Mathematics - 2021-04-28 (page 1 of 3)

Ted Shifrin

00:01

You're right, as we said, but you need to say “likewise for $b$”.

barista

OK OK

Ted Shifrin

And the sum of positive and nonnegative is positive. Hard to know how much to say without knowing the level of the course.

Or the level of you and your friend.

Howdy @Thor

Thorgott

hi Ted

Clarinetist

D2L is horrible, and I expect Canvas is going to take over

Or has already. The place I teach at is just behind the times. Massive bureaucracy.

It's interesting to me that students at large universities complain about a bloated administration and then use community colleges as examples of institutions that don't need as many levels of administration yet manage to get things done. What students fail to realize is that community colleges don't all operate independently.

Jade Vanadium

00:22

Hi

Jade Vanadium

00:33

Anyone want to take a look at a question I'm working on and give opinions?

Ted Shifrin

Unless you link to the question, we can't answer that.

Jade Vanadium

Oh yeah lol. It's this one
https://math.stackexchange.com/questions/4094288/does-a-closed-curve-exist-for-which-a-square-cannot-intersect-it-8-or-more-times/4116607#4116607

Ted Shifrin

Oh, I already linked that in here yesterday. I do not have the fortitude to read — let alone think about — an answer that goes on for what feels like ten pages. Sorry.

Jade Vanadium

O ok

Ted Shifrin

I'm sure you've spent dozens of hours working on it, and good for you for that!!

Jade Vanadium

00:42

Oh, yeah I spent most of yesterday I suppose.

It seemed at first like the kind of thing that could be solved in a day.

I do not have that impression anymore.

Ted Shifrin

I think I believe that 8 is the minimum. Because the game allows rescaling the square to force intersections of each of its edges with the curve, then the minimum each edge will intersect is 2 when I make the square transverse to the curve (i.e., not intersecting tangentially). So that makes 8. I'm satisfied.

Unless I misunderstand the rules of the game, I think that's the proof.

Jade Vanadium

That only works if the curve is convex.

Ted Shifrin

Nope, not at all.

BigSocks

So are there some standard results of what you get when you take $\Pi_i^{n!} \sigma_i$ for $\sigma_i \in S_n$? I think order will matter for $n > 3$ so maybe something interesting happens when we pick some "natural" order on permutations?

Ted Shifrin

I'm just using transversality (non-tangential intersection) plus obligation for each edge to intersect the curve.

Jade Vanadium

00:45

It's not necessary for the edge to intersect twice. it's possible to have the edge intersect only zero or one times.

Ted Shifrin

There is no natural order, @Big. You can declare one arbitrarily, I suppose. I don't understand your question. You pick $n$ distinct elements of $S_n$? Any $n$ in any order?

Jade Vanadium

In the case of one, the curve can cross in one edge and then out the other.

Ted Shifrin

@Jade That is not the rules. If so, the answer is 0.

BigSocks

@TedShifrin My bad, meant all of them

Jade Vanadium

@ted

Oops

Ted Shifrin

00:47

@Big So you can cancel out the elements of orders >2 if you want to, or you can try for as big a mess as you hope to get.

Jade Vanadium

The rules are stated so that the square is chosen second

The person choosing the square is meant to maximize the intersections

BigSocks

@TedShifrin No, I hope to get something nice, but I fear I get a mess

hope and fear I guess are dual

Ted Shifrin

So obviously choosing the square so that each edge intersects the curve nontangentially is the game plan, not matter what curve you draw. This is why I didn't bother any further with the question. I find it either obvious or ill-posed.

Jade Vanadium

Choosing the square is the hard part though?

Ted Shifrin

@Big What is "something nice"? You want the shortest possible simplified word(s)?

BigSocks

00:50

nah, I want to know if you could somehow pick orderings that cancel out to a single transposition/the identity

I guess equivalently that is what I want

Ted Shifrin

I don't see why, @Jade. I can make it tiny (no intersections) or make it huge (no intersections). In between, most configurations will be transverse intersections with all the edges.

BigSocks

but specifically those 2 so not just it being shortest will be the best I think

Ted Shifrin

Hmm, so you want to pair up most $k$-cycles ($k>2$) with their inverses, but leave a few out to try to cancel products of $2$-cycles.

Have you played with actual examples with $S_3$ or $S_4$?

BigSocks

yeah in a way

Jade Vanadium

You need all four edges to be transverse simultaneously. It's not obvious how to force that to happen. It's not as simple as picking a square and then linearly scaling it up.

BigSocks

00:53

and I was just looking at some stuff with $S_3$, but I'll take a while

I was mostly wondering if there were obvious results I was forgetting

Like I remembered there were some idempotents you got in similar ways in knot theory, so maybe this happened for $S_n$

Ted Shifrin

You can rotate the square, translate it, and resize it. As long as the curve is piecewise-smooth (and not a wild continuous space-filling curve), I believe I am correct.

I'm not going to think further about either of these questions. I'm out.

BigSocks

take care

Jade Vanadium

Bye

vitamin d

Please upvote. Important post from a moderator, XanderHenderson.

Enforcement of Quality Standards.

Ted Shifrin

Done.

Quin

01:10

@vitamind sounds good!

HereToRelax

will be interesting to see who they go after first

Ted Shifrin

I'm the one commenting for the OP to show his work and say where he's stuck, maybe a hint, no more.

leslie townes

01:27

i like low quality posts. i like giving the answer to a freshman calculus question that someone spent longer mathjaxing than thinking about. it's how i got fifty thousand internet points.

Ted Shifrin

You shall be shackled and punished with gruel.

leslie townes

ted is just envious of my fifty thousand internet points.

did i say fifty? i meant six.

HereToRelax

with six you arent trusted

You need 25 to be a trusted responsible bulwark and bastion for the site

leslie townes

i'm more of a bastion than all of them will ever be.

copper.hat

i just want the t-shirt & mug

leslie townes

01:32

bulwark, too.

copper.hat

you are a bit of a bastion

leslie townes

laughed out loud.

copper.hat

i think mse should commission a klein like object with covid like things sticking out. get get one when you hit some number of points.

Ted Shifrin

Ion … ard … a few typing fingers off.

Would a porcupine do?

HereToRelax

only if it has covid

leslie townes

01:36

the shirt ought to say 'ask me how to do your homework.'

alternatively, 'ask me about ted shifrin's differential geometry book'

i miss going to cafes.

Jade Vanadium

This is a counter example to the transversality argument by the way.
https://i.sstatic.net/260eX.png
You can get eight intersections on a square, but it's impossible to do it with two on each edge. The problem is nontrivial.

leslie townes

one time in law school i went to a cafe and right behind me in line i heard two very distinctive voices, it was NPR's the car guys. i went back a little over a year ago and the cafe had closed. it's impossible for anybody other than starbucks to afford cafe rents in cambridge, MA anymore.

Ted Shifrin

So I don't get to move the square however I want, then. As I said, I don’t get the rules. Annoying.

Jade Vanadium

You can move the square however you want.

Ted Shifrin

Then that is not a counterexample.

HereToRelax

01:44

There is a youtuber called Louis Rossman that repairs macbooks and he has a whole series looking at new york real estate

Jade Vanadium

Go ahead, draw a square on that crescent. Any angle, any size. If you can get two intersections on every edge, I'll eat my hat.

HereToRelax

apparently a Popeyes rented an eggregiously expensive place he looked at and he's convinced it can't be profitable

leslie townes

there's stuff like that i saw in long beach, pre-pandemic, that didn't make any sense.

Ted Shifrin

Happy hat eating.

leslie townes

i've often thought that every popeyes should have a cannon on the roof that chooses a random direction and angle every couple of minutes and fires off a piece of chicken. that's how i'd market the business.

Ted Shifrin

01:46

Or a canon?

Paging Bach.

HereToRelax

would the distance be constant?

disregarding wind and height and stuff

Jade Vanadium

@TedShifrin I'll wait for your drawing.

leslie townes

that's exactly the kind of thing that the customer base would congregate around the store to find out.

HereToRelax

would probably be good publicity

you would get people outraged

leslie townes

i would get stuff from popeyes when i lived in oakland. here there is roscoe's house of chicken and waffles, so i have not attended popeyes in some time.

HereToRelax

01:48

like the guys at burger king

I have never attended Popeyes or Chipotle

I hear people my age are supposed to have their Chipotle order ready for when people bring it up in conversation

leslie townes

chipotle is pretty good. there was one in iowa city that i never attended because they had a local analogue of the same type of thing, pancheros, across the street. i went there instead.

i lived on chipotle in law school, which by the way is another business that could not afford to continue operating in harvard square.

HereToRelax

pancheros is the most mexican non-mexican thing I have ever heard

it's very good ngl

leslie townes

my favorite mexican place is cactus taqueria, right next to the rockridge bart station in oakland. they have other locations. the quality has gone down somewhat over the years but it is still good.

HereToRelax

One of my favorite tacos is a night place that opens near the curve two blocks from my house

it gets full of taxi drivers parked on the sidewalk blocking half the street

that's how you know it's good

copper.hat

chipotle have great bacteria.

cactus have tiny soccer mom burritos.

"organic"

HereToRelax

02:03

how known is it that chipotles are dried smoked jalapeños

copper.hat

in the albany berkeley area? toddlers know that.

HereToRelax

o.o

copper.hat

they compare aromas and fight over how sustainably sourced they are

just kidding. i am in a foul mood today.

HereToRelax

I actually believed you

Jade Vanadium

If you find that curve too easy, you can stretch it out to make it harder. One of the cases we looked at was this one.
https://i.sstatic.net/jINLF.png
This one shows how to get eight intersections, but it only does it by having four on just one edge.

HereToRelax

02:10

what kind of curve is that?

Jade Vanadium

Just some kind of weird crescent. It was designed to attempt to minimize the number of crossings any square can have

There's a question on Math.SE asking if there exists a closed curve such that no square intersects it eight or more times

Some people were trying to make counter examples. None of the attempts worked, and I think there aren't any counter examples, but the proof is very hard, if it is true.

Some of the counter examples, like this one, are very good at showing why the problem is hard

HereToRelax

So it doesn't exist?

Jade Vanadium

The counter example curve? We don't know. I think it doesn't exist, but I'm not entirely sure

It's comparable in difficulty to the inscribed square problem, which is still open

I figured it would be easier than inscribed square, but after having tried it and seen their differences, I'm not sure.

The question on Math.SE doesn't have any affirmative answer yet. I have a partial answer solving a bunch of subcases, but the general problem is really hard.

The crescent up there, with the square through it, shows a really novel configuration to produce a square giving eight intersections. I'm pretty sure that, for the crescent curve at least, that sort of configuration is the only way to get eight crossings. It's not the kind of thing you would think of immediately.

HereToRelax

oh, the concavity shows why the inscribed square problem doesn't imply this one?

Jade Vanadium

Yes exactly

That was part of my original answer actualy

Someone suggested that inscribed square would solve this, but that's not necessarily true, since we can have weird concavity

Inscribed square does solve the convex subcase really easily, though.

HereToRelax

02:20

cool

Jade Vanadium

Yeah it's pretty neat. It's very nice, I think, when something you think will be easy ends up being really hard.

Since that improves your intuition

leslie townes

soccer mom burritos, laughed out loud again.

the burrito mejor used to be very large, but they have cut it down.

copper.hat

02:40

picantes used to be the goto, but they started downsizing their tasty but pricey burritos and they stopped the curbside pickup. gordos was never a fave, but when the opened online ordering i warmed to them. and they are ginormous. my fave that i am aware of in the area for burritos is sinaloa on telegraph in berkeley

just checking that the organ helicopter is not nearby.

leslie townes

a lot of people in my school swore by gordos and another one in that area. i think they were swayed by a variety of salsas, which to me is gimmicky.

i had a cactus burrito as my friday treat almost every week from 1998 to 2007

copper.hat

i used to bring the kids there. but they are not big enough for my 6'3" 17yo.

leslie townes

the burrito mejor used to be huge, but they did cut it down. they also reduced what they put on it. i wonder if it was a change in management or ownership.

copper.hat

dunno. its gordos for the foreseable.

no wait

there is talaveras. they are awesome.

just not as convenient for ordering.

what was i thinking.

and equidistant to us.

leslie townes

i miss solano ave. if i could drag-drop one street into long beach it might be solano.

copper.hat

02:46

their chimichangas are incredible

one place i liked that has gone was liu's kitchen

and there was an iranian run place on solano across from colusa (i think, if i recall correctly)

they did great breakfasts

salmon omelette with a chilled mimosa

hard to believe i just polished off 0.5lb fresh salmon and 3 spuds worth of mash

the temptation to hit the kona is strong but i need to do some thinking work.

leslie townes

and now i'm hungry again.

copper.hat

software development is, for the most part, an incredible drudge.

i should not complain, i have a nice setup at the moment. and my customer likes our stuff.

rare to have an eda customer say that, i am proud to say.

Andrew Micallef

@TedShifrin ex 1.1.2. I think

leslie townes

before you click through that looks as if it's written on a refrigerator.

i was mildly disappointed to see it was a clipboard and paper setup.

Andrew Micallef

Oh

Why?

copper.hat

02:53

is $v$ a constant?

hardly in a helix. Maybe $vt$?

Andrew Micallef

I tried to hold my hands steady, but im v shaky

Yeah v is the derivative of vt wrt t

copper.hat

sry, really slow. part of my aging process.

why are you reparameterising the curve?

Andrew Micallef

Np im reaally slow at this stuff

Because it was the excercise

leslie townes

my handwriting is far worse. what is the goal, where are we going. don't make me click through to ted's book. i understand he is paid exorbitant royalties every time somebody does that.

copper.hat

i see. mechanics.

Andrew Micallef

02:56

Sorry leslie, i will leave you hanging. My lunch is over. Maybe I will rememver to tell you tonight :)

copper.hat

twist and frenetic ferret curves.

life would be much simpler if we just outlawed any rotational acceleration.

TNB sounds like a crude description of something.

Ted Shifrin

@Andrew That is totally illegible.

Kevin Zheng

Hey, I don

Hey, I don't have any reputation, so I was wondering if anyone could direct me to more information about the first answer in the following post math.stackexchange.com/questions/2281246/…

More specifically, I was wondering if anyone has any information about (1 + D)^-1, and if there's a generalization like (1 + D)^r, possibly using Newton's general binomial theorem

copper.hat

i have no idea, sry.

leslie townes

03:12

it isn't even clear to me what D is. the operations look familiar. rota had some good papers on the calculus of finite differences but this seems continuous. somehow.

writing $\mathrm{d}t$ before the integrand will never not seem wrong to me.

kevin there is a notion of a continuous functional calculus which can be applied to sufficiently nice operators (bounded and self adjoint is ideal, normal is OK, unbounded is fine but requires a lot of work). you can write out integral expressions for the functions of the operators and even use software to approximate them.

anything called D might be unbounded, which makes me nervous.

Ted Shifrin

$D$ is the usual $d/dx$, of course.

leslie townes

okay yeah, unbounded. and away we go.

Ted Shifrin

But, this is formal Taylor expansions. Assume real analytic functions.

shintuku

03:44

ah, I finally understood limit proofs

Rover

why in Ellipse $h^2-ab<0$ ?
If the general equation is $ax^2+by^2+2hxy+2gx+2fy+c=0$

$ax^2+2x(hy+g)+(by^2+2fy+c)=0$
Then, it has two roots for any y=k
Therefore, delta>0
4$(hy+g)^2-4a(by^2+2fy+c)$>0
$4y^2(h^2-ab)+By + C $>0
delta>0 and $h^2-ab>0$ ?
So, $h^2>ab$

leslie townes

x^2 + 3.

Jade Vanadium

@KevinZheng Sometimes yes, but also your question is complicated. If the exponent r is fractional, there's not always a well defined meaning for a fractional exponent of an operator. In the case r is an integer however, then yes the standard binomial theorem applies, when defined

Rover

@leslietownes you got $h^2-ab<0$

I want to know how?

shintuku

yes, x^2 + 3 is under control now

leslie townes

03:45

i did. i'm regretting it, but i did get that.

shintuku

i erased any geometry from my mind, and all was fine

leslie townes

if the discriminant of that (regarded as a quadratic in x) is a polynomial in y with a positive coefficient of y^2, the real square root in which it appears will be defined and produce solutions for an unbounded set of y. that's not what happens with an ellipse.

hence, i think, my desire for a < instead of an >

Rover

@leslietownes Common..didn't we did same thing?

shintuku

he called you a commoner!

leslie townes

there were a few hours this morning where i thought we were in the same page. then i had some phone calls with some annoying attorneys and i don't remember where we left off.

Rover

03:49

@shintuku hey what's commoner?

@leslietownes okok

shintuku

a person who has rights over another's land, and thus, makes said object of alienated rights suffer greatly from the tragedy of commons

leslie townes

i'm not a commoner, i'm royalty.

copper.hat

what is wrong with a commoner?

shintuku

nothing, they are the vanguard of true, radical change. bless the commoner, but not the one with implicit rights over the commons, the other one

"bless" here, is, of course, a figure of speech

Rover

I guess it's 9:30 pm there now ?

leslie townes

03:54

we have some weird 15 minute difference. it is 8:54 PT.

i don't think it's 9:30 anywhere.

there are countries that offset standard time zones by 30 minutes but i do not live in one of them.

shintuku

i guess it is for people working at CERN

Rover

@leslietownes Oh alright , to be exact it's 9:24 here.

leslie townes

so there is a 30 minute difference. fascinating.

shintuku

@Avra stackedit.io is nice to preview your mathjax

leslie townes

avra even an approximate version of your question would be appreciated.

copper.hat

03:57

Nepal has a 15min offset, India 30 min.

leslie townes

i forgot about that.

Avra

Hello. I have simple question about algebraic simplification of the following quantity:

$\frac{4n^2+4n+1}{4n^2+8n+4} < \frac{n}{n+1}$

The final answer I got is
$1<0$

But the book solution is

$n+1 < 0$

I am no sure how the quanity $n+1 < 0$ was got!

shintuku

$1<0$ is false

copper.hat

How did the order creep into the setup?

leslie townes

the quantity is just a quantity, you can divide it using known methods for division by polynomials. the inequality is something else.

copper.hat has asked the question more pointedly.

oh i think i was responding to an unmodified version of something that has been edited. let me think.

shintuku

04:00

huh, wolframalpha agrees with you, there's no solution

are there any other restrictions? @Avra

Avra

@leslietownes. Sorry. I edited it.

shintuku

according to wolframalpha, $1<0$ should be fine if there is no other restriction

Avra

@shintuku. There is no solution, but I am asking about the final form, how the book got $n+1 <0$ please?

copper.hat

I get $n+1 <0$. You just need to be more careful.

Avra

@copper.hat. Thanks. I will double check again.

copper.hat

04:04

There is no magic, just multiplying & adding.

Note that $\frac{4n^2+4n+1}{4n^2+8n+4}= 1-\frac{4n+3}{4n^2+8n+4}$ and something similar for the rhs.

shintuku

$n+1<0$ iff $n \neq -1$

copper.hat

That is not true.

$n=1$

Avra

@copper.hat.

This is how I did it,

$\frac{4n^2+4n+1}{4(n+1)^2} < \frac{n}{(n+1)}$
$\frac{4n^2+4n+1}{4(n+1)} < n$
$4n^2+4n+1< n(4(n+1)) $
$1< 0 $

copper.hat

How did you even do the first step???

Why did you decide that $(n+1)$ is not negative???

You need to be careful.

Avra

multiply both sides by $(n+1)$ given that $n \in mathbb{Z}$

copper.hat

04:13

if $-5 < -1$ is $5 < 1$?

Avra

$n$ is only positive integers $(1,2,3, \cdots, n)$

shintuku

ah, there you go

copper.hat

Why, was that part of the problem statement?

Avra

@copper.hat. Yeah!

shintuku

@copper.hat the original statement, if true, implies $\frac{n}{n+1}$ is undefined if $n = -1$

copper.hat

04:15

I did not see that above.

Avra

@copper.hat. I apologize.

copper.hat

Then the book solution $n+1 <0$ makes no sense.

Which is true for any $n$ for which the equation makes sense.

shintuku

$1 < 0$ is a nonsequitur if $n$ is in the natural numbers, but not if it is unrestricted

Avra

@copper.hat. So my solution is true $1<0$?

copper.hat

I would not write it like that, I would say there are no solutions satisfying $n \ge 1$.

Or $n >1$.

shintuku

04:21

That the original equation implies $1 < 0$ is true if $n \in \mathbb{R}$

but the original equation does not imply $1 < 0$ if $n \in \mathbb{N}$

Avra

@leslietownes. @copper.hat. @shintuku Thank you.

copper.hat

You can write the equation as $1 - {4n+3 \over 4 (n+1)^2} < 1 - {1 \over n+1}$. THis is equivalent to ${4n+3 \over 4 (n+1)^2} > {1 \over n+1}$. Multiplying both sides by $(n+1)^2$ (which is positive for $n \neq -1$) we get $4n+3 > 4(n+1)$ which reduces to $0 > 1$.

Rajorshi Koyal

04:46

66÷12÷6

How to solve this?

Which divides which

Please help..

leslie townes

parentheses, please.

that's the kind of thing that will go viral on social media.

Rajorshi Koyal

The question doesn't have one..

shintuku

just assume ÷ is commutative

leslie townes

if an operation BLERP is associative, which division is not, it is considered okay to write a BLERP b BLERP c.

Rajorshi Koyal

So what should be the next step?

leslie townes

04:48

absent associativity, parentheses are needed.

Rajorshi Koyal

It is not there in the original question what do I do..??

shintuku

the expression has no meaning

leslie townes

ask for a better question. (66/12)/6 and 66/(12/6) are different things. the question is not well posed.

you could go viral on twitter with it.

Quin

I am thinking of buying a few volumes of Landau and Lifshitz' books on theoretical physics. Does anyone here know if the volumes build on each other or are they relatively self-contained (unfortunately, the topics of interest are not in order in their series!)?

Rajorshi Koyal

@shintuku What does @shintuku with this?

Please explain this to me..

shintuku

04:51

i was joking about the commutativity, but that expression has no determinate meaning

as pointed out by leslie

what's the context?

leslie townes

quin, i do not. just tossing that in there so you know you are not being ignored.

copper.hat

i thought you had accidentally posted here, realised that this was maths not physics and gone elsewhere...

leslie townes

i always experiment with copyright infringement before buying volumes of anything. i don't suggest doing that.

Rajorshi Koyal

cube root of (46656)÷square toot of(144)÷square root of (36)

that is the whole question..

Maybe it could revolve around some assumption

leslie townes

when x is not associative, as division is not, a x b x c has no meaning.

you might as well ask for 'the division of all the numbers on the chalkboard.'

shintuku

04:54

@RajorshiKoyal is it a math book?

or what sort of book/

maybe we can get the meaning from the context

Rajorshi Koyal

No from my study material

Quin

Lol, no, I'm partial to math, so I post here. Just throwing it out there in case anyone knew. Googling around for such things didn't reveal anything. Yeah, I think I might not infringe a bit :P

Rajorshi Koyal

No book a handout

leslie townes

sometimes people do adopt a left to right convnetion or something else.

Quin

yeah, that division question is nonsense as far as i can tell

Rajorshi Koyal

04:55

ok.. So I guess it will be a left to right convention

leslie townes

but this is higher order stuff than the basic premise which is that if you don't tell me whether to to (a x b) x c or a x (b x c) i don't know what to do.

Rajorshi Koyal

ok..

leslie townes

if you have a twitter account, create a poll and have your followers vote on the outcome. that is how most things are decided these days. [i am joking, please do not do this]

shintuku

definitely need to assume a convention. but if someone will correct it, make sure to write both and explain why you wrote both possibilities

if 66÷12÷6 means (66/12)/6, then blablabla. if 66÷12÷6 means 66/(12/6), then blablabla

leslie townes

that seems like the only opportunity.

shintuku

04:59

i'm an econ student and i've done this multiple times on my homework sunglasses

leslie townes

it is also the lawyerly response.

Silent

Where formula (3) is $\ln z = \ln |z| + i \text{ Arg } z \pm 2n\pi i$

I can't understand what would happen if, say for $n=0$, we include negative real axis as well? I mean, derivative is obtained to be $1/z$ even at negative real axis, hence in particular it is continuous there, right?

Rajorshi Koyal

@leslietownes The answer reads 0.5

How would that happen?

leslie townes

i'm beginning to wonder whether $\div$ means something completely new.

Rajorshi Koyal

:P

leslie townes

05:01

i don't mean to be sarcastic; any way i can think of interpreting that expression, which i think we agree is not unambiguous, does not lead to 0.5.

Rajorshi Koyal

ok..

shintuku

is it a programming language question?

Akiva Weinberger

Quin

@Silent If you used the integral formula for the log and integrated to the negative real axis via two different paths (ccw vs. cw) you would get different answers

Akiva Weinberger

^Reminds me of @BalarkaSen

leslie townes

05:07

clank

Rajorshi Koyal

Please check problem 24 and problem 30..

copper.hat

did i miss some big room arrangement?

can i dump my problems here?

Rajorshi Koyal

I am not dumping they are tricky indeed.

Please understnad.

leslie townes

i need parentheses to make sense of $\div$.

Rajorshi Koyal

ok try 30 once may be I am wrong..

Please check this problem out

leslie townes

05:18

whoever is posing these has a grudge against parentheses

Rajorshi Koyal

:P

copper.hat

i'm saving up my internet bits so i have some if it breaks again in the morning.

smart, huh!

satan 29

hi

copper.hat

lo

satan 29

does anyone here know about the frobenius method

for DEs?

shintuku

05:32

@RajorshiKoyal we can safely say none of these

copper.hat

only the application

Rajorshi Koyal

ok

shintuku

@RajorshiKoyal we can safely say that neither of the two possible meanings result in either 0.5 or 0.51

copper.hat

it is just matching Taylor series coefficients to get a difference equation.

Rajorshi Koyal

ok..

shintuku

05:33

for problem 30, there's no such ambiguity, so you should be able to find the proper solution no problem

Euler 2

Frobenius method

In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a second-order ordinary differential equation of the form z 2 u ″ + p ( z ) z u ′ + q ( z ) u = 0 {\displaystyle z^{2}u''+p(z)zu'+q(z)u=0} with u...

Rajorshi Koyal

Those are not plus signs divisions only

It was misprinted

shintuku

can you give me the possible interpretations of the question?

with what we've shown up to now, you should be able to tell how many different ways there are to interpret question 30

you should also be able to list the proper answer to each interpretation, and see if any of the multiple choices match it

Euler 2

what is question 30? the picture is a bit blurry

is it $175616+56^3+8=?^2+7$?

Rajorshi Koyal

No no those are divisions @Euler2 I am posting a fresh question

Please ignore that

copper.hat

05:46

jeez

shintuku

06:35

there is $\longrightarrow$, is there a long \mapsto somewhere?

robjohn

Oh, let me see

shintuku

wait

$\longmapsto$

$\mapsto$

yeah, I guess that is a long mapsto

robjohn

\longmapsto = $\longmapsto$

shintuku

thanks!

robjohn

@shintuku When you want to find LaTeX characters, try Detexify

shintuku

06:49

an excellent resource, thanks!

Euler 2

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$Mathematics$ Mathematics