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.
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.
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.
@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.
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 $\Bbb{R}$)
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
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.
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)
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 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
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 *
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.
@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?
@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?
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.
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
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.
So , 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...
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.
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.
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.
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 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
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.
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 $(Y,B) \rightarrow (Y,B)$
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.
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?
@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.
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 ?
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..
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.
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).
