prithu the idea is to use properties of the squaring function and clever choices to show that if m^2 is bigger than 2 you can find something smaller than m that also squares to something bigger than 2. this turns out to contradict one of the properties of m.
same with the other case, but with 'smaller' and 'bigger' interchanged. the arithmetic and choices behind these formulas are not very illuminating. they involve somewhat arbitrary choices, and leverage the specificity of the squaring function.
the choices that guide those arguments are from people playing with algebraic formulas to see what they can simply prove is bigger than 2, or smaller than 2, using a small number of order hypotheses.
this is not a useful life skill.
yeah lesliecoin is going to blow up. everyone who bought early is going to retire to their own island.
where we will defend ourselves from people who bought late, and the authorities.
also my job has a constant-changing list of like 5000 stocks i can't own, or have to disclose owning, and it's just simpler not to. but yeah the AT&T thing was scummier than usual.
prithu some methods of computing sqrt(2) rely on sequences generated by a specialization of newton's method, which at least generalizes beyond squaring and polynomials and gives some intuition for where the method 'comes from.' if i recall correctly, the sequences you get are not hard to analyze, but maybe slightly subtler than the ones that other proofs choose. maybe worth checking out.
I did always warn my students about that. Nor did I act like high school teachers and take off points for missing $+C$. However, when they had to solve an ODE with an initial value, I nailed them if they didn't have the $+C$.
well as ted noted, we can basically assume that any of our internet/phone/whatever providers are powering everything we use by throwing orphans and puppies into a wood chipper. so maybe AT&T is not a big deal.
@leslietownes Now I am kind of lost at this point about where should I study real-analysis ? should I study from point-set toplogy ? which book:abbot? ruding? spivak? Are there a hierarchy of theorem I can prove?
prithu it is an interesting question with no single right answer. analysis-oriented books vary a lot in how much topology they put in. and in how much they focus on counterexamples and higher levels of generality vs. merely justifying what you might see in a 'calculus' book.
i don't think rudin is a great introduction. it has some good chapters and some bad chapters and an introductory student won't be able to tell which is which. some of the proofs are legendarily inscrutable.
for a newer book, i like thomas koerner's "companion to analysis." published by AMS in the USA. if you look on the internet archive, there are some very-close-to-final drafts on the author's website.
meaning, chapters that are a distraction from what most people look to in an introduction to real analysis, or are just poorly written. people don't agree on which chapters are bad, but a lot of people feel rudin has this issue
@Koro You might enjoy the short story Unreasonable Effectiveness by Alex Kasman, which is set on an island. Kasman is a mathematician & author. He maintains a very comprehensive list of mathematical fiction, which has descriptions & reviews of over 1000 stories, and links to free versions, when available.
you didn't ask, but my opinion of rudin's PMA is that the chapters 2 and 8-11 are bad and best replaced with other material (maybe a tiny bit of the topology of 2???) or skipped (everything else)
chapter 2 is almost an orphan from another text, so little of the generalities introduced there are used in the rest of the book
also everything about constructing reals in chapter 1
what's a nice constant that's very very big? e.g. i need a very big constant to justify my sketch of the isocline of $y = \frac{-x}{y'}$ on the x-axis by setting $y'$ equal to that constant
sketching an isocline on the x-axis of a differential equation whose slope on the x-axis is visually vertical, i.e. setting $y' = \textrm{biiig constant}$ in $y=\frac{-x}{y'}$
@leslietownes Slower than Newton's method, but slightly simpler, so a little easier to work with is $x_{n+1} = \frac{x_n+2}{x_n+1}$, which comes from the continued fraction of $\sqrt 2$. Newton's method doubles $n$ on each loop.
that or newton's are exercises in rudin's chapter 3. i do like those exercises, although it wouldn't have killed him to say where the sequences were coming from. :)
i once had a phone number that was prime. i sometimes use that when i need a big but not too big number.
When I was a kid, we had 7 digit phone numbers in my city (Sydney). One day, I dialled the first 7 digits of pi. I got a shock & immediately hung up when the phone was answered.
PM: i feel like you missed an opportunity there, although i can't imagine what it would have been. ask them for the next 7 digits of pi?
we were very limited in my hometown, there were only a handful of local exchanges you could dial without running long distance charges, and none of them were very good alphabetically.
when i was a grad student in the ee building in cory hall in berkeley, the phone number of a nearby student spelled michael. to confuse things, another student in the cube next to me was named michael, and to add further confusion, his last name was, you guessed it, jackson.
i forget how i did this the last time i did it. i don't think it would have done it in 2 seconds. well, maybe with playing some games about what a 'normal' computer is. but whatever online judge i was feeding it to did not give me that many cycles.
Here's that infinite descent proof of the irrationality of $\sqrt2$. $$1<x \implies 1<x<x^2$$ So $$1<\sqrt2<2$$ Assume $\sqrt2$ is rational, i.e., $\exists \frac pq =\sqrt2$ Firstly, $$1<\frac pq<2$$ $$q<p<2q<2p$$ so $$0<p-q<q$$ and $$0<2q-p<p$$ Now $$p^2=2q^2$$ $$p^2-pq=2q^2-pq$$ $$p(p-q)=q(2q-p)$$ Both $q>0$ and $p-q>0$, so we can divide. $$\frac pq=\frac{2q-p}{p-q}$$ So we have a new fraction, where the numerator & denominator on the RHS are positive and lower than the numerator & denominator on the LHS. We can repeat this process indefinitely. But that leads to a contradiction, since an…
you can start with simple stuff, like compute M^powers of 2 via iterated squaring (however you multiply two matrices), and then group the ones you need to get M^n for any n.
strassen is a possibility, although the last time i remember looking at this, you needed M or n or both to be pretty big to really notice a difference
i think there is stuff provably better than strassen asymptotically, but the implicit constants might be so big that in practical terms it's not worth it
if M has structure or is not that big, the stuff that is asymptotically good in general might not be as good as more specific approaches
avra: it's something you can do with any multiplication algorithm to reduce the number of multiplications from the n that one might at first think are needed to compute M^n. it doesn't speed up individual multiplications.
another consideration is if M is going to be changing over time (and how), or if you will need lots of different powers of the same matrix over time, or just one of them. something that seems slow because it requires a certain up-front investment in precalculation might be better if you have to do it a lot with the same (or similar) ingredients.
Functional analysis is the study of linear operators on vector spaces. Most of the work is done in infinite dimensional spaces (e.g. the space of all continuous functions on a compact set, or the space of square integrable functions on $\mathbb{R}$). However, matrices are linear operators on finite dimensional spaces, so all of the theory from funky anal applies.
Indeed, if you are willing to imagine infinite dimensional matrices, you can get a pretty good understanding of lots of spaces (e.g. an analytic function is "really" just a series, i.e. an element of a space with countable dimension, so linear operators on the space of smooth functions are "just" infinite dimensional matrices).
So a lot of the intuition one one gets from matrices carries over.
Also, I think that Kreyzig's book is a very gentle introduction to the topic, with some emphasis on spaces of sequences (which are, perhaps, slightly more tractable to the neophyte).
Suppose we have matrices $A_0, \cdots, A_{n-1}$ (you can say n matrices). Matrix $A_i$ is with dimension $d_i \times d_{i+1}$. If we would like to find all possible permutations to find the best parnthesization possible, that would take $4^n$ time complexity, any idea here please?
For example, B is 3x100, C is 100x5, D is 5x5, the best one is (BC)D = 1575 ops while B(CD) takes 4000 ops.
I don't think his discussion of contrapositive is flawed. (NOT B) => (NOT A) is equivalent to (NOT B) AND (A) => (NOT A) AND (A), a fallacy. So equivalently you can assume A and NOT B and try to get a contradiction
I have seen this strategy before but I am not sure it's called "contrapositive". I might be missing out on terminology.
I don't think he should have said "which means". It's an equivalent strategy.
I want to make a generalization of Catalan numbers, so I make an $n\times m$ (where $n$ and $m$ are coprime) grid and try to find a number of paths which is not crossing the diagonal.
I think it will be ${2\over n+m} \binom{m+n}{n}$, but I can't find proof of it.
Please help me. Thanks.
if the rules are you can only move up and right, you've clearly got to do n ups and n rights and it's a question of how to cleverly count the ones where the ups don't stack up too much ahead of the rights. thinking in terms of symbol strings maybe easier than paths.
if C(a,b) denotes the count of allowable paths from (0,0) to (a,b) it is often possible to identify recurrence relations that compute C(a,b) in terms of adjacent/previous C(x,y)'s and then solve those relations.
i assumed 'more pedantic logical sorts' excluded me. it looks OK but it is not what people normally call proof by contraposition, and i think in some logics (intuitionistic logic?) that deny PEM it is not the same as controposition.
i'm hesitant to enter such discussions because once i pointed out that some people do not assume PEM or the existence of infinite objects and a high-rep set theorist curb stomped me like three hundred times for what felt to me like a mild, ignorable comment
if people tell me 'hey, you forgot inseparable hilbert spaces' i let it slide. but i think set theorists see the world differently because every other post on SE is initially tagged 'set theory' if it involves a set
It's equivalent to $\lnot(\lnot Q \vee P)$, is his point. But he's messing around, point (2) is "proof by contradiction of the contrapositive statement", not "proof by contraposition".
i was joking earlier about wiley and sons being imperialist pigs but it made me wonder if copyright law has something to do with dover's republication of so many soviet-era classics.
my understanding of soviet law is that after some early point in the republic, all international copyrights in soviet authors belonged to the state. us copyright law explicitly denied the validity of these transfers.
so either there were no us copyrights, or nobody who could enforce copyright claims in soviet-authored works.
which is why we can all enjoy the sweat of the workers' brow.
us law does generally defer to foreign law in questions of ownership in foreign-authored works, which is kind of weird. in theory you might need an american court to interpret a french will or italian bankruptcy proceeding to figure out who can sue whom but in reality it doesn't get litigated much.
on the bright side, now that the us is sliding into authoritarianism, i guess everyone can choose their favourite type so you might eventually have communism
when the postal service lost about half of my math book collection (which was not large, but very high quality and included rare items), i took to the dark corners of the internet to 'replace' the works with electronic copies. i would do it again tomorrow.
you'd think that the main competitor to the usa in the international sphere is bound to be brimming with intellectual activity, so where are the translated works of people
just, known in the general academic public to non-specialists
interesting point. there may be something of a language barrier and not a history of translating too much in either direction, that is less in e.g. french or german or even russian works.
of course a lot of chinese mathematicians publish in english. although maybe not as much "public-facing" stuff. even america and england don't really have a tradition of that the way russia did. or even france.
i'm fairly ignorant of the history of chinese mathematics but an enormous amount of stuff that is given european names in western countries was there first. "gaussian" elimination comes to mind although even gauss's contemporaries would have recognized it as already known.
india too. maybe some of this is a legacy of colonialism and a response to colonialism.
with stuff like that, i dunno. i think there is a level of political control over a lot of fields of inquiry that doesn't exist elsewhere. although with a fairly strong state, researchers may have access to data sets that people in other countries could only dream of.
and there's certainly levels of indirect control from powerful interest over social inquiry just about anywhere.
I want to make a generalization of Catalan numbers, so I make an $n\times m$ (where $n$ and $m$ are coprime) grid and try to find a number of paths which is not crossing the diagonal.
I think it will be ${2\over n+m} \binom{m+n}{n}$, but I can't find proof of it.
Please help me. Thanks.
