Mathematics - 2017-04-14 (page 2 of 5)

Anonymous

10:00

Can the functional equation be solved?

DHMO

@blue in general a functional equation has 1,000,000 solutions

@blue is that your whole question?

Anonymous

@DHMO Well I have been given that f(1)=1

Anonymous

@DHMO No

Anonymous

I'm showing the whole question

Anonymous

One sec

DHMO

10:01

You should have shown your whole question in the beginning

Anonymous

Anonymous

@DHMO Really sorry

DHMO

that isn't the whole question

Anonymous

Uploading the options

Anonymous

10:03

Yep

Anonymous

That's it^

Anonymous

@DHMO

DHMO

can you find another value of x such that f(x)=1?

I guess not

Anonymous

@DHMO f(1)=f(2(1)-1)=1

Anonymous

So

Anonymous

10:04

no

Anonymous

How to deduce the continuity stuff?

DHMO

give me a minute; i am thinking

SteamyRoot

I'm guessing the question is to give the answer that is correct for all such $f$ ?

Well, I know at least one function $f$ that satisfies $3$ of those conditions :P

DHMO

@SteamyRoot you can't satisfy 3

A and B are contradictory

C and D are contradictory

SteamyRoot

Oh, wait, $2$ of them, yeah

forgot the "only" bit

DHMO

10:07

@blue can you fill in the blank? f(2) = f(___)

BAYMAX

me too

Anonymous

f(2)=f(3)..not useful

Anonymous

If 2x-1=2

Anonymous

Then x=3/2

DHMO

yes

now can you work on f(3/2)?

essentially, f(x) = f((x+1)/2)

Anonymous

10:10

Oh right

Anonymous

f(2)=f(3/2)

DHMO

and f(3/2) = ?

Anonymous

No

Anonymous

Ok lets see

Anonymous

If we use f(x)=f((x+1)/2)

Anonymous

10:12

x=3/2

Anonymous

Then we get f(5/4)

BAYMAX

$f(3/2) = f(5/4) = f() = f() = f()$

DHMO

@blue if we continue like this, what would we get?

Anonymous

Doesn't help

DHMO

@blue it does

Anonymous

10:13

Oh

Anonymous

It tends to 1

DHMO

so?

SteamyRoot

Hmmm

Anonymous

And f is continuous...

DHMO

bingo

@SteamyRoot Was?

Anonymous

10:14

So at infinitely many points f(x)=1

Anonymous

But not at all points (?)

DHMO

we don't need all points

SteamyRoot

It's a shame... with a slightly different problem I could've probably made a nice function that was only continuous at $1$.

DHMO

@SteamyRoot tell us

SteamyRoot

It just uses the idea that you can make a function that is continuous in a single point

Like, the indicator function on the rationals is continuous only at $x = 0$.

But, it doesn't work here. oh well

DHMO

10:16

@SteamyRoot surely there are a lot of functions that are continuous in a single point

Anonymous

So only option A is correct? @DHMO I'm not sure about the continuity part

DHMO

@SteamyRoot not that

@blue the continuity part cannot be decided

do you understand why f(2)=1?

SteamyRoot

Well, yeah, but that's one of the standard examples

Which is nice enough to modify easily

Anonymous

@DHMO Yes, that I understood

DHMO

@blue why?

And I'm thinking why can't I use the same argument to argument that f(x)=1 for all x?

Anonymous

10:17

@DHMO Because f is continuous at x=1

DHMO

@blue alright

I could prove that f(x)=1 for all x>1

no

for all real x

@SteamyRoot can you make up a function satisfying the conditions but is not constant?

@blue that would make D correct...

SteamyRoot

Not easily, unless I'm missing something

Anonymous

@DHMO How?

DHMO

@blue using the same method really

Anonymous

Say we start with $x=\sqrt{2}$

DHMO

10:20

yes

Anonymous

Yes

Anonymous

It is working for sqrt(2)

DHMO

yes it is

so your question is quite fxcked up

Anonymous

Cool

Anonymous

=P

DHMO

10:23

would you consider putting to the main to have them verify?

Anonymous

@DHMO Okay. But your solution seems correct!

DHMO

"Let $f: \Bbb R \to \Bbb R$ be a function which satisfies $f(x) = f(2x-1)$ for all $x \in \Bbb R$ and $f$ is continuous at $x=1$. Is it true that $f$ must be constant?"

@blue ^

Anonymous

0

Q: Is it true that $f$ must be constant?

"Let $f: \Bbb R \to \Bbb R$ be a function which satisfies $f(x) = f(2x-1)$ for all $x \in \Bbb R$ and $f$ is continuous at $x=1$. Is it true that $f$ must be constant?"

Anonymous

Ok?

DHMO

thanks

Anonymous

10:35

Anonymous

Does $g'(0)$ exist or not?

Anonymous

$g'(x)=xf''(x)+f'(x)$

Anonymous

$f''(0)$ doesn't exist

Anonymous

Is it okay to say $0.f''(0)=0$ and $g'(0)=1$

Anonymous

Any idea @DHMO ?

DHMO

10:37

wait a minute

i am thinking

$g'(0) = \displaystyle \lim_{h\to0} \frac {g(h) - g(0)} {h} = \lim_{h\to0} \frac {hf'(h) - 0} {h} = \lim_{h\to0} f'(h) = f'(0) = 1$

Anonymous

@DHMO Yes, by that it is 1

DHMO

by what?

Anonymous

By that method

Anonymous

But from $g'(x)=xf''(x)+f'(x)$

Anonymous

It isn't so evident

Balarka Sen

10:40

You cannot apply chain rule.

DHMO

@BalarkaSen it was product rule?

Balarka Sen

Product rule, yeah.

Product rule is applicable iff the factors are differentiable.

Anonymous

@BalarkaSen Oh I see

DHMO

@blue do you want a question from me in the same vein?

Anonymous

@DHMO Sure

DHMO

10:46

@blue see this answer

so your question is pretty much done

@blue let f(x) = x^2 sin(1/x), and f(0) = 0.

is f'(x) continuous at x=0?

Anonymous

f'(0)=0

Anonymous

ok after that...

DHMO

@blue how?

Anonymous

(h^2sin(1/h)-0)/(h-0) as h->0

Anonymous

So it becomes $h\sin(1/h)$

DHMO

10:49

yes

Anonymous

So that is 0

Anonymous

as sine lies between -1 to 1

DHMO

agreed; continue

Anonymous

And h->0

Anonymous

Then we differentiate f(x)

Anonymous

10:50

And check for continuity

Anonymous

$2x\sin(1/x)-\cos(1/x)$

Anonymous

lim x->0

Anonymous

The cosine is oscillating

Anonymous

So cannot be continuous

DHMO

nice

Anonymous

10:52

good question :-)

DHMO

Consider $f(x) = x^2\sin \left(\dfrac{1}{x}\right)+\dfrac{x}{2}$

(a) Find f'(0)

(b) Prove that f is not increasing in any interval containing 0

Anonymous

Not increasing or not decreasing?

Anonymous

That seems like an increasing function

Balarka Sen

it is not.

Anonymous

desmos.com/calculator/ziojrj8uo2

Anonymous

11:01

Anonymous

@BalarkaSen How does it seem decreasing/non increasing ?

Balarka Sen

It's not strictly increasing. There are "ups" and "downs" on any interval containing 0.

DHMO

@blue your eyes deceive you

Anonymous

@BalarkaSen Wait a second. What does interval containing 0 mean ? You mean like [0,2) ?

Anonymous

Yes it is not strictly increasing

Balarka Sen

11:03

(0, 2) does not contain 0.

(-1, 1) does.

Or (-epsilon, epsilon) for any epsilon >0

Anonymous

[0,2) ?

Balarka Sen

Yes but in general half open intervals are dumb intervals. Usually when one says "interval around 0" one means an open interval containing it.

Anonymous

Yeah, so taking the derivative does the job

Anonymous

It is an oscillatory function

Anonymous

And so obviously not strictly increasing

Balarka Sen

11:07

I can give you oscillatory functions which are strictly increasing.

Anonymous

I mean the derivative is oscillatory

Balarka Sen

Right :)

Anonymous

It takes both positive and negative signs

DHMO

@BalarkaSen you can?

Thorgott

f'(0) isn't defined, is it

DHMO

11:09

@Thorgott it is

Thorgott

In which sense

Anonymous

x+sin(x)

DHMO

@Thorgott f'(0) is 0.5

Balarka Sen

@DHMO For sure. Think of something which looks like $x^3$ near each integer.

DHMO

@BalarkaSen how do you define "oscillatory"?

Balarka Sen

11:11

I don't. How do you? :P

DHMO

functions which are increasing and then decreasing and then increasing etc

Balarka Sen

By that definition trivially there can be no strictly increasing oscillatory functions :P

BAYMAX

actually it is wiggles

DHMO

agreed @BalarkaSen

Thorgott

Do you derive that as a limit?

DHMO

11:13

@Thorgott no, f'(0) is really 0.5

Thorgott

How do you arrive at that result?

1/0 isn't defined

Anonymous

@DHMO The variation of a function which exhibits slope changes, also called the saltus of a function. A series may also oscillate, causing it not to converge. mathworld.wolfram.com/Oscillation.html

Anonymous

So even $x+\sin(x)$ should be oscillatory

BAYMAX

some kind of functions of bounded variation

Anonymous

It is a vague term though

Thorgott

11:18

$f(x)=x+sin(x)$ has derivative $f'(x)=1+cos(x)$ which is always non-negative, so the function is monotonically increasing, yet not strictly

Anonymous

@Thorgott The question is: Can it be called oscillatory?

Thorgott

It exhibits periodic slope changes, so I could bring myself to say so

Anonymous

@Thorgott I'd agree

Thorgott

Anyways, $[x^2sin(1/x)+x/2]'=2xsin(1/x)-cos(1/x)+1/2$, so how exactly would $f'(0)$ be derived when $1/0$ isn't defined and even the limit $lim_{x\rightarrow0}cos(1/x)$ doesn't exist.

Anonymous

@Thorgott Use the basic definition of derivatives

Anonymous

11:32

The limit definition

Anonymous

It is not a continuously derivable function

Thorgott

How can I do that when $f(0)$ isn't defined either

Studentmath

look at the limit from the left + limit from the right

I've got a rusty question. I want to show that $\pi =2+4\sum_{n=1}^\infty \frac{(-1)^n}{(1-4n^2)}$ using the Fourier series of $cos(x/2)$.
So I managed to get to $\pi^2=4^2(1/2+\sum_{n=1}^\infty |\frac{(-1)^n}{(1-4n^2)}|^2)$ And for some reason I get stuck there. I forgot how it works with infinite sums..

Would appreciate a tip

DHMO

11:57

@Thorgott I forgot to put f(0) = 0. It is my fault.

Thorgott

Oh, alright, then $f'(0)=0.5$ makes sense

Secret

Random and weird question: Does there exists nonlinear operators that cannot be represented as a power series of linear operators?

i.e. suppose $A$ is a linear operator. A power series of linear operators will be something like: $\sum_{i}^{\infty}a_iA^{i}$ where $a_i$ are the coefficients. Of course, the power series has to converge for it to be well defined

Studentmath

Or maybe I should've approached it differently, but it looks like one step away from the right answer

Secret

0

Q: How to represent non-linear operators computationally?

I have a finite dimensional vector space V, and want to compute a non-linear operator $R: V \rightarrow V$. I want to have a "general" form of this operator R. I think of the following series expansion: $$ R = a_0 I + a_1 T + a_2 T^2 + a_3 T^3 + ... $$ where T is a linear operator over V. But ...

functional-analysis numerical-methods operator-theory nonlinear-system

Ok nvm, false in general

DHMO

@Secret I thought power series of linear operators must be linear?

Secret

12:07

@DHMO yeah, that's what the MSE said (and hence why what I said previously is false)

DHMO

Alright

Is $\{a+b\sqrt2:a,b \in \Bbb Z\}$ dense in $\Bbb R$?

@Secret

Secret

$1<\sqrt{2}$ but no $x$ such that $1<x<\sqrt{2}$, thus no

DHMO

I don't understand

Secret

A set $S$ is dense if $\forall x,y\in S, x<y, \exists z\in S, x<z<y$?

DHMO

that is a necessary but not sufficient condition, but go on

Secret

12:20

wait, not sufficient? Ok I think I have a knowledge gap here...

DHMO

For all intends and purposes, $[0,1) \cup \{2\}$ satisfies that

Secret

12:41

Hmm, the topology of $\Bbb{Z}[\sqrt{2}]$ is the subspace topology in $\Bbb{R}$.

Since I can find e.g. $\{1\}=(0,2) \cap \{1\}$ and $(0,2)$ is open in the standard topology of $\Bbb{R}$, therefore all singletons in $\Bbb{Z}[\sqrt{2}]$ are open wrt $\Bbb{R}$...

DHMO

Is that to me?

Secret

not quite, because it is not an answer

DHMO

None of your statements whatsoever make any sense to me.

Are you saying that the only element of $\Bbb Z[\sqrt 2]$ in $(0,2)$ is $1$?

Secret

No, I just pick $\{1\}$ as an example to illustrate that all singletons are open wrt $\Bbb{R}$ in the subspace topology

DHMO

Read the first line of your link.

Secret

12:48

Yes, $\Bbb{Z}[\sqrt{2}] \subset \Bbb{R}$ thus it is a subspace of $\Bbb{R}$. Thus its most natural topology is the subspace topology (the topology induced from the topology of $\Bbb{R}$, which is usually assumed to be the standard topology if it is not mentioned)

DHMO

Continue proving that $\{1\}$ is open.

Secret

The open sets in this topology is then given by the intersections of one of the open sets in the larger set (in this case, open intervals in $\Bbb{R}$) with the subspace (in this case, $\Bbb{Z}[\sqrt{2}]$). Now $\{1\}=(0,2)\cap \{1\}$ where $(0,2)$ is open in $\Bbb{R}$ and $\{1\} \subset \Bbb{Z}[\sqrt{2}]$ hence $\{1\}$ is open in $\Bbb{Z}[\sqrt{2}]$ wrt this topology

DHMO

You are supposed to intersect one of the open sets in $\Bbb R$ with the subspace $\Bbb Z[\sqrt2]$.

Not with $\{1\}$.

Secret

ok sorry, using (0,1.5) instead. Then $(0,1.5) \cap \Bbb{Z}[\sqrt{2}]=\{1\}$

DHMO

Not really.

What do you think $\Bbb Z[\sqrt2]$ means?

Secret

12:57

$\Bbb{Z}[\sqrt{2}]=\{a+b\sqrt2:a,b \in \Bbb Z\}$ ?

DHMO

Yes.

And what does $\Bbb Z$ mean?

Secret

Ok, I missed out the element $2-\sqrt{2}$, let me try again

nope...

$2-\sqrt{2}$, $3-2\sqrt{2}$, $5-3\sqrt{2}$ ... something strange happens...

Ok, I am not sure how to prove it, but it seems $\lim_{(a,b)\to (a',b')}a-b\sqrt{2} =0$

Kasmir Khaan

Hello all !

DHMO

@Secret what the hell is $a',b'$?

Secret

The expression $a-b\sqrt{2}$ can get arbitrarily close to 0, However, once again, I have no idea how to test all neighbourhoods of $0$ (because the open sets in $\Bbb{Z}[\sqrt{2}]$ are countable sequences of singletons) and thus showed that $0$ is a limit point

put it in another way, the observation of the elements $a-b\sqrt{2}$ suggest $\Bbb{Z}[\sqrt{2}]$ is dense in $\Bbb{R}$, but I am not sure how to prove it

DHMO

13:09

@Secret do you want hints?

Secret

I think so

DHMO

Show that there are infinitely many elements in $[0,1)$.

nbro

13:55

Hi!

in which course(s) are usually tensors introduced in a mathematical curriculum?

Semiclassical

In physics, they typically make an appearance once you start doing relativistic physics as well as upper-level electrodynamics or mechanics e.g. multipole moments.

(More upper-level relativistic physics, though. You don't need tensors to discuss length contraction/time dilation.)

Secret

@DHMO There are no continuous bijection $f: [0,1) \to \Bbb{R}$ thus we consider the following discontinous bijection: Let the countable sequence $(x_n)=\{0,0.9,0.99,0.999,....\}$. Define $f(x)$ such that $f(x_n)=x_{n+1}$ and $f(y)=y-0.5$. Next define $g(f(x))=\tan (\pi f(x))$. This thus give a bijection to $\Bbb{R}$ and hence $[0,1)$ has $\mathfrak{c}$ many elements

Semiclassical

In math, I think the main place you'd start seeing it is in differential geometry?

i mean, there's a little bit of it in linear algebra to the extent that matrices are rank 2 tensors. But that's a bit too cheap.

DHMO

@Secret oh I see we have a misunderstanding

I meant show that there are infinitely many elements of $\Bbb Z[\sqrt2]$ in $[0,1)$.

Semiclassical

Is there a reason to have that as [) rather than ()?

DHMO

14:05

@Semiclassical no

Semiclassical

mmkay.

So you want to show that there are infinitely many integers $a,b$ such that $0<a-b\sqrt{2}<1$. (probably said this above somewhere.)

Secret

Well I guess that's what I am stuck at, because there seemed to be no nice pattern between $a_n,b_n$ such that a sequence $(a_n+b_n\sqrt{2})\to 0$

I mean, yes, I can see how $2-\sqrt{2}$, $3-2\sqrt{2}$ are getting towards zero, but $4-3\sqrt{2}$ overshoots it

Semiclassical

One cheap way, I think, would be to obtain a sequence of such $a,b$ via the continued fraction of $\sqrt{2}$.

DHMO

@Semiclassical no need to do that

there's a much more elegant proof

@Secret proofs don't have to be constructive

Semiclassical

Hence why I said cheap :)

DHMO

14:07

look for the wider world

Semiclassical

I like constructive proofs, though.

DHMO

macroscopic view

not microscopic view

Semiclassical

I would presume that this holds for any $a+b\sqrt{q}$ with $q>0$ square-free.

DHMO

@Semiclassical that would hold for any $a+br$ with $r$ irrational.

Semiclassical

Yeah.

nbro

14:10

@Semiclassical ok... please, next time tag me ;)

Semiclassical

Woops.

I think you could still exhibit such an infinite sequence via the continued fraction for $r$, but you wouldn't be able to give an explicit formula.

Whereas you could for $\sqrt{q}$.

DHMO

whereas my macroscopic non-constructive argument would still work

Semiclassical

Right.

DHMO

$q$ doesn't have to be square-free though.

Semiclassical

You're right, it just can't be a perfect square.

But without loss of generality it's square-free, I think?

Anyways. If you use the continued-fraction expansion, the sequence for $\sqrt{2}$ starts as

$-1 + \sqrt{2},3 - 2 \sqrt{2},-7 + 5 \sqrt{2},17 - 12 \sqrt{2},-41 + 29\sqrt{2},99 - 70\sqrt{2},-239 + 169 \sqrt{2}, 577 - 408 \sqrt{2},-1393 + 985 \sqrt{2},3363 - 2378 \sqrt{2}$

bah

(one has the alternating signs because the continued-fraction convergents for sqrt(2) are alternatively over/under estimates.)

Amusingly, if you take $r$ to be the golden ratio $\phi$, you get the sequence $\phi-1,2-\phi,2\phi-3,5-3\phi,5\phi-8,\ldots$

So that in this case the coefficients are, up to signs, just the Fibonacci numbers.

Not surprising if you know much about Fibonacci numbers, but still fun.

More broadly you can express the coefficients via an appropriate recursion relation and thereby express them in powers of $r$.

Fawad

14:33

How to show $\cos^4 x +\cos^2 x =1$ when $\sin x=\dfrac{\sqrt5 -1}{2}$ ?

Semiclassical

Write $t=\sin^2 x$, rewrite the first equation using this, and find the roots of the resulting quadratic in $t$. Only one of these will have a corresponding real value of $x$.

Jack Lam

hey can someone help me with a "simple" step in this paper:
http://fermatslibrary.com/s/a-proof-that-euler-missed
Page 2, column 2. I can't seem to make the substitution work. How does it work? u is a function of both variables, so how is it valid to treat x as constant?

Secret

@Semiclassical Ok, I think I have a proof by contradiction (but it is nice to see the sequence you have wrote down via continued fractions (which I have not thought of because I am not that good at continued fractions yet):

Suppose there are finite number of elements of $\Bbb{Z}[\sqrt{2}]$ in $(0,1)$. Then since this set is finite, it must have a minimum. Let the smallest element be $a+b\sqrt{2}$. Therefore $0<a+b\sqrt{2}<1$. Now if we sqaure this element, the new element $0<(a+b\sqrt{2})^2=a^2+2b^2+2ab\sqrt{2}<a+b\sqrt{2}$, contradicting the assumption that the set is finite. Now since $\…

DHMO

Nice

Secret

In fact, this proof by contradiction happens to be constructive, since the sequence $((a+b\sqrt{2})^n)\to 0$

Semiclassical

14:38

One hole I see: A finite set need only have a minimum if it is nonempty.

DHMO

@Semiclassical very nice.

Semiclassical

So you should also argue that there's at least one element in (0,1).

Secret

Ah right

Semiclassical

Not that that's really a challenge.

This also reproduces the same sequence I gave above, interestingly.

Fawad

@Semiclassical sorry but I don't want to "solve" that quadratic equation. I need to "make" quadratic equation from that $\sin x=...$

Semiclassical

14:41

What?

DHMO

@Fawad 1. find $\sin^2 x$.

Secret

My first guess to your observation might be perhaps there's a connection between powers of elements in $\Bbb{Z}[q]$ and its continued fraction expansion.

DHMO

2. Substitute it by $t$

3. Plug it in to verify.

Semiclassical

@Secret Yeah, almost certainly.

DHMO

Is anybody interested in my proof?

Semiclassical

14:42

Sure.

DHMO

Fix $b$. There must be an $a$ such that $0 < a + b \sqrt 2 < 1$.

Since the possible values of $b$ is $\Bbb Z$, it is countably infinite.

Semiclassical

What's the domain of $a,b$?

DHMO

$\Bbb Z$

Semiclassical

Hmm, okay.

Secret

How do we show there must be an a obeying that inequality that corresponds to any $b \in \Bbb{Z}$?

Semiclassical

14:45

Because $b\sqrt{2}$ is a real number.

DHMO

@Secret $b \sqrt 2 \pmod 1$

Archimedean property

Semiclassical

So it has a definite floor.

Yeah, that's cute.

I like the constructive aspect of @Secret's approach, but this is definitely better in generality.

Fawad

@DHMO you didn't understand 😓 I know $sin x=..$ and need to make quadratic equation in terms of $cos x$

Secret

I am in general terrible at non constructive proofs (though after reading it I can often understand it) because of how I am used to have something "tangible" to manipulate around with when solving problems

DHMO

@Fawad $\sin ^2 x + \cos^2 x = 1$

Meow Mix

14:48

Hi

Jack Lam

cos⁴x+cos²x=1
(1-sin²x)²=sin²x

Semiclassical

^

Hence what I said about $\sin^2 x$.

Jack Lam

now

can we move on to my integral?

Semiclassical

"Just ask, don't ask to ask."

DHMO

what is your integral?

Fawad

14:49

@JackLam cool,thanks

Semiclassical

Oh, the Apostal proof.

Jack Lam

hey can someone help me with a "simple" step in this paper:
http://fermatslibrary.com/s/a-proof-that-euler-missed
Page 2, column 2. I can't seem to make the substitution work. How does it work? u is a function of both variables, so how is it valid to treat x as constant?

thankgoodness for my clipboard memory

Semiclassical

Think about the Jacobian of that transformation.

Jack Lam

I tried that

Semiclassical

Then it should be apparent that $\int f(u,x)\,dudx=\int f(\cos\varphi,x)\,d(\cos\varphi)dx$.

Jack Lam

14:52

...um

I'm know how to get from that line to the conclusion.

Secret

@DHMO ok, so by repeating our proofs by translating the inequality up or down any element in $\Bbb{Z}[\sqrt{2}]$, we thus showed that all elements in $\Bbb{Z}[\sqrt{2}]$ are limit points. But how to pick sequences in $\Bbb{Z}[\sqrt{2}]$ such that it approaches to real numbers not in $\Bbb{Z}[\sqrt{2}]$ ?

Jack Lam

I don't know how to get from the original form of P+Q, to the intermediate integral.

in terms of u and x

because I can't get rid of a factor of 1/(1-xy)

also the region of transformation isn't even nice

Semiclassical

oh, I thought you meant the next change of variables.

DHMO

@Secret when did we show that all elements in $\Bbb Z[\sqrt 2]$ are limit points?

Semiclassical

you said page 2 column 2, but the initial integral is right at the bottom of column 1.

But it should still be a matter of the Jacobian

Jack Lam

14:54

apologies, but I was referring to the given substitution

the jacobian gives me the same result as a naive change of integrals

unless you would care to demonstrate otherwise?

Secret

@DHMO We have showed 0 is a limit point by the 3 different versions of the proofs that we have done above. There's no reason that the same argument cannot hold for arbitrary (a,b)? (well except for my proof, as the squaring argument will fail for any interval that is not (0,1))

Akiva Weinberger

@DHMO You have a sequence of thingies that converge to $0$. Add your favorite element of $\Bbb Z[\sqrt2]$ to each element of that sequence.

Semiclassical

Well, the Jacobian is enough to explain why you only care about $\partial u/\partial y$ in said transformation.

DHMO

@Secret The group is additive. If 0 is a limit point, it must be dense everywhere.

@Secret My proof only showed that there are infinitely many elements of $\Bbb Z[\sqrt 2]$ in $[0,1)$..

Meow Mix

Hi @Akiva

DHMO

14:59

@AkivaWeinberger I know. I'm teaching him.

Akiva Weinberger

Ohh

Derp

Sorry

Semiclassical

Hence the substitution would give $dx\,du = (\partial u/\partial y)\,dx\,dy = (1+x y)\,dx\,dy$.

DHMO

And it's called the pigeonhole principle @AkivaWeinberger

Akiva Weinberger

No it isn't

I mean, you can prove it with that

SBM

Um, no

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