But @leslie no longer speaks standard English. That's what leaving Iowa or some parts of WI does to people: their speech gets distorted. I imagine robjohn also believes Don and Dawn are homonyms?
@robjohn Don and Dawn in particular, goes back to when my family spent Christmas in Claremont with my aunt and uncle and cousins. We went out to a place serving "family style dinners" (not a family diner) and among us cousins, we were surprised how we each spoke. Don is one of the cousins, who the other cousins called Dawn!
@robjohn DON is not Down, nor DAWN. DON, to say correctly in Spanish, would be Dan, ahhhhhhhhh Dahhn. Dawn is like D-awe-n. Down rhymes with Town, and Brown,
the issue is that a lot of the transliterations read the same too. 'awe' and 'ahh' are exactly the same. they don't help me to differentiate don and don (which are the same)
in law school i got a lot of guff from people for saying 'defendant' in a way that i was told sounded like 'defindant'. guilty as charged, it's the same sound. the fin in a fish is the same as the fen in fen-phen.
@leslietownes If it wants to be part of Europe, I wouldn't say it's pretending, at least Western Ukraine. But I do not know what's all at stake. I just know a two hour conversation between Biden and Putin took place today precisely about the Ukraine.
Not sure how to go about showing this. $f(x) = \frac{1}{1-2x} + \frac{1}{1-5x}$ looks like the sum to infinity of two geometric series. For $f(x) - 7xf(x)+10x^2f(x)$, I've got that this equals to $\sum_{n=0}^{\infty} a_n x^n(1-7x+10x^2)$. But not sure how to proceed. Would appreciate some hints
$7x f(x) = \sum_{n=0}^{\infty} 7 a_n x^{n+1} = \sum_{n=1}^{\infty} 7 a_{n-1} x^n = 7 a_0 x + \sum_{n=2}^{\infty} 7 a_{n-1} x^n$, and $10x^2 f(x) = \sum_{n=0}^{\infty} 10 a_n x^{n+2} = \sum_{n=2}^{\infty} 10 a_{n-2} x^n$ so if you collect the x^0 and x^1 terms separately from the sums, $f(x) - 7x f(x) + 10x^2 f(x) = a_0 + a_1 x - 7 a_0 x + \sum_{n=2}^{\infty} (a_n - 7 a_{n-1} + 10 a_{n-2}) x^n$. maybe some typos there.
and because of that recurstion the series goes away.
this gives you $(1 - 7x + 10x^2) f(x) = L(x)$ for some linear polynomial $L(x)$ and presumably algebra shows you that $L(x)/(1 - 7x+10x^2)$ is what it's supposed to be. that would be a partial fraction decomposition.
look at my chatjax, everybody. this is rare stuff from me.
experience. if you think of a power series as just a sequence of coefficients, multiplication by x is a shift. (a,b,c,...) goes to (0,a,b,c...) etc. and if you equate linear combinations of those you get linear recurrence relations in the coefficients.
Is anyone familiar with the definition of continuity when it is written as 'a function $f$ is continuous on $X$ iff $f^{-1}(U)$ is open for each open set $U$'? $f^{-1}(U)$ is the functions inverse image $\{x\in X|f(x)\in U\}$, not the inverse function.
This definition is meant to be a generalisation of the $\epsilon$-$\delta$ definition, but for the life of me, I can't figure out if the function, $f(x)=1$ if $x>0$ and $f(x)=0$ if $x\leq 0$, is continuous under the new definition or not. It is definitely not continuous under the epsilon delta definition, but for the new definition there are no open sets in the range of $f$ to look at.
you get to pick the open sets in the range to look at. it does require some mental rearrangement to think in terms of this definition. some books prefer one view to the other, to the point of not quite acknowledging that the other exists.
might help to work that out in detail. e.g. if $\lim_{x \to a^{-}} f(x)$ is less than, or greater than, $f(a)$, how do you grab open sets that will show the discontinuity in the open set sense
huh, now I realised why I was confused. The full definition was 'A function $f:(X,T_X)\rightarrow(Y,T_Y)$ is continuous on $X$ iff $f^{-1}$ is open for each $U\subset Y$'. I was thinking of $Y$ as the target set, which would be $\{0,1\}$ in my example, but then since every $U$ needs to be taken out of that, I wasn't able to get that larger subset of Leslie's $(\infty,\frac{1}{2})$. Come to think of it, is it even compatible with the definition?
In the definition given, the domain and range are not specified. Unless you are referring to my example in which case yes I didn't write out the domain and range.
I am trying to code multivariate polynomial division, and I wonder if, by following the long division algorithm, is it possible to get terms with repeated monomials in the quotient, or terms out of the chosen monomial ordering.
I suppose the discrete topology works too, either of them would make the problem the simplest. I don't know what a/the subspace topology is. The discrete topology to me is all the subsets of $\{0,1\}$.
@TedShifrin Let's say the domain is just R, with the metric topology on it.
I feel kinda bad that I wound up dropping his Top 2 class. Though I might have felt worse staying in, considering that was my, uh, catastrophe semester.
I talked to him about it; my impression is that he doesn't. He warned the class early on that it was, for reasons at least partly outside his control (as I remember), going to turn into quals prep.
Not sure the nature of it. I know it was going to be Munkres part 2 before that, so roughly, yes, but I don't know about after because I dropped very early.
I lasted through that material for a little while, and then his announcement + my dropping, which came just a bit before all the cool theorems (FTA, Borsuk-Ulam, etc) and Seifert-Van Kampen, IIRC.
Or whatever the theorem is called. It has been too long since I looked at AT.
@Fargle Elements of the topology are called 'open sets' of the space. So $\{0\}$ is an open set in this context. Since $f^{-1}(\{0\})$ is $(-\infty, 0]$, the function will not be continuous.
> There are three paths from $\mathrm{A}$ to $\mathrm{B}$ each consists of one or more semi-circles of unknown radii. The paths $\mathrm{AXB}, \mathrm{AYB}$, $A Z B$ are called I, II and III respectively.
The middle figure in AZB doesn't look like a semi circle :-/
Is it possible to calculate a primitive, even if it does not exist? Let me explain: I want to calculate $\int \frac{1}{\sqrt{1-u^{2}}}\sqrt[4]{\ln u}{\rm d}u$ but it seems that over $[0;1]$ this integral does not exist. Still, is it possible to calculate it?
@copper.hat I came across a question where they asked to calculate that indefinite integral. However, it seems that this integral does not really exist.
I tried to show if $\int_0^\infty (\cos^2x)^{x^4}\,dx$ converges but nothing seemed to work. @robjohn
Any hints on that please?
I tried considering $exp (x^4 \log (\cos^2x)$ and then $x^4 \log (\cos^2x)\le x^4 e^{-x}$ and integrating both sides gives RHS as $\Gamma {5}$
But then I realized that LHS is negative and I couldn't conclude convergence of the integral. And even if the convergence could be concluded still convergence of $\int exp (x^4 \log (\cos^2x)$ could not be established, I think.
Background: this integral came up while discussing one day: if there exists a continuous function $f$ such that that $\int_0^\infty f(x)\, dx$ exists but $\lim_{x\to \infty} f(x)$ does not exist.
Hi, i am getting a very serious problem, because suddenly i cant comment, neither answer nor ask a question, also i cant bote and i got more than 500 in reputation, i dont know if it is compatibility issues on my phone, any suggestion?
@S.M.T The presupposes that mathematics is a whore, who only gains value by satisfying the needs of other disciplines. In reality, mathematics is a beautiful queen, who serves her own purposes, and occasionally deigns to grant a favor to the dirty sciences.
jay: yeah, that use of "formal" means "following an expected or established form" and not formal in the sense of "formal proof" in a proof writing context, or formal reasoning. and in context is a signal that what follows probably isn't rigorous at all
very common e.g. to write the so-and-so transform is 'formally' given by some expression, when it maybe the expression only makes sense for a subset of the domain it is being considered on
i just skimmed one of my old papers where 'formally' i deduced f(z) = Re g(z) using a comparison of power series that diverged for a lot of z i wanted to use that equality on, where the real proof would have detoured through some integrals and limits
if you want to introduce an inner product you can think of it as the orthocomplement of the null space.
again, that's not quite as directly tied to the transformation as its image is.
i am a huge fan of matrices but unless you have to calculate something i think it is better to think in terms of properties of the maps themselves. matrix entries, whether rows or columns, aren't themselves very interesting.
yeah. this is sometimes or maybe even usually the definition of 'nonsingular' (note that it makes sense for rectangular matrices that are not square).
although some books define 'nonsingular' only for square matrices by matrix algebra (meaning: one of a long list of properties equivalent to invertibility) in which case this is maybe a theorem and not a definition.
@leslietownes It's a typo. This is the correction: if $x_{1}\in (0,1)$ and $x_{n+1}=x_{n}-x_{n}^{2}$ so $\sum x_{n}$ is divergent series? I think we can prove that $a_n \not\to 0$ as $n\to \infty$ so we can conclude that $\sum x_{n}$ is a divergent series. But I'm no sure as to work here.
@Alex It would be nice to learn how to use the comment links. See how this message is linked to your last message? I did that using the arrow at the far right of the message:
@Alex if $x\in (0,1)$ then $x-x^2=x(1-x)\in (0,1/4)$. if you go one step further the interval shrinks further, and so forth. so i'm not sure on what basis you're concluding that $x_n$ isn't converging to zero
@Alex If $f(x)=x(1-x)$ then the sequence can be recursively defined as $x_{n+1}=f(x_n)$. Now if $x_1\in (0,1)$, it follows that $(x_n)$ is a sequence of positive terms and decreases. Therefore $\lim x_n$ exists and can be shown to be equal to $0$.
@Koro Sorry and I just thought I could use the convergence theorem. I should have thought a bit more about the limit of the terms. So how could you prove that the series diverges?
@Alex Ah, I have also used monotone convergence theorem -$(x_n)$ is decreasing and bounded hence must converge. About convergence: I think it's easy to see that $\sum x_n^2$ converges. I'll think about convergence of $\sum x_n$ though.
No, no, I wanted to teach it. I used Strang's brand new book, and it made a fabulous course, although I had to fill in all sorts of proofs and application derivations ... and make up a lot of better homework.
I actually don't think stationary phase was in that book, but a lot of good stuff is in Bender/Orszag (a course I should have taken at MIT, but I was too busy doing grad pure courses).
@robjohn The typical applied course was taught out of Kreyszig or some such and was just a compendium of unrelated stuff. Strang's book was very wonderfully thematic.
@TedShifrin I never looked at it. I was teaching at UCLA 86-88, so it would have been nice to have a good book to teach from, but I was not in any position to create a course.
@Alex I used that to conclude that $\lim x_n$ exists. Then note that $x_n-x_{n+1}=x_n^2$ so telescoping it will give you insight into convergence of $\sum x_n^2$
Functional analysis, no. This was a year-long course in applied mathematics, so various things from linear algebra, ODE, PDE, transforms, complex variables, asymptotics, ...
Integrating over the interval $[k\pi,(k+1)\pi]$ makes it a lot easier than doing separate uniform estimates when the function is $<1-\epsilon$.
@Koro Yes, robjohn has a certain style :P ... I don't know if you'll recognize me so easily ... but you don't look at many questions I answer anyhow :D
I have by far the greatest number of answers in differential geometry and differential topology. For multivariable calculus there are some obvious people who try to answer every question.
@TedShifrin yes, they fixed one of the bugs that I reported, but another still remains. The scores for your answers should be correct, but the acceptance state is the acceptance state of the question, still.
Well, the five that show up on my summary page are correct. I had to click to get more answers to see more, and then I ended up where you don't want me :P
No, no. I'm not criticizing. What is interesting, I suppose, is that we only see the top five. To see more, it takes us to the page with more complete information.
And I hate that to see more detailed information, we cannot just click on the "Reputation" title, which is simply text, but we have to scroll back to the top and click the "Reputation" button.
@TedShifrin more information, framed by a lot of empty space
If S is a subset of $\mathbb R$ such that $S'\ne \mathbb R$, here S' denotes set of all limit points of S in $\mathbb R$. Is there an example of S such that $S'\ne \mathbb R$ but still S is dense in $\mathbb R$?
if y is in R \ S' there is a punctured interval around y containing no points of S, and hence subintervals of R containing no points of S, so S is not dense.
I have a sample of size d from each of n normal populations with common unknown variance bur possibly different unknown means $X_{ij} \sim \mathcal{N}(\mu_i, \sigma^2)$
Does it mean my data looks like this:? https://media.discordapp.net/attachments/554278933554659328/918573644081422336/unknown.png
@Koro ok, I think it works: since $x_{n}\to 0$ so $\frac{1}{x_{n+1}}-\frac{1}{x_{n}}=\frac{x_{n}}{x_{n+1}}\sim 1$ because $x_{n}-x_{n+1}=x_{n}^{2}$. Now, by Cesaro we have $x_{n}\sim \frac{1}{n}$. Hence $\sum x_{n}$ is a divergent series.
