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12:48 AM

1:38 AM
it is now

1:58 AM
Well now, this is funny. In writing the paper, I realized something about an older identity that I had found for quotients which appears to be even cheaper than I initially had realized, but my mind was too weak to notice it at the time.
So, obviously, a reciprocal $\frac{1}{y}$ can be represented in base $y$ as $(0.1)_y$. Great.
Converting it to binary is as trivial as computing $2^{k(n)} \bmod y$ where $k(n) = k(n) + 1 | k(0) = 0$. We do this up to $n$ for $n$ bits of precision. If we then evaluate $2(2^{k(n)}\bmod y)$ and it is greater than $y$, then the value of that bit is one, and zero otherwise.
But me and my slow brain just realized that this describes a number $2^{n+1} - 1$.
It's almost possible to just use $(2^{n+1} - 1)\bmod y$ to do this, but it's off by one step. We recognize that $(a + b) \bmod c = ((a\bmod c) + (b\bmod c)) \bmod c$.
In place of the latter, what we have is, for $n$ bits, an expression of the form $(a_0\bmod c) + (a_1\bmod c) + ... + (a_n\bmod c)$.

2:44 AM
Wowwww ok so I can literally just compute in hexadecimal to get more throughput (or even base $2^8$ or $2^{16}$ depending on how high we can divide) since the corresponding digit in that base is apparently just the floored quotient $\lfloor\frac{b(b^n \bmod y)}{y}\rfloor$. Using base $2^{16}$ with four execution units, you could get insanely good latency and throughput with integers.
It also would appear that $\frac{1}{y} = \sum_{n=1}^{\infty} (y+1)^{-n}$ just at a glance, but that could be wrong.
Looks like Wolfram agrees. wolframalpha.com/input/…
This makes sense given what I learned as well about infinite sum representations of a reciprocal in its corresponding integer base, viz. that $(0.1)_y = (0.0(y-1)(y-1)(y-1)(...))_y$ where $AB$ and $(A)(B)$ here specifically represents the concatenation of two integers.

3:09 AM
Huh... ok, welp... looks like I'm archiving the paper draft and starting a new one.

4:07 AM
The ode $y’+\frac 1x y=1$ is non-linear.

koro, huh. up to you. i think most books would call that a linear equation

Hm, now that I think about it, I don't know what the formal definition of linear is.

different books define it differently. at the most informal you will see stuff like, an equation is linear if it "can be put in" some particular form.
or others will get more particular and say it's only linear if it's in some particular form.
and what that form is varies on the application.

I'm going to predict that Ted's diffgeosense is tingling and he's going to [say something charitable about] Koro's statement or something...

you see this even in algebra. is x + 2 = 1/x a quadratic equation? well, probably not to most people, but you can certainly rewrite it as one.

4:20 AM
Yeah, but where's a rigorous and universal definition that holds for all real-valued algebraic polynomials?
"Degree of at most 1" fails for negative exponent unless you say "degree of at most magnitude 1".

i don't know what rigorous and universal means, but that's a good example. what is a polynomial equation. well, you can pretty much write out what that is.
but other equations that don't fit that definition might be capable of being turned into polynomial equations.

Hi all

o/
@leslietownes Right, well, a bit of philosophy here is handy. The substance of some mathematical entity is accidental to the form it takes within a potency, formally speaking. Applied to this particular case, that is to say, formally in the language of mathematics, "If an algebraic function $f:\Bbb{R}\to\Bbb{R}$ can be written as a polynomial of at most degree one [has at most one real root], then it is linear." as just one possible definition.

the complication with differential equations is that to get good thoerems you often need to make at least some assumptions on the functions you're working with. but the obvious sets of assumptions either don't work, or are far too restrictive in terms of what you want to do, or don't give you good theorems.
which is why the papers on the differential equations oriented parts of the arxiv are littered with different function spaces, chosen for whatever the people interested in them are working on.
rb: does it have to be good advice?

4:32 AM
@leslietownes Leslie: I took a look at the linear ode definition again and again and concluded that. We say that $y’+a(x)y=b(x)$ is linear if a and b are continuous.
If it’s not linear, then it’s called non-linear.

koro: OK, that certainly is a definition. the next question i guess would be, continuous where. everywhere?

Yes. So in the example: not continuous happens at 0 so that way.

I've created a simulation of a ball bouncing around in a triangular billiards table.

OK. i think a lot of books would work with a less restrictive definition. but that is a fine definition.

4:34 AM
Problem is: it's really slow. It takes quite some time to plot just one path.

And I concluded that the ode was linear.
But I was wrong.

Does anyone have any advice for speeding things up?

it's sort of a 'classical' definition, where people wuld classify equations according to what they looked like and maybe the abstract notion of linear map didn't even really have cachet yet

Yesterday, I discussed that with teacher :). He clarified the linear/non-linear confusion.

Here's how the ball bounces:

4:35 AM
If I had a piecewise differentiable function with discontinuities at arbitrary points, am I to say that it is simply non-linear, or rather, might we say that it's better to say "The function is linear everywhere except ..."?
I would say the latter is perhaps more useful under that definition of linear.
Makes continuous synonymous with linear.
(In this context)

AMDG: no new definition is being introduced. It’s the universal definition, I think. I have misunderstood the definition for a long time.

Fair, but I'm just saying that I believe it is non-trivial to mention which parts of a non-linear function are linear or non-linear.

AMDG, the use of 'linear' to characterise ODE is not the use of 'linear' that people normally use with functions.

Hi folks, I'd really appreciate any input on my problem.

the word 'linear' is a little overloaded.

4:39 AM
It seems misleading in the case of 1/x to say that 1/x is non-linear without stating that the only discontinuity is at x=0.

1/x is not discontinuous at 0
as per the definition I use.

here we go

:-)

@leslietownes Also fair.

Any suggestions are greatly appreciated.

4:41 AM
I follow Rudin’s definition that defines discontinuity only at points in domain of definition of the function.

rb this sounds like a coding thing. ie one might have to know what language you are writing in, how you are storing this stuff in data structures, how you are displaying it to screen or writing it to a file. are you running it on your phone with 20,000 apps open. the math might not be the half of it.

@rb3652 1) implementation details 2) algorithm choice. There's a problem in one or both of these if your software is running slow.

Best guide to proofs and logic I have ever seen, they should make these notes into a book admissionstesting.org/Images/…

koro: there's something related to this that bourbaki does differently than most analysis books do, i forget exactly what.

Yeah, Leslie has the right idea. The first thing I'd check is cache misses or data transfer.

4:42 AM
@leslietownes No, no, I'm just thinking about a good algorithm. The ball's path is governed by the law of reflection, so that's how I'm doing it right now. I define an initial $(P_0,V_0)$ and once the ball hits a wall, I just make its velocity vector reflect. I'm wondering if you all have any better ideas.

rb: oh, at the level of the algorithm, i'd just do that. :)

GPUs are not well-suited to recursion. If you're writing anything iterative in a GPU shader, that could be a cause. Still, this is all just very poor speculation without anything more concrete about how you are performing these computations.

i mean, from a starting P,V one thing you could do is generate the sequence of points that it hits the walls at. so don't think about the path, just think of the points that would be drawn on the walls if the ball had ink.
computing that sequence of points might be as easy, or easier, than animating the thing. you know where that ball is going. not a lot is happening to that velocity.

If you're going to have an iterative process in the GPU, convert it to a function of time instead, or do the iterating on the CPU and send the result to the shader.

@AMDG I understand that even “solution” of an ode has a meaning in the sense that it must be defined on an interval. So we may restrict non-linear/linear definition to an interval, I guess.

4:45 AM
then if you want to do a visual, traverse the set of points, being careful to tune your stepping through of them consistent with 'constant velocity.'

@leslietownes Thank you for the suggestion. The thing is that I actually do care about the path ... I'm looking for a periodic path and so the ball's trajectory does indeed play a role, at least for me.
Essentially, the ball behaves like a beam of light, reflecting off the walls just like a light beam would.

rb: sorry. i didn't mean "you don't care about the path" so much as, if you're modeling this thing with short time increments and in most of those increments nothing happens to the velocity, another way to do it would be to precompute just the set of points where things do happen with the velocity.

Here's an example of a periodic path in triangular billiards:

then if you want to animate it, have your imager zip through that path.

@leslietownes Hmm, interesting.
"if you're modeling this thing with short time increments and in most of those increments nothing happens to the velocity" -- That's exactly my issue
"precompute just the set of points where things do happen with the velocity" -- hmm ... I'm thinking how I'd do this.
I guess I can just solve for the intersection between the velocity vector and the table's wall?

4:49 AM
yes, roughly.
intersect lines. compute new velocity. intersect lines. compute new velocity. etc.

Hmm, let me sketch this out while I'm at it.

It's simple enough that you could compute the intersections analytically.

Please bear with me for a second

this'll be tricky in finite precision, eventually the math errors will pull you away from the ideal triangle.
i will be afk for a bit but back later.

You can get around finite precision using symbolic encodings and/or arbitrary precision.

4:52 AM
@AMDG I think arbitrary precision would be fine for me. Going the symbolic route may be a hassle
OK I've just sketched it up.

It would seem that your problem which you intend to solve is, "Given an initial point and initial vector, and a set of surfaces which transform this vector, is the path traced by this vector over time cyclic?"

@AMDG Precisely, yes.

iirc, that's NP-Complete.
if not stronger

@AMDG Well, I'm trying to see if I can get any insights by just numerically simulating the problem.

You know, symbolic computation isn't as much of a hassle as you might think. You do it every time you evaluate expressions by hand.

4:57 AM
Haha, ok. But I don't want to use SymPy and whatnot.

Ah, you're writing this in Python?

That's right.
Seems like I've got my intersection work cut out for me ...
658

There’s a nice approach to this problem that uses vector cross products. Define the 2-dimensional vector cross product v × w to be vx wy − vy wx. Suppose the two line segments run from p to p + r and from q to q + s. Then any point on the first line is representable as p + t r (for a scalar pa...

Writing optimized implementations in Python is not trivial.

No kidding. But it doesn't have to be state-of-the-art or anything. I'm just looking to get it sufficiently quick.
At the moment, it's hair-pullingly slow.

Still, I have a hard time believing that the interpreter is introducing that much overhead.
You give me the impression that your program is running incredibly slow.

5:01 AM
It's not the interpreter -- it's my algorithm, which, as @leslietownes nicely put it, spends a lot of time modeling the ball even when nothing's happening to its velocity.

What are you doing, anyways? Allocating and initializing data every frame?
Is it stutters or long pauses? I've got nothin'. You haven't given many symptoms.

I'm running a while loop that checks (every 1/1000th of a second, I believe) if the ball is hitting a wall. If not, keep going. If it is, reflect using $\theta_i=\theta_r$

Ah, so possibly cache misses with a gazillion branch mispredictions. No wonder it's slow, but there must be more that you're doing...

That's more or less the gist of the algorithm right now. I'm working on changing it.

If you see a significant speed up after precomputing the bounces, the issue was probably the one or more if-statements in your loop.

5:05 AM
Hm, for sure for sure.

It would be easier to tell if I could see the while loop for myself.

Sure.
Let me put it in a gist.

Is this the original unmodified before you started refactoring?

Right

Note that Python is well-known to be 100-1000x slower than equivalent C++ code for such things
CPython interpreted

5:10 AM
@PenAndPaperMathematics No way I'm doing C++. I don't even know C++.

That's not where its strength lies. Think of a slow moving python to remember this

Just use GNU C and don't worry about UB.

You could try maybe speeding it up with NumPy, seems like lots of vector ops

@PenAndPaperMathematics Yes, but as far as I've looked into it, np has nothing for reflecting a vector across a line/vector

You have to code some part of it :P
You do as much as you can in vector ops such as scalar mult
If you did numpy with Cython then you'd have C-like speed, but Cython made it hard to debug Cython code
Or even just Cython alone would work
For this simple app gist, a Cython solution is nearly a 1-1 translation

5:13 AM
Thank you for the suggestion, but nevertheless, I've got to refactor my current algorithm.
Anyway, thank you everyone for all the help and suggestions. I'm going to refactor now!

@rb3652 I see a problem with your code

try parametrizing past, now; incomingVector appears to be instantiating an object every frame even though you are only modifying pos and axis; same issue with perp and reflectedVector. Refactor those and try again.

See how you're adding to xPlaces and your only looking back -50. Make it so that you initialize array with xPlaces = [0] * 50
I mean cancel that
Remember to pop the first element in xPlaces yPlaces
Every time you append to it
You have a growing list that might be dragging your app memory-wise
Ever-growing I should say
@rb3652 did you see our suggestions ?

I doubt it's a memory leak, but memory leaks rarely are the cause of bad performance from cache misses unless you're doing something insane like creating a new page every time an object is instantiated and the heap is so fragmented that you get at least one cache miss every frame.

It is in fact a memory leak - it's an ever growing list that doesn't need to be more than 50

5:19 AM
Yes, but I assume this is prototype code. No need to diagnose a memory leak here without necessity. It's just going to cause confusion.

Usually people do the same thing in their prototype in their real design
We both suggested some fixes and you're like "oh mines relevant but not yours".

The occasional call to realloc/calloc due to array resizing isn't going to cause issues literally every frame unless it's being called every frame.
@PenAndPaperMathematics No one is pointing any fingers here...

K, just checking :)

Then don't accuse me of such. Anyways...

Yours barely made sense by the way

5:25 AM
I don't see how. There's very little information available, especially given that I don't know Python very well at all (in terms of library calls). The best I have is some speculations about what is going on here, such as whatever the arrow function is doing. It has the form of an initializer, so I assume it is an initializer; if the interpreter reads to and from the array all the time, then parametrizations ensure everything stays in cache longer; etc.

5:40 AM
@rb3652 Just as a rule of thumb, the more algorithmic and pure your function is, the more it should be parametrized in order to keep things neater and hopefully stay in cache more instead of dereferencing globals. This will also make debugging much easier since you can just look at the function itself without having to scroll to where you defined something.
It also helps to keep procedures decoupled from particular implementation details.
In addition, pay attention to wherever memory is being allocated and data objects are being initialized. Don't make copies without necessity. Use as little memory as possible.

Hi @AMDG and @PenAndPaperMathematics Thank you very much for all the suggestions.

Any time :)

I've just taken a look at them and I'll be thinking about them as I refactor the code.

Basically optimization today is just about memory and little else. If you want to optimize an algorithm, the latency is still important and must not be neglected, but memory is the primary bottleneck in most applications.
(This is for CPU-side on contemporary x86 microarchitectures, not GPU-side code. GPU is different.)

no, today it is about vector/parallel.

5:46 AM
vectorization is trivial

muahahahahah, tell me how long have you been working?

Little bottleneck there. You get it for free practically with (partial) loop unrolling.
@copper.hat Working how?
You mean at a job or working in the field of IT?

it? no, i mean in writing vector/parallel code

Not much. I've spent much of my time in software design and studying contemporary hardware, however, as opposed to writing software which is the easy part and makes up much less than half the time spent in making software.
The majority of software production is design and error correction.
The most hands-on I've got is with writing Minecraft shaders.
But it isn't like it's difficult to grasp vector algorithms and operations...
Sequential execution is one dimensional; parallel execution is two dimensional.

i am not young enough to know everything

5:52 AM
I'm not sure what that's supposed to mean.

parellising code is far from trivial. in the real world.

Then my metacompiler will change that.
I really don't see how identifying temporal dependencies is non-trivial.
Call it lack of experience, but in principle, it's incredibly straightforward. Give me an example of something difficult to parallelize.
I want to see what you're talking about.

i don't really want to pursue this particular convo. good luck with your metacompiler.

$(Z/{nZ})^*$ is same as $U_n$? 😱😮😮

Alrighty then...

5:56 AM
Depends on what $U_n$ is

$U_n$ is multiplicative group of integers modulo $n$.
and is of order $\phi(n)$.
I didn't know the former symbol $(Z/{nZ})^*$ earlier.
I am familiar with $U_n$ though.
Any idea on how to show $\sum_{r=1}^{p^n-1}\binom{p^n-1}{r}p^r$ is divisible by $p^n$, where $p$ is an odd prime and $n\ge 1$ is a natural number? Induction is not allowed.

I have seen $(\mathbb{Z}/n\mathbb{Z})^\dagger$ and similar, presumably the $*$ means the elements with a multiplicative inverse.

Instead of dagger there is x (cross sign) in the version that I am familiar with now.

ah, it might have been a cross. Ted & Leslie would be better on this.

Aight, it's late here, and I can't make up my mind on whether or not to go to bed, but I have two words to say before I leave for now: deterministic threading.

6:11 AM
About the question: I say that $p+1\in U_{p^n}$

Now that I think of it, I should really change my profile picture to a pig because I find myself more dumber than a baby. Also Now I realized that I am too braindead to study pure mathematics and computer science&programming. The only thing I seem to be capable of is following robotic instructions and pumping my bicycle.

So $(p+1)^{\phi(p^n)}\equiv 1 \pmod {p^n}$

@AMDG I hope you can publish your paper =)

nope that doesn't seem to work :(
the question seems very difficult. :(
feels like an open problem.

6:34 AM
Hello to everyone!
$-1<x+y+z+w<1$

$-2<2x-4y+5z-3w<3$

$0<10x+3y+8z+7w<5$

$0<-x+2y+4z+6w<5$

$b_1<Av<b_2$

where $A$-matrix,$v=[x,y,z]$

$b_1=[-1,-2,0,0]$

$b_2=[1,3,5,6]$
There is a system of linear inequalities bounded on both sides. Are there methods for analytically solving such inequalities? Is it possible to use the inversion of the coefficient matrix A in the solution process?

2 hours later…
9:02 AM
@Jakobian what's that?

How to show using bionomial theorem that if p is odd prime, n is $\ge 1$ a natural number then $(1+p)^{p^{n-1}}\equiv 1\pmod {p^n}$?
I wrote $\color{blue}{(1+p)}^{p^{n-1}}-1=(\color{blue}{1+p}-1)(1+\sum_{j=1}^{p^{n-1}-1}(1+p)^j),n\ge 2$

9:23 AM
@Koro You are not using the binomial theorem
The general term in the binomial expansion is $\binom{p^{n-1}}{k}p^k$. Then use Kummer's Theorem.
which, if you need it, is proven in this answer.

I was going to use binomial inside the ()^j term inside parentheses but it was only complicating things. :(
But I don't know p-adic :(

why does that matter?

9:38 AM
Because Kummer's formula mentions "carries" in base p representation.
And I've never worked with numbers in another base than 10.

base $p$ does not involve $p$-adics

@Koro the $p$-adic valuation is just the number of factors of $p$
I don't know why they used that term.

But I know Legendre formula (de-Polignac's formula) that you mentioned in your answer, although in a different form.
Different in the sense that the version that I know doesn't involve $\sigma$.
The version that I'm familiar with: If $p$ is a prime then $\sum_{i=1}^\infty \lfloor \frac n{p^i}\rfloor=:a$ is number such that $p^a$ divides $n!$ but $p^{a+1}$ does not divide $n!$.

@Koro there might be another way to show that $\left.p^n\middle|\binom{p^{n-1}}{k}p^k\right.$ for $k\ge1$, but that is the way I know how.

9:49 AM
I'll try understanding your way also professor Rob. Thanks. Meanwhile, I'm posting the question on mse also because the question has a group theoretic flavour to it. Also, the question is from Dummit and Foote's.

If $V$ is a finite dimensional vector space and $V = U\oplus S_1$ and $V = U\oplus S_2$ then $S_1 = S_2$. Is this correct?

what if $V=R^2, S_1=$ line y=x through origin, and $U=$ y-axis and $S_2=$ x-axis?

2 hours later…
11:50 AM
@AlessandroCodenotti a closed base satisfying certain properties. What you do is define a space of ultrafilters using this base, and then topology on the space of those ultrafilters
It's what you can find online about Cech-Stone compactifications, but the method is more refined, and actually gives you all possible (separable, metrizable) compactifications

12:22 PM
hi all
0

Let $p_1 = 1, p_2 = 3, p_3 = 5,...$ where the sequence $p_n$ is $1$ and the odd primes ordered by size. Let $Re(s),Re(z) > 0$. define $g(z,s) =\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{z^n}{p_n^s}$ Notice this sequence diverges for $Re(z) > 1$ but we use analytic continuation to define it there. What...

any ideas ?

hello math chat

1:01 PM
@mick hi mick nice to see you

1:33 PM
Hi All..
$i^2 = i\cdot i = \sqrt{-1}\times\sqrt{-1} = \sqrt{-1\times-1} = \sqrt{1} = 1$
$i^2 = (\sqrt{-1})^2 = -1$
What is wrong with $i^2 = 1$. What i did wrong there?

the rules for complex exponentiation of complex numbers are not the same as the ones for real numbers
think about it this way: you have to be careful with what you mean by $\sqrt{-1}$ because $i^2=(-i)^2=-1$

What's the fastest way to improve mathematical maturity? Learning harder and more advanced mathematics or learning to solve harder and harder problems? I feel like my mathematical maturity is improving, but not at a fast enough rate.

in the real numbers we get rid of that problem by requiring the square root to be non-negative, and there's a way of doing that for complex numbers too, but then it isn't always true that $x^ay^a=(xy)^a$

@Derivative No. high school. Starting university in 7 months

@Derivative How $i^2 = (-i)^2=-1$
@Derivative Yes. But i found the above relation by rules of complex numbers. and found 2 different results.

1:44 PM
It is true that $i. i$ is a square root of $-1\times -1 = 1$. Simply it isn't the square root of $1$ as you commonly know it. Indeed any complex number has two square roots. When dealing with positive real numbers, there is a consistent choice to make, by taking the positive square root. And it turns out that this choice is consistent with the product rules $\sqrt{a}\sqrt{b} = \sqrt{ab}$

@LearningCHelpMeV2 pick up a book on a topic and read it. I suggest Spivak's Calculus. You'll be better prepared than most of your classmates. Try focusing on learning techniques
@123 I have to leave now, but write out which part you're having trouble with

I am having trouble how $i$ have two results 1 and -1 using laws of exponents.

Those theorems only apply to real numbers.

What is the name of law which proves $i^2 = -1$ not $i^2=1$

@123 There isn't a "law" which does that.
It is the definition of the imaginary unit $i$.
$i$ is an object which, by definition, satisfies the property that $i^2 = -1$.

2:11 PM
It means we don't take $i^2=1$ only we allowed to take $i^2=-1$

I don't understand what you are trying to say.
By definition, $i^2 =-1$, not $i^2=1$. That is either precisely the definition of $i$, or follows immediately from however you choose to define the complex numbers.
This isn't a law or theorem---it is simply the definition.

@XanderHenderson Okay . Thank you. I was confused due to two results.

2:25 PM
Yeah, exponentiation is not commutative for the complex numbers, meaning that it is not necessarily true that $(z^a)^b=(z^b)^a$. This has always broken my mind, too, because it does respect multiplication. Like $(z^a)^b=z^{ab}$, iinm, but it's just that $z^{ab}$ can have multiple possible answers.

2 hours later…
3:59 PM
Any experts here on limits?
I'm looking at the radius of conv. of a Taylor series but can't figure out how to get this done

4:10 PM

Sorry, yes
Looking at a series of where this is $a_n$
Any ideas how I can tackle the limit of $n\rightarrow\infty \abs{a_n/a_{n+1}}$?
It's a... limit of a fraction of sums I guess
To be clear I also thought about how to post this on the main site, but I've not yet passed the staring phase so I have very little work to show meaning the question will not be very good
The $lambda_{kj}$ should be $lambda_\ell - \lambda_j$, the $k$ shouldnt be there at all (to avoid confusion with the index of the sum)

5:00 PM
I got it now. For $n=1$, $(1+p)^{p^{n-1}}\equiv \pmod{p^n}$ is true. So suppose that $n\ge 2$. Now suppose that our induction hypothesis is: $(1+p)^{p^{n-2}}\equiv \pmod{p^{n-1}}$.
It follows that $(1+p)^{p^{n-2}}=1+k p^{n-1}\implies ((1+p)^{p^{n-2}})^p=(1+p)^{p^{n-1}}=1+\sum_{j=1}^p\binom {p}{j}(kp^{n-1})^j$
(By binomial theorem)
$\ddot\smile$
So we conclude that $(1+p)^{p^{n-1}}\equiv 1 \pmod{p^n}$ for every odd prime $p$ and $n\ge 1$. $\ddot\smile$

5:28 PM
I have posted the answer here.

Hello.
What does this symbol $z>>d$ mean?
Is it greater or equal to?

Hi, I think that's used in Physics and means that 'z is very very large compared to d'.
For example: radius of the earth >> height of a human.

Yes it's used in physics. Thanks for help.

on the wikipedia page for ramsey's theorem (https://en.wikipedia.org/wiki/Ramsey%27s_theorem#2-colour_case) they state: It is clear from the definition that for all n, R(n, 2) = R(2, n) = n.

shouldn't it be R(n, 2) = R(2, n) = **2**? Because any coloring of the complete graph with two vertices K_2 yields a subgraph (K_2 itself) where exactly 2 vertices are either red or blue?

5:55 PM
hello, any idea to solve this :math.stackexchange.com/questions/4369160/…

vrouvrou just a word of warning. problems like these are fairly difficult to approach from first principles. they often appear in textbooks after some big theorem about existence and uniqueness has been proved. the hypotheses of these theorems vary, so it's possible that someone may answer with a technique that doesn't quite match what you are familiar with.
if you do know of some big theorem on existence and uniqueness, it might help to state what it is, and point out where you encounter difficulty applying it. that might lead to more focused answers.

i need any methods or any idea

@Koro Also used in math!
Hi @leslie … Is lesliecoin hanging in there with all the financial instabilities?

good evening, is it possible for two vectors to span $\mathcal{R}^3$? It seems to me we need at least three vectors.

i sure hope not.
vrouvrou: sites.oxy.edu/ron/math/341/10/ws/05.pdf ("existence of a unique solution") is the kind of theorem i am thinking of. often these problems are solved by using something like that as a black box.
your problem can be solved in this way, but it would help to know the context of the problem. i.e. what 'can you use' and what would you have to prove.
ted: it's doing better than ever. it is more valuable than ever in times of uncertainty.

6:10 PM
I hope it’s immune to all the havoc wreaked on us by the “unbiased” SCOTUS.

Thanks.

i'd love to, koro, but my time machine doesn't go back to questions from 2011.
ted: that reminds me, i tried 'wroke' in wordle the other day and was surprised it was in the dictionary.
i mean, i knew it was a word, i was just surprised it was in wordle's limited dictionary.

Yeah, a bit archaic, methinks.

Leslie, I’ve been at that question for a long time (background: an exercise problem from Dummit and Foote). I even posted my question on mse on that but it seemed to that that I was not going to get an answer to my question so I deleted that.

koro: i was kidding. :) i do think it's helpful for people to provide new answers to the same question, even if the question is very old. it avoids duplicates and cross-talk between near-duplicates.

6:17 PM
how can I type and upload image in the same comment in here?

Why I ask for review/verification is because the comments to that linked post suggest Kummar theorem and Robjohn also suggested the same.
But the answer I posted uses only induction and binomial theorem :).

@CroCo I don't think you can.

@TedShifrin any feedback about my question above. Is it possible for two vectors to span $\mathcal{R}^3$? thank you.
If I try to do the following
z will always depend on x and y.

please, how to write this problem into a system of differentiel equations of order one
y''+f(x,y)=0
y(0)=y_0, y'(0)=z_0

@Vrouvrou

6:34 PM
@CroCo Why are you doing the computation? Do you know about dimension? If not, the column space of a $3\times 2$ matrix can never be all of $\Bbb R^3$. If the augmented matrix $[A|b]$ has rank $3$, that happens only when the equation $Ax=b$ is inconsistent.

@TedShifrin yes this is the way I understand it but I'm trying to understand the point of this question which is by the way from A Modern Introduction to Linear Algebra by Henry Ricardo.
The book introduces the spanning sets before dimensions.
even before matrices

In my linear algebra text(s), we talk about constraint equations for $Ax=b$ to be consistent before we get to dimension.
Oh, so you haven't got to the algorithms for solving inhomogeneous equations?

the book starts with vectors.

If so, your brute-force solution is fine.

but I'm aware of consistent systems

6:43 PM
If you haven't done any matrix algorithms yet, your solution is fine.
Two vectors span a plane, as you've shown. You have the equation of the plane if you have $z$ in terms of $x$ and $y$.

but my question is how two vectors span $\mathcal{R}^3$, I thought we need at least three vectors.

Doesn't all that robotics stuff assume a lot more sophistication than this?
Well, yes, of course. Two vectors span at most a plane.

I'm refreshing my linear algebra course so I don't ask silly questions. :)
But what is the point of this question if I may ask?

I don't know the book, so it's hard to say. They want students to develop intuition for dimension before getting there? I have questions like this in my book(s), but mostly after the discussion of $Ax=b$ and consistency.

6:48 PM
One linear algebra book, one multivariable mathematics book (with linear algebra and multivariable analysis/calculus combined).
The 112 YouTube videos are for the course with the latter.

that sounds bulky. Usually linear algebra is an independent textbook.
what is the title?

You can google and find it.

You wanted just linear algebra, so not that one.

I think this one Linear Algebra: A Geometric Approach

6:55 PM
Yes.

A bit expensive though :). I hope I can get it from my university's library.

We tried to make the publisher keep it cheaper. That failed after one year, decades ago.

they are greedy.

Well, all the illegal uploading of books to illegal websites doesn't help.

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