Mathematics - 2019-09-08

user76284

00:08

What's the state-of-the-art in global optimization with access to gradients?

user193319

What does it mean for a function $f : [0,1] \to [0,1]$ to preserve the orientation of $[0,1]$?

Ted Shifrin

What would you guess?

user193319

00:25

Not sure. I read it in a paper on Thompson's group and it sort of confused me.

My initial guess was that the function is increasing, but something about that doesn't seem right.

Ted Shifrin

That's right. Everywhere positive derivative.

Leaky Nun

00:45

@TedShifrin hey

Ted Shifrin

howdy Leaky

usukidoll

Okkk I read the example wrong. It was going backwards D;

But now I know that well-founded has an R-Minimal element

Like the smallest element in a set

Ajay Mishra

03:17

If we have to maximum error or maximum value of lagrange form of remainder in two variables, say for R3, in some range like |x-1| < 0.1 and similar for y, we have to compute f_xxxxx and f_xxxxy.......... to find the critical points, as the maximum error depends on all fourth order derivate in this case.

Is there any shortcut?
Because this is obviously very long method, even longer when the maxima is not in the given so in that case we have to check allllll the boundaries and corner points.

Shine On You Crazy Diamond

04:12

Let $A$ be a commutative ring with $1$ and consider $C = \{$ functions $A \to A\}$. Let $c \in C$. Then since every element of $A$ is integral over $A$ via $X - a = 0, a \in A$, does it hold that every element of $C$ is integral over $A$? Embed $A \subset C$ as the subring of constant functions. What conditions would make every element of $c$ integral over $A$?

For example $A = \Bbb{Z}/(2)$. Then $|C| = 2^2$. If $c\in C$ and $c(0) = 1$, $c(1) = 0$ then we want to find $f \in A[X]$ such that $f(c) = 0$. Can this even be done? $c^2(x) = x, \ \forall x \in A$, so that $c^0(x) = x$ an…

Harshit Gupta

11:31

I have a question

that if i answer to a bountied question and the person who put up the question doesnt tick my answer so will i get boutny after bounty peiod is over ?

Mary Star

11:59

Hello!!

I am looking at the following question:
Which are the most important concepts that are included at the pythagorean theorem and how can we convince someone (the students) for their importance?

Are the concepts meant to be right angled triangle, hypotenuse, etc?

For the importance, do we have to say the use of the pythagorean theorem?

If we know the lengths of two sides of a right angled triangle, we can find the length of the third side.

user193319

13:27

Claim: Let $f : [0,1] \to [0,1]$ be continuous and differentiable almost everywhere on $[0,1]$. If the derivative of $f$ is positive wherever it exists, then $f$ is increasing. Proof: By way of contradiction, suppose there exist $x < y$ in $[0,1]$ such that $f(x) \ge f(y)$. I think I can say by continuity (intermediate value theorem?) that $f(x) \ge f(z)$ whenever $z \in [x,y]$.

Now, if for every $z \in (x,y]$ we had that $f$ wasn't differentiable on $(x,z)$, then this would contradict the fact that $f$ is differentiable almost everywhere. Hence, there must exist a $z \in (x,y]$ such that $f$ is differentiable on $(x,z)$. By the mean value theorem, there is a $c \in (x,z)$ such that $f'(c) = \frac{f(z)-f(x)}{z-x} \le 0$, which is a contradiction. Hence, $f$ must be strictly increasing.

I'm a little skeptical about the proof I just gave...

Ryan Unger

14:11

@user193319 I don't see why $f$ should be differentiable in any open interval

user193319

@RyanUnger Hmm...I guess that is one error. Is there anyway of fixing the argument, or is the claim false?

@RyanUnger Is there continuous function whose set of points of differentiability is the fat cantor set?

user193319

14:35

0

Q: Differentiable A.E. with Positive Derivative Implies Increasing

Claim: Let $f : [0,1] \to [0,1]$ be continuous and differentiable almost everywhere on $[0,1]$. If the derivative of $f$ is positive wherever it exists, then $f$ is strictly increasing. Here's my fallacious proof: By way of contradiction, suppose there exist $x < y$ in $[0,1]$ such that ...

jserv

15:04

Afternoon, all. This statistics question is driving me insane, I'm just working out some sample size and power problems. I have the question and solution right here, but I just don't know where that 0.10 is coming from? "This leads to"... what leads to it, exactly? i.ibb.co/c1g0rXD/Q2-Question-and-solution.png

Ryan Unger

@user193319 it's false

see Remark 5.5.1 in DeBenedetto's Real Analysis

you need either absolute continuity or bounded variation as an additional hypothesis

Oops I think BV isn't enough

you need AC

in which case it's trivially true

Érico Melo Silva

@Ryan are u going to any of the scheduled stuff for orientation

Ryan Unger

yeah I'm about to head out

are you?

Érico Melo Silva

15:20

im def going for food but idk how worth it like the scheduled talks and stuff are

Ryan Unger

ok are you going now

Érico Melo Silva

ye im heading that way in a sec, i’m already on campus tho

@Ryan i’m out in front of butler college if u wanna meet me on ur way there

or just at frist

Secret

16:21

iep.utm.edu/math-inc

user193319

18:26

If $f : B(a,r) \to \Bbb{C}$ is an analytic function and $\alpha = f(a)$, what does it mean to say that $f(z) - \alpha$ has a zero of order $m$ at $z=a$?

ALannister

@Semiclassical you around? I wanted to pick your brains about something.

Semiclassical

@user193319 Same sort of things it would mean for a polynomial: 1) the first $m-1$ derivatives of $f(z)$ vanish at $z=a$ (but not the mth), 2) $(f(z)-f(a))/(z-a)^m$ is nonzero and behaves like a polynomial near $z=a$

@ALannister sure, though I’m on mobile atm

ALannister

Remember that question I asked a couple of weeks ago about Gauss-Newton?

user193319

@Semiclassical Oh, I see. Thanks!

ALannister

Basically that Cesareo guy's algortithm did the same thing mine did, didn't it?

Semiclassical

18:38

I think so?

ALannister

So I just need to change the number of iterations? Maybe the tolerance?

He has a different way of approximating the Hessian than I did.

Semiclassical

Hmm

ALannister

Unless he just calculated his all at once and I calculated each component separately.

Semiclassical

I’d have to look at that more carefully to say

It might be useful to test on a simpler example tho

ALannister

I honestly don't understand his code. It's mathematica.

Semiclassical

18:41

for instance, f(x,y)=x^2+y^4 should present similar behavior as it iterates towards the origin

Actually, that’s probably too simple an example

Coming up with good examples is annoying

ALannister

Anyway, I can't use his code. I'm supposed to use my Newton's Method code and just feed into it a function with my f's Jacobian approximation of the Hessian worked out rather than the Hessian itself.

Semiclassical

I will say, though, that iirc I got the same numbers as C did by following the Wiki page on Gauss-Newton and using C’s seed values

ALannister

but is $H\approx J^T*J$ the same thing as $H(i,j)\approx 2\sum_{i=1}^{3}[J(i,j)*J(i,k)]$?

I guess that's what I'm asking.

If it is, there was probably nothing wrong with my code in the first place and I just needed to change the number of iterations or something.

or the tolerance

Semiclassical

Hmm. Aside from a factor of two, yes: the definition of matrix multiplication is $$C=AB\iff C(i,j)=\sum_k A(i,k)B(k,j)$$

ALannister

Why do I have the factor of 2 in there though? That's weird

there must be a reason

Semiclassical

18:49

And transposing a matrix is equivalent to swapping the index order.

Yeah, that worries me too

ALannister

The only thing I can think of is to offset the factor of $\frac{1}{2}$ in my original $f$

Crap. I cannot find a whole bunch of my work either.

Semiclassical

I thought of that too, but since H is like J^2 I’d expect a factor of 4 more than a factor of 2

ALannister

Oh according to the notes Itook to prepare for this problem, I got that 2 from Wikipedia

Semiclassical

It could still be something silly tho

Did you document where on Wikipedia you found that?

ALannister

Gauss–Newton algorithm

The Gauss–Newton algorithm is used to solve non-linear least squares problems. It is a modification of Newton's method for finding a minimum of a function. Unlike Newton's method, the Gauss–Newton algorithm can only be used to minimize a sum of squared function values, but it has the advantage that second derivatives, which can be challenging to compute, are not required.Non-linear least squares problems arise, for instance, in non-linear regression, where parameters in a model are sought such that the model is in good agreement with available observations. The method is named after the mat...

where it says derivation from newton's method

I'm really concerned I can't find my work for one of the earlier problems on this asignment i did though. I"m starting to panic

Nevermind found it. Its ok

Also, is it okay to use points with negative coordinates?

Semiclassical

19:04

Had to step away, back now @ALannister

Negative coordinates should be fine, yes

But note that their objective function is $S=\sum_i r_i^2$

And I think yours had 1/2 overall?

ALannister

Yeah it did

I just tried to run mine with his tolerance and #of iterations. Mine aren't like that at all

Semiclassical

Ok. Then that would cancel the 2

So I suggest trying your algorithm without the 2

ALannister

I get to 1 much faster and it keeps going way past 1 afterwards

let me try it.

wait a minute

I don't know what initial point he used to generate that outcome

Oh I guess just enter as my initial point whatever value was there in his 0th iteration

well that might change the outcome a little here

I'll try that with my original function with the 2 still in it

Okay, I'm getting VERY different numbers.

It's not converging in 20 iterations at all. Not even close.

Okay even for the one with no two it is not converging.

expletive

function[val,g,H]=givenfGN(x)
%givenf() modified to output the Hessian approximated by the Jacobian, as
%required by the Gauss-Newton Method

val=(1/2)*((2*x(1)-(x(2)*x(3))-1)^2+(1-x(1)+x(2)-exp(x(1)-x(3)))^2+(-x(1)-2*x(2)+3*x(3))^2);
if(nargout>1)
g=[5*x(1)+2*x(2)-3*x(3)-2*x(2)*x(3)+(exp(x(1)-x(3))+1)*(x(1)-x(2)+exp(x(1)-x(3))-1)-2; x(1)+5*x(2)-6*x(3)-exp(x(1)-x(3))+x(3)*(x(2)*x(3)-2*x(1)+1)+1; 9*x(3)-6*x(2)-3*x(1)-exp(x(1)-x(3))*(x(1)-x(2)+exp(x(1)-x(3))-1)+x(2)*(x(2)*x(3)-2*x(1)+1)];
end
if(nargout>2)

That is my code. I already have a Newton's Method function I can feed that into. One that I know works.

Maybe it's how I'm doing the calculations here.

I'm doing them by brute force and maybe that's why they're not working.

Instead of letting Matlab do the work for me and calculate the Jacobian approximation to the Hessian, I am attempting to tell it how to calculate each component of the approximated Hessian and that is probably why it is not working.

Semiclassical

19:22

Yeah, when I did it in mathematica I did in terms of vectors/matrices rather than three separate functions

ALannister

I'm basically copying the original function our teacher gave us that for some reason she called a banana function (?) and in the code it has the nargout >1 thing taking the gradient and the nargout >2 thing taking the Hessian, but here we're supposed to modify it to take the Jacobian approximation of the Hessian. When I tried just calling the Jacobian function in Matlab at first, I wasn't getting the right outputs so Idecided to just brute force it.

Then it worked.

But I was able to because that was a very simple function. This one is not and so it took me a VERY long time to come up with the approxiations by hand and then put them in brute force

Semiclassical

@ALannister the name “banana function” has a simple explanation: the level sets around the optimal point look like bananas

ALannister

lol makes sense then ;)

But basically we were told just alter the banana function so that it uses the approximation. Then, have the Newton's Method code she provided us with call it.

Semiclassical

Hrm

I don’t have much help to share there: I ended up just implementing Gauss-Newton in Mathematica rather than having to use someone else’s code

ALannister

19:39

Okay it doesn't like that

I just did this

function[val,g,H]=givenfGNM(x)
%givenf() modified to output the Hessian approximated by the Jacobian, as
%required by the Gauss-Newton Method

val=(1/2)*((2*x(1)-(x(2)*x(3))-1)^2+(1-x(1)+x(2)-exp(x(1)-x(3)))^2+(-x(1)-2*x(2)+3*x(3))^2);
if(nargout>1)
g=gradient(val, [x(1), x(2), x(3)]);
end
if(nargout>2)
J=jacobian(val, [x(1), x(2), x(3)]);
K=transpose(J);
H=mtimes(K,J);
end
end

And it said Undefined function 'jacobian' for input arguments of type 'double'.

schn

If $\textbf{u}_1,...,\textbf{u}_n$ form a basis in a linear space, how does one determine the dimension of the span $\textbf{u}_1-\textbf{u}_2, \textbf{u}_2-\textbf{u}_3,...,\textbf{u}_n-\textbf{u}_1$? Since $\textbf{u}_1,...,\textbf{u}_n$ form a basis, they're linearly independent. If one finds the number of vectors that make up the basis of the span $\textbf{u}_1-\textbf{u}_2,...,\textbf{u}_n-\textbf{u}_1$, then that number is also its dimension.

To find the basis of the span, I check the linear independence of the vectors:
$x_1(\textbf{u}_1-\textbf{u}_2)+x_2(\textbf{u}_2-\textbf{u}_3) +...+x_n(\textbf{u}_n-\textbf{u}_1)=0$ (1)
Rearranging terms:
$(x_1-x_n)\textbf{u}_1+(x_2-x_1)(\textbf{u}_2)+...+(x_n-x_{n-1})\textbf{u}_n=0$

From the linear independence of the vectors $\textbf{u}_1,...,\textbf{u}_n$, $(x_1-x_n)=(x_2-x_1)=...=(x_n-x_{n-1})=0$, which means that $x_1=x_2=...=x_n=k$ for some $k\in\mathbf{R}$. (1) can thus be written $k((\textbf{u}_1-\textbf{u}_2)+(\textbf{u}_2-\textbf{u}_3) +...+(\textbf{u}_n-\textbf{u}_1))=k(\textbf{0})=0$. Thus they're all linear dependent. How does one proceed from here?

Tobias Kildetoft

@ALannister Well, what type is it meant to be given?

ALannister

I think I see my mistake.

@TobiasKildetoft vector input

I think it might be becaue I used commas and not semicolons

Tobias Kildetoft

Well, you gave it two values, one of which is a double and the other being a list

ALannister

Nope. Still doesn't like it.

Okay so what I need to give it is a vector valued function and then a vector

not a double and a list

So how do I fix that?

Tobias Kildetoft

19:47

Not sure. I suck at Python

ALannister

This isnt Python it's Matlab

Tobias Kildetoft

Ahh, I such at that even more

ALannister

I could post a question but I doubt anyone will answer it

Tobias Kildetoft

(it did look like very strange Python)

ALannister

Everybody wants to answer with what they know (which is either Python or Mathematica)

neither of which will help me

Problem is, I need a teacher. Like a real one. I need to be able to go to my professor's office hours and ask her qustions. But I am in such a sticky situation right now where I don't know if that wil lbe an option for me.

nbro

20:12

Hi

Anyone familiar with interpolation methods?

northerner

What is the fastest way to see if two integers have any prime factors in common?

Gabriel Romon

@nbro ye

nbro

@GabrielRomon Are you in particular familiar with Catmull-Rom splines?

Gabriel Romon

No, I don't know much about splines

nbro

I think it is not necessary to be familiar with any spline though

Essentially, a piecewise Catmull-Rom spline is a C1 function (it has a continuous derivative). I am trying to explain this with an argument. My argument is that the tangents at the control points (which are used to approximate the derivative at those control points) are calculated equally for all Catmull-Rom pieces, which implies (?) that the values of the derivatives at the control points that are shared by two consecutive pieces are equal.

Do you think that this is a valid argument and good explanation?

MatheinBoulomenos

21:02

@ÍgjøgnumMeg nein kp

ÍgjøgnumMeg

@Mathein kann mich nichtmal dran erinnern was ich gefragt hab :D

MatheinBoulomenos

ob ich wisse, ab wann man sich immatrikulieren kann

ÍgjøgnumMeg

ahhh okey

egal in dem fall :)

krauser126

21:40

Anyone good with combinatorics? Having some troubles

johnny09

22:53

can we turn an inequality problem to a max/min problem? for example turn $\sum_i (x_i^*-y_i)^2 \leq \sum_i (x_i-y_i)^2$ into $\min_{x_i} \sum_i (x_i-y_i)^2$?

usukidoll

Quick question. If set A is an empty set and set B is also an empty set, wouldn't taking $A \times B$ also result in an empty set?

Transcript for

$Mathematics$ Mathematics