Comic Sans is famously the font you use when you don't want to be taken seriously. Algerian and Hobo I'm not familiar with, but Papyrus is reserved for people who want to make movies about blue aliens
Can someone please explain to me why I'm getting these rather convenient values using the function listed in cell 18 as an argument to circular and inverse circular functions? Why are the results "nice" numbers like approx. 1 or approximately the original function even? desmos.com/calculator/zkfagfdxwx
This doesn't make too much sense to me other than the somewhat vague connection between $e^{x+iy}$ and all of these functions here.
@robjohn You will find the reciprocal approximation in cell 17 and the recursive formula to improve its accuracy at the very bottom in cells [39, 41]. I believe it will be of use to you.
I honestly have never come across something like this before, but hopefully someone else here has.
Also, I have nothing more to say concerning reciprocals now apart from recreation :)
Could this actually be related to the hyperbolic functions since $\frac{1}{x}$ is a hyperbola?
Thanks for your question.
I will continue from inductive step.
Inductive step: Assume $P(k)$, then we want to show it holds for the inductive step $P(k+1)$:
$$\bigcup_{j=1}^{k+1} A_j \subseteq \bigcup_{j=1}^{k+1} B_j = \left(A_1 \bigcup A_2 \bigcup ... \bigcup A_k\right) \bigcup A_{k+1} \subseteq...
i was driving around with expired registration on my license plate. i'd paid to update the registration, but forgotten to put it on/in my car. i fixed that today.
when i mentioned this to my wife, i learned that her registration expired the month before mine, and that she hadn't even received an updated registration (although she paid for it, probably stolen in the mail). this troubles me more than it troubles her.
i am generally a more anarchistic person than my wife, but i do keep track of the bureaucratic stuff. things like tickets for random driving infractions, fines for overdue library books, etc. bother me. they don't bother everyone.
i am holding the universe together with my regard for rules and regulations.
yeah, isn't it basically the lebesgue decomposition of a measure?
i guess some people might be interested in it for generalizations of the FTC, independent of any general concern with measure theory. those people should buy their own books.
ted, we don't know if it's the flu. might just be a long-running sinus infection. whatever it is, definitely from day care.
i got it too, as a bonus. had a splitting sinus headache that did not go away with OTC meds for about two days. made it difficult to sleep. it is gone now.
we know it's from day care because they haven't sent her home, so i'm guessing all of the kids have it. the only hard-and-fast rule with those folks is you can't run a temperature and be at day care. munchkin had a temp last weekend but it was gone by monday.
daughter is finally asleep. cat is curled in a tight little ball at her feet.
hey, so what's a good place to self learn algebraic topology very well? hatcher seems to have very less details for a beginner, so that isn't working for me
twink, suffices to check for any fixed power of x. implied by knowing that x^n/(e^(x^2)) goes to 0 as x goes to infinity. could prove that e.g. by induction on n using l'hopital's rule.
twink: another way to prove that x^n/e^(x^2) goes to 0 would be to first prove that x < e^x (by whatever method you like) and note that it implies x^n e^(-x^2) < e^(nx - x^2) which again goes to 0 because nx - x^2 goes to -infinity no matter what n is.
lotsa other methods and tricks exist, no doubt, based on whatever tools feel simple or are readily at hand.
ohhh i can get a measure of the economic power of an agent by taking the inner product of his economic activity with that of the total economic activity on a $\Bbb R^l$ space of $l$ commodities. isn't that the niftiest of things
in my generation and before, irish folks were very observant of matters bureaucratic because, as the saying goes, you can't fight city hall. job applications would be denied because your letters strayed outside the box, etc. i have receipts back to the 80's, etc. someday i will have to let go.
spice was the name of a popular analog circuit simulator.
So f and all its derivatives are bounded. Hence, for every $l$ there exists M such that $|f^l(x)|\lt M$. Consider $g(x)=|x^k||f^l(x)|$. If $g(x)$ does not have limit $0$ as $x\to \infty$ , then g(x) is larger than a non zero value d for infinite distinct values of $x$ (let's say they make a set S). Then by choosing larger and larger $x$ in S, we see that $g(x)$ is unbounded as $x\to \infty$ (x are in S). This contradicts the given hypothesis. @Twink
I think this should work, but I may be wrong also.
if sup |x^2 g(x)| is some finite C, then |x^2 g(x))/x| <= C/|x| allows you to evaluate lim |x g(x)| = 0. nothing too special about the powers 2 and 1 here. a finite sup assumed on a higher power should gives you a 0 limit on any lower power.
@Twink $g(x)$ is greater than $d$ for x in S. Hmm, I think it should also be proven that $f^{l}(x)$ is away from zero (I mean my argument works if $|f^{l}(x)|\gt m$ for some $m\gt 0$.
the lim finite implies sup finite direction is easier because if you assume lim finite for f, you immediately get sup finite for f without considering other functions. the other way does require you to hop around a bit in the schwarz space. e.g. if you only make those assumptions for 0 <= k, l <= some fixed N, the lim finite implies sup finite for all k, l implication still works, but the other one would break down.
feels like it ought to work, both in the sense of those two conditions still being equivalent to one another, and also in the sense that it shouldn't make a difference in the definition of the underlying space.
what makes this all fairly uncomplicated is that these things are being assumed for a wide range of parameters (all powers/polynomials of x, all k, all l). it gets tricky if you begin placing limits on those things.
How is the problem called? You want to cross a river with a boat, but it has a current. So you have to travel at a specific angle to cross it the fastest way possible.
We all know that a circle is exactly defined by three distinct non-collinear points. But I need a way to solve the following problem:
Given three points, calculate a circle with all three points on its border if it exists, else calculate a circle with minimum radius which has two points on its b...
A question for the analysis people around (which I'm thinking about making a proper question on the site): Is there such a thing as a maximal nonmeasurable set under the Lebesgue measure? I'm thinking the answer is "no," because my attempts at applying Zorn's lemma have encountered the issue that the union of nonmeasurable sets might not necessarily be nonmeasurable
can you send me the original statement, I can't read like this. I forget the exact statement myself. It should be the mean value theorem for integrals, anyway
Is there a method that has a name which might be derived from Newton's method which finds peaks in functions instead of roots? (i.e. $f'(x) = 0$) You could use a similar method by computing the perpendicular of the derivative.
@robjohn, well but some points sit at a distance $> \log(n)$ of one another. hence, the question is what fraction of points sit within, and without $\log(n)$ out of the total number of possible pairwise distances I guess? Idk what I'm looking for here.. the way I just asked could be just $\sim \frac{\log(n)}{n}$ maybe, just by looking at a single row and hoping hard it might stay that way after considering all points
what is the difference between two points on a plane connected by 4 curves and two points in $\Bbb R^3$ connected by 4 curves, s.t. no curves are co-planar
the original question is: let $(X_n)_{n \ge 1}$ be rvs on the same probably space with $\mathbb{E}(X_i) = \mu$. Suppose $Cov(X_i, X_j) \le f(|i - j|)$ where $(f(n))_{n \ge 0} \rightarrow 0$ as $n \rightarrow \infty$. Prove $n^{-1} \sum_{i=1}^n X_i \rightarrow \mu$ in $L^2$.
so you get $\mathbb{E}(n^{-1} \sum_{i=1}^n X_i - \mu)^2 = Var(n^{-1} \sum_{i=1}^n X_i - \mu) = Var(n^{-1} \sum_{i=1}^n X_i) = \frac{1}{n^2} \Big[ \sum_{i=1}^n Var(X_i) + 2 \sum_{1 \le i < j \le n} Cov(X_i, X_j) \Big ]$
and so for that last covariance term I need to know that there is only a "small" amount of indices $(i, j)$ (looks like no more than $O(n \log(n))$) that are "close" to one another (hence I'm thinking at most of distance $O(\log(n))$ of one another). and the rest can go to to $0$ since you can pick $n$ large enough so that $f(\log(n)) < \epsilon$
or, well, ok. maybe we can fix $k \in \mathbb{N}$. and let $n \rightarrow \infty$. then the number of pairs increases as $n \rightarrow \infty$, but the proprtion still goes to $0$.
I'm going to squint, pretend that this works, and hope for the best.🤞
mm, no, the thing i was thinking of wasn't it. hypotheses way too different and with operators and not random variables. it's some idea like that, though, joe.
just doing that, all by itself, is a form of mild illness. they are lucky if the only harm is that they annoy people. some of those people waste a whole lot of personal time and become very, very unhappy if they feel that nobody is listening to them.
you never hear about campaigns to ban actual math textbooks from the classroom. i exclude K-12 stuff with fluff inside from this, because i'm sure that's been done. someone should pick a standard undergrad book and wage a scorched earth social media campaign against it.
yeah, anyone can troll. i think the more intense case that joe was describing is something beyond that. the word 'crank' is sometimes used. i can choose to be irritating and at least amuse myself for a minute. many of those folks aren't deliberately choosing to do that. they can't shut it off and they aren't having fun.
or maybe they are. let's ask them! can i get an email address, or better yet, a telephone number? i want to discuss the validity of cantor's diagonal argument.
I know that {nt} is dense in [0,1] , where t is a fixed irrational. Notation abuse here purely for brevity. {nt} may be treated as a set of all {nt} such that n is any natural number. Since {$\sqrt 3n$} is a subset of {$\sqrt n$}, and the former is dense in [0,1] so is the superset.
Why I want to know about correctness of the same is because then knowing $\{nt\}$ dense in [0,1] gives denseness of {sqrt n} as a corollary, we might say.
Given: $\{nt\}$ is dense in [0,1], where t is a fixed irrational. This can be proven to be true. Now question is to show: $\{\sqrt n\}$ is dense in [0,1] using the given statement.
I say yes because let S={{$n\sqrt 3$}: n is a natural number}, T={{$\sqrt n$}: n is a natural number}. Now S is a subset of T (because for every $\sqrt 3 n$ in S, we note that $\sqrt 3 n=\sqrt{3n^2}$ and RHS is in T). S is dense in [0,1] by the given statement.
Yes @Leslie, n runs over natural numbers and {.} represents fractiional part function.
ok, yes, i agree. note that it is sometimes easier for people to follow and read these types of arguments if you don't use the same dummy variable in both set definitions
it might be a tiny bit easier to follow if you wrote a chain of set equalities like {m sqrt(3): m in N} = {sqrt(3m^2): m in N} subset {sqrt(n): n in N}. your point is that the extreme left hand side is dense in [0,1] by some theorem, so the extreme right hand side is too. but i get a little dizzy if i see n sqrt(3) and sqrt(n)
