« first day (3912 days earlier)      last day (59 days later) » 
00:00 - 21:0021:00 - 00:00

12:12 AM
@leslietownes i knew it, you two are becoming one
 
this is non mathematical but i had my first pfizer shot today. that felt good. i celebrated with a cappuccino.
 
it was a placebo
the cappuccino, i mean
 
12:31 AM
the cafe was empty obviously so the barista was taking his time doing art in the top of the foam. he drew a mouse. i would have settled for having the drink a little sooner, but he seemed to be enjoying himself.
first time i'd been in a cafe since march 2020. very weird. i also may have been within 6 feet of a coworker. also weird.
 
i definitely have the last century irish/uk perspective on food. drawing on or playing with edible items was asking for a slap on the wrist growing up. never quite got the food as entertainment thing.
gimme potatoes of gimme death
 
there's a certain pour you can do with a cappuccino where you get a fern-like picture. i don't mind that. but this guy was actually drawing on the surface of the foam with an implement. anything to pass the time i guess.
 
12:46 AM
@leslietownes They draw those very fast. I've never learned how, despite having made cappuccini for 50+ years.
 
my sister can do the fern pour but nothing more complicated.
i'll be honest, the guy's mouse was very good.
 
at the risk of offending, i have a gdp per capita vs food as entertainment thesis.
 
how dare you.
:)
 
all this travel and intermingling has spoiled it a bit
 
all these foreigners coming over from ireland and spoiling our native californian culture. i agree.
 
1:01 AM
Drunkards all!
@AndrewMicallef I promise not!
 
started with st brendan the navigator of course
 
 
1 hour later…
2:42 AM
So close dammit
The term on the left must equal zero because the term on the right is the answer
That sounds like a physics level proof to me
 
i've certainly seen worse
improved focus, too!
i can almost see the grain of the paper.
 
(Still the same priblem i have been on all week)
Yeah im trying to work on my photography, seeing as I cant do tex during the day
I feel like the solution is going to be the derivative of psi is zero at infinity...but I dont think I can say so from what has been given to me thus far
@leslietownes ive done that, it feels like cheating. Ferns arent so hard, ita all in the quality of the milk though (smaller air bubbles gives you better drawing resolution)
Also its a sign both of good well brewed coffee and milk to have fancy art (again because of the size of the bubbles)
@TedShifrin have good coffee + well steamed milk, hold the cup at an angle, start your pour on the rim, and swish as you move to the opposite edge
 
i wondered if that was true or if it was just me being snobby. when we lived in michigan we'd go to a place and my main criticism was that the bubbles had the size and consistency of sea foam. or bath bubbles. just all wrong.
 
I feel there is a good description for that with differential geometry....
 
sea foam?
 
2:54 AM
Its almost time for this cinderella to leave
 
one of the double bubble guys was on my thesis committee.
 
No ferns in coffee
Also you probably are a coffee snob for caring about the art in the milk :)
Times up
 
you're a pumpkin now. goodnight
 
@AndrewMicallef Didn't you say that $\Psi$ and $\Psi^\ast$ vanished at infinity? Probably their derivatives do, too.
 
I think it is sin(exp(x)) for the motion
 
2:57 AM
it certainly isn't automatic, but it stands to reason. see if the resource quantifies the decay at infinity in any way beyond 'vanishes'
 
@AndrewMicallef I think that is definitely the case, but I don't know what they've given you to show that.
 
3:15 AM
So compact support? Schwartz class?
 
these things solve a differential equation, right? that ought to help. although i don't know what V is.
 
its physics, it doesn't matter
 
the particles know, that's all that matters.
 
the amazing bezon prime particle
 
 
3 hours later…
6:00 AM
Looking at twin primes through linear independence of characteristic functions of ideals: math.stackexchange.com/questions/4109254/…
$$\tau(n) = \chi_2(n) + \sum\limits_{a,b = 1 \\ \text{rad}(ab) = ab \\ ab \neq 4}^{\sqrt{n}} (-1)^{\omega(ab)}\chi_a(n) \chi_b(n+2) \tag{1}$$
Is the arrived-at characteristic function of twin primes $(n, n+2)$
@copper.hat lol
 
6:16 AM
Looking for a detailed answer (answer eligible for a 500 bounty) math.stackexchange.com/questions/4106641/…
 
@BooleanWick nice question. Does it relate to twin prime research?
I upvoted it
^_^
 
@StudySmarterNotHarder Thanks, I don't think so, my question relates to Riemann zeta function asymptotics and it's closed form in terms of nested sums.
 
Nice, I'm new to zeta functions
and complex analysis
@BooleanWick take a gander at the formula I came up with above
what are your first thoughts?
 
I see, even I'm not much good at complex analysis (especially using residue theorem)
Sure, let me see. (give me couple of minutes.)
 
Great minds think alike (lol)
 
6:22 AM
@StudySmarterNotHarder You really came up with all those formulas?
 
@BooleanWick yes, took about 5 hours of study, but I've messed with those divisor characteristics before so I knew about lcm, etc.
So it took longer than 5 hours, but just today when I started on that description in particular, I ran into many troubles, then it hit me.
 
Your question probably seems out of my league lol
 
It's not
What's a characteristic function of a set?
It's just $\chi_A(n) = 1 \iff n \in A, $ else $0$.
I'm saying, let $A = (a)$ an ideal
so it's iff $a \mid n$.
 
Yeah but your last question, it seems difficult to grasp
 
by definition of integer ideals
Oh linear independence?
Well, I'm thinking about tackling it using simple methods
In the case of $\chi_a$'s linear independence, basically if you have a linear sum of $\chi_a$'s with coefficients in (any field I guess, not just $\Bbb{R}$)
 
6:26 AM
Your lemma 6 and lemma 7, do you have rigorous proofs for them? I'm quite doubtful on your statement of lemma 6 lmao
 
@BooleanWick you just expand the product, just like Euler product formula for Riemann zeta, same thought process
Since you have for instance $(1 - \chi_2(n))(1 - \chi_3(n))(1 - \chi_5(n))$
 
I see, also what's the significance of your work with twin primes? does it give some sort of function to evaluate them?
 
Each prime occurs uniquely as a factor of the subscript
That's the big question. It's hard to prove twin primes, but this is certainly related, since it gives a characterstic function formula in terms of something else
 
Hmm I think I get it now, I might be able to contribute to your question after spending significant time working on it lol
 
It gives a function $\tau(n)$ such that $\tau(n) = 1$ iff $(n, n+2)$ is a twin prime pair, else it equals $0$. So it's technically a char. function for the set of twin prime pair firsts.
 
6:30 AM
Yeah I got it when I read it for the first time lmao
 
first $(p,q)$ = $p$.
Well, it doesn't give you a formula for computing the $n$th twin prime, but that's not neccessarily a requirement to prove infinitude of them
 
I don't think so if there's any formula which gives us nth twin prime (there might be a few but they probably give approximations like PNT and it's extensions)
@StudySmarterNotHarder btw where are you from?
 
San Diego right now
But from all over US and two years in Thailand when I was in first and second grade
 
Ah nice, I'm from India.
 
Nice!
That's the best place to be from because Buddha lived there.
Its a Holy Land.
 
6:35 AM
haha yeah lol (not better than US though)
 
We bombed your bay, that's why it's called Bombay, so no... we're not proud of such atrocities.
We melted 100's of thousands of people in Japan, in an instant
That's probably worst thing that ever happened on Earth
Not melted, vaporized. Then we went on building megaton warheads like it was okay
 
You're in college? (If yes, then which?)
 
Nope, but I did go to NAU (Northern AZ university) and ASU at a time.
But I'm going back to school (Open University, online) soon
 
I see, I'm planning to go to US uni's for my undergrad
next year for class of 2026 (I'm a junior right now)
Please go through my question once https://math.stackexchange.com/questions/4106641/closed-form-of-2-sum-n-0m-1-1n-zeta4m-2n-1-zeta2n-2

I have placed a 500 bounty on it. Thanks.
 
Would I be asking for too much if I ask for faster mathematical techniques in this group?
 
6:45 AM
Nah it's completely fine. go ahead
Just make sure that your questions are worth everyone's time (that is they shouldn't be too elementary)
 
I am actually preparing Quantitative Aptitude...I would like to have the questions collection in a single place..It will be extremely difficult for me to have a recourse to all the questions at a single place. So I am thinking of creating a group..If there are a few more people who would agree to assist me there that would be extremely generous of one and all..
@BooleanWick Are you ok with the proposal..Would anyone like to join him..
 
I'm not getting what you mean by ~Quantitative aptitude~
 
Plain and simple mathematics..In a separate name.
 
What do you mean by group, do you mean chatroom?
@RajorshiKoyal
 
Yes I do..I created one.Please join if you wish to.
 
6:50 AM
send me invite first lmao
 
I am an unemployed person preparing for some job entrance exams..
So need some assistance.
 
how come you're unemployed being a math enthusiast?
 
COVID is driving me crazy..Anyways..
 
You're from India?
@RajorshiKoyal
 
6:53 AM
which part?
 
Please take an entry into the room I created @BooleanWick and @StudySmarterNotHarder
 
also where did you go to college? (unrelated)
 
I am not able to sent you a request..
@BooleanWick Jadavpur
@BooleanWick Kolkata
And what about you?
 
ISI?
High school student still
 
You are from ISI?
ok..
 
6:54 AM
In my final year
No
I'll be going to Princeton University though
 
Which place?
 
USA
I can't join your room, you need to send a request, you know how to right?
 
Not really..
 
Click on my icon and then you can see the invite user option
in the actions tab you can see the option
 
7:36 AM
I'm not sure, I'm going through the text once more, trying to see if I can glean anymore info know I have tried to solve the problems.
For one thing there is this sentence which I don't know how to parse:
> "Physically realizable states correspond to the
square-integrable solutions to Schrödinger’s equation"

I also know that $\int\limits_{-\infty}{+\infty}{|\Psi^2|}dx = 1.$
Oh and there was this footnote that I overlooked. I think this helps me!
> A competent mathematician can supply you with pathological counterexamples, but they do not arise in physics;
> for us the wave function and all its derivatives go to zero at infinity.
Done!
@leslietownes Also I thought more about coffee pouring, and I now realise the reason you tilt the cup is not to pour the milk under the coffee surface layer, but actually to maximise the canvas that you are working on. When you tilt the cup back the image get's squished into the bounds of the smaller circle. So you can get more fancy details in your fern image
@robjohn Yeah robjohn, it looks like that was the case, I should have paid more attention to the text after all
 
8:34 AM
In Geometrical transformations, we can transform by either transforming geometrical objects or transforming coordinate systems .According to my book , Euclidean transformations come under transforming geometric objects, but $Euclidean transformations$ are $translation of axes, rotation of axes, or reflections$ so, this should be under transforming coordinate systems, isn't?
Euclidean transformations
are translation of axes , rotation of axes , or reflections *
 
@Rover why are you adding $s in your comments? They really make things look weird with ChatJax running.
@BooleanWick If you look at their profile, you should see a list of the rooms they own. You should be able to click on one and go there.
I don't think they can create a room that you cannot go to.
 
9:03 AM
@leslietownes You explained it very well
I came with same contradiction by the definition with constant function
 
@robjohn Thanks, btw did you go through my question?
 
@BooleanWick I am looking at it now
 
@leslietownes Fogot the say thank you
 
Thanks, just to make sure, you're looking at this one right? math.stackexchange.com/questions/4106641/…
@robjohn
 
yes
 
9:11 AM
@robjohn Thanks. What's your intuition though? Does it have a nice closed form?
 
Dunno, but I can give you the sum for any $n$...
I know its generating function
 
Cool, you can just answer the question once you're done with it (you can also just give the generating function but a closed form would be highly appreciated. Thanks.)
 
@TedShifrin I got it! The epsilon doesn't tell me anything about the x because when my epsilon is near the that 1^2 I have two values. That's why they use delta inequality implies epsilon inequality.
 
@robjohn Also if you get the closed form then also put the generating function in the answer.
 
@TedShifrin Thanks for the example.
 
9:17 AM
@robjohn Do you have any idea on how to prove @Cluade's claims (in his answer)?
 
9:41 AM
Hello! Can I ask if 'area under the curve' applies to net signed area?
 
@soupless yes
 
Oh, okay
Thanks, @robjohn
I was confused because 'under' will be 'above' if the function is less than zero
 
@BooleanWick I'll have to get back on that. I noticed that it's odds times evens. I think at least one of the other answerers thought as I did that it was evens times evens (or he would not have been talking about the Bernoulli numbers)
I think both may have now that I look at Claude's statement
@BooleanWick were you really looking for a closed form, or an asymptotic approximation?
 
Hey! Can someone help answer a basic doubt related to provability in logic?
 
10:15 AM
@robjohn @robjohn I'm looking for a closed form but after Cluade's answer I changed I also added asymptotic approximation, did you get the closed form?
I mean what did you get so far?
 
@BooleanWick No. I, and I think the others, were thinking of $\zeta(4n-2k+2)$ instead of $\zeta(4n-2k+1)$ in the sum. I am working on that now.
I can see how Claude remembered that answer assuming he was thinking about $\zeta(4n-2k+2)$
and the other answer with Bernoulli numbers was definitely thinking about $\zeta(4n-2k+2)$
 
No actually the other answer wasn't, If you read the other answer carefully then it's $\zeta(4n-2k+1)$. I'm not sure about Claude's statement coz he didn't give any reasoning or justification to his claims.
@robjohn
 
10:35 AM
The (underdamped) Langevin equation describes the position and velocity of a particle under an external force, friction, and a diffusion. In the introduction to this paper see equations 5a-5c they introduce a third coordinate, they call this a "coupling to a heat bath", does anyone know an introductory source for this kind of thing... im confused
thats a math paper...
 
@robjohn you there?
 
@BooleanWick No, pisoir's answer definitely does NOT consider even zeta times odd zeta
they are only considering even zeta times even zeta
There is no nice formula like that for odd zeta
 
Alright, so you're working on even zeta time odd zeta?
 
yes
The hard part is that the generating function for $\zeta(n)$ is $-xH_{-x}$, and that has a nice even part, but not so nice odd part
 
why're we concerened with the generating function though? I mean why do we need it?
 
10:51 AM
Because the generating function of the sum you want is the square of the generating function for $\zeta(n)$
or can be derived simply from that
 
Hi guys :)
I came across the definition of an open set Ω having locally finite perimeter if $\nabla\textbf{1}_{\Omega}$ is a locally finite $\mathbb{R}^m$-valued measure. Is the notation with the $\nabla$-sign something that mathematicians are usually familiar with? Not sure, if this is specific to the material that I read or not. I have very much of an engineering background for that matter.
 
wait, really?
@robjohn should we talk about the problem in a different room? maybe personal room?
 
 
1 hour later…
12:10 PM
if a norm of a function converges to anything but 0, does it mean that the function doesn't converge in this norm?
 
12:28 PM
you mean a sequence of functions? L_2 norm, or euclidean norm of function evaluated at $x$?
 
$\sqrt{\sum_{i=1}^n(x_i-y_i)^2}\leq\sqrt{\sum_{i=1}^n\max_{i=1,\dots,n}(x_i-y_i)^2}\\
$ Can someone explain why this inequality is true?
 
@Flows. Yeah, I got a sequence of random variables, which are functions. I calculated the limes of their L2 norm
which is euclidean distance
and it converges to 1
can I say that this function (random variable) doesn't converge in L2?
 
12:47 PM
ok, sorry, that was a stupid question
 
-5
Q: How is value of slope = 1 here?

Srijan M.TSo , I are going to represent equation of boyles law. I will provide the math . It’s ok if you don’t know Chem. So , PV = nRT ( Ideal gas equation ) y= p , x=V and nRT = K. Yx = K Therefore , graph of hyperbola we got. Let us do Y= K/x Now , Putting log on both sides. Log P = Log K + Log(1/V) No...

 
1:06 PM
@MadSpaces each term in the left sum is no bigger than the max of all the terms, which is each term in the right sum
 
wklm if a sequence of vectors converges in a normed space, then the sequence of norms is a convergent sequence of real numbers. depending on the space, you may be able to rely on a weaker form of convergence than norm convergence + convergence of the sequence of norms to deduce norm convergence.
not quite what you're asking, but related.
 
Pls check my Q too.
 
srijan, i'm not sure i understand the question. if P = constant/V, then log P is a linear function of log(1/V) with slope 1, or a linear function of log(V) with slope -1. it's just the properties of the logarithm. why P has that form is a matter of chemistry, i presume (which may well depend on additional assumptions, e.g. T being held constant)
to put it another way, it's not a situation where you're fitting a linear model to some data. the assumption of the law is that P actually is that
i got the first vaccine shot yesterday and my arm really, really hurts. i guess i know it's working.
 
2:07 PM
Hi guys im reading some lecture notes : "Open Quantum Systems II The Markovian Approach". The first line of the introduction says : "Open systems are usually understood as a small Hamiltonian system (i.e. with a fi- nite number of degrees of freedom) in contact with one or several large reservoirs." Can anyone give me a model example to supplement this statement, specifically what are the degrees of freedom and what are reservoirs ?
The Hamiltonian system I imagine as a distinguished particle following Newtons Laws.
 
2:21 PM
hi, if I have the following graph (1-dimensional CW complex) imgur.com/a/cfOCe4S , with the boundaries of the circles identified according to the arrows, is my space homotopy equivalent to a wedge of two circles?
or do the identifications somehow change that
and another question, if I fill in the interior of the graph , does the resulting space deformation retract onto the graph ?
i.e. does the disk with two holes , and with the same identifications of the boundary of the disk and the boundary of the holes, deformation retract onto the graph above?
im asking because it shouldnt be possible, but for sure without those identifications the disk wtih two holes deformation retracts onto that graph (without the identifications), and im not sure why we cant use the same deformation retraction just applied to the quotient space
 
@porridgemathematics you can take the inner two circles out of the picture and nothing changes
 
right, but then why doesn't the disk with two holes with those identifications deformation retracts onto that graph?
 
@porridgemathematics I don't know what "interior" is, but a disc doesn't retract onto its boundary circle and that should answer the question in the negative
 
like why can't you pull radially inwards from the center of each hole, and leave all the points on the boundary in the same place
yeah i know a disk doesnt retract onto its boundary circle
oh okay because with the identifications that retraction isn't continuous
yeah I see it
but without those identifications that retraction is continuous
 
2:38 PM
yeah, that sounds correct
 
i guess its a good lesson for me, quotienting things requires a lot of care
 
belt and suspenders
 
i guess what confused me a little is that CW complexes are locally pretty nice, e.g. every subcomplex has a little neighbourhood that deformation retracts onto it
but what i was trying to do was too global in a sense, and the quotient structure said no, pretty much
 
the problem is that too many things are representable by CW complexes. nothing can be too nice globally about them or math would be too easy. and we live in a hostile universe.
 
2:48 PM
math.stackexchange.com/questions/4109732/… the pruefer p-group raises its ugly head. for the second day in a row.
 
another sort of related topology question, if I know that $f : X \rightarrow Y$ is a homotopy equivalence, and $f(A) \subset B$ for subspaces $A \subset X, B \subset Y$, and I know $f \restriction A : A \rightarrow B$ is a homotopy equivalence, is $f : (X,A) \rightarrow (Y,B)$ a homotopy equivalence?
i.e. isthere a map $g : (Y,B) \rightarrow (X,A)$ s.t. $f \circ g$ is homotopic to $id_{Y}$ through maps of pairs $(Y,B) \rightarrow (Y,B)$
(and similarly for $g \circ f$)
 
take A=B=point
then you're asking whether a homotopy equivalence is automatically a pointed homotopy equivalence, the answer is no
you get this from an example of a contractible space that does not deformation retract to a point
which you may or may not have seen in ch0 of Hatcher
 
@Flows. that's an SDE for the noise-term
 
3:06 PM
Many people thought the idea of complex numbers is silly but still it helped in the sweeping development of mathematics blyat
 
i still think it's silly, but i'll use them.
 
@Thorgott ah okay, that makes sense
 
Why do you think it's silly
 
this came from the question of whether these conditions mean $$f_{\ast} : H_n(X,A) \rightarrow H_n(Y,B)$$ is an isomorphism
 
any1 good with representation theory of finite groups?
wanna talk about cliffords theorem proof
 
3:13 PM
it's not silly, i take that back. real numbers are silly. nothing squares to -1? give me a break.
manolis a question would be welcome. i would not volunteer myself as 'good' at anything, but might have something to offer. or not. only the question itself will let me know.
 
Everything is silly
 
i would like to go through the proof of cliffords theorem in Gordon James & Martin Liebecks book
 
i do not have this book, but others may have.
man, my arm hurts.
 
@porridgemathematics they do, but you have to argue differently
 
hm, im trying to prove that $f_{\ast}$ is injective, but im probably doing something stupid, if $\alpha$ is a $A$-relative n -chain, and $f \circ \alpha $ is zero in $H_n(Y,B)$ then does that mean $f \circ \alpha = \beta + \partial b$ where $\beta$ is some $B$-relative n-chain and $b$ is some $n+1$ chain?
is that how im supposed to argue
 
3:27 PM
@robjohn ya , it's looking weird, but I thought it may help in highlighting words and I edited but some part was too late to be edited..
In Geometrical transformations, we can transform by either transforming geometrical objects or transforming coordinate systems .According to my book , Euclidean transformations come under transforming geometric objects, but Euclidean transformations are translation of axes, rotation of axes, or reflections so, this should be under transforming coordinate systems, isn't?
 
rover this is a well known thing. i forget the nomenclature. whether you operate on the thing itself, or the labels that describe the thing. it can be either way. focus on the formulas and just do what they do.
 
$\beta$ is a chain in $B$
but yes
fwiw, this fact is completely algebraic in nature
 
ok suppose i have a e=< x,y> where x is the character of module X and y irreducible the character of a module Y where y is also a constituent to x . Why there exist a submodule of X with character ey ?
 
oh right, $\beta$ is a chain in B
 
if you know the five lemma, use it, if not, prove this by hand, then look up the five lemma and convince yourself you've accidentally proven it
 
3:31 PM
oh i do know it
wow the five lemma makes it really clear
all the vertical maps are just induced by $f$ so everything commutes
and the rows are from the LES of a pair
 
yuh
pure homological algebra
 
i still wanna write it out in terms of the chains for some reason
although of course the five lemma makes this super clean and effortless
brilliant solution
 
@porridgemathematics you'll get over that
 
@Rover It would be better to use * italicized * or ** bold ** text for highlighting purposes.
 
3:46 PM
does stackexchange chat support strikeout? that's the one i most want to use.
a commonly neglected form of expression.
 
I think it does not
 
It could be bolder, though; that line is pretty weak
===test=== I guess not
 
it also seems higher than it ought to be. on my screen, anyway. i'm viewing in chrome.
 
it's probably the same height and is right for capitals
STRIKEOUT
----test---- darn, that worked for a second and then reverted. It looked darker and then got remarked up
 
3:49 PM
i'll just type in all capitals from now on. problem solved.
 
how did u do dat
QnllIGJ5ZQ==
 
hm, it seems pretty unclear without the five lemma, once I have $f \circ \alpha = \beta + \partial b$ then I get $f \circ (\partial \alpha) = \partial \beta$, which tells me that $\partial \alpha$ is the boundary of an $n$-chain in $A$, so $\partial(\alpha) = \partial(c)$ for $c$ a chain in $A$, but now I don't know how to see that $\alpha - c$ is a boundary..
 
is this measures stationary for the hamiltonian system?
ignore that^
 
4:05 PM
@Euler2 QmUgd2VsbA==
 
@porridgemathematics well, you will have to use the LES at some point
 
@robjohn TG9yZW0gaXBzdW0gZG9sb3Igc2l0IGFtZXQsIGNvbnNlY3RldHVyIGFkaXBpc2NpbmcgZWxpdCwgc2VkIGRvIGVpdXNtb2QgdGVtcG9yIGluY2lkaWR1bnQgdXQgbGFib3JlIGV0IGRvbG9yZSBtYWduYSBhbGlxdWEu
 
none of you pass the turing test
 
Tm9yIGRvIHlvdQ==
 
brb donating to Leakys bitcoin wallet
 
4:17 PM
-- .-. -.. .-- -.-- ..... -.-. ..--- -.- -. -... -.. .- -.-- - --.. .. .--- ...- --. -.-. .-. ..--- --. --- ...- -. -.. . ...- .-.. .... -- .--- -- --. .-- --.. ..--- ..--- -.- ....- ..--- .-- ..- -.-- ...-- --- -. .-. ...-- .-- .. .-. ...-- -- --- --.. .-. --. ... --.- - ..--- -- ...- -- . ....- -- -.-. ..--- -.- ....- -.-- -.. ..--- -...- -...- -...-
 
oh for heaven's sake.
as i say to my daughter (who is two and a half, what excuse do YOU have?), you might be having too much fun right now.
none of you are people, you're all bots, you're my CPU come to life. F- on the turing test.
 
@leslietownes Wikwqi qk a ccm
Really, Wikwqi qk a ccm
 
nice try, computer
 
Dlme ct
 
what can i help you with today?
 
4:33 PM
global thermonuclear war please.
 
Dmdwg gwhpff
 
he's speaking in welsh for some reason now. i don't know. the AI has gone off kilter.
 
we have a special of the day, defcon 1.5 for $999.99
 
DLME CT
Well bye bot has to power off
 
goodnight or morning or whatever it is, wherever you are.
 
5:00 PM
@leslietownes how about a nice game of chess?
 
open the pod bay doors, hal
 
@LeakyNun U2VtcGVyIHViaSBzdWIgdWJp
@leslietownes I'm sorry, Dave. I can't do that.
 
 
2 hours later…
7:10 PM
good afternoon
 
do u guys have electric bikes?
 
7:56 PM
i have an ebike
 
i don't
 
just wanted that bit of knowledge or any more questions :-)
 
i have an electric toothbrush
if that helps
 
8:25 PM
it helps a lot
 
I use pedal power
Whats up with your bike though?
Did you forget to charge it?
 
you can borrow my skateboard
 
@HereToRelax is the problem wind resistance? I have a spreadsheet for calculating battery lfe of a ebike under various conditions
 
8:41 PM
I was considering buying one
I don't have one
 
i have a folding e-bike. if you get similar, do not get one with wire spokes. get one with 'solid' spokes if you get my drift.
 
I dont have an electric tootbrush either
Yeah I understand that
 
in the first year with mine there were 5+ warrantee issues, mostly the throttle lever breaking, but also the batter had to be replaced.
 
did a wire one break?
 
many times.
 
8:43 PM
Oh, that's great advice thanks
 
i need to replace them at the moment, its just a pain because of covid.
really depends on your need. i used to take mine on the train from berkeley to santa clara many times.
 
I would drive it like 3 miles a day on roads
pavement roads
 
perfect application. i like to ride my normal bike, but if i am formally dressed do not want to get them dirty plus i sweat easily. e-bike is perfect for that.
 
Also 3-5 kW are more than enough. (Any more is probably dangerous overkill bordering on motorcycle territory)
 
i bought an e-joe epik se a few years ago, around $1.6k if i remember. i would not get a folding bike unless it was necessary. (if i knew what i know now i would have bought a magnum premium 48 or whatever the equivalent was back then).
 
8:59 PM
Did you bend a wheel @copper.hat?
 
00:00 - 21:0021:00 - 00:00

« first day (3912 days earlier)      last day (59 days later) »