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Ted Shifrin
Ted Shifrin
12:08 AM
I didn’t just fall off the culinary turnip truck!
 
PM 2Ring
PM 2Ring
12:39 AM
@robjohn I'm sure you're correct. :) I haven't really thought about that puzzle since it was posted, and I don't remember the details. At the time, I found you only need a few Euler-Maclaurin terms to reduce the error for 86 to around 1/1000000, but of course that's a bit tedious without a calculator. ;)
 
 
1 hour later…
copper.hat
copper.hat
1:46 AM
i love turnips
i like Spanish tortillas. I discovered the difference when buying food in a rural town outside Barcelona. Never occurred to me that tortilla would mean something different there.
 
user10478
user10478
2:40 AM
From the en.wikipedia.org/wiki/Moving_average Wiki page, "This is also why sometimes an EMA is referred to as an N-day EMA. Despite the name suggesting there are N periods, the terminology only specifies the α factor. N is not a stopping point for the calculation in the way it is in an SMA or WMA." This is very shocking to me. Does this mean an N-period exponential moving average on, i.e., candlestick charting software is not actually averaging over just the last N periods? If the
EMA weights don't stop at N, but are presumably infinite in number, how do you even convolve the EMA? Can you take the convolution of an infinite sequence of weights with a finite amount of historical price data?
 
robjohn
robjohn
2:58 AM
@PM2Ring I've posted the answer I wrote here. I didn't post there as I think it may have too much detail.
 
PM 2Ring
PM 2Ring
3:36 AM
@robjohn Nice. I think it'd be ok on Puzzling, but maybe inside a spoiler.
@Mohammed Here's a Sage program you can run that tests your twin prime pattern. It starts out with lots of twins fitting your pattern, but after a while most pairs fail.
 
robjohn
robjohn
@PM2Ring: I am not familiar with that site. What typically goes into spoilers there?
 
PM 2Ring
PM 2Ring
I did a test upto 100 million. The result was total 293796 bad 249146 0.848
@robjohn Well, they don't like solutions to be visible, especially when the puzzle is active. But once people start openly discussing the fine detail, spoilers aren't really necessary. But they're still appreciated, for the benefit of future readers.
I must admit I'm not a frequent reader of Puzzling. I mostly only browse it when I see something interesting on the HNQ. Sometimes I've worked on answers for my own enjoyment but didn't bother posting because my solution was already posted by someone else.
 
robjohn
robjohn
@PM2Ring so there are active puzzles. I'll look for where those are.
 
Arief
Arief
 
robjohn
robjohn
@PM2Ring Yeah, my solution is essentially the accepted solution, just more detailed, with an estimate of the error.
 
PM 2Ring
PM 2Ring
3:51 AM
@robjohn Adding an answer with more details is quite acceptable.
 
copper.hat
copper.hat
@Arief why are you inviting me?
should i repeat?
@Arief if you have something to ask, please ask it here. otherwise, please stop pinging me.
 
PM 2Ring
PM 2Ring
@Arief I guess you're trying to promote your YouTube channel. But appearing in a chat room & pinging a bunch of strangers to join you in separate rooms is a bit annoying, and also a bit creepy. Mentioning your channel is ok, I guess. But what you're doing is bordering on spam, IMHO.
 
robjohn
robjohn
4:06 AM
@PM2Ring Bordering?
 
PM 2Ring
PM 2Ring
I was trying to be diplomatic. :)
 
leslie townes
leslie townes
BUY LESLIECOIN N O W ! ! ! YOU WILL NEVER FORGIVE YOURSELF IF YOU DON'T lesliecoin.leslie.biz.leslie.ru.biz
 
Koro
Koro
that's exactly how spam looks like, with some bold words, a link and all. :)
 
PM 2Ring
PM 2Ring
@Mohammed BTW, all primes>5 must be coprime to 2×3×5=30, so are congruent mod 30 to one of $\{1,7,11,13,17,19,23,29\}$. Twin primes p,q with pq not divisible by 5 must therefore be congruent to (11,13) or (17,19), so pq+26 is congruent to 19.
 
Alex
Alex
Hi
I'm here
 
leslie townes
leslie townes
4:15 AM
koro it is not spam it is an INVESTMENT OPPORTUNITY !! ! ! MEGABUCK$S leslie.leslie
 
PM 2Ring
PM 2Ring
Greetings
 
leslie townes
leslie townes
hi PM
 
Alex
Alex
We can find the closed form to $\displaystyle \int \frac{e^{\tan^{-1}(x)}}{\sqrt{1+x^{2}}}{\rm d}x$?
 
copper.hat
copper.hat
speaking of spam :-)
 
Alex
Alex
is about my question?
 
leslie townes
leslie townes
4:21 AM
yeah, nice try. i'm not buying those NFTs
 
Alex
Alex
wolframalpha.com/input/… seems not to find a good form for the primitive
 
leslie townes
leslie townes
ignore wolfram alpha and try a substitution
2
there are a lot of complex identities around inverse trig functions that symbolic packages love to apply even if there is something real that they could also be doing
 
Ted Shifrin
Ted Shifrin
@PM2Ring Diplomacy forbidden here.
 
PM 2Ring
PM 2Ring
My first response to Arief was rather harsh, but I deleted it before posting. I guess I went too far in the opposite direction. ;)
 
Alex
Alex
@leslietownes I understand. I will try a substitution.
 
leslie townes
leslie townes
4:27 AM
alex: i don't guarantee success, but it will be more informative than WA can be. for example, it might explicitly reveal the form of an integral that is known to require a special function
the fact that WA just spits something out is not always a guarantee that it's required; it can be easier to see by hand
if there were specific bounds on the integral and the antiderivative is not the full focus of attention, that may also be helpful to indicate
 
EM4
EM4
hello.
 
leslie townes
leslie townes
i tried to post something on my daughter's instagram account the other day, and it somehow found spam-ish characteristics in the comment i applied to the photo, and warned me about this and asked if i wanted to post it
since when does instagram care about spam anyway, spam is the whole purpose of the app
salut EM4
 
copper.hat
copper.hat
whoa, a question i just answered just picked up 5 votes in a minute. is that suspicious? i mean, this is calculus of variations (with convex flavour, of course), not batman!
hopefully i have not inadvertently flattened a grade curve somewhere...
 
leslie townes
leslie townes
oh, was it you who gave us question no. 5 on our final exam
many thx
i find it's either feast or famine, with my answers. either a lot of upvotes immediately, or nothing
probably because like you, i too seek out only the juiciest PSQs around exam time
 
PM 2Ring
PM 2Ring
Funny voting patterns are to be expected during Winter Bash. Also editing of ancient posts.
 
leslie townes
leslie townes
4:36 AM
oh is that what is going on.
i was wondering about that. all of this stuff from 2013 suddenly on the main page.
i voted to close something and then realized (1) it had an answer (2) from over 5 years ago
and then wondered how it had even bubbled up to the surface
now i know
 
PM 2Ring
PM 2Ring
I first noticed it on Physics, with OPs editing their own 5+ year old questions.
 
copper.hat
copper.hat
well, i fell a little bad as well, there was an existing answer but it was incorrect, so i knocked that out of place.
 
leslie townes
leslie townes
oh haha copper i think i voted on that
i didn't even realize it was you
 
copper.hat
copper.hat
hey, how can you vote twice :-)
 
leslie townes
leslie townes
well i meant i voted on the Q with an accepted answer that looked wrong
 
copper.hat
copper.hat
4:39 AM
oh, good news to report
 
leslie townes
leslie townes
can they dis-accept an accepted answer?
 
copper.hat
copper.hat
jewel lake is full again, from 0 to full in a few days
 
Koro
Koro
Leslie: yes, I think.
 
leslie townes
leslie townes
that's awesome. apparently the last system got california basically to 'normal' with hopefully more to come
 
copper.hat
copper.hat
had a wonderful ride. i am getting old, or california is making me soft, but i was a little chilled at times!
 
PM 2Ring
PM 2Ring
4:40 AM
Sure. The OP can un-accept at any time.
 
leslie townes
leslie townes
math.stackexchange.com/questions/820981/… brand new edit on an obviously close-able post from 2014
(i taught out of eccles once, it was not a horrible book)
 
PM 2Ring
PM 2Ring
Here in Sydney, we've had a really slow start to summer. We've had a huge amount of rain since the end of winter. Last week, it got so cold in the middle of the night that I had to use my heater. But in the last 3 days its been rather warm, and very humid.
 
copper.hat
copper.hat
see, there are lots of targets for reasonable closing, but i think people pick on new, low ranked askers because they are easy meat.
 
leslie townes
leslie townes
we've had a wintry week here. got into the 30s one night!
 
copper.hat
copper.hat
i can feel a suspension coming on
 
EM4
EM4
4:44 AM
global warming @PM2Ring .
 
Ted Shifrin
Ted Shifrin
Most of those books are horrible,
 
leslie townes
leslie townes
i struggle generally with the idea that it is possible to teach people how to 'do proofs' outside of a specific setting where there is substantive math that you are also learning
and all of those books, to varying degrees, fail at teaching anything subtantive
eccles does do elementary number theory
 
copper.hat
copper.hat
yep, bay area had 30s last night. right now, dublin (the real one), walnut creek, albany & oxford all have about the same temperature
while it was engineering, i really did not get the value of proofs until my first semester at berkeley
 
PM 2Ring
PM 2Ring
 
leslie townes
leslie townes
i had to teach out of solow's 'how to read and do proofs' once, and while i know many people have had a lot of success with that book, it made me want to walk into the ocean
 
EM4
EM4
4:47 AM
yes!
 
Koro
Koro
Is there an example of a norm of $\mathbb R^2$ which is not obvious?
 
leslie townes
leslie townes
i mined it for homework exercises and completely diverged from everything else in lecture
 
copper.hat
copper.hat
i think the idea of a formal proof too early on is pointless and dispiriting
 
Koro
Koro
the norm which is tricky to be proven to be a norm on R^2.
 
copper.hat
copper.hat
@Koro $\|x\| = \sup_{t \in [100,101]} |x_1+t x_2|$.
 
leslie townes
leslie townes
4:49 AM
koro: i don't think the question is well formed.
 
Koro
Koro
the question is not well formed. true.
 
leslie townes
leslie townes
you might ask, given the following general recipes for constructing norms on R^2, is there any norm that can't be realized as one of those recipes
even that though might involve norms that are very easy to prove are norms
R^2 is only R^2
 
copper.hat
copper.hat
i guess that rules out minkowski then
 
Koro
Koro
By recipes, I had "L_k$ norms only in mind.
@copper.hat thank you! copper.
 
leslie townes
leslie townes
ok. one way to disitnguish L^p norms from one another, and from other norms, is via the modulus of convexity
 
copper.hat
copper.hat
4:51 AM
squirrel
 
Ted Shifrin
Ted Shifrin
Acorn
 
leslie townes
leslie townes
NUTS
 
Koro
Koro
peanuts
 
copper.hat
copper.hat
chest nuts
sorry, didn't mean to put a space there
 
leslie townes
leslie townes
what is with this trend of people using subscripts for L^p spaces instead of superscripts
i've seen it all over MSE today and now here, too
 
copper.hat
copper.hat
4:53 AM
the world is going upside down
it will be $l^1, l^2$ next
 
Koro
Koro
Haha
 
leslie townes
leslie townes
i even answered a question involving this. i thought about saying something salty about what i understood the subscript to mean, but surprisingly, better judgment prevailed
 
copper.hat
copper.hat
i need some of that
better judgement
 
robjohn
robjohn
Yeah, but all most of us can afford is poor judgement.
 
copper.hat
copper.hat
i'm saving my judgement for the pearly gates. i'm really looking forward to seeing the hell demo.
 
robjohn
robjohn
5:56 AM
break's over; back on your heads.
 
copper.hat
copper.hat
my convex job is done for today
 
 
4 hours later…
Koro
Koro
10:24 AM
copper et leslie: If $a<b$ and $f(x,y):=\sup_{t\in (a,b)}|x+ty|$, then I claim that $\lim_{(x,y)\to (0,0)} f(x,y)=0$.
Proof: Since all norms on $R^2$ are equivalent, there is a constant $c>0$ such that $f(x,y)\le c\sqrt{x^2+y^2}$. Since $f$ is non negative, given any $\epsilon\gt 0$, the result follows by choosing $\delta=\epsilon/c$. Proved. Is my understanding correct? Thanks.
 
 
2 hours later…
Jam
Jam
12:22 PM
@Thorgott So can you help me see how $O(1)$ and $O(-1)$ as bundles/sheaves and how the divisors of X play a role is their definitions? i cant understand from wiki.
 
Alex
Alex
Hi
\begin{align}
\int \frac{e^{\tan^{-1}(x)}}{\sqrt{x^{2}+1}}{\rm d}x&\overset{x\mapsto \tan u}{=}&\int \frac{e^{u}}{\sqrt{x^{2}+1}}(x^{2}+1){\rm d}u\\
&=&\int e^{u}\frac{(x^{2}+1)\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}{\rm d}x\\
&=&\int e^{u}\sqrt{x^{2}+1}{\rm d}x\\
&=&\int e^{u}\sqrt{\tan^{2}(u)+1}{\rm d}u\\
&=&\int e^{u}\sqrt{\sec^{2}u}{\rm d}u\\
\end{align}
I don't know how to move forward from here.
 
Mad Spaces
Mad Spaces
12:44 PM
fUnderstanding the Proof of: A Compact set is only then a lesbegue meassure zero when its volume "jordanian" is zero"
The Right direction
Since A is lesbegue meassure zero then we know that
$ A \subset \bigcup_i^\infty W_i $ with $ \sum_i^\infty W_i < \epsilon $
However since compact then according to lesbegue - Borel there exists a finite coverage of $W_{i_1} ... W_{i_k}$
Then some inequalities are written. Here is the one which i do not understand.
for some partition $P$ we have
$Vol \bigcup _{W\in P,_W\cap \bar A \neq \emptyset} W \leq Vol(W_{i_1})+ ...+Vol( W_{i_k}) $ where as we are …
Why is this inequality true? Say why is the sum of the volumes bigger or equal to that of the sum on the left side, since the left side also considers the closure, where as the right side only includes the set. i fail to see the truth behind it.
 
Thorgott
Thorgott
@Jam I don't know anything about divisors, though I also don't think that's required to understand $\mathcal{O}(k)$, but I also don't really know what precisely you're asking for
 
Mohammed
Mohammed
1:03 PM
@PM2Ring thanks to made a codo
@PM2Ring, I mean, this is a range-bound guess and it always excludes all twins that end in 9 and 1
 
Sebastian Nielsen
Sebastian Nielsen
1:31 PM
I am trying to explain the difference between a hyperplane and manifold in simple words. Is this correct?

Assume we are in 4 dimensions, then a hyperplane will in 3 dimensions be your "flat" hand pointing in any direction.
And a manifold would then be your hand folded in some weird way pointing in some direction.
 
Thorgott
Thorgott
2:01 PM
I think these are syntactically different concepts. Manifolds are particular spaces. Hyperplanes are particular subspaces. A hyperplane is not just the plane itself, but also the way it is embedded in ambient space.
So, if we're in a four-dimensional vector space as you want to assume, the right thing to compare a hyperplane to is a codimension one submanifold and then the former is simply a special case of the latter.
in either case, I don't understand your points about "directions", a direction is effectively something one-dimensional
 
Sebastian Nielsen
Sebastian Nielsen
Okay, thanks. I clearly don't understand what I am talking about.
 
robjohn
robjohn
2:43 PM
@Thorgott If the hyperplane has codimension 1, then I had assumed the "direction" was a unit vector in the 1 dimensional subspace orthogonal to the hyperplane, but then maybe I don't understand what I am talking about either :-)
 
Thorgott
Thorgott
oh yeah, that's something you can say of course, but I didn't understand the message that way
the notion of the flat hand pointing in a direction confuses me, a flat hand strikes me as two-dimensional anyway
 
robjohn
robjohn
but then we're not higher dimensional beings who may have higher dimensional hands.
When I visit a four dimensional friend, the food they give me refuses to stay inside my stomach.
 
soupless
soupless
I want to ask how to solve for $(30! - 1) \bmod 930$. What I did was guess that since 93 and 10 are relatively prime and that $(30! - 1) \equiv 9 \bmod 10$, then the last digit of the remainder mod 93 should also be 9 since 10 is smaller than 93. Also, since the question is a multiple-choice type, the only choice there is 29. Is this a 'wrong solution, correct answer' situation or 'correct solution, lacking details' one?
Sorry if this is an easy question I asked. I really don't know how to use the Chinese Remainder Theorem.
 
antonio
antonio
3:21 PM
I always meet partitions at most countable. I wonder if a partition can be uncountable as a set.
 
Alessandro Codenotti
Alessandro Codenotti
@antonio Sure, why shouldn't they be
 
antonio
antonio
@AlessandroCodenotti: any notable example?
 
Alessandro Codenotti
Alessandro Codenotti
Any equivalence relation with uncountably many classes partitions the set its defined on in uncountably many pieces. Say $x\sim y\iff x-y\in\Bbb Q$ as a relation on $\Bbb R$ which is an important counterexample in analysis (its transversals are Vitali sets, which are not measurable)
 
antonio
antonio
@AlessandroCodenotti: thanks
 
leslie townes
leslie townes
3:41 PM
the partition $S = \coprod_{s \in S} \{s\}$ of any set $S$ (aka the equivalence classes of the equivalence relation $=$) has the same cardinality as $S$ itself :)
 
Fargle
Fargle
3:58 PM
@soupless Shouldn't work, I don't think. Note that, say, 98 is 8 mod 10, but 5 mod 31, even though 10 is smaller than 31.
Better yet, replace 31 with 93 in my example. What a coincidence.
@soupless To use CRT, you'd also be best served to completely prime factorize for this problem, so you want to find remainders for 10, 31, and 3. 10 and 3 aren't too bad, and 31 is an application of Wilson's theorem.
(Not correct that you have to completely factorize, CRT only requires coprime. Don't math in the morning. Still, it makes this problem easier.)
 
Xander Henderson
Xander Henderson
@Fargle As a resident of Trump's 'Merika, I have been told that CRT is the devil.
 
soupless
soupless
4:20 PM
@Fargle Thank you very much.
 
Mad Spaces
Mad Spaces
@Fargle we look the same. :D
 
soupless
soupless
Yes, I laughed. That's true: don't math in the morning
 
Fargle
Fargle
We sure do.
@soupless You are correct that the remainder mod 930 would have to end in 9, by the way. I missed this a bit ago.
 
Koro
Koro
If $a<b$ and $f(x,y):=\sup_{t\in (a,b)}|x+ty|$, then I claim that $\lim_{(x,y)\to (0,0)} f(x,y)=0$.
Proof: Since all norms on $R^2$ are equivalent, there is a constant $c>0$ such that $f(x,y)\le c\sqrt{x^2+y^2}$. Since $f$ is non negative, given any $\epsilon\gt 0$, the result follows by choosing $\delta=\epsilon/c$. Proved. Is my understanding correct? Thanks.
 
Fargle
Fargle
Just not mod 93. I think it winds up being 92 mod 93.
 
leslie townes
leslie townes
4:31 PM
koro: yes. in a classroom setting, possibly missing from this argument would be a verification that f is in fact a norm.
with this particular norm, it's also not too difficult to write down a 'c' without appeal to the general equivalence of norms. but i do like using the general result as a sledgehammer.
 
Koro
Koro
Thanks Leslie for the response. I thought about this following suggestion by copper today and your suggestion yesterday. I could show that f is a norm as follows: 1) f is non-negative and $f(x,y)=0\implies |x+ty|=0\; \forall t\in (a,b)$ so for $t_1\ne t_2$ we have $x+t_1y=0 $ and $x+t_2y=0$ so $(x,y)=(0,0)$ and converse is trivial. 2) Triangle ineq. follows by $\sup (A+B)\le \sup A+\sup B$., 3) scaling: $f(kx,ky)=|k| f(x,y)$
Yes, we can do it without using norm but I wanted to try this new approach :)
When I become a prof., I'll put this question in exam/test paper :)
 
Fargle
Fargle
@soupless And a cute observation for the problem, not using CRT per se: it's -1 mod 10, -1 mod 3, and -1 mod 31. So, it has to be one less than a multiple of 10, of 3, and of 31; thus it would have to be -1 = 929 mod 930.
 
UnderMathUate
UnderMathUate
Hi all.
 
leslie townes
leslie townes
undermathuate will now lecture us on all of the new math they've learned since the end of finals. you have had no other distractions, so we expect a full report.
start with the theory of linear partial differential operators.
 
UnderMathUate
UnderMathUate
Ha...
So you see, I've actually been coding.
I've also just witnessed the steepest curve known to man.
I had a C- in my programming course and I somehow ended up with a B.
I feel bad about it, so I'm working ahead on my next programming course.
 
leslie townes
leslie townes
4:48 PM
it's a christmas miracle.
 
UnderMathUate
UnderMathUate
I know. ;-;
 
leslie townes
leslie townes
the grinch realized that all the whos in whoville would happily program regardless of what grades they were given, so he brought a big bag of Bs back
 
UnderMathUate
UnderMathUate
Honestly, that might have actually been his plan.
But yeah, apart from that and Ted's vids, I haven't done anything new yet.
 
adamaero
adamaero
General math question. To factor inflation into the compound interest formula with reoccurring contributions...
...is it as simple as just subtracting speculated inflation rate from the speculated return rate?
So if I assume 8% return, and 3% average inflation, just use 5% for "r"?
P(1+r/n)^(nt) + PMT((1+r/n)^(nt)-1)/(r/n)
Example. Assume 8% return and 3% average inflation. Monthly contributions are $1000 for five years and $3500 for 15 years. Using a rate of return of 5% , that's over $1M.
 
Fargle
Fargle
5:04 PM
@UnderMathUate What language, out of curiosity?
 
UnderMathUate
UnderMathUate
@Fargle We learned the basics of programming using C++, but we're switching to Java to cover OOP.
 
Curio
Curio
I do not understand why this integral does not converge: https://bit.ly/3J8UdTq
It's asymptotic to -4logx/x^(3/2) for x-> + infinity. As 3/2 > 1, it should converge
 
Fargle
Fargle
@UnderMathUate Gotcha. Been a while since I touched either language, but I had to take similar courses back in undergrad.
I don't hate either, but I do have various idiosyncrasies that wind up drawing me to different languages and paradigms.
 
UnderMathUate
UnderMathUate
Ha, I'm in a similar position. I enjoy programming in relation to personal projects but struggle to muster the same enthusiasm for my courses.
 
Fargle
Fargle
I think it's at least partly a problem of the pedagogy not really being caught up to the present. I know that, for example, I took my data structures class in 2016, and we were still using C++98.
OOP is also falling out of favor in a lot of contexts where it was previously industry standard, as far as I understand. Though GUI programming is still a good case for it.
I'd like to see more functional programming stuff and other recent developments make their way into the standard pedagogy, but that feels about as far off as "The Year of Linux on the Desktop".
 
UnderMathUate
UnderMathUate
5:18 PM
@Fargle Yeah, one of the major topics of the Java course is gonna be GUI's.
 
Semiclassical
Semiclassical
very random empircal math observation: apparently, the nth Pisano period $\pi(n)$ is much more likely to be a bit larger than $2n$ than a bit smaller than $2n$.
 
Fargle
Fargle
It's a model that definitely makes sense there.
 
Semiclassical
Semiclassical
same seems true for 3n
(could be a matter of sampling size: i'm using the first 8k Pisano periods)
 
 
1 hour later…
jay
jay
6:36 PM
Question : Let $A$ and $D$ be two matrices ( both invertible ), is the product $A-DA^T(D^{-1})^T$ anti symmetric ?
oh and D is symmetric
so is $A-DA^TD^{-1}$ symmetric ?
sorry anti-symmetric*
 
leslie townes
leslie townes
6:54 PM
not necessarily
 
jay
jay
sorry it is zero when $A$ is symmetric
 
leslie townes
leslie townes
are you asking if A = DAD^{-1} when A and D are symmetric? again, not necessarily
it'll hold if and only if A and D commute
 
jay
jay
Sorry can I phrase my question again
Question : Let $A$ and $D$ be two matrices ( both invertible and symmetric ), what can we say about $A-DA^T (D^{-1})^T$
 
leslie townes
leslie townes
as phrased this makes sense, but it's confusing. if A is symmetric then A^T is just A and if D is symmetric then (D^(-1))^T is just D^(-1)
 
jay
jay
sorry yeah
$A-DA (D^{-1})$
is this guy anti symmetric or we cant say?
 
Thorgott
Thorgott
7:04 PM
@leslietownes should be $A$ and $D^2$ commute, no?
 
leslie townes
leslie townes
"what can we say about __" is a different question from "is __ antisymmetric"
thorgott, i was thinking along the lines of A - DAD^{-1} = ADD^{-1} - DAD^{-1} = (AD - DA) D^{-1}
jay: in general it does not need to be anti symmetric
 
jay
jay
:(
 
leslie townes
leslie townes
it's anti symmetric in dimension 1 :)
 
jay
jay
pffffffff
xd
 
Thorgott
Thorgott
$A-DAD^{-1}=(A-DAD^{-1})^T=A-D^{-1}AD$ iff $DAD^{-1}=D^{-1}AD$ iff $AD^2=D^2A$
 
leslie townes
leslie townes
7:07 PM
oh i see you were answering his question
thorgott, you and i were answering different questions. i was answeing "it is [sic] zero when A is symmetric"
which i provisionally took to mean "is it zero when A is symmetric"
 
jay
jay
I just wanted
$A-DA (D^{-1})$ to be anti symmetric when A and $D$ are symmetric
:(
 
leslie townes
leslie townes
jay, well in general, it isn't. do you know more about A and D? why is it so important that this hold?
 
Thorgott
Thorgott
asking it to be anti-symmetric seems even more unnatural than asking it to be symmetric
 
leslie townes
leslie townes
you get a goofy equation instead of a nice one
 
Thorgott
Thorgott
I guess the reasonable observation is that if $D$ is symmetric or antisymmetric and $A$ is symmetric (resp. anti-symmetric), then $A-DAD^{-1}$ is symmetric (resp. anti-symmetric) if $A$ and $D^2$ commute
 
leslie townes
leslie townes
7:15 PM
i wanna hear the unreasonable observations
 
jay
jay
Let $f$ be some real valued function. Given a matrix $D$ (symmetric invertible) and a general matrix $A$, I want to write $div(fAx)=div[fD(Sx)]+div[f Qx]$ where $S$ is symmetric and $Q$ is anti symmetric. this is why I was asking
so I tried $div(fAx)=div(fD(D^{-1}Ax))$
then started trying to decompose
 
Ted Shifrin
Ted Shifrin
7:43 PM
@jay Forget about the $f$, since there are rules for that. So you're asking if $\text{div}((A+B)x) = \text{div}(Ax) + \text{div}(Bx)$?
Where did $D$ come from?
Oh, this is nuts.
Note that $\text{div}(Ax) = \text{tr}(A)$. Just work with that?
I see there was a long discussion before I returned.
 
leslie townes
leslie townes
hi ted :)
 
Ted Shifrin
Ted Shifrin
Howdy munchkin's toy.
 
leslie townes
leslie townes
this morning, one of munchkin's imaginary friends, who sometimes is malevolent, was antagonizing munchkin by making false claims of ownership of a stuffed animal toy, and an orange cup, and by knocking on the door of munchkin's room.
 
Ted Shifrin
Ted Shifrin
Aggressive grinch action.
 
leslie townes
leslie townes
i now have the stuffed animal and cup on my desk. they were brought to me during a work meeting this morning.
 
Ted Shifrin
Ted Shifrin
7:52 PM
I'm sure they made a grand official entrance.
 
leslie townes
leslie townes
along with part of the story. "so-and-so was saying my bird was her bird. she can't take it."
yes, thank you. mute. thank you again. please go now, i'm not on winter break.
then someone on the meeting asked if i had two children and i had to say, no, i have one child with an imaginary friend who claims her stuff.
 
Ted Shifrin
Ted Shifrin
Well, remember that Christopher Robin had all sorts of imaginary friends. Well, more precisely, his father did.
 
leslie townes
leslie townes
munchkin has approximately five imaginary friends. approximately because the characters of some of them are not very well developed.
their names are persistent, however. there's something to it.
 
Ted Shifrin
Ted Shifrin
Well, she's getting close to Pooh, Kanga, Roo, Eeyore, Owl, Rabbit, and Tigger ... but then there are all of Rabbit's friends and relations.
OMG, I left out poor Piglet.
 
 
1 hour later…
PM 2Ring
PM 2Ring
9:20 PM
@Mohammed No worries. As Ted mentioned, all primes of the form 4n+1 can be written as a unique sum of 2 squares. Composites which are the sum of 2 squares can have even powers of prime factors of the form 4n+3. The product of 2 numbers that are the sum of two squares can also be written as the sum of 2 squares. That's the special Fibonacci form of the Brahmagupta identity. $$(a^2+b^2)(c^2+d^2)=(ac\pm bd)^2 + (ad\mp bc)^2$$
I wrote another small Sage program related to this. For each odd prime p<hi, find the first odd q: $(p+1)^2+q^2$ is prime.
 
PM 2Ring
PM 2Ring
10:03 PM
@Mohammed Here's a version that graphs q (on the Y axis) vs p (on the X axis).
Here's the plot for p<15000
 

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