Mathematics - 2021-12-13 (page 2 of 2)

Koro

09:20

@runway44 I think you meant $(A-\lambda I)^nv=0$ where v is a generalized eigenvalue of T.

runway44

I said I was restricting to a generalized eigenspace

but if you want to think about it that way then sure

Koro

But still, having known $R_i$‘s I don’t understand how R is a square root of T. :(

I agree that we can construct a square root viz. $\sqrt{\lambda (I+\frac N \lambda)}$ and divisibility by lambda is allowed as T is invertible.

Here is the proof from the book: google.co.in/books/edition/Linear_Algebra_Done_Right/…

runway44

the problem with your question is that you never defined the $R_i$s

actually, no

you specifically said $R_i$ is a square root of $R$ restricted to the generalized eigenspace

by that very description, the eigenspace is invariant

Koro

I did. $R_i\in L(G(\lambda_i, T))$ is such that $R_i^2=T|_{G(\lambda_i, T)}$ etc.

runway44

that's not a definition

you say "the proof goes on to show there exists [...]" but never mention any of that proof

that proof presumably gives an actual construction, one which shows how it's a square root of (R restricted to the eigenspace)

Koro

09:26

Ok, so here's one small background to $R_i$.

runway44

in other words, $R_i$ is a linear operator defined on that eigenspace, so by definition its range is within that eigenspace

Koro

For every nilpotent operator $N$ a square root of $I+N$ exists over complex vector space.

And its proof is same as as you were suggesting earlier. (binomial expansion of (1+x)^(1/2))

@runway44 So here square root of $A$ (your notation) exists and that is R_i in my notation.

Do we have to explicitly write $R_i$? I thought no, so didn't write it in the question explicitly :(

@runway44 I agree. But in the question: my problem comes from $R_iR$. Had it been $R R_i$, then yes I would apply what you said here and be happy.

runway44

I mean, what's wrong with $RRe_j=RR_je_j=R_jR_je_j=Te_j$? Dunno why they wrote $R_jR$ instead of $RR_j$, but whatever.

Koro

The problem, I think is the definition of $R$. First we are writing v as a sum of elements $u_i$'s from generalized eigenspaces (this is possible and has been proven: that V can be written as decomposition of generalized eigenspaces) and these $u_i$'s are unique (thanks to direct sum) so we have $v=\sum_{i=1}^k u_i$ then $Rv:= R_1u_1+...R_ku_k$.

runway44

ok, and?

Koro

09:35

So to prove the statement: I want to show that $R(Rv)=R^2v=Tv$ for every v

runway44

which I just shows in my comment, no?

Koro

I said since this is going to be complicated, I'd show equality only at $e_i$'s which are basis of V consisting of generalized eigenspaces and that's the idea I had. So I must have $Re_i=R_ie_i$ which means $R_i^2 e_i=Te_i$ (by definition of R_i) and LHS is simply $R_iRe_i$ and that's how I end up with $R_i$ first.

@runway44 The problem is how do we know that $R_je_j\in G(\lambda_j, T)$?

that's what I don't understand.

runway44

you said $R_j\in L(G(\lambda_j,T))$ and $e_j\in G(\lambda_j,T)$, so $R_je_j\in G(\lambda_j,T)$ by definition

Koro

Ohh yess!!

You are so right.

Thank you so much. :)

runway44

yee

Koro

09:41

You beat my question to death :)

If you answer it there, I'll accept it and the question will go away from unanswered query.

runway44

nah, you answer it

Koro

Thanks a lot, I've been thinking on this question since today morning.

(morning as per IST)

porridgemathematics

does anyone know why the following operator is continuous? Apparently its by holders inequality, but I don't see why $\frac{1}{z-w}$ is in $L^{p'}$ for any $p'>1$.. $L^p(K) \rightarrow \mathcal{C}(K)$, $K = B(0;R) \subset \mathbb{C}$ for some $R > 0$, $f \rightarrow \int_{K} \frac{f(w)}{w-z} dw_1 dw_2$, and $p > 2$

the target space is continuous functions on $K$, and the norm on the target space is the uniform (supremum) norm

here $w = w_1 + iw_2$ and $dw_1dw_2$ is the lebesgue measure

the presumption is we have $|\int_{K} \frac{f_1(w)}{w-z} - \frac{f_2(w)}{w-z} dw_1dw_2| \leq ||f_1 - f_2||_{p} ||\frac{1}{w-z}||_{q}$ where $q$ is $p$'s Holder conjugate, but then why is $||\frac{1}{w-z}||_{q}$ even finite?

Koro

@runway44 done!

Thanks a lot :)

porridgemathematics

much less, for the same reasons, why is this operator even well-defined? Shouldn't we require $f \in L^{\infty}$ for this to make sense?

runway44

09:50

np

robjohn

@porridgemathematics If it were Holder, then $\frac1{|w-z|}$ would also be bounded, but it is not.

porridgemathematics

im referring to the accepted answer in this post mathoverflow.net/questions/307713/…

its the remark made in the first sentence of that answer

well Holder will work if $f \in L^{\infty}$, because $\frac{1}{|w-z|}$ is integrable on bounded subsets of the plane

robjohn

Singular integral

In mathematics, singular integrals are central to harmonic analysis and are intimately connected with the study of partial differential equations. Broadly speaking a singular integral is an integral operator T ( f ) ( x ) = ∫ K ( x , y ) f ( y ) d y , {\displaystyle T(f)(x)=\int K(x,y)f(y)\,dy,} whose kernel function K : Rn×Rn → R is singular along the diagonal x = y. Specifically...

They are bounded, but there is cancellation. It is not simply Holder.

@porridgemathematics wait.. you're on $\mathbb{R}^2$?

porridgemathematics

yes, $\mathbb{R}^2$

robjohn

well, $\frac1{|w-z|}$ is not in $L^1$ in any case, so how is it bounded on $L^\infty$? How is it even defined on $1\in L^\infty$?

porridgemathematics

10:02

it is in $L^1$ on bounded subsets of the plane

robjohn

so?

porridgemathematics

its locally integrable

im not saying its bounded on $L^{\infty}$

robjohn

but you're convolving over all of the plane.

porridgemathematics

im saying if you assume $f \in L^{\infty}$, you can directly apply Holders to see that $\int_{K} \frac{f(w)}{w-z} dw_1dw_2$ makes sense

no, over a ball

not the entire plane

$K = B(0;R) \subset \mathbb{C}$ for some $R > 0$

robjohn

Ok, then $\frac1{|w-z|}$ is in $L^q(K)$ for $q\lt2$, is it not?

and how do you define $\frac1{w-z}$ on $\mathbb{R}^2$? how do you divide by $w-z$? I usually see $\frac{w-z}{|w-z|^2}$

like the Hilbert Transform on $\mathbb{R}^n$, which actually has a higher exponent in the denominator

runway44

10:20

presumably it's supposed to be interpreted with complex numbers

which will end up being the same as what you said you usually see, up to complex conjugation

porridgemathematics

@robjohn oh doh.. yes, it is in $L^q(K)$ for $q < 2$, and since $p > 2$, this is exactly the situation im in.. sorry!

robjohn

@porridgemathematics yes

porridgemathematics

@runway44 yeah everything is assumed to be a complex number as you say

Dan Donnelly

@shintuku can you explain to me the notation:

28

A: Predicate logic: How do you self-check the logical structure of your own arguments?

Truth tables are not enough to capture first-order logic (with quantifiers), so we use inference rules instead. Each inference rule is chosen to be sound, meaning that if you start with true statements and use the rule you will deduce only true statements. We say that these rules are truth-preser...

"⇒sub ⇒restate (We can create a conditional subcontext where A holds.)"

I'm not sure if that belongs to the LaTeX above or below

it

Also, what good is $\implies \text{sub}$ if you don't also include $A$: $\implies \text{sub}(A)$?

shintuku

10:36

here's a quick proof of the transitivity of the material conditional

Dan Donnelly

thx

shintuku

1 A -> B
2 B -> C

3 If A:	(-> sub)
4	A (from 3)
5	B (from 1, 4)
6	C (from 2, 5)
7 A -> C (-> elim)

-> restate doesn't do anything, it's a superficial rule for readability that allows you to state things you've said previously

Dan Donnelly

I see, so those are "descriptions of applied rules"?

The stuff in parentheses?

And what does the horizontal concatenation of the contexts mean? Is that just for compactness of their text?

putting figures adjacent

shintuku

whoops, I didn't cite the correct rules. give me a sec

Dan Donnelly

thx :)

shintuku

10:41

1 A -> B
2 B -> C

3 If A:	(-> sub)
4	A (from 3)
5	B (-> elim with 1,4)
6	C (-> elim with 2, 5)
7 A -> C (-> intro)

Dan Donnelly

In the post, does the $(

$\implies \textbf{sub}$ go to the diagram above or below that?

shintuku

i don't understand the question

Dan Donnelly

shintuku

below

Dan Donnelly

okay, they could use some <hr>s

lol

shintuku

10:46

heheh

Dan Donnelly

So what's an algorithm for applying these natural deduction rules, elims, intros, etc?

Do you just take the obvious / naive way in order to match the rules hypotheses?

shintuku

there's no algorithm, the correct matching consists in thinking about a correct proof

Dan Donnelly

Then output the attached conclusion with proper variables, etc

thx

shintuku

there are a couple of exercises listed by user21820 over in the basic mathematics chatroom

and solutions by a couple of users

Dan Donnelly

Thanks for your help @shintuku

shintuku

10:59

np

Dan Donnelly

11:43

@shintuku are we better off (as a computer program, not human) representing $M_1, M_2, M_3, \dots M_9$ as a map from $\{1,2,3,\dots, 9\} \to X$ or treating $M_1, M_2, \dots, M_9$ as variables themselves?

shintuku

i don't understand what those mean

Dan Donnelly

In the first case, variables are sort of limited to single latin letters, and you create the more complex integer-subscript (say for a 3x3 grid of $R$-modules $M_{i}$ in homological algebra

I mean

You know how syntactically you have these objects on paper called variables?

To a human, the distinction between variable and something else is done automagically, but for a computer system you have to be very specific about that

So I'm asking, would a computer program used for proof checking rather have simplest variables (single letters, latin or greek) and complex variables not actually variables but say maps from $\{1..9\}$ into some other space

With a base variable $M_{\cdot}$

shintuku

no idea heheh, i haven't used proof checkers yet

Dan Donnelly

where $\cdot$ shows where to put the number syntacticall

@shintuku, ok, thx

I will ask on a CS forum

actually the Lean Zulip chat exchange is best

I think I answered my own question

In homological algebra, things can get even wayyy more notationally complicated: $M_{1,2,3}$ etc

and you don't want to overcomplicated what we call variables in the system

Thus $M_{1,2,3}$ should be implemented always as a function $M_{-,-,-} : \Bbb{N}^3 \to R-\textbf{Mod}$ (e.g.)

or from some finite index set

This allows "made up / complex" variables to take any complicated form say in LaTeX at the end user point

while letting your formal notation of variable be just the base: $M$

I'm only going to allow single-letter variables in my code

greatly simplifies things I think, and mathematically that's what we do anyway!

Prithu biswas

yesterday, by Prithu biswas

@shintuku can you show me that proof?

shintuku

11:53

busy atm heheh

Prithu biswas

@shintuku Oh ok.

robjohn

@Prithubiswas are you asking to see a map from the natural numbers to the integers, or the particular one they mentioned?

Dan Donnelly

@Prithubiswas link?

shintuku

@robjohn they want a proof that the metric space axiom $d(x,y) \geq 0$ is implied by the others

Dan Donnelly

I would then just give a coursest topology st that holds

You can define $d$ however you'd like as long as triangle, positivity, symmetry

and $d(x,y) = 0 \iff x = y$ (maybe there's 4 axioms)

Or that's how a human should do it

Not sure about in PA

shintuku

12:13

@Prithubiswas

1 d(x,y) = 0 <-> x=y
2 d(x,y) = d(y,x)
3 d(x,y) ≤ d(x,z) + d(z,y)

4 d(x,x) ≤ d(x,y) + d(y,x)
5 d(x,x) ≤ 2d(x,y)
6 0 ≤ 2d(x,y)
7 0 ≤ d(x,y)

2

Prithu biswas

@shintuku This is what I meant.

yesterday, by shintuku

since the day I had to write 70 lines to prove there exists a function between the natural numbers and the integers, i got tired worrying about axioms

I am interested in "that" proof.

shintuku

oh lol

pastebin.com/02h30nWG

Arief

(Putnam 1953) infinite nested radicals problem and the general case

►

Prithu biswas

@shintuku thanks.

shintuku

np. the proof consists in showing that the defined set is a well-defined function

love_sodam

13:24

Let $f:(X,x_0)\to (X,x_0)$ be a homeomorphism and $M_f$ be a mapping torus of $f$. Let $i:X\to M_f$ by $x\mapsto (x,0)$ be an inclusion map and $\gamma:I\to M_f$ by $t\mapsto (x_0,t)$ be a loop. Let $\Phi_\gamma:\pi_k(M_f,x_0)\to \pi_k(M_f,x_0)$ be a base changing homomorphism. In this setting, I'm trying to show $\Phi_\gamma\circ i_*([h]) = i_*\circ f_*^{-1}([h])$ for $[h]\in \pi_k(X,x_0)$.

i.e., $[f^{-1}\circ h] = [\gamma h]$.

How can I show this?

S.M.T

16:11

What would you say are the ways a teenager I.e 18 can earn money ? Please share your experiences as well & what did you do. Person who hasn’t done undergraduate yet but only high school.

Xander Henderson

@S.M.T This has nothing to do with mathematics, and is off-topic in this room. Why are you asking it here?

S.M.T

@XanderHenderson Sir , I totally agree with you that we only talk about maths here. I have also seen a little bit we do talk about off topics things as well & here are many experienced people & I have spent a lot of time here as well. That’s why I though I could ask here. I’m sorry as well. Could you recommend which site should I ask it at ?

Thorgott

@love_sodam your last formula is wrong, you want conjugation by $\gamma$

as for how to show this, understand geometrically why it ought to be true

picture the cylinder $X\times I$ and then pass to the quotient

leslie townes

SMT i think the answers would vary extremely based on location, particularly in view of covid. maybe a location-specific forum? a lot of info you might get from older people, even near your location, may be extremely out of date

Xander Henderson

@S.M.T The classified section of your local newspaper?

leslie townes

16:26

that would be an example of very out of date information :D

2

Alex

16:51

Hi! I have a question: Suppose that a primitive $I$ there's not exists on $\mathbb{R}$, so it is possible that the definitive integral associated there exists? For example we have the following integral on $\mathbb{R}$, $\int \frac{x}{a^{2}-\cos^{2}x}{\rm d}x$ with $a>1$ there is not exists on $\mathbb{R}$ but $\int_{0}^{\pi} \frac{x}{a^{2}-x^{2}}{\rm d}x=\frac{\pi^2}{2\alpha\sqrt{\alpha^2-1}}$ why?

leslie townes

why do you say that that function would not have a a primitive?

i'm not sure i understand the relation between the first function and the second, but that's the first question i have

Alex

@leslietownes hi, wolfram give is: wolframalpha.com/input/… and the primitive is in $F\in \mathbb{C}[x]$ or is this not correct?

leslie townes

it's possible that all of the complex stuff cancels out. you see this in its polynomial root finding algorithm (e.g. wolframalpha.com/input/?i=solve+x%5E3+-10x+%2B+1+%3D+0 where the polynomial has three real roots but WA expresses them in terms of complex radicals)

also, the fact that WA is producing a nightmarish symbolic formula doesn't mean that the function has no primitive, only (at most) that it might not have a primitive with a nice formula

from the fundamental theorem of calculus, int 0...t x/(a^2 - cos^2(x)) dx is a real valued function that is a primitive of t/(a^2 - cos^2(t))

a lot of methods of evaluating specific definite integrals (e.g. specific bounds) do not involve finding a nice formula for a primitive and then evaluating that primitive at the end points

a separate issue, i'm not sure how int x/(a^2 - cos^2 x) dx relates to int x/(a^2 - x^2) dx. a trigonometric substitution x = cos(t) in the latter integral would transform both the numerator and the denominator of the integrand

shintuku

17:49

@S.M.T programming

wherever you are in the world

robjohn

18:23

@Alex The second integral is wrong. Did you mean $a^2-\cos^2(x)$ in the denominator?

Curio

18:36

I was wondering why, while solving an integral, I can assume that sqrt(cosh^2 (x) - 1) = sinh(x)
Why not |sinh(x)|?

leslie townes

often there are domain restrictions relevant to the integral that guarantee sinh is nonnegative in the relevant places

in general you'd need that

Alex

18:54

@robjohn Yes, I'm sorry. It was a typo. The correct should be $\int_{0}^{\pi} \frac{x}{a^{2}-\cos^{2}x}{\rm d}x=\frac{\pi^2}{2\alpha\sqrt{\alpha^2-1}}$.

@leslietownes :-( I'm trying to understand

Ted Shifrin

morning @leslie @robjohn

robjohn

@TedShifrin hello. Just finished clearing out our gutters for the upcoming rain. Last I looked, we were in for 1.91"

that's 4.85cm for those who need it ;-)

Curio

Consider integral of x^3 * sqrt(2x^2 + 1)
u^2 = 2x^2 + 1
sqrt(2x^2 + 1) = sqrt(u^2) = |u|
But |u| became u for some reason o_O

Ted Shifrin

Whoa. Well, no question that we need it. We're due for less than an inch, however.

robjohn

@Curio It is NOT |u| it is often written that way, but it is either u or -u. it won't change sign in the midst of things.

Ted Shifrin

19:06

@Curio Better to just say $u=\sqrt{2x^2+1}$ from the beginning and realize $u>0$.

leslie townes

i don't see anything wrong with taking u to be the positive square root of 2x^2 + 1 there

Curio

@robjohn Why? Is |u| incorrect?

robjohn

because that gives the impression that u can change sign

since $|u|\ge1$ from your condition, it never even gets close to $0$

Curio

OK

What about this last example?
Integral of sqrt(x^2 - a^2) dx
x = a*cosh(t)
...
a^2 * integral ( sqrt(sinh^2(t))*sinh(t) dt )

So for the same reason it becomes sinh^2(t) and not |sinh(t)|sinh(t)

robjohn

19:51

@Curio Sorry, I had to go afk for a while. In this case, $\sqrt{\cosh^2(t)-1}$ can approach $0$, and $\sinh(t)$ can change sign as $\cosh(t)$ passes through $1$.

but away from that point, it is either $\sinh(t)$ or $-\sinh(t)$

@TedShifrin I'm sorry it's not more. Our forecast had been for 2.4", then dropped to 1.66", then went back up to 1.91"

we'll see what it actually ends up being (usually less than predicted).

copper.hat

20:18

hercules (a little north of albany) has over 2" in the last 2hrs.

a friend has his own station (part of the wunderground setup).

leslie townes

they're guessing maybe a half inch around here.

Ted Shifrin

Accuweather tells me 0.01" here today and 0.74" tomorrow.

Nemanja Vuksanovic

Hello, I have a curious question.

"In a model for change in the amount of individuals in a
population, where N(t) is the amount of individuals in the population to the time t (calculated in days).
Below is shown a graph for N.
Use the graph to decide N'(10), and write what this number tells you about the change in the population."

imgur.com/Y7o4hon

robjohn

@copper.hat not so strange for NY, for SoCal, it is more extraordinary to have 2" of rain.

@TedShifrin I hope we get the 1.9" we are forecast.

Nemanja Vuksanovic

How would one go about and solve this? This is an old exercise from 2011

@robjohn I would love if it rained in my country!

robjohn

20:32

@NemanjaVuksanovic $\frac{\Delta N}{\Delta t}(10)$ is approximately $100$.

Nemanja Vuksanovic

@robjohn Sorry, your text is kinda wierd looking, I can't really read it sorry ://

Semiclassical

i'm curious about finding an elementary way to see the following: Given $a,p\in(0,1)$, the expression $\dfrac{a\sqrt{1-p}+(1-a)\sqrt{p}}{\sqrt{a(1-p)+(1-a)p}}$ is largest when $p=1/2$

Nemanja Vuksanovic

weird*

robjohn

@NemanjaVuksanovic have you looked at link for installing ChatJax in the sidebar?

I bet a lot of things here look weird if you don't have that installed.

Nemanja Vuksanovic

@robjohn No, I have not, pardon me. Could you perhaps point me where to do that?

robjohn

20:35

@NemanjaVuksanovic The link is in my last comment.

Semiclassical

i can verify that the first $p$-derivative vanishes at $p=1/2$ and the second is negative there, so that's indeed the local maximum

but it feels like there should be a better way of doing it

Nemanja Vuksanovic

It all looks so cryptic

I did bookmark it, but nothing happens?

Semiclassical

have to click the bookmark while you're in this room

the 'start chatjax' bookmark to be specific

Nemanja Vuksanovic

Ahhh, it worked!

@robjohn Can I ask where you knew this was the way to go?

Ted Shifrin

@Semiclassic It looks like convexity stuff. Page @copper.

Nemanja Vuksanovic

20:37

Anyhow you can link me to further reading perhaps? :)

And remember, I am supposed to find the differentiated to N(10), so N'(10)

robjohn

@NemanjaVuksanovic the derivative is the limit of the ratio of differences. The plot is not very precise, so just dividing $\Delta N$ by $\Delta t$ would give an approximation of $\frac{\mathrm{d}N}{\mathrm{d}t}=N'$

Semiclassical

the more symmetric version of this is of course $(a\sqrt{q}+b\sqrt{p})/\sqrt{aq+bp}$ where $a+b=1,p+q=1$ and all variables are positive

if i square that, i guess i have $1+\dfrac{2ab\sqrt{pq}}{aq+bp}$

Ted Shifrin

I just approached it a little differently. Let $x=\sqrt{1-p}$, so $p=\sqrt{1-x^2}$ and it looks like $a\cos\theta + (1-a)\sin\theta$.

I haven't put in the denominator yet.

Nemanja Vuksanovic

So if I choose, let's say, to calculate the difference between 500 and 100 which is 400, what would the difference in t then be?

Semiclassical

yeah, that seems sensible

Ted Shifrin

20:43

The denominator seems to be $\sqrt{a\cos^2\theta+(1-a)\sin^2\theta}$.

So maybe now it's natural to square the whole fraction and play.

Semiclassical

my 1+ statement was nonsense, so ignore that

Ted Shifrin

I was ignoring it all because I was working on my approach :)

Semiclassical

heh

Ted Shifrin

So, if I didn't mess up, $a$ is fixed and we're looking at $$\frac{(a\cos\theta+(1-a)\sin\theta)^2}{a\cos^2\theta+(1-a)\sin^2\theta}.$$

It's sorta surprising that $a$ is irrelephant.

Semiclassical

well, it does need to be 0<a<1 for this to work: if a is outside that interval, the second derivative ends up being positive so this becomes a local minimum

i guess a simpler observation is that $\theta=0,\pi/4,\pi/2$ respectively give $a,1,1-a$

so $1$ can only be the maximum when $0<a<1$

leslie townes

20:50

or so they would have us believe

copper.hat

@robjohn i meant albany CA :-)

Ted Shifrin

So now I can get the natural variable $m=\tan\theta$ and look at $$\frac{(a+(1-a)m)^2}{a+(1-a)m^2}.$$ Is this any clearer?

Semiclassical

it feels like it should be

Ted Shifrin

So that is optimized at $1$. Symmetry somehow?

robjohn

@copper.hat Oh... I didn't even know there was an Albany in CA.

copper.hat

20:52

its a pimple beside berkeley

Ted Shifrin

@robjohn Just up the bay from Berkeley a hop.

copper.hat

my kids call it smallbany

leslie townes

for some reason they decided that solano avenue should have a mayor

Ted Shifrin

@copper We paged you because things looked convex.

robjohn

@TedShifrin Ah. I will add that to the geographic database that is loosely packed between my ears.

copper.hat

20:53

@TedShifrin thanks, am taking a look :-)

robjohn

@Semiclassical Ah, the square root in the denominator is longer than when I looked last.

Ted Shifrin

@robjohn Does my last formula (with $m$) seem more tantalizing?

robjohn

@TedShifrin Actually, the original looks a lot like Jensen...

Ted Shifrin

Yeah, it all sorta does.

robjohn

with square root being concave

Ted Shifrin

20:55

I got rid of the square roots.

robjohn

and $a+(1-a)=1$

Ted Shifrin

And used the natural nature of the circle.

Semiclassical

@TedShifrin hmm. mathematica says $$1-\frac{(a+(1-a)m)^2}{a+(1-a)m^2}=\frac{a(1-a)(1-m)^2}{a+(1-a)m^2}$$

robjohn

That would indicate that the quantity is less than $1$

Semiclassical

which is nonnegative for all $m$ and zero only if $m=1$

Ted Shifrin

20:57

Aha. Done. :)

Semiclassical

yeah

Ted Shifrin

This all looks rather tantalizing.

Semiclassical

ya

Ted Shifrin

But I like my modification :)

Robjohn will soon have the slickest interpretation.

Semiclassical

this is a better solution than "compute derivatives at p=1/2$, but still feels like it's not as simple as possible

Ted Shifrin

20:58

Well, the introduction of the slope as the natural variable seems right.

I'm gonna make a sandwich and check back in a bit.

Semiclassical

sure. but it'd be great if Jensen were enough

robjohn

$$
\begin{align}
\left(\frac{a\sqrt{1-p}+(1-a)\sqrt{p}}{\sqrt{a(1-p)+(1-a)p}}\right)^2
&=\frac{a^2-a^2p+p-2ap+a^2p+2a(1-a)\sqrt{p(1-p)}}{a+p-2ap}\\
&=1+\underbrace{\frac{a(1-a)\left(-1+2\sqrt{p(1-p)}\right)}{a+p-2ap}}_{\le0\text{ and }=0\text{ only when }p=\frac12}\\
\end{align}
$$

The $p$ in the denominator might throw a wrench in the works

Otherwise, the numerator is maximized at $p=\frac12$

copper.hat

@Semiclassical since $f(t)=\sqrt{t}$ is concave, we have $ af(x)+(1-a)f(y) \le f(ax+(1-a)y)$ so the quantity above is bounded above by $1$. If you substitute $x=p={1 \over 2}, y = 1-p$ you get $1$ so this is the $\max$.

i'm only on my first pot of tea, so take the above with a bag of salt

robjohn

at $p=\frac12$ the square of the whole thing is $1+2a(1-a)$

copper.hat

oh, i misread then

i still get $1$, let me put on the kettle

robjohn

21:10

Oops, I missed an $a^2$ for an $a$...

Semiclassical

there we go

copper.hat

wonder why the caffeine time constant seems to get longer as i age? metabolism, i guess

robjohn

now it is correct

Semiclassical

there is some geometric context to this: this is arising as the dot product between two specific unit vectors

copper.hat

i am frequently wrong, but i think what i wrote above holds...

Semiclassical

21:16

so i guess in context a simpler argument is: "you're taking a product betweeen unit vectors, so of course 1 is the largest you could get. now verify it works when p=1/2"

robjohn

@Semiclassical: yes, at $p=\frac12$ it is $1$

and Jensen says it should be less than or equal to $1$.

as long as $a+p-2ap\gt0$

which it is as long as $a\ne 0$ or $p\ne0$ (divide by $ap$ and it is clear)

Semiclassical

yeah

copper.hat

the simpler argument is that $\sqrt{\cdot}$ is concave :-)

robjohn

@copper.hat That is where we used Jensen

Semiclassical

for context, the vectors here were $\sqrt{a} e_1+\sqrt{1-a}e_2$ and $$\frac{\sqrt{1-p} \sqrt{a} \, e_1+\sqrt{p} \sqrt{1-a}\,e_2}{\sqrt{a (1-p)+p(1-a)}}$$

copper.hat

21:24

i think of jensens as the extension to probability measures

Semiclassical

it's easy to see that both vectors are unit vectors, so Cauchy-Schwarz guarantees that their inner product can't exceed 1 and should only be attained when $\sqrt{p}=\sqrt{1-p}\implies p=1/2$

copper.hat

for me the concave bit makes it easier to understand.

robjohn

Okay. I have removed the need for Jensen ;-)

Ted Shifrin

@Semiclassic I agree that Cauchy-Schwarz is the optimal approach.

Semiclassical

yeah. i know why i missed it---the solution for the HW I'm grading assumed they'd compute the inner product and grind out the derivative by hand---but Cauchy-Schwarz is much simpler

robjohn

21:39

@Semiclassical we are assuming that $e_1\cdot e_2=0$? That is, basis vectors.

Semiclassical

ya

copper.hat

have my nice new SE t-shirt & fancy socks :-)

no idea why

Ted Shifrin

21:54

They must be mine!

copper.hat

22:11

:-) i have two now,but the MSE one tore the first time i put it on

Ted Shifrin

You’re just too muscled and strong!

copper.hat

those days are long gone :-(

leslie townes

22:28

they cut him off at the bar and he hulked out (readiness to anger is a characteristic of his people)

goodbye, mse shirt

they should send out swag for maintaining a friendly, approachable, decent but by no means stellar amount of reputation

copper.hat

they're not getting it back he said, clutching his socks tightly

and listening to the rain pelting the glass

amWhy

@copper.hat I have two too to wear. (homonyms). I love my zip up hoodie, almost seven years old, but it was better made than the T-shirts. I do love the socks I recently got!

@copper.hat No email asking for your size?

copper.hat

22:43

@amWhy i did get an email asking for size, but still have no idea what prompted the second round. the 100k ones took forever to arrive, and i am not, nor am likely to ever be, near the 250k mark which i think is the next boundary...

my goal was reached when i answered a question for my daughter :-)

amWhy

@copper.hat When I passed 200K, I got the email. Something about having been backlogged. Maybe they couldn't confirm that we were sent swag for 100K, so just to be sure, they sent us swag again?

leslie townes

copper: sounds like reputation fraud to me

stop the t-shirt steal

amWhy

Because like you, I assumed "250K" was required.

copper.hat

i replied saying that i already had the first round swag but they sent it anyway

amWhy

@leslietownes Hah!

copper.hat

22:48

i would like to have an MSE specific t-shirt that lasts :-)

amWhy

I also won swag from a Christmas challenge on SE from years ago. It took them 2.5 years to send it. I had given up on it after a couple of months....

copper.hat

you know, when i work on the driveways it always opens up the convo

i would even pay for (gasp) a reasonable quality MSE shirt.

they could have a pay for rep thing going that gets printed on the shirt :-)

what, you are a 500k MSE user, omg

yeah, only cost me $20

you want gravel on top of the tarmac as well?

amWhy

@copper.hat Hah!

copper.hat

i do feel bad for travelers, much as i implicitly make fun

that's for you @leslietownes

it is difficult to escape a cultural anchor

leslie townes

:D

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$Mathematics$ Mathematics