Mathematics - 2022-01-22

i don't have ethical issues with eating animals. i don't like seeing animals mistreated of course.

How about my cat, who mistreats me?

your cat should have ethical issues

Indeed.

copper: greenqueen.com.hk/octopuses-lobsters-crabs-sentient/….

5:56 AM

We need to focus on treating other people properly first...

Hi there, I'm trying to find the kinematic singularities of a robot. The book says these postures occurs when the robot resists motion which corresponds to the nullity of the jacobian matrix. I've computed the nullity but I'm not able to find such angles. See the following picture.

Is it possible from the nullity and the matrix's transpose to find such angles?

Places where the Jacobian drops rank, not necessarily nullspace!

@TedShifrin how to check when the rank drops?

Likely Non-manifold points of its configuration space is what I said

So max rank is what?

3

for this example

6:10 AM

Right. So when does it drop to 2?

Think about columns.

when at least one column becomes dependent, I think.

But look at how that can happen here.

to be honest, I'm not sure.

Throw away the zero columns. There are three columns left. Can you use the column of 1s to get a nontrivial linear combination?

Equal to 0, I mean, of course.

one moment.

I think that is not possible.

You mean this, right?

6:23 AM

Presumably the $L_k>0$ and $s,c$s are $\sin, \cos$?

yes

One way it can drop rank is if the last two columns are multiples of each other.

@copper.hat true but in this example at which angles this occurs? I'm trying to see it but I couldn't

That’s what I wanted Croco to figure out.

or may be at 45

6:28 AM

No, constant multiple allowed!

But the same constant on every row.

What ratio?

It looks like a mess, though.

Oh, do row operations !

row echolen form?

Sure. Just the two columns we’re talking about.

@TedShifrin Matlab gives this

[1, 0]
[0, 1]
[0, 0]
[0, 0]

You can’t use matlab.

You have to do it by hand.

sure

6:44 AM

perhaps you could think in terms of $\tan (\theta_1), \tan(\theta_1+\theta_2),...$?

gn

Nighty night!

@TedShifrin I think this is it.

what next?

love_sodam

7:04 AM

Matrix group is interesting although I don't know it well.

You’re not thinking smart, Croco. You’re proceeding mechanically.

Think about using the first row to simplify the others. Then think ratios.

@TedShifrin sometimes I can't think out of the box. I admit it.

but I like your way by pushing your students to their limit.

@TedShifrin the first row is zero. what do you mean by simplifying the others.

7:37 AM

Now I'm thinking smartly.

@TedShifrin

Do I need to equate ratios with zero so these angles are my target?

Q: Has there ever been a case where someone wished a theorem or important result wasn't named after them? Has it happened more than once?

8:40 AM

Let S be a parallelogram not parallel to any coordinate plane. Let S1, S2,S3 be projections of the parallelogram on the three coordinate planes. Then the area of the parallelogram is....?

How do I find this area using surface integrals?

uhoh

9:08 AM

3

The current answers to the Academia HNQ Should I ask for permission to name a mathematical theorem? can be loosely paraphrased as "go for it" and "it would be bad if you didn't", e.g. There is no need to ask permission, and mostly likely they will be happy to have the theorem referred to with th...

hah, that's a very interesting question.

i think sometimes there is disjointness between what a person regards as important, and what gets named after them. usually the people have no control over what is named after them. it will be interesting to see the responses.

10:04 AM

I am loser abd mentally retarded person I am worthless piece of poop and I will never be a successful mathematician.

10:16 AM

You don't have to be a mathematician. It doesn't dictate your value either

It's a normal thing to have those kind of feelings when you're trying to push yourself too hard, I guess?

You might try to push forward, but you'll have to change your mentality

it isn't helpful to characterize oneself or any person as "mentally retarded." these phrases say more about the people who employ these terms than the people they are applied to. even if they're the same people.

10:34 AM

Sometimes I feel I am not creative enough.

Can't think outside the box.

10:48 AM

Is that really the issue though

I think, in math, you'll meet lots of situations like that. You'll be stuck without progress. That's normal

But you're saying, you have low value in yourself because of those things. That's abnormal

I'm not saying such things are easy to not get frustrated about

But you shouldn't take it out on yourself, or at least try to

11:07 AM

How should I parametrize a surface?

I understand that $z=f(x,y)$ can be parametrized as $r(u,v)=(u,v, f(u,v))$.

But what about parametrizing that region of a plane p that gets cut off by a cylinder $x^2+y^2=a^2$?

Here is the question that I am trying to solve: math.stackexchange.com/questions/3250821/…

Since I'm confused at parametrizing the region, here's a solution that I came up with:
$z=a-x-y$ so $z_x=-1=z_y$
The required area $a(S)= \int\int_T \sqrt{1+z_x^2+z_y^2}dxdy$.
What is $T$? $T$ is basically the projection of the plane on the cylinder. So $T$ is the region: {(x,y):x^2+y^2<=a}.

Leaky Nun

@MethNoob what's wrong with never being a successful mathematician?

7

Using polar coordinates: $a(S)=\int_{0}^{2\pi}\int_{r=0}^a\sqrt 3 r dr d\theta=\sqrt 3\pi a^2$

Is that correct? Thanks.

11:25 AM

Ah I finally solved the problem I am the best

@LeakyNun doesn't it sound regrettable when you spend so much time on math and then you fail to contribute to mathematics before dying?

Leaky Nun

11:42 AM

@MethNoob depends on why you spend time on maths

if you had fun and enjoyed it, then there is nothing to regret about

and also depends on what you mean by "contribute to mathematics"

if you have shared a piece of maths that you like with your friends, you have already contributed to mathematics

@MethNoob the mind can often be trapped in its own negative thoughts and as a result view the world negatively; and the solution is to take a step back and view your thoughts as opinions that you generate, and analyse those opinions as they are

"Most people don't think about things. Thoughts appear in their head and they believe them."

About products of Baire spaces again. Well, I didn't prove it myself, but I found an article by Oxtoby which seems to prove this, and I will probably analyze it.

eudml.org/doc/213586

This one

12:06 PM

He basically shows that a product of arbitrary amount of second-countable Baire spaces is Baire

Alessandro Codenotti

1:37 PM

@Jakobian that's a nice result

For finite products of second countable spaces it is an easy corollary of Kuratowski-Ulam of course

anyone know how to reduce the size of an image in a post on main ?

Mhm ^^
According to the article, to prove the arbitrary product case, you first have to deal with the countable product one by using Kuratowski-Ulam repeatedly.

A: How to maximize area bounded by thread and a fixed line?

Specifically in this answer I would like the first image to be much smaller and - if possible - centered

1

You are asking the same as maximizing the surface area between the thread and the thread and the diameter, imposing that the thread stays below the half circle. The solution to this problem is the arc circle: indeed we know that the circle has optimal area/perimeter ratio. Consider the circle of...

There's example <img src="https://example.com/sample.png" width="100" height="100"> here stackoverflow.com/editing-help, I think it works the same

Ah ok thanks

Jonas

2:05 PM

Is there any "intuitive" relationship (say geometric, for example) between a matrix and its adjugate/adjoint? I have an assignment which asks me to "determine and calculate" the adjugate of a given matrix. I'm confused regarding how I would "determine" it.

Maybe they mean for you to provide the definition

determine usually means the same thing as calculate so maybe they just mean for you to calculate it

Jonas

maybe yes

it's probably best to ask some of my fellow students on monday and see how they understood it :)

robjohn

2:48 PM

@Astyx try <img src="https://i.stack.imgur.com/5lAfH.png" width="350">

or whatever width you like

I did a test edit and it works (I didn't save)

This works even better: <a href="https://i.stack.imgur.com/5lAfH.png"><img src="https://i.stack.imgur.com/5lAfH.png" width="350"></a>

That will let them click on the smaller image and get the full size image

Alessandro Codenotti

@Jakobian that's reasonable, failure if Bairness should be witnessed on a countable product. Iirc if $X^\kappa$ is not Baire, then $X^\omega$ already isn't. Don't quote me on that though, I haven't looked at those things in a while

4:52 PM

Any idea how to run this applet? mathinsight.org/applet/…

5:16 PM

@robjohn Thanks I ended doing this. for some reason Jakobian's example didn't work for me

well, it's not my example, I only googled it

I had googled it, but I thought MSE used BBcode and not Markdown

I don't know why it didn't work though -- maybe I made an error when copying it

thanks for your help either way

Hawk

@Jakobian so I think I know what happened. Basically defining smooth functions on aren't defined on open sets can lead to not so well-defined derivatives. I used a one-dimensional case before like f(x) = x on [0,1), but that I found out it was very difficult to see what was wrong. But a good example math.stackexchange.com/questions/2843163/… can be found here when we add 1

more dimension, the problem becomes clearer. So I guess that's why we compare against functions f on arbitrary sets against smooth functions thet are by default defined on open sets

5:55 PM

I don't know if I am asking dumb question but I have blindly believing proof of pyramid is true

The proof was for a pyramid with square base

which is obviously 1/3l^3 for length l

What proof?

but when it is rectangular one can you show me a geometric proof that the volume is 1/3bh where b is base and h is height

nrich.maths.org/1408

I have enough tools to prove it but I think this method doesn't obviously show volume of pyramid with rectangular base

how do you know that the pyramid inside the rectangular prism is exactly 1/3 of it's volume if show that way in the link

when not using calculus

The link you just gave justifies the formula for every rectangular base and triangular base (and from that you can justify any base if you allow limiting processes).

all three pyramid will have different base and different height

In general, you cannot, I believe, fit three congruent shapes inside a box. But you use similarity arguments as the person in that link does.

6:05 PM

oh shoot I didn't read that proof

I read another website

proof

Just shows you should take your own medicine before giving it to us :P

lol

6:33 PM

what's bothering me with this link proof is how they knew what scale factor was without knowing the volume of pyramid with height h and base a^2

scale factor should be (volume of pyramid with height h)/(volume of pyramid with height a)

6:49 PM

but instead if it is treated as very thin slice of rectangular prism

oh now i can see thanks

i hope you enjoy the moosic

I am a very good bad boy

►

8:02 PM

Problem: Compute the area of the region cut from the plane $x+y+z = a$ by the cylinder $x^2 +y^2 = a^2$.
Since I'm confused at parametrizing the region, here's a solution that I came up with:
$z=a-x-y$ so $z_x=-1=z_y$
The required area $a(S)= \int\int_T \sqrt{1+z_x^2+z_y^2}dxdy$.
What is $T$? $T$ is basically the projection of the plane on the cylinder. So $T$ is the region: {(x,y):x^2+y^2<=a}.
Using polar coordinates: $a(S)=\int_{0}^{2\pi}\int_{r=0}^a\sqrt 3 r dr d\theta=\sqrt 3\pi a^2$
Is that correct? Thanks.

8:16 PM

hello, what is the idea to prove that $f(x,y)=x^y^4$ is lipshitzien over y on $\{(x,y), |x|\leq1,|y|\leq 3\}$

$f(x,y) = (x^{y})^{4}$?

no sorry $f(x,y)=x^2+y^4$

Are there equivalent conditions for being Lipschitz?

I vaguely recall there being something about a bounded first derivative

i don't know

Also is this Lipschitz in $x$ or $y$?

8:27 PM

in y

According to Wikipedia
| We have the following chain of strict inclusions for functions over a closed and bounded non-trivial interval of the real line: Continuously differentiable ⊂ Lipschitz continuous ⊂ α-Hölder continuous

Looks like a shortcut?

@copper Yikes!

It's easy to prove with the Mean Value Theorem that any $C^1$ function is Lipschitz on a compact connected subset.

8:43 PM

@TedShifrin Scarily early start to fire season!

ugh, awful. fire season is just every day now.

weirdly i think the bixby bridge is a different family of early-out-west folks than the bixbys who have so much named after them in long beach.

growing up in the west means growing up knowing somebody who has the last name of the street or neighborhood they live in. people ran out of ideas pretty quickly.

albany used to be named ocean view. apparently changed because there were so many ocean views that mail was misdelivered frequently. mild irony in choosing the name albany.

@leslietownes I loved Bill Bixby!

8:59 PM

he had to have been from one of the two families.

el cerrito used to benamed rust

people of my generation, but maybe not yours or anyone here, would know bixby from the movie ferris bueller's day off. in which ferris lived in a house in bixby knolls.

in the script he lived near chicago but the house is in long beach.

i think of the hulk

the ang lee hulk was filmed in berkeley.

i like ferrigno, he comes across as a decent human being

9:03 PM

and of course the graduate.

the graduate said stanford, so when i watched it i was surprised to see telegraph avenue

yeah, durant and college is prominent too.

they also drive the wrong way over the bridge.

in "la la land" there is a scene with emma stone filmed a few blocks from my old house. i remember thinking "that's a cute apartment building" prior to the movie. maybe i have a second career as a location scout.

third career i guess.

educating rita was supposed to be in a uk university but was filmed in trinity, dublin. they painted all the mail boxes red (instead of the irish green). was a bit odd watching it and slowly realising that it was dublin (my birthdplace).

in one of the thin man movies nick and nora drive over the newly constructed bay bridge and you can see the hill in albany and they visit the race track. i think they visit the long beach airport (whose front facade is unchanged) in the same film.

worth a look.

@leslietownes yes, the car goes into Lake Michigan, allegedly.

9:09 PM

well if anyone's in the area i can drive us over to the bueller house. and the la la land apartment building.

my intermittent working on abstract algebra is not consistent with good progress. however, i realised that i am making mild movement when i encountered some $x^2+y^2=z^2$ problem and the first thing i did was take $\pmod{4}$.

Mazltov!

must be the hafner winery trip yesterday.

All my yelling at Under paid off.

a bit depressing, i have realised how barren my social life has become

:-)

9:11 PM

Social life?

exactly

my wife and i were married in a spot that dw griffith used for filming one of his non racist movies. there are scenes literally where we were married. i didn't know this until later, i am something of a film nerd and i was watching it and then it's like "wait isn't that where we cut the cake?" it was.

we got married at the brazilan room on tilden, all the irish folks ended up getting sun burned.

my social life is texting. i had dinner with one of my work friends a few weeks ago. but only one time in a year. no general socializing.

my sort of interaction style doesn't really work remotely, unfortunately. people either get mildly offended or think wtf

9:16 PM

i'm on the wtf side, not mildly offended.

besides, there is a huge difference between chatting with a cute girl on the train and texting some girl.

Good day all -

Your wife will be glad to hear.

Hi Clarinet.

yeah, basically still on the wtf side.

I suppose you've probably heard about Nature moving to open access and Nature Neuroscience opening it up with a $11,528 USD open-access fee?

9:18 PM

clarinetist, i had not, although there is something mildly hilarious about a $10k+ "open access" fee.

@TedShifrin i chat with anyone who will listen. my wife knows i am a loyalty freak.

Here you go: nature.com/articles/…

> Beginning in January 2021, we and our sister journals, including Nature, became ‘transformative’ journals. This means that authors of accepted primary research articles can now choose to have their papers published under a ‘gold’ open access model... the cost of publication is covered by an Article Processing Charge (APC)... The APC for Nature Neuroscience in 2022 is €9,500/US $11,390/£8,290

I think the additional $138 USD comes from something else, can't remember

does that mean to publish a single paper you need to pay $11k?

journals don't do a thing except provide their names. i am no longer an academic so i don't care, but if i had professional security, i would not review for any journal that bled libraries dry. i understand that people without professional security have to do about anything.

@copper.hat That is how I am interpreting that, yes, for Open Access

9:22 PM

god, i wasted so much of my graduate time doing detailed reviews

i would read references, etc

how stupid i am

I've only done one journal review in my lifetime. Got about $240 for it and only spent 2 hours.

i did a lot of reviewing on behalf of my advisor. none of it felt exploitative but it was unpaid work and i don't do unpaid work anymore.

well, i have great respect for my advisor, but he certainly plumbed me for value.

so many stories, not really to be shared here

maybe over a pint at ben and nicks.

he is a concentration camp survivor.

any time you want :-)

i wonder if i listen to johnny nash's "i can see clearly now" enough times if it will improve the mood...

9:25 PM

i'd need a dirty big glass of porter, 8% minimum.

:-) i usually stick with their plonk.

the journaling industry does depend in large part upon people feeling that they have to do stuff.

being raised catholic certainly helps

Got $240? Every journal article I had to review was part of my professional obligation. When I've reviewed books pre-publication for publishers, I've usually gotten a few books out of it.

i never got paid for reviews

wth???

9:26 PM

Yeah, I've never heard of such a thing.

oligopolies. the only way to do business.

i was also never paid for anything. sometimes the publisher would send a book gratis and then ask for a review, which felt a little bit like pressure, but i ignored it.

@copper.hat Yeah, one of the former faculty I used to work with is the associate editor of a journal and was looking for a statistical reviewer to settle the debate between two reviewers about the appropriateness of a statistical method.

cool!

i really am in a truly foul mood.

A number of my editor friends knew they could count on me to be tough and reject stuff, so I got mostly things I rejected. One time I was sent an article by one of my teachers and someone I admire, so I was very flattered, but didn't have the time to do it justice.

calls referee on copper

9:28 PM

:-)

the people who advertise textbooks have utterly no connection to the subject matter. they would visit and it would be like, here are three or four recent graduates of charm school here to bother you.

life would be so much simpler if people (including myself) were up front

this is how they sold opiate pills everywhere. you get young people to bother people relevant to the delivery of the product, and just push, push, push.

maybe i'm cynical. in my business if you make a guess at the worst reason why something is happening it is usually the right one.

"Maybe" you're cynical?

unfortunately i think that is true

9:37 PM

:D

i mean @leslietownes'sd comment

although, i have to say, that when i was at my most successful it was because i trusted people to do what they said they would do

When I was associate department head for 8 years, I did lots of things I should have assigned to a secretary to do, but I figured I was more efficient and more likely to do them correctly the first time.

same here. formally i am assigned a secretary but any request is weighted by, would this actually be faster, and usually the answer is no.

it can take a whole lot of time to tell someone how you want something to be done in that time you can just do it.

there was line in What They Don't Teach You at Harvard Business School that took me years to understand, "spend 5 hours to save 5 minutes"

3

we have this all the time with huge corporate clients. some policy somewhere will say, you need to do X via Y form. ok, filling out the form will take 5 hours, and in 5 minutes i can tell you what you are not going to like if i filled out the form.

9:45 PM

:-)

spoiler alert, we fill out the form.

And charge $400/hr to do so.

it would be ungentlemanly to do otherwise.

But of course.

Yet another interview to do in a few. It's a good thing I'm getting towards the end. After a dozen or so I'm sorta sick of it ... although the last few have been quite fun.

10:06 PM

strange. a 2 word text from my son changed the sign on mood. becoming emo as i age

11:30 PM

@copper.hat $100 prize for guessing the two words.

:-)

I have to compute the following limit $$\lim_{n \to{\infty}} \begin{bmatrix} 1+x & -x & 0 & 0 & \cdots & 0 & 0 \\ \frac{-1}{2} & 1+\frac{x}{2} & -x & 0 & \cdots & 0 & 0 \\ 0 & \frac{-1}{3} & 1+\frac{x}{3} & -x & \cdots & 0 & 0 \\ 0 & 0 & \frac{-1}{4} & 1+\frac{x}{4} & \cdots & 0 & 0 \\ \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1+ \frac{x}{n-1} & -x\\ \\ 0 & 0 & 0 & 0 & \cdots & \frac{-1}{n} & 1+ \frac{x}{n}\\ \end{bmatrix}$$

In what world are you computing that limit?

I can see the limit is gonna be $e^x$, and with some computations I got that $A_n=\left(1+\displaystyle\frac{x}{n}\right)A_{n-1}-\left(\displaystyle\frac{x}{n}\right)A_{n-2}$

Huh?

Are you taking a limit of the determinant or something?

11:36 PM

Yeah sorry

So $A_n$ is the determinant of the $n\times n$ matrix?

Yeah

The determinant is continuous.

So I want to prove by induction that $\displaystyle A_n = \sum_{i=0}^{n}\dfrac{x^i}{i!}$ with the recurrence I found

Not when you're varying the sizes of the matrices, @copper.

11:37 PM

Ahh, I did not see that.

No det in that sentence, @Odes.

I've seen the cases $A_1$, $A_2$ and such, but meh I can't see how to proceed.

So you need to say $A_{-1}=0$ and $A_0=1$, I guess.

I noticed that $A_n = \frac{x^n}{n!}+A_{n-1}$ but not sure how to get this from my recurrence relation

Well, does induction do it?

11:40 PM

Uhm, not sure how to proceed. What I am trying to do is:

Since I noticed $A_n = \frac{x^n}{n!}+A_{n-1}$, then the first recurrence relation $A_n=\left(1+\displaystyle\frac{x}{n}\right)A_{n-1}-\left(\displaystyle\frac{x}{n}\right)A_{n-2}$ could be rewritten somehow

I tried to see if its "telescoping" but no luck :(

No, you're confusing things. You didn't notice $A_n$ is that formula. You want to prove that.

Right

Well, that's my problem, I don't know how to use $A_n=\left(1+\displaystyle\frac{x}{n}\right)A_{n-1}-\left(\displaystyle\frac{x}{n}\right)A_{n-2}$ to prove that

I tried to compute $A_{n-1}$ and substitute it in that relation but meh

Do you know how to do proofs by induction?

Yeah, I know I want to compute $A_{n+1}$

But should be the same than doing the $n-1$ case implies $n$ case, shouldn't it?

Assuming your conjectured formula is correct for $A_{n-2}$ and $A_{n-1}$, plug those in and do the algebra in your recursion formula and hope it gives you your formula for $A_n$.

It does work, and it's not too hard.

11:47 PM

Do you mean to plug it here? $A_n = \frac{x^n}{n!}+A_{n-1}$

No.

You conjectured that $A_n = \sum_{i=0}^n x^i/i!$.

Yeah

Prove that by (complete) induction as I indicated.

But I don't see how to get a sum from my recurrence relation

Did you read what I wrote?

11:49 PM

Yeah but I guess I didn't understand it. Let me re-read it and think.

Put in the conjectured formula for $n-1$ and $n-2$ into your recursion formula.

Ah!!

God. I was trying to derive the sumation from my recurrence going backwards until $A_1$.

Induction is a powerful technique.

Hoping to see some kind of "simplification".

Wrong latex line :D

$\displaystyle A_n=\left(1+\displaystyle\frac{x}{n}\right)\sum_{i=0}^{n-1}\dfrac{x^i}{i!}-\left(\displaystyle\frac{x}{n}\right)\sum_{i=0}^{n-2}\dfrac{x^i}{i!}$

You meant that right :)

Yes!

11:53 PM

And yes now its telescoping

Thanks :)

Do the algebra intelligently and economically. You don't even have to write out the telescoping. Just cancel.

You're welcome.

Yeah the entire sumation cancels after introducing the fraction into the sum.

You don't even need to do that!

Just look at $\sum_{i=0}^{n-1} = \sum_{i=0}^{n-2} + (n-1)\text{-term}$.

Ah!! right, thanks :)

I guess what I was doing is stupid. I was trying to rewrite the recursion formula $A_n=\left(1+\displaystyle\frac{x}{n}\right)A_{n-1}-\left(\displaystyle\frac{x}{n}\right)A_{n-2}$ in the $A_n = \frac{x^n}{n!}+A_{n-1}$ form.