argh, each time i learn a new thing and i ask why it is so i have more math to do

i'm on a schedule man

i think you need to go slower and develop some intuition in the fundamentals. intuition comes before formalising.

sorry, that was meant to be general advice, not specifically for you in particular.

intuition is a myth there exists only geometry

it is getting used to something :-)

von neumann said that.

he was very understanding

some version of it. in math you never understand things you just get used to them. funny remark coming from him.

If there exists only geometry, that is proof incontrovertible that @leslie doesn’t exist.

i like the presumably apocryphal story about some dinner problem involving a bird flying back and forth between two trains and how long it would take before they crashed (tells you something about the mathematical mindset).

he answered immediately, someone said "ahh, you know the trick". he replied "yes, what could be easier than summing an infinite series".

i like that story.

even if apocryphal it has a point. if you have a machine for solving problems and you can run it faster than being clever there is no point in being clever.

Tricks are harder than geometric series. Oops — there’s that word again.

that is my line :-) don't think, compute

thinking uses a lot of energy

they should call those something else, because i acknowledge their existence and use them.

Shame on you!

i refer to them as best practices.

Is this mathematically correct? $f_X: R \to N$. I want to say for all $a\in R_+ \cup \{+\infty\}$, $f_X=f_{X+a}$. (sorry to interrupt the conversation)

no

why

Huh?

unless you tell me what $f_X$ is?

And the rest of the symbols.

perfectly fine, this is mostly interruption

@Copper It was a fly, not a bird :)

wonder what peta has to say about insects?

Most importantly, $f_X(x):=\sum_{n\ge X}g(x,n)$ (g is a nice function I don't care about now) My main concern is if it's correct to include the $+infty$ in the definition of $a$.

somehow the bird image seems a little nicer for after dinner convo :-)

@vitamind you are not giving enough information. why don't you explain what you are trying to do?

birding is really good right now. most of the nesting season is over but you still see some very aggressive behavior from birds who still have nests.

is $g(x,\infty)$ defined (apart from the shiftiness of it all)

i watched a group of crows, probably a family, mob a cooper's hawk this morning. crows are like humans in that when young ones grow up they stay around and you might see two or three years of crows defending one nest.

yes, g is defined at infty

really hard to comment on correctness without knowing what i would be commenting on.

But why should $\infty$ be in the sum?

Too f****ing vague, despite copper’s polite pleas.

I mean can I work normally or rigorously with +infinity when it's (1) an upper bound in the sum and (2) added with pos. real numbers

if $f_X = f_{X+a}$ for all such $a$ then it would seem to mean $f_X = f_\infty$.

which seems a waste of symbols. perhaps you mean something else?

@TedShifrin Sorry.

oops, gotta go. wife's gone to bodega bay, so its just myself & my son & heart healthy king of the phillys and large curly frys.

Bye, thanks for your help!

i am entirely inconsistent with my moods & politeness. i like Ted's consistency.

later :-)

Ted’s a consistent bitch.

i'm a sassy california bitch who lives for drama

We won’t discuss mental state.

@TedShifrin Sorry again. After thinking a little bit I am now able to put it clear. If I have $f(X)=\sum_{n\le X}g(n)$, would it be mathmetically correct to say $f(\infty)$ if I mean $\sum_{n\le \infty}g(n)$ or should I write $\lim_{x\to\infty}f(x)$.

I solved it as follows that: at each levels we need $2^{k-1}$ at level $k-1$ as for each vertex we do one comparison given that $k$ level is the level of all leaves in a binary tree.

However, the solutions I got says we need to make one comparison at each level starting from $k-1$.

Is it still to vague? If that's the case I don't what else I can do. Maybe let $g(n)=1/n^2$, not important.

@vitamind these are not the same, if $g(\infty)$ has a nonzero meaning. But $\infty$ is a symbol, not a number.

$\infty$ is not in the domain of the $g$ you just wrote.

I see. So lim is the right way to go.

Most likely, yes.

@TedShifrin I just realized this is not what I want. I'll try to be more specific. I have a conditionally convergent double series $f(X)=\sum_a\sum_{n\le X} g(a,n)$. Suppose there exists an $X_0$ so that for all $X\ge X_0$, $f(X)=f(X+a),$ where $a$ is a real positive number. Can I say that $f(X_0)=\sum_a\sum_{n\le \infty} g(a,n)$?

X>1

You have no business writing $\le\infty$.

You also cannot reuse the letter $a$.

I have a conditionally convergent double series $f(X)=\sum_a\sum_{n\le X} g(a,n)$. Suppose there exists an $X_0$ so that for all $X\ge X_0$, $f(X)=f(X+b),$ where $b$ is a real positive number. Can I say that $f(X_0)=\sum_a\sum_{n< \infty} g(a,n)$?

@robjohn Did you have the chance to take a look at this?

@epsilon-emperor the whole approach is wrong, for the reason you've pointed out.

We can find a sequence $f_n\in C(T)$ so that $\|f_n\|_\infty\le1$ and $\int_Tf_n(t)D_n(t)\,\mathrm{d}t\ge C\log(n)$, right?

Think about approximating the signum function of $D_n$

@robjohn Thanks, I'll try this out

Consider $\sum\limits_{k=1}^\infty 2^{-k}f_{2^{5^k}}$

and the fact that a sequence is unbounded is NOT the same as the sequence tending to $\infty$.

any hints to find this limit?

$\lim \frac {n}{2^{\sqrt n}}$

would that be the same as $\lim\limits_{n\to\infty}\frac{n^2}{2^n}$?

@robjohn no I think

why not?

because $\frac 1{2^{\sqrt n}}\le \frac 1{2^{\lfloor \sqrt n\rfloor}}$

So $\frac n{2^{\sqrt n}}\le \frac {\lfloor \sqrt n \rfloor +1}{2^{\lfloor \sqrt n \rfloor}}$

@robjohn that's why I think no

but i am not sure

I was also thinking if I could show that $n^2\lt 2^{\sqrt n}$ (eventually) then also I am done.

but for this I guess I'll have to use mean value theorems

if it is kitten then why is it not kat

but that seems complicated due to power of 2 on right hand side of the inequality

@robjohn ahh yes, it will be same indeed.

$n$ replaced by $n^2$

Then $\lim\limits_{n\to\infty}4\left(\frac{n/2}{2^{n/2}}\right)^2$

You could even back substitute to get $\lim\limits_{n\to\infty}4\left(\frac{\sqrt{n}/2}{2^{\sqrt{n}/2}}\right)^2$

and you know what the limit of $\frac{n}{2^n}$ is?

and if you're worried about the fractional parts, you can look at $\lim\limits_{n\to\infty}\frac{n+1}{2^n}$

Rob, there's also another question on which your thoughts would be helpful. imgur.com/a/J3Naerx

I'm looking for a necessary and sufficient condition.

I asked this on MSE sometime back, but the answer assumed more than just $\mu$ being a positive measure.

For all $f\in L^\infty$, if $1/f\in L^\infty$, then $M_f$ is surjective. I'm struggling with the converse. If the converse is not true, then we'll have to look another condition that is both necessary and sufficient.

@robjohn I got that professor Rob, thank you so much. I was trying to avoid log completely

@robjohn yes of course, now I can show this limit to be 0 without using log :)

great

thank you so much :) @robjohn

Here's my attempt at the converse:

This didn't help a lot since I got a finite lower bound.

If the converse is true, we want $\int_X |g|^2$ to blow up for some $h \in L^2$

@robjohn So pulling out a factor $x^m$ for any $f(x)\in o\!\left(x^m\right)$ is not possible. Now, equation 4 claims though that one can pull out a factor of $x^{m-1}$ for any $f(x)\in o\!\left(x^m\right)$, right? Is this the least factor one can pull out?

@robjohn I meant largest.

Amazing. Apparently, a poorly-made game has manifested an apparent hardware design flaw in 3090 GPUs.

Gone are the days of "Will it run Crysis?" Behold, the time has come: "Will it run Amazon New World?"

@schn you can pull out a factor of $x^m$. You were trying to pull out a factor of $x^{m+1}$

@robjohn do you know whether or not the difference of a number and its bitwise negation for all significant bits is equal to the upper portion of the square of the number?

how is this website called where it renders latex online and then I can send a link?

There's mathbin, but if you want to generate a latex as an image (which is more portable than plain latex or mathbin) or just want to ensure a latex input gives the result you want, I've always used editor.codecogs.com

They recently improved it to this editor version which is significantly better than what they had before.

(5)Has both Propositional and Predicate Calculus.

@AMDG Thank you

You're welcome!

i hate when a result you think is straightforward turns out to use a lot more machinery that you intended.

Same

Is there a good alternative to mathematica?

It's kind of annoying how limited I am right now in terms of analysis.

AMDG do you have a particular application in mind? mathematica is a very good general purpose tool but many of its functionalities are separately available in other software. it is probably the best source where a large number of things are in one place.

Both algorithmic optimization and binary analysis.

i do not know of good tools for that.

my 2.5 year old daughter just said "workers got footprints on the base of our umbrella." referring to something on our patio. she comes up with these perfectly constructed sentences sometimes and it's scary.

I know that $\{\cos n: n\in \mathbb N\}$ is dense in $[-1,1]$

this is true. i think there are a few short proofs on math.se

Hi Leslie. The proof follows easily from the fact that the set $\{a+b \alpha: a,b\in \mathbb Z\}$, where $\alpha$ is irrational, is dense in $\mathbb R$

Now let’s think of the set A=$\{\sec n: n\in \mathbb N\}$

I claim that this set A is dense in the set $U=(-\infty,-1]\cup [1,\infty)$

that seems likely.

Supposing that we are given that that cos n set is dense in $[-1,1]$, we can prove that sec n set is dense in U. Let me present my proof for this.

Let $u\in U$ be arbitrary. Let $t=\frac 1u$ so that $|t|\le 1$ and we know that there exists a sequence $x_n=\cos p_n$ where $p_n \in \mathbb N$, such that $x_n \to t$

Noting that $x_n\ne 0$ for any $n$, we can safely define $y_n =\frac 1{x_n}=\frac 1{\cos p_n}=\sec p_n$ and by limit rules, it follows that $y_n \to \frac 1 t=u$ hence we have shown that $U\subset A’$, where $A’=$ set of all limit points of A.

To show that $A’\subset U$, we say that let v be a limit point of A and if $v\in (-1,1)$ then since $(-1,1)$ is open, we must have a $\delta \gt 0$ such that $(v-\delta,v+\delta)\subset (-1,1)$ but this is a contradiction as range of sec function is $\mathbb R\setminus (-1,1)$

Hence $v\in (-1,1)^c=U$. It follows that $A’\subset U$ and hence in totality we have $A’=U$ and since $A\subset U$, it follows that $ cl(A)=A\cup A’=U$. This proves my claim.

Leslie, what do you think of my proof ?

A possible generalization: if a set S(not containing 0) is dense in an open bounded interval, then the set $\frac 1S$ (defined as set of all $\frac 1s$, where $s\in S$). I have not yet proven this though but I’ll try.

math.stackexchange.com/questions/1146931/… related.

@leslietownes that’s one of exercises from Rudin’s PMA. continuous functions are determined by their values on dense sets.

but I don’t know how that is related to above.

oh i was only referring to the first half of it. take E to be {cos(n): n in N} and f to be 1/x. is what i was thinking

Ahh, that’s a great suggestion. I’ll think on this and revert.

@AMDG the bitwise negation of x is -1-x

For two's complement, right?

that's what I thought you meant by bitwise negation

Yeah, logical NOT.

But restricted to the significant bits of x.

e.g. $\neg 000001011 = 000000100$

I'm asking because it would seem that $x^2$ has some of the higher bits of $x^2$ equal to $(x - \neg x) 2^n$.

If so, this might be useful for computing reciprocals since it took me long enough to realize that the reciprocal of a number is related by multiples of the number itself. Using this principle, I could perhaps reverse engineer a solution from the result of a reciprocation to get back to x, generalize it, and get an algorithm for computing reciprocals.

In case it wasn't obvious, my reasoning is that $\frac{x}{x^2} = \frac{1}{x}$, $\frac{1}{x}\cdot x^2 = x$.

Unfortunately, I can't just use some app and type in some expression in C to evaluate apart from a compiler and tell me "does this equal the upper bits of $x^2$ for all values in [some range]?"

printf is great and all, but graphs are easier to visualize the relationship with.

@leslietownes I thought about it. According to this, set A should be dense in $\mathbb R\setminus \{0\}$. What did my proof miss?

No no, my proof seems correct. A should be dense in complement of (-1,1)

Let $f: [-1,0)\cup (0,1]$ be defined as f=1/x then let $C=\{\cos n:n\in \mathbb N\}$ and hence $f(C)$ must be dense in $f([-1,0)\cup (0,1]$= U

yes, i think that's true.

Thanks a lot Leslie for suggesting me this powerful application of that exercise.

Wow! My life became so much easy now. Now I say: Let $f:[0,1]\to \mathbb R$ be defined by $f(x)=\log \sqrt {x^2+1}$ then clearly range f=$[0,\frac 1{\log 2}]$ and hence, $\log \sqrt {\{\sqrt n\}^2+1}$ is dense in $[0,\frac 1{\log 2}]$

Typo: it should be $\frac{\log 2}2 $ in place of 1/(log 2).

Can anyone explain the move in this answer (math.stackexchange.com/a/4199761/109355) where $\tilde g(x)$ is defined as $g(-x)$? This isn't a regular, "Introduction to Calculus u-substitution" for $-x$, because $g$ is given the new name, not $x$, but if it is just saying "consider some new function which happens to be the same as $g$, but where $-x$ is replaced with $x$," then it isn't clear how the answer justifies applying the chain rule in the second equation that follows.

their $\tilde{g}$ is the composition of $g$ and the map $x\mapsto-x$. That's what the chain rule is being applied to.

How is that different than making the substitution $\tilde x = -x$ and writing it as $g(\tilde x)$?

@user10478 there isn't any difference, but to showcase the chain rule, they are doing it their way

Okay, that seems way less obvious to me that the chain rule would apply using their notation, but I guess that's subjective.

they said "defining $\tilde{g}(x)=g(-x)$" and that is what that means

Like, if $g(-x) = (-x)^2$, then $\tilde g(x) = x^2$, and $\tilde g'(x)$ should be $2x$, whereas their approach would produce $-2x$?

@user10478 No, their approach still produces $\tilde{g}'(x)=2x$.

but $g'(-x)=-2x$

it's different notations for the same thing, $g(\tilde{x})=\tilde{g}(x)$

yes, but $g'(\tilde{x})\ne\tilde{g}'(x)$

indeed

@robjohn So I understand how I would check that claim by applying the chain rule to the LHS. How would I apply the chain rule to the RHS, being that $\tilde g$ is not a function of $g(x)$, but directly a function of $x$?

So the function is

$\tilde{g}(x)=g\circ\nu(x)$ where $\nu(x)=-x$

The chain rule then says $\tilde{g}'(x)=g'\circ\nu(x)\nu'(x)$

awesome cycle in tilden this morning

Okay, ty

Hello, is the map $f:(X/\sim)\times I\to X\times I$ that sends $([x],t)\mapsto (x,t)$ continuous? [x] is an equivalence class of x. Thanks!

Does anyone have some hints?

The signum function of $D_n$ is in $L^1(T)$ right

and by density of $C(T)$ in $L^1(T)$, we can approximate it by continuous functions on $T$

@epsilon-emperor it is $L^\infty(T)$

@robjohn That too. It is in both

It is the $L^\infty$ nature that we are using here.

The signum function's magnitude is bounded by $1$, and we're integrating over a finite interval, $[-\pi,\pi]$

So it's in $L^1$ as well

@robjohn but, C(T) is not dense in $L^\infty(T)$ right

so how do you approximate hmm

Also I have no idea how that sum $\sum\limits_{k=1}^\infty 2^{-k}f_{2^{5^k}}$ is even related

@epsilon-emperor did you see the comment i left on the g=0 ae problem?

hi copper

you don't need denseness, you just need to approximate a particular function.

@copper.hat Yes! I've linked another SE post to my comment which solves the problem

@copper.hat Okay, I'll think more about this

I feel like I've hit a dead-end though and Idk how doing this stuff will help in the bigger picture of the problem at hand

@pritchard that looks like an incredibly non-well-defined map to me

Let $f_n(x)=1$ where $D_n(x)\ge1$ and $f_n(x)=-1$ where $D_n(x)\le-1$. Then make them as close as needed to the signum of $D_n$ in the remaining intervals. Then $f_n-\text{signum}(D_n)$ only needs to be controlled where $|D_n|$ is less than $1$

@epsilon-emperor certainly easier than invoking the inversion theorem.

Then we can use the fact that $f_n$ is close to $\text{signum}(D_n)$ where $|D_n|\lt1$ and $f_n=\text{signum}(D_n)$ where $|D_n|\ge1$

@Thorgott Sorry I meant the other way, i.e. $(x,t)\mapsto ([x],t)$. Is that defined?

@robjohn but what do we gain from all this

@robjohn Yes and I guess there is no largest factor since all functions $x^q$ such that $q\geq{m+1}$ are also in $o\!\left(x^m\right)$.

Then we can use the fact that $D_n\in L^\infty(\{|D_n|\lt1\})$

@pritchard sure. now, when is a map into a product space continuous?

@epsilon-emperor You get functions that exploit the large $L^1$ norm of $D_n$

@robjohn Right. The $L^1$ norm of $D_n$ goes to infinity as $n\to\infty$

Then I can apply the argument I had thought of before with this small modification?

@schn all functions $x^q$ where $q>m$ are in $o\!\left(x^m\right)$.

@Thorgott If the individual component maps are continuous?

indeed, now what are the components in this case and can you tell why they are continuous?

@schn even more, all functions $\frac{x^m}{\log(x)}$ are even closer to $x^m$, yet are still in $o\!\left(x^m\right)$.

@Thorgott The first one is the canonical projection map which is continuous from definition of quotient topology, and the second is just id. on I. Is that correct?

almost, the components maps should have the form $X\times I\rightarrow X/\sim$ and $X\times I\rightarrow I$

Does this operation (:=op) have a name: op(f)=\lim_{\epsilon\searrow0}(f(x+\epsilon)+f(x-\epsilon))/2

Thank you @Thorgott !

@robjohn Yes :) I guess one can construct infinitely many. With respect to your clarification in your answer, would you still say that equation 3 is correct, in particular using $f(u,h)$ instead of $f(hu)$?

I'm not saying that $f(hu)$ is correct. I'm just interested in hearing why $f(u,h)$ instead of $f(hu)$.

Never mind my question. You have motivated this very well in your answer, but it is still somewhat hard to digest, i.e. little-o with two variables.

@vitamind did you want an \epsilon in the denominator as well

Huh, is $x - \neg x =$ x >> 1 where >> means right circular shift to msb?

@Thorgott No. The op(f(x)) only differes from f(x) if we have a discontinuity at x

ok then

Getting closer to an answer. $\frac{1}{x}=\frac{1}{\lfloor \log_2(x)\rfloor - \neg x}$

Actually a bit misleading, though (no pun intended). We're operating under the assumption that $x - \neg x$ multiplied by some unknown power of two, $2^n$, is a subpattern of $x^2$. If it is, then $2^n = \lfloor \log_2(\frac{x}{y})\rfloor = \lfloor \log_2(\frac{x^2}{x})\rfloor$

Or at least that some number of bits at the head of $x$ are a subpattern of $x^2$.

Unfortunately, I can't just graph that on desmos because it doesn't support boolean operations (yet), and emulating them mathematically is inconvenient and cumbersome.