it was not feeling bad, it was questioning the fabric of reality, my place in the world. the logical connection between the social structures that gave birth to my understanding
it was the birth of a revolutionary passion
which would have burned for years to come
but thankfully it was just a minor mistake and the fervor quickly went away
I felt so bad that day that I couldn't watch where I was going and walked into the closed door of a manga store slamming my face against it and hurting my nose
@TedShifrin why's this such a recurrent pattern? When I look at my own lab report and writing, as well as my peers' who are also STEM majors, its usually pretty mediocre at its best. Then you look at the written work on language or literature majors and its completely out of this world
@TedShifrin yeah it comes back to bite me later on, when I don't write down proper steps and a bit of explanation on what I did, especially in math problems and while writing code (i.e. leaving comments)
I look at code I wrote a week ago and wonder, why did I do this?
@leslietownes out of this world compared to what I read on the daily. I mean, of course its still nothing compared to, say, a manga I'm reading. But that manga is written by an extremely gifted author so, that's an unfair comparison I think
You call showing $\frac{x^p}{p} + \frac{y^q}{q} \geq xy$ a challenge problem?....I laugh at thee....I do actually see why you would make $xy = c$ a constraint....interesting...
Find a filtration $1=M_0 \leq M_1 \leq \cdots \leq M_{n(n-1)/2}=U_n(\FF_p)$ so that $M_i \trianglelefteq M_{i+1}$ and $M_{i+1}/M_i \tilde{=} C_p$ (the cyclic group of order $p$). Any hints.
@leslietownes plenty of restrictions, but I was feeling myself and got carried away........now the next question which involves geometry has me quite humbled on the other hand.....
I'm not even sure why I referred to that. It's a very obtuse reference. At the end you need to think about the geometry of angles inscribed in a circle.
But, yes, the geometry in 1.2.14 is one of my favorites, and it does show up when you want to know the polar coordinates equation of a circle like $(x-a)^2+y^2=a^2$.
So working on the quadrilateral question I've arrived at some intermediary questions that I have to work out to gt to the final result. Right now I have to find the quadrilateral which provides maximum area. Just based on experience I'm assuming it is the square, but I want to prove this.
Right now I'm thinking of an idea of showing that the area of the triangles from a square would be larger than the area of triangles from any other quadrilateral. But I'm having trouble relating the lengths of all quadrilaterals.
What condition on angles is necessary and sufficient for a quadrilateral to be inscribed in a circle? Why? ......Clearly it must be that the angles sum to $2\pi$?
@peek-a-boo thank you i understand now. I got confused because I thought the table was talking about the D.E. decreasing/increasing. Makes more sense now
what happens to opposite angles in a square (which you know can definitely be inscribed in a circle)? taking a huge leap of faith, what would you dare to guess?
@TedShifrin So that is how to characterize the constraint. I'm going to have to ponder more on the reasoning of it, but it makes sense. Now I have to figure out the area of a quadrilateral using the idea of angles.
this is perhaps a not well enough defined question, but is graph theory useful in finite degrees of freedom settings?
say I want to map connections between an infinite number of nodes (for simplicity represent each node by a distinct natural number); does graph theory have the machinery to meaningfully work in this situation?
for some people range and image are synonymous and there is no difference
for these people, "codomain" is sometimes used for what you might be thinking of, roughly, the declaration of the set that the function is regarded as a function to
can you say what you mean by "range"? do you mean codomain, or do you mean the set {f(x): x in domain}? i'm not trying to be annoying. both definitions are in use, and the first is fairly common in a lot of calculus books
the set {f(x): x in domain} absolutely does depend on the domain
figuring out the range of a function, in rudin's sense, can involve a whole lot of work that might be unrelated to what you want to do with the function
for continuous functions from [a,b] to R for example, it requires computing the maximum and minimum values of f on [a,b]
many people find it preferable not to build that into the definition of a function, and use the codomain concept instead
so you can say "here is a function f from [a,b] to R, and its range is [something that might be smaller than R]"
instead of "here is a function on [a,b] and we just don't know what its range is until we do a bunch of work"
no. it is to separate the concept of "the set that you define as part of defining a function, that you regard the function as taking values in" from the concept of "the set of the values taken by a function on its domain"
it is to say "hey, it might be useful to distinguish these"
instead of to say "we only have one concept, range, and codomain is redundant"
again, whether you care at all about any of this depends on what you're actually doing or hoping to do with a function
which may be why calculus books don't always introduce the concept of codomain or distinguish it from the range
i know that this question has been asked and answered many many times on this site :P but i've read several answers to no avail. does anyone have an explaination for rudin's proof of theorem 2.12 (countably infinite union of countable sets is again a countable set)
I don't understand at all how to create a bijection from $\mathbb{N}$ to the sequence (17)
and also how to understand if this proof is rigorous or not :P it doesn't really seem that rigorous at all
I'll try to make something that's closer to what you want. Give me a little while, I'm having a slow brain day...
We're doing a triangular scan through that array, and we can do that via triangular numbers, $T(n)=n(n+1)/2$. The core trick is to invert that quadratic.
Rudin's saying we can make a bijection from the naturals to the cells of the array, by traversing it on those diagonals, since we know that each diagonal has finite length. But we may need to drop some cells because they contain elements that we met earlier in the scan.
I was talking about this yesterday when I said that Cantor used a simple diagonal pattern.
@SillyGoose What's wrong with it? Dropping cells is ok because it can't increase the size of the set. But if we were adding items to the set, that would be bogus, because it might make the set uncountable.
Do you want me state the definition of contravariant functor that I am referring to above?
Suppose that $C$ and $D$ are categories. Suppose that $T: C\to D$ is a function which does the following: 1) For every c in obj C, Tc is in obj D. 2) If f is a morphism in Hom (a, b), a,b are in obj C, then Tf is in Hom (Ta, Tb). 3) Suppose fog is defined, f is in Hom (a,b), g is in Hom(c,a), then T(fog)= Tg o Tf, 4) identity goes to identity. Such T is called a contravariant functor.
3) does not appear in the definition of a functor.
Like it requires c=0 at x=0, but why is that considered a singular solution and not particular? In general are singular solutions specific values of x,c, and particular solutions are functions with specific values of c?
pndas: for particular examples of normed spaces you might be able to give an example without the axiom of choice (or something equivalent) but for an arbitrary infinite dimensional normed space you will need something like AC
pndas: its equivalent to showing the existence of a nonzero linear functional on the space that is not continuous (given such a functional, f, then ker(f) turns out to be a maximal subspace that is not closed, and given such a subspace, S, the canonical map into the quotient of your space by S turns out to be such a functional)
sure. using the axiom of choice as needed, convince yourself that there's a countable linearly independent subset (c_n) of unit vectors in the space and a functional f satisfying f(c_n) = n for all n
@leslietownes I'm sorry. But I don't see why the existence of non-zero unbounded linear function would imply that ker(f) is not closed. Does ker(f)= closed imply that f is continuous?
pndas: the main site has all of this stuff (you do have to watch out for answers that unwittingly use finite dimensional results that dont hold in the infinite dimensional setting). see e.g. math.stackexchange.com/questions/2327189/…
you're not wrong to be worried about it, though. there's an analogous technique for partial differential equations (i.e., more than two variables), and there you would get more than one integration constant
there actually is one sorta weird thing with this example: @PNDas is right to point out that, for isntance, $y=-e^x$ is a solution
but in that case we'd seemingly need $\log y=x+\log(-1)$
i forget what the best way to deal with that is tbh
@leslietownes yeah, though that can only really bite you when you want to do an integral from negative to positive y. which you shouldn't for the $\int dy/y$ example
@Obliv well, lets consider the case of interest to see what goes wrong
ignoring the singular solution for now, you correctly had that $2y^{1/2}=x^2/2+C$ is the general solution
suppose we evaluate this equation at $x=0$. if we write $y(0)=y_0$, then the equation becomes $2y_0^{1/2}=C$. so you can write $y^{1/2}=x^2/4+y_0^{1/2}$.
if you now try to take $y_0\to 0$ from above, this is $y^{1/2}=x^2/4\implies y=x^4/16$. but the singular solution $y=0$ has $y(0)=0$ as well
so the "general" solution doesn't include $y=0$
which is to say, yes: your singular solution is one which cannot be obtained by varying the integration constant
even not-very-good ODE books will sometimes slip examples of this kind of stuff around their statement of existence and uniqueness results, even if they don't include the proofs of those results
continuing: if i require $r=R$, $v=0$ at time $t=0$, this yields $C=-(2/3)R^{3/2}$. hence $v^2=4/3(r^{3/2}-R^{3/2})$. in particular, as $R\to 0$ this is $v^2=(4/3)r^{3/2}$
Did some reading on the quadrilateral problem @TedShifrin and I think I almost have the setup. So given our fixed sides I was going to use Bretschneider's formula as the function to maximize. But I'm having trouble creating a constraint. What I expect to happen is based on the correct constraint I will get values $\theta$ and $\gamma$ for my opposite angles and those values I could sum together to show that the quadrilateral of maximum area is inscribed in a circle.
In geometry, Bretschneider's formula is the following expression for the area of a general quadrilateral:
K
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
(
s
−
d
)
−
a
b
c
d
⋅
cos
2...
So to at least explain how I came to that idea. I was asking myself "if a general expression for the area of a quadrilateral exists?" and so happens this did that corresponds to my opposite angles
