@TedShifrin I think that using the given assumption, I can find a smooth curve $\gamma$ between $p$ and $p'$ where $\pi(p)= \pi(p') = q$. So I can find a vector field on $\gamma(I)$. Extend this vector field to $M$? Is this correct?

What kind of curve? Be specific. You don’t need a vector field on $M$, do you?

Since $\pi^{-1}(q)$ is a connected embedded submanifold of $M$, there is a smooth curve segment between any two points in $\pi^{-1}(q)$. Since $\omega\in\Omega^k(M)$, to use Cartan's formula, I thought I need a vector field on $M$.

@user10478 Some care is needed in interpreting the result. The formula I have above is $R_n=\Pi_k (2f X_k+(1-f))$, where $X_k \in \{0,1\}$ is the $k$th coin toss value. $R_n$ is the return on a unit investment after $n$ bets. Since $X_k$ is binary you can rewrite as $R_n=(1+f)^{\sum_k X_k} (1-f)^{n-\sum_k X_k}$. The geometric 'return per bet' is then $\sqrt[n]{R_n} $, and so the CLT tells us that almost surely this converges to $(1+f)^p(1-f)^{1-p}$.

But it does not really make sense to look at the absolute return after a fixed $n$ bets and expect it to match the latter value. For example, $E R_1 = 2pf + (1-f)$ from the formula above, but this 'tells' you to bet everything if $p> {1 \over 2}$ and nothing otherwise, which is a lot different than the Kelly bet. The Kelly bet only applies if you are betting lots of times..

@onepotatotwopotato I want you to choose a path in the fiber, of course. Certainly you do not need the vector field defined on all of $M$.

@Jakobian The spoiler always can choose either

and the duplicator does the other one

I have a question.

I was doing this problem and left and right limits match.

what's 9 - 12 + 5 again

2 LOL.

OK, and same for the other one? don't make me do these calculations

yes.

we are so far, provisionally, in agreement that the left and right limits match

then I was wondering, there isn't x=3 does it mean is not defined at g(3)?

I'm stunned that Leslie could get to 9-12+5

Of course, EM4.

yes, assuming that this is all that we know about g, then g is not defined at 3.

So don't you dare say it's continuous at $3$.

This is a Calc 1 question. Why are you doing this sort of thing now?

I am just bored and refreshing my 28 years old brain.

Wow. Truly bored.

hehehe.

generally, g(x) given by a recipe bleh(x) for x > a and blah(x) for x < a would fail to be defined at a, without examination into what bleh and blah are doing near 3

I was thinking like hmmm should I watch a movie or do some math.

watch a movie!

And don't forget the calculus textbook vs mathematician debate about whether $f(x)=1/x$ is a continuous function.

I vote for movie.

now its late, I want to do more math HAHAHA.

Well, just so long as you do it and not us.

i'll watch a movie for you.

if you do the math for me

I'm behind on movies, too.

yay! please do.

your math is harder than mines haha.

akiva knows about all of the best new movies thanks to his brother.

@leslie The current issue of Bon Appétit has recipes from an LA (WeHo) pizza chef that look just amazing.

i'd be tempted just to drive to wherever that is. it would probably take me as long to source the ingredients.

The ingredients aren't that esoteric. But, yeah, the person who wrote the article came from the Grapevine to WeHo to get pizzas.

Why don't you pick up a few and I'll meet you at your house!

i was over there last week visiting a friend of mine. she is a walk from this place (looked it up).

Lucky her!

there's great stuff along that stretch of beverly although we tend not to go to it. takeout sometimes but not in the actual restaurants.

PIZZA!!!!

One of my favorite pizza places here closed their on-the-street shop and now only do food truck business wherever ...

a lot of the best tacos in long beach are like that. you sometimes have to find them.

Now that I have to watch my weight and triglycerides, I'll be skipping some of these fine foods for a while.

I wish Boston has good taco places.

i slummed it at kensington pub

Surely Boston must have good taco places?

not in the city, but in Salem yes.

The SF Bay Area used to have a lot of great, reasonably priced food. Not any more. Gentrif*cation.

its sad, gentrification isn't cool at all.

Boston has some things better than most other places … or did when I lived there centuries ago.

now it has changed a lot :( .

@TedShifrin I think I roughly understand what you're saying. If $\gamma$ is a smooth curve between $p$ and $p'$ contained in $\pi^{-1}(q) =: S$, it gives a vector field $V$ on $S$ along a $\gamma$. Since $S$ is a level set of smooth submersion, $v\in T_xS$ for $x\in S$ is contained in the kernel of $d\pi_x$. Hence by the given assumption and Cartan's formula, $\mathcal{L}_V\omega = 0$ so $\omega$ is invariant under the flow of $V$ which shows the independence of $p$.

Right. You don’t need that level set of submersion stuff, I think.

I think it's needed because the assumption is that $i_v\omega_p = i_vd\omega_p =0$ for every $p\in M$ and every vertical vector $v\in T_pM$.

I don’t follow you. It’s $\pi$ that is a submersion, no?

yes

So that says the fibers are submanifolds.

I forget the original post. We need connected fibers.

The statement ensures that $v\in T_xS$ is vertical so the assumption can be used.

Hi chat!

If $N=7$, $\beta_{1} = b_{1} + b_{4}+ b_{7}; \beta_{2} = b_{2} + b_{5}; \beta_{3} = b_{3} + b_{6}$

If $N=4$, $\beta = b_{1}+b_{4}; \beta_{2} = b_{2}; \beta_{3} = b_{3}$

I want to generalise $\beta_{1}, \beta_{2}, \beta_{3}$ for $N$ of the form $3n+1$

I am stuck here $\beta_{j} = -\delta_{3,j} + \sum_{k=0}^{\frac{N+2}{3}-1} b_{k}$

I am trying to think of the summation but puzzled

$j=1,2,3$

any ideas

Good morning mr shifrin

So the first derivative of a multivariable function is a linear map, while the second is a symmetric bilinear form am I correct? the derivative at v is <v,v>.. can this analogy be extended to higher derivatives as well? The third derivative is multilinear so <v,v,v>.

How do I write the answer of: $\binom{6}{2} + 2\binom{6}{3} + \binom{6}{2}$ in the form $\binom{n}{k}$?

nvm, I got the answer

Wait, is $\binom{n}{r-2} + 2\binom{n}{r-1} + \binom{n}{r} = \binom{n+2}{r}$?

In the notation $df/dx$, if I substitute $y=-x$, then I must evaluate the derivative with the rules of derivation "seeing the $-y$ as the independent variable"? I mean, it is not correct to say, after substituting $y=-x$, that $d(e^x)/dx=d(e^{-y})/dy=-e^{-y}$, instead it is correct to say $d(e^x)/dx=d(e^{-y})/d(-y)=e^{-y}=e^x$, right?

Intuitively, I must derive as I do with the usual rules treating the $-y$ quantity as the usual $x$, hence I must write my function as a function of $-y$ and apply the usual rules?

Another example: letting $y=-x$ in $x^2+x^3$ I get that $d(x^2+x^3)/dx=d(y^2-y^3)/d(-y)=d[(-y)^2+(-y)^3]/d(-y)=2(-y)+3(-y)^2=2x+3x^2$; basically it is s chain rule, but I have to modify the variable of derivation in the formal notation of $df/dx$ as well when I substitute like this, correct?

Is there any meaning to "which infinity is bigger?" I am getting this question because of the Hilbert's paradox

Could I infer than the number of irrational numbers are much more than the rationals?

@JaiSriKrishna mathworld.wolfram.com/CardinalComparison.html

Also f(x) is defined as x if x is rational and 1-x if x is irrational, Evaluate the $$\lim_{x \to 1/2} f(x)$$ equals

@feynhat Aah Thankyou!

@JaiSriKrishna Can you guess what it is?

For the limit? How could I know if the number nearest to 1/2 is a rational for sure?

You can't, but the way the function is defined, it shouldn't matter.

Oh yeah It would be 1/2 in either case right?

Yes, now try to prove this.

What if I am given Some other number in place of half ?

@feynhat The limit being half?

@A.G Yes, but you have to be careful. The function must be twice continuously differentiable (class $C^2$) for the second derivative to be symmetric. If it is not, then the second derivative may not be symmetric even if all the partial derivatives exist.

@A.G I am not sure what you mean by derivative at v being <v, v>.

@JaiSriKrishna Then the limit won't exist.

@JaiSriKrishna Yes.

Limit won't exist for sure? or can't we comment on the limit?

Yes, the limit won't exist. For example, in any neighborhood of 1/3 the function takes both 1/3 and 2/3 frequently.

Is there any way of comparing the frequency of occurrence of 1/3 and 2/3 in the smallest neighborhood?

I meant which would repeat the higher number of times?

This might seem vague.......But I just had this idea because I had once read that there are more irrationals than rationals, So I could I make a claim that the frequency of occurrence of 2/3 is more than 1/3?

Why is the limit called "undefined" when it tends to infinity?

Hi everyone, I've been struggling with a problem for some time.

Can I ask it here?

@rb3652 Just ask; don't ask to ask.

@HelpMeToUnderstandContours OK

Find the equilibrium points of the system given by $x_{n + 1} = ax_{n}^2 + c$

My first thought was

@rb3652 have you tried solving $x=ax^2+c$?

@copper.hat Hm, thank you for the idea, but wouldn't that just be $ax^2+x-c=0$?

how do you define an equilibrium point?

As I understand it, an equilibrium point is some solution to a differential equation that is "steady-state"

In other words, it's invariant over time, but I'm not really sure if that's correct.

that would be a good starting point. the above is not a differential equation.

Right, so would I just solve it as a quadratic as such: $x=\frac{-1\pm \sqrt{1+4ac}}{2a}$?

Assuming that is what you mean by an equilibrium point, then yes.

Ah, so would I solve $x_{n+1}=ax_{n}^2 + c$ the same way?

Just as a quadratic equation?

Hello @copper.hat?

Yes?

Is that correct?

No.

Would I set $\frac{d}{dx}(x_{n+1}-ax_{n}^2-c)=0$?

Sorry, you really need to review some stuff. It is a difference equation. I have no idea what $x$ is about or why you are differentiating.

What are you trying to do?

Oh, ok

No, a friend asked me this problem and thought I'd give it a shot, but I'll just forfeit this one.

Although this page (eric-roca.github.io/courses/math_app/steady_state) does look promising

Well, thanks for your help.

Are you looking for equilibrium points or finding a general solution?

He just asked for an equilibrium point, but if it's too hard, I don't think it's worth my time

Are you actually reading what I wrote?

Equilibrium points

You want to find a point such that $x = a x^2 +c$, which you did above. Then you went and did something else.

I solved that equation as a quadratic, but you said that $x_{n+1}=ax_n^2+c$ couldn't be solved the same way.

The latter equation describes the evolution of $x_n$. if you start at an equilibrium point (hint in the word equilibrium) then you will stay there, but if you start somewhere else then $x_n$ will not be constant.

Ah, I see. Well, I'll try this a little more on my own, and come back if I need further help. Thank you for your time @copper.hat

Good luck. Just to illustrate with a simpler example that has an explicit solution, $x_{n+1} = {1 \over 2} x_n$ has exactly one equilibrium point, but the solution to the difference equation is $x_n = {1 \over 2^n} x_0$.

How do I show that $GL(n, \mathbb R^+)$ is path connected?

I tried to connect a matrix A in GL(n, R+) to I by f(x)= xA+(1-x)I

where $x\in [0,1]$

The problem is det f(x) can be zero for some x.

So I think that linear ansatz for f won't work.

But apart from linear f, I don't know what f I should consider so that det f(x)>0 for all x in [0,1].

How can it be path connected, $I$ and $[0 \ 1 ; 1 \ 0]$ are there but their determinants are $1, -1$.

copper: [0 1; 1 0] has determinant <0 so it doesn't count.

@Koro the + should be outside the bracket

What do you mean by $GL(n, \mathbb{R}^+)$???

$GL(n,\mathbb R)_+$

:(

invertible matrices with positive determinant

nono, the + is a superscript, not subscript

The set of all n by n invertible matrices with positive determinant.

that is, $GL(n, \mathbb R)=GL (n, \mathbb R)_-\cup GL(n, \mathbb R)_+$

anyway wasn't this discussed in detail not too long ago already

That was, for GL(n, C)

and for GL(n, R)

Then I thought of $GL(n, R)_+$, which is path connected but I don't know how to prove it.

nah, I discussed this case with Ted not too long ago

you first reduce to $SO(n)$, then I suggested an inductive approach and Ted suggested the isometry normal form

@Thorgott Gram Schmidt

Or Ounce Schmidt in non-metric countries.

haha

yes, that's the idea

The reason is that there are no real numbers called $\pm\infty$. It's just a symbol.

Also, the $[a,b]$ notation is used for closed intervals. The interval $(-\infty, \infty)$ is certainly not closed.

I think Help wanted to close it.

@rb3652 You have a sign error (or two) there.

@rb3652 You also need to determine if the equilibrium is stable. If not, the sequence may not converge.

A couple of French guys investigated $z \leftarrow z^2 + c$...

Julia sets?

Mandelbrot, too

Oui

Here's a more extensive plot:

@PM2Ring I saw the plots. My first guess would be that the error bounds would be on the order of $\frac{n}{\log(n)^2}$. I'd have to look into it further to verify and get a handle on the constant.

That seems to be what you are using there.

Yep

I suppose there might be something nicer involving the logarithmic integral &/or zeta functions. I'd prefer something involving elementary functions, but I can easily graph fancier stuff with Sage.

The data seem to indicate there is a better bound. The $\frac{n}{\log(n)^2}$ arises from the $\operatorname{li}$ function.

I found an expansion I posted a while ago.

Perhaps li is better than that error. Have you tried plotting the same thing with error bounds of $\frac{n}{\log(n)^3}$?

@robjohn Nice!

@robjohn Not yet. I'll take a look at it.

I should write a simulation in Mathematica and see

sagecell.sagemath.org/…

That's my plotting script. It lets you look at sub-intervals.

@HelpMeToUnderstandContours "Infinity" has no particular value; so, you can't treat ♾️ like a real number with all the real number properties.

If you want to look at bigger numbers just edit hinum in the script.

Of course, Sage has various ways of creating lists of primes, but my sieve is pretty fast. ;)

I didn't put any error estimates

I just tried (x / ln(x)) * (1 + u * e^2 / ln(x)^2) fir the boundary curves, with u in [-1, -1/2, 1/2, 1]. It looks pretty good.

$\frac{3n}{\log(n)^3}$ bounds pretty well near $50000$

$\frac{6n}{\log(n)^3}$ is the outer bound

Looking good! Thanks, robjohn.

running $100000$ intervals now

I remember when just sieving primes upto a million was a big deal. :) Of course, RAM was more expensive in those days...

The bounds look pretty nice up to $100000$

They do, but one always has to be cautious with patterns involving the distribution of primes. ;)

Indeed, but the plots are interesting.

I like how they illustrate they convergence due to the prime number theorem but also show the inherent orneriness of the primes.

They seem to follow a smoother distribution than I thought they would.

I'm kind of surprised that I haven't seen this kind of plot before. I guess it's considered more "natural" to look at geometrically increasing intervals, due to the log in the PNT, Bertrand's theorem, the logarithmic integral, etc.

commenters on the post have said that the two integrals are not, in general, the same. it seems to have evolved into some discussion of what an image on wikipedia means. i don't know about any of that, but the integrals don't need to be the same

i don't see the truncation to [-R,R] appearing in any of the general formulas on wikipedia. if you see that kind of truncation on a graph, i would not assume it's also being used in formulas used to generate the graph. other than that i dunno.

many images on math wikipedia are trash, even if the pages themselves are not

i have no idea why people are downvoting, but it often helps to keep a post on main to a single question. the second sentence of the post[ perhaps with hypotheses on f and g!] would be a single question. the wikipedia image thing might be a separate question.

a background question might be, if f and g are supported on [-R,R], is the convolution also supported on [-R,R]. the general answer to that is no, although you can relate the supports of f, g, and f*g. you might click around and look at other questions about that.