Mathematics - 2019-04-13 (page 1 of 2)

12:59 AM

Anyone know another way to prove the series $\sum_{n=0}^\infty \tan(1/n)$ diverges other than the LCT with $1/n$?

2:10 AM

Suppose G is group of order 30. Then G cannot have both six subgroups of
order 5 and 10 subgroups of order 3 (for then G would have more than
30 elements).

How to see that in above case group will have more than 30 element?

Leaky Nun

2:40 AM

@Jeff $\tan x$ is the argument of the complex number $\exp(ix)$, and $\sum_{n=0}^N \tan(1/n)$ is the argument of the complex number $\prod_{n=0}^N \exp(i/n) = \exp(i \sum_{n=0}^N 1/n)$

in a way the two series you mentioned are connected with each other

Karl Kronenfeld

3:17 AM

@Silent Any pair of distinct subgroups of order $5$ intersect trivially. So, e.g., six subgroups of order 5 account for $6\cdot 4$ nonidentity elements.

Can you continue this train of thought?

Silent

So, 10 subgroups of order 3 account for $10\cdot2$ nonidentity elements, because 3 prime and each order three subgroup cyclic,

But order 5 and order 3 subgroups can all be overlapping, right? So, how do we see that this creates problem?

Oh! because each order 5 subgroup's nonidentity element has order 5

and each nonidentity element of order 3 group is of order 3, this is what prevents from overlapping, right?

Thanks!!

Rithaniel

So, for the semidirect product $H\rtimes_{\phi}G$, do you need any special criteria for how you select $\phi$? Are there $\phi$ for which the semidirect product does not end up being a valid group?

Jeff

3:37 AM

@LeakyNun That is waaaay beyond me.

Leaky Nun

@Jeff complex numbers are fun!

Jeff

Mmmm.... yes. But how is $e^{ix}$ a "number" and what is it's "argument"?

Leaky Nun

@Rithaniel as long as $\phi$ is a homomorphism from G to Aut(H) then you're fine

@Jeff $e^{ix}$ can be defined as $\cos(x) + i \sin(x)$

its argument is the angle it makes on the complex plane

from the positive x-axis, counter-clockwise

Jeff

I get that.

Leaky Nun

you can prove that $e^{i(x+y)} = e^{ix} e^{iy}$

which basically summarizes the "cosines of sum" etc you learnt in high school

Jeff

3:39 AM

OK

you mean the trig identities for adding subtracting sines and cosines

Leaky Nun

the left hand side becomes $\cos(x+y) + i\sin(x+y)$

the right hand side becomes $(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))$

Rithaniel

Alright, then an associated question: If two groups are isomorphic, is the isomorphism between them unique? If I had $H\rtimes_{\phi}G$ and $G\cong\text{Aut}(H)$, with $\phi$ being an isomorphism, then is $H\rtimes_{\phi}G$ unique?

Leaky Nun

which you can FOIL to become $\cos(x)\cos(y) - \sin(x)\sin(y) + i[\sin(x) \cos(y) + \cos(x) \sin(y)]$

Jeff

@LeakyNun OK.

Leaky Nun

if you compare the coefficients you get $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$ and $\sin(x+y) = \sin(x)\cos(y) + \cos(x) \sin(y)$

@Rithaniel if G and H are isomorphic then there are as many isomorphisms as Aut(G)

Jeff

3:43 AM

@LeakyNun So far so good.

Leaky Nun

and the argument of $e^{ix} = \cos(x) + i \sin(x)$ is $\sin(x)/\cos(x)$

which is $\tan(x)$

Rithaniel

It's not $G$ and $H$ which are isomorphic. I'm thinking of if $G$ and $\text{Aut}(H)$ are isomorphic.

Jeff

That last part loses me.

Leaky Nun

draw a picture! :P

Jeff

maybe i don't understand the term "argument"

Leaky Nun

3:45 AM

argument is the angle it makes from the positive real axis, counter-clockwise

Rithaniel

Imagine a line segment drawn from the origin to the point that the complex number occupies. Argument refers to the angle that line segment makes with the positive real line.

Jeff

@LeakyNun (my gosh you're quick, I've only barely googled the term)

Semiclassical

in physics you tend to use the word "phase" instead of "argument"

Rithaniel

Leaky is indeed quick.

Silent

I am having hard time digesting that suppose $X$ and $Y$ homeomorphic metric space, and still one of them may be complete and other may not be. (e.g., $\Bbb R$ and $(0,1)$.)

I think that this is because, although homeomorphism maps each open set of $X$ to that of $Y$ bijectively, it does not say that 'small' open set won't be mapped to 'large' open set. Am I looking at it right?

Leaky Nun

3:47 AM

@Silent right

"complete" isn't a topological property

it's a metric space property

Silent

thank you!

Jeff

OK, so the argument is the angle. But $\tan(x)$ is the slope. Surely we can get the angle from the slope. But i would use the $\arctan$ for that. leaving $x$, which is the argument.
i'm confused

Leaky Nun

oh wait...

someone help me

Semiclassical

What exactly is the confusion?

Leaky Nun

I think I said something nonsensical

way before this

I was trying to connect the divergence of $\sum \tan(1/n)$ with the divergence of $\sum 1/n$ by using complex numbers

Jeff

3:51 AM

well don't panic, i took a screenshot so it can live into posterity! :D

Leaky Nun

I'll just hide in a hole

Jeff

hehe... imagine how I feel sometimes.

Semiclassical

seems like the most obvious connection is via the limit comparison test

Jeff

Today, a student asked me if there was another way to prove $\sum \tan(1/n)$ diverges other than comparing it to $1/n$ using the LCT (scroll up if you're wondering how that went).

Semiclassical

haah

hmm

Jeff

3:54 AM

@Semiclassical You must have come in after this: chat.stackexchange.com/transcript/message/49909065#49909065 :D

Semiclassical

yeah, totally didn't see that

Jeff

@LeakyNun Does this invalidate the whole discussion? I only ask because it was kind of interesting.

Leaky Nun

@Semiclassical help me

Semiclassical

i got nothing

Rithaniel

You can probably use the root test for that particular sum. All you have to do is show that the limit approaches 1 strictly from above.

(That is $\text{lim}_{n\rightarrow\infty}\sqrt[n]{tan(1/n)}$, if it wasn't clear.)

Jeff

4:04 AM

that limit's probably not true. $\tan(1/n)$ is approaching zero, and the $n$th root of that is approaching either 0 or 1, but if it's approaching 1, it's probably from below.

WolframAlpha says the limit is 1

Rithaniel

Yeah, plugging in a few numbers it does appear to be approaching from below.

(Though, this is just approximation. Trying successive powers of 10)

Jeff

Oh well. Interesting thought though: Is it true that if the $n$th root limit = $1^-$ that it still passes?

Rithaniel

Nah, in that case the root test is inconclusive.

Well, as far as we know

Jeff

I mean equals $1^+$.

Rithaniel

So, you mean it equals 1 and is approaching that limit from above?

Jeff

4:10 AM

yes

which is what you said

Rithaniel

In that case it's my understanding that the root test implies the sum diverges.

If the root test gives a limit less that one, the sum converges absolutely. If it's greater than one or equal to 1 and approaching from above, the sum diverges. Otherwise the test is inconclusive.

Jeff

gotcha.

Isiah Meadows

Is there a term for directed graphs where all cycles have length at most 1 (looping back directly to their start node)?

Jeff

thanks for thinking about this guys (and gals). I'm gonna go to bed now.

Isiah Meadows

Or to put it another way, a graph that's effectively a DAG + possible single-edge loops.

Rithaniel

4:17 AM

I'm not too familiar with graph theory, but perhaps what you are looking for is a case of a direct acyclic graph?

Isiah Meadows

No, a specific superset of directed acyclic graphs.

Just wondering if there's a term for them.

Rithaniel

Not that I know of, but perhaps you could describe it as "a graph whose strongly connected components only consist of single points"

Isiah Meadows

Okay. (I'm even less familiar with graph theory - I'm a web dev who happens to have a significant partially-formal background in CS, so I'm certainly not the traditional math person.)

Rithaniel

Well, all the stuff I'm giving you is from a very shallow wiki dive. I can link the pages, if you'd like to read these things for yourself.

Isiah Meadows

Sure.

I've basically been doing the same just now myself. ;-)

Rithaniel

4:23 AM

https://en.wikipedia.org/wiki/Strongly_connected_component
https://en.wikipedia.org/wiki/Directed_acyclic_graph

Isiah Meadows

Just a little less shallow, but I know a lot less about graph theory (and so it's harder to really know where to look).

Rithaniel

There you go, the two pages where I got the terminology from.

Isiah Meadows

I was aware of DAGs (they come up surprisingly frequently in the context of regexps, code optimization, state machines, and similar), but I had not even heard of "strongly connected component" until now.

My main reason for looking at it is I was also curious if there was anything about how that particular set of graphs relates to the subgraph isomorphism problem (as in, if there's a special case hidden in there somewhere).

Mainly for programming language type-oriented applications.

Rithaniel

Well, I don't even know what the subgraph isomorphism problem is, so that's another thing for me to google.

Isiah Meadows

Great...(I also only discovered that today, even though I knew the related graph isomorphism problem subset for much longer).

Subgraph isomorphism problem

In theoretical computer science, the subgraph isomorphism problem is a computational task in which two graphs G and H are given as input, and one must determine whether G contains a subgraph that is isomorphic to H. Subgraph isomorphism is a generalization of both the maximum clique problem and the problem of testing whether a graph contains a Hamiltonian cycle, and is therefore NP-complete. However certain other cases of subgraph isomorphism may be solved in polynomial time.Sometimes the name subgraph matching is also used for the same problem. This name puts emphasis on finding such a subgraph...

There you go

Save you a search. :-)

Rithaniel

4:29 AM

Already searched it, but thank you.

Isiah Meadows

No problem.

I was just dropping in to see if there was anything interesting (and useful) known about that subset. I guess I can post a question on the main site about this, in hopes someone much more well-versed in graph theory than either of us can answer.

Rithaniel

A fairly simple statement, but if finding isomorphisms between graphs is anything like finding isomorphisms between groups or rings, I doubt there's going to be a general solution.

Also, definitely. There is a better chance to find someone with the expertise you're looking for one the main site.

Isiah Meadows

Graph isomorphism is slightly simpler than that, but only slightly. It's relatively straightforward to code for directed acyclic graphs (and it's also reasonably fast), which is the primary subset I usually care about in practice. This just happens to step just past where my knowledge ends on graph theory.

user21820

@IsiahMeadows Not that I'm aware of, and it seems such a concept is not very useful?

Isiah Meadows

4:49 AM

I also took a step back and realized my actual problem is that of a bipartite graph, but a different special case of them (where specific clusters of 2^16 nodes are connected to all others in that particular cluster with edges, but no other cycles exist).

Really, where I'm coming from is trying to figure out what graph arises (indirectly via the recognizing FSM) from a regular language where the max star height is 1 and the Kleene star can only be applied to the union of all characters in the alphabet. Using the syntax of most regular expressions, it'd be where you can only do the empty pr…

Balarka Sen

There seems to be a polynomial time algorithm for the graph isomorphism problem for graphs of bounded genus. I'd like to understand why, though.

@IsiahMeadows I'd expect that there are polynomial time algorithms in general if the graph doesn't have "too many cycles". This is analogous to eg saying that the word problem is solvable in certain small cancellation groups.

Precisely because if, say, I have a C'(1/6)-cancellation group (basically it has a finite presentation with certain technical conditions on the size of how much things cancellation if I merge two relators), then it's Cayley graph doesn't have too many cycles.

Isiah Meadows

@BalarkaSen By any chance, could you explain what "genus" means in the context of graph theory? Wikipedia explains it in terms of spheres, but that doesn't really make sense to me coming from a more directed background (formal language processing, state machines, etc.).

I only learned a small subset of graph theory, just enough to find how it relates to automata theory and tree processing.

(I maintain a front-end virtual DOM framework, so I had to learn some about sequences, trees, and related, but I hadn't yet needed to learn much more generalized than that like graphs.)

Balarka Sen

So basically it's a measure of how nonplanar the graph is. Planar graphs have "genus 0". If you have for example, the K_{3,3} graph, it's nonplanar, but if you wanted to draw it on a coffee mug without the edges intersecting each other you'd be able to do it.

The coffee mug has "one handle", so the genus of K_{3,3} is 1.

In general if you can draw your graph with edges not intersecting each other in a coffee mug with "g handles" but not on a coffee mug with "g-1 handles" then the genus of the graph is g.

Rithaniel

So genus counts the number of holes in a torus, in an indirect sense?

Balarka Sen

Ya, it's the minimum g such that the graph can be embedded on a torus with g holes (a surface of genus g)

Isiah Meadows

5:02 AM

So with my case, would I need 2^16 - 1 holes (where it's 2^16 nodes all interconnected with edges)? Or is my intuition here incorrect?

Balarka Sen

your graph is a complete graph on 2^16 vertices?

I looked up the formula, and apparently the complete graph on n vertices has genus = the smallest integer larger than (n-3)(n-4)/12. So much less than what you anticipate.

Isiah Meadows

If you look here, I have clusters of that size, namely U+0000 to U+FFFF.

Yeah, I just went by random guess - I didn't actually look up a formula for that (graph genus).

Rithaniel

So, this stuff can be associated to topology. That makes sense, but it is always cool to see how two fields of math interconnect with each other. (On a related note, I want to take some algebraic topology in the fall. Do you have any recommended books on that topic, Balarka?)

Balarka Sen

@Rithaniel Hatcher is the canonical reference. Munkres has a famous point set topology book which contains a fare share of introductory algebraic topology

@IsiahMeadows Er, I meant much more than what you anticipate. The genus of K_n is a quadratic in n.

Rithaniel

I've been through a course in point-set, so while some review would probably be good, I probably shouldn't go for a book specializing in it. Also, Ted recommended Hatcher before. The only issue there is that I'm setting up a reading course with a friend, and he is dubious of Hatcher's book after reading some negative reviews of it.

Isiah Meadows

5:09 AM

@BalarkaSen Ooh, nice.

Balarka Sen

Hatcher is very good, I don't know why people hate it so much.

Isiah Meadows

I did see that much on Wikipedia when looking at the graph isomorphism problem.

But my main curiosity is around the subgraph isomorphism problem, specifically the special case where the subgraphs start from the same root node as the larger graph itself.

Rithaniel

Supposedly the hate is coming from people who want more symbolic explanations of the math? I can't say for sure because I've not read the book.

Isiah Meadows

That's my other big curiosity.

Balarka Sen

@Rithaniel Yeah, I don't think it's a wise complaint.

But depends on the reader's taste of course

Rithaniel

5:11 AM

Me neither. Though, if this is the source of the complaints, then my friend's reaction makes even less sense, because he wanted a less theory-heavy text.

Balarka Sen

Strange!

Skies burn

hatcher is for nerds

Rithaniel

I think he's being reactionary, myself.

Isiah Meadows

Okay, off my topic: people are complaining that a math book is too symbolic? That's a very weird complaint about a higher-level math book.

Rithaniel

He saw negative reviews, so he's thinking of it in a negative light.

Isiah Meadows

5:12 AM

Confirmation bias, maybe?

Rithaniel

Nah, that the book is not symbolic enough.

The book relies a lot on geometric interpretations, as far as I understand it.

Balarka Sen

Ask him to give it a try at least. If it suits, all's fine. If not, you can look for other books

Isiah Meadows

Oh, then never mind. I wasn't fully following this conversation.

@Rithaniel

Rithaniel

I'll run it by him again.

Daniel ML

P-adic Numbers the geometric representattion is in fractals?

Skies burn

5:15 AM

@DanielML What do you mean by geometric representation?

Daniel ML

Why they cannot be represented in the real líne

Skies burn

How?

Real numbers are one equivalence relation on cauchy sequences, p-adic numbers are another

Isiah Meadows

@BalarkaSen Oh, and many thanks for the help here. Very deeply appreciated. Here's the underlying context for that question: github.com/Microsoft/TypeScript/issues/…

Rithaniel

You're gonna have a difficult time representing p-adic numbers geometrically at all, I think. In that metric, imagine you have a circle. It's fairly easy to show that every point not on the circle is the center of the circle.

Skies burn

Depends what one means by geometrically, and whether one means the elements, or the p-adic numbers themselves

Isiah Meadows

5:18 AM

Gotta love how often category theory just finds ways to invade programming language design and how often graph theory just finds ways to invade programming language implementation...

Rithaniel

Similarly, given any three points, the triangle with those points as it's vertices is always isosceles.

Well, distinct points, at least.

Balarka Sen

@Isiah I hardly helped; I know very little graph theory.

Isiah Meadows

@BalarkaSen You know a lot more than I do, though.

Balarka Sen

Just saw an interesting conversation happening so piped in

Isiah Meadows

I know virtually nothing outside my little niche.

That being mostly automata theory, computability, and the like, with a hint of DAGs at the worst I usually end up having to deal with.

Balarka Sen

5:21 AM

I know nothing about those, on the other hand :)

Isiah Meadows

Graph theory is something I rarely touch, and when I do, I don't generally care about equality. It's more often just graph normalization and, on rare occasion, rewriting.

I come from a web dev background and just picked up a bunch of CS and related math out of necessity.

Balarka Sen

Gotcha. Pretty cool stuff though

Daniel ML

Yes a friend told me that one can see p adic numbers as a fractals you have the triangle and inside you have circles and another circle and another but i dont understnd that

user131753

Let me share one of my experiences regarding Hatcher @BalarkaSen. One of my friend who is onto logic had to take an Algebraic Topology course. His teacher suggested Hatcher as a good book on Algebraic Topology. His review was very much close to what @Rithaniel said. It is not enough symbolic and detailed. It is very much weird that I have never found any logician (with whom I have met, of course), who liked Hatcher and suggested it as a good book for Algebraic Topology.

Isiah Meadows

I maintain a front-end web framework that takes trees and tries to patch the DOM using as few as pragmatically possible, so I necessarily have to care about trees, lists, and the occasional sequence bit (my framework uses longest increasing subsequences for patching aware of identity), but graphs are a bit beyond my usual stuff.

Rithaniel

5:29 AM

Perhaps it would be good to acquire two books. Hatcher being one and a more symbol-heavy book to supplement Hatcher.

user131753

@Rithaniel Like?

Rithaniel

I don't know, yet. These are plans for the fall semester, so I have some time ahead of me to work things out, but it's good to have a plan.

Perhaps Rotman?

user131753

I was thinking about Tammo tom Dieck.

Rithaniel

Oh, haven't heard of that book before.

user131753

@Rithaniel This one.

Rithaniel

5:34 AM

Danke schon.

Was just getting ready to google it, myself

user131753

@Rithaniel Bitte schön.

Alessandro Jacopson

7:01 AM

Hello, do you know of a technical bookshop in London? some second hand bookshop

I am interested in computer science related books or math/stats books, in the Central London area.

Secret

8:48 AM

[Random]

Towards The Ambigurity: Absolutely undefinable numbers, definable numbers using indeterministic procedures and quagmire space

Recall a number $a$ is undefinable in a language $\mathcal{L}$ if there is no formula $\phi$ in $\mathcal{L}$ such that $\phi (a)$ is true

Normally, however, $\mathcal{L}$ can be extended into some $\mathcal{L^+}$ such that $a$ becomes definable, the simplest being to include $a$ as a constant in $\mathcal{L}$

Hence, a number $a$ is absolutely undefinable in $\mathcal{L}$ if there exists no extension $\mathcal{L}^+$ such that there exists a formula $\phi$ such that $\phi (a)$ is true.

Thus such $a$ cannot be a real number because there exists models of set theory such that all reals are definable

One possible way to make such $a$ will be to ensure that any attempt to include $a$ into the language $\mathcal{L}$ will trigger a contradiction or similarly undesirable phenomenon

For example, $a$ could be something such that $\phi(a)$ satisfy the Inclosure schema, despite $\phi (x)$ for anything $x \neq a$ is well defined

Now, onto the next item, quoting from: rudyrucker.com/infinityandthemind/#calibre_link-294

> The point of the mind recipe is that we specify a complex machine M with a name of the form, say, “Seed your randomizer with the number 1946, create an initial population of a million random programs, and evolve the population through a billion generations, using the mutation and crossover genetic operators, and gauging fitness according to the following specific tests: answer the enclosed quizzes on the following books and movies, do better than the other programs at
the following list of games, score well in the programs’ mutual rankings of each other, etc.” And this name can be transpa…

A number $b$ is indeterministically definable if there exists some indeterministic procedure $P$ such that $P(b)$ returns true as one of its outcomes.

The details of generating $b$ however cannot be written down because of the indeterministic nature of $P$, however it is a subset of definable uncomputable numbers similar to how the Chaitin's constant, despite no procedure can exists to give its digits deterministically, it can be defined as "the probability that a randomly constructed program will halt"

With these defined, it is time to move to the last item:

Inspired from the April Fools arxiv about the Marshland Conjecture

Secret

9:23 AM

A space $(Q,d(\cdot,\cdot))$ with semimetric $d(\cdot,\cdot)$ is called a quagmire space if for all $x,y$, $d(x,y)$ is an indeterministically definable real number

Intuitively, such space have a desolated notion of distance between points such that the distance between any two points (except itself) is uncomputable, and cannot be defined using any deterministic procedure

This result in that as you step outside of the point you are standing, you can potentially end up anywhere in the space but you cannot knew in advance where you ended up

suffice to say, its topology is also pretty indescribable in the algorithmic sense

In the future, we will explore how much structure is still present in $(Q,d(\cdot,\cdot))$ despite the disordered nature of the semimetric in this space

yellow_flash

10:09 AM

I'm having some difficulty wrapping my head around the concept of units in ring theory. One such troubling instance is - In a ring $R$, if $p$ is a prime such that $p = am$, where $p, a, m \in R$ and $a$ is a unit, then $(p) = (m)$ (the ideals generated are equal). How do I go about reasoning this ?
Thanks in advance!

Alessandro Codenotti

Think about $\Bbb Z$, why would the ideals generated by $3$ and $-3$ be different?

yellow_flash

they should be the same
This is the main problem which I am facing - I feel that units in ring theory are somehow correlated to 1 and -1 (in $\Bbb Z$) ; there are also statements like - factorization in UFD are unique upto units

I'm not able to draw the connection b/w the definition of units (existence of multiplicative inverse) and the statements above

Alessandro Codenotti

Sure, $1$ and $-1$ are the units of $\Bbb Z$

The point is that $a\mid b$ iff $\epsilon a\mid b$ when $\epsilon$ is a unit, they are irrelevant for divisibility

RaphX

10:27 AM

Hi everyone I need a help regarding a PnC problem

There are 'X' boxes containing notes of denominations 10, 20, 50, 100 and 500 in infinite number. Exactly 9 notes are randomly taken from each of the boxes. What is the minimum possible number of boxes needed to make sure that atleast 8 notes of the same denomination are taken in out in total?

RaphX

10:38 AM

Can anyone help please?

ÍgjøgnumMeg

12:14 PM

Really nice little A-Level proof question: Prove there are no $m, n \in \Bbb N$ such that $m^2 - n^2 = 6$.

Chebotarev Density

lol just factor

Rithaniel

Well, we know that $m^2=(n+1)^2=n^2+2n+1$, and so it $m\neq n+1$ as 6 is not odd. Next consider $m^2=(n+2)^2=n^2+4n+4$ then $m^2-n^2=4n+4$ Which cannot equal 6 if $n\in\mathbb{N}$. Any larger difference between $m$ and $n$ will just increase the constant term.

Probably could do it a little more cleverly with quadratic reciprocity, though.

ÍgjøgnumMeg

:) It's nice for an A-Level question to have a couple of steps, often there is one step and it's quite transparent

Rithaniel

Yeah, you've gotta identify the cases you need to check, that's true.

It is a pretty neat question.

vesii

12:36 PM

Hi guys!, is it true to say?: ${n \choose \frac{n}{2}}=\frac{n!}{(0.5n)!^{2}}=\frac{n!}{[(0.5\cdot1)\cdot...\cdot(0.5\cdot n)]^{2}}=\frac{n!}{0.5^{2n}\cdot n!^{2}}=\frac{4^{n}}{n!}$

user21820

@Rithaniel Uhh... why bother with high-powered tools? Many simple tools suffice: (1) Sufficiently large consecutive squares have difference more than 6. (2) Mod 4. (3) parity of m+n = parity of m−n.

Rithaniel

Well, check a couple of numbers. $\binom{10}{5}=252$. Meanwhile $\frac{4^{10}}{10!}=\frac{4096}{14175}$, so you've probably made a mistake somewhere.

(Consider whether you can really expand a factorial in the way you're thinking)

@user21820 Because the high-powered tools are cool, man. (You're entirely right, though)

user21820

@Rithaniel Hahaha. Let's see what we can whack it with... =)

ÍgjøgnumMeg

1:00 PM

it's a pell-like equation :P

user21820

1:21 PM

@ÍgjøgnumMeg Yea I thought of that too, but didn't get any nice idea.

Leaky Nun

1:38 PM

Supreme Court Revokes Annoying Man's Free Speech Rights (Season 1 Ep: 3 on IFC)

►

anakhro

2:31 PM

@LeakyNun what are you up to today?

Leaky Nun

@anakhro thinking about why sample variance of normally distributed random variables follows a chi-squared distribution with one less degree of freedom :P

anakhro

I love the idea of statistics.

But never got to friendly with it.

A great little book is Rosenthal's little intro to rigorous probability theory.

And I read little bits and pieces.

But I should go through it formally.

anakhro

2:57 PM

I need to pick some $\varepsilon > 0$ so that $\|(x,y) - (p,q)\| < \varepsilon$ implies $\|(x,y)\| > \|(p,q)\|/C$, for a fixed $C\in\mathbb R$ and $(p,q)\in\mathbb R^2$.

RockDock

while reading proof of that x^4-y^4=z^2 has no solution - let us assume it admits a solution x1,y1,z1 where we have chosen x1 to be least positive integer .Then x1 must be odd . why ?

Joe Shmo

friends, i want to clarify something -- Peter Lax (I forget on what page) writes that $\langle \cdot \rangle,\ \| \cdot \|$ are independent of the choice of basis as geometric quantities. I buy his proof over there. However, there's a subtlety --

mm.. maybe not

thanks y'all!

Rubber duck debugging

In software engineering, rubber duck debugging is a method of debugging code. The name is a reference to a story in the book The Pragmatic Programmer in which a programmer would carry around a rubber duck and debug their code by forcing themselves to explain it, line-by-line, to the duck. Many other terms exist for this technique, often involving different inanimate objects. Many programmers have had the experience of explaining a problem to someone else, possibly even to someone who knows nothing about programming, and then hitting upon the solution in the process of explaining the problem. In...

ÍgjøgnumMeg

@RockDock I guess if $x$ is even then all of them are so you can remove a factor of $2$

and then it wouldn't be a minimal solution

James Groon

4:04 PM

How would you solve cos(z)=cosh(z) without using sum-to-product identity for cosine in cos(z)-cos(iz)=0 ? Since that formula is hard to remember and also I believe it is hard to derive it.

anakhro

4:14 PM

@JamesGroon do you mean you don't want to be able to use $\cos(iz) = \cosh(z)$?

Or what are you not wanting to use?

James Groon

@anakhro I want to use that

I don't want to use trigonometric identity for cos(x)-cos(y)

anakhro

So if we set $z = x+iy$, what do we get?

James Groon

I've tried that way also

I get some wierd equalities involving exp, cosine and sine

Which I cannot solve

Have you solved it and trying to give me a hint ? @anakhro

anakhro

Well what do you get for solutions for $\cos(z) = \cos(iz)$?

James Groon

If I used that unwanted trigonometric identity I get : z=(1+i)$\pi$k, z=(1-i)$\pi$k, k$\in\mathbb{Z}$

which is correct according to WA

anakhro

4:22 PM

Which trig identity are you concerned with?

James Groon

I wouldn't say concered , I simply don't want to use it because I know I will forget it eventually.
It goes like: cos(x)-cos(y)=-2sin((x-y)/2)sin((x+y)/2) .

I'm curious if there is another way

anakhro

Well what do you do when you solve for $\cos(z) = \cos(iz)$

James Groon

I use the definitions of cos(z) for complex z , then use the definition for exp(z) for complex z

After which I equate real part of the left side with the right side , same with the complex part

I would say all standard stuff

anakhro

Well $z=(1+i)\pi n, n\in\mathbb Z$ is easy to get without doing anything hard.

James Groon

hmm that's weird, what about the other set of solutions ?

btw thank you for your time

anakhro

4:32 PM

I think I am missing something.

if $z = iz + 2\pi n$, then you get those solutions.

That's the easy part.

James Groon

how do you get that ?

anakhro

Well certainly cos(z) = cos(iz) if z = iz + 2pi n

Periodic in 2pi

James Groon

just like that ? woah

anakhro

I feel like I am missing something else obvious that makes it easy to get the other one.

Something with conjugates?

James Groon

I seriously don't know

Till now I didn't know sine and cosine are periodic

I haven't known*

anakhro

4:37 PM

They are periodic because of e^z being periodic.

James Groon

shouldn't the period be 2$\pi$i then ?

anakhro

$\cos(z) = (1/2)(e^{iz} + e^{-iz})$, so $\cos(2\pi) = (1/2)(e^{2\pi i} + e^{-2\pi i})$

Si?

James Groon

yup I see now

anakhro

Well half your problem is solved.

I don't know what I am missing for the other half. :(

James Groon

Maybe that's the most you can get out of the peridiocity

anakhro

4:49 PM

Well it means that I am missing something about the naive solution to cos(z) = cos(iz)

It's not just z = iz + 2pi n

OH

@JamesGroon

James Groon

yah

anakhro

z = 2pi n + i z

and z = 2pi n - iz

The first yields z = pi n(1-i). The second yields pi n(1+i)

James Groon

you got it from the fact that cosine is even function ?

anakhro

Yes.

James Groon

how would you prove there aren't any more solutions ?

anakhro

4:56 PM

It should be an iff for the cos(z) = cos(iz)

So cos(z) = cos(iz) iff $z = 2\pi n \pm iz, n\in\mathbb Z$

James Groon

and how do we know that ?

anakhro

Probably will need some trig identity for that.

Like yours for cos(x) - cos(y). :P

James Groon

it's not trivial?

aah

prolly yeah

anakhro

"trivial" is a very subjective thing.

To most working mathematicians, it's probably "trivial" on the basis that they know that cos is 2pi periodic and even.

James Groon

regardless on that basis , if you need a trig identity to prove it , it certainly is not trivial for anybody

anakhro

5:04 PM

Heh.

James Groon

It could be easy , but trivial? Don't think so.

so in conclusion , cos(z)=cos(w) iff z=2$\pi$n (+-) w , n$\in_mathbb(Z)$

anakhro

Yes.

So this is easy to remember, and helpful.

James Groon

I could agree with that

Are you sure though ?

anakhro

Positive.

James Groon

Thank you very much :=)

anakhro

5:07 PM

I agree that relying on obscure trig identities is equivalent to memorizing a bunch of tricks.

However, trig identities aren't all that bad.

And you slowly learn them well.

Ultradark

How would I take the first step to solve this pde? $$ \frac{\partial}{\partial s} \phi(x,s)\frac{\partial}{\partial t}\phi(x,t)=\frac{\partial}{\partial I}\phi(x,I)$$

James Groon

My "trick" was to remember identities sin(x+-y) and cos(x+-y) and then most of the other identities are fairly easy to derive .

anakhro

Yeah, those two are quite fundamental.

James Groon

But regarding my problem , I believe the identity for cosx-cosy is hard to derive from those fundamental ones

Lemme try it now for fun

Nah I can't do it now since I've remembered the actual identity haha

But it is hard imo

Ted Shifrin

5:27 PM

@James: You have to "know" the idea. Write $x=u+v$ and $y=u-v$ for the appropriate $u$ and $v$. So you were on the right track.

James Groon

5:39 PM

@TedShifrin where exactly was I on the right track ?

in proving trig identity or the problem I asked at the beginning ?

Ted Shifrin

I thought you were going to use the usual addition/subtraction formulas to derive $\cos x-\cos y$.

Ajay Mishra

What does it mean that an equation has complex roots? I mean, Something corresponding to the meaning of the fact that, in real system of equation the roots correspond to value of abscissa of intersection of graph of function and the line y =0?

Ted Shifrin

@Ajay: I don't understand your question. What do you do with the equation $x^2+1=0$?

Ajay Mishra

I solve it algebraically, and solutions are $i$, $-i$. But what does it meant?

Ted Shifrin

There is no real picture for what it means. If you think of $z$ and $w$ as complex numbers and graph $w=z^2+1$, then the graph crosses $w=0$ at $z=\pm i$.

I have no idea what you're looking for.

Ajay Mishra

5:47 PM

I dont see it crossing. haha

Ted Shifrin

You don't know how to draw the complex graph. That requires more real dimensions than we have.

Ajay Mishra

Actually, there is a problem in my text, which say the roots of the given cubic equation constitutes equilateral triangle, I cant figure out what it meant?

Ted Shifrin

They're graphing the points in the complex plane.

The equation is something like $z^3=1$ or $z^3=8$?

You need to draw pictures in the $xy$-plane with $z=x+iy$.

Ajay Mishra

Okay, How one verifies that this system of equation is giving the right answers?

Without using the same system to verify.

Ted Shifrin

There is no other way to make sense of what you're asking.

Ajay Mishra

5:52 PM

how do I verify that $z^3 = 1 $ has solution $\omega , \omega^2 $?