Mathematics - 2021-12-30 (page 1 of 2)

copper.hat

01:51

If someone changes their username it is possible to determine the old one(s)?

leslie townes

copper, i don't know if there's a formal way to do it without higher powers. old posts are associated with the new name, while comments on old posts sometimes reference the old name.

Odestheory12

Maybe internet archive?

I guess if you can find an old post and that post is archived, you would find the old username assuming he changed it only once.

Ted Shifrin

02:22

Some of us are downright boring and have a unique identity!

waits for inevitable math humor, so-called

Odestheory12

You only need to prove the existence then :P

Ted Shifrin

I prefer not to exist!

Fargle

Dark.

robjohn

03:06

It's too bad we don't have an emeritus professor from UGA in this chat.

@copper.hat I don't think so, but the chat transcripts hang onto the names used when posting, I believe.

sometimes sleuthing through comments on main can give some information, as well.

copper.hat

@robjohn Thanks, yes, that was what prompted the question.

too many joes in the world, i'll stick with scupper.hat for now

robjohn

@copper.hat changing identities?

copper.hat

i mean copper.hat :-)

robjohn

ah

cucap

copper.hat

i am torn between anonymity and its opposite, whatever that may be labeled

robjohn

03:21

fame?

copper.hat

that was a 70s show, right?

robjohn

notoriety

@copper.hat was it that long ago?

copper.hat

i think so, unfortunately

robjohn

1982-1987

copper.hat

for no particularly good reason, i bought \$99.01 worth of btc 30 mins ago and its now worth \$98.70.

robjohn

03:30

leslie will have something to say about that.

copper.hat

70s, 80s,.. so long ago

i could not find lesliecoin on robbing hood

robjohn

80s were not so long ago, but the 70s are. They are unearthing creatures from the 60s for fuel

copper.hat

i used to cut turf for fuel back in those years

i think elementary number theory is my mathematical kryptonite

robjohn

why is that?

copper.hat

i find the results, or rather their proofs to be unintuitive

i mean, i eventually find a path, but it is hours & days, not minutes

and this is basic stuff.

im working through some exercises in Ted's Abstract Algebra

modular stuff

i think any intuition i have comes from geometric stuff, and modular does not have any obvious (to me) connection

robjohn

03:36

I'll have to look again, but there has to be some intuition there, or it seems like pure memorization.

copper.hat

well, it may reflect my way of looking at things, but yes, for me it is close to memorization

not my forte

robjohn

I knew early on I would never be a chemist.

copper.hat

same for me

i love chemistry, but it needs too many facts at hand to be productive

i still finger through my pauling's general chemistry wistfully

robjohn

The two paths that were most alluring were astronomer and mathematician.

copper.hat

i like working on the boundary of hardware and software

i like hardware, but for my impatient self i like the software because i can change things quickly and retry

embedded is fun, i should have stuck in that direction

robjohn

03:41

I am a software person, but I have written code for embedded software

copper.hat

there were too many interesting things

AMDG

Good evening and Merry Christmas!

robjohn

hello

copper.hat

i like software, but really only with expressive languages like lisp, etc

if i write 1+1=2 in c++ it takes three pages a .cc, .h, and a few other classes

AMDG

Three pages?

wat

copper.hat

03:43

i am exaggerating every so every so slightly

don't believe everything you read

AMDG

If I write 1 + 1 = 2 that will be 9 bytes Kapp

copper.hat

kapp?

AMDG

external-content.duckduckgo.com/iu/…

robjohn

@AMDG who is that?

AMDG

I don't remember the exact story. Look up the Kappa emote. Some guy who worked at or still works at twitch or something.

copper.hat

03:47

for example, with elementary NT, there is a beautiful result about the power of a prime that can divide a binomial coefficient math.stackexchange.com/a/1941501/27978

but the proof took me ages.

i forgot about it, but encountered a related problem in Ted's text

AMDG

But can it divide generalized binomial coefficients for complex arguments tho. That is the real question.

copper.hat

i gave up last night and hit MSE to find a result

and discovered an answer (elaboration really) by myself

that was disturbing

leslie townes

there are some unsolved problems in multinomial coefficients related to that, copper.

you're always a step away from the unknown.

AMDG

past you answered future you's question.

leslie townes

also, i pity the person who buys BTC instead of LC.

copper.hat

03:49

yeah, i know, but this thing is i have zero intuition regarding the proof direction

hey, its \$98.86 now, only \$0.15 loss in less than an hour

leslie townes

see e.g. arxiv.org/abs/0806.0607 within that is the usual stuff about divisors of binomial coefficients, p-adic carries, etc.

still unknown i think, and mentioned in some edition of conway and guy

copper.hat

how do people even think about this stuff???

AMDG

They pass it to their intellects and get an answer Kapp

copper.hat

volumes, integrals, convex stuff is all geometrically obvious

there must be some ah-hah moment that i have yet to come across

i hope

robjohn

@copper.hat Kummer's Theorem?

AMDG

03:53

Usually takes as long as it takes for the intellect to properly abstract something and give a universal many-to-one representation of it. That's why humans are still better and faster to learn than a machine learning algorithm.

copper.hat

@robjohn yep, all the results that i have been looking at for current trip-me-up problem seem to revolve around Kummer's

@AMDG i think we are hard wired for certain sorts of problems

robjohn

@copper.hat yeah, that seems to be very intuitive to me

copper.hat

@robjohn serious?

robjohn

I am

AMDG

@copper.hat I have to go attend to my friend who is waiting on me, but that would ring true with the fact that each person has a unique vocation in life.

Night guys!

copper.hat

03:55

night

robjohn

l8r

Koro

Good mor9g :)

copper.hat

good morning @Koro

i think i'll stick with convex things

leslie townes

said the actress to the bishop

copper.hat

i would guess that the bishop had his eye on certain convex objects

the relationship between de Polignac's formula and Legendre's formula is interesting, thanks @robjohn

i never thought to look at the base $p$ representation

we're back down to \$98.64 now

is lesliecoin this volatile

robjohn

04:02

@copper.hat are they not the same?

copper.hat

i always knew one as an infinite sum and the other as a statement of carries

or the explicit formula you have with $\sigma_p$

robjohn

@copper.hat Hmm, the carries one is Kummer, I thought.

copper.hat

i meant the $\sigma_p$ formula.

robjohn

@copper.hat That I think of as Legendre

copper.hat

yes, and the infinite sum i think of as de Polignac's

robjohn

04:05

You mean the infinite sum with finitely many non-zero terms?

copper.hat

yes

robjohn

ah, okay

copper.hat

$\sum_k \lfloor {n \over p^k} \rfloor$.

i suspect i am not spending enough consolidate time to build up my elementary NT muscle to allow some progress, piecemeal is not working for me

unfortunately there are other minor distractions.

i wish one could bookmark answers in addition to questions

Koro

Given $f$ defined on interval $I$, define $f^*(t):=\sup_{x\in I}|tx-f(x)|$. The set of values of $t$ for which $f^*(t)$ is real is either empty or consists of a single point or is an interval. How do I show this?

leslie townes

try showing that if a is in there and b is too then so is ta + (1-t)b

for t in [0,1]

copper.hat

04:13

how can $f^*(t)$ be anything other than real?

leslie townes

if f is unbounded

it's a poorly written expression of that

copper.hat

thx

the function is convex

leslie townes

we both honed in on that immediately

maybe i have a future in selling stuff to apple

copper.hat

:-)

Koro

So if $a$ and $b$ are values of $f^*$ then so are the values in interval [a,b].

copper.hat

04:15

i still harbour resentment towards tim cook

leslie townes

koro i just mean if the sup is finite at a and b then it's also finite between them

a and b here are in the domain of f* , not values of f*

copper.hat

each $t \mapsto |tx-f(x)|$ is a convex function.

hence the $\sup_x$ is also

Koro

@leslietownes Ah, I was erroneously analyzing range $f^*$

copper.hat

hence if $f^*(t), f^*(s)$ are finite then so are values in $[s,t]$.

Koro

thanks a lot. :)

copper.hat

04:19

it looks like the absolute value of the convex conjugate

Koro

This exercise has many subparts. I'll try writing the complete answers to them.

copper.hat

woo hoo, its worth $99.46 now, i'm off to buy my private jet

Ted Shifrin

@copper.hat what is worth a jet?

copper.hat

hobbylinc.com/…

Ted Shifrin

Ah

copper.hat

04:23

i used to love making polystyrene models when i was a kid (and not so kid)

my daughter & myself put together a kit of the space shuttle

when she was around 5

Ted Shifrin

A project for munchkin

leslie townes

could be. she's begun building with normal-sized legos. she's not too bad at it

huzaifa abedeen

Hi

How to proof these two theorem

??

copper.hat

imgur.com/xfXHX8h

@leslietownes here is a Lego project to do with your daughter reddit.com/r/Damnthatsinteresting/comments/rrg2zl/…

leslie townes

you know, i think i had that as a child

the shuttle

those decals were a pain in the ass

Ted Shifrin

04:35

@huzaifaabedeen These are formulas, not theorems. What have you done?

copper.hat

the crew are all present and painted

@huzaifaabedeen if you pick a permutation of $n$ objects, and $p$ of the objects are the same then there are $p!-1$ other permutations that end up with the same result.

AMDG

@copper.hat Personally, I like thinking about concave surfaces.

Namely about how to put them in a compact format so that games are not over 100GB as the norm.

:P

copper.hat

why concave surfaces? surely convex are nicer

huzaifa abedeen

How to derive those formulas then Ted shifrin. I did some problems based on those formulas from exercise 7.1 from NCERT maths textbook.

AMDG

Convex are easier to work with, but things in nature tend to be concave. Convex is practically solved. Just define a convex model as a function of radius and theta.

Alex

04:47

Hi good morning mathematicians

AMDG

We can map a flat plane onto a sphere and define offsets, then use spherical interpolation so that there aren't any "holes" in the surface.

The question I still have yet to answer for both convex and concave models, however, is the manner in which the curvature is represented compactly.

Fun challenge!

copper.hat

@huzaifaabedeen why don't you consider my comment

Alex

@huzaifaabedeen Did you read the comment of cooper?

copper.hat

bingo :-)

AMDG

If only it were as easy as taking a derivative of a 4D function to get a vector field, but it's a chicken and egg problem: what's the 4D function? :L

Here's an interesting question concerning the development of models: what is the minimal surface which passes through a discrete set of points in R^3?

With the knowledge of such surface, we can do some advanced transforms such as minimizing aliasing in polygonal models without changing the number of vertices, or rotate a set of points as travel along the surface... of the surface, given any arbitrary vector.

copper.hat

04:53

you would have to define minimal before that makes sense

AMDG

"In mathematics, a minimal surface is a surface that locally minimizes its area."

Alternatively, see mathworld.wolfram.com/MinimalSurface.html

Semiclassical

so, here's something wacky which i discovered that i hadn't seen before

first, something familiar. one way to write the fibonacci sequence is by iterating $M=\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}$ starting with $(1,0)^T$

leslie townes

withdrawn

Semiclassical

feh

leslie townes

it would have been funnier if you hadn't fixed it

:)

Semiclassical

04:58

alternatively, i could've just said to have act $M$ on the right :P

leslie townes

i can't let you do that

Semiclassical

lol

suppose we think of this as acting not on $\mathbb{Z}^2$ but on some $(\mathbb{Z}/p)^2$

i played around with a few choice of $p$, and noticed something interesting when i pick $p=5^n$

leslie townes

p is an odd label for a power of a prime, but continue

haha i'll stop s--tposting and listen, i promise

Semiclassical

had to pick something

here's what i get if i plot the orbit for $p=5^2$ (five is too small to be that interesting)

leslie townes

said the actress to the bishop

Semiclassical

05:01

that is a lot more symmetric than i would've expected

leslie townes

if you work mod p, is the sequence polynomial? i.e. if D is the forward difference sending x(n) to x(n+1)-x(n), is there some k for which D^k on that is 0?

Semiclassical

there's 5^3

copper.hat

sorry, are you using $p$ for a non prime?

wtf

leslie townes

i tried to stop him, copper

copper.hat

@Semiclassical if you hear a noise upstairs don't worry, its just the wind

Semiclassical

05:04

and 5^4

copper.hat

we operate under the prime directive

Semiclassical

@copper.hat quiet, you

copper.hat

look at the eigenvectors

Semiclassical

it's not divisible by two, therefore it must be prime :P

@copper.hat is that still useful if we're acting modulo a power of five?

copper.hat

im searching for the other even primes

Semiclassical

05:05

i guess it should be

leslie townes

semi do you know the answer to my Q above ^^^ unlike most of what i say it was not a shitpost

copper.hat

my sh*tcoin has gone negative again

Semiclassical

i did, but i didn't understand it at first glance. lemme try

copper.hat

i should have sold when i was up $0.15

leslie townes

copper this never owuld have happened if you had invested in the world's only cryptocurrency that only goes up

copper.hat

05:07

up modulo 10?

modular arithmetic where every direction is up

leslie townes

no, it just goes straight up

said the bishop something something

copper.hat

did the parish priest pay you a visit?

leslie townes

i don't even know who my parish priest is

i know my mom's priest, he is a cheerful old irish guy

Semiclassical

well

leslie townes

when i was visiting my mom once i saw him at the grocery store and i had that moment, like you have when you see a teacher in a grocery store

like "i didn't know you were allowed to go to normal stores"

copper.hat

05:12

unfortunately the only parish priest i know (knew, he said the service for my mom) was run over & killed by a bus last august.

and he was a decent fellow, not your usual sort

leslie townes

that's awful. sorry to hear it

Semiclassical

i don't get any D^k which annihilates it, but i do seem to have D^100 gives back the original sequence in the mod 25 case

copper.hat

joking aside, he was a real loss

Semiclassical

but i think that's just b/c the sequence is 100-periodic

copper.hat

not like some of the tossers

leslie townes

05:13

semi interesting. one next step would be to look for polynomial relations in various powers of D. i guess factors of x^100 - 1

copper.hat

you can see the eigenvectors in the picture

leslie townes

i block all pasted images in the chat, on the off chance that they might reveal something of geometric significance

even if it's related to eigenvalues

copper.hat

in my old age i use a hot water bottle when it is a little chilly at night

it leaked last night unfortunately

i tried to get a booster at cvs today but they were so miserably disorganised i left after 10 mins

Semiclassical

the eigenvectors don't seem to match those directions unfortunately

leslie townes

you should get an electric blanket. they're great. not the firetraps of yore

copper.hat

05:17

i only need the bottle for 5 mins then i'm all good.

leslie townes

i don't even know what yore is, but maybe ted could tell us

copper.hat

can't have the 5g electric blanket waves radiating my body

used to be lots of yores on san pablo

the bigger the hoops

leslie townes

my cat has a heated bed that functions like an electric blanket when she steps on it. it's pressure activated and very comfy

copper.hat

i know a lot of otherwise seemingly sane people who lose it with cell towers & 5g

wow, in socal?

leslie townes

so comfy that my daughter figured out how it worked and sometimes sits in it

cell towers and 5g woo woo has some overlap with general hippie suspicion of technology. we had that in the bay area from way back.

yeah, even here. every once in a while, it gets into the 30s at night :)

Semiclassical

05:20

i take that back, partially. the orientation of the eigenvectors does seem to match the crosses that show up

but they they don't match the lines

so, this is sorta neat

the four centers in the figure in the 5^2 picture are (15,5), (20,15), (10,20), (5,10)

which form a four-cycle via the action of M

Semiclassical

05:48

okay, apparently the lines i'm seeing are all of slope 2 (or of slope -1/2 for the downwards ones)

so that's...something

copper.hat

how are you creating the trajectories? $M^n x_0$?

cabmetric

06:49

Can I have an uncountable dimensional subspace W of the vector space V = C[0,1] such that dim(V/W) is countable ?

I am taking the field to be R

leslie townes

with the axiom of choice, could you take a basis {v_i: i in I} of V, with I necessarily uncountable, and then choose a subset J of I with I \ J countable or finite, and let W = span {v_i: i in J}

copper.hat

Let $\phi(f) = f(0)$ and $W= \ker \phi$?

leslie townes

yeah, if you are OK with dim V/W finite (and not 'countable' meaning countably infinite) that's the way to go

copper.hat

ima simple guy

leslie townes

why did the tarmacking rinse off in tonight's rain

cabmetric

06:57

@copper.hat , Is phi a function from C[0,1] -----> R ?

copper.hat

i think my friend did that one, he's in kentucky right now

@cabmetric yes

Koro

I have exactly the same question as here math.stackexchange.com/questions/1383105/…. It is still unanswered. Can anyone please help me understand why "as a consequence of theorem 12.11 and 12.12" is true?

cabmetric

@copper.hat We have C[0,1] (R) / W (R) is isomoprhic to R(R) ? Right ?

copper.hat

yup

leslie townes

copper: yup (R)

copper.hat

07:05

an example of element of the quotient is the set of all functions such that $f(0) = \alpha$ for some $\alpha \in \mathbb{R}$.

i r(R)efr(R)ained fr(R)om adding extr(R) r(R)s

Koro

I s(S)e(E)

Here also, the same question is still unanswered: math.stackexchange.com/questions/1402689/…

leslie townes

is there some kind of skull hat for people who resurrect these old questions

AMDG

Apparently the error in one iteration of Newton-Raphson over a linear approximation of $\frac{1}{x}$ can be used to approximate $\csc(x)$. This makes sense considering the relationship to the Gamma function between cosecant and 1/x.

Koro

Or it may also be true that "as a consequence of theorem 12.11 and theorem 12.12" above is like saying "Taylor series expansion of functions is a consequence of mean value theorem".

copper.hat

la la la

leslie townes

07:16

ba ba ba

Koro

ha ha ha

AMDG

Spaniards be like ja ja ja

Anyways, by error I of course mean that if $f(x) \approx \frac{1}{x}$, and we apply one iteration of NR to f(x) which we will call $f'(x)$, then $\frac{1}{1-f'(x)} \approx \csc(1-f'(x))$.

copper.hat

la la la$^2$

ok, i'm going to sleep

AMDG

And if you want to make it sound more impressive, we can say $\frac{1}{1-f'(x)}\approx \frac{\Gamma(1 - f'(x))}{\Gamma(2-f'(x))}$. :P

Koro

"As as consequence of..." , I think is like saying "it's very easy to see that".

It's as easy as showing that $f(x)=\sum_{n=1}^\infty\sin (\frac 1{n^2}) I(x-q_n)$,(where $q_i'$s are enumeration of all rationals and I is unit step function defined as 1 on non negative numbers and zero elsewhere) is continuous from right at every real number.

copper.hat

07:28

if $D_{k,r} f, D_{r,k} f$ exist and are continuous at $c$ then $D_r f, D_k f$ are differentiable at $c$ by 12.11.

then the result follows from 12.12

it is straightforward, but the author should point out that it is being applied to $D_r f, D_k f$.

Koro

@copper.hat I'm afraid, I don't understand that. In 12.11, I think that n is the dimension of $R^n$ so to use that we need n partials. How can one conclude only on the basis of partials w.r.t. r and k?

copper.hat

the other partials are immaterial

Koro

i thought that only one partial is immaterial (in the sense that it doesn't need to be continuous but existence is required in 12.11) and that we do require continuity of the rest of all n-1 partials, don't we?

Am I misinterpreting theorem 12.11?

copper.hat

no, but if both are continuous as opposed to one continuous and the other exist then the consequence still holds

don't get fixated on the $n-1$ part. $n=1$ in this case.

Koro

I’ll think about this. Thanks copper. :)

copper.hat

07:44

only the $k,r$ indices matter here

RaMathuzen

07:55

I have a doubt on this proof that is How proving W as non-empty based on the existence of zero vector tells it has vectors $\alpha,\beta$ ?

user688539

What's theorem 1?

I think the issue is that the "let alpha and beta be.." thing is senseless if W is empty. If W actually has elements ( if we are given that it does), then directly strike out the existence part and then directly continue from the previous arguement into the part right after the existence part

love_sodam

08:11

Is there any Lecture notes or books that writes linear algebra in a group theoretical sense?

Ignore the above chat

love_sodam

09:05

@RaMathuzen I know that book its HK LA

RaMathuzen

@love_sodam Hoffman Kanze

love_sodam

yes

RaMathuzen

@user688539 Ok

Is there any other good book for linear algera?

@love_sodam What's your recommendation?

love_sodam

@RaMathuzen Well I first met LA via HK too.

RaMathuzen

09:20

@love_sodam Oh

RaMathuzen

09:42

@user688539 Is the Theorem actually correct?

user688539

09:59

Why do doubt that it may not be true?

cabmetric

10:38

Can we have this map M: C[0,1] ------> R^N defined as
M(f) = (f(1), f(1/2), f(1/3), f(1/4), ...) onto ?

I think we can. Considering (x_1, x_2, x_3, ...) in R^N I should be able to obtain a function g in C[0,1] such that g(1) = x_1, g(1/2) = x_2, g(1/3) = x_3, so on..

Further, we can define an injective map from C[0,1] to R^N.

Is C[0,1] and R^N bijecitve.

Here R is the field of real numbers and N is the set of all natural numbers.

maths student

0

Q: Inequality to show monotonus increasing function

$$ -\frac{\partial P / P}{\partial y}=\frac{C\left\{-(n / y)+\frac{1}{y^{2}}\left[(1+y)^{n+1}-(1+y)\right]\right\}+n F}{C\left\{\left[(1+y)^{n+1}-(1+y)\right] / y\right\}+F(1+y)}=\frac{C A(y)+n F}{C B(y)+F(1+y)} $$ To show that $-(\partial P / P) / \partial y$ increases monotonically if $y>0$ as ...

robjohn

11:00

@Semiclassical How many points did you plot there? I Plotted 1000000 and mine came out less dense.

RaMathuzen

@user688539 The doubt I ask is somewhat like "reading against the grain"

What if the subspaces of V are disjoint?

i.e They can have zero vector common but not some other

robjohn

@RaMathuzen they all contain the zero vector

oh, I see.

then the intersection has only the zero vector

Koro

@copper.hat I thought about this and got stuck here: For simplicity, let's consider f defined on a subset of $\mathbb R^2$. Then, $f_x$ and $f_y$ exist at c. $f_{xy}$ is continuous at c. That is, $f_x$ and $(f_x)_y$ both are continuous at $c$ so to use theorem 12.11 to conclude that $f_x$ is differentiable at c, one also requires existence of $f_{xx}$, which is not given. :(

Koro

11:23

@robjohn: professor Rob, can you please help me understand what I am missing?

I looked up on mse and found the questions (still unanswered) here math.stackexchange.com/questions/1383105/… and here math.stackexchange.com/questions/1402689/…. Since the question has already been asked twice (atleast) without an answer, I thought of not posting my question.

The equality of the two partials (theorem 12.13) can be shown independently without referring to theorems 12.11 and 12.12 but I want to understand how theorem 12.13 follows "as a consequence of theorem 12.11 and 12.12".

RaMathuzen

@robjohn @user688539 But its fine to accept the theorem as zero vector alone can form a subspace

robjohn

@Koro these proofs rely on what has been shown before and I don't have access to what they have shown before. $f_x$ is definitely differentiable in $y$, but as you say, it may not be differentiable in $x$.

Koro

In one of the above posts, the proof image has also been posted. I see, so you mean that "as a consequence..." also takes stuff from "proof" and not just from the statements of the theorems .

robjohn

$f(x,y)=x^2y\sin(1/x)$, would be something to consider. $f_x$ and $f_{xy}$ exist, but are not continuous where $x=0$

Koro

@RaMathuzen if W is singleton, then also W is a subspace. If not, then suppose that there is some $a$ in W. Then $a$ is in $W_i$ for all $i$ and since each $W_i$ is a subspace, each of these contains $-a$ too and therefore $W$ contains $-a$ too. This in a way justifies taking "let $\alpha$ and $\beta$...". :)

maths student

11:36

1

Q: Inequality to show monotonus increasing function

$$ -\frac{\partial P / P}{\partial y}=\frac{C\left\{-(n / y)+\frac{1}{y^{2}}\left[(1+y)^{n+1}-(1+y)\right]\right\}+n F}{C\left\{\left[(1+y)^{n+1}-(1+y)\right] / y\right\}+F(1+y)}=\frac{C A(y)+n F}{C B(y)+F(1+y)} $$ To show that $-(\partial P / P) / \partial y$ increases monotonically if $y>0$ as ...

real-analysis derivatives inequality

robjohn

@Koro rather than say "let...", a better way to have said it is "suppose..."

"let" supposes existence, where "suppose" does not

2

Vrouvrou

hello, someone know "strophoide" ?

robjohn

@mathsstudent please stop repeatedly posting that to chat. Once is enough, unless you are going to go over something that is confusing you. A better approach would be to put a bounty on the question.

@Vrouvrou you mean this?

Koro

@robjohn $|f_{xy}(x,y)|=|x^2\sin \frac 1x|\leq |x|$ showing that $\lim_{(x,y)\to (0,0)} f(x,y)$ exists. And since $(0,0)$ is not an interior point (because not in domain), how does that give counterexample?

Vrouvrou

@robjohn yes

Koro

11:41

In the theorems (stated above in linked post), we want $c$ to be an interior point. :(

robjohn

@Koro that wasn't my function, but $f_{xx}$ does not exist

and I didn't say it was a counter example. I said it was something to consider

and what is the domain?

I am assuming that $f(0,0)=0$

Koro

By domain here: I meant $\mathbb R^2\setminus y-$ axis. But that can be fixed if I define $f(0,y)=0$ etc.

Vrouvrou

i have this question: calculate the area of the ball of the right strophoide with polar equation $r=cos(2\theta)/cos(\theta)$

i don't know how to do

Koro

And I think I understood your point: existence of $f_x$ and $f_{xy}$ does not give $f_{xx}$

maths student

@robjohn can you give me hint with that question ?

robjohn

11:46

$f(x,y)=x^{4/3}y$ is another thing to consider $f_x$ and $f_{xy}$ are continuous, but $f_{xx}$ does not exist

Koro

That works.

$\ddot\smile$

RaMathuzen

@Koro Your approach is more good than that

Koro

So "as a consequence of..." is probably a typo or is like saying "L'Hospital's rules' are consequences of mean value theorems.

PM 2Ring

@Semiclassical There's an open problem related to that Fibonacci stuff.

RaMathuzen

@Koro thanks

PM 2Ring

11:53

Pisano period

In number theory, the nth Pisano period, written as π(n), is the period with which the sequence of Fibonacci numbers taken modulo n repeats. Pisano periods are named after Leonardo Pisano, better known as Fibonacci. The existence of periodic functions in Fibonacci numbers was noted by Joseph Louis Lagrange in 1774. == Definition == The Fibonacci numbers are the numbers in the integer sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, ... (sequence A000045 in the OEIS)defined by the recurrence relation...

Vrouvrou

can you give me a hint please ?

PM 2Ring

> two primes are anomalous. The prime 2 has an odd Pisano period, and the prime 5 has period that is relatively much larger than the Pisano period of any other prime.

Koro

@RaMathuzen you're welcome. :)

PM 2Ring

There are a few questions about this stuff on the main site, eg math.stackexchange.com/q/1610374/207316

robjohn

@mathsstudent I don't think that your simplification works to show that that function is monotone. You probably have to use other properties of the original functions.

Koro

12:02

@Vrouvrou Did you plot the graph?

maths student

The sensitivity of the percentage price change to changes in interest rates measures price volatility. We define price volatility by $-(\partial P / P) / \partial y$. The price volatility of a coupon bond is
$$
-\frac{\partial P / P}{\partial y}=-\frac{(C / y) n-\left(C / y^{2}\right)\left((1+y)^{n+1}-(1+y)\right)-n F}{(C / y)\left[(1+y)^{n+1}-(1+y)\right]+F(1+y)}
$$
where $n$ is the number of periods before maturity, $y$ is the period yield, $F$ is the par value, and $C$ is the coupon payment per period. For bonds without embedded options, $-(\partial P / P) / \partial y>0$ for obvious rea…

Vrouvrou

i don't know how to plot the graph i don't know where is $\theta$

Koro

You can start by taking $\theta$ in $[0,\frac \pi 2)$ for example.

PM 2Ring

@Semiclassical And here's a fast mod Fibonacci Sage / Python script:

sagecell.sagemath.org/…

Koro

When $\theta =0, r=1; \theta=\frac \pi 4, r=0; \theta=\frac \pi 2, r:=-\infty$

and then also note that $r(\theta)=r(-\theta)$ so above x-axis is mirror image of the graph below x-axis.

Vrouvrou

12:12

i don't know how to plot it

Koro

Without plotting it, I don't see how you'll find the area.

Vrouvrou

12:31

Thorgott

@RaMathuzen "Let $\alpha,\beta\ in W$" does not specify any two elements of $W$ to exist, so your question is not well-posed. What the phrase means is that we take two arbitrary elements of $W$ and just give them the names $\alpha$ and $\beta$. This is to confirm one of the subspace axioms, which has the same syntax: if you take two arbitrary elements of a subspace, their sum is in the subspace. This is what's being checkd.

PM 2Ring

@Vrouvrou Here's a plot from -pi/4 to pi/4

sagecell.sagemath.org/…

Vrouvrou

i forget x

Thorgott

@cabmetric No, the image lies in the subspace of convergent sequences, for if $g\in C[0,1]$, we have that $g(1/n)\rightarrow g(0)$ as $n\rightarrow\infty$ by continuity.

the answer to your second question, $C[0,1]$ and $\mathbb{R}$ having the same cardinality, is that it's true

Vrouvrou

then the area is $\int_0^1 \int_{-x\sqrt{\frac{1-x}{1+x}}}^{x\sqrt{\frac{1-x}{1+x}} dx dy$

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