I used to be the best player on the soccer team back in the States

Here, I'm mediocre at best lol

Still the fastest tho

so for $y'' - k^2y = 0$ we have the fundamental set of solutions $y = c_1\cosh(kx) + c_2\sinh(kx)$ and for $y'' + k^2y = 0$ we have $y = c_1\cos kx + c_2\sin kx$ I'm told these are important in applied mathematics

linearly independent so they span the solution set or something

I am trying to not use the language you used to explain the hint on the handout, but it is very simple and explanatory. So I will try to expand on it to express the idea. The good rectangles of $P$ are the rectangles which are fully contained in each rectangle of $P'$. The bad rectangles of $P$ are those that are in multiple rectangles of $P'$. Either way I will still have the relationship of $U(f,P') - L(f,P') < \epsilon/2$

Not in multiple. They overlap two or more. So, no, you don’t have the last sentence. For the good rectangles, you understand why the $U-P$ estimate for $P’$ controls that?

The point will be that $P’$ control won’t help on bad rectangles. So you can only use $\sup f-\inf f$ times volume of those. The point is to relate that volume to $\delta$. Whence all the other verbiage.

@TedShifrin You interested in giving Eucl geometry another go? Found a pretty interesting question on Main, also about squares

For the good rectangles the $U - L$ estimate will control $P'$ because $U_{g} - L_{g} \leq U - L < \epsilon/2$ where the $g$ indicates just the "good" portions

The words at the beginning are totally wrong, but because all the good rectangles are contained in rectangles of $P’$, the good $U-L$ is at most $U(f,P’)-L(f,P’)$.

That's what I was trying to convey, but my use of language was poor.

So I'm working on relating the $\delta$ part. For the bad rectangles I will have something of the sort: $\sum (\sup f_i - \inf f_i) v_i < \sum (\sup f_i - \inf f_i) A$. The $v_i$ are the volume of the bad rectangles, at most as a bound the sum of those $v_i$ could be is the total area $A$.

But it will never reach that because there are less $v_i$ than would make up the total area $A$

Volume versus area? Look at the 2D picture for area in terms of lengths if the darkrsus area? Look at the picture for area in terms of lengths of the dark line ssgments.

darksus?

Ugh … hate typing on phone. Lengths of the dark segments. Read the second version.

Your right I need to be discerning and use the proper words. WHat could be said though it may not make much difference is: $\sum (\sup f_i - \inf f_i) a_i < \sum (\sup f_i - \inf f_i) A$, where the $a_i$ would be the area of the rectangles that overlap rectangles of $P'$

But in higher dimensions, you want volume in terms of $\delta$ and the $A$ I described in the problem. You want the bad rectangles to contribute $<\epsilon/2$ to $U-L$.

So there was a picture I was fiddling with that had to do with the diagonal.

imagining I have my rectangle projected down, this drawing represents the "3rd dimension"

I've bothered you enough for today ted, I'll think about things and hopefully come back to you tomorrow with positive results..... 😓

My algebra professor was talking about the importance of $G$-set or $G$-action on a set and then suddenly talked about infinite Galois theory.

Does this have a discord server? Bc this is really slow damn

no offence tho

peep

It was downvoted and I'd like suggestions on how to improve it.

new question :)

for $y'' + y' + y = g(x)$ $g(x) = e^{5x}$ the particular solution should be $Axe^{5x}$ no?

says in my book $Ae^{5x}$

No, it looks wrong.

I was down with fever, pain in legs, cough, headache and loss of sleep too for last two days. After consulting with doctor, I feel better now. I think it was because of me taking cold drink (weather is changing here) and coffee consecutively for many days (this caused loss of sleep).

oh the assumption in the text was that there is no duplication in the complementary solution

Hope you feel better Koro

@Obliv I mean neither $Ae^{5x}$ nor the other solution seem correct. You can check this by putting for example $Ae^{5x}$ in the equation.

Actually, the answer should involve some oscillating terms (sin, cos etc.) also.

@Obliv thanks :-).

oh yeah that was a mistake, I should have wrote it as $a_ny^{n} + a_{n-1}y^{n-1}+...+a_0y = g(x)$

and what is g(x)? Is it given that y=g(x) the solution of this equation?

it's just some function to make it non homogeneous

we're given a table of particular solutions for such g(x)

(I think I know this book).

So what is your question now?

It's Zill and Cullen diff eq with boundary value prob

@Obliv in view of this comment, it seems that you already know the answer.

no question, I see that it said "taking for granted that no function in the assumed particular solution y_p is duplicated by a function in the complementary function y_c" makes sense now

alright.

it's just in the example prior it had $y''-5y'+4y=8e^x$ which had $y_p = -\frac{8}{3}xe^x$

but that is a "glitch" in the method

Suppose that X and Y are homotopy equivalent, then their nth homology groups are isomorphic. That is, $H_n(X)\simeq H_n(Y)$.

I understand why the above is true.

Can we also say that nth reduced homology groups are isomorphic?

For n>0, the answer is yes for sure.

For n=0, $H_0(X)=\tilde H_0(X)\oplus Z$

i.e., $\tilde H_0(Y)\oplus Z= \tilde H_0(X)\oplus Z$.

Can we 'cancel' Z here to get the desired equality?

@Obliv those are trial particular solutions. There are situations where they might not work. For example: say f(x) and x f(x) both are complementary solutions.

How much about Vector spaces do I need to know in order to understand Pedoe properly? Is a standard calc 3 and linear algebra course enough or do I need to go further? I just received the book @TedShifrin

@Obliv: You may find this interesting math.stackexchange.com/questions/4350045/…. Note the method here.

Nope, we can't cancel Z for my question: math.stackexchange.com/questions/2178841/…

I wonder how to get the equality then.

Hey, shouldn't it be possible to just determine the pair of coprime integers representing a rational in the form of a numerical approximation? From the questions I've found on SE, it seems to be related to the Stern-Brocot tree, but I'm wondering if there's perhaps a closed form means of doing so for any arbitrary rational's numerical approximation.

I'm trying to find a closed form expression for computing the power of two coefficient of an integer divisible by two.

If we restrict the rationals to all rationals representable as $2^y x = z$ for integer $y$ and positive integer $x$, then the number of bits required to represent $z$ is $\lfloor\log_2(x)\rfloor$, but the question is how to compute $x$ and $y$ given only $z$.

I'm trying to find a way forward by encoding $z$ into an added coefficient $2^{\frac{1}{2} z(z + 1)}$, or perhaps some similar value just greater than $z$, to solve this.

@Koro In this question, the resulting subspace are still isomorphic (both $\mathbb R$)

@冥王Hades That should suffice. He does some stuff in the early part of the book, as I recall, by way of background.

@Astyx oh right, thanks.I understand. So the answer is wrong?

And could you please give a hint on to how to prove the 'cancellation'?

I tried to have dinner but felt dizzy while doing so could not have it. I need more medication, it seems.

My taste is also off due to taste of medicines :(.

@Koro There is no cancellation, other than up to isomorphism.

The answer is correct.

isomorphism as 'groups'?

As vector spaces.

Ahh, I'm sorry, I did not see the = sign in the OP there.

So we can cancel upto isomorphism as groups.

The OP there has asked about equality.

But how will one prove it?

Right. I would mod out by the common factor and use $(U\oplus Z)/Z \cong U$.

Or you can do it with bases. Start with a basis for $Z$ and extend ...

Oh cool. Thank you so much!

Define $f: U\oplus Z\to U: (u,z)\mapsto u$. So ker f= $\{0\}\times Z$. Hence, by FIT, we have the isomorphism :-).

I am trying to understand the formula for covariance of two random variables. $$\mathrm{cov}(X,Y)=E[(X-E[X])(Y-E[Y])]$$ I understand that if there is a positive linear dependence, then the product $(X-E[X])(Y-E[Y])$ will be positive and negative for a negative linear dependence. This indication makes sense. However, I don't understand why one takes the expectation of the product $(X-E[X])(Y-E[Y])$. What is the purpose of that?

@schn Why are you talking about positive/negative linear dependence? What in the world?

@TedShifrin positive/negative slope, if you like

@Koro No, the answer is correct, the subspaces are not equal

Oh Ted already answered

Salut @Astyx

yes, indeed. I understood that, thanks :-).

Hello

@schn It still makes no sense to me. Neither is a function of the other.

It's important to understand that if the covariance is $0$, this does not mean that the events $X$ and $Y$ are independent.

(I think you said linearly dependent out of linear algebra habit.)

If you take a foliation of punctured planes embedded in $\Bbb R^3$ where the punctures are along the z-axis and you require all the punctures to be at the origin you get a neat foliation of $\Bbb R^3.$

I am a little confused about some representation theory. Say we have a vector space $V$ of dimension n and the special unitary group SU(n). To my understanding, an element of SU(n) is simply a matrix. It is not a linear transformation. How do we then formally define the action of SU(n) on V? In particular, how do we transmute some abstract matrix (element of SU(n)) into a linear transformation over V? I assume from there a natural group action would to represent elements of V as vectors,

make sure the basis is common between the elements of SU(n) and V and then use matrix multiplication

hm well from wikipedia it seems like the generic general linear group is isomorphic to the group of all linear transformations over a finite dimensional vector space so that answers that. but is there any analogous result for infinite dimensional vector spaces?

@SillyGoose pick a basis. extract the coefficients, make a column vector, multiply it by the matrix, extract the coefficients from the new column vector and go back to $V$. this action is obviously base dependent

this is why representation theory usually works with $GL(V)$, the group of all invertible linear transformations $T:V \to V$. this is not base dependent by any means

also, if $V$ is an inner-product space, you can work with unitary transformations because they are a closed subset (w.r.t. composition) of the group of all invertible transformations. but I doubt that an analogous result for infinite dimensional spaces (i.e. "$SU(V)$ is unique up to group isomorphism"), because there are weird spaces like non separable Hilbert spaces. I think that maybe you could get a nice representation of $SU(V)$ applying Riesz's but that's just speculation.

@PM2Ring: Voyager says that the Sun is 4.0" south of the equator at 2:17:40, and is moving about 1" north per minute.

Voyager says that the Sun has crossed the equator

Greetings, @robjohn.

Happy Spring!

Is that today?

With all our deluges all over the state, it's hard to keep track.

It was just minutes ago

Ah, no wonder you're examining Voyager so meticulously.

Happy Equinox!

And to you, too!

Anyone that could help me compute a value numerically?

I have a graph of the RA speed (& a few other graphs) here, @robjohn astronomy.stackexchange.com/a/48594/16685

desmos.com/calculator/xpr7va8yzg see plotted

Happy autumn down under, @PM2.

@PM2Ring Very interesting.

Thanks, Ted. It's overcast this morning, but we've had some very warm summery weather over the last week or so, some of the warmest weather of the year. And thankfully, not much rain. We had too much rain last year. :(

Well, good thing there's no climate change.

@PM2Ring: Oh, I didn't realize that you were in the southern hemisphere. Happy Autumn!

:D

@AMDG I can't tell from that what you are trying to compute.

@robjohn Aight, gimme like two minutes to write this in chat

@robjohn While you're bored and waiting for AMDG, there's this.

@Koro yes

@robjohn $$\frac{\ln(C(2x, x)\land (C(2x, x) - 1))}{x}$$ where $C(n, k) = \frac{n!}{k! (n-k)!}$

it's not like $H_0(X)=\tilde{H}_0(X)\oplus\mathbb{Z}$ is just some abstract isomorphism. it comes from explicit maps that are natural in an appropriate sense, and that's at the heart of the answer to your question.

@AMDG What does the $\land$ mean in computation, bitwise and?

@robjohn Yes

@Thor Maybe you'll like this one?

@robjohn The innermost expression here comes from oeis.org/A000120 , formula from Gary Detlefs, Jul 10 2014

@AMDG I recognize it as $2^\text{number of factors of $2$ in $x$}$

So the whole expression is the number of factors of $2$ in $x$ times $\frac{\log(2)}{x}$

What, did you look harder than I did on that page, or you just know this?

@AMDG Related: oeis.org/A007814

I didn't look at that page

@PM2Ring Yeah that's what I'm trying to compute... algebraically in closed form

Oh, wait. I am using the exclusive or...

One of the rabbit holes leads all the way to A000120

The $\land$ just peels off the bottom 1 in the binary representation

Right, and that's trivially cnttz, but the question is what that looks like as an algebraic expression (factors of 2 in $x$) without requiring some sum or product (finite or otherwise).

Well $2^{\operatorname{cnttz}(x)}$, but anyways

Path I took is A007814 -> A001511 -> A000120

$x\land(x-1)=x-2^\text{number of factors of $2$ in $x$}$

cries

The closest I am to A007814 is $2^{-\lfloor\log_2(n)\rfloor - 1} n$

$$2^{-\lfloor\log_2(xy)\rfloor} xy = 2^{-\lfloor\log_2(x)\rfloor} x = 2^{-\lfloor\log_2(y)\rfloor} y$$

Yes my rep is now $\pi(1000)=168$

I take it you don't like any of the standard popcount algorithms on stackoverflow.com/q/109023/4014959 FWIW, we had a question asking about the name of this operation a couple of days ago: math.stackexchange.com/q/4661812/207316

@geocalc33 Nice. I know that number well, from numerous hours testing prime sieves. :)

I used to know $\pi(1000000)$, but my memory is getting a little fuzzy. ;)

$1000\pi = 168$??!!

@PM2Ring needs to be algebraic closed form expression for my needs... unless you want to solve a recursive formula involving $a' = 2^{\operatorname{cnttz}(\neg a)} a + a$ for odd $a$ ;)

Honestly, I'd just shift in a loop until the lsb isn't 0.

that would defeat the purpose.

@TedShifrin Prime counting function

@PM2Ring 78498 :)

Yeah, that looks right. :)

Also $\neg x = 2^{\lfloor\log_2(x)\rfloor +1} - x$, so half the battle is won.

what is the sum of the first 100 natural numbers go

iff $xy = 2^p (2^q - 1)$ is true, then the recursive sum terminates in a Mersenne number (then starts over).

5050

yes

Okay now what is the largest prime gap less than 25,000?

@PM2Ring I knew ;)

i think it's like 52

i used Cramer's heuristic

@geocalc33 Ask DLeftAdjointtoU, the twin primes guy.

@TedShifrin I suspected you were just being obtuse about number theory stuff. ;)

@PM2Ring So the solution to the repeating part of this recurrence relation is in fact $2^p (2^q - 1)$, but those intermediate shift offsets must by definition be the bits of $y$. ;)

i.e., it's a division algorithm.

@AMDG A few weeks ago, I was messing around with dividing by Mersenne numbers using multiplication and bitshifts.

@PM2Ring Nice. You might like this, then (explanation of this equation in my comment): oeis.org/A204983

Listed conjecture because I can't formally prove it myself

And since any odd number is a divisor of a Mersenne number, the method could, in principle be used for any division. But it's generally not practical. And of course, to use it as a general integer division algorithm, you have to separate the powers of two out first. But it can still be useful in special circumstances.

The equation literally gives the repeating integer $y$ in the binary expansion $\frac 1 x$, but getting a closed form for cnttz or the recurrence relation I gave would permit efficiently computing $y$ given $x$.

or at least $p$ and $q$ for sure

@AMDG Sorry, I suspect there's no shortcut. Otherwise, the crypto people would have found it by now. en.wikipedia.org/wiki/Multiplicative_order

@PM2Ring You saw the A numbers in the formula, right?

desmos.com/calculator/eoylhqkder

No

;;;;;;;;;;;;;;)

I generally try to avoid Desmos.

I don't have anything else besides Desmos for research, sorry; but the A numbers are in my formula listing on A204983

@AMDG I think that tends to $\log(4)$

@robjohn yeah it's presented as something weird but close as 3.94 or smth using Desmos' low precision Gamma implementation.

@AMDG As $x\to\infty$, $\frac{\log\left(\binom{2x}{x}\right)}{x}$ tends to $\log(4)$

dropping the lowest order bit in the binary representation should not alter this much

but hence the quest for boolean isomorphism to get $\{ +, -, \cdot, \div \} \mapsto \{ \land , \neg , ...\}$

If I could space bits apart (equivalent to squaring in GF(2)), it would be quite easy.

ez as usual with Mersenne numbers

For all else, the formula I found doesn't generalize to all integers. :L

If only it were that easy...

$\frac{1}{\left(\frac{1}{x}+1\right)^{2}-1}$ works for $x = \frac{1}{2^n - 1}$

All this does is spread the bits by 1, but it isn't hard to see how that works for any Mersenne number reciprocal.

@TedShifrin what is the $\tilde\beta$?

is it just another variable?

Predel'nyy ugol ataki

kakoj yazyk

@robjohn Also, thanks

Hm, that would explain why it's like 3.94, but probably some irrational... in which case a closed form is certainly desirable to obtain eventually.

I bet as $x\to\infty$ it gets closer to $4$

Well I mean the shape of the expression is certainly $e^x$ in some fashion...

Probably something out of the identity you gave for $x \land (x - 1)$

@robjohn It's the dummy variable for the first integral. Stupid letter choice, but ... you know engineering types :P

The only reason it's plotted modulo a $2^n$ is because my bitwise operation implementation is brute force and requires specifying how many bits you want for representing $x$

@TedShifrin Ah, when I first looked at it, it looked like a conjugate. Very bad letter choice

with cost increasing as more terms are used. Basically a sum of square waves.

Is this the correct shorthand notation for $a_2(x)y'' + a_1(x)y' + a_0(x)y = g(x)$, $L(y) = g(x)$ where $n = 2$ ?

also LHS can't be reduced to have constant coefficients unless a_2=a_1=a_0 (x) right

not sure why my book assumes $y_p = u_1(x)y_1(x)$ for $L(y)=g(x), n=1$ and $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$ for $n=2$

I mean for $n=1$, $a_1(x)y' + a_0(x)y = g(x)$ we have $y_c = ce^{-\int \frac{a_0(x)}{a_1(x)}dx}$ and by multiplying both sides of the equation by $\frac{-y_c}{c}=e^{\int P(x)dx}$ where $P(x) = \frac{a_0(x)}{a_1(x)}$ we can get this in form $D[yy_c] = e^{\int P(x)dx}f(x)$ where $f(x) = \frac{g(x)}{a_1(x)}$, so I see for n=1 but not sure for 2nd order

Welp, I basically use Desmos as a latex renderer for my work, but FWIW, here's my workspace and what I mentioned earlier about using $2^{\frac{1}{2} x(x + 1)}$ to encode $2^y x = z$ desmos.com/calculator/ztf5hhzs8l

@AMDG If I set $x=1000000$, I get the value to be $\log(3.99997)$.

Oh nice

Huh, I just thought I was dealing with approximation error in the shape looking hyperbolic.

It just looks like a straight line for small values is why

I guess I get it, it's because you need two independent solutions

Come 20 years from now and the power of hindsight will tell me the answer for an optimal division algorithm and boolean isomorphism smh. On the one hand, this is relatively close to my domain (to no one's surprise), but the algebra causes the solution to be just beyond reach for me for lack of study and intuition in plenty of things here.

Now is the best time to plant a tree @AMDG

Don't have to do it all now but if you stay interested you will find answers

Like, yes, I fully understand $xy = 2^p (2^q - 1)$ via its numerical representation--that's how I intuit it and came to find it in the first place--but the relation of binomial coefficients and AND, etc., to this equation are not immediately obvious.

I understand $n!$ as the product of integers in $[1, n]$, and $\Gamma(n)$ through factorial. Floored division is an imperfect coefficient calculator. The logarithm is an imperfect factoring algorithm. The only thing missing to compute all of this (for the most part) is just a perfect factoring algorithm (solve $z = b^y x$ for $b$ and $x$ given only z).

Knowing whether or not $y$ is zero is the only easy thing to compute about this for all $b$.

Every $b$th integer $n$ is necessarily divisible by $b$, $\frac{n}{b}$ times.

Right, well I think this calls for a harmonic analysis break.

Bass guitars are underutilized and underrated smh youtu.be/p4425BfWv2c

At least in this way

lol does that just mean a music break :P

Yep XD

BUT IF YOU INSIST

is this legal: $\frac{d[y\cdot e^{\int P(x)dx}]}{dx} = f(x)e^{\int P(x)dx} \to d[y\cdot e^{\int P(x)dx}]=f(x)e^{\int P(x)dx}dx$ then integrating both sides

seems messed up but i can't think of any other way it's done

oh wait that is totally fine actually $\frac{dy}{dx} = g(x) \to dy = g(x)dx$

So you just do this variation of parameters stuff for higher order derivatives but instead of $y_c = ce^{-\int P(x)dx}$ you have $y_c = ce^{-\int P(x)dx} + cxe^{-\int P(x)dx}$ maybe.