from the Riemann's series theorem, we can rearrange the terms of a non absolutely convergent series in a way that we can make it converge to any real number. So, how can we make sense of the sum of such series? For instance, I know that $\sum_{n=1}^{+\infty} \frac{(-1)^n}{n}=-\log 2$, but this series is not absolutely convergent. So why is $-\log 2$ considered the sum of this series, when we can make it converge to any other real number/diverge?

@TedShifrin You don't rememebr exactly where do you?

Not answering.

You need to focus on exactly what it is you need here.

Gwyn, my analysis chops are weak, but the series is by definition the limit of the partial sums, which of course depends on the order in which the terms are presented. Playing around with Python code, it seems clear the series converges to $-\log 2$. If you change the order of the terms then you're not working with the original series anymore, I don't think.

nudge in the right direction?

If $Ax\cdot y = Bx\cdot y$ for all $x,y$, why does $A=B$?

Well the only explanation I came up with is that $(A-B)x\cdot y = 0$ so either $x \cdot y = 0$ or $A-B = 0$ so $A = B$.

I will ignore this.

I chuckled.

This is a day-1 mistake.

How about $(Ax - Bx) \cdot y = 0$, then $Ax = Bx$, but this means the columns of A and B are the same.

Or reduce to $Cx\cdot y=0$ for all $x,y$. Why is $Cx=0$ for all $x$?

@D.C.theIII this is one reason i like $(Ax)^\top y$ better than $Ax\cdot y$. the notation in that case makes it evident that you have $[(A-B)x]^\top y$ not $(A-B)(x\cdot y)$

@Gwyn as series are usually defined, -log(2) is definitely the sum of that series. its helpful to remember that the notion of the sum of a series includes, in part, an ordering of the terms. "sum n=1..infty a_n" refers to a limit of a specific sequence of partial sums, which depends not just on a bag of numbers, but how the numbers are indexed as a_1, a_2, a_3, .. etc.

tho in quantum physics we'd use dirac notation i.e. $\langle Ax|y\rangle$ and thus $\langle (A-B)x|y\rangle$

@Semiclassic In defense of my notation, only symbol pushing with zero brain results in reading $Ax\cdot y$ as $A(x\cdot y)$. I admit I corrected that mistake on homeworks over the years, but anyone who passed the course soon stopped making it.

@leslie I've been cleaning out year-old emails today and came across some old pictures of Munchkin. I imagine she's now as tall as I am.

eh, it depends for me. the point of good notation is to make it so you don't have to spend time thinking about stuff like that

@Gwyn when sum a_n is convergent and the a_n are nonnegative, it turns out that the choice of order doesn't matter at all. and, maybe less evidently but still importantly, some of our intuition about sums in general (in particular, "rearranging" sums with "...." in them by regrouping them in various ways) implicitly assumes that the way you add up those a_n's doesn't matter.

Well, $\langle Ax,y\rangle$ removes that mistake, but it's just a pain to type. And standard calculus, physics students are used to dot product.

counterpoint: physics students -should- see the $\langle Ax,y\rangle$ notation, because it's a good prep for Dirac notation in QM

@Gwyn riemann's result is just a reminder that order generally matters, how you index a sum generally matters, you can't always arbitrarily move around pieces of an infinite series and expect the same result. so when you see proofs about series that do use regrouping sums, or switching order, etc., there is probably more going on there than just "i have a convergent infinite series so i can do this."

hi @copper ... why did your son's driving to Santa Cruz upset you? Surely he's done plenty of 2-hour drives by now. I do recall my driving to Santa Cruz in my 1971 Saab back in 1977 or so and the master brake cylinder started leaking. I had to make the return trip with almost no brakes. Scary (but standard shift, at least).

tho tbh what i mostly find frustrating is the use of i,j,k for basis vectors

Well, you folks started that, too :D

hmm

i assumed they were i,j,k because Hamilton and quaternions

@Gwyn a helpful point of comparison, not directly related to riemanns theorem, is how there is no contradiction in sum_n (-1)^n diverging, sum_n (1 - 1) being 0, 1 + sum_n (-1 + 1) being 1, and so on. different sums yield different results and there is nothing surprising about that. it's only surprising if you expect your intuition to be able to "evaluate" 1 - 1 + 1 - 1 + 1 - ... by grouping elements appropriately. the convergence (and what the sum is) will be affected by that.

i know there was a period where people wrote physics textbooks using quaternions not vectors, which is sorta hard for me to wrap my head around

@TedShifrin she's tall enough to be very dangerous.

@Semiclassical I think they arose around the same time, so probably yes.

@leslietownes Not that she needs another reason to be dangerous.

Yeah, and now geometric algebra is trying to push exterior algebra and classical vector analysis out the door.

oh. here's a question that someone else asked me: what's the simplest way to explain why determinants correctly give the volume of parallelpipeds?

@TedShifrin ugh....the only way that can occur is if $C = 0$ in this case matrix $0$

@leslietownes thank you very much for these insights :)

@TedShifrin Oh it was the physics folks that started the $i,j,k$ nonsense?........and as an aside I'm all for $\langle, \rangle$ as I am an erudite now...

i'm a bit more skeptical of that

which part? Physicists' fault? my "erudition"? or my claim a line above?

first of those

defending your tribe is admirable.

Any idea how I can show a limit of the type $\lim \frac{1}{N} \log(Z_N)$ exists. Where $Z_N=\int \exp( \sum_{i=1}^N V(x_i)+\frac{1}{N}\sum_{i,j=1}^N U(x_i-x_j))dx_1,\ldots, dx_N$. Here $x_i\in \mathbb{R}$. I guess I need to show this $Z_N$ is of order $\exp(N)$

what context is this? it looks like something out of stat mech

yeh it is

neat. tho that doesn't help much

geometric algebra i.e. the decadence of society

maybe laplace method

doing some digging, it looks like ijk in vector analysis is best attributed as being in continuity with ijk for quaternions

whaddaya got against ijk? is it the implication of the closedness of that world? the implication that you can only add a few more letters before physics ends at z?

@leslietownes math >> physics

i guess all of those letters sometimes want to stand for other things in physics too

hence the sometimes-boldfacing or hats or other tomfoolery

@Semiclassical Depends on the audience and the dimension. Do we know cross product?

i asked the prof, he said he didn't want to introduce it

@TedShifrin what symbol do you use for cross product? $\times$ or $\wedge$?

using anything but $\times$ for cross product is weird to me, since only \times is a cross

@Semiclassical in italian it's called "vectorial product"

that makes more sense

and dot product is called "scalar product"

scalar product comes up in english too

Wedge is wrong, actually, because the bivector is not a vector (or pseudovector).

As a geometer who uses exterior algebra and differential forms and wedge heavily …. Nope.

on principle the derivation i'd prefer is that the multilinearity of the determinant captures exactly the properties you'd need for a volume

but that feels like like the wrong argument for this audience (more of a physics elective aimed at people outside of science and engineering)

@Semiclassical you can explain the geometric meaning of the determinant using Gram matrix

but it's a bit technical

By the way, @Sine, there was a mistake in that problem. Certainly not a basis for a topology on $\Bbb R$.

that route is more or less where we're trying to end up, i think, but it unfortunately means we can't -start- from the Gram matrix

You need a lot more background to use Gram.

right

My usual approach is Cavalieri and multilinearity, but for that audience I would do scalar triple product.

one idea i had was to start with the 2D version where you can prove it by diagram

@TedShifrin I've checked in fact and the correct one seems to be $B=\{\emptyset, \Bbb{Z},(a,a+1)_{a\in \Bbb{Z}}\}$

and then figure out a way to bootstrap that to 3D

I'll never talk to a med school major again

i guess Cavalieri gives the bootstrapping

@ペガサスSeiya why

@Sine Now it makes sense. You should have this almost like yesterday.

@SineoftheTime huge misunderstanding. We brought up vectors, and I was talking about the usual vectors in math and she was talking about "vectors" that carry disease from one organism to another

lol

@TedShifrin but in this case $(a,a+1)$ is connected right?

@ペガサスSeiya sad

@ペガサスSeiya i work with a guy with a lingering Dutch accent, and therefore his v's sound like f's. hence at one point i couldn't distinguish between him saying vectors and factors

Yes, and closed.

@Semiclassical same here for Japanese people. Their L's and R's are a little weird

The world = Za Warudo

@TedShifrin I tried to think like yesterday, I considered $f:I\to X $ s.t $f(0)=f(1)=1/2$ and $f(I)$ is connected

and I was reasoning about the open sets that contain $1/2$

And?

What if the base point is $0$?

@TedShifrin for example $f(I)$ can't be $[-1,1]$ since it's not connected

What are the path components?

$(a,a+1)$?

What’s wrong with that answer?

so the only open connected set that contains 1/2 is (0,1)

Why no larger?

But take for example $(0,2)$, you can see it as $(0,1]\cup (1,2)$ no?

and all points of $\Bbb{Z}$ are open

Yes, so all those sets are path components

yes, right

But after identifying these sets I can't make progress

So, what is $\pi_1$ with base point $0$?

@TedShifrin Maybe $\Bbb{Z}$ ?

This problem is annoying:

@SineoftheTime why?

@TedShifrin I was thinking about the paths with basepoint 0 but probably I'm wrong

What paths are there?

the path must be "inside" the path component

Old, old saw.

@SineoftheTime So?

Very old. But I figured that some of the younger members might not have heard it.

@TedShifrin thinking about the nature of the open sets $(a,b)$ with $a,b\in \Bbb{Z}$ and $0 \in (a,b)$ I'm inclined to say $\pi_1(X,0)=\Bbb{Z}$

You are making guesses without any justification. Come on.

What is the path component of $0$?

@PM2Ring wouldn't be surprised if the conversation had gone that way

@TedShifrin sometimes I forgot you're 70

Here's to another 70 though

No thanks.

there's no interval $(a,a+1)$ that contains 0

Why not? Seen enough?

so is $\Bbb{Z}$

What are you thinking, @Sine? Explain that.

Maybe I misread your definition of the topology.

I'll rewrote it

Oh, I see.

$\Bbb Z$ is the open set. I was thinking each integer was an open set.

$B=\{\emptyset, \Bbb{Z}, (a,a+1)_{a\in \Bbb{Z}} \}$

OK, so any function $[0,1]\to\Bbb Z$ is continuous. Therefore?

I can't conclude

@ペガサスSeiya so L is a point such that DAL=45 degrees and the perpendicular bisector of DL passes trhough B

Do you agree with what I just said?

i'm a scrub at synthetic geometry now so i'd probably just go for analytic

@Semiclassical Or you move $L$ along the ray from $A$ until the line from $D$ to the midpoint is perpendicular.

@TedShifrin yes, because $\Bbb{Z}$ is open

hm, yes

It's the only open set contained in those points.

yes

So if every function from $[0,1]$ is continuous, you can't tell me $\pi_1(X,0)$?

i do wonder how you'd construct this setup. "move L until something happens" is fine as reasoning but it doesn't lend itself to compass-and-straightedge

Why do you care? You only need the intermediate value theorem to know this situation exists :)

maybe you'd start from DL instead shrug

because i like being able to make things in geogebra :P

@TedShifrin can't tell

Ah.

@Sine This concerns me.

I guess

Aren't any two crazy functions homotopic?

yes

so all paths are equivalent

Very unintuitive, but maps to the indiscrete topology aren't interesting.

Yes, all paths are equivalent.

an in particual are equivalent to the constant path

Yes.

so the group is trivial

So this base point is done.

Now back to $1/2$. You told me any path had to be contained in $(0,1)$.

ok, so in general if the basepoint is in $\Bbb{Z}$ we're done

@TedShifrin yes

@Semiclassical It's still painful analytically, depending on how you attack it. And to figure out the best way to attack it, a bit of synthetic geometry is useful. ;) This problem (or something very similar) came up a year or so ago on the main site. I wasted a few hours on it, without solving it. At first, it seems like you don't have quite enough info to get a unique solution.

And what functions from $[0,1]$ to $(0,1)$ are continuous?

yeah, it's confusing me

@TedShifrin constant maps

They always are. Anything else?

@TedShifrin no because they're not homeomorphic

That is a terrible thing to say. So there are never continuous maps that are not homeomorphisms?

the most obvious points I get are A(0,1), B(1,1), C(1,0), D(0,0). i can use the 45 degree assumption to get L(-d,1-d) and thus G(-d/2, (1-d)/2)

Anyhow, @Sine. You have to pay attention here. What is the topology on $(0,1)$?

@TedShifrin you're right, I must be hitting myself. there are continuous maps that're not homeomorphism

@TedShifrin is the topology induced by the topology on X

oh, i see my error

There are zillions of continuous maps in the usual topology. But we don't have the usual topology. Induced from what?

So what is that topology, given your basis?

the open sets of $(0,1)$ are the empty set and $(0,1)$

Doesn't this sound familiar?

yes, but I don't know the name of the topology in english

shouldn't be that bad now. compute (D-G).(B-G) and require it to vanish

I just said it above.

Once again, what are the continuous functions $[0,1]\to (0,1)_{\text{indiscrete}}$?

every map from $[0,1]$ to $(0,1)$ is continuous because of the indiscete topology

tho i'm getting d=-1/2 here which makes no sense, so i'm doing something wrong

Right.

Therefore ... ?

So all paths are equivalent and the fundamental group is trivial

Right. Follow-up question: Even if we use the usual topology on $(0,1)$, what is $\pi_1$?

I think is trivial because with the usual topology the interval is contractible

@Semiclassical This may be useful, IIRC. en.wikipedia.org/wiki/Ceva%27s_theorem But I'm getting sleepy, and I don't want to think too hard about it. ;)

Yes, of course.

In general, to show that the group is trivial, it's enough to show that all paths are equivalent

@Semiclassical here's how I did it

Pretty annoying problem

Sometimes, it's easier to show what you need to: that every path is equivalent to the constant path. Then the general case follows by transitivity.

that makes sense

@PM2Ring problems like these require that you treat them like puzzles. Therefore, treat every individual shape as a puzzle piece. For example, to me $\triangle DAL$ is a piece of the puzzle, which I can and did move around and placed it at the top of the square to arrive at the answer

That's a good description, Seiya. As I said, you're far cleverer at this stuff than I ever have been.

You should easily see that $\angle ABG=\angle DAL=x$, since $AEDG$ would form a cyclic quadrilateral

@TedShifrin thanks Ted, nice to be complimented for something that I used to be told that I'm really weird and strange for

Well, I still think you should broaden your geometric horizons significantly.

I am and will. I'm learning affine geometry. That stuff really seems to click with me

Geometry vs Algebra. Who wins?

I do not understand why $\frac{dy}{dx} = {3y \over x }$ is not $y = {x+C \over 3}$

Welp, time to review more calc 3. Seiya later everyone!

is it not $\int \frac{dy}{3y} = \int \frac{dx}{x} \to \log 3y = \log x + C \to 3y = x + C$

If you take the derivative of $y = {x+C \over 3}$ you get $y'=1/3$. But that's not equal to $\frac{3y}x$

@Obliv that'not correct

I thought you could do e^{} on both sides

yes, or rewrite $\log x +C $ as $\log(x e^C)$

wait you can just bring constants into a log like $\log a + b = \log ae^b$

makes sense actually

$\log a +b=\log a+\log e^b=\log ae^b$

But note that in this case the constant is not arbitrary, if you rename $k=e^C$ you have $k\in \Bbb{R}^+$

the choices still don't include something of $y = \frac{xe^C}{3}$

the closest thing I think is y = Clnx

but not sure how that's achieved

what are the choices?

Oh, maybe you missed a $3$ when computing $\int \frac{dy}{3y}$

$\log 3y = e^{3y}$ though

?

$e^{\log 3y} = 3y$ ?

@Obliv yes

so we have $3y = e^{\log xe^{C}}$

I think is B

Doesn't make sense to me :(

@Obliv this is wrong

It's $\frac{\log y}{3}$ not $\log 3y$

B is the correct answer. Keep 3 on the right.

yes is B

$\ln y = 3\ln x + C$

Right, so take e^{} both sides?

You’ve got to master basics.

Yes.

@PM2Ring welll, doing it analytically I reduced it to $\tan x =d/(1-d)$ where $d(d+1)=1/2$. There may be a simpler way but from there I can solve to get $d=(sqrt{3}-1)/2\implies \tan x=1/\sqrt{3}$, hence x is 30 degrees

@Obliv first bring the 3 inside the log

my calc1&2 capabilities are like that of an old car who has spent too many winters in a parking lot.

Which, yay

@Semiclassical Nice

You can bring the 3 inside the log? @SineOfTheTime

So yeah, analytic geometry sometimes beats synthetic :P

$a\log b = ?$

@Obliv $3\log x=\log x^3$

Oh my, that is quite an important step

@PM2Ring in retrospect, picking D as the origin was fortunate

is there something like induction done the other way around, i.e., from n to zero? seems like I'll need that to do this

i think I also need the fact that an $n$ degree polynomial function has at most $n$ roots

if you give me a small $n$ i can do this proof manually, but for an arbitrary $n$ i have no clue how to do induction the other way around

the shape of the proof would basically be: use (1) then (2) exactly $n$ times, then we get our result

Induction is fine. First decide what $w_0$ is and reduce to the case of leading coefficient $1$.

Do induction on $n$ in the usual way.

@TedShifrin I've heard some professors saying that every proof of the fundamental theorem of algebra uses topologial concepts, is it true?

@TedShifrin oh! fantastic, I got it, thanks!

No. Gauss did that, but one can avoid it.

@SineoftheTime math.ucdavis.edu/~anne/WQ2007/mat67-Ld-FTA.pdf

this one only uses only real analysis theorems

Right. I have that proof in my algebra book.

Ok I'll take a look after the exam

Ted I have a time management question.........I've done the whole p-set, except two questions the challenge questions on interpreting the lagrange multiplier and the second derivative test for constrained extrema. Should I do them now or will these ideas come up again in my calc/analysis journey once I have stronger footing?

Bare with me, $e^{-y} dy = \frac{dx}{x^3} \to -e^{-y} = \frac{1}{-2x^2} + C$

that is correct so far?

@TedShifrin it was a last minute change (my wife was going to drive down and bring the car back) and it took my helicopter mind a while to adapt to the idea of him driving the hill on his own. the 1971 Saab is a beautiful car.

if so, why can't I do $\frac{-1}{e^y} = \frac{-1}{2x^2} + C \to e^y = 2x^2 + C$

then $y = \log 2x^2 + C$

@Obliv just rewrote $\frac{1}{2x^2}+c$ as a unique fraction and then take the reciprocal

I'm looking for something of the form