces the Fundamental Theorem of Calculus (and it is fairly standard).
we are all biologically designed to feel good after winning, for a very short time, then that passes. If we felt good for longer than a short time this would be an evolutionary disadvantage
@Jakobian I just added to his opinion that he could also help the loser feel better by encouraging them to do better next time. I don’t know if it’ll help a lot but it’s worth a try
@sunny: Note that if all points in $[a,b]$ are in the interior of the domain of $f$, then we could define "differentiable on $[a,b]$" to mean $f$ is differentiable at every point of $[a,b]$. That would be a different definition to the one I just gave, and I am fairly sure I have seen yet another different definition of "differentiability on $[a,b]$". I think it is best practice to explain what you mean by "differentiability on $[a,b]$" before using it.
@冥王Hades Alternatively, I can choose to engage in activities where there are not winners and losers. I can choose to do things where individual or team success aligns with group success.
@sunny: One of the issues with differentiability and continuity (in the context of introductory analysis) is that continuity on closed sets like $[a,b]$ is arguably a more natural notion than continuity on $(a,b)$. I believe the abstract reason for this is that closed intervals are compact, and there are a lot of theorems about continuous functions on compact sets. On the other hand, differentiability is a more natural notion in the context of open intervals (or at least open sets).
@sunny The way things are defined in elementary calculus classes, it would generally be wrong to claim that a function is differentiable at the boundaries. The language just isn't developed.
@sunny: Also, one-sided differentiability doesn't generalise particularly nicely to higher dimensions. There are directional derivatives, but just knowing that the directional derivatives in all directions exist doesn't tell you much. For instance, a map $f:\mathbb R^2\to\mathbb R$ could be discontinuous at a point, even though its directional derivatives in all directions exist. So I think you are right to find intunintuitive
It might also be worth noting that one of the more important theorems of introductory calculus is the mean value theorem (because the MVT is used to build so much theory). The hypotheses of the MVT are that the function being considered is continuous on a closed interval (the function has to be defined at the endpoints), and differentiable on the interior of that interval.
@geocalc33: Take $f:[0,1]\to\mathbb R$ given by $f(x)=x$ for $x\in[0,1)$ and $f(1)=0$. Then, there is no $x\in(0,1)$ such that $f'(x)=0$. So the hypothesis of continuity on endpoints is needed.
@Jakobian On the contrary, Toei animation dropped the idea of adapting the manga any further because the 2016 movie did horribly in terms of sales and merchandise
I don't see why you couldn't just use a limit argument to conclude that although the curve is not defined precisely on the boundary points of the closed interval, it gets arbitrary close to being defined on those points
are we arguing about differentiability at the endpoints vs differentiability as being extended to a smooth function?
@Joe ah, yes, it is what you were arguing
@geocalc33 So you mean that $f(a^+)$ and $f(b^-)$ are still defined? Wouldn't that be the same theorem just applied to a new function $g(x) = f(x)$ for $x\in (a, b)$, $g(a) = f(a^+)$ and $g(b) = f(b^-)$
I found a theorem weakening the differentiability assumption
If $f$ is continuous in $[a, b]$, differentiable from both sides on $(a, b)$, then $\frac{f(b)-f(a)}{b-a} = pf'_+(c) +qf_{-}'(c)$ for some $c\in (a,b)$ and $p+q = 1, p, q\geq 0$
I think studying by myself is good and worked for me so far. But people say that it's better to study together with colleagues. I tried once long ago but I felt like it was wasting time. I mean understanding concepts by reading various books seems to me more effective in the long term than discussing with colleagues. If I'm planning to solve some research level problem then I guess discussing with colleagues would be helpful, but not for solving some exercise problems or homework problems.
@robjohn do you know of any theorems that would generalize mean value theorem for a function $f:[a, b]\to\mathbb{R}$ ? The standard way of expressing MVT could be thought of as the limits $f(a^-)$ and $f(a^+)$ existing. I'm wondering if this can be weakened
You can already make statements about that integral (as in improper Riemann integral, for example) without needing some nebulous generalization of MVT.
So, I think, this might hold: If $f:[a, b]\to\mathbb{R}$ is differentiable in $(a, b)$ and $a, b$ are Lebesgue points of $f$, then $\frac{f(b)-f(a)}{b-a} = f'(c)$ for some $c\in (a, b)$
But, even then, I would imagine that $f$ needs to be a pretty nice function. The theorem pretty crucially relies on the intermediate value property.
(This would be for a theorem which gets you a $c$ such that $f(c) = \frac{1}{b-a} \int_{[a,b]} f(x)\,\mathrm{d}x$, or whatever the right statement is).
the point isn't to introduce integrals for the sake of having MVT about them, but to reduce the assumption of continuity at the endpoints by replacing it by a weaker condition
like Lebesgue points above
it's about being able to apply MVT to a larger amount of functions
@Jakobian Okay... but the statement I suggested is a version of MVT for a larger class of functions (i.e. functions defined by integrals of Darboux functions).
@user10478: I can't give you a definite answer, but my understanding is that random walks on $\mathbb Z$ are already very well understood mathematically, and so is there is not so much need for simulation. On the other hand, I would presume that in a more complicated set-up, simulation would play a larger role
I don't know much about the study of continuous-space discrete-time random walks. How do you define such a thing?
That said, my guess is that, in the Euclidean case, the limiting behaviour is still going to be classical Brownian motion. There's a theorem whose name I can't remember right now which suggests that things must converge to the usual Brownian motion.
Donsker.
Basically, if I recall correctly, Donsker tells you that if you have any kind of reasonable notion of "random walk", as you scale the space and time increments to zero (in an appropriate manner), the process will converge (in a meaningful sense) to Brownian motion.
I am interested in a setup in which voters have fixed, numeric preferences over multiple issues, and face a sequences of policy proposals comprised of random values for each issue. The proposals are discarded from the sequence unless they pass a majority rule vote.
@Jakobian Sure. I wasn't making a general statement. I was simply saying that I am interested in the mathematics of random walks (particularly in non-Euclidean spaces, e.g. on fractals, or in the $p$-adics).
