Mathematics - 2016-11-17 (page 2 of 5)

arctic tern

16:00

it is a k-algebra

Alessandro

With the definition I was given extensions can be just rings, weird

Mike Miller

@Balarka did you see the question I posted

arctic tern

ring extensions can be rings, field extensions are fields

Alessandro

You don't need the extension to be a field to talk about minimal polynomials though

Balarka Sen

@MikeMiller I did. I didn't read your answer though.

Mike Miller

16:05

ok

thought you'd like it is all.

DHMO

Is it true that (real-coefficient) polynomials form rings while power series form fields?

Alessandro

(I believe we just call both of them extensions in Italian, I wasn't aware that field extension implies that the extension is a field in English)

Balarka Sen

@MikeMiller It's a neat question. I don't have much experience with symmetric products so didn't understand the line of approach (I know the statement of Dold-Thom theorem, but that's it).

Mike Miller

I just got that fact from the comments.

Balarka Sen

Heh. Ok, I blackboxed it and read a bit; that looks really nice at a glance. How did you compute $H^k K(\Bbb Q, 2n+1)$?

Oh, just think of $\Bbb Q$ as a direct limit of $\Bbb Z$'s, and compute termwise.

Mike Miller

16:29

That's not going to work, at least not without serious pain. First prove that the integral cohomology agrees with the rational cohomology. This follows because the space is rational (all its homotopy groups are rational vector spaces); its homology with $\Bbb F_p$ coefficients is always zero. From this and some algebra you can conclude that its integral homology is also always rational vector spaces.

Eh, I guess I don't think that's necessary.

It's just an inductive spectral sequence argument on the fibration $K(\Bbb Q, n+1) \to \mathcal PK(\Bbb Q,n+1) \to \Omega K(\Bbb Q,n+1) = K(\Bbb Q, n)$.

Sir Cumference

@arctictern Yes, but people these days are becoming too dependent on calculators

Not even being able to add on their own

That's not an exaggeration

Balarka Sen

@MikeMiller Ah, ok, I see.

Sir Cumference

Calculators can prevent people from understanding the mathematics

arctic tern

@SirCumference there is a difference between 2-3 (I've seen my students put that into a calculator) and computing sin(39) by hand (the thing that sparked this discussion).

Sir Cumference

@arctictern Yes, but learning how to compute sine on one's own can be helpful

In general it just exercises the mind

arctic tern

16:33

one could make the same argument for multiplying ten digit numbers mentally. but not being able to do so is not something to lament. failing to multiply one digit numbers mentally is something to lament.

Sir Cumference

@arctictern Perhaps it's not lamentable, but learning it should be encouraged

arctic tern

encouraged to multiply ten digit numbers mentally? waste of time.

party trick

Sir Cumference

@arctictern So give people a calculator instead of exercising their mind?

That's like giving crutches to people who can walk

Their legs will get weak

DHMO

encouraging people to multiply ten-digit numbers is like forcing them to run every day

waste of time

Sir Cumference

@DHMO Don't you think that'll make them stronger?

DHMO

16:35

@SirCumference yes, but it is pointless

arctic tern

Walking is basic. Multiplying ten digit numbers mentally or computing sin(39) accurately by hand is not. Your exaggerations are so ridiculous I think I'm wasting my time talking to you, either that or you're trolling me.

6

JeSuis

@MikeMiller Hello, do you know something about the following theorem: if $f$ is an application from a compact submanifold $M$ to $\Bbb{R}$, and if $f^{-1}([a,b])$ has no critical points then $f^{-1}(a)$ and $f^{-1}(b)$ are $C^1$-diffeomorphic ? thanks

Sophie

multiplying 10 digit numbers isn't doing actual mathematics

Sir Cumference

Multiplying ten-digit numbers might not be particularly useful, but learning how to do tricks in your head can help you improve your ability to do mental calculations

Mike Miller

@JeSuis yes.

DHMO

16:36

@SirCumference so we're in agreement

Balarka Sen

@JeSuis That's a theorem of Morse theory, yep.

Astyx

There are better exercises than multiplying 10 digits numbers, sure you should know how to do is (and if you can multiply 2 digit number, you have the method to multiply 10 digit numbers) but spending time speciffically on 10 digit numbers is only tiring and does not bring much improvement

Balarka Sen

But I should chicken out because I don't know Morse theory...

Sir Cumference

@DHMO I never advocated for specifically encouraging people to multiply ten digit numbers. I'm talking about mental calculations in general.

DHMO

3 mins ago, by Sir Cumference

@arctictern Perhaps it's not lamentable, but learning it should be encouraged

Sir Cumference

16:37

@DHMO In general, mental calculations should be encouraged

JeSuis

@MikeMiller@BalarkaSen do you a reference for that ?

Balarka Sen

Milnor, "Morse theory"

Sir Cumference

I never once said ten digit numbers

DHMO

@SirCumference it was a response to this message:

4 mins ago, by arctic tern

one could make the same argument for multiplying ten digit numbers mentally. but not being able to do so is not something to lament. failing to multiply one digit numbers mentally is something to lament.

Sir Cumference

@DHMO You know what I'm trying to say.

DHMO

16:38

No, I don't.

I'm just a harmful DHMO molecule

3

Sir Cumference

The point is about mental calculations vs. using a calculator

Often the former should be encouraged

Mike Miller

@JeSuis You flow along the gradient vector field.

DHMO

only if there are actually tricks

learning how to do squares of X5 (e.g. 65^2) should be encouraged

but not reciting the first 100 squares

Krijn

@DHMO You can do the first 100 squares easily if you know the tricks

Sir Cumference

@DHMO The point I'm making is that people are becoming too dependent on calculators

DHMO

16:39

@Krijn but not memorizing

JeSuis

@MikeMiller ??

Sophie

I like to know I could multiply 10 digit numbers of calculate sin(39) or pi by hand if I had to, I just think that's boring

Sir Cumference

Practicing calculating sine by hand a few times isn't necessarily a waste of time

Could help them understand the function

Mike Miller

that is how you prove the result

Sophie

devising the method to, say, compute pi and then get a computer to do it is interesting

Mike Miller

16:40

@SirCumference Why don't you go calculate sin(39) by hand and come back to this discussion when you're done.

2

Balarka Sen

lol

Semiclassical

hi chat

Sir Cumference

@MikeMiller I think you're missing the point. We're teaching people how to plug numbers into a calculator, not use mathematics

JeSuis

@MikeMiller ok

Semiclassical

I think the point is that there are interesting calculations, and there are boring calculations.

Astyx

16:42

I think maths is more than just multiplying ten digit numbers

DHMO

@SirCumference not everyone wants to know how to compute sin(39) by hand

Balarka Sen

@Sir Well, that's precisely what he's asking you to do: use the mathematics and calculate that instead of plugging it into the calculator.

DHMO

and not everyone needs to know how to compute sin(39) by hand

Semiclassical

If the calculation amounts to just grinding out an algorithm through N steps, that's pretty boring.

Now, understanding -why- that algorithm works is valuable.

Sir Cumference

@DHMO Not everyone has to.

Semiclassical

16:43

But having a human being act like a computer is just silly.

Sophie

@SirCumference Taylor's theorem is mathematics. Plugging numbers into it is just arithmetics

teadawg1337

Calculators are a replacement for performing needlessly tedious calculations. While I agree that using one to find out what 7*13 is (it's 91) would be ridiculous, it's also equally as ridiculous to expect people to be able to perform complex operations on the fly.

DHMO

@SirCumference You said "We're teaching people how to plug numbers into a calculator, not use mathematics" meaning that the educational system is teaching people to depend on calculators

Semiclassical

@teadawg1337 On the other hand, recognizing that 7*13=(10-3)(10+3)=10^2-3^2=91 is rather nice.

Sir Cumference

@DHMO Essentially

You're using a specific example, with sin(39)

DHMO

16:44

@SirCumference so I responded by saying that not everyone needs to know how to use mathematics to calculate sin(39)

Sir Cumference

I'm talking generally

DHMO

for example?

Semiclassical

To the extent that mental math is used to build an appreciation for algebra, I'm fine with it. To the extent that it tends to elevate arithmetic over actual reasoning, I find it silly.

Astyx

But essentically this is how society has been evolving, there is no point in being good at calculating 10 digits numbers any more

Sir Cumference

@DHMO Do you think we should teach children all the derivative rules in calculus? Or just show them how to plug it into a calculator?

DHMO

16:45

@Semiclassical wow, i just memorized it all these years...

Astyx

Since any computer can do it for you

DHMO

@SirCumference derivative rules are not "numbers"

Sir Cumference

@DHMO I have never once seen a teacher explain how the calculator figures out the sine of an angle

Semiclassical

Derivative rules != arithmetic.

DHMO

> We're teaching people how to plug numbers into a calculator, not use mathematics

Astyx

16:45

Because being able to compute such numbers is not something that is required to make maths advance. Being able to improve algorithm so that machines can do it faster is

DHMO

@SirCumference because that is what calculator industry teachers teach

Sir Cumference

@DHMO I mean plug formulae into calculators, press a button, and get the derivative

Astyx

^

DHMO

the general public is not supposed to know how a calculator works

Sir Cumference

@DHMO Don't we lose something from that?

Semiclassical

16:46

"not supposed to know" makes it sound rather conspiritorial :/

DHMO

@Semiclassical i don't care

Semiclassical

I think it is valuable to appreciate the huge role that algorithms play

DHMO

but it is not valuable to know how calculators work

knowing the taylor series for sine is good, but knowing how a calculator calculates sine is pointless

Sir Cumference

@DHMO I'm not talking about how to build a darn calculator

Semiclassical

Eh. "valuable to whom"

DHMO

16:48

2 mins ago, by Sir Cumference

@DHMO I have never once seen a teacher explain how the calculator figures out the sine of an angle

@Semiclassical the general public

Sir Cumference

@DHMO What do you mean by "how calculators work"?

Semiclassical

General public is, by definition, a broad phrase

2

Sir Cumference

It seems like you're talking about how to electrically build a working calculator

Semiclassical

Very little is meaningful to the entire 'general public'

DHMO

@SirCumference exactly what you mean by "how the calculator figures out the sine of an angle"

@Semiclassical but education is for the general public

@SirCumference firstly they mod the angle by 2pi (this also needs explanation)

Mike Miller

16:49

This is extremely tedious. I agree with anon.

Sir Cumference

@DHMO I'm saying we should learn how to calculate sine angles for ourselves

Mike Miller

As a final note I guarantee everyone you're arguing with as thought about pedagogy significantly more than you have.

DHMO

@SirCumference that isn't what you said

Sir Cumference

@DHMO I obviously meant that

That's what this discussion is about

Relying on calculators vs doing calculations on one's own

Semiclassical

I'm a bit unsure.

DHMO

16:50

@SirCumference but you obviously didn't say that

@MikeMiller who is anon?

Sophie

@SirCumference Isn't this usually done in introductory calculus courses? $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$

Semiclassical

The thing is, there's no god-given algorithm for how one calculates sine.

Sir Cumference

@Sophie Yes, but you don't need to delve too deep into calculus to explain that to algebra 2 students

DHMO

@Sophie but we do not usually plug numbers in

Sir Cumference

@Semiclassical It's approximated through the Taylor series

Semiclassical

16:52

One can use the Taylor series, but I dunno if that's the algorithm that a calculator uses

That's not necessarily the algorithm it uses.

Astyx

I think it is

Sir Cumference

@Semiclassical Regardless, it's still useful to teach

DHMO

@Astyx have you built a calculator?

Sir Cumference

Right now kids are left in the dark regarding how to calculate sine

Astyx

Not really

As in : no

Semiclassical

16:53

My point more broadly is that there can very well be a difference between a simple but inefficient algorithm and a more subtle but also more effective one.

Sir Cumference

@DHMO Imagine you were never told how to take the derivative, and never learned all those rules because they were "a waste of time"

Semiclassical

Now I'm curious, though.

Sir Cumference

You just learned how to plug a formula into a calculator and press the derivative button

Astyx

Anyway when doing math who the hell ever computes $\sin(39)$ ?

Sir Cumference

That's the equivalent of not explaining the sine Taylor series

Semiclassical

16:54

HMM: qc.cuny.edu/Academics/Degrees/DMNS/Faculty%20Documents/…

Sir Cumference

@Astyx Don't be so specific

Semiclassical

Looking at the first sentence of that, I think that article could be interesting

Astyx

I'm not being specific, but all the examples you give seem silly to me

Semiclassical

"Almost every calculus teacher I ask answers the question, “How do calculators compute
sines and cosines?” with the words, “Taylor Polynomials.” It comes as quite a
surprise to most that, though it is reasonable to use Taylor polynomials, it is really
a method known as CORDIC that is used to compute these and other special functions..."

2

Astyx

Why would you want a specific value of $\sin$ ? Why would you want to compute the product of 2 10 digit numbers ? And so on

Sir Cumference

16:56

@Astyx Just so you can use your own head when you don't have a calculator.

Sine is pretty specific, but there's a point about learning how to do these things on your own

DHMO

CORDIC

CORDIC (for COordinate Rotation DIgital Computer), also known as Volder's algorithm, is a simple and efficient algorithm to calculate hyperbolic and trigonometric functions, typically converging with one digit (or bit) per iteration. It is therefore also a prominent example of digit-by-digit algorithms. CORDIC and closely related methods known as pseudo-multiplication and pseudo-division or factor combining are commonly used when no hardware multiplier is available (e.g. in simple microcontrollers and FPGAs), as the only operations it requires are addition, subtraction, bitshift and table lookup...

Astyx

I don't see it honestly

Semiclassical

For special functions in general, I think it's more important to know what kinds of approximations one makes.

Sir Cumference

@Semiclassical The calculator probably approximates sine with the Taylor series

Semiclassical

e.g. $(1-x)^p\approx 1-p x$ if $x$ small is something any math student should know

No, it probably doesn't. See the article I just linked; that's the point of it.

Astyx

16:58

Any physics student too

DHMO

@SirCumference did you read the text above?

3 mins ago, by Semiclassical

"Almost every calculus teacher I ask answers the question, “How do calculators compute
sines and cosines?” with the words, “Taylor Polynomials.” It comes as quite a
surprise to most that, though it is reasonable to use Taylor polynomials, it is really
a method known as CORDIC that is used to compute these and other special functions..."

Balarka Sen

@SirCumference Nobody disagrees on learning to do things on their own. I don't even understand what your issue is.

Maks

Any good technique to graph in $ R^3 $ ? Its fucking my mind

Sir Cumference

@DHMO Missed that

Semiclassical

@maks Depends on what you're trying to graph.

DHMO

16:58

@Maks then enjoy it

Balarka Sen

@Maks Depends on what you want to graph.

Sir Cumference

@BalarkaSen The fact that people never learn how to figure out the sine of something, for example.

Semiclassical

jinx (almost)

Balarka Sen

@Semiclassical You beat me by a second.

Astyx

I have never ever felt the need to compute the sine of something other than ${\pi\over2}, \pi, \dots$

Semiclassical

16:59

@SirCumference Well, given that we've proven our ignorance (mine included) of how calculators -actually- compute sine...

Maks

@DHMO I dont like it hahaha @Semiclassical @BalarkaSen simple vectors (x,y,z) with real numbers

Sir Cumference

@Astyx Sine is plenty useful

teadawg1337

@SirCumference I'd need to reconsider my life decisions if I ever found myself being forced to calculate $\sin(39)$ by hand. In any practical setting, everyone would pick up a calculator for that. Why spend a good 30 seconds (at the very least) approximating it by hand if a calculator can give you the first 12 digits in less than a second?

Balarka Sen

@Sir Knowing how to compute sin(0/30/45/60/90/etc) is important. Other than those, nobody ever needs to know anything like computing sin(e) or something.

Sir Cumference

Astronomers use it all the time, for example

DHMO

17:00

@BalarkaSen heh, sin(15) and sin(75) are good

Semiclassical

I think it is reasonable to ask, though, if a math student can, in a reasonable amount of time, estimate a sine function from arithmetic alone.

4

DHMO

knowing sin(36) is also good

Semiclassical

i.e. addition/subtraction/multiplication etc.

DHMO

@Semiclassical I think it is possible

Semiclassical

I think so as well.

Balarka Sen

17:01

Those can be obtained from the addition formulas etc from what I said.

Astyx

It's useful, but not at specific values. At least it never has been to me

Semiclassical

It's a bit like asking about digits of pi.

DHMO

@BalarkaSen not sin(36). it's a special angle that one forgets

Semiclassical

Do I really care what the 10000th digit of pi is?

Astyx

I fully agree with @Semiclassical on this

Sir Cumference

17:02

@Semiclassical I've asked a bunch of high school students to give any definition of pi they can

Semiclassical

I don't care what the nth digit of pi is. Do I care that algorithms exist which -can- compute it? Yeah, that's interesting.

Sir Cumference

None were able to

DHMO

@SirCumference i don't expect high school students to say "the period of sine"

Astyx

^

Sir Cumference

@DHMO It could be as simple as "the ratio between the circumference and diameter for any circle"

Semiclassical

17:03

@SirCumference I do find that a bit disappointing.

DHMO

@SirCumference then you also are not able to

it's the old definition

Sir Cumference

Just anything would be nice

Vrouvrou

@DHMO have you an idea about this : math.stackexchange.com/questions/2017179/…

Semiclassical

Or the proportionality factor between the area of a circle and the square of its radius.

Mike Miller

@DHMO I sure as hell don't know sin(36).

DHMO

17:03

@Vrouvrou no idea, sorry

teadawg1337

Isn't sin(36) from a 3-4-5 triangle?

Semiclassical

It's related to the golden ratio, if memory serves.

Sir Cumference

@DHMO Enlighten me?

DHMO

@teadawg1337 no

Semiclassical

@teadawg1337 Nooo.

DHMO

17:04

@SirCumference look above

teadawg1337

Lol

Sir Cumference

@DHMO Oh, huh

DHMO

@MikeMiller then derive it

Sir Cumference

Actually would be nice to know

Can you elaborate?

Balarka Sen

sin(2pi/10), and solve a polynomial. I don't know how to do it simply.

Semiclassical

17:05

I will speak in defense of sine(36) for one reason: 36 degrees shows up in the interior angles of a pentagon.

DHMO

@SirCumference sine is (re)defined as the series x-x^3/3!+...

Balarka Sen

I don't care much either.

Sir Cumference

@DHMO 'Course

DHMO

then using real analysis we can prove that it has a period

Astyx

is it 36 degrees or 36 radians ?

DHMO

17:05

and then we define the period as pi

Semiclassical

degrees

DHMO

@Astyx radians

Sir Cumference

@DHMO Yeah

Astyx

Right :p

Sir Cumference

@DHMO Is that the formal definition?

Semiclassical

17:05

36 degrees, I meant.

DHMO

@SirCumference I believe so

Semiclassical

That's the real analysis definition.

Mike Miller

@DHMO No.

Balarka Sen

You'll get a solvable quintic that you can solve with some tricks. I remember working this out before at some point of time.

Sir Cumference

@DHMO I don't see anything saying the "circumference:diameter ratio" definition is inacurate/outdated

DHMO

17:06

@MikeMiller then what is it?

Mike Miller

I don't care

DHMO

@SirCumference it is circular

Sir Cumference

@DHMO How?

Balarka Sen

@DHMO He's referring to the "then derive it" message

DHMO

@SirCumference how do you find its circumference?

@BalarkaSen oh, lol

Sir Cumference

17:07

@DHMO No pun intended?

Semiclassical

Too many conversations at once.

DHMO

@SirCumference no pun intended

@Semiclassical people who don't use reply tags

Sophie

can we please stop discussing arithmetics and go back to mathematics?

DHMO

we are discussing mathematics

Alessandro

And most of them not really worth having, imho @Semiclassical

Semiclassical

17:07

contradiction in terms :P

DHMO

"the old definition of pi is circular" is related to mathematics

Mike Miller

@Balarka The part of that answer I was really proud of was the argument that the group had to be divisible and the calculation for Prufer groups.

Semiclassical

I think that sine(36) is a reasonably interesting question, but only because it can be interpreted in terms of a regular pentagon.

Balarka Sen

No, we're mostly discussing semantics and wrong-headed pedagogy.

Sir Cumference

@DHMO Even if you can't measure the circumference, it still has a definite value

Pi is the ratio between it and the diameter, isn't it?

Semiclassical

17:08

Sin(37), by contrast, is just not that interesting.

DHMO

@SirCumference in mathematics we don't measure (except in topology). we find using integration.

Sir Cumference

@DHMO Yeah, realized

DHMO

@Semiclassical you can find sin(1) from sin(36) by solving cubic equations

Semiclassical

yes, but you wouldn't.

DHMO

sure

Sir Cumference

17:09

Regardless, there's probably not a single definition of pi

DHMO

but there is a non-circular definition of pi

Astyx

Of course there is

Semiclassical

for context: cut-the-knot.org/pythagoras/cos36.shtml

Sir Cumference

@Astyx Care to share?

Vrouvrou

@robjohn please have you an idea about this :math.stackexchange.com/questions/2017179/…

Astyx

17:09

Why would one have two definitions of pi ?

DHMO

@Semiclassical thanks

Sir Cumference

@Astyx Ok, how would you define it?

DHMO

@Astyx there are 5 definitions of e^x

all of which are equivalent

Semiclassical

Different definitions needn't be incompatible.

3

Sir Cumference

Have to go

Later

Semiclassical

17:10

Or, more to the point, seemingly-different definitions can in fact be equivalent.

Astyx

But all are the same intrinsically

Bye @SirCumference

Semiclassical

Sure. But if you're going to write down a definition, you need to pick one.

Astyx

Not necessarilly

If you prove they are equivalent before defining it

DHMO

pointless

Astyx

Why ?

Semiclassical

17:12

Depends on the context. Some important theorems hinge on showing that two seemingly different definitions are equivalent.

DHMO

you can establish one as definition and the other as properties

it would be less confusing

Balarka Sen

@MikeMiler Ya, I was looking at the computation for the Prufer p-group. Lots of neat ideas, but I wouldn't claim to understand it in detail.

Astyx

But at least you would have one, unified definition

Mike Miller

Pi is, of course, the first positive root of the unique solution to f'' = -f, f(0)=0, f'(0)=1

Astyx

Anyway, not that interresting a discussion

Semiclassical

17:13

Two definitions being equivalent can be boring, or it can be interesting. Problem is, it's sometimes not at all obvious which definition is the better starting point.

DHMO

@MikeMiller that is an equivalent definition

Semiclassical

If someone was to ask for a formal, rigorous definition of pi, I might give the one Mike just gave. But if someone was to ask me for intution, I'd give the geometric one.

DHMO

@Semiclassical I would just say "the period of sine"

and give the formal definition of sine

Semiclassical

At which point an introductory geometry student would stare at you and say...okay?

Balarka Sen

That's exactly the definition Mike gave

Semiclassical

17:14

A definition can be true and useless, depending on the context.

DHMO

@BalarkaSen Mike gave f''=-f,f(0)=0,f'(0)=1 as the definition of sine; I gave a taylor series

they are equivalent but not equal

Alessandro

You probably want "the smallest period of sine" or something like that

DHMO

@Alessandro period in english is automatically the smallest

Semiclassical

The definition of it as the period of sine would, in all likelihood, not be useful for someone in high school

Alessandro

Not every periodic function has a smallest period

DHMO

17:15

i agree

Semiclassical

Because they'd then ask why on earth you'd define it that way.

DHMO

@Alessandro then it is not periodic

Semiclassical

Yeah, I'm not seeing how you'd have a periodic function without a period.

Well, aside from a constant function.

Balarka Sen

I don't know what it means to say two definitions are equal but ok. I am not going to discuss semantics anymore

Alessandro

The indicator function of the rationals surely is periodic, f(x+q)=f(x) for every rational q

DHMO

17:16

@Alessandro right

Semiclassical

Hmm, fair enough.

Continuous periodic function, then.

Mike Miller

@Balarka Do you know what a cohomology operation is?

DHMO

@Semiclassical math.berkeley.edu/~kpmann/Lakatos.pdf

oops, too long

Balarka Sen

@MikeMiller A functor between $H^n(-; G)$ and $H^m(-, H)$, I suppose?

Semiclassical

yeah, I've seen that (though not read it in detail)

DHMO

17:17

basically they just keep restricting the domain until the theorem holds

Balarka Sen

Sorry, natural transformation, I meant.

Mike Miller

Yes, but as sets. Doesn't need to be a homomorphism.

Balarka Sen

I see, gotcha.

Mike Miller

So if I tell you that every element of $H^*(K(\Bbb Z/p^h, n);\Bbb F_p$ is 1) an F_p-cohomology operation applied to the fundamental class, or 2) a special element Qi, we see that if you kill the fundamental class, you kill everything of type 1.

That was my point. So the colimit is sero ezcept possibly along the Qis.

Balarka Sen

@MikeMiller I suppose by the fundamental class you mean the generator of the cohomology of the specific grade you are looking at? (1) is an interesting fact.

But actually it should not be surprising. This is just the Yoneda lemma or something.

Astyx

17:34

Does anyone know about Knuth's algorithm for the 5-step mastermind game ?

Balarka Sen

Which is to say $H^m(K(\Bbb Z/p^h, n); \Bbb F_p)$ is $[K(\Bbb Z/p^h, n), K(\Bbb Z/p, 1)]$, which is once again in 1-1 correspondence with natural transformations between $H^n(-, \Bbb Z/p^h)$ and $H^m(-, \Bbb Z/p)$ I guess.

Astyx

https://en.wikipedia.org/wiki/Mastermind_(board_game)#Five-guess_algorithm
The algorithm here states that the first guess has to be 1122 to minimize the number of steps required. I do not understand why that is, since when I computed the "score" of each combination 1123 had a better score than 1122. What do I miss ?

DHMO

17:48

Who cares about topology? (inscribed rectangle problem)

►

Mike Miller

@BalarkaSen Right, but I'm saying most of them come from natural transformations after the "canonical one" coming from the homomorphism Z/p^h -> Z/p.

Balarka Sen

@MikeMiller I think I buy that.

Mike Miller

All but one of them, even.

Obliv

why is the gradient theorem like so: $\int_a^b \nabla f \cdot dr = f(r(b)) - f(r(a))$ I see $\int_a^b \nabla f \cdot T~ds \to \int_a^b \nabla f \cdot r'(t)~dt \to \nabla f \int_a^b r'(t)~dt \to \nabla f [r(b) - r(a)]$ I think I am missing something.

Semiclassical

$\nabla f$ isn't independent of $t$, for one.

Obliv

17:57

i think $\nabla f = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial x_2} + ...$ for all components of the vector $f$ is a function of but does that change $\int_a^b \nabla f \cdot r'(t)~dt$

Semiclassical

No, that's not the definition of $\nabla f$.

Balarka Sen

$\nabla f$ means $\nabla f(r)$.

Obliv

hmm so it's more like $\nabla f = \frac{\partial f(r(t))}{\partial x} + ... $, then?

Maks

Hey, who can help me with this ??
I have the points on $ R^3 $ $(3,1,1) (-1,2,1) (2,-2,5) $ which are the vertices of a triangle, and I have to find the angles of it.
I saw that $ (3,1,1) $ and $ (-1,2,1) $ were orthogonal, so I figured it would be a rectangle triangle, but $ a^2 + b^2 \neq c^2 $ as $ 6 + 11 \neq 33 $
What am I doing wrong ??

Balarka Sen

Look at SemiC's message

Semiclassical

17:58

$$\nabla f = \frac{\partial f}{\partial x}\hat{x}+\frac{\partial f}{\partial y}\hat{y}+\frac{\partial f}{\partial z}\hat{z}$$

need those vectors in there, because the gradient of a scalar field is a vector field.

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$Mathematics$ Mathematics