00:00 - 19:0019:00 - 00:00

12:00 AM
I suspect the Irish clustering had more to do with the abundance of warmth in Irish Februaries...

... can't wait for spring ;-)

my dad's birthday was yesterday. i shouldn't have forgotten him.

has he forgotten you

no, he's just gone out for cigarettes. he's coming back.

12:17 AM

one of my dad's admin's jobs was to keep him supplied with cigarettes.
those were different times.

1:22 AM

1:34 AM
@copper.hat Sad. Cigarettes be cringe. Cigars are better and don't capitalize on addiction to nicotine.
@user178758 You wouldn't happen to know if there is a similar method of argument reduction for computing logarithms as there is for computing the exponential, would you?

my dad used to smoke cigars too. he inhaled them, so i'm not sure i agree that they are better.

Inhaled them? Bruh...
You puff cigars.
You're not supposed to inhale them, so I can't say using them wrongly can validly contribute to the statistic here.

@J.Doe The map $(t,r,\theta) \to (r, \theta-t)$ is smooth.
@AMDG you are not supposed to drive over the speed limit.

Exactly
It's just something you don't do, albeit of a different gravity morally than going over the speed limit.

wow, you are judgemental.

1:39 AM
Nah, I've only stated what is.

i am not young enough to know everything.

Inhalation will be of great detriment to health.

yes, but if you do not inhale that would be a bigger detriment.

I don't follow. If you're just puffing the cigar, i.e. letting it linger in your mouth, then blowing out, you're not letting tar build up in your lungs or let nicotine directly into the blood stream.
Now, I haven't yet researched it, but it would seem reasonable to me to presume that the exposure to the tar (which is the carcinogen) would be like exposing one's self to the char on your grilled steak.

my da used to inhale a pipe as well.

1:47 AM
Honestly, I have no clue how that's even possible, either for cigars or pipes, but I'll take your word for it.
I inhaled cigar smoke by accident and it made me nauseous.
10/10 would smoke again, but my mom is asthmatic and I have yet to convince her to let me smoke cigars again so I can get some quality recreation.
I might be able to change her mind on this vacation trip.

i had enjoyed an evening of port & cigars with friends a few days ago. very relaxing.

You ever smoke pipes before? I've only had two cigars, but I'm thinking I'll probably get into it enough to want to enjoy pipes.

i hate smoking, both parents were chain smokers. i love the smell of nice cigars and pipe smoke however.

2:03 AM
That's fair, I suppose.

My father used to smoke cigars. He said it calmed his nerves.

It is indeed a very relaxing activity.

i had an officemate my last year of grad school who was always quitting smoking. i'd walk into the office and i could tell by the look on her face that she would tear my head off if something upset her.
she didn't quit that semester.

2:20 AM
It's a shame, really. Had the companies existed for their true purpose, to create good products which people need, we wouldn't have such problems like cigarettes and nicotine addiction.

my advisor also smoked. sometimes our meetings would just be 20 feet away from the math building, or whatever the required distance was.
i'm preparing my lawsuit against the university.

That is reasonable.

iowa lost to purdue today. purdue! you wanna talk addiction, i wonder what some of those referees were on. iowa still would have lost. they are dead to me now.

What was the case?
Oh, sports?

yes. i even wore an iowa shirt to the park today. before they humiliated me and all that iowa stands for.

2:29 AM
lol
So what sport is this? Football or something?

yes, football.
i used to live within a short walk of the stadium. could usually estimate who was ahead, and sometimes even the score, from the crowd noise.

Ah, yes, the one where they kick the checkered hexagonally tiled spheroid into the trapezoidal open polyhedra.

yeah. the other one.

Yes, the other one. :P

2:33 AM
Just think of all the topological and geometrical leaps and bounds we can make analyzing sports!
Oh yeah, can't forget about car racing.
Especially the one where they drive on a circle. They make a left turn... and another left turn... followed by two more. Riveting...
Sometimes they even make a right turn which then becomes a left turn.

as mitch hedberg said. "i want to be a race car passenger. just a guy who bugs the driver. 'say man, can I turn on the radio?' 'you should slow down.' 'why do we gotta keep going in circles?' 'can i put my feet out the window?' 'man, you really like Tide.'"

ded
The best part is when they ruin the likely several-grand logo in the grass when they win.

there's stuff to appreciate in almost any sport. except for rowing, which is really dumb.

The real question, though, is how's the beer on the winning car taste after a year and can I get a swab to make that winning homebrew.
I can imagine, but I mean, if I'm being honest, I'm heavily biased against sport because I'm not into it in the least, and so if there's anything to appreciate, then in my totally objective and unbiased opinion, the appreciation requires a microscope to really get.

something like driving, there's a lot of strategy. it's not just go fast because most of those cars can go the same speed, or if they can't there's little the driver can do.
i joke about rowing being dumb because my wife does it. she has some kind of subscription to a rowing magazine. as i understand it, the way to win a rowing competition is to be faster than the other competitors. that's it.
they often don't even race at the same time.
running is pretty dumb too, although as a hobby i understand it. stupid sport.

2:41 AM
Like, the car racing thing is just a bunch of cars whose performance appears to vary within margin of error of each other unless it's the more interesting street races with real racing cars as opposed to mere stock cars.
Like the twisty circuits as opposed to the flat circle.

stuff like basketball, football, and even baseball is much smarter than people give it credit for.

Those definitely require skill.

but rowing, if you find that you're losing, row faster. get to the end before the others.
pull the oars more, or whatever they're called.
in some configurations, there's even a person on the boat who doesn't row. that's hilarious to me.
it's said that they steer. have you seen what a rowing stadium looks like? i'll draw a picture. ================

@leslietownes His friends let him tag along to get the experience points so he could level up.

you'd think, maybe their job is to be as light as possible, but it isn't even that. if they're too light they put weights in the boat.
i usually flush out at least one fan of rowing when i go on this rant, but it isn't working, so i give up.

2:46 AM
Ok, so I'm assuming it's a matter of ensuring the center of mass is, well in the center of the boat; the alternative is there's weight balancing among teams and they try to normalize the weight that all teams must deal with to make the sport more fair.
Can't think of other reasons

my wife's going rowing tomorrow and my daughter is psyched for the day that she goes rowing. i'm for it, as long as she realizes that it's pointless.

@leslietownes You didn't insult them enough. You have to treat them the way you do mathematical proofs: you stare them down until they comply to assert dominance over your adversary.
Tell them their coffee was brewed poorly or smth idk

i would never treat a mathematical proof that way.
my family gets the insults. not my proofs.

Can't decide which of these movies to watch first: either The Man Who Knew Infinity or A Brilliant Young Mind (X + Y).
I think I'll watch the latter

3:25 AM
daughter went to bed 20 minutes ago, just shouted "don't look at me like that." i hope the cat is in there.

3:42 AM
the cat was in there

4:01 AM
said the Narrator

4:52 AM
here's a permutations question. suppose i have a set of permutations on $n$ objects which i know generate the full $S_n$. how would one go about finding the smallest (by number of factors) number of permutations required to generate a particular element?
(in my case of interest, i've got a set of 9 permutations on 8 objects, and there's another permutation I'm trying to write in terms of these 9 as simply as possible. so far the smallest product i know uses 20 factors, and that seems...excessive.)

5:40 AM
@copper.hat But I don't need the partial derivative with respect to $t$. I need to prove that the partial derivatives of $p_r(\theta-t)$ with respect to $\theta$ and $r$ are also continuos with respect to $t$.

found a brute-force approach via mathematica that got me down to 5 permutatins (and showed that that's the minimum possible in my case)
(looks like there's 6 such factorizations total. huzzah for trying every single possibility)

@J.Doe you need to do some work. if $f(r,\theta) = P_r(\theta)$ is smooth and $g(t,r,\theta) = (r,\theta-t)$ is smooth, then so is $f \circ g$.

huh. turns out that my answer ends up such that the last three factors in my permutation product all commute
so aside from that, the permutation product is actually unique

6:21 AM
@Semiclassical what do you mean by "generate the full $S_n$" and "finding the smallest number of permutations required to generate a particular element"?

suppose you want to generate any permutation on n elements. then it's enough to use transpositions, for instance. more precisely, we could restrict ourselves to adjacent transpositions swapping k and k+1
so, for instance, the permutation (1324) on four elements may equivalently be written as (23)(12)(34)(23)
so the transpositions (12), (23), (34) are enough to express any permutation

you could just use swaps with 1, even

now, my case isn't adjacent transpositions or (1k) transpositions

6:35 AM
pick a goofy set of generators, get a goofy game to play for minimal representation

right

i seem to have walked into the end of this, what was the beginning

pretty much what you just said
2 hours ago, by Semiclassical
here's a permutations question. suppose i have a set of permutations on $n$ objects which i know generate the full $S_n$. how would one go about finding the smallest (by number of factors) number of permutations required to generate a particular element?
if you have goofy generators, how does one find minimal representations?

oh. that seems tough in general.

yeah, i figured
i had to brute-force my problem in mathematica. (i said it was some permutation on 8 elements, but i should've said 7)
what makes it extra complicated: my generating set has 9 elements
when you only need 6 for S_7

6:38 AM
if you abstract this to groups (not full groups of permutations) i think it's not nice at all
but hrm

the brute-force way was to write out all sequences of n elements on the integers 1 through 9
say, for n=3
then have mathematica compute all products $\sigma_1\sigma_1\sigma_1,\sigma_1\sigma_1\sigma_2,\ldots$
and check if any of them were the one i wanted
had to go to n=5 to get a result

6:56 AM
hi, im struggling to apply DCT to show that this map is continuous: Let $R$ be a fixed closed rectangle in $\mathbb{C}$, and let $f \in L^p(R)$ (complex valued) for some $p > 2$, the map is $R \rightarrow \mathbb{C}$ , $z \rightarrow \int_{R} \frac{f(w)}{w-z} dw$
anyone see a way forward?
the issue im having is that i dont know how to majorize $\frac{1}{w - z_n}$ in $R$, where $z_n \rightarrow z_0$ in $R$
basically since $\lim_{n \rightarrow \infty} \frac{|w-z_0|}{|w-z_n|} = 1$ does not happen uniformly in $w$, and we need larger and larger $n$ for $w$ closer and closer to $z_0$ in $R$
sorry, that integral should read $d\lambda(w)$ at the end where $\lambda$ is the two-dimensional lebesgue measure, rather than $dw$ (its not a line integral)

@copper.hat I get it but I don't want to prove that the partial derivatives with respect to $t$ are continuos in $t$, I want to prove that the partial derivatives with respect to $r$ and $\theta$ are continuos in $t$.

if they are all smooth i am not sure what the issue is.

i have a hacky method in mind, namely prove this for $f \in C_{0}^{\infty}(R)$ (by rewriting it as $\int_{H} \frac{f(w' + z)}{w'} d\lambda(w')$ where $H$ is some large rectangle containing all of the $R - z_n$ and $R - z_0$ in its interior) using uniform continuity of $f$ (so DCT is easily applicable), and then apply a density argument to get the general result.. but im hoping there is a direct proof
the $p > 2$ here is pretty essential (for the hacky argument) or else we wont get to use holders

7:13 AM
Why does this problem sound NP-complete?
(RE permutations)
In fact, from the manner in which it was solved, it sounds like the definition of NP-complete
I don't know if this is relevant, but if factors here refers to products of composite integers, I've learned a bit in this regard through modulus, but I don't think it is of much use as-is.
I would consider it maybe a sort of gateway or food for thought, but I learned for the most part that you can list factors of a number in the form of $2^n x$ as all power of two factors of $x$ relative to some modulus. The only thing missing then is all the factors in-between.
It's something about equivalence classes and the fact that you can end up getting powers of two from sums and differences in those equivalence classes that you get something easily (relatively speaking) manipulable.
I just thought it might be interesting given that I heard something about factorization of integers not having an efficient algorithm. I believe it was prime factorization or something similar in the same category.

7:41 AM
if $p > 2$, and I have a continuous linear functional $L^p(K) \rightarrow \mathbb{C}$ where $K \subset \mathbb{C}$ is a closed rectangle, why is this functional necessarily of the form $f \rightarrow \int_{K} \psi f d \lambda$ where $\lambda$ is the two-dimensional lebesgue measure, for some $\psi \in L^p(K)$?
for $p = 2$ I can see this via riesz representation, but im not sure why it follows as well if $p > 2$

where did the Lebesgue measure come from? surely $\psi \in L^q(K)$?

oh sorry, I mean $\psi \in L^q(K)$ yes
oh nvm, i see that it follows from a riesz representation theorem for $L^p$ spaces

1 hour later…
9:08 AM
@TedShifrin the kind of diagrams that are in geometry books eg dover books on projective geometry

1 hour later…
10:26 AM
yesterday, by Ted Shifrin
@BannedUser What sort of geometry? Like high school geometry? Use Geometer’s Sketchpad.
@BannedUser There is a lot of software out there. Did you look at Geometer's Sketchpad?

@robjohn Yes I saw it. I will try it thanks :)

2 hours later…
12:39 PM
Hello does someone know what this symbol << means in math or physics?
Is it greater or equal symbol?

12:56 PM
You sure that wasn't a CS context?
@BannedUser In case it happened to be so, it usually indicates left logical shift traditionally in most programming languages. This is all I have to contribute.
In otherwords, multiplication by $2^x$ for $x\in\mathbb{N}_0$ usually.

1 hour later…
2:14 PM
@BannedUser $x << y$ means that $x$ is MUCH, MUCH lesser than $y$. Think $10^{-23} << 1$

it's usually less extreme than that tho
but it depends on context

yeah its not a well defined order relationship
en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm might be a starting point, which is more or less the same thing as grobner bases
so actually maybe not np-complete. idk..

2:29 PM
interesting

1 hour later…
3:41 PM
@user178758. Thank you. Appreciated. Yep I found that indeed means "mean, median, etc." ~_~
You can compute the indices for the probe sequence more efficiently by using the recurrence relation
$k_{i + 1} = (k_i + 2i + 1)$ modulo n for i ≥ 0 and $k_0 = k$
Derive this recurrence relation.
I found that it's $jk_i+2ji+j$!

(Law of the iterated logarithm) let $\lambda > 1$. He claims here that I can pick $\lambda > 1$ so big, so that
$$\dfrac{(1-\dfrac{\epsilon}{10})\sqrt{2(\lambda^{k+1}-\lambda^k)\log\log(\lambda^{k+1}-\lambda^k)} - (1 + \epsilon) \sqrt{2 \lambda^k \log\log(\lambda^k)}}{\sqrt{2 \lambda^{k+1} \log\log(\lambda^{k+1})}} \ge 1$$
which is categorically false, because already
$$\sqrt{2(\lambda^{k+1}-\lambda^k)\log\log(\lambda^{k+1}-\lambda^k)} < \sqrt{2 \lambda^{k+1} \log\log(\lambda^{k+1})}$$
I think I know what he meant to say. and it's not it ^

4:03 PM
It would seem my modular identity holds for several values; I think it is worthwhile to try and prove it. desmos.com/calculator/kivfq3qjvx
I'm terrible at proofs. Where do I begin?

@JoeShmo. Thius is for me :0

How we can replace module with one comparison and subtraction?

Assuming you mean modulus, then for $x\bmod y$, $\lbrace x,y \rbrace \in \mathbb{N}$, and $y\leq x\lt 2y$, you can conditionally subtract $y$ from $x$ to compute the modulus. This obviously requires reducing $x$ such that it satisfies the inequality if $x$ as-is does not satisfy it.

I was looking at your demo
OMG
Looks crazy

Yes, my demo is an algorithm for modulus. You can repeatedly apply it recursively to reduce $2^x$ to compute $2^x\bmod y$.
It's only useful for $2^x \gt y$

what applications does this have anyway!
It seems this is moslty theoricial

4:15 PM
Computing quotients efficiently in what I believe is O(n log n).

Do you have a purpose in mind playing with these demos :0

It can be used to compute a reciprocal very quickly and then all you need is a multiply.

I don't get it :( sorry
But good that you have something in mind
keep going

This should explain everything: rapidtables.com/convert/number/…
The base is $y$ and we want to compute $\frac{1}{y}$.
To convert to base two for an arbitrary amount of precision, you begin with 1 and repeatedly double it in base $y$. Double it once. If the result is greater than or equal to 1, then the corresponding bit is a 1; 0 otherwise. Take the fractional portion of the result and repeat up to $n$ times for $n$ bits of precision.
In case you didn't notice, we only have to double and take the fractional portion which means that for $x$ doublings, the fractional portion is $2^x\bmod y$.
So, find a way to compute $2^x\bmod y$ fast enough and you can compute reciprocals quite quickly this way for each bit in parallel.
Exploiting the properties of commutative rings, then we can represent $2^x$ as the product of several smaller powers of two, therefore, on average, the cost to compute modulus should be O(log n). We wish to do this for n bits, therefore the cost to compute the reciprocal should be O(n log n).
And, obviously, you can compute multiplies if you can compute reciprocals by dividing by the reciprocal of the divisor, so you could probably make something in hardware that computes multiplication and division in the same execution unit.

wow
:0

4:25 PM
:)

does this improve along any dimension the usual algorithms for binary gcd with bezout coefficients

No clue, honestly. I presume it does assuming you're trying to find an integer in some modulus like you would be with the method presented here: stackoverflow.com/a/48482313/2950711
If you look at the number there which multiplies $x$, it is actually a multiple of $\frac{1}{3}$.
Power of two multiple, specifically

@leslietownes. Can you perform the same modulous operation only using one comparision and subtraction?

the line of thought i'm considering is, if a is invertible mod n, then gcd-with-bezout will produce P and Q with Pa + Qn = 1 and then P is an inverse of a mod n

I mean can we replace the modulo operation with one comparison and an occasional
subtraction

4:29 PM
people have spent a lot of time optimizing gcd although i don't know if they also optimize the size of the bezout coefficients

Yeah I consider gcd algorithm slow. The only way to compute gcd efficiently is with division, and this is a fast division algorithm as you can see.

avra, i'm not going to wade too deeply in these waters, but division with remainder of positive integers can be implemented as iterated subtraction

(Supposedly. I still want a proof.)

to the extent you want anything more subtle than that i have no idea

@AMDG. Proof!:0
Then good luck man haha

4:30 PM
A proof is composed of many small proofs.

You need leslie, Ted, robjohn and so on

Proving this now that I look more at this does not seem too difficult even for me.

Mathematicians have endless ways to prove you are right or wrong

I would love to just wait around and let someone else prove this, but I think in the mean time, I'm just going to implement this and hope someone else can prove it for me in the mean time. It is quite reasonable to believe that I should begin implementing this now.

@leslietownes. I am just trying to show like if the module is given by (x+1) mod n, then how we can replace the mod with one subtraction and comparision? 1) check if (x+1)>n is the first step I see, then we can subtract (x+1) - n if the comparision is true

4:33 PM
@robjohn You wouldn't mind verifying my algorithms once again, would you?

However, if we have let us say n=17 and (x+1) = 3333, following the previous strategy won't work

Also, I'm pretty sure this is more the nature of the sort of thing to which you told me to mention if I found anything.

This is anyway the solution I got

avra if i'm understanding things correctly, in general, you may need iterated subtraction, not just one subtraction. that's the end of it. what is the end game here? is this a textbook problem or an applied problem or what?

Demonstrate that you can replace the modulo operation in Part b with one comparison and an occasional
subtraction.
This is the question :/ So occasional subtraction does not mean one !

4:38 PM
"an occasional subtraction" does not appear to limit to one

:0
OMG
Thanks
I got it then, so we keep subtraction by $n$? However, it's required that we do only one comparision

as a matter of prose, i don't like "occasional" in that, if an algorithm is designed to subtract, over and over without interruption, until [condition] is satisfied, i would not personally describe the subtraction aspect as "occasional"
i don't have much insight into this problem, it seems poorly written and i am a terrible mind reader

100%
Welcome to my world :(
Most people who taught those courses probably not even mathematicans
I will skip the question then

mathematicians may not be models of clarity, but non mathematicians writing math is the worst of both worlds
2

Yeah, take me for example

4:44 PM
@AMDG. Have you did bachelor of in math haha?

I do not have any degrees in math

wow great
respect

Yeah... the downside is the tedium of learning math on your own while having to also figure out new terminology on your own
and notation for that matter

5:02 PM
@AMDG which algorithms?

@robjohn $2^x\bmod y$ via a recursive algorithm for which the first iteration is found here (blue-colored graph): desmos.com/calculator/kivfq3qjvx
If you don't beat me to it, I will create an implementation in C and send a godbolt.org link like last time.
However, it has been a long month and a half. I need a break.

@AMDG I don't see an algorithm there. I see calls to mod, and if you add one of those for each bit of a general number, I don't see that it will be any more efficient than simply repeated shifted subtraction.
perhaps there is more buried in desmos code that I don't see

Will respond when done eating.

that sounds like a machine code instruction
rwde

5:29 PM
lel
nope, it ain't
not in x86 at least
@robjohn There is more buried in it that isn't immediately obvious. The first iteration of the algorithm which is used to reduce the input, $2^x$ to be $2^{x\prime}\leq 2^x$, is in there as $2^{n}\left(2^{\operatorname{floor}\left(\log_{2}\left(y_{0}\right)\right)+1}-y_{0}\right)-2^{n-1}y_{0}$.
The starred message three up from the bottom is an example of just one iteration of the algorithm.
$2^7$ was reduced to $-2^3$
We can then repeat this process once more, taking into account the number of factors of $-1$ that we accumulate per iteration.
And just plug in, for this example, $2^3$ again to reduce it further, but it cannot be reduced further for $2^x\bmod 11$ since $2^3 \lt 11$, so we now multiply by -1 to get $-2^3$ again and finally compute the modulus as a single subtraction (or addition).
It could probably be reduced to a constant-time implementation if we could efficiently compute addition-subtraction-chains and/or the number of iterations required to reduce $2^x$ which is always a finite number of iterations.
Also I'm quite certain the cost decreases as $y$ increases.

6:17 PM
However, is this an iteration that must be performed once per bit of the number being modded? I assume so since you are only doing this on powers of 2.
to apply the mod to a general number, you will need to break it into powers of 2
This doesn't seem more efficient than simply subtracting and discarding the result if a carry is performed.

It's not for computing $x\bmod y$, only $2^x\bmod y$, and that, for determining the value of a single bit for the reciprocal of $y$.
$x$ is the index of the bit we wish to compute for the reciprocal of $y$. Then the value of the bit using Iverson bracket notation is $[2\left(2^x\bmod y\right)\geq 1]$.
From there, just use the resulting division algorithm to compute the modulus of any two values rather than just powers of two.
Oh and of course $x$ is an integer beginning with $0$ as the index of the least significant bit.
Just make sure to compute using fixed point.
So actually it's $[2\left(2^{|x|}\bmod y\right)\geq y]$ for $x\lt 0$.

@AMDG It might only compute the highest order bit for $1/y$ since any lower bits would be masked by other bits of the "numerator". I guess I am just not getting how this computation helps in general.

No, it computes an arbitrary number of bits for $\frac{1}{y}$.

How many bases of $(\Bbb F_p)^n$ are there
unordered

equipping the cycle graph $C_{3}$ with the graph topology, $C_{3}$ then becomes homeomorphic to $\mathbb{S}^1$?

6:32 PM
yeah

cool. Thanks

"Graph topology" is just "edges are lines and vertices are points", right?

yup
that topology coincides with the toppology of a CW complex, right?

I believe so

akiva, (p^n - 1)(p^n - p)...(p^n - p^{n-1})/n! ?

6:34 PM
@AkivaWeinberger since any two elements of a base are distinct, we can just divide the number of ordered ones by $n!$, no?

thorgott and i might be the same person

maybe this is fate

@leslietownes @Thorgott Yeah
So follow-up question
Is there an easy way to see that this is an integer, without knowing this argument
i.e. without knowing it counts based in a vector space
Hmm maybe write it in base p...?
Then it's 11110*11100*11000*10000
(for p=5 for example)
Wait no
I meant 44440*44400*44000*40000 if p=5 in base 5

Howdy, @Thor, @leslie, @robjohn, DogAteMy, @monoidal. Whew.

Incidentally, vocalizing math is an interesting topic

6:39 PM
Hi Ted

someone had a system for vocalizing binary numbers. they haven't been around in a while.

for example, "$f'(x)>0,~f''(x)<0$" is pronounced "going up but slowing down"

Reminds me of the infamous math songs of yore.

and yet, not slowing down. that's a challenge. easing off the gas pedal, maybe.

Huh?
The speed is decreasing, is it not?
you're quibbling over the up/down, I know

6:42 PM
yes, that's all.

inconsistency in English idioms

we should speak in byte sequences.
my daughter does this stuff. she would appreciate the logic, or lack thereof, in slowing up.
particularly when you ask her to calm down

@robjohn @leslie DogAteMy ... Take a look at this classic sort of mis-proof. The OP thinks it's salvageable. I think not.
She's acting up

oh, this is interesting.
this is one of the things my analysis instructor blew. there's an exercise in rudin relating to this, the instructor offered a 'simplification' that actually wasn't a simplification.

"Phrasal verbs"

6:46 PM
i should have to pay a dollar to some charity every time i complain about my first analysis class.

are the difference between throwing a potato, throwing out a potato, and throwing up a potato
If you did all three at once perhaps you could say you "threw it and it up and out"

that example seems unusually well attuned to the context of dispute resolution in irish-american families.

Are you summoning @copper so soon?

i know, he's still sleeping it off.
i don't see how to salvage this either. if it's any consolation, someone who got a postdoc at berkeley had a similar idea 23 years ago.

in German we have two verbs "umfahren" and "umfahren" which have the stress on different syllables. The meanings are basically opposites. If you stress the first syllable, it means "drive over", if you stress the second syllable, it means "drive around"

6:48 PM
No, I think it's total garbage, but nice try for someone teaching high school :)
Oh, right, @Lukas. I remember that vaguely. One instance where English is not the most f***ed-up language.

Hi all

OOM-fahren. oom-FAHRen. got it. i may need this, i have a german friend who visits sometime, and long beach has a roundabout between our house and the airport.

Also notice "look at the potato" -> "look at it" but "throw out the potato" -> "throw it out", not "*throw out it"

Hi @Balarka

Howdy, a @Balarka.
@leslie I have no recollection of a related exercise in Rudin. Where he does Taylor?

6:50 PM
@leslietownes Is that Olga?

I recently learned that the kanji (Chinese character) for "medicine" literally means "fun grass"
medicine = 薬 kusuri
fun = 楽しい tanoshii
grass = 艹 (radical, meaning it only has this form when it's a component of other kanji)

The integration proof of Taylor, if he does it, has to come in the integral chapter?

lemme look it up.
amWhy: it isn't! although if olga had been born at the another time i think she would have been born in germany.

(It's actually a phono-semantic compound, which means the word for "medicine" just sounds like the word for "fun" in Old Chinese)

@leslie You make me wonder how many math students at MIT complained about that darn postdoc who messed things up.

6:51 PM
(and the "grass" is the semantic compound (used in lots of characters to do with plants)

@leslietownes dark times

On the other hand, music (音楽) does literally mean "fun sound" or "joyful sound"

(the same is also true of Chinese and Tibetan, apparently)

See also (and I only realized the connection recently): the phrase "to play music" in English

@Ted: Just saw the guy who proclaimed your comment on that logic question was flawed. Hilarious!
Can't stop laughing.

He sure put me in my place.
We're now going to prove everything by contradictory contrapositive. It's a pact.

Hahaha

6:53 PM
@TedShifrin Is it reigning in CA? Maybe knot, you seam two get nice whether their!

It hardly ever reigns in southern CA. Don't you know the song?

my postdoc had some harebrained solution to that, which is weird because rudin, somewhat uncharacteristically, spells out what to do.

Oh, @leslie, nothing whatsoever to do with the integral proof.

yeah. my memory was focused on taylor only.

6:54 PM
integration form of Taylor's theorem is not there in Differentiation chapter in Rudin's.

I like the explicit proof that I ultimately put in the 3rd and 4th editions of Spivak's book. I guess I came up with it by myself.
It's probably equivalent to Rudin's proof, but more motivated.

the CV of the postdoc i had indicates that they never taught analysis before or since my ill-fated class. good job, math departments.

Well, they needed LOTS of bodies to teach 104 and 113.

@TedShifrin How many wrong spelling did I have, though I'd be perfectly understood, if I spoke them

@amWhy: You are never going to outdo GHOTI.

6:56 PM
@TedShifrin I surrender. That is brilliant!

It's hard for us mere mortals to beat G.B. Shaw.

people would strategically flee from 104/113 if taught by actual faculty. the word on the street was the faculty were 'too hard' and the postdocs were 'easy because they can't fail too many people.'
not true as a factual matter, but true in practice.

Often postdocs are better teachers than tenured faculty. Even teaching as a grad student I was a better teacher than many of the faculty. Yes, and modest, too.
They let me teach my own Guillemin & Pollack course, which then ultimately became a course in the curriculum.

here's a fun fact about ancient greek. All these forms have different meanings (sometimes multiple ones per form): ἡ, ἤ, ἥ, ἦ, ἧ, ᾗ, ᾖ

there were a few tenured people at iowa who were incompetent but had exclusive jurisdiction over certain classes.

6:57 PM
@TedShifrin That is true.

I don't remember having to twist too many arms, either.

@LukasHeger lol

i don't think it's an accident that at many schools, non-tenure-track faculty are disqualified from winning university-wide teaching awards. it would embarrass them.

@LukasHeger Did they have reading glasses back then?
@leslietownes Hah!

Oh, at UGA we got college-wide teaching awards for non tenure-track folks who clearly deserved them.
But the Regents in GA are about to abolish tenure. So it's not just pro-COVID now.

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