In the manufacture of chocolates of a specific variety with a nominal weight of 20.4 grams, there may be fluctuations in the actual weight of a praline. We describe the weight of a praline of this type using a random variable X and assume that X has mean 21.37 and variance 0.16. One box contains
...
i will crawl out on a limb here and suggest that this problem has never actually come up in the context of chocolate manufacturing.
i don't know what their tolerances are but i imagine that decimal points are not involved.
we did work for a biotech startup whose main thing required manufacturing lots of stuff with diameters varying less than 2 microns. pralines were not an input to that process.
hi ted. there was a weird ibis-like bird at the duck pond again today. and the northern shovelers were back.
math.stackexchange.com/questions/4358730/… interested to hear other views of this. a lot of my anti-visualization thing is an act i do for fun, but in some cases i think insisting on a visualization inhibits actual analysis.
venn diagrams are great examples of this, actually. they absolutely do not scale.
I would like to ask as I was confused: From Abbott's Understanding Analysis, about the construction of $\mathbb{R}$: Why is it that the upper bound of $A \subseteq \mathbb{R}$ is defined as $b \in \mathbb{R}$ instead of $b \in A$ such that $a \leq b$ for all $a \in A$?
@leslietownes What I am trying to ask is that, why say that the upper bound is a real number when we still don't know other properties aside from the inherited properties from the rationals?
simpler examples would be $A = \{1 - \frac{1}{n}: n \in \mathbb{N}\}$ which also has an upper bound that is not in the set. this example does not truly test the order completeness of the rational numbers.
soupless if you have a page in abbott i can look at it. in general, the least upper bound of a set $A$ will not lie in the set $A$. it might lie instead slightly outside of it.
ok. you could do definitions 1.3.1 and 1.3.2 for any ordered field or perhaps even for any ordered set. the only issue is that sets that are bounded above might not have least upper bounds.
they seem to be specifying to $\mathbb{R}$ because that is a context in which bounded sets do always have least upper bounds.
abbott is my mother's maiden name, i wonder if we're related.
if we're not doing anything yet, it's not clear in any context that an ordered set does, or does not, contain least upper bounds of bounded sets.
you can define the notion in any ordered set, and then it is just a set-dependent question of whether the least upper bound exists or not.
the example of $\mathbb{Q}$ is helpful. the positive stuff that squares to something less than $2$ is bounded. the least upper bound would be something that squares to something that is exactly $2$. this takes some work, but if you assume it, it helps.
Proof let $g(x)= f(b) - [f(x)+\frac{b-x}{1!}*f'(x)+\frac{(b-x)^2}{2!}f''(x)+...+\frac{(b-x)^{n-1}}{(n-1)!}*f^{n-1}(x))]$ Then is $g'(x)= \frac{(b-x)^{n-1}}{(n-1)!}*f^n(x)$ and $g(b)=0$
So ... what did we prove here? Deriving it gives that but i am not sure how to understand this
Naturally for $f$ diff n times at interval [a,b] with x_o in that interval..
Google that. you will see something that helps you
Question. If we have a function $h: R \rightarrow R^n $ and $f: R^n \rightarrow R$ then the function $g:=f \circ h$ have derivative of $D(g) = D(f(h))*D(h)$ suppose the function $h : t \rightarrow t*x, x \in R^n$ then simply we have $dg/dt = \sum_{i=1}^n \partial f(h)/ \partial x_i * x$ right?
where as $x$ is a vector, but why i have in my notes the following written $dg/dt = \sum_{i=1}^n \partial f(h)/ \partial x_i * x_i$ Ok this is bad notation from my side where as the first x_i is a variable and the second x_i is a vectorcomponent nevertheless why are we taking the singular components of the vector? it doesnt add up!
Okay sorry i think i reliazed why it is like that... it is because $D(f(h)$ is basically gradient. and then $x = D(h)$ so we have a multiplication of vectors.. we need to write it in that order.. never mind.
I always get the derivatives mixed up when it is in R^1 to R^n and R^n to R^1
confusing
@robjohn Hello Prof Rob i hope you are doing good!
Yes Prof i have figured that out. One just needs to be a bit careful with the indicies and the dimensions and it will work out quite fine.
Dimensional Analysis is very useful! In physics as also in mathematics (IE comparision of dimensions or units and figuring it out how th eresult should be like)
Great, so if you have to start making choices for one of these subsets, how many different choices do you have for a first element of one of these subsets?
Actually it doesn't because like you said, we could have {1,2,3,4} = N and then for r = 2 we pick orders of R_A = {(1,2),(2,1)} and R_B = {(2,4),(4,2)}
Are there any analogs to $\int \frac{1}{x} dx = \ln(x)$ for the exponential?
I want to see if there's another way to speed up convergence for approximating the exponential by using the sum I found for division.
Multiplication is still relatively expensive, let alone using multiplication and exponentiation by squaring to do exponentiation for arguments that have been reduced for a given $e^x$ approximation.
Taking a simple approximation that first computes $(e^{2^{-64}x})^{2^{64}}$ for 64-bit precision would be expensive even with a fixed procedure that uses an addition-subtraction-chain.
How to solve this double integral?
$$f(a,b)=\int _{x=0}^{\infty }\int _{t=-\infty }^{\infty }\exp \left(\frac{-a t^2+i b t}{3 t^2+1}+i t x\right)\frac{x}{3 t^2+1}\mathrm{d}t\mathrm{d}x$$
$$\text{with }a>0,b\in \mathbb{R},i^2=-1$$
Known special solution for ${\bf b=0}$
$$f(a,0)=\frac{\pi}{\sqrt{3}...
I like how Weisstein included a joke as part of the page on contour integrals.
> Renteln and Dundes (2005) give the following (bad) mathematical joke about contour integrals:
Q: What's the value of a contour integral around Western Europe? A: Zero, because all the Poles are in Eastern Europe.
> As a result of a truly amazing property of holomorphic functions, a closed contour integral can be computed simply by summing the values of the complex residues inside the contour.
Hey wait a sec
So you're saying I can compute indefinite integrals by applying a (convenient and easily manipulable) transform that makes a function holomorphic and summing the residues?
I have a map $f:X\to Y$ between compact metric spaces which is irreducible. I'm trying to prove that for every non-empty open $U\subseteq X$ there is open non-empty $U^\#\subseteq Y$ such that $f^{-1}(U^\#)\subseteq U$ is dense in $U$.
Irreducible means that it's a continuous surjective map such that for every closed proper $A\subseteq X$ we have $f(A)\neq Y$
I've tried setting $U^\# = Y\setminus f(U^c)$ which is a non-empty open subset such that $f^{-1}(U^\#)\subseteq U$ but I have trouble showing if this is dense in $U$.
Taking a non-empty open $V\subseteq U$, and assuming that density fails for this open set, I have that $f(V)\subseteq f(U^c)$. I don't really see any contradictions from this.
@TedShifrin I agreed with you that it could be better written, but holomorphic doesn't necessarily mean "no singularities". Depending on how, precisely, an author is defining the term, I suppose.
So if $f^{-1}(U^\#)\cap V = \emptyset$, then $f^{-1}(U^\#)\subseteq V^c$ so that $U^\#\subseteq f(V^c)$ but this means that $Y = f(V^c)$, in particular $V$ is empty. I think the above is correct
yep, pretty sure this solves the problem of density
You want to show $$ \frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}\tag1 $$ You know that if $a\ne b$, $$ 2ab\lt a^2+b^2\tag2 $$ Take reciprocals $$ \frac1{2ab}\gt\frac1{a^2+b^2}\tag3 $$ Multiply by $2a^2b^2$ $$ ab\gt\frac{2a^2b^2}{a^2+b^2}\tag4 $$ Substitute $a\mapsto\sqrt{a}$ and $b\mapsto\sqrt{b}$ $$ \sqrt{ab}\gt\frac{2ab}{a+b}\tag5 $$ Divide $(2)$ by $2$ and take square roots $$ \sqrt{ab}\lt\sqrt{\frac{a^2+b^2}2}\tag6 $$ Putting $(5)$ and $(6)$ together gives $$ \frac{2ab}{a+b}\lt\sqrt{ab}\lt\sqrt{\frac{a^2+b^2}2}\tag7
@Vrouvrou For future reference, you can always use Lagrange multipliers to prove these sorts of inequalities. Just state it as a minimization subject to a constraint.
@anakhro That's the inverse matrix, of course. But if you have an orthonormal basis, then the dual basis is declared to be orthonormal. You can work out what that means in terms of matrices. Mumble mumble inverse transpose ...
@Vrouvrou: it is not immediately obvious that you have that:
You have $$ \left(\frac{2ab}{a+b}\right)^2\leq \frac{a^2+b^2}{2}\iff8a^2b^2\le\left(a^2+b^2\right)(a+b)^2=\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right) $$ The questionable step is $$ 8a^2b^2\le\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right)\iff8a^2b^2\le2\left(a^2+b^2\right)^2 $$
If you can show that last biconditional, you are good, but then you should start at the end and go backwards.
starting with a true statement and deriving what you want is always more convincing
It's clear why sums won't work, but I don't see why scalar multiplication would be a problem. If $c_n\to c$ (in the usual sense), doesn't $c_n F_n$ $\Gamma$-converge to $F$ if $F_n$ $\Gamma$-converges to $F$?
The issue with general sums is that the optimizers won't be related to one another at all. But if you take $F$ and $cF$, they have the same optimizer for any $c$.
So, if $u_n$ optimizes $F_n$, it also optimizes $c_n F_n$ with optimal value $c_n F_n(u_n)$. Maybe it's better to think about $c_n F_n - c F = (c_n-c)F_n + c(F_n-F)$. I see. So if the optimal $F_n(u_n)$ is unbounded as $n\to\infty$, we have a problem. But if it stays bounded, I think I'm right.
@TedShifrin Just to confirm that I have the right idea: basically the metric g creates an isomorphism $f\colon TM\to T^*M$ and so for covectors $\alpha_p,\beta_p\in T^*_pM$ we define $g_p(\alpha_p,\beta_p) = g_p(f^{-1}(\alpha_p),f^{-1}(\beta_p))$.
I don't like thinking of it that way, @anakhro, but it's fine ... You need $\tilde g_p$ or something. I think it's worth understanding how to induce a metric on associated bundles in general. The easiest way is to work with orthonormal frames.
Well, if $F_n\to F$ (in the $\Gamma$ sense), then certainly $cF_n \to cF$ is fine.
@TedShifrin Is this something I could expect to come up with myself, or is there a good reference on this? I am not super informed when it comes to Riemannian geometry.
I think it's pretty much in every book, @anakhro. Certainly the more modern geometric analysis books, like Jost and the French guys' ... I'm sure it's in Spivak somewhere and probably even in Warner.
@jay I think you should work with the specific $F_n$ and $F$ you have. This stuff is not going to work in generality; that much is clear.
I've been taking it a bit slow during the winter break, just before that... I was slowly reading about quantum unique ergodicity
And I had given a presentation about mapping class groups which required learning about mapping class groups :P
And that was a lot of fun
This was in dynamics, and a friend of mine wants to do his masters thesis with the prof for that class, probably in translation surfaces and billiard stuff. I very much intend to learn some of the stuff with him
Also we're reading through some scheme theory together which is coo. And trying to learn some more rep theory
Not making it very far in any endeavor tbh but it's fun :P
What sort of stuff are heading towards thesis-wise?
I find it hard to say I have been learning any math, but apparently I have been doing basic Riemann geometry computations while trying to follow along a few papers in geometric analysis.
