Mathematics - 2021-09-01 (page 1 of 2)

Ted Shifrin

00:20

@robjohn Thanks, but doesn’t save on phone neither! :)

Avra

01:12

!!

I did not see I deleted some lines

leslie townes

hoovered them right up, you did. and just after chopping them up.

Avra

Given $r= (ak +b) \bmod{b}$, $s=(al+b) \bmod{p}$. Professor Ted always tell me that I have something missing!

I see now

How we can get b from r,s please above?

$b=(r−ak) \bmod{p}$

leslie townes

the + b in (ak + b) mod b strikes me as unusual. it's the same relation as r = ak mod b.

Avra

sorry!

OMG!!

leslie townes

recovering the modulus b from the other congruence also strikes me as unusual, but modular arithmetic was never my strong suit and i have thought about this for about five seconds only.

is there an application or problem in the background that boiled down to this?

Avra

01:16

$ r=(ak+b) mod ~ p$

s=(al+b)modp

$k≠l$

$a≠0$

I should delete everything and start over

This is a complete mess more than chaos theory

Given, $r=(ak+b) \bmod{p}$, $s=(al+b) \bmod{p}$. Also given that $k\ne l$ and $a \ne 0$. Also $a,b, k$ are integers and $p$ is prime. How we can get $b=(r-ak) \bmod{p}$ please?

@leslietownes. You did not study number theory before?

This is very popular topic in number theory I guess please?

leslie townes

i studied it and even taught it, i just have a difficulty wrapping my mind around systems of these sorts of equations after too long a time. i had no intuition for it, it was symbol fiddling for me.

throw some geometry in there, that's how you spark my interest.

Avra

haha

fine. Yeah every mathematican likes certain area and not all

My brother's wife loves differential equations the most...

leslie townes

eww. there's no accounting for taste. i guess there's no law against that.

Avra

what was the most annoying math topic you have ever studied if any please!

leslie townes

ooh, tough one.

if i'm being honest, beyond the basics, i didn't like algebra very much. once we were doing quotients of multivariable polynomial rings and drawing icky diagrams, time out. this is not a criticism of algebra, just my reaction to it.

i liked the stuff i took in undergrad but then in grad school realized that i was probably going to not like what came next. thankfully it was optional. i took one class 'for culture.'

Avra

01:30

hahaha

Ted Shifrin

@leslietownes WHAT T F !?

leslie townes

this isn't leslie, ted. this is the guy who broke into leslie's house when he wasn't home. the computer was on.

Avra

hahaha

It seems that this is Prof Ted favourite subject?

leslie townes

avra, yes. my general 'brand' is to lightly discourage the use of geometry. i am mostly being ridiculous, but it's rooted in something real. maybe high school. i should therapize these feelings and why i am so sarcastic about them.

Ted Shifrin

@leslietownes That makes more sense.

Probably Olivia.

Avra

01:36

You are talking complex analysis high level language now that I am not able to understand

Ted Shifrin

Huh?

Avra

@TedShifrin. I am just saying @leslietownes language is too complex :/ It needs math expert to understand

AMDG

01:55

Geometry is useful for immediately grasping the intuition and application of algebra. Algebra is useful for manipulating your intuition and application of algebra.

leslie townes

i'm not sure that 'therapize' is even a word, let alone a word for something that i could do to myself.

avra, i try to speak more clearly when discussing math. much of the above is just me making jokes.

AMDG

02:12

It would seem English has no verb for therapy, but German does.

It could probably pass as a sort of inverse of gerund as noun to verb. "I can therapy you."

Edward Evans

@Avra This is a weird question

Semiclassical

@AMDG wiktionary does list 'therapy' as a verb, but only rarely

Edward Evans

@AMDG What's the German word?

Semiclassical

there is also therapize, but i can't say i've ever seen it used

AMDG

verbformen.com/conjugation/therapieren.htm

Edward Evans

02:16

eh, I've never seen that used

I'd just use behandeln

Semiclassical

So i'd say English does have verbs for therapy, but they're altogether uncommon

AMDG

/shrug

Edward Evans

I'd admit I've only seen the word therapise as an English verb, but that I've rarely seen used

AMDG

Well, it isn't like there's a committee for the <native> language lol

Edward Evans

Aye, not like in France

Semiclassical

02:19

for demonstration, compare the Google ngram viewers:
[therapy](https://books.google.com/ngrams/graph?content=therapy&year_start=1800&year_end=2019&corpus=26&smoothing=3&direct_url=t1%3B%2Ctherapy%3B%2Cc0) vs [therapize/therapise](https://books.google.com/ngrams/graph?content=therapise%2Ctherapize&year_start=1800&year_end=2019&corpus=26&smoothing=3&direct_url=t1%3B%2Ctherapise%3B%2Cc0%3B.t1%3B%2Ctherapize%3B%2Cc0)

Edward Evans

Weird that the correct spelling sees less usage

Semiclassical

hrm. i don't get why those embeddings didn't work

AMDG

I keep forgetting what a radix is

Semiclassical

i think American english tends to use -ize endings whereas British uses -ise

Edward Evans

Yeah sorry, I was being silly

Semiclassical

02:22

what i find interesting is the early blip of 'therapise' in the mid 1800's

AMDG

British tends to use French spellings. I'm not sure where z popped in for America.

Edward Evans

Would be interesting to see if the British committed some heinous war crime around that time

Semiclassical

there's actually a whole wikipedia article on this one issue

Oxford spelling

Oxford spelling (also Oxford English Dictionary spelling, Oxford style, or Oxford English spelling) is a spelling standard that prescribes the use of British spelling in combination with the suffix -ize in words like realize and organization, in contrast to use of -ise endings. Oxford spelling is used by many British-based academic/science journals (for example, Nature) and many international organizations (for example, the United Nations and its agencies). It is common for academic, formal, and technical writing for an international readership (see Usage). In digital documents, Oxford spelling...

with the logic apparently being that the British use of French spellings is actually the more modern one

displacing older use of -ize

Edward Evans

That's brilliant

Typical of the Brits to try and distinguish themselves from other organisations

AMDG

I don't see the point of innovating that. It's the same either way is not any clearer.

I think I'm starting to see numbers more properly in terms of what they're made of, especially after having contemplated more about division.

Semiclassical

02:26

oh, wait, i misunderstood that

leslie townes

it isn't in some american english dictionaries although a lot of online dictionaries have it in one form or the other.

i'm glad that this began with my need to therapize away my fear of geometry.

Semiclassical

the one oddity i ran into lately was "focussed"

which looks just wrong to my eyes

but it's regarded as a variant spelling in most places

leslie townes

a lot of double letters in English words fell off the boat on the trip across the Atlantic

while travel[l]ing, i should say.

Edward Evans

the double-s makes more sense to me, shortening the u

AMDG

¯_(ツ)_/¯

Semiclassical

02:28

yeah

and, say, programming vs. programing? first one > second for me

Edward Evans

yeah I agree there too

leslie townes

we kept the double m in that.

Semiclassical

even as we say program not programme

leslie townes

but removed it and the e from programme

jinx

AMDG

Based on my understanding of grammar, if a vowel is intended to be pronounced according to its modification by the presence of two or more vowels, then there should be no more than one consonant separating the two of them.

Edward Evans

02:30

Strange, we write Dsuinks instead of jinx

Semiclassical

the table they have comparing British, Oxford, Canadian, and American spelling conventions is neat

Edward Evans

Analogue vs analog

Xander Henderson

analoogie.

Edward Evans

Gross

Semiclassical

Only American uses "behavior" rather than "behaviour", and only British uses "organisation" rather than "organization".

AMDG

02:32

And then there's anagogy.

Xander Henderson

@AMDG That's not grammar. That's orthography / spelling.

AMDG

Yeah, that's what I meant.

Edward Evans

It's a common joke in the UK that people from the US are too stupid to spell things the "proper" way

The reality is that we need the extra letters to get the words past our bad teeth

Semiclassical

the weird one to me is aluminum vs aluminium

Xander Henderson

@EdwardEvans I think that Webster felt very much the same about the Brits some 300 years ago when he wrote his dictionary.

@Semiclassical But the words are pronounced completely differently in the US and the UK.

AMDG

02:34

The British never find me humorous.

Semiclassical

yes? that's why i spelt them differently

what's weird isn't that we say them differently, though

Edward Evans

humorous is actually correct in both orthographies

Xander Henderson

There are historic reasons for the differences. I'll see if I can dig up a reference.

AMDG

Well the joke is that they spell it humour.

Semiclassical

it's that both versions of the word came out almost simultaneously, and there's no evident reason why America got one vs. the other

AMDG

02:35

Vanity I guess idk

Semiclassical

The Oxford English Dictionary reports that in a lecture he delivered in 1809 and published in 1810, Davy does not use the term alumium, but refers only to good old alumina as alumine. By 1812, Davy had revised his coinage, opting instead for aluminum. But the previous year another scientist, in a review of another Davy lecture, had coined aluminium, with the nice -ium that was so familiar in potassium and sodium (which, incidentally, Davy had also coined).

Edward Evans

Nice

Xander Henderson

merriam-webster.com/words-at-play/aluminum-vs-aluminium

In any event, I need to sleep.

Later.

Semiclassical

yes, that's where i'm quoting from :>

night

Edward Evans

Good night

AMDG

02:37

Well I'd think if it were to be more like potassium and sodium, the stresses would be more like a-lu-mi'-ni-um.

Good night Xander

Semiclassical

my guess is that the usage probably reflects pre-existing usage patterns for other words

so that aluminum slotted into the American vocab whereas aluminium into the British

AMDG

Meh. It would seem to me that these things are just conventions the same the programming languages have conventions, some of them even idiomatic, or more properly called, idiotic.

Semiclassical

So to my mind the M-W article doesn't really answer the question. It says where both words came from, but it says nothing about why American English got one and British English got the other. (Especially because both usages arose within a year or so.) Hence, weird.

Edward Evans

Aye strange indeed

Seems like the British spellings were made to artifically preserve etymology

Semiclassical

right

whereas the American ones just reflect usage

Edward Evans

02:41

That ends up being a big culprit for the non-phonetic-ness of other languages

A good example is Faroese, where the official orthography was only actually written down in the 1800s some time and the orthography was completely based on old Norse etymology, making the language more or less entirely unphonetic

Only about 10 people speak it though, so who cares

Semiclassical

today's XKCD slipped by me at first. had to look at the alt-text to get it: xkcd.com/2509

leslie townes

i love that. xkcd is very hit or miss for me but that one hit.

Semiclassical

yeah, that one was delightfully subtle

leslie townes

03:02

reminds me a bit of cliff stoll's video on measuring the area of a piece of paper. youtube.com/watch?v=9yUZTTLpDtk

Euler 2

03:22

but sometimes xkcd deserves r/iamverysmart

Ted Shifrin

So many removals.

Akiva Weinberger

@Euler2 That one seems to be parodying r/iamverysmart-type people, no? Like, each person in that comic thinks they're smarter than the others, but they're actually all capable of thought

Euler 2

yeah

i don't like people who think they are very smart
what makes them think so

Semiclassical

03:40

if there's an element i don't like about it, it's the notion that everyone in the car feels like that

leslie townes

maybe the car is going to an xkcd fan convention

:)

Euler 2

socrates could be there

Semiclassical

there's certainly a proportion of the population that does think like that, mind

but that could be just as well conveyed by, say, having only three of the five thinking that

Euler 2

relevant

dc3rd

04:10

Somebody want to talk adjoints briefly with me? :)....I'll give you some candy....

leslie townes

04:30

what kind of adjoints? the word is a tad overloaded.

dc3rd

I don't even think it is even talking about the adjoint yet, this is the idea prior to mentioning the adjoint:

So is this theorem saying that for every linear transformation I can characterize it in terms of an innner product?

robjohn

what kind of candy? let's talk about the important stuff.

dc3rd

you should know better than to take candy from strangers.....

robjohn

I should know better than to discuss adjoints with strangers, too

leslie townes

more specifically, for any linear transformation from V into the field F, which a priori is just a linear map satisfying the axioms, you can grab a vector y in V and indeed write the linear transformation as "to compute the value of the transformation, just take the inner product of the input with the vector y"

dc3rd

04:35

well played...........this was a display of superior thinking...lol

robjohn

@dc3rd define it on a basis first

I'm an alien Im an eagle alien

adjoints are a social drug

leslie townes

i would prefer not

dc3rd

the proof of the theorem uses an orthonormal basis

leslie townes

i was instructed never to take adjoints

I'm an alien Im an eagle alien

04:37

the poof of the adjoint

robjohn

@leslietownes take them, but don't inhale

I'm an alien Im an eagle alien

roll up and smoke adjoints where invented by Grothendieck

leslie townes

using an orthonormal basis is one way to do it. there are coordinate-free routes but probably overkill in the finite dimensional context. e.g. they use topological notions

and depending on what F is maybe you're stuck with just algebra anyway

I'm an alien Im an eagle alien

We're computers in a giant computer making more computers

robjohn

@dc3rd (assuming an o.n. basis) for $w=\sum\limits_j V(e_j)\,e_j$, note that $V(x)=\langle w,x\rangle$ is satisfied by each $e_j$. By linearity, it is satisfied by all $x$.

dc3rd

04:41

baby steps first, baby steps first......finite dimensional and defining over a basis first. THen we can get to the abstract stuff.

I'm an alien Im an eagle alien

Just write $v_j e_j$ via Einstein notation

leslie townes

and arguably some of the topological proofs use g-om-t-ical notions, which are always best avoided.

I'm an alien Im an eagle alien

I'm coding a forex expert advisor now for MetaTrader5. I've encoded a moving average cross as an expression tree. The idea is to try random such expression trees and see if one does better than the other at making profit

This falls under genetic algorithms.

If anyone codes in MQL5 I can share the code with them. Having some trouble with it

There's a bug, where it doesn't even make one buy or sell over an entire year of backtest

Though the moving averages are visually crossing

Edward Evans

Ergh damn, I was just about to press enter to write to Ted

I'm an alien Im an eagle alien

You can still do it, they'll get an email

leslie townes

04:45

he's a reverse valdemort.

Edward Evans

Gave birth to lily and james?

leslie townes

possibly. i never actually read the books but know he's one of those villains who appears if you say his name too much, or something. ted will disappear if you say geometry.

Edward Evans

s'aink like that

dc3rd

so this vector $y$ or from robjohn's example $w$ is a chosen fixed vector for our specific linear transformation?

Edward Evans

Hi @Ted, I have a silly sheafy question when you're around

leslie townes

04:47

yes, each transformation V gets a single fixed vector associated with it.

or i guess your reference uses g for the transformation and V for the space. but it sounds like you get the point

dc3rd

Yes. I get the idea. Thanks for the elucidation (if that could be considered a word in that form, but I like it anyways)

Edward, just say something sacrlegious to him like "geometry doesn't exist without algebra" or "you let garlic fry in your pan before adding your other ingredients" and he will for sure return

Edward Evans

lol

dc3rd

maybe even saying Thor is always right might send him into a tizzy.....

Ted Shifrin

What, sir Edward?

Edward Evans

Hail!

Here goes: a sequence of sheaves $0 \to \mathcal{F} \to \mathcal{G}, \to \mathcal{H} \to 0$ is exact if and only if $0 \to \mathcal{F}(U) \to \mathcal{G}(U) \to \mathcal{H}(U)$ is exact and some other condition whose proof I understand

in the proof my prof shows that $\mathcal{F}(U) \to \mathcal{G}(U)$ is injective (this is easy from some diagram) and then goes on to start proving that $\mathcal{F} \to \mathcal{H}$ is the zero map and thus that $\mathcal{F}$ is contained in its kernel, and hence that $\mathcal{F}_x \cong \mathcal{H}_x$ for all $x$ (the stalks)

but I mean, this is vacuously true because the sequence is assumed to be exact, right?

oops, I mean $\mathcal{F}_x \cong \operatorname{Ker}_x$

Ted Shifrin

05:01

Huh?

Edward Evans

idfk

the formulation in the notes is really confusing, is it confusing to you too?

Ted Shifrin

Oh, you corrected the nonsense.

Edward Evans

right

Ted Shifrin

So you mean the kernel of the map $\mathcal G\to\mathcal H$. But are we doing this on $U$ or with stalks? No, that's not right, either.

What kernel?

And what are we actually trying to do?

Edward Evans

So we're assuming $0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$ is exact and we want to show that $0 \to \mathcal{F}(U) \to \mathcal{G}(U) \to \mathcal{H}(U)$ is exact

In the notes my prof writes "The injectivity of $\mathcal{F}(U) \to \mathcal{G}(U)$ follows from diagram. One shows analogously that the composition $\mathcal{F} \to \mathcal{H}$ is zero, so that $\mathcal{F} \subset \ker\{\mathcal{G} \to \mathcal{H}\}$."

Ted Shifrin

05:07

Seems like he's being sloppy with sheaves versus sections on $U$.

Edward Evans

It's very confusing lol. The next line he says "since taking stalks and taking kernels commute, the above inclusion shows that $\mathcal{F}_x \cong \ker\{\mathcal{G} \to \mathcal{H}\}_x$ for all $x$."

oops

Ted Shifrin

Sounds like a hot mess to me.

Edward Evans

Time to switch to a book then

You'll be pleased to know I'm taking complex geometry next semester :P

Ted Shifrin

I don't know what your prof is using for definitions, either.

Edward Evans

Sheaf exactness is exactness at the level of stalks, if that's what you mean

Ted Shifrin

05:11

But then you need to use the definition of stalks and the sheaf axiom.

Edward Evans

Right, he goes on to say "Exactness at $\mathcal{G}$ follows from the next lemma: let $\psi : \mathcal{F} \to \mathcal{K}$ be a sheaf injection with $\psi_x : \mathcal{F}_x \cong \mathcal{K}_x$ for all $x \in X$. Then $\mathcal{F} = \mathcal{K}$" and uses the sheaf axioms to prove this

kinda difficult to follow what he's doing

Ted Shifrin

I'm still confused about the overall logic here. But I don't have the time or patience to sort this all out now.

This is standard stuff in lots of books.

Edward Evans

Don't worry, I'll switch to some book and work it out

Thanks anyway

Ted Shifrin

You're not welcome.

Edward Evans

lmao

Ted Shifrin

05:14

Since I didn't do anything.

Edward Evans

You listened to me rant

You confirmed that it's a "hot mess", which is what I was looking for I guess

Ted Shifrin

Well, it's not that bad if you just look at a decent text.

Edward Evans

I know, I was just looking for confirmation that I'm not just being stupid and the formulation genuinely is confusing

love_sodam

06:47

0

Q: Classify the groups of order $20$

Classify the groups of order $20$ (there are five isomorphism types). $\newcommand{\semi}{\rtimes}\newcommand{\aut}{\operatorname{Aut}}$ Proof. If $G$ is abelian, $G\simeq\Bbb Z/20\simeq\Bbb Z/4\times\Bbb Z/5$ or $\Bbb Z/2\times\Bbb Z/2\times\Bbb Z/5$. Suppose $G$ is not abelian. Then by Sylow...

need a little help here

LearningCHelpMe

09:25

Did Stack change it's user profile interfaces or did I click on something that I shouldn't have? It seems I can no longer stalk people and find out when they were last online... Can I revert back?

mick

11:28

0

Q: $\sum_{n=1}^{oo} (-1)^{n+1}(ht(n)-ht(n-1)) =^? \frac{31}{60}$

Let $ht(n)$ be the n th harmonic triangular number : $$ht(n) = \sum_{i=1}^{t(n)} \frac{1}{i}$$ $$ht(0)=0$$ where $t(n)$ is the n the triangular number ( $t(1) = 1, t(2) = 3 , t(3) = 6,...$ ). Now consider $$T= \sum_{n=1}^{oo} (-1)^{n+1}(ht(n)-ht(n-1))$$ Thus $T = 1 - 1/2 - 1/3 + 1/4 + 1/5 + 1/6 ...

calculus sequences-and-series closed-form

Avra

12:49

@EdwardEvans. Thanks for the reply. Given $
r=\left( ak+b \right) mod\ p,\ s=\left( al+b \right) mod\ p,\ r\ne s
$,and $ a=\left( \left( r-s \right) \left( \left( k-l \right) ^{-1}mod\ p \right) \right) mod\ p,\ b=\left( r-ak \right) mod\ p$, then why why this is a one-to-one correspondence between pairs $(a,b)$ and $(r,s)$ given that $r \ne s$, $a\ne 0$ please?

robjohn

13:08

@Avra Never mind, I found the question

Avra

13:36

@robjohn. Thanks. How do you rate this question please from 1 to 10? Do you think it's easy, medium or hard?

Wolgwang

@LearningCHelpMe No

shintuku

14:09

say I have a square matrix $A$ in $\mathbb{R}^n$, and say $e_j$ is the $j$th basis vector and I want to designate the $j$th row vector of $A$. is the standard way to do this $(A^Te_j)^T$?

Avra

14:33

@robjohn. Can you please in your own words say what this question is asking for:

"How can we apply the
division method $h(k) = k \bmod{m} $to compute the hash value of the character string $r = r_0 r_1 \cdots $ without using more than a constant number of words of storage outside the string itself?"

PM 2Ring

14:50

@Avra It means that you need to create a string hashing algorithm based on mod. Your algorithm has to work on a string of any length. It has to use a fixed amount of RAM to do the calculation, it cannot use an array for temporary / partial values whose size depends on the length of the input string.

Avra

@PM2Ring. Thank you! Do you think wording is clear please?

PM 2Ring

@Avra It's clear to me, but I'm a native speaker of English, and I've been programming for decades.

They aren't asking you to write an excellent string hashing function. They just want you to write something that makes a reasonable hash.

That's pretty easy, and there's an example on Wikipedia if you get stuck.

Avra

@PM2Ring. Thanks appreciate it. I understand the solution, but it was only that part

Node.JS

I am wondering can someone help with this question: math.stackexchange.com/questions/4237866/…

Flows.

I never do geometry :/

Avra

15:02

@Node.JS. Is this a question from Linear Algebra please?

Flows.

Im confused by the definition of geodesic on wikipedia en.wikipedia.org/wiki/Geodesic#Metric_geometry

Avra

if,

$$
h\left( u \right) =\left( \sum_{i=0}^{n-1}{u_i2^{ip}} \right) mod\left( 2^p-1 \right)
$$
$$
h\left( v \right) =\left( \sum_{i=0}^{n-1}{v_i2^{ip}} \right) mod\left( 2^p-1 \right)
$$

How please by Euclidean algorithm, we can get that,

$$
0\le h\left( u \right) \le 2^p-1,\ 0\le h\left( v \right) \le 2^p-1
$$

PM 2Ring

15:19

@Avra I think you've left out some important context information.

Avra

@PM2Ring. What please?

PM 2Ring

@Flows. Which part is confusing?

Avra

It's given that $u$ is a sequence of numbers and $v$ is another sequence where both $u$ and $v$ differ by only 1 number.

So, if $u=<1,2,3,4>$, then $v=<2,1,3,4>$.

Pair of numbers are swapped between $u,v$.

PM 2Ring

@Avra Well, in programming, if we say r = b mod m then that normally means that $0 \le r < m$. But in mathematics if we say $r \equiv a \mod m$ then $r$ might not be restricted like that, all we know is that $m | (a - r)$

Avra

$u,v$ differ by interchanging any two indices from $v$ and then give that to $u$ as a result. $v=<1,2,3,4>$ then we interchange $(2,3)$ of $v$. $u=<1,3,2,4>$.

@PM2Ring. Great! Thanks. This is where I got confused. It says by Euclidian algorithm we have,

$$
0\le h\left( u \right) \le 2^p-1,\ 0\le h\left( v \right) \le 2^p-1
$$

why $2^p -1$ is included above please?

Is this a typo?

Why Euclidiean algorithm is there anyway please? I don't see what rule it plays

PM 2Ring

15:30

@Avra That seems a little strange to me. But it's not wrong, since any number $<2^p-1$ is also $\le 2^p-1$. Although of course the converse is not true. ;)

Avra

Then it proceeds by saying that,
$$
-\left( 2^p-1 \right) \le h\left( u \right) -h\left( v \right) \le 2^p-1
$$

@PM2Ring. By the function $h(k) = k ~mod~2^p-1$ will give less that $2^p-1$ and not equal to it

PM 2Ring

@Avra The Euclidean algorithm usually means the algorithm for determining the GCD of two numbers. However, Euclidean division just means finding $q, r$ that solves $a = q×m + r$, with $0 \le r < m$.

Avra

residue system of $h(k)$ is $[0, \cdots, 2^p-1]$

PM 2Ring

@Avra Of course. But, for example, every number less than nine is also less than ten. Right?

Avra

Ohhh! I got the Euclidean part, it does not add anything then?

@PM2Ring. Yes!

PM 2Ring

15:34

:)

Avra

Probably, it was added for convenience of proof

PM 2Ring

Probably

Avra

Finally please, how to get

$$
-\left( 2^p-1 \right) \le h\left( u \right) -h\left( v \right) \le 2^p-1
$$

It's same strategy

if it's greater than 0, then it's greater than $-(2^p-1)$ you mentioned with 9 and 10?

I see now! Since h=

$h(v)$ residue system is $[0, \cdots, 2^p-1]$...

we can get $h(u)=0$, so we get $-(2^p-1)$

@PM2Ring. Thank you!

PM 2Ring

@Avra Correct

You need to be smart to do good mathematics. But it's even more important to be careful. Make sure that what you're reading really says what you think it does. And when writing things, make sure you don't leave out anything important, or accidentally change something.

robjohn

@Avra note that $2^{ip}\equiv1\quad\left(\!\!\bmod 2^p-1\right)$

PM 2Ring

15:44

A lot of the time when you post problems in this room, you leave something out, or make some little error. And so it ends up taking much longer to solve your problem than it needs to. I'm not scolding you, I'm just trying to give you some helpful advice.

Avra

@PM2Ring. Thanks.

@PM2Ring. I will keep that in mind.

PM 2Ring

I can't imagine how hard it must be to do mathematics in a foreign language. It's hard enough doing it in my native language.

AMDG

Truly awful I'd guess.

PM 2Ring

@robjohn Oh, yeah. I was going to mention that earlier, but I got distracted...

robjohn

@Avra That $h$ is in those ranges seems to be by definition, though $a\bmod b$ is an integer in $[0,b)$ where $a\equiv x\pmod b$ talks about an equivalence class.

you seem to be mixing up these concepts in your questions

PM 2Ring

15:49

There's a discussion in the room archives about how in mathematics PhD programs it used to be standard to have requirements for multiple foreign languages, eg, French, German, and Russian. Some programs still do require at least some ability to read mathematics papers in one or more foreign languages.

robjohn

@PM2Ring I had to read passages from a French and a German math book.

nothing more than that, though. I had taken Latin in high school and German in college, so that got me through.

Avra

@PM2Ring. Honestly, modulous is the worst topic I studied in math

@PM2Ring. It confuses me

Similar to how implicit theorem in calculus confused me

robjohn

$\{1,4,7,10,13,\dots\}$ are all equivalent to $1$ mod $3$

PM 2Ring

At least with modulus you can play with small examples and see what's going on. With more abstract mathematics, that can be difficult, or simply impossible.

Avra

@PM2Ring. Not to mention, abstract math and functional analysis are nightmares for me

PM 2Ring

15:56

Modulus is very important. And it gets used a lot of use in coding. You've already seen that it gets used in hashing. It's also used in many encryption algorithms.

copper.hat

think of clocks

Sidvhid Hsinynjad

@robjohn Do you have a PhD in mathematics?

shintuku

we use modulus all the time when dealing with trigonometric functions @avra

copper.hat

i need concrete examples to learn

shintuku

since $\sin(x) = \sin(x+2\pi)$ @avra, and the same holds for all trig functions

PM 2Ring

16:01

I don't exactly need concrete examples, but I do find it harder to learn and absorb stuff without concrete examples.

AMDG

If you just graph modulus, you can see what it does pretty quickly and easily.

In terms of the number, imagine the number as a string with length equal in magnitude to the number. Consider that in $x\bmod m$, all values are reduced to the range $[0,m)$. Wrap the string around after traveling a length of $m$ along the string. Continue to do this until you reach the end of the string. Measure the length of the tip to the last wrap's crease or bend. That is the modulus.

Bruh I can't do braces

copper.hat

orthodontics?

AMDG

Unfortunately no

shintuku

suppose you see $x^2 - 6x + 2$. any fast ways to see the roots are $3 \pm \sqrt{7}$ or do even math people use the quadratic formula?

AMDG

I learned a method called "tictactoe" but I forgot how to do it.

shintuku

16:10

i'll look it up, thanks for the tip!

copper.hat

the quadratic formula is surely fast enough?

PM 2Ring

I almost never use the quadratic formula. If the factorization isn't obvious, I prefer to complete the square. I only use the formula if I'm trying to solve a quadratic where the coefficients are messy expressions rather than numbers.

Also, I like using simple rational approximations for square roots, to make quick mental estimates. Eg, using $8^2 -7\cdot 3^2=1$ we have $\sqrt 7\approx 8/3$, and using the Brahmagupta identity, it's easy to generate better approximations. Of course, we can also use Newton's method, which converges faster.

See en.wikipedia.org/wiki/Brahmagupta%27s_identity

shintuku

nifty, but how come you work with approximations? your answers don't depend on exact values?

PM 2Ring

16:30

@shintuku I like to know the value of numbers I'm playing with. ;) Sure, I could use a calculator, but doing mental arithmetic is good brain exercise.

Also, rational approximations can be useful when you're doing geometry on a grid of pixels

shintuku

cool!

btw, might anyone know why wolframalpha thinks this is true

$\int \frac{1}{x^2-6x+2} dx = \frac{1}{2\sqrt{7}}\int(\frac{1}{x-3-\sqrt{7}}-\frac{1}{x-3+\sqrt{7}})dx$

but thinks this is false:

$\int \frac{1}{x^2-6x+2} dx = \frac{1}{2\sqrt{7}}(\int\frac{1}{x-3-\sqrt{7}}dx-\int\frac{1}{x-3+\sqrt{7}}dx)$

?

copper.hat

16:55

indefinite integrals always seem like a recipe for ambiguity to me...

shintuku

hm, well $1/(x^2-6x+2)$ does go weird places between 0 and 8

robjohn

17:39

@shintuku as definite integrals on $(0,\infty)$ they are not equal, though you do need to take principal values all around.

As principal values on $(-\infty,\infty)$ all the integrals are $0$

shintuku

huh, up until now I thought the integral was distributive with respect to substraction

thanks a lot!

first time I hear about principal values too. time to do some analysis

robjohn

17:54

@PM2Ring umm... the quadratic formula is pretty much just completing the square.

@shintuku definite integrals are, but indefinite integrals have constants of integration to worry about.

That is why people get confused about $\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\sin^{-1}(x)+C_1=-\cos^{-1}(x)+C_2$ among many others.

shintuku

hm.. neither rudin nor my goto analysis book have anything about principal values, though i found Cauchy principal value on wikipedia

any clue where I could read up on it?

robjohn

Wikipedia and cauchy-principal-value are places to start

shintuku

oh swell, thanks!

Avra

@robjohn. You are superhuman by how many questions you follow and answer

robjohn

@shintuku $\mathrm{PV}\int_{-\infty}^\infty\frac{\mathrm{d}x}{x}=0$ The principal value can be used at a singularity and at infinity.

shintuku

18:03

I guess it makes sense if you suppose opposite infinities, which increase at exactly the same but opposite rate, are zero when added together

robjohn

@shintuku it is technically $\lim\limits_{\substack{\Lambda\to+\infty\\\epsilon\to0^+}}\left(\int_{-\Lambda}^{-\epsilon}\frac{\mathrm{d}x}{x}+\int_{\epsilon}^{\Lambda}\frac{\mathrm{d}x}{x}\right)$ and those integrals are negatives of each other.

shintuku

noted, thanks for the help!

PM 2Ring

@robjohn Of course. I just find it more satisfying to complete the square explicitly. ;)

PM 2Ring

18:23

A few hours ago, the network was vulnerable to XSS attacks for 44 minutes. Fortunately, almost nobody noticed. meta.stackoverflow.com/q/411177/4014959

Avra

Given two sequences,

$$
u=\sum_{i=0}^{n-1}{u_i2^{ip}},\ u_i\in \mathbb{Z},\ p\ is\ prime
$$
$$
v=\sum_{i=0}^{n-1}{v_i2^{ip}},\ v_i\in \mathbb{Z},\ p\ is\ prime
$$

If we interchange two positions in sequence $u$ at positions $k, l$, such that $k>l$. All other elements in the two sequences in $u$ and in $v$ are same except the one interchanged at $k, l$ of sequence $u$. Then if we subtract the two sequences, $u-v = (u_k2^{pk}+u_l2^{pl})-(v_k2^{pk}+v_l2^{pl})$. Any hint please here?

I know that since all other terms are similar, then they get cancellled from $u,v$, but how we got as result …

copper.hat

18:53

are you saying that $u_i = v_i$ for $i $ different to $k,l$?

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$Mathematics$ Mathematics