Mathematics - 2022-02-19 (page 1 of 2)

copper.hat

00:06

well, that was exciting

geocalc33

where was exciting?

copper.hat

i was being facetious

geocalc33

I'm trying to prove that a function is even

well actually that it's symmetric about a vertical line

Koro

02:18

main website is down.

LearningCHelpMeV2

something happen?

Koro

Probably the site is down for maintenance.

leslie townes

02:37

i saw it in 'read only' mode, but not down, when i checked 20 min ago.

Ted Shifrin

Doesn’t seem down, but no new questions.

leslie townes

so yeah maybe maintenance and not the usual ddos crazies.

Koro

The site’s back now.

rb3652

03:08

Hey @TedShifrin, @leslietownes, @copper.hat, and @PM2Ring -- first off, thanks a lot to all of you for spending almost two hours (I think) trying to help me fix my problem.

Turns out it all came down to a minus sign:

2

Q: Collision of Ball in Triangle

A ball with position and velocity $(P_0,V_0)$ is in a triangle. Which side of the triangle it will hit? Calculations: The ball's motion is $$L( t) = P_0+t*V_0 = \ ( V_{0_{x}} t+P_{0_{x}} ,V_{0_{y}} t+P_{0_{y}}) \tag{1}$$ The sides of the triangle are given by $A_1x+B_1y=C_1$, and we can represen...

leslie townes

hah. you tricked someone into actually reading the equations.

well done. :)

mohan10216

What are p-adic numbers?

ive been reading on wikipedia for like an hour and I don't get it...

leslie townes

subtle stuff. textbooks probably a better resource than wikipedia, which isn't all that great for real numbers.

are you familiar with any construction of real numbers? it might help to start there.

the stuff with formal series i'm seeing on wikipedia is confusing. it's just "here are these crrrrazy numbers!" kinda like looking at real numbers from the point of view of decimal expansion. suddenly there's all this work to do, and infinite sequences, when the concept is simpler than that but also subtler than that.

you would want to do the construction of R from Q as the completion of a metric space. not the order theoretic route.

soupless

03:41

@leslietownes Question for now. Rest assured that this is not from the contest as it will start at 12:00, UTC+8. When simplifying $a \bmod c$ where $c$ is composite, how should CRT come into play?

leslie townes

i dunno what you mean by 'simplifying' a mod c. the CRT relates the value of a mod c to the values of a mod c_k where c_k are pairwise coprime factors of c.

or at least some version of CRT does that. i think there are a family of results called the 'CRT.'

i'm probably going to be sued in florida just for saying this much

soupless

I mean, something like $1897 \bmod 48$.

I am referring to the Chinese Remainder Theorem. Sorry about that

leslie townes

if you wanted to find the n with 0 <= n < 48 with 1897 = n mod 48 i would just do it? it's division, no CRT necessary.

i understand. but i think there are a family of results, all referred to as 'the chinese remainder theorem.' i'm not seeing where you would use it here.

soupless

I am trying to solve problems of the form "last n digits of $x$"

And I found this problem, specifically problem 9.

leslie townes

if you just do division with remainder you see 1897 = 39*48 + 25. 25 is the "simplified" version of 1897 mod 48.

soupless

03:48

Find the last three digits of $2003^{2002^{2001}}$.

leslie townes

i would use euler's theorem for that. en.wikipedia.org/wiki/Euler%27s_theorem that theorem. there have to be worked out examples.

i should warn you, i never did contest math, and the solutions i find may not be the ones that work 'quickly.'

soupless

Yes, I was typing that I should use Euler's theorem. Then, I should solve $2002^{2001} \bmod \varphi(1000)$

leslie townes

math.stackexchange.com/questions/2186887/… is an example

soupless

Let this value be $p$. Then, $2003^p \equiv 1 \bmod 1000$

leslie townes

i guess you can use CRT to work out what it is mod a power of 2 and then mod a power of 5.

once you know that, you know what it is mod 1000.

why anybody would want to do this, that's what i'm interested in. :)

soupless

03:51

Yes, that is what I am talking about. How can I use CRT here? Is it that I should factor out the modulus to its prime factors, then use CRT on that?

Wolgwang

@soupless What's the answer?

soupless

I still don't know for now.

Anyway, the contest will be starting soon, I should go for now. Thank you very much!

Wolgwang

@soupless All the best! :-)

Ted Shifrin

Is a contest like a war?

user2236

04:09

if you want to win at all costs, then yes

soupless

The contest is from a particular group, which is affiliated to the school I am applying to. So yes

I keep in mind that failing to reach the national round = no scholarship

user2236

Win and you're in.

soupless

@TedShifrin So yes. I need to win the war, even if I lose the battle

user2236

04:23

Setting up 🍯 honey pots to catch cheaters is part of that "war"

Prithu biswas

I am waiting for the final proof of the Collatz conjecture by PenAndPaperMathematics.

Sort of feels like a giant cliffhanger.

user2236

don't hold your breath

The landscape of mathematics has many cliffs :-)

Always remember, we live for the climb.

Prithu biswas

And I am also waiting for the "An Infinite Sum Suitable for the Efficient Computation of Quotients" by AMDG.

user2236

A Network of Fake Test Answer Sites Is Trying to Incriminate Students

Aka, entrapment.

The JEE has been doing it for years.

Random Variable

04:41

@robjohn If $f$ and $g$ are functions in $L^{2}(\mathbb{R})$, does $$\int_{-N}^{N} f(x)\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \left( g(t) \boldsymbol{1}_{[-N,N]}(t) \right) e^{-ixt} \, \mathrm dt \, \mathrm dx $$ converge to $\int_{\mathbb{R}} f(x) \hat{g}(x) \, \mathrm dx $ as $N \to \infty$? That would seem to provide a way to justify switching the order of integration.

user2236

05:43

@rb3652 ouch, live and learn

copper.hat

05:55

@rb3652 i added a non answer to your question illustrating how it can be done using inner & cross product.

SMC not Sven Magnus Carlsen

09:57

Any hint on how to prove the following integral equality? I'm getting lost because it seems to me that we are mixing some Cauchy integral formula with some Fourier Series but I've never seen anything similar in class

I can just "copy" the proof that the coefficient of a Fourier series are in that form and adapt it to Homogeneus Expansion for holomorphic function?

I can use the classic Cauchy-inequality to prove the theorem but I would like to understand from where that integral cames up

William John

I want to prove that there exist sequence $x_n<k$ such that $x_n>k$ as n approach infinity x_n and k are real numbers and n is positive integer

user123456789

Hey

is anyone there ?

I have a stupid question to ask

helloooooo

SMC not Sven Magnus Carlsen

10:13

@user123456789 Just ask; don't ask to ask

user123456789

There is something I am missing here and I cant figure out what

SMC not Sven Magnus Carlsen

the "from this" we can say it's just false

you can't assume that from what you have wrote

that thing is 0 because a=sqrt(b) but you already know that, it's literally your first line

user123456789

Oooooooo

so in cases like this at least one of them is zero.

I don't know how to explain but when should we be concerned about more than 1 anwers ?

SMC not Sven Magnus Carlsen

10:28

in that case you can have more than 1 answers only if sqrt(b)=-sqrt(b) as you wrote hence b=0

Friends lie, Girlfriends lie, Algebra does not

you wrote it yourself

user123456789

X^12 = 49^(1/124)

Would this have 1 solution or more than 1 solutions ?

Wolgwang

91

Q: Why can't you square both sides of an equation?

Why can't you square both sides of an equation? I've been asked this many times and can never quite give a good, clear, concise answer (for beginning algebra students) in plain language. I just searched the web and still couldn't find a simple-to-understand answer for why squaring both sides giv...

algebra-precalculus

@user123456789 Two solutions...

user123456789

How do you know ?

William John

Why is limit point of circle not outside circle. Intutively it makes sense but how I prove it? Is there a proof?

Wolgwang

@user123456789 (I am assuming you meant real solution)By solving... $x^{12}=7$ Take square roots then eliminate one equation then again take square root.

William John

10:43

or may be say why limit point of set S=(0,1] is not 2?

William John

10:53

It seems I got a bit vague definition from text.

My text says x is limit point of S if there exist sequence x_n such that x_n approach x as n approach infinity

x_n is element of S except x

sorry it seems that I can solve it I am being too vague

It's gonna contradict the definition of limit

Jakobian

@TedShifrin I meant it more as... do they hold the same amount of importance, kind of. It's an informal question.

user123456789

@Wolgwang How do you know that it has 2 solutions just by looking ?

Typo

11:12

can someone point me towards the solution to 5^x - 2^x = 117 utilizing modular arithmetic? it's clearly x=3 but I want to get the answer without inspection.

I got to $125(5^{x-3}-1)=8(2^{x-3}-1)$, which tells me $2^{x-3}-1$ is a multiple of 125, so $x-3=100k$ for some integer k as $\text{ord}_{125}(2)=100$. similarly for the other side, $x-3=2m$ for some integer m. I don't know where to proceed from here

Typo

11:31

Actually, I'll ask on the main site

user 85795

@user123456789 have you read through the answers in the posted question?

1 hour ago, by Wolgwang

91

Q: Why can't you square both sides of an equation?

Why can't you square both sides of an equation? I've been asked this many times and can never quite give a good, clear, concise answer (for beginning algebra students) in plain language. I just searched the web and still couldn't find a simple-to-understand answer for why squaring both sides giv...

algebra-precalculus

You know it has two solutions because if two numbers have the same sign their product is positive.

soupless

12:11

@leslietownes I didn't get to use CRT at all.

Xander Henderson

12:35

@soupless Here in Arizona, they are trying to make CRT illegal.

Prithu biswas

3=3 ⇒ 3^2 = 3^2 ⇒ 3^2-3^2 = 0 ⇒ (3+3)(3-3) = 0 ⇒ 3+3 = 0 ⇒ 3 = -3 ???

Xander Henderson

No.

If $a,b\in\mathbb{R}$ and $ab = 0$, then either $a=0$ or $b=0$. It is possible that only one is zero.

Prithu biswas

@XanderHenderson Isn't the problem division by zero?

Xander Henderson

12:50

Not really, no.

You could, I suppose argue that the problem is a cancelation by zero. But then the appropriate response is that $ac = bc$ if and only $a=b$ and $c\ne 0$.

But I guess that a lot of students are taught that if $a=b$, then you can divide both sides of the equation by the same thing. This is generally the effect of cancelation, though I would argue that the actual property is a bit more subtle.

PM 2Ring

Shouldn't that be $ac = bc$ if and only if $a=b$ or $c=0$?

Prithu biswas

So $ac=bc$ if and only $a=b$ and $c≠0$ is a true statement?

soupless

@XanderHenderson Cathode-ray tubes?

Xander Henderson

@PM2Ring Not if you want an iff statement.

I could clean it up a little, I suppose.

PM 2Ring

I have implied parentheses: $ac = bc$ if and only if ($a=b$ or $c=0$)

Xander Henderson

12:59

Yes, I know.

But that doesn't capture the idea of "dividing both sides". That captures the idea of multiplying both sides.

Prithu biswas

What about $ac = bc$ iff $a = b$

user 85795

c is not equal to 0

because 0 times any number is 0

Prithu biswas

Like $a=b$ does imply that $ac=bc$ . But $ac=bc$ doesn't imply $a=b$

Xander Henderson

In any event, my original statement was the most relevant: $ab = 0$ implies that $a=0$ or $b=0$.

PM 2Ring

There are a zillion questions on SO where the code doesn't behave as expected because the OP hasn't applied De Morgan's laws properly.

user 85795

13:02

aka the zero product property

PM 2Ring

@XanderHenderson Most certainly.

Prithu biswas

@XanderHenderson Is there another example where division by zero is not the main problem?

user 85795

Division is multiplication by the reciprocal.

Xander Henderson

Again, I don't think that division by zero is really the problem here. You aren't really dividing by zero. You are attempting to apply the law of multiplicative cancelation in a situation where it doesn't apply.

soupless

@user85795 Multiplicative inverse.

user 85795

13:06

Yes.

0 has no multiplicative inverse

Xander Henderson

It might be best to state the "rule" as "If ($ac = bc$ and $c\ne 0$), then $a=b$."

user 85795

👍🏼

Xander Henderson

As I said above, this amounts to more or less the same thing as "dividing both sides by $c$" (or multiplying by $1/c$), but I think that there is a distinction here which is relevant in more general settings, e.g. rings with zero divisors.

user 85795

Not the reals.

Quoting the original poster:

> I've been asked this many times and can never quite give a good, clear, concise answer (for beginning algebra students) in plain language. I just searched the web and still couldn't find a simple-to-understand answer for why

are they ready for a more general setting :P

PM 2Ring

Eg, $1\cdot 5 \equiv 3\cdot 5 \pmod{10}$ but $1 \not\equiv 3 \pmod{10}$

Xander Henderson

13:19

@PM2Ring Right, which was why I originally specified that if $a,b\in\mathbb{R}$ and $ab=0$, the either $a=0$ or $b=0$.

It is a theorem about the reals (or, more generally, rings without zero divisors).

Ditto multiplicative cancelation. If $ac = bc$ and $c$ is not a zero divisor (or some similar hypothesis), then $a=b$.

PM 2Ring

Understood

Prithu biswas

@XanderHenderson In the standard axiom of reals, there is no division ?

Just multiplicative inverse?

user 85795

and additive inverse

Xander Henderson

I don't know what is "standard"---there are many ways axiomatizing the reals---but, generally speaking, we don't define "division", except by way of multiplication by an inverse.

Prithu biswas

@XanderHenderson Oh ok.

user 85795

13:33

has your confusion been cleared up?

55 mins ago, by Prithu biswas

3=3 ⇒ 3^2 = 3^2 ⇒ 3^2-3^2 = 0 ⇒ (3+3)(3-3) = 0 ⇒ 3+3 = 0 ⇒ 3 = -3 ???

Prithu biswas

Yea.

user 85795

coolio 😎

user21820

13:59

@XanderHenderson I just want to comment that for ℝ specifically, at the beginner level, it seems best to understand all equation manipulation as doing the same thing to both sides as far as possible. This includes dividing by non-zero reals, and the above error is forbidden because we simply are not allowed to divide by zero (it is simply not a defined operation so you cannot do it to both sides).

This viewpoint is supported by the fact that even at a foundational level ℝ is constructed either by Cauchy sequences or Dedekind cuts from the field ℚ, and it is again crucial that we have division by non-zero in ℚ.

However, you are of course correct that once we move to non-fields such as division rings, then we need to realize that although the field axioms do not hold (and hence division by non-zero may not be a supported operation), the ring may still support cancellation of non-zero factor.

PM 2Ring

If we don't define "division", except by way of multiplication by an inverse, then we don't need to worry about division by zero. You can't multiply by the inverse of zero because it doesn't exist. ;)

user21820

So I would say that there is nothing wrong if a student identifies the error (for real a,b) in ( a^2 = b^2 ⊢ a = b ) as due to a division by something that may be zero, even though in practice every student who actually gets to that step via ( a^2 = b^2 ⊢ (a+b)·(a−b) = 0 ) will not make that mistake intentionally precisely because it is clear that they cannot divide by (a−b) unless they know it is nonzero.

@PM2Ring Of course, nobody disagrees with that, but you must remember what level you are teaching at. Everyone who learns abstract algebra is taught that division is just multiplication by inverse. But for ℝ specifically, at a level below abstract algebra, students know division and actually the cancellation property is at a higher level than necessary for ensuring correct understanding.

Formally, cancellation is ∀a,b,c∈ℝ ( a·c = b·c ∧ c ≠ 0 ⇒ a = b ). Right? But that's not in the same form as the "do the same thing to two equal things and you get two equal things", which is a simpler and sufficient principle for ℝ.

You can also imagine teaching a very young child. Dividing a cake by 2 is not quite the same as multiplying a cake by the inverse of 2.

Xander Henderson

14:15

@user21820 Sure. If a student in a precalc or calculus class were to day "we can't divide by zero", I would not tell them that they are wrong.

But I think that the issue is, in fact, somewhat more subtle.

And I think that students in a calculus or precalculus class are sufficiently mathematically mature to deal with some subtlety.

user21820

@XanderHenderson Haha well, if they make that kind of mistake and cannot understand their mistake, then I wouldn't go anywhere else until they can get it solidly down that they can only claim two things are equal if they can justify it, and that they cannot just move things over the equal sign, which is a very common pedagogical bogosity.

PM 2Ring

@user21820 Fair enough. OTOH, when we learn division, we're also taught that you aren't allowed to divide by zero. But at that stage we might not have a solid understanding of why we can't divide by zero.

Xander Henderson

@user21820 Exactly.

user21820

@PM2Ring Ah, that needs to be taught (well). There's a reason why in many countries the standard education syllabus puts fractions in a year by itself after multiplication. =)

PM 2Ring

Agreed.

user21820

14:23

@XanderHenderson So yup I think we agree. I just wanted to emphasize that we can do a lot of rigorous mathematics even at elementary school level without touching logically complicated notions like cancellation (which would seem like magic if the foundations are lacking). Something like 16/64 = 1/4 by cancellation. =D

user 85795

lol

PM 2Ring

This wasn't an issue for the ancient Greeks, because they didn't have zero. Or algebra. ;)

user21820

@PM2Ring LOL.

PM 2Ring

That's the Romans.

user21820

@PM2Ring Greeks have χ.

=P

But their alphabet that corresponds to our x is ξ, which in my subjective opinion is a really ugly letter.

PM 2Ring

14:29

Greek numerals were more painful. They used the whole alphabet, including a couple of extra obscure letters. So you had 9 letters for 1 to 9, another 9 letters for 10 to 90, and another 9 letters for 100 to 900. I can't remember how they did bigger numbers, but Archimedes came up with an exponential notation scheme in The Sand Reckoner.

user21820

@PM2Ring One would guess that any mathematician would come up with exponential notation sooner or later.

As long as they lived long enough. =P

PM 2Ring

From a modern perspective, it's kind of surprising how long it took for negative numbers to be accepted as legitimate. Sure, they got used in algebra, but they were seen as a bookkeeping device or computational trick, not as proper numbers. Somewhat ironically, it wasn't until complex numbers were accepted as valid mathematical entities that negatives became grudgingly accepted. Even by the time of Euler, negatives were still a bit suss.

OTOH, multiplicative inverses have been accepted & studied for a long time.

user21820

14:45

@PM2Ring Yes I also do not understand it. I believe I would accept it immediately based on geometrical motivations, since position on a line clearly has a direct symmetry across a given reference point. But it's hard to tell whether there was social or political suppression of the idea of negative numbers. If everyone around us (imply that they) despise negative numbers, would we go against that publicly?

One could argue that water can be divided but cannot be negated, but I believe the geometric position on a line makes it untenable not to accept negative numbers.

However! There is an interesting modern mathematical point that may be worth mentioning, which is that we can get to ℝ in two somewhat distinct ways! The first is ℕ → ℤ → ℚ → ℝ. The second is ℕ → ℚ+ → ℝ+ → ℝ. Funnily enough, the second way avoids negation all the way until we are literally at the end... And when you really look at the actual technical constructions involved in both ways, you will find both benefits and drawbacks of each way due to the choice of when to introduce negation!

PM 2Ring

@user21820 It seems obvious to us, because we're used to it. But it was a huge conceptual leap to unite magnitude and direction into a single entity.

On a related note, people were doing computations with complex numbers for several centuries before the geometrical interpretation of the Argand plane was devised.

user21820

@PM2Ring That's another funny thing. I would have thought that somebody cared enough to use the ancient euclidean geometry to find a representative of their complex numbers.

But again this is hindsight, as you said.

user 85795

Negative numbers are a tough sell to very young learners.

They don't know about debt.

Temperature 🌡️ is a good motivator.

Except you can't talk about "what is twice as cold as 0°?

PM 2Ring

15:07

Yeah. Zero degrees is a fake zero, unless you're using an absolute temperature scale, like Kelvin.

John Baez briefly mentions that in Torsors Made Easy

user 85795

👍🏼 even 0 as a number meaning "nothing" can be problematic.

user21820

@user85795 What about position as per my geometric motivation? No need for a continuous line; even just seating in a line. +1 means move to the right 1 seat. +2 means move to the right 2 seats. What about not moving? What about moving to the left?

The seating motivation also supports multiplication as an action (but you don't tell students that). 2 × +1 is repeating +1 twice.

I actually say much more about this here:

11

A: What makes negative numbers different from positive numbers other than their being (almost) opposite?

None of the answers so far has addressed the underlying question, which is why it seems that multiplication is asymmetric and satisfies "positive times positive is positive" while "negative times negative is positive". The reason is that multiplication is not as simple as it seems. $\def\nn{\math...

Under Viewpoint 1.

user 85795

The number line is a great motivator.

Turn the thermometer on its side :-)

+1

user 85795

15:31

Negative vs Minus

The problem here^ is kids hate subtraction already.

user21820

@user85795 Ahahaha... I personally read "−" as "minus" in all cases, and see no problem since to me "−n" is the same as "0−n" where "0" is the appropriate zero element of the appropriate ring.

But I also have no problem with people who want to distinguish the unary and binary "−".

geocalc33

16:38

$$e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$$

Where $K$ is a modified Bessel function of the second kind.

Then letting $x=e^{-n^{-s}}$ we have:

$$\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$

Does anyone see the error here?

$0<x<1$

unit 1991

17:21

Are there general methods for graphing functions given by parametric form?

Xander Henderson

17:42

@user85795 That thread makes me sad. So much opinion, so little of which is truly relevant. :(

Ted Shifrin

17:58

@unit1991 Computers love parametric form. They just plot lots of points. For curves like the cycloid there is no way to write them explicitly (globally). For curves in higher dimensions, it’s almost always hopeless.

@XanderHenderson I tend to use negative for the additive inverse and minus for the operation.

rb3652

Hi @TedShifrin, just wanted to give a quick update: it all came down to a minus sign. And then everything worked flawlessly.

Ted Shifrin

OK. Still hope you understood our geometric preference.

rb3652

Yep. I saw @copper.hat's answer. Something I didn't think of (although I tried a similar approach).

Wietlol

18:19

what have I done?

also, I wrote a function for which I am not sure I know the name

fun interpolate(origin: Vector2d, target: Vector2d, factor: Double): Vector2d =
	origin + (target - origin) * factor

iirc it was something about interpolation, but not sure if it is called just that

if factor = 0, the result is origin
if factor = 1, the result is target
anywhere between 0 and 1, it shifts linearly from origin to target

is this just called "interpolate" or does it have a different name?

leslie townes

18:41

you can call it whatever you want. pretty common to have a family of things indexed by a parameter in [0,1], where at 0 you get one thing, and at 1 you get something else, and if you vary in the middle it moves continuously from one to the other.

Ted Shifrin

“Homotopy,” “deformation,” …

leslie townes

sometimes people use 'interpolation' to mean more specialized or specific things but usually context makes it clear.

Wietlol

hmm... I guess I'll keep the name then

copper.hat

18:59

interpolopy

Wietlol

intertopyation?

robjohn

@TedShifrin does that include the Homotopy of Notre Dame?

Ted Shifrin

19:17

@robjohn I have a hunch not.

Xander Henderson

@TedShifrin Me, too. But it doesn't really matter, so long as one is clear and consistent.

The tone of so many of the answers is "If you don't say [x], you are wrong!"

leslie townes

this is the tone of the internet.

there oughta be a feature where you can automatically block seeing all questions with more than N answers. where N is something like 4.

robjohn

@TedShifrin I had best to not use a quasimodal verb (nor split an infinitive)

leslie townes

minus and negative are both what happen to your net worth and your worth as a person if you don't invest in lesliecoin. maybe i should add an answer that explains this, with links to relevant resources.

amWhy

@leslietownes No, no, sorry, amwhycoin is outperforming lesliecoin! ;P

Ted Shifrin

19:30

Maybe I should switch to amWhycoin, as lesliecoin makes big promises but follows not through.

Wietlol

im still not entirely sure how I always end up with so many "1 - X" situations

leslie townes

wiet: it's a pretty good example of a function that is 1 at 0 and 0 at 1 and continuous in between. good for turning things on and off in a continuous way.

Lukas Heger

19:54

I think I'd prefer something like $\begin{cases}1,& x <0\\
e\exp\left( -\frac{1}{1 - x^2}\right), & x \in (0,1) \\
0, & \text{otherwise}
\end{cases}$ for that

copper.hat

i suspect crypto will go the way of dotcom, or dotgone as they say

Ted Shifrin

@copper.hat Dotgonit!

copper.hat

:-)

found a convex psq, but don't want to deprive a learning experience :-(

Ted Shifrin

20:12

What makes you think there might be a learning experience?

copper.hat

i need some glimmer of hope

robjohn

@copper.hat Yeah, my first instinct is to answer the question to help, but then I have to change my thinking and realize that if they put little effort into the question, even though the question might be a good question, it is probably just a homework grab.

copper.hat

@robjohn you are right, but i hate to see folks in pain

not the case in the above situation

robjohn

@copper.hat it might be pain or skating

Xander Henderson

@copper.hat I've been saying that for more than a decade.

Ted Shifrin

20:16

I was proud of the success here last night!

Xander Henderson

Hasn't happened yet. :(

robjohn

@TedShifrin are you referring to the comments?

Ted Shifrin

Well, ultimately, the OP figured it all out correctly and posted the answer, as I suggested he do. To me, that is success.

copper.hat

yes, that is a success

robjohn

It is.

Ted Shifrin

20:19

No one jumped in with the answer too soon; no one closed the question because we were having learning in the comments. Rare!!

copper.hat

to me it is a success if the gain is $>1$

Balarka Sen

I see Deane Yang approved the thread.

copper.hat

i think much of what is missing in modern education is the narrative

Ted Shifrin

Yes. Deane and I are old friends, however. :) He actually took a class from me when he was in grad school. :)

Wietlol

@leslietownes the linear formula is easy though... the sin wave formula was a bit more difficult, the offset sin wave formula is a pain... which is the one I posted above

Balarka Sen

20:20

Ah

Ted Shifrin

We're also mathematical brothers (he was a student of Phil Griffiths, too — an official one).

Balarka Sen

Gotcha

copper.hat

my latest helpful comment: "To be honest, I give a crap what WA computes."

Ted Shifrin

I agree.

robjohn

Yes, but I want to know how it computed it

Xander Henderson

20:22

@SMCnotSvenMagnusCarlsen Does Lie lie?

Do Lie algebras lie?

Ted Shifrin

a @Balarka: Isn't this citation from wiki totally inaccurate? When does Whitney ever make reference to a Riemannian metric?

copper.hat

the truth about Lie algebras, many folks hurt during push forward

Balarka Sen

@TedShifrin It's badly written. What wiki means is that first you can embed your compact Riemannian manifold $M$ of dimension m in R^2m, then scale it down so that its embedded in a very very very small ball in R^2m -- which means the embedding is short, $f^* g_{Euc} < g_M$ -- then approximate it by Nash-Kuiper h-principle by C^1-isometric embeddings.

One uses Whitney to produce a short map, afterwards Nash to approximate the short map by isometric ones.

Ted Shifrin

Yes, that's the solution that was just posted. And I knew that. But the statement is horribly misleading.

Balarka Sen

Yeah, badly written

Ted Shifrin

20:28

Worse than that.

Balarka Sen

No mention of "compact" either, which I did include above (paused for 1 second and put it in the hypothesis), and Deane Yang is just giving the answerer a hard time about that :)

Ted Shifrin

Good.

Balarka Sen

Precision is important!

Ted Shifrin

Speaking of imprecision, @copper, note that we didn't pay attention. $\int_\gamma \Re(f(z))\,dz$ is certainly not $\Re\left(\int_\gamma f(z)\,dz\right)$.

copper.hat

Yep, I already ate crow on that one

Ted Shifrin

20:32

Was it cooked rare?

copper.hat

Not sure why crow gets such a bad rap

I do like corvids in general

Balarka Sen

Was it a raven rapping at your chamber door?

copper.hat

one for sorrow, two for joy, three for a girl, four for a boy

Ted Shifrin

Our analysts should relish this one.

Balarka Sen

I'm so Poe at analysis

Ted Shifrin

20:36

Of course, someone just gave the OP an example. Grrrr.

copper.hat

i think the student i helped most in mathematics really had little to do with mathematics. he (like my son) had a 9th grade teacher whose approach to starting high school maths was a weeder class and he ruined maths for many, including my student. he saw how i approached a problem and made lots of mistakes until i got it right and i guess he figured if joe can do it so can i. he's on his way to becoming a mech e.

Ted Shifrin

And now another one posts an answer. It's hopeless.

I worked with two twins long-distance while the AP calc teacher of one (female) was making her feel stupid (even with remarks in class, I gathered). All it took was a little fun and encouragement. They both ended up with As in their calculus courses.

copper.hat

hopefully i will not encounter the 9th g teacher in a dark alley :-)

Balarka Sen

@TedShifrin And it was wrong.

copper.hat

there may be some unbounded variation invlolved

Balarka Sen

20:40

As pointed out by OP :-)

Ted Shifrin

Yeah, I went back and looked at the answer.

The person who gave it away in the comment did remove it. That was nice of him.

leslie townes

can someone re-post an answer, it's due in 20 minutes

thanks

Balarka Sen

This answer is hidden. This answer was deleted and converted to a comment 3 mins ago by Xander Henderson♦.

LOL

@XanderHenderson So you converted it to a comment, deleted it, and then deleted the comment?

Xander Henderson

I converted the answer.

Then saw the comment thread.

Then deleted a bunch.

Balarka Sen

Nice!

Xander Henderson

20:51

And now I see that there was discussion here. :/

copper.hat

is $[n]$ usually taken to mean $\lfloor n \rfloor$?

leslie townes

it's iverson bracket notation. returns 1 if n is the one true number and 0 otherwise

seriously though i wouldn't think it is 'usually' taken to be anything

robjohn

@copper.hat Often it does when people don't know the $\lfloor\dots\rfloor$ notation

copper.hat

i think i hate ambiguity, but am not sure now

Xander Henderson

Depends on context. In number theory, I think it is "usually" the integer part.

Or the gif.

robjohn

20:55

@copper.hat I'm not decided.

Balarka Sen

$[n] = \{1, \cdots, n\}$

Xander Henderson

Depends on context. :D

copper.hat

i think i should go back to bed, gray matter is slower than usual

Xander Henderson

Go for it.

Ted Shifrin

@copper.hat It's often the equivalence class of $n$ mod something.

I thought where you're from it's grey matter, @copper.

Xander Henderson

20:57

It is just a grouping symbol.

Balarka Sen

grey matter is greying matters.

copper.hat

Ahhh, those beautiful gray skies...

Ted Shifrin

soon copper will be grey and blue.

Balarka Sen

I came up with a good tongue twister recently

Baron on a barren desert eating dessert on a baren

copper.hat

my spelling is all over the place since i communicate regularly with various branches of the language

Ted Shifrin

20:58

What's a baren, a Balarka?

copper.hat

fortnight

Ted Shifrin

How do you eat dessert on a baren?

Balarka Sen

its what they use for pressing woodcuts

Ted Shifrin

I thought tongue-twisters had to have consonance with only one consonant.

Ah. Esoteric.

Xander Henderson

Baren (printing tool)

Baren (馬連、馬楝) listen is a disk-like hand tool with a flat bottom and a knotted handle used in Japanese woodblock printing. It is used to burnish (firmly rub) the back of a sheet of paper, lifting ink from the block. == Construction == A traditional (hon) baren is made of layers. A flat coil of braided cord forms the core. This is placed on a disk (ategawa) consisting of 30–40 sheets of high-grade long-fibred hosokawa paper, wrapped in tissue and black lacquer. This is covered by a thin bamboo sheath (takenokawa) twisted in such a manner as to form the handle on the top. According to Hiroshi...

Who knew...

copper.hat

20:59

red leather, yellow leather,

Ted Shifrin

Balarka knew, evidently.

Transcript for

$Mathematics$ Mathematics