Mathematics - 2022-05-12 (page 2 of 3)

copper.hat

6:00 AM

dava sobel had a nice paperback on said subject

Akiva Weinberger

Have you read the book Longitude? Great book on the subject

Dammit, beat me to it

copper.hat

@PM2Ring somehow mathematics and mathematics educators are becoming separated

Akiva Weinberger

I read a different book which suggested that (a) there were no technical barriers to inventing radios, only knowledge barriers and (b) radio could have solved the longitude problem

so if you ever find yourself with a time machine, go back to 1600s and teach them how to build a transmission tower

copper.hat

@AkivaWeinberger i have a student sextant which i play with from time to time for giggles

Akiva Weinberger

(pun)

PM 2Ring

6:02 AM

Newton realised that you could calculate longitude from the Moon, if you had good observations, and good Moon tables. But it's not easy to get the required precise observations at sea, and calculating good Moon tables is really hard.

Akiva Weinberger

I have no idea how to use a sextant

Yeah, the book Copper mentioned is basically a race between Mr. Build Better Watches and Mr. Get Better Moon Tables

copper.hat

its not hard in principle, in practice tricky, especially when the surface you are on is weaving wildly

Akiva Weinberger

You'd think that the one that was an infinitely replicable bit of data would win over the physical object, but apparently…

Of course, nowadays we all rely on the good graces of the US government to keep their GPS satellites free to access

copper.hat

forms of loran are making a comeback

Akiva Weinberger

(several nations have since launched alternatives I think?)

copper.hat

6:05 AM

apparently there is an underwater version used for tracking sea creatures

Akiva Weinberger

Loran?

copper.hat

long range navigation, since we are on the subject of navigation

Akiva Weinberger

How does that work?

copper.hat

one cute fact is that one minute of latitude is 1nm

loran, like most systems, relies on timing

Akiva Weinberger

Took me a bit to realize you meant nautical miles and not nanometers

It's a small world, after all

@copper.hat Like comedy

copper.hat

6:09 AM

there was that movie a long time ago where they shrunk a space ship thingy and injected it into a human

gps ruined it all

PM 2Ring

A century ago, the world relied on astronomy for timekeeping & navigation. Now we have atomic clocks & GPS. Ephemerides are still produced, but they no longer have such a central role. And of course, we use UTC, not TAI, so astronomical observations are needed to determine the difference between atomic time & mean solar time and to determine when we need leap seconds.

copper.hat

inertial navs were all the rage for a while

Akiva Weinberger

GPS ruined Fantastic Voyage?

copper.hat

no, it means people who don't know which way is north on a map appear in places in the back country that they shouldnt

not that i can reach the back country any more, unfortunately

the irony of it all

Akiva Weinberger

I try to keep track of where north is but I generally lose track when I'm in a building

If I build a building I'll paint all north walls the same color so you can tell

and/or install windows

PM 2Ring

6:13 AM

We occasionally get questions on Astronomy.SE from people who are amazed when they discover that the Moon is often visible during daylight hours. :)

copper.hat

well, i'll allow a compass :-)

i can barely remember the name of the big dipper

Akiva Weinberger

I should learn the constellations at some point

copper.hat

there is something a little spooky about looking at stars at high altitude

they seem unreal, maybe less twinkling or something, or just a darker background

PM 2Ring

I've never seen the Big Dipper, but I can see Alpha Centauri any night, weather permitting.

Akiva Weinberger

I suppose the atmosphere causes the twinkling. Well, that and astigmatism

PM 2Ring

6:17 AM

Definitely less twinkling at higher altitude, both due to less air, and less inhomogeneity caused by water vapour.

Akiva Weinberger

With crappy eyes anything is possible

copper.hat

understanding always takes the fun out of stuff

but i like to understand, so what does that make me?

a kill joe

Akiva Weinberger

Feynman has some talk called "Ode to a Flower" on how understanding can enhance beauty

copper.hat

i like feynman, albeit a bit full of himself

Akiva Weinberger

Not a fan of the Manhattan Project thing tbh

copper.hat

6:24 AM

knowledge is neither good nor bad

Akiva Weinberger

In that case, can I know your MSE password

copper.hat

i don't have one

PM 2Ring

@AkivaWeinberger goodreads.com/author/quotes/1429989.Richard_P_Feynman

copper.hat

some companies have a little hardware device and when you login to a computer you need to scan your fingerprint on the little nubby.

Feynman used to say, "physics is to sex as mathematics is to masturbation".

PM 2Ring

Allegedly, Feynmann had his sleazy moments. OTOH, I dont think it's totally fair to judge him by modern standards.

copper.hat

6:31 AM

yes, i agree, not just with feynmann

different times

generally i try not to judge unless i need to

PM 2Ring

Indeed

copper.hat

except for ice cubes in wine

jk

Akiva Weinberger

It's Mirzakhani's birthday

copper.hat

would have been

PM 2Ring

en.wikipedia.org/wiki/Maryam_Mirzakhani the International Council for Science has agreed to declare Maryam Mirzakhani's birthday, 12 May, as International Women in Mathematics Day in respect of her memory.

Akiva Weinberger

6:42 AM

Neat

I know some people like that

copper.hat

i have met very few female mathematicians, unfortunately

you mean females?

Akiva Weinberger

…Women in mathematics

Several classmates and two professors (that I've taken classes from)

copper.hat

wow, its sad that i cannot recall a single female mathematics professor.

even socially.

lots of ees, physicists, etc.

Akiva Weinberger

I read "Category Theory in Context" by Emily Riehl, but have not met her

PM 2Ring

Another female mathematician into hyperbolic geometry: Latvian Daina Taimiņa en.wikipedia.org/wiki/Daina_Taimi%C5%86a

copper.hat

6:47 AM

that's gotta be a thrilling read

Akiva Weinberger

Ah yeah, I love her hyperbolic crochet TED talk @PM2Ring

I've shared that one here a few times

PM 2Ring

I've mentioned her here before too.

Akiva Weinberger

@copper.hat It started to lose me towards the end, but the parts I understood were interesting. Eventually I'll probably read another category theory book like Seven Sketches

copper.hat

i have never mentioned here anywhere, i feel left out

that is a level of mathematics that is beyond me

Akiva Weinberger

Jessica Purcell wrote a book on hyperbolic knot theory that I might check out at some point

copper.hat

6:49 AM

too many little arrows

Akiva Weinberger

At one point I felt that differential equations were beyond my limit, and then I learned them, so ever since then I've felt invincible

That's my origin story, lol

It's just gonna take some time

copper.hat

i'll stick with brion tosses knots

odes i understand. groups i don't.

i mean i have zero intuition.

Akiva Weinberger

Knot portal

Trefoil knot in KnotPortal

►

Calvin Khor

I attended stochastic analysis lectures from li xiaomei who is the wife of fields medalist martin hairer. didn't understand a thing (presumably 100% my fault)

copper.hat

specific groups, $SO(3)$, etc, I have familiarity with, but in general

Akiva Weinberger

6:51 AM

I shared a Visual Group Theory lecture series on here not too long ago, but I have not yet watched it

copper.hat

like the class i took from steven smale

Akiva Weinberger

(youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv)

copper.hat

i thought i might learn something about horses

Akiva Weinberger

Just their shoes

Calvin Khor

lmao

copper.hat

6:53 AM

it took me a week to just make a dent on the first lecture. i gave up after a few

you really don't understand something until you can use it fluidly with 10 other concepts at the same time

so combining measure, manifolds, implicit function theorem and pdes in one sentence was a bit much

Akiva Weinberger

Von Neumann once said "In mathematics, you don't understand things, you just get used to them"

I never agreed with that statement, but I've gotten used to it

copper.hat

yeah, i have often quoted that to my kids

on that note, its wine o' clock.

Akiva Weinberger

What time zone are you in

copper.hat

i have answered my obligatory convex questioh

i'm in california, pst

Akiva Weinberger

psst

Gonna sleep

copper.hat

6:56 AM

:-)

Akiva Weinberger

That's a great sentence^ because now my two options are lying or lying

(geddit)

copper.hat

there are 10 types of computer scientists in the world

PM 2Ring

Those that dont understand binary, and those that do.

And those that also understand tri-state logic.

Balanced ternary is interesting. You can use it with Tower of Hanoi.

copper.hat

you can do cute things with tristate

Koro

7:14 AM

@robjohn oh yes :(

is there a fix for that?

robjohn

Not sure if there is a fix for that approach.

you might be able to show some uniform bound, but the dependence throws a wrench in the works.

Calvin Khor

@Koro yeah I'm not confident it can be fixed. sorry for not checking carefully

Koro

7:31 AM

😬

I propose the following: I choose d in (0,1/2).

robjohn

@AkivaWeinberger I have one of those.

Koro

instead of choosing d in (0,1).

hmm, that should work.

@robjohn et @CalvinKhor

robjohn

does that remove the dependence of $d$ on $a$?

Calvin Khor

how does that help?

Koro

The following holds: \begin{align*}
|\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} | & \leq |\int _{0}^{d}\left( a^{x} -a\right) x^{a} \ dx|+|\int _{d}^{1}\left( a^{x} -a\right) x^{a} \ dx|
\end{align*}

The second term is $<\epsilon \int_d^1 x^a dx\le \epsilon (1-d)$.

The first term is $\le \int_0^d (a-a^x) x^a\le \int_0^{1/2}(a-a^x)x^a$

robjohn

7:43 AM

but $d$ so that that is true depends on $a$.

Koro

and then $\le \int_0^{1/2} (a-a^x) (1/2)^a $

@robjohn it does but we are bounding it by 1/2 and we can do this.

I think you pointed to the second line: fix for that is $\le \epsilon(1-d)\le \epsilon$.

robjohn

$a-\sqrt{a}$ is not bounded

It is the part where $x$ is near $1$ that is the bad part

Koro

For that, I can use continuity of $a^x$, i.e., that is $|a^x-a|<\epsilon$ for x very close to 1.

isn't that right?

robjohn

But then, the $d$ needed for that is dependent on $a$.

Continuity is fine for a single $a$, but $a$ is tending to $\infty$.

Can you bound $\left|a^x-a\right|\le\epsilon$ for $x\ge d$ where $d$ is independent of $a$?

Koro

yes, but why can't this be true: Fix a >1 and choose $d_a$ (recording that d depends upon a) so that $d_a<1/2$ and that $|a^x-a|<\epsilon$ for x in [d,1]. The following holds: \begin{align*}
|\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} | & \leq |\int _{0}^{d}\left( a^{x} -a\right) x^{a} \ dx|+|\int _{d}^{1}\left( a^{x} -a\right) x^{a} \ dx|
\end{align*}?

and then I use the steps I used above. Yes, d depends upon a but we are also showing that the fluctuations due to this dependence are bounded by a diminishing quantity (i.e., (1/2)^a)

@robjohn: should I post it on mse so that it remains at one place?

robjohn

7:57 AM

@Koro If you post the question on the main site, someone may answer.

Koro

you may answer too :).

robjohn

that is sort of what I meant ;-)

Koro

:-)

Calvin Khor

If you choose d so that $|a^x-a|≤\epsilon$ for all $x\in[d,1]$ then in particular $\epsilon >a - a^d>a^d(a^{1-d}-1)$. If $d<1/2$ then $a^{1-d}-1≥a^{1/2}-1\ge100000$ once $a$ is large enough. so you do need d close to 1

robjohn

indeed

Calvin Khor

8:04 AM

>s should be ≥s but whatever

robjohn

The idea of breaking up the integral is good. I use that in my answer. However, I need to use $d$ close to $1$.

Calvin Khor

@robjohn ahhh so tantilizing.....don't pull me back to the integral now i need to work lol

Koro

0

Q: Finding the value of $\rm \lim_{a\to \infty} a^x x^a \,dx$

Let $\rm\epsilon >0$ be given. $\rm x\mapsto a^{x}$ is continuous at $\rm 1$ so there is a $d_a\in ( 0,1)$ such that $|a^{x} -a|< \epsilon $ for all $ x\in [d_a,1]$. WLOG, let $d_a<1/2$. $ |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |=|\int _{0}^{1}\left( a^{x} -a\right) x^{a} \ dx|...

@CalvinKhor O_O

S.M.T

Hii everyone

I have a Q regarding vectors

b = |b| cos θ i + |b| sin θ j

Let’s consider this as case 1.

Calvin Khor

@Koro 😩

@S.M.T hello

S.M.T

8:13 AM

Diagram for this case is :

@CalvinKhor Hii

On left is the diagram & on right is the coordinate value

I hope it is clear till here

My question 1 is : What will be the value of θ for situation 1 ?

Calvin Khor

@Koro if @robjohn suddenly has to walk about 200 more dogs and MSE's servers were taken off for everyone but me for the next few years....i might beat robjohn to the answer

@S.M.T what's a v value?

vertical value?

S.M.T

There are more follow questions

Koro

@CalvinKhor 😁

Calvin Khor

oh ok

S.M.T

Sorry

45 right ?

Calvin Khor

8:16 AM

yes, if b = (1,1) then theta is 45º

S.M.T

Great. Let’s move to case 2 now

Consider we want to find the |b|

Calvin Khor

calculated as arctan(1/1)

S.M.T

Only way to convert the vector b into its scalar form I.e |b| is by using dot product. Correct ?

Calvin Khor

|b|, given what information?

S.M.T

What other information will u need ?

Koro

8:19 AM

@robjohn with mod powers, can you know who upvoted your comment?

Calvin Khor

well if you know b=(b_1,b_2) then yeah $|b| = \sqrt{b_1^2 + b_2^2}=\sqrt{b\cdot b}$

S.M.T

@CalvinKhor i don’t know about this

robjohn

@CalvinKhor I've already put the answer on the site, about 15 hours ago.

S.M.T

@CalvinKhor This seems to be important. Can we like come onto it in the next follow question I have

Calvin Khor

@robjohn ahaha

@S.M.T sure

S.M.T

8:22 AM

So , the case 2 I meant to talk about is this:
b.b I.e b*b cos θ I.e |b| =( |b| cos θ i + |b| sin θ j ) . ( |b| cos θ i + |b| sin θ j)

Agree with this statement ?

Calvin Khor

no

S.M.T

K , where is it wrong ?

Calvin Khor

well.

are we still talking about b = |b|costheta i + |b| sintheta j

which is not usually the case

S.M.T

Yes

Calvin Khor

ok then I agree. sorry

S.M.T

8:25 AM

K.

robjohn

@Koro do you understand my comment to the question?

Calvin Khor

oh. it is usually the case, i'm just silly

S.M.T

Here will u agree that θ = 0 ?

Calvin Khor

where is theta=0? is there another picture coming?

S.M.T

No but I am thinking of I like in terms of dot product

Because we wrote b . b

Calvin Khor

8:27 AM

|b| =( |b| cos θ i + |b| sin θ j ) . ( |b| cos θ i + |b| sin θ j) is correct

S.M.T

We meant to project the same vector I.e b onto b

Calvin Khor

but idk what this means b.b I.e b*b cos θ

oh ok. i see, yes, 0

S.M.T

So , then θ will be 0

robjohn

@Koro oops... someone answered your question.

Calvin Khor

yes, but this theta and the other theta are different

S.M.T

8:28 AM

Yes absolutely

Calvin Khor

@robjohn i can't believe you've done this

robjohn

what, posted the answer?

I said I would if he posted on main

Calvin Khor

joke fell flat, ignore me lol

robjohn

Sorry, I missed it.

Calvin Khor

was trying to riff off your oops 😬

robjohn

8:33 AM

@Ted you're up late

S.M.T

@CalvinKhor b.b = ( $|b| cos^2θ i^2 $+ $|b| sin^2θ j^2$

U will get this in end

Agree ?

$|b| cos^2θ $+ $|b| sin^2θ$

Calvin Khor

well...where $i^2$ means $i\cdot i$ and $j^2$ means $j\cdot j$, and you used that $i$ and $j$ are orthogonal

S.M.T

i.i = 1

Calvin Khor

missing a square on the |b|s?

S.M.T

Ohk

$|b|^2 cos^2θ $+ $|b|^2 sin^2θ$

Better now

Calvin Khor

8:36 AM

yeah

S.M.T

@CalvinKhor I have a Q. We cannot find the value of |b| only from 1st case without going to 2nd case , right ?

26 mins ago, by S.M.T

b = |b| cos θ i + |b| sin θ j

PM 2Ring

@S.M.T In MathJax, you should use \theta, not θ. Are you using the ChatJax bookmark?

Calvin Khor

I suppose so? "Another way" is to rotate b until it lines up with say the x-axis. Then the x component is |b|

S.M.T

I used a google engine. It is because on my tablet , it doesn’t convert latex to normal form of reading. So , I use these symbols for better clarity to chat.

I do use latex on my questions on site

@CalvinKhor Can u tell how exactly ? I didn’t understand clearly this part

Calvin Khor

physically, you rotate the ruler until it is parallel to b. then you measure

but better to stick to the dot product

S.M.T

8:41 AM

How will u write the equation ?

Calvin Khor

with a rotation matrix

S.M.T

Ohk , I don’t know what that is now

Calvin Khor

sure, don't worry about it. if you can get the dot product way you're doing good i think

S.M.T

Ohk. What is this “ $|b| = \sqrt{b_1^2 + b_2^2}=\sqrt{b\cdot b}$”

Calvin Khor

@S.M.T its exactly case two that you typed out

I wrote $b=(b_1,b_2)$ instead of $b = |b|\cos\theta i + |b|\sin\theta j$

S.M.T

8:45 AM

K

Calvin Khor

but other than this notation change its exactly what you said

S.M.T

Why did u write a square root ?

Koro

@robjohn oops 😬

Calvin Khor

well your computation computed $|b|^2$, which is the norm squared. if you want $|b|$ then you need to take a square root

S.M.T

K

U mean , what I did is |b|*|b|

Instead of finding |b| , right ?

Calvin Khor

8:47 AM

11 mins ago, by S.M.T

$|b|^2 cos^2θ $+ $|b|^2 sin^2θ$

S.M.T

Ohk , I get that.

Calvin Khor

$|b|^2 \cos^2θ+|b|^2 \sin^2θ = |b|^2$

S.M.T

Yes , got it.

Calvin Khor

good

user1027216

Hi! I have a question

In this answer math.stackexchange.com/a/4448563/1027216

I'm given an explanation about a question subject to a bit of interpretation depending on the definition of logarithm. However, the user below indicates that my proof is circular. Why? I think that if he interprets this as the definition of a derivative then he is correct, but my interpretation of the question was just about the limit. I would like to know if my answer is incorrect to proceed to eliminate it if it is the case. Thanks you so much!

robjohn

8:51 AM

@Koro I'd posted it in my room a while ago. I just moved it.

Koro

I’ll take a look at the answer shortly. :-)

one potato two potato

10:08 AM

Apr 29 at 12:43, by one potato two potato

If $\{f_n:\Bbb D\to\Bbb D\}$ is a sequence of analytic function on unit open disk $\Bbb D$, then $\{f_n\}$ is equicontinuous on each compact set $K\subset\Bbb D$. How can I prove this?

Is there any example that $\{f_n\}$ is not equicontinuous on $\Bbb D$ itself?

Calvin Khor

@onepotatotwopotato maybe something like $f_n(x)=\sum^n_0 x^k$?

one potato two potato

@CalvinKhor $f_n:\Bbb D\to\Bbb D$.

Calvin Khor

oh didnt see that

one potato two potato

Something like $f_n(z) = {1\over 1-z^n}$ but also in this case, image does not lie in $\Bbb D$.

Hmm.. $\{z^n\}$?

porridgemathematics

10:28 AM

hi, anyone know why this is true math.stackexchange.com/questions/4448800/… ? Its about kahler manifolds

Koro

10:47 AM

@robjohn yes.

:)

Violet Flame

11:39 AM

Is the set of all numbers with only 0, and ones in decimal expansion an uncountable nowhere dense set?

Calvin Khor

does unital algebra just mean an algebra with unit?

Violet Flame

@CalvinKhor is my above comment correct?

Calvin Khor

@VioletFlame i do not recall the definition of nowhere dense

Violet Flame

A set which isn’t dense in any open set @CalvinKhor

Calvin Khor

ok...its clearly uncountable, im gonna guess already closed

for any point p in the set, every ball around p contains points outside the set

so the interior is empty. does that sound right?

@VioletFlame do you know what a unital algebra is?

oh i found it on wikipedia nm

Ajay

12:02 PM

Nvm

Hello @Prithubiswasleftmse

Prithu biswas leftmse

Hello

Ajay

I have a question

Grayson’s Groupcycles rents bikes with multiple seats for large groups of people.
They rent 7-seater, 13-seater, and 25-seater bikes. A group of 14 people could fit
on two 7-seater bikes, however a group of 15 people could not fit exactly on any
of the bikes since no combination of bikes have exactly 15 seats.
What is the largest group size that cannot fit exactly on any combination of
bikes from Grayson’s Groupcycles?

That I just felt like posting here for no apparent reason...

However, do feel free to try it out.

ioch

1:33 PM

is there a formula for the inverse of the (exponential matrix plus the identity) ?

$(e^{tB}+I)^{-1}$

Semiclassical

2:08 PM

Is there a way to articulate the following? I have a differential equation $Ly=\lambda y$ subject to periodic boundary conditions (PBC). Is there a way to say that PBC are part of the definition of $L$?

(I ultimately want to consider the projection of $L$ onto even/odd functions, which amounts to supplying additional boundary conditions.)

I guess it amounts to the choice of domain of $L$, rather than how it acts on this domain

ioch

mb via the domian

yh I was going to say

where does $y$ live

Semiclassical

2:25 PM

right

user1027216

3:29 PM

Hi! Is there any remark about my question?

7 hours ago, by user1027216

Hi! I have a question

Violet Flame

@robjohn do you think chat.stackexchange.com/transcript/message/61110139#61110139 is correct?

@Koro do you think it’s correct

Koro

@VioletFlame It is uncountable but I doubt if it's nowhere dense.

Let's first note that such a set is contained in the closed interval $[0,10/9]$.

why do you think that it's nowhere dense? @VioletFlame

hi @copper.hat!

robjohn

3:50 PM

@VioletFlame @Calvin Khor's assessment seems correct.

Violet Flame

@koro why is it dense in that interval

am i missing something or @robjohn is saying the opposite

robjohn

It is closed and it's complement is dense in it, so it cannot contain a non-empty open set

@VioletFlame Koro said that he doubted it was nowhere dense. That means not certain (no proof).

leslie townes

4:06 PM

semi: very late to this but as noted it is often done with the definition of the domain (which can make a huge difference for properties of the operator, such as our favorite, self adjointness)

or its uglier cousin, symmetry

robjohn

or the painful-to-see double-jointness

leslie townes

demonry

Ted Shifrin

smacks robjohn for practice

robjohn

practice to smack me later, or to smack someone else?

Ted Shifrin

Yes.

leslie townes

4:13 PM

sometimes in operator theory people would define stuff so the domain was arranged as to make the operator self adjoint and then the eventual proof of self adjointness would have the flavor of a tail wagging a dog

Ted Shifrin

"the domain was self adjoint"? Huh?

leslie townes

you can often do a thing where, assume the domain of something that is a priori defined in a whole lot of places is assumed to contain only X, and play the game of, what else does it need to contain to make the thing self adjoint and not worse than that

raising the question of whether the eventual proof of self adjointness is even a theorem or a hypothesis

it's like my belt holds my pants up, but my belt loops hold my belt up, who's the real hero, what's going on down there, etc

Ted Shifrin

gives up

leslie townes

i did a version of this with a technical consultant once and he didn't find it remotely funny

Calvin Khor

@robjohn knew it all along 😎

Ted Shifrin

4:19 PM

Word to the wise: Stand-up comic is ok for Munchkin, not for you.

leslie townes

it wasn't exactly standup. there was a turbine for a power plant that had a fan, whose purpose was to feed air into a chain of compressors and then a turbine, which powered the fan, and it actually mattered who the real hero was

but that part is boring

Calvin Khor

@VioletFlame why are you not convinced?

leslie townes

there was a starter that was not written on any of the schematics we had been given that was the real hero

robjohn

@leslietownes well, for a perpetual motion machine, what use would be a starter? if it is really perpetual

leslie townes

"what isn't in this picture" is a good question for engineers although they are often so accustomed to not depicting irrelevant things that you have to ask it five or ten times before the point gets through

i think i broke through with an analogy relating to a lawn mower

sadly the actual power plant was not started by tugging a rip cord

rob: hahaha it just is

the turbine has always been powering the fan, and the fan has always been feeding the turbine

Violet Flame

4:41 PM

@CalvinKhor i was convinced untill @Koro said im wrong,

copper.hat

4:56 PM

@Koro Hi!!

Ted Shifrin

Greetings, @copper.

copper.hat

Hi Ted!

robjohn

@VioletFlame Did Koro say you were wrong, or did they just doubt and ask you why you thought it was so?

Violet Flame

5:46 PM

@robjohn if such sets are so easy to construct what’s special about the cantors set?

robjohn

6:06 PM

@VioletFlame Is Cantor's Set difficult to construct? That amounts to a similar construction. The Cantor set represents the base 3 fractions with only 0 and 2 in them.

leslie townes

cantor had a pretty good hairstyle if you have male pattern baldness that wipes out most of the front of your head. i used him as a reference for the first part of the pandemic.

eventually i decided the mustache/beard thing was too much and shaved that off.

Akiva Weinberger

6:34 PM

Remind me: is Munchkin Leslie's kid?

Semiclassical

hope so, otherwise these conversations would take on a darker meaning

Vio Ariton

7:12 PM

How did he get from the first part to the second

Semiclassical

7:26 PM

@VioAriton the notation isn't great, but it seems ilke he meant d(x^2)

it's a shorthand way of doing u=x^2 subsitution

in which case the integrand is (u+y^2+25)^(3/2) which does have 2/5(u+y^2+25)^(5/2) as indefinite integral

Vio Ariton

Oh I see, thanks

ZaWarudo

7:54 PM

Talking with a friend, he said that "I can define cosine as the derivative of sine, and sine as the anti derivative of cosine", however I think there is a logical mistake in this, because it is circular (defines cosine using sine and define sine using something else related to cosine -> defines cosine using cosine); so I said that we must use functional equations, differential equations (y''+y=0) or power series.

But he didn't agree with me, saying that his reasoning is equivalent to the differential equations one; I don't agree with this latter affirmation, because both sine and cosine satisfy y''+y=0 and so there is nothing circular about that in my opinion. Who is right?

Akiva Weinberger

9:32 PM

@ZaWarudo If you're defining f as the derivative of g, and g as the antiderivative of f, then that can be satisfied by any derivative-antiderivative pair

If you're defining f as the derivative of g, and g as the derivative of -f, then… it's still kinda circular, but at least this way you know f and g are linear combinations of sine and cosine

The best way (related to what you've done) is probably to define sine as the solution to y''+y=0 with initial conditions y(0)=0 and y'(0)=1, and cosine as the derivative of sine

It's a fun exercise to show that y''+y=0 implies that y^2+y'^2 is a constant (hint: find the derivative of y^2+y'^2)

robjohn

10:03 PM

@AkivaWeinberger or multiply by $y'$ and integrate.

leslie townes

10:18 PM

@AkivaWeinberger yup

Ted Shifrin

10:52 PM

Semi-affectionately named by Ted ... although soon it'll be changing.