Mathematics - 2024-06-08

EE18

00:26

I want to be able to show that for a sequence defined by a seed $x_0$ and the recurrence relation $x_{n+1} = \frac{a}{1+x_{n}}$ that for all $n$, $x_{n+2} \geq x_n$ and $x_{n+3} \leq x_{n+1}$. (Leslie we were discussing this a couple days ago)

I tried to proceed by induction but couldn't make much headway with the inequalities

leslie townes

02:15

well, using that recursion it looks like x_n - x_{n+2} is (x_n^2 + x_n - a)/(a + x_n + 1) (see wolframalpha.com/input?i=x-a%2F%281%2Ba%2F%281%2Bx%29%29 for details)

you didn't say anything about the sign of the "seed" or of "a" but assuming they're nonnegative, this fraction would suggest that the sign of that difference is determined from what x_n^2 + x_n is in relation to a

which may suggest a more informative thing to induct on or at least some ideas about how to analyze that issue

user 85795

03:06

@robjohn or an ingratitude

@ZaWarudo learning is hard work.

@leslietownes political correctness strikes again.

Xander Henderson

@copper.hat I never make missteaks!

EE18

03:29

@leslietownes Will work on this tomorrow, thanks Leslie. (I'm also starting concrete mathematics tomorrow :) )

Thomas Finley

@leslietownes Thanks for your assistance, but the commenters and answerers (only 1 till now) seems either not to understand my question or they are providing vague answers. I don't know what's unclear with my post.

leslie townes

well, your saying that three separate commenters and one answerer either do not understand you, or are 'vague' does not inspire any confidence in me that what i say will satisfy you. what about the accepted answer is "vague"? if you were to apply your formulation of theorem with F = C instead of F = R (which my earlier comment suggested doing), does it at least get you any closer what you want?

Thomas Finley

@leslietownes yes, that not only makes me closer but also gives me what I want.

But the problem is, as I mention that $F\neq C$ but $F=\Bbb R.$

leslie townes

OK, and the answer you received begins with the assertion that for a real matrix, the minimal polynomial will not depend on whether you take F = C or F = R. that's literally the first sentence of the answer. the remainder of the post sketches a proof. maybe it isn't how i would have written it, but, what about that is "vague"?

user 85795

03:45

> Anyone who has never made a mistake has never tried anything new.

:-)

Thomas Finley

03:59

The proof of that surely looks vague. Take for instance, the line "This is because taking the real and imaginary parts of the minimal polynomial over C gives two real polynomials which annihilate the matrix"--- What it meant the real and imaginary parts of the minimal polynomial?

leslie townes

i would think of that as a miniature homework assignment for you to do. instead of "that's vague, so i'll stop here," try to figure out what they might mean. one candidate interpretation would be: the pair of polynomials given by the lists of coefficients that are [respectively] the real and the imaginary parts of the [complex] coefficients of the given polynomial. would that make sense? would it have any relation to what they are saying?

and if you aren't willing to take the time to investigate something like that, why should anyone else? i mean no offense, but this issue comes up a lot on MSE where people focus way too much on the exact form or phrasing of the feedback they are given, and less on what they would do if they were stuck on a desert island with only that feedback, and forced to think about where it would lead them.

2

i'm not saying that every answer or comment on MSE is written carefully, or is correct, or is coming from a perfect place, and we certainly spend a lot of time on the chat dealing with violations of those expectations. but, at least check first to see if this is one of those situations, or isn't it

Soumik Mukherjee

@SineoftheTime XD

Thomas Finley

05:44

@CyclotomicField My definition for eigenvalues: Let $A\in M_{n\times n}(F)$. If $\exists v(\neq 0)\in F^n$ such that $Av=\lambda v$ for some $\lambda\in F$ then, $\lambda$ is called an eigenvalue of $A$ corresponding to the eigenvector $v.$ — Thomas Finley 2 hours ago

The commentator disagrees with my definition saying it's incorrect.

I really don't have a clue on what to do now.

Milad Jangjan

In this link, https://en.wikipedia.org/wiki/Tridiagonal_matrix the Similarity to symmetric tridiagonal matrix has been shown. What happens if the element of tridiagonal matrix would be 2×2
blocks matrix?

Thomas Finley

@leslietownes All the answers and comments seems to be using a definition that's different from mine. I think that's the crux of all the issues and that's precisely the reason why none of them were making sense to me.

Now, I don't know how to proceed any further.

leslie townes

i don't have much to add to my very first comment on this, which is, you seem to be limiting your view of this problem to taking the ground field to be R in every theorem that you have ever seen, i guess because the matrix involved in the theorem is real. you are not forced to do that

neither are you forced to engage with every defect in exposition of every comment on your problem. you can independently take the comments for what they are worth, and see if they help you (meaning you, not following every sentence that someone says as if it is a black box) get closer to an answer

maybe put this problem aside and come back to it in a week

Thomas Finley

Why not? If the base field is R, why will I think the problem in C? Ofc the theorems are valid in C, but why should I consider C when the given field is R.

Jakobian

Because you want real matrices to have eigenvalues

leslie townes

05:54

you're also changing the subject from a project i suggested earlier, which was to see if you can understand the post that someone submitted as an answer to your question

one answer to "why should i consider it this way" is that "because when you asked the question, someone considered it that way, and purported to solve the problem"

you have the option of trying to see if there is anything worthwhile to take from that, even if you don't see why they would have thought to look at the world that way

user 85795

> if they were stuck on a desert island with only that feedback, and forced to think about where it would lead them

Learning is hard work.

Jakobian

Well its not really about learning. More about the willingness to work with something even if it doesn't give you answers on a silver platter

@MiladJangjan how about you try assuming $b_ic_i$ are positive definite and proceeding similarly

user 85795

06:23

Sure, "spoon feeding, in the long run, teaches us nothing but the shape of the spoon."

Jakobian

07:42

Yeah but this is not even about learning, its about personal development

psie

Reading Spivak's proof of the inverse function theorem. Quite soon into the proof he makes an assumption that we can consider the linear transformation $\lambda=Df(a)$ to be the identity linear transformation. All this is summarized here. In the answer in the link, shouldn't $z\in\mathbb R$ be $z\in\mathbb R^n$? If $f$ is a function from $\mathbb R^n$ to $\mathbb R^n$, then so is the derivative, right?

Jakobian

08:37

@psie for the first question yes, for the second question... the derivative is a function from $\mathbb{R}^n$ to $L(\mathbb{R}^n, \mathbb{R}^n)$, technically. A derivative at a point is a function from $\mathbb{R}^n$ to $\mathbb{R}^n$.

psie

ah right, thanks!

psie

11:11

Given a bijective linear map $\lambda:\mathbb R^n\to\mathbb R^n$, is the following observation correct? Since $\lambda^{-1}$ is also a linear map and thus continuous, then $\lambda(W)$ is open for any open $W\subset\mathbb R^n$. (I'm asking because this is another statement done in the link above I posted about some hours ago.)

Feels elementary, but a confirmation would be greatly appreciated.

Jakobian

11:27

@psie yes

$\lambda$ and $\lambda^{-1}$ are continuous means that $\lambda$ is a homeomorphism

in particular, $\lambda(W) = (\lambda^{-1})^{-1}(W)$ is open

this is preimage of the inverse

(of course, continuity follows since we are in a finite-dimensional space, linear maps over infinite-dimensional Banach spaces need not be continuous)

psie

ah, ok 👍

Jakobian

the property that $\lambda(W)$ is open for open $W$ is called an "open map"

in general, an invertible map $\lambda$ is open iff $\lambda^{-1}$ is continuous, by the same argument

Note that we also have so called open mapping theorem, which is stronger, when talking about surjective maps in general, instead of bijective ones

Open mapping theorem (functional analysis)

In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem (named after Stefan Banach and Juliusz Schauder), is a fundamental result that states that if a bounded or continuous linear operator between Banach spaces is surjective then it is an open map. == Classical (Banach space) form == This proof uses the Baire category theorem, and completeness of both X {\displaystyle X} and Y {\displaystyle Y} is essential to the theorem. The statement of the theorem...

that is if $\lambda:X\to Y$ is a continuous linear surjective map between Banach spaces $X, Y$, then $\lambda$ is an open map

psie

ah yeah, I've heard about this

Jakobian

of course "continuous" is redundant for finite-dimensional spaces

yeah the open mapping theorem is more like a curiosity to relate this to something more advanced

@Jakobian this is all that you need

psie

ok, thanks for the overview though :)

copper.hat

14:06

@XanderHenderson you meshed up your speuling

Xander Henderson

@copper.hat Oh, no! I has dun a missteak!

copper.hat

:-)

robjohn

14:32

Stahp et yoo too

Serilena

I have the following equations:

$L_1 = \sqrt{A_2^2 + A_3^2 - 2A_2A_3 \cdot \cos(\alpha_1)} \\$
$L_2 = \sqrt{A_3^2 + A_1^2 - 2A_1A_3 \cdot \cos(\alpha_2)} \\$
$L_3 = \sqrt{A_1^2 + A_2^2 - 2A_1A_2 \cdot \cos(360 - (\alpha_1 + \alpha_2))}$

how do I know if $L_1 + L_2 + L_3$ can ever exceed $A_1 + A_2 + A_3$

I assume my latex syntax did not work?

Jakobian

15:06

@Serilena yes, it didn't

\begin{align*} L_1 &= \sqrt{A_2^2 + A_3^2 - 2A_2A_3 \cdot \cos(\alpha_1)} \\
L_2 &= \sqrt{A_3^2 + A_1^2 - 2A_1A_3 \cdot \cos(\alpha_2)} \\
L_3 &= \sqrt{A_1^2 + A_2^2 - 2A_1A_2 \cdot \cos(360^\circ - (\alpha_1 + \alpha_2))} \end{align*}

fixed

@Serilena this seems to be some kind of cosine law, but its weird in the sense that the angles add up to $360^\circ$ and not $180^\circ$

Serilena

Yes it is for points inside a triangle

I guess my idea was wrong, I was trying to find.

if sum of lines from the tips of a triangle to any interior point, can never exceeds the perimeter of a triangle. and vice-versa if the point is exterior, the sum maybe greater thus using this as a test method for points inside a triangle or not.

I know the common way is to use areas and sum them up if they form the same size as a the original triangle then it is inside

@Jakobian actually at least on my screen, it does not to render your latex equations

Jakobian

it does on my screen

it doesn't render your latex equations though

Serilena

should I use a plugin for chrome or something?

yes, both set of equations are not rendered.

Jakobian

are you using the bookmark?

it doesn't automatically render in the chat, you need to use a bookmark that you click

Serilena

bookmark? what is that? I usually enter from going through chat.stackexchange.com and then select the math room

Jakobian

15:16

LATEX in chat: tinyurl.com/cfqcvpc

tinyurl.com/cfqcvpc

see upper right of the chat description

if you're on a phone it will be harder, although technically possible

Serilena

I have seen two links before and clicked them, they don't work

I just checked the link you posted, it also does not open

Jakobian

@Jakobian this doesn't open?

Serilena

saying tinyurl refused to reconnect. But I am from Iraq, and I know many US websites don't work properly so if tiny url is from a US provider, it might be the issue.

Jakobian

math.ucla.edu/~robjohn/math/mathjax.html

does this direct link work

Serilena

yes this opened. Let me see if I can follow the instructions

Yes this worked @Jakobian Thank you

I assume I only need renderMathJax one

Jakobian

15:21

> "render MathJax" installs MathJax and renders LaTeX once per execution.
This is intended for use on web pages where the contents of the page are static.

Serilena

Cool thank you!

Jakobian

this means that with render MathJax option you need to click the bookmark to see LaTeX every time someone writes something in LaTeX

> "start ChatJax" installs MathJax and starts a loop that renders LaTeX as needed.
This is intended for use in chat, where the contents of the page are not static.
Reloading the page will stop the loop, so the bookmark needs to be run again.

but if you use the start ChatJax option, you only have to click it once until you refresh the web page

Serilena

Got it.

Jakobian

so I recommend using start ChatJax

alright, well, coming back to the question, now I see why do the angles add up to $360^\circ$

one can easily come up with an example where $A_1+A_2+A_3 < L_1+L_2+L_3$ yet the point doesn't lie in the interior

so $A_1+A_2+A_3 > L_1+L_2+L_3$ is not equivalent to lying in the interior of the triangle

but I do think that every point lying in the interior needs to satisfy $A_1+A_2+A_3 < L_1+L_2+L_3$

$L_1, L_2, L_3$ being the sides of the given triangle, and $A_1, A_2, A_3$ being lengths of segments from the vertices of the triangle to the given point

What I think is that $A_1+A_2+A_3 < L_1+L_2+L_3$ might be indicative of lying in the interior of some bigger triangle in which the given triangle is included

Jakobian

17:11

could someone help me understand this proof? I'm not very good at matrices

i.e. there exists $y^k\in T_k$ in general position

leslie townes

baire category theorem? each of those determinants being nonzero specifies a dense subset of R^n?

Jakobian

how do you see that $\det [\frac{y^1, ..., y^{n+1}}{1, ..., 1}] = 0$ has finite amount of solutions?

(determinant of one of those (n+1) x (n+1) submatrices)

Adhway

17:28

Hello everybody here !!! I am new to maths.. and I was facing one problem.. I apologise if the question that I am going to ask is very elementary........... I have a question that , let us say that we have two functions f(x) and g(x) , and their ranges are R1 and R2 respectively, then we have to find the range of f(x) + g(x)... Now normally we cannot add the ranges of both the functions to obtain the range of f(x) + g(x)

For example , in sinx + cosx , the range is from - root 2 to + root 2

So the question is that , what is the condition that f(x) and g(x) must satisfy , so that we are able to obtain the range of f(x) and g(x) as R1 + R2?

Oh wait.. I am sorry. I interupted.... were you guys discussing some other question?

Jakobian

@Adhway If $f, g:[a, b]\to\mathbb{R}$ are continuous then one such condition could be $\max_x (f(x)+g(x)) = \max_x f(x) + \max_x g(x)$ and $\min_x (f(x)+g(x)) = \min_x f(x) + \min_x g(x)$

Adhway

Oh okk.. and just one more question , we require both the functions to have the same range for this?

Jakobian

no

Adhway

Like R1 = R2 is not required right??....

Yeah thankss

I understood this

Jakobian

@Adhway yes

Adhway

17:35

I was searching all over the net for this but couldn't reallty find anything

Thanks a lot man!

Jakobian

I could be a woman (or other), technically, but yeah, no problem

Adhway

Sorry.

nvm

Thanks a lot

Xander Henderson

17:56

I think I'm funny.

I don't care that no one else does. :D

Soumik Mukherjee

lol

I have never seen this $\eth$ before

Jakobian

oh I think I understood what the author in the screenshot is saying

robjohn

@Serilena I fixed the latex in your comment

Jakobian

that if you have some fixed parameter $x$, and you have a matrix with $x$ in it, such that every submatrix not involving $x$ is non-degenerate, then this can have only finitely many solutions for it to be degenerate

and this is because if we do Laplace's expansion on it, then it will be some linear function $ax+b$ with $a\neq 0$

and so since there is finitely many subdeterminants involving $x$, there is finitely many choices which we don't like

so we can definitely complete the row of $1$'s with $y_i^k$ satisfying the inequality, by going over $i$ and $k$

Jakobian

18:26

this argument is nice, it generalizes to arbitrary infinite fields

Xander Henderson

@SoumikMukherjee It's an eth.

(In the IPA, it is used to denote some kind of dental fricative).

leslie townes

18:43

4 out of 5 dentists recommend eth for your dental fricative

Soumik Mukherjee

Just heard the sound in wiki, and tbh sounds like something one would hear in a zombie apocalypse

here

smth

19:09

how to compute int_{-\infty}^{\infty} e^{x-x^2} dx ?

leslie townes

completing the square would suggest a change of variable that turns it into a multiple of the integral of e^(-u^2) du over the same region

namely, x^2 - x = (x - 1/2)^2 - 1/4

May 31 at 14:05, by leslie townes

by completing the square and a change of variable it is a multiple of e^(-u^2) du, which does not have an elementary expression, but is well tabulated and known. if you were taking a definite integral, that might have an elementary expression

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