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00:09
I hate iced coffee. I hate it so much
00:27
@冥王Hades Cool beans. More for the rest of us.
I'm a hot coffee person, myself
I drink cappuccino and espresso. In seasons other than winter, my second cup is usually iced.
@Rithaniel Me, too. I work my way through a couple of macchiatos every morning.
But in the summer, cold brew is nice.
I actually just made myself a dessert coffee (since it's Friday and I don't have to get up tomorrow)
Help, I fell down a rabbit hole asking myself "Is it ice coffee or iced coffee"
:-/
00:39
I used some of this "Mandarin orange" flavored syrup that came in today, and some coconut-flavored cream. It's a bizarre, but not unpleasant, combination
(Every time I take a sip, I find my bewilderment at the taste renewed)
@user4539917 iced, as one puts ice in it :)
@Rithaniel I don't usually like sweet coffee all that much. I sometimes will make a mocha, but I think of that as hot cocoa, not coffee.
Not a speck of sweetness in mine, thanks.
The rapper was "ice tea" not "iced tea," I think :-)
Yeah, if you go the chocolate route, that's hot cocoa
00:42
2 oz espresso, 1 oz steamed milk (as straigt coffee has, in the last several years, started to hurt my poor tum-tum).
I don’t determine my syntax by names of rappers.
@user4539917 No, he was Ice-T.
Ah, indeed!
Yeah, straight coffee tastes bad to me. I have to drown it in additives, but I've realized that this opens up whole avenues for what coffee can be
00:44
Ice, ice baby...
Like, hazelnut, peppermint, mandarin orange (still confusing), irish cream, honey, chocolate. So many options
STOP collaborating and listen :D
@TedShifrin We need to rework all of mathematical syntax to mirror Snoop Dogg's naming conventions
Hammer time.
Can't touch ths^
00:46
this is 2 Legit
@user4539917 (I believe that the verb you are looking for is "touch".)
To quit
"Ay Ayyyyy"
Rithaniel, I reject your taste in beverages … perhaps in food altogether.
@XanderHenderson thank you
00:48
So... does anyone want to give me $400 to buy Legos? Anyone?
Bueller?
SPeaking of food, I was surprised that you were able to relate a vending machine sandwich to a shape....as if you have experience with them.....I would not expect this with your food sophistication Mr SHifrin
@XanderHenderson Playmobil is better (may be aging myself there)
@D.C.theIII Lies.
Playmobil sucks.
Sweet little lies.
I'm not sure if they still make sets
@D.C.theIII They do.
00:50
I did subscribe to the reddit page for it....
Or did, as of a few years ago.
They also attempted a Playmobil film recently, if I recall correctly.
Actually I do remember the film being a thing
Playmobil: The Movie is a 2019 English-language French computer-animated adventure comedy film based on the German building toy Playmobil. The film is directed and conceived by Disney veteran Lino DiSalvo, in his directorial debut, written by Blaise Hemingway, Greg Erb, and Jason Oremland, and produced by On Animation Studios. The film stars the voices of Anya Taylor-Joy, Jim Gaffigan, Gabriel Bateman, Adam Lambert, Kenan Thompson, Meghan Trainor and Daniel Radcliffe. The film follows a girl named Marla who tries to save her brother from a Playmobil world that the two are sucked into and becomes...
sheesh they got some big names to voice
@D.C.theIII That is usually a bad sign.
Hire real voice actors. :P
Like Joe Dimaggio.
Or Hank Azaria.
00:52
Until you mentioned it I never really thought of that angle, with regards to any anime or other types of media I consume
Yeardly Smith.
@DCthe You missed my cute skunk surface from calc 2 :)
it does make a difference
@TedShifrin That actually sounds interesting.
That said, my very favorite Batman ever is Will Arnett.
There really is no other Batman....the way he played it still brings tears of laughter to my eyes
00:53
And it turns out that Mark Hamil is a pretty excellent voice actor, even if he is only okay as an actor actor.
Only real acting gig he get's to live off of is the original one that made him famous..
Also, how cool is it that Lego gave Nic Cage the opportunity to play Superman (a role that he has wanted to play for at least 30 years).
Build a solid whose cross-sections perpendicular to the $x$-axis are squares with one edge a chord of the unit circle. Draw.
@TedShifrin It's okay to be incorrect, Ted. I accept you, regardless
I'm going to give it a quick go
00:56
@Rithaniel Oh, dear...
You do realize that if you and Ted both make a statement of fact about the universe, and those statements disagree, you are wrong, right?
@Rithaniel You obviously don’t live in deSantis land!
That's president elect to you...
Or right, wrong? Xander
@TedShifrin Left.
Can we all just agree to disagree.
00:57
@user4539917 NO!
On what? I’m lost.
Ice tea
@TedShifrin Lol, you have me pegged
00:59
@user4539917 Iced tea is a drink. Ice-T is a rapper. Ice tea is tea which has been frozen in an ice cube tray.
T-cubes.
Does that mean that ice coffee is frozen coffee?
@Rithaniel Yes.
And, for the record, when I make iced coffee, I typically cold brew about a gallon of coffee, then freeze two ice cube trays worth. Then, when I want to have some iced coffee, I pour it over the frozen coffee cubes.
01:00
I suppose that means iced ice coffee would be a coffee ice cube that you have somehow gotten pockets of pure ice to form inside of
01:19
@XanderHenderson no. No iced coffee for anyone
I'll erase it from existence
Should've just had some cola with my pancakes instead
01:38
@冥王Hades They only did that to make it interesting. We all know he is the Best.
@D.C.theIII is he? I can't recall Ikki ever sacrificing himself like Shaka either
Set of complex number.
02:14
I have been studying complex analysis and visual complex analysis is the best book.
Especially the set theory combined with complex number gives surprising result.
Mods I will not disturb you today I am going back to study. Cya mods I will be actively watching you. Your sincere fan - John Connor
02:43
Do we have a similar "formula" for Stirling numbers of the first kind?
I'm trying to justify changing the order of the summation symbols in $\sum_{n = -\infty}^{\infty}\sum_{m = -\infty}^{\infty} a_{m, n}$, where $a_{m, n} \geq 0$ for all $m, n$. I am aware that some version of Tonelli's theorem allows the interchange for $\sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty} a_{m, n}$. How hard is it to argue that Tonelli's theorem applies in the former case?
Maybe it would be more obvious if I actually understood how Tonelli's theorem applies in the latter case
03:07
Apply Tonelli in the same way.
I don't understand how Tonelli's theorem applies in the known case, unfortunately
I think this would solve my problem, at least it would if I knew how to prove it
 
1 hour later…
04:22
Hello. Where can I see a proof to\[n^n\left(1+\frac1{4(n-1)}\right)\le\sum_{k=1}^nk^k?\]I often see it, but not any proofs. Nor can I prove it myself.
04:56
@youthdoo Seems straightforward by induction. You need to recognize that $e=\lim (1+1/n)^n$.
Professor @TedShifrin hello sir.
How are you?
What’s up?
05:15
The standard condition on ax+by=c to represent the graph of a line is a and b not both 0. Wouldn't it be more general to say ax and by not both 0?
Thus include the possibility of x and y both 0.
No. We’re talking about equations with $x,y$ as variables. If you want the line $x=0$, you take $a=1$, $b=c=0$.
If y is a variable why can't y=0?
(0,0)
5
Q: In the formula $Ax+By=C$, is it true that $A$ and $B$ can't both be zero? If so, why not?

ReliableMathBoyI read in a math book that in the formula $Ax+By=C$, I read that $A$ and $B$ can't both be zero. I think C will also be zero because anything times zero equals zero and on a graph, the x- and y- intercepts will both be zero meaning the two points will be at ($0$, $0$), so then we wouldn't be abl...

05:43
We”re talking about equations of lines, not of points.
$x=y=0$ is a point given by intersecting two lines.
06:41
When do we say a Riemannian metric is/has a finite area?
 
3 hours later…
09:52
How is it derived that $p = \frac{2\pi}{b}$, where $b$ is the factor of $x$ in $\sin(x)$?
The greater $b$ is, the smaller the period, so $p \cdot b = k$ for some $k$, $p = \frac{k}{b}$.
Since $p \cdot b = k$ holds for any $p$ and $b$, take the sine function with $p = 2\pi$ and $b = 1$, thus, $k = 2\pi$ and $p = \frac{2\pi}{b}$.

Like that?
10:19
Given a certain formula, then reasoning by analogy, one can calculate a zero to be 4003/3001 of the analytic continuation of the polynomial $2 x^2 - 2 x - 2$
Is there such a thing as the analytic continuation of a polynomial?
Here is the Mathematica code for the zero:
(*start*)
n = 100;
s = 0;
f[x_] := 2 x^2 - 2 x - 2;
s + 1/n +
1/(1 - Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n], {k, 1,
n}]/Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/f[s + k/n + 1/n], {k, 1, n}]);
N[%, n]
N[4003/3001, n]
(*end*)
I claim to know the zero, not the function, of the analytic continuation of $2 x^2 - 2 x - 2$.
I meant $x^2 - x - 1$ instead of $ 2 x^2 - 2 x - 2$, but the result from the program is the same.
10:40
This is the other example:

(*start*)
(*Mathematica 8.0.1*)
n = 30;(*set n=200 for more digits*)
s = 3/2 + 14*I;
s + 1/n +
1/(1 - Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/HarmonicNumber[10^10000, s + k/n], {k,
1, n}]/Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/
HarmonicNumber[10^10000, s + k/n + 1/n], {k, 1, n}]);
N[%, n]
(*end*)
which gives:
0.4999999999999999999956000526 + 14.1347251417346937904557731916 I
10:55
where the first non-trivial Riemann zeta zero is the zero of the analytic continuation of f[x]:=HarmonicNumber[Infinity,x].
although no analytic continuation is given by f[x] itself.
The seed point s = 3/2 + 14*I is also outside the analytic continuation area where the root is found.
11:33
Regarding Riemann zeta I think I have resolved my confusion now, since:

Clear[x, s]
Normal[Series[HarmonicNumber[s, x], {s, Infinity, 2}]]
returns:
s^-x (1/2 - s/(-1 + x) - x/(12 s)) + Zeta[x]
which has Zeta[x] in it.
11:43
I don't understand CW complex construction in Hatcher's.
12:03
0
Q: Understanding CW complex construction

KoroHatcher's construction of a CW complex is as follows: (1) Start with a discrete set $X^0$ , whose points are regarded as $0$ cells. (2) Inductively, form the $n$ skeleton $X^n$ from $X^{n−1}$ by attaching $n$ cells $e_{\alpha}^n$ via maps $\phi_{\alpha}:S^{n-1} \to X^{n-1}$. This means that $X^n$...

 
2 hours later…
13:53
Could you please explain the last equality? How do you get $X^{n-1}$ in the last line? I think it has something to do with 'as sets' but I'm not quite understanding how. Thanks. — Koro 6 mins ago
I have this confusion because I think that 'attaching maps' changes the 'cardinality' of the set to which we are attaching. Here, we are attaching $D^n_\alpha$ via identification map $\phi_\alpha$ and this identification is happening on the set $X^{n-1}\sqcup \partial D^n_\alpha$, not on $X^{n-1}$ so I don't understand how it can be concluded that $X^{n-1}= X^{n-1}\sqcup \partial D^n_\alpha$. — Koro 58 secs ago
Can anyone please explain these to me? Thanks.
14:04
@MatsGranvik You might find this answer related.
ok I will have a look.
and formula $(11)$ of this answer
In this answer, the expansion is carried out further.
I have more along those lines, if you're interested.
@Ted good morning.
Koro, every point of $\partial D_{\alpha}^n$ gets identified with a point of $X^{n-1}$, namely its image under the attaching map
14:38
@Thorgott yes, I understood that. Thanks.
 
1 hour later…
15:53
Can anyone please help me with this : math.stackexchange.com/questions/4651990/… ?
16:12
here's one direction, showing that alpha + 0* is contained in alpha. fix t in alpha + 0*. by definition of the addition operation of cuts, there are x in alpha and y in 0* with t = x + y. because y is in 0*, y < 0, and hence t = x + y is less than x. because x is in alpha, and t is less than x, and alpha is a cut, t is in alpha.
these are definition chases. the definitions are long and complicated, but there is nothing more to it than that.
there is a "forall" quantifier in the last equation display of your question that should not be there.
16:29
@Franklin you've an equality between sets so proceed showing the two inclusions
hi @leslie
16:49
I saw ant man today. It s
so weird
how can he shrink smaller than quantum object
that means we can't see antman
how are we looking at it
and if it shrink that small
than his density should be conserved
how did those ant carry antman?
and how is he able to breathe when he take out helmet?
is it because ant man is bug movie or is it?
shit I forgot to do proper vsaucd intro
Anyway mods you guys are doing great job thank you for your efforts to have people like me fired :) It makes me heart warmed because of your contribution
have a great day
Roy
Roy
17:07
is there a bound relating the eigenvalues of a matrix A and the normalised matrix A'
where A' is normalised over the columns or rows
17:24
I remember eigenvalues is related to something with scaling I forgot it sorry can't help I am a loser
@Koro modular arithmetic problem for you: math.stackexchange.com/questions/4652035/…
It's an open problem in the sense that I'm unaware of any research into it or any formula for it
17:45
I really hope one day I will be back to do math but I gotta sustain myself for food and also have fun chatting here. It is really good complaining all day. Yeah I gave up math after I know I cannot even buy Ramen and have to save my money for internet to watch meme. What a loser have I become. I could instead answer question in mse but look how worthless have I become. I apologize to do this and not contributing to this community. My sincere apology. Take it or leave it.
Dude i wish i was as smart as Siong Thye Goh God he looks so smart.
18:11
also I am not mocking siong thye goh I am just really fan of his so I renamed myself I hope nobody feels offended and I hope to entertain you guys with my mathematics
18:39
What is that squiggly e
greek epsilon
Thank you!
0
Q: Smash product of X with $S^1$ (Example 0.10 in Hatcher's)

KoroMy first question/confusion is regarding the definition of reduced suspension $\sum X$ and suspension $SX$ of a space $X$. Usually, we denote the quotient space of $X$ with subspace $A$ identified to a point as $X/A$. I understand that suspension of a space is like a 'double-cone': we take a spac...

Can anyone please help me this one? Thanks.
koro i do not envy you :) i took an algebraic topology class, two semesters in length, that used a lot of hatcher. it is a struggle of a book. good exercises, but, there is a lot to do on almost every page.
two semesters?? 👍👍
I'd say that's a good amount of time.
my mid semester exams are over. So I'm trying to understand stuff in my few days holiday.
I expect nothing from the 'teachers' at my college.
18:52
it was a very high tech class, most of it was beyond me.
someone emailed the teacher a question he was stuck at, and the reply was 'it's very easy'.
I see the ubiquitous world famous Paul's Notes have made an appearance on MSE
@Koro well, that's helpful. now they know.
@Koro 😄😄😄.....It is just intuitive..don't you see it?
but the functional analysis teacher here is good to talk to.
he has lot of knowledge and is a very nice guy.
18:55
that's cool. i got into functional analysis because i liked the people who did it. there are definitely some rude folks in the field, but they are few and far between.
@D.C.theIII emailer should have put the head of the department in cc in that email.
I learnt this from my boss(es) at the company I worked at earlier. If one is not doing their job, email them cc to their boss.
then they may get back in line.
Very smart
but sadly this doesn't work here.
they all are one group. 😅
I had done this here for something but nothing happened.
It sounds like your masters is not as enjoyable as you imagined. Sounds like the profs are gatekeeping
It's fine as long as I don't expect anything from professors here.
19:00
😂😂😂 Sensational.
@Koro the thing you say is not true is true
I just go to class for attendance, see what concert is going on in class. If it makes sense, fine, else I just do my self study.
@D.C.theIII I meant if I expect that yeah this teacher here will teach, and he just wastes time, then it would make me sad.
10100001010000101.... (period $p = 7$)
10100101001010010.... (period $q = 5$)
If you or'd these two bi-infinite sequences together, what is the maximum length of any run of $1$'s? Is my question.
So better is don't expect anything from them.
AT least you are adjusting and making the best of things
19:05
1
Q: Local flows and diffeomorphism flow map

MathematicallyInterestedIn Loring Tu's book introduction to manifolds, the definition of a local flow, page 223-224, the following is said: For a smooth vector field $X$ on $M$ and $p$ in $M$, there exists an open set $U$ of $p$ and an $\epsilon>0$ and a smooth map $\varphi: (-\epsilon,\epsilon)\times U \rightarrow M$ s...

Hi everyone
yeah, one of my classmates is very depressed, said that his exams didn't go well and that he'll leave the course etc.
I'm looking for some textbooks for studies after baccalaureate
I said here knowing things and scoring well in exams are two different things. Look, I memorized about 20 solutions in algebraic topology, 2 showed up in exam. I wrote them.
In order to learn after that fast Fourier transform
Any recommendations?
19:07
there were total 4 questions in the exam. Pretty much everyone in the class did the same 😅
I already have a master degree but I forgot everything about maths and physics
Unfortunately that happens sometimes. You don't get to actually "learn" the material for the exam it is just memorize and survive
I want to get a strong background in maths before starting fft
I think now that college is not the best place for me to learn.
The best place for me to learn is while being in industry.
at least I understand what I want now. So that's a good thing.
it probably wouldn't have happened without coming to college here.
$\ddot\smile$
19:45
Can we define smash product using pushout in Top?
Can we define wedge sum using pushout in Top?
not a very well-defined question, nor a useful one, I think
the wedge sum should be understood as coproduct in the category of pointed spaces, the smash product as tensor product
@Koro imagine if the teacher is in the chat
@SineoftheTime I don't care if the teacher is here or not.
20:02
@Koro just kidding :(
20:25
@TedShifrin can I ask you something to an answer you gave on MSE?

It's about the following question:
https://math.stackexchange.com/questions/1469521/curves-with-constant-curvature-and-constant-torsion
I got that $n(s)=C_1\cos(\sqrt{K^2+T^2}s)+C_2\sin(\sqrt{K^2+T^2}s)$, but I don't know how to continue.
Could you help me futher?
20:44
1
Q: Are fields $Q[i\sqrt[4]{5}]$ and $Q[\sqrt[4]{5}]$ isomorphic?

romperextremeabuserAre fields $Q[i\sqrt[4]{5}]$ and $Q[\sqrt[4]{5}]$ isomorphic? I tried to prove they are not simiralry like one can show that $Q[i\sqrt{5}]$ and $Q[\sqrt{5}]$ are not isomorphic because if they were then $(f(i\sqrt{5})^2=f(-5)=-5$ but in $Q[\sqrt{5}]$ there is no element such that it's square is n...

I think that OP's conclusion is correct.
@SineoftheTime :-)
that OP should explain their notation
@user123234 How are you interpreting that equation? What are $C_1$ and $C_2$?
@Koro So what is the correct statement and proof?
20:59
So I mean i solved the differential equation using Euler approach and I left with two constants C_1 and C_2 but normally I would have some "initial problem" to find out what the constants are. And about n(s), I know that it is orthogonal to t(s) and points into the center of the osculating circle, i.e. shows the direction where the curves turns?
@pourjour Math is a huge and vast subject. For FFT you need a bit of complex numbers and linear algebra. It's hard to know what to suggest without knowing your background. Why don't you just pick up some book(s) that treat FFT and read? Strang does them in various of his books.
But you haven't quite answered my question, @user123234. You certainly can start with $n(0)$ as a particular vector, etc. But I wanted you to specify that $C_1$ and $C_2$ must be vectors here.
@TedShifrin ah yes because n(s) is a vector but cos and sin are only real numbers right?
Right.
but do I need to find C_1, C_2 explicitly to finish the exercise?
Why don't you just make an arbitrary choice for $n(0)$, like $(-1,0,0)$ or something?
You're going to have to integrate the Frenet equation(s) to get the curve.
21:13
I don't know why I can take an arbitrary choice for $n(0)$
No one has told you how the curve is sitting in $\Bbb R^3$. You are free to translate and rotate it. You will have other choices to make.
ah, but then I also need to take for example n(pi/2) to be for example $(1,0,0)$ to get C_2 or how do I get C_2?
You should think of specifying $\alpha(0)$, $T(0)$, $N(0)$, not messing with different values of $s$.
because at the moment I know that $C_1=(-1,0,0)$ with your choice of n(0)
Leave $C_2$ alone for the moment and proceed. What should happen next?
21:20
Next I would write $t'(s)=K n(s)$, $b'(s)=T n(s)$ and integrate both to get t and b
Why do you need $b$? You're trying to get to $\alpha$.
ah no sorry it is enough to know $t$ because then I can integrate again and get my curve right?
Right. So you're going to get more constants of integration, and then you can stop to think about them.
why can I stop think about them?
Just work on it.
21:27
@TedShifrin thanks for answering, what's Strang, any books from McGraw for analysis linear algebra and probability that's what I wanna learn?
Any textbooks from university from undergraduate to graduate
Gilbert Strang has written lots of books on linear algebra and applied mathematics.
It's impossible to list textbooks for you. This is nuts.
Why it's nuts?
Because there are thousands of books and we know nothing about you and your knowledge.
@TedShifrin so what I get is that $\alpha(s)=\frac{K}{K^2+T^2}\pmatrix{ -1 \\ 0 \\ 0 }\cos\left(\sqrt{K^2+T^2}s\right)+\underline{C_2}\sin\left(\sqrt{K^2+T^2}s\right)+Ds+E$
That doesn't look right, @user123234.
21:31
oh shit okey.
yet another unexplained downvote. some answers i can understand why someone might have a perspective, but minimisation in reflexive banach spaces does not seem like one of them.
I blame leslie.
Yeah, must be Leslie
Lovely cycle to the Oakland docks today, view of The City is incredible, got caught in a downpour on return.
I haven't been to the Oakland docks in many, many decades.
But I don't see my error, I got that $t(s)=-\frac{K}{\sqrt{K^2+T^2}}\pmatrix{ -1 \\ 0 \\ 0 }\sin\left(\sqrt{K^2+T^2}s\right)+\underline{C_2}\cos\left(\sqrt{K^2+T^2}s\right)+D$
21:34
Nope.
Why are you being careful with only one term?
with which one you mean?
Figure it out.
is $t(s)=\frac{K}{\sqrt{K^2+T^2}}\pmatrix{ -1 \\ 0 \\ 0 }\sin\left(\sqrt{K^2+T^2}s\right)-\underline{C_2}\cos\left(\sqrt{K^2+T^2}s\right)+D$
Take the derivative and find out.
You need to work on your own here and be on top of it.
@TedShifrin in my opinion my t is true now
21:42
@TedShifrin well I have a master degree in signal analysis I found a book called elementary analysis by Kenneth and it's level is fine as a start
21:56
@copper.hat Better than the snowstorms here.....
22:27
@D.C.theIII are you on the East Coast?
Or in LA?
LOL.....at LA........In Toronto
:-) A 72 yo friend stayed with us for a few weeks and just returned to Winnipeg, where it was a balmy -10c on return
but some parts of socal are snowed in atm. my wife's friend's new generator in tahoe had cold weather issues (intake froze) and they were out for a few days
Yeah I've read about the Cali winter weather........and that truly is balmy for Winnipeg....guess I have that to be appreciative of...
22:45
hello, if $f\in L^{\infty}(0,1)$ does $\frac1f\in L^{\infty}(0,1)$ ?
Prove that $\angle BEA=30^\circ$
Any ideas?
$AC=DC$ and $BD=EC$
23:01
@Vrouvrou You need to try basic stuff before asking. How about $f(x) = x$?
23:44
@copper.hat coldest it's been here recently is -1°C (34°F).
that was the low that day, not the high
@robjohn about the same here :-).
@robjohn hope all is well with your recent activity
It's been okay. Got more treatment down the road (hormonal and radiation), but the doctor thinks things are good.
The surgery seems to have been pretty successful.
@Vrouvrou As @copper.hat says, try simple examples. Constant functions; the identity function, maybe an exponential (since it is its own derivative / antiderivative, funny things can happen). You might also thing about how things can fail to be $L^1$: essentialy, there is either a big spike, or a heavy tail.
Since you are on the interval $(0,1)$, there is no heavy tail.
@robjohn good luck with recovery & future
Can you think of an $L^1$ function which has a spiky reciprocal?
What feature of a function would cause it to have a spiky reciprocal?
23:51
i think the question was in $L^\infty$, but your remarks are still useful
Oi... indeed.
But the same basic examples typically work for any $L^p$ space.
The bigger $p$ is, the more spikes matter, and the less tails matter.
Anywho... time to go make dinner.
@copper.hat Thanks!

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