"Show that for every real number x there is exactly one integer n such that n <= x < n + 1." Where to begin?

i have reals defined in as limits of cauchy seqs

Perhaps the fact that no two integers are closer than 1 unit apart is relevant

Integers are recursively defined with a successor operator, so yeah i think your claim follows inductively.

I heard recursively

>.>

*returns to lurking*

ψ_0 <--- "YARR! SHIVER ME TIMBERS!"

ok so suppose you start with n=0 and keep iterating while n<x: n++

then ill end up with n==floor(x), like i want

if there were 0 such n, that'd mean you never get to x. but that's clearly not gonna happen

if there were more than 1 such n, that'd mean there are >=2 n such that n<=x<1

so suppose you do it once, and get some n meeting the criteria

where's the other one gonna come from

well, maybe you'd encounter one somewhere else below x, maybe you'd encounter one somewhere above x

well one above x isn't gonna work because it has to be <x

what about one below x

it has to be either above or below the floor(x) that we already found

if its above it, we need to show its not gonna be an integer

if its below it we need to show n+1 !>x

if its below, we wouldve found it in the original algorithm, so that's not gonna happen

so the remaining place it could be is above the previously found floor(x) and below x

n+1 is the smallest integer above n, so if x isnt an integer, there's nothing there

if x is an integer, then we wouldve found it n=x, and there is no room between n and x to fit another one

so there's exactly one, the one we found

@AkivaWeinberger qed. how's that?

Lurkers gotta lurk @SimplyBeautifulArt

:P

anybody mind looking at that ^? is there a nicer way to Show that for every real number x there is exactly one integer n such that n <= x < n + 1."?

I feel like you can probably manage something inductively

@SimplyBeautifulArt do you have something in mind, or you just have faith that i can come up with something?

Hopefully both

what's your inductive hypothesis?

Let us show there exists at least one integer such that n ≤ x < n+1 first?

Then show that if n implies n ≤ x < n+1, then this implies every integer greater than n or less than n does not satisfy this property by transitivity

i.e. n+1 ≤ k > x < k+1, so it fails for any k ≥ n+1

I assume we can't just say that n = floor(x) does the job?

@Hatshepsut Assuming x > 0?

@SimplyBeautifulArt yeah

this is the proof that floor() exists

Oh, okay lol

Well, I imagine its equivalent to showing that there exists int n for every real x s.t. x-1 < n ≤ x

Not entirely sure how to prove, got to head to bed. G'night

cya, thanks

I found a flaw with my big proof that I was hoping to publish... :( I feel like strangling someone

:(

depressing

Life goes on pal :(

but I still believe it is true. Just gonna take a lot more work

Well, there you go. All is not lost.

maybe I will need to combine forces with a wizard

Nah, become a grand master wizard!

99% perspiration, right?

Maryam (RIP) said its like "torture."

who is that?

The female Iranian fields medalist who recently died.

(only 40 years old)

One of her quotes was about how patient you have to be to see the beauty of mathematics.

+ the hard work is like torture

[Random] Let a sequence $(a_n)_{n\in \Bbb{N}}$be: 5,0.5,4,0.25,3,0.125,2,0.0625,1,0.03125,1,0.01,1,0.001,1,0.0001,1,0.00001,1,0.00‌​0001,...

Then $\lim_{n\to \infty} a_{2n} = 0$ and $\lim_{n\to \infty} a_{2n+1} = 1$

so $\lim_{n\to \infty} a_n$ does not exist

Also note how $(a_n)_{n}$

$(a_n)_{n\in \Bbb{N}} \supset (a_{2n})_{n\in \Bbb{N}}$ despite both have the same cardinality of $\aleph_0$

“You’re torturing yourself along the way,” she would offer, “but life isn’t supposed to be easy.”

Does anybody have any experience taking the math subject GRE?

I have 1.6 months or so to prepare for it and am wondering how to spend my time. I have covered introductory analysis, linear algebra, abstract algebra; I have been working through the basics of point set topology for fun mainly

Just in case there are some people who might be interested - it seems that some folks want to start some kind of course on general topology tomorrow:

Thanks for the info.

So I've just seen something which I think gets glossed over quite a lot in complex analysis but is worth pointing out

Usually when you're learning that holomorphic functions are black magic, you're doing stuff like Cauchy's theorem and the immediate knowledge of analyticity

And those are definitely crazy

But also, in real analysis, if you want to do term-by-term differentiation, you sorta have a problem

Because to know that $f_n' \to f'$ when $f_n \to f$, you sorta want to have that the $f_n'$ converge to something uniformly

So like, uniform Cauchy-ness is a precondition for being able to differentiate term-by-term

In complex analysis, you can say that if $\sum f_n = f$ and $f_n$ are analytic, you know that $f$ is and that $\sum f_n' = f'$, automatically

Which is pretty insane

Hey @Alessandro, @Tobias, and @s.harp!

@Daminark Hi

How's it going?

Good. Looking through some lecture notes trying to figure out which ways I want to change them for the course I will be teaching this fall

Hmm, which class is it?

Algebra (i.e. introduction to groups, rings and such)

Hi @Dami

That's one of the many magics of complex analysis

I see, that's neat

And yeah @Alessandro

Though the notes I am looking through are just for the final part of the course which is on representation theory. These are fairly new, so they still need some work

Have you seen Cauchy's integral formula yet in the course? @Dami

I see, and that's pretty cool that you're getting to some rep theory!

Yeah @Alessandro

That's some magic too imho

Which version did you see?

@Daminark Yeah, not quite sure how much I can realistically cover. I was hoping to get to show that the degree of an irreducible character divides the order of the group, but that seems to need a bit too much algebraic number theory

@Tobias Is this over the course of a quarter, semester, year?

@Daminark A semester, but the rep theory part is only the very end, and this is the first course the students take in abstract algebra (they are as far as I recall going to be second year undergraduates)

@Alessandro that $f^{(n)}(w) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{(z-w)^{n+1}}dz$

What's $\gamma$ and where's $w$?

So, let's say $f$ is holomorphic on some domain $D$ containing $w$, and $\gamma$ is some closed contour in $D$

Let's say $D$ is simply connected, so this is any contour

In general you can use $\gamma$ null-homologous and $S$ not necessarily simply connected

True, my issue is that I don't yet know much about homology in order to pull off that witchcraft

I mean not null-homologous in this case, presumably homologous to the circle of appropriate radius, no?

@Tobias that sounds like quite a lot in that time, for sure, but at least you won't run out of stuff to do

We didn't know any homology either, but if you pick a chain $\gamma$ in some domain $\Omega$ you can say it's null-homologous if every point outside of $\Omega$ has winding number $0$ wrt $\gamma$

(Chains are formal sums of curves here)

What does winding number with respect to a chain entail? Is it like, degree in $\mathbb{C}\setminus\gamma$?

I don't know what the degree is here

Oh wait degree might not work nicely in that case anyway

Ignore the latter half of that sentence

What is the probability that a random n-by-n binary matrix is linearly independent?

So, what does winding number with respect to a chain involve?

$\text{Ind}_\gamma(z)=\frac1{2\pi i}\oint_\gamma\frac1{w-z}\mathrm{d}w$ is the winding number of $z$ wrt a curve $\gamma$

Okay

It's always an integer, it counts how many times, with sign, $\gamma$ turns around $z$

Okay so it's the same as what I saw but backwards :P

We had defined $\text{Ind}_z(\gamma) = yada yada$

A chain is a formal sum of curves $\Gamma=\sum\limits_{i=1}^n\alpha_i\gamma_i$ and you define the winding number wrt $\Gamma$ as $\sum\alpha_i\text{Ind}_{\gamma_i}(z)$

(For a point outside all of those curves)

@LeakyNun Hmm, that seems tricky and seems to boil down to counting $0-1$-vectors in a given subspace

Which seems to be badly behaved as it depends too much on the subspace

@TedShifrin it would be nice to find a criterium to say about that. Are there any functions that are monotonically increasing and go through $0$? Sure. How many among them have a gradient in $t$ that is exactly a square root from $y(t)$? Intuitivly - only one. But intuition is not mathematics - there should be an equation that says it. For the initial value $1$ there is something to do with $\ln$. For the value $0$ - there are some until I have proven that there are none.

@TedShifrin I know ya only the basic-basic separation of variables, nothing else.

@TedShifrin on the other side, there can be discrete defined functions that are continuous on their domain, partly differentiable somewhere and will suit into the conditions. I have just thought about continuous functions on the whole $\mathbb{R}$ that are differentiable in every point.

@Kirill You mean $y'(t) = [y(t)]^2$?

And do note that the singular of criteria is criterion :)

@LeakyNun oh, German mistake. Sure, criterium is about bicycles, thank you :)

@Kirill is this what you want?

@LeakyNun It might be that, but I do not know the notation on the right side. The question was about the first derivative on the left, square root on the right: $y'=\sqrt{y}$.

Oh... sorry, I reversed it

@LeakyNun Mr. Shifrin is leading me through the basics of differential equations :)

@Kirill is this what you want?

@LeakyNun The solution is not correct

@LeakyNun I have had $\frac{t^2}{4}$ for $t \ge 0$. Maybe, I am thinking.

@Kirill differentiation in discrete spaces is often... ill-defined

Sorry, I forgot the square

I meant, $y = \left(\dfrac{t+C}2\right)^2$

The answer isn't as important as the steps are.

@LeakyNun The steps are also lacking some things. Such as dealing with the fact that the solution has $y=0$ at some points which causes the intermediate steps to be problematic

@LeakyNun I have the same, involving the initial of $y(0)=0$.

@TobiasKildetoft I just wanted to present the general idea to @Kirill :)

@LeakyNun still, with separation of variables I cannot say if there other functions that suit into the equation.

@Kirill why not?

(not to mention that technically, you can't just put an integral sign on both sides of an equation, even integrating with respect to different variables. Nor is line 3 technically meaningful).

@TobiasKildetoft please.

Separation of variables is a bit of a miracle that is intuitively clear except the steps are all wrong

@LeakyNun you will maybe laugh: Fourier methods are perfect, but maybe not perfect enough, so that there some solutions that are not covered by the separation of variables.

@TobiasKildetoft our professors told us about that. It was only said to do it, as it were normal at that moment.

@Kirill And indeed, one does end up with the correct answer

@TobiasKildetoft is this better?

@TobiasKildetoft I will take the differential equations the next year in order to look at that more precisely.

Integration is, in a simplified view, just summation, so summation of both sides seems fine to me

@micsthepick you can't summation both sides either

I had a doubt in Differential equations which i posted here a day ago but how do i find it now ?

@BAYMAX search your message

@LeakyNun So why did you do it in the first place?

@micsthepick I was wrong

Is there a specific case where taking dt to the other side and then putting an integral on both sides specifically creates a paradox, or the wrong result?

@micsthepick "putting an integral on both sides" just isn't a well-defined notation; @TobiasKildetoft on the other hands, differentials are perfectly well-defined, so could you elaborate on the problem of multiplying both sides by a differential?

I fell like $\mathrm{d}x, \mathrm{d}y$ and other guys with $\mathrm{d}$ before a variable are seen like an obsolete notation piece that one should write every time. I have some books where a great attention is paid to the differentials. Why don't I have them in my courses? Are differentials obsolete as idea, can they come later in study or other versions? How do you think?

Two things to note

1.

The first is that when you get to stuff like differential topology, the d has meaning

@LeakyNun How do you define $dy$?

Specifically, $dx_i$ is the projection to the ith coordinate

@TobiasKildetoft proofwiki.org/wiki/Definition:Differential

The other side is that you have measure theory to worry about

Or, you can interpret the $d$ as an infinitesimal :P

@SteamyRoot non-standard analysis, aka heresy! :D

three days pass like a day to you :D

A measure is basically a way to assign "sizes" to certain sets, and if you have a few different measures floating around in a given context, you need to know which you're integrating against

@LeakyNun With that definition, you get that the two sides are now functions of different variables

@SteamyRoot Uh oh...

Yeah time flies :)

If solution exists to the above ODE in $[0,L)$ then range of $L$ is ?

@TobiasKildetoft $\begin{array}{rcl} \dfrac{\mathrm dy}{\mathrm dt} &=& \sqrt y \\ y'(t) &=& \sqrt y \\ \dfrac{y'(t) h}{\sqrt y} &=& \operatorname{id}'(t) h \end{array}$, right?

now putting an integral sign to both sides no longer has any meaning :D

@Daminark so, in common one could say: differentials is not the thing one should study in basics, right?

@Kirill I would say so

now do you understand my solution?

@LeakyNun I have made the same solution the day before, so yes.

@Kirill then what is your question? whether there are some functions we missed?

@LeakyNun we were speaking with Mr. Shifrin, he offered the equation and we were discussing it

@Kirill so you don't have a question?

@LeakyNun :) I am obligated to have questions being a mathematician. It is something different from being a musician.

@LeakyNun let me think, I will definitly post one soon :)

@LeakyNun and thank you for the explanation

We can discuss this statement

I don't think your statement is true

@LeakyNun sure. The thing is - I am given the method they say it works. I just believe that it works and get the result. But then it is religion, not science. So, how much solutions are obtained with the separation of variables? Why?

It isn't a religion.

Each step presented here is an implication

meaning that what satisfies the previous step must satisfy the next step

Overall, this says if $y$ is a function such that $y'=\sqrt y$, then we must have $y=\left(\dfrac{t-C}2\right)^2$.

So there can be no functions missed.

@BAYMAX if $x>0$ then how can solution exist in $[0,L)$?

so you are telling we cannot take $x > 0$ as $y(x = 0) = 2$

@LeakyNun

I'm saying that if you specified $x>0$ then $0$ is obviously not in the domain of $y$ it would make no sense to say that the domain of the solution is $[0,L)$ which includes $0$.

@LeakyNun ok. Saying it another way - I see that you have proven the existence of a solution. I do not see, that there no other ones.

@Kirill I have proven that whatever satisfies $y'=\sqrt y$ must be $y=\left(\dfrac{t-C}2\right)^2$.

Or, if we take the contrapositive: if $y \ne \left(\dfrac{t-C}2\right)^2$, then we cannot have $y'=\sqrt y$.

@LeakyNun according to the method you have chosen. Why there no other methods to get other results? Or, why other methods will lead to the same result?

@Kirill because each step is an implication

each step is $A \implies B$.

what satisfies $A$ must satisfy $B$

as $-\frac{1}{y} = x + c$ when $x = 0$

so $0 \notin $ domain of $y$

?

@LeakyNun

I'm simply pointing out the contradiction between $x>0$ and $[0,L)$.

ok I was thinking when $x = 0$ then why $y \neq 0$

?

never mind, it was a pointless distinction. Ignore what I said so far.

ok

@LeakyNun I will read some logic again, as I do not think implications say all the truth. Say I want to find the smallest number in the world. 8 is the smallest number is the world. Lets take 9. 9>8 $\Rightarrow$ the statement ir true. The first one was false, the second one is right, but the formal implication is still true.

"If $9>8$, then $8$ is the smallest number in the world"

@LeakyNun maybe I mix formal and normal logics, not sure

"$9>8$" is true and "$8$ is the smallest number in the world" is false

so the formal implication is false

and I do not see your point.

@Kirill should we deal with a more simple example?

Each step is an implication

What satisfies the first line must satisfy the third line

but you can see that what satisfies the third line does not necessary satisfy the first line

@LeakyNun You have a sqrt that should not be there

@TobiasKildetoft where?

the first line

@LeakyNun is an equivalent transformation always an implication?

@TobiasKildetoft that was stupid

@Kirill equivalent ($\iff$) is even stronger than ($\implies$), so yes

@LeakyNun what does stronger or weeker means here?

@Kirill it means yes

@LeakyNun no, when is the statement $A$ stronger than $B$?

@Kirill when $A \implies B$

ok

@LeakyNun so, setting the initial value implies the constant $C$ in our example?

implies fixing the constant $C$.

the constant $C$ isn't a statement

i.e. I don't want to test each choice one by one

@LeakyNun you have an indefinite integral. But as I see, setting $y(0)=0$ implies that there is only one function, as you say. I still try to understand why.

too complicated

@LeakyNun i have $y(t)=\left(\frac{t-t_0+2\sqrt{y(t_0)}}{2}\right)^2$. So, your constant is $C=t_0 - 2\sqrt{y(t_0)}$, right?

@Kirill yes

@Kirill the indefinite integral generates many solutions, only one of which satisfies the constraint $y(0)=0$.

@LeakyNun that was it what I meant. Giving a starting value, you get the only one $C$, so only one solution.

so?

@LeakyNun so that was the question I was asking.

what is the question?

@LeakyNun What was or what is?

whatever

are you still not convinced that there cannot be other solutions?

@LeakyNun still.

Let's look at this example.

The first line implies the second line; and the second line implies the third line.

As a result, whatever satisfies the first line must satisfy the third line.

@LeakyNun I am ok with $C$ now. Whats about $t$? Is there nothing to substitue inside to get another solution?

[Super random] To be expanded: a mathematical object that has well defined building blocks but it's constructs cannot be explicitly written down

However, it is not true that whatever satisfies the third line satisfy the first line.

@Kirill $y$ is a function in $t$. $t$ is not what you can control.

@LeakyNun yes I got it from the first time.

Then the same with this.

Whatever satisfies the first line must satisfy the last line.

Do we agree? @Kirill

@LeakyNun can $t$ be a function itself?

@LeakyNun yes, we agree.

@Kirill so there are no other solutions!

@Kirill it doesn't matter

@LeakyNun you could probably make a variable change somewhere that would help (I'm on my phone so I'm no too sure)

@Astyx where?

However you can definitely change these to inequalities and go from there

can't we use dynamic programming?

However you want an algorithm that solves it right ?

@Astyx yes

What are the parameters ?

googles parameter

well, there would be $n$ variables

wait a minute

What are the variables ? You can solve this one by hand probably

Plugging 1 and 3 in 2 gives you a set of values y can take

and there can be way more than $10$.

In fact, there are $80$ in the real version of my problem

And I need to solve such equations $80$ times.

Right, harder to do by hand

Are the x_i only integers ? Naturals ?

positive integers

It's actually an MCFP, but I haven't read through the algorithms yet...

I just thought that maybe dynamic programming would help.

I don't have the time right now but I'll give it some thoughts, it looks interesting

@Astyx thanks

I guess you'd do it recursively by making hypothesis on x_0

Hello.

That is suppose x_0 = b_0 (which gives a lower bound for x_1) and you're left with the n-1 case plus some condition on x_1

Otherwise you have x_0 =x_1 +C_1 and then again with some work you can get the n-1 case plus some condition

(Are the constants the same throughout the equations ? I feel like you could generalise to different constants for each equation with complicating things too much)

@LeakyNun

@Astyx The $x$ in this set of equations becomes the $b$ in the next set of equations

I'd appreciate some help..

and the $c$ is completely different

@Mahmoud $(x-y)^2 \ge 0$

I need to prove the Schwarz inequality

It's hard to explain on my phone, it's doable via recursion

$x_1 y_1+x_2 y_2 \le \sqrt{{x_1}^2+{x_2}^2} \sqrt{{y_1}^2+{y_2}^2}$

@Astyx so basically we try the first value of $x$, reduce the problem, and then try the second value?

No no, you get the solutions from the n-1 case and apply you conditions to get the n case

but... the last equation in the n-1 case is different from the second-last equation in the n case

Can we discuss this in 6 to 7 hours ?

@Astyx bien sur.

@AkivaWeinberger tienes alguna idea?

(The c_i change from one line to another)

Hello, can you help me for this question math.stackexchange.com/questions/2365917/…

Nevermind, I figured it out

@Astyx wouldn't two iterations of dynamic programming be enough?

I mean, set $x_0$ to $b_0$, and then dynammic program once in the correct order

and then starting from the last value, dynamic program once in the reverse order

this is what I mean

Dynamic programming doesn't seem to be adapted for this unless I'm mistaken

(Can't try your link online on my phone)

@Mahmoud the harder version of that is to prove that it works regardless of how many real variables you choose (your case being the 2-variable one).

How do you (specifically you) prove Cauchy Schwartz btw ?

That's what I'm trying to remember if I'm honest

ams.org/mathimagery/…

Imagine carrying the image of a kaiedoscope around. It is roooooouuuunnnnndddd and it morphs

I think one may be able to use induction to get the cases with $n=2^k.$ one can then obtain any n less than 2^k by setting an appropriate number of x,y to zero

There should also be a geometric approach, since the dot product is related to the scalar projection of y onto x.

I think one may also be able to use Lagrangian multipliers

@Astyx right, it didn't work for the 80x80 matrix

(I tested it with a 5x5 matrix provided and it worked)

So a bunch of ways

@Semiclassical would you have any idea?

Ew

Which is to say, no

@Semiclassical the proof I've learned uses a trick by exploiting polynomials

I feel like I've seen that, but not recently

That is, consider $t\mapsto ||x+ty||^2$

(I hope this renders correctly)

@Astyx nope

Use \| which gives $\|$

@Astyx \Vert

thumbsup

Oh right, V, I need to revise my latex commands ...

@LeakyNun btw, to what mathematical field refers your problem?

Compsci

right, computer science

@LeakyNun your equation set :)

@Astyx you're right, dynamic programming is wrong here

it merely returns the local minimum

I'm pretty sure recursion is the way to go

The 1 equation case is trivial and you can divide the probleme into two subcases

> probleme

Meh

@Astyx and then?

However this would lead to an exponential complexity

oh, never mind then

I might be saying crap, I'll take pen and paper when I get home

Still waiting for my plane ...

@Astyx bon voyage

Merci

@LeakyNun my thinking would be to consider the n=3 case first, and see what insight can be gleaned from that.

@Semiclassical I can't even solve n=3 o_O

well, in that particular instance the situation can at least be visualized

$x_0=\text{min}(b_0,c_1+x_1)$ means that $(x_0,x_2,x_2)$ lies somewhere on the boundary of the region defined by $x_0\leq b_0,x_0\leq x_1+c_1$

visualizing does not reduce the complexity :p

in this manner you get a total of ...2+3+2=7 inequalities which together define some region in the positive orthant of $(x_0,x_1,x_2)$ space

yeah.