I messed this up in my head yesterday. You need either the derivative condition or $|f(z)|=|z|$ somewhere to get what we were thinking. Surjectivity gives neither.
Yeah, I think this method works. Just generate ten numbers between 1 and 10 without replacement, then interpret that sequence as the second line in two-line notation. Works for me
a sometimes useful fact about cycle notation is that if you write the cycles so the first number in each cycle is the largest, and you order the cycles so the first numbers increase, you can drop the parentheses. e.g. (412)(53)(6) is unambiguously written 412536.
i'm not sure if that automatically leads to another algorithm.
reminds me of a problem i saw today (not formally stated but implied by a puzzle)
you're trying to arrange 1,2,3, ... n in some sequence. however, you know that each number can only occur at some of the positions (like, only first or fourth)
how do you determine whether such a sequence exists?
This is an exercise from Linear Algebra by Friedberg, Insel.
The question asks to determine if True or False: Any polynomial of degree $n$ with leading coefficient $(-1)^n$ is the characteristic polynomial of some linear operator.
I know that it is true after scouring the internet and trying to u...
I computed a determinant for a smaller companion matrix and got that it was the characteristic polynomial of the restriction $T_w$ of my linear operator to a cyclic subspace.
@TedShifrin this question is in chapter 5, canonical form isn't until chapter 7 :(
Until the suggestion I had never heard of the companion matrix, but I see how it fits in ....even if not fully yet. Ok...well in terms of true or false it is true. I'll work through the details when I come up to them gradually.
Thanks for the help....I hadn't realized how much I burnt out in the second half of the year until I saw how long ago that question I posted was......well I'm refreshed now.
Should've been done Insel by now.......such is life...shrug
i also drank about a gallon of tea today, so i may need to expel some discourse on or in somebody's desk. you know, the way the founders debated the constitution.
@robjohn If $f$ and $g$ are square-integrable functions on $\mathbb{R}$, is it true that $$\int_{\mathbb{R}^{2}} f(x) g(t) e^{-ixt} \, \mathrm dt \, \mathrm dx = \int_{\mathbb{R}^{2}} f(x) g(t) e^{-ixt} \, \mathrm dx \, \mathrm dt?$$ I vaguely remember seeing this stated somewhere, but I don't remember where.
The conditions don't guarantee that Fubini's theorem is satisfied. Take, for example, $f(x) = \frac{\sin x}{x}$ and $g(t) = \frac{1}{t} \boldsymbol{1}_{[1,\infty)}(t)$.
whatever (Z/Z)* is it is an uninteresting object. Z/Z has one element. if you decide to allow that thing to be invertible then yes it is its own group of units. hooray for the zero ring.
used to be engineers got themselves twisted in a knot over positive or negative logic. but they sorted that out. mathematicians can't even decide if its rng or ring.
i wear a ring. it took me about six months to get used to wearing it. i still notice it when i pick up something and hear it clink against glass or metal.
in cardano's memoirs, at one point he is bragging about his health, and he says, i still wear the same rings. this was a measure of health, that age 70 or whatever he had not wasted away.
my wedding ring is some kind of chromium alloy. it was brushed to not be shiny, and countless interactions with metal have made it even less shiny. i'm beginning to think i may not be able to return it to the store and get my $100 back.
@robjohn How would you use Plancherel's theorem to argue that $$\int_{0}^{\infty} \int_{0}^{1} \frac{\sin (x)}{x} \frac{\sin(xt)}{t} \, \mathrm dt \, \mathrm dx = \int_{0}^{1} \int_{0}^{\infty} \frac{\sin (x)}{x} \frac{\sin(xt)}{t} \, \mathrm dx \, \mathrm dt? $$ The function $\frac{1}{t} \boldsymbol{1}_{[0,1]}(t)$ is not square-integrable on $\mathbb{R}$.
Could the theorem be modified to state that if $f(x)$ and $g(t)\sin(xt)$ are square-integrable on $\mathbb{R}$, then $$\int_{\mathbb{R}^{2}} f(x) g(t) \sin(xt) \, \mathrm dt \, \mathrm dx = \int_{\mathbb{R}^{2}}f(x) g(t) \sin(xt) \, \mathrm dx \, \mathrm dt?$$
I'm still stuck at this: if $T:E\to F$ is a continuous surjective linear map between Banach spaces, can we always find a continuous linear map $T':F\to E$ such that $T \circ T' = \text{Id}_F$?
Note that we can always pick such continuous, but not necessarily linear map by Micheal selection theorem
@Rithaniel The standard algorithm is the Fisher–Yates shuffle. It's essentially building the permutation from its index (aka rank).
On a related note, if you want to iterate through all permutations lexicographically, see the algorithm of Narayana Pandita, which efficiently handles duplicate items. There are various other permutation ranking & unranking algorithms which are a little more efficient, although they generally aren't lexicographic.
@Koro If you're still interested in generating & counting partitions & compositions, I've got some useful links here: math.stackexchange.com/a/1287927/207316
@robjohn Wikipedia seems to state that $f$ and $g$ need to be in both $L^{1}(\mathbb{R})$ and $L^{2}(\mathbb{R})$ for Plancherel's theorem, but other sources state can it be extended to functions only in $L^{2}(\mathbb{R}) $.
@robjohn because isn't it so that if the integral converges, then there exists some $h(x)$ that bounds the inner function and that the integral of $h(x)$ also converges. Because $\int_0^{\infty}|\frac{sin(ay)}{y}| = \pi$
@RandomVariable So the integral surely converges, but how to find such h(x), or if we can't find it why is that so ?
@VLC If $a>0$, the integral $\int_{0}^{\infty} \frac{\sin (ay)}{y} \, \mathrm dy$ converges to $\pi$, but it does not converge absolutely. See this question.
OEIS is like the cheat codes for a ton of enumerative math
whenever you find a sequence that isn't in OEIS, the best thing to do is back away slowly, don't make eye contact. if you have a jacket, hold it up so you look big and the sequence is less likely to attack
Hi. I have the limit $\displaystyle \lim_{x\to\infty}\frac{-3x^4+x^3+x}{-2x^3+x+1}$. Dividing everything by $x^4$ yields me $\displaystyle \lim_{x\to\infty}\frac{-3+\frac{1}{x}+\frac{1}{x^3}}{\frac{-2}{x}+\frac{1}{x^3}+\frac{1}{x^4}}$. This last limit goes to $-3/0$ which is $-\infty$. But clearly the limit is $+\infty$. Where did I mess this up?
@koro Here's a little Sage program that prints partition numbers. I think it uses Pari to do the computation, so it's quite fast. If you want to code it yourself, I recommend using generalized pentagonal numbers, as explained on oeis.
note that the wrong application of limit rules results in funny results: for x>0, $x=\frac{x^2}x=\frac{-x^2}{-x}=\frac{-1}{-\frac 1x}$ and now do what you did with the limit in your question. @Odestheory12
@PM2Ring For now, I require partitions of n for small n only by hand calculation so not a problem :).
@Koro Ah, ok. Producing partitions (and compositions) by hand is a good exercise. It helps you understand how the algorithms work. :) FWIW, here's some plain Python that generates partitions of a fixed size. stackoverflow.com/a/40220262/4014959
i heard through the grapevine that knuth is not well although i do not know the reliability of this information. he is basically gauss in my book. his work influences almost everything.
@robjohn If $f(x)$ and $g(t)$ are real-valued functions, can Plancherel's theorem be modified to state that if $f(x)$ and $\color{red}{g(t) \sin(xt)}$ are in $L^{2}(\mathbb{R})$, then $$\int_{\mathbb{R}^{2}} f(x) g(t) \sin(xt) \, \mathrm dt \mathrm \, dx = \int_{\mathbb{R}^{2}} f(x) g(t) \sin(xt) \, \mathrm dx \mathrm \, dt?$$ This would be useful in situations where $g(t)$ by itself is not square integrable due to a non-integrable singularity at $t=0$. I'm come across this a few times.
random i have been looking for my euclidean harmonic analysis notes on the basis of this. perplexingly it is the one class (!) that is not in the box of notes from my grad era. i don't know where they could be.
in the box that i do have, my notes on undergrad differential geometry are 1/3 incomplete. a kid was sick for two weeks and asked for my notes and never returned them. i should track him down.
there was a question on math.se within the last six months that was easily dealt with via a result on 'almost orthogonality' and i discovered the loss of my notes when i went to look up its precise statement, which of course i had forgotten.
if you have a sum of operators O_j and O_i O_j is small if nonzero for distinct i, j then you can estimate the norm or do other things with the sum that otherwise would not be permissible.
weirdly i think i understand a lot of what i studied better now, because i have forgotten the details. i remember the basic moves.
i really wish i had my notes on this stuff. the instructor had a draft book (unpublished as of 20 years later) which contained refinements of the stuff in the literature. also lost.
The Knuth XKCD comic is number 163, the largest Heegner number. I've sometimes wondered if that's a coincidence, or if Randall intentionally gave Knuth an interesting number.
The difference between being a manifold and not is that for a manifold, in the nbhd of every point in the set there exists some way of being able to represent that nbhd as a $\mathcal{C}^1$ function?
Yes, that is a trouble point, but sometimes a trouble point is not really bad. Most of the time it is. But what's the correct statement in your question?
I had 4 x-ray sources disguised as monitors attached to my computer there. One for QuickMail, one for display of the output, one for code and one for debugging.
well if we say our manifold is of dimension $k$ for concreteness, then domain is of $\mathbb{R}^k$ and codomain $\mathbb{R}^{n-k}$. Where our implicit equation $F$ would've been $F:\mathbb{R}^n \to \mathbb{R}^m$
I only had two displays at one point in my life ... when I went from external monitor to iMac and kept the external monitor. It was extremely helpful for editing my books.
@dc3rd The important point is that the $\Bbb R^k$ is one of the standard coordinate $k$-planes, not a random $k$-dimensional space. Also, domain is only open set blah blah blah, of course.
Given a non constant continuous function $f:[0,1]\to [0,1]$ with the property that $fof=f$, I want to show that the set $S=\{x\in [0,1]: f(x)=x\}$ is an interval.
No, the sophisticated answer is the one you already have. If you do some searching on MSE, you'll see that I've used that as an answer for half a dozen questions.
It's worth understanding the proofs when you get to section 6.3.
That's true too I shouldn't get bogged down by this cause this was only an appetizer for the main dish. I've done the questions, but I'm revisitnig them all since I'm a bit fresher and it is clearer thank god ( or your deity(s) of choice if you have one or some))
Amateurs love proof by contradiction, even though it's unnecessarily clumsy.
Proof by contradiction can be useful for developing intuition, but usually the contradiction is totally unnecessary. Here's one of my standard examples of this: Suppose $x\cdot y = 0$ for all $y\in\Bbb R^n$. Prove $x=0$.
I assigned this in the first assignment every time I taught my multivariable math class and linear algebra. Students almost always did it by contradiction.
@TedShifrin that's precisely what I thought and at that time, in order to present it here to you, contradiction seemed like the best/easiest way for communication.
Hello, if $f: [0,1]\to [0,1]$ is increasing and consider $A=\{x\in [0,1], f(x)\geq x\}$ I know that A is not empty and bonded from above by 1 how to prove that sup(A) is in [0,1]?
