given $y' + x^{-1}y = 1$, since it is a first order linear DE, we can solve it using the integrating factor $e^{\int x^{-1} \ dx}$. the lecturer says it is $e^{\ln x}$ (and this works). why not $e^{\ln x + C}$? or should i just assume we've set $C = 0$?
answer: you can just cancel out $C$ as soon as you multiply the equation by the integrating factor
yeah. the way to think about it is that the integrating factor in that method just needs to be some antiderivative of the thing in that method. you don't need the most general one, or any particular specific one.
which is maybe somewhat at odds with how something like "$\int P(x) \, dx$" is usually interpreted, but, thats what it's standing in for, in that formula.
i'm procrastinating entering my time for march into my work's timekeeping software. those dollar signs gave me one more half second of procrastination.
dc: i think the only people who do that at my work are very old and sufficiently inactive that their secretary can easily keep track of how they spend time.
Show that for a space X, the following three conditions are equivalent: (a) Every map $S^1→X$ is homotopic to a constant map, with image a point. (b) Every map $S^1→X$ extends to a map $D^2→X$. (c) $\pi_1(X,x_0) = 0$ for all $x_0 ∈ X$.
I didn't know until now that it's holiday here today. :)
So for (a) to (b): Let $f: S^1\to X$ be any map. It is nullhomotopic by (a) so there exists a homotopy $f_t:S^1\to X$ and a constant map c from $S^1$ to X such that $f_0=f, f_1=c$
I define $F:D^2\to X$ as follows: $F(x)=f(x)$ if x is in S^1 else F(x)=c(x).
@leslietownes Because if the interval is from $[a,b]$, we can always find a number that is halfway between each successive interval? $a+\frac{a+b}{2},a+\frac{a+b}{2}+\frac{a+\frac{a+b}{2}+b}{2},...$
koro: what if you do a kind of polar coordinates thing. identify D with complex numbers of mod <= 1, and C with the unit circle and do F(r e^(it)) = f_r(e^(it)), where i'd choose f_0 to be the constant map (so this is well-defined, i.e., independent of t, when r = 0) and f_1 to be f, or otherwise fiddle with the formula.
The true meaning of Jordan Holder theorem is a well definedness of a finite group $G$ to a power set $\mathcal{H}$ of finite simple groups up to isomorphism.
In mathematics, a hyperelliptic surface, or bi-elliptic surface, is a surface whose Albanese morphism is an elliptic fibration. Any such surface can be written as the quotient of a product of two elliptic curves by a finite abelian group.
Hyperelliptic surfaces form one of the classes of surfaces of Kodaira dimension 0 in the Enriques–Kodaira classification.
== Invariants ==
The Kodaira dimension is 0.
Hodge diamond:
== Classification ==
Any hyperelliptic surface is a quotient (E×F)/G, where E = C/Λ and F are elliptic curves, and G is a subgroup of F (acting on F by translations). There are seven...
It's different from the hyperelliptic surface I know in Riemann surface
We may assume that each $f_j$ is of degree $3$? $f_j$'s are meromorphic functions, in particular degree = 0.
I think it should be the order of poles is 3 counting multiplicities.
meromorphic function with simple pole cannot exists on compact RS so it starts with two poles. If $f_j$ has two poles then $D_j$ can be chosen to be degree $2$ so $M$ is hyperelliptic.
Oh degree here is not a map from divisor class to Z... it's a degree as a map. so the same thing
i dunno if anyone calls it a 'theory,' results like the ones mentioned on that page are often what people somewhat ambiguously refer to as 'the peter-weyl theorem'
i can think both of reasons why you might include it, and reasons why you might not include it, alongside other stuff in functional analysis, so, who knows. is it in a textbook that the instructor likes?
So our syllabus on college website includes compact operators and there is not much detail available as to what extent the operators are to be covered.
@leslietownes No, he does not follow any textbook.
Also, please note that this is an introductory functional analysis course so I think that the course structure should follow Kreyszig's or Conway's.
In algebraic geometry, a hyperelliptic curve is an algebraic curve of genus g > 1, given by an equation of the form
where f(x) is a polynomial of degree n = 2g + 1 > 4 or n = 2g + 2 > 4 with n distinct roots, and h(x) is a polynomial of degree < g + 2 (if the characteristic of the ground field is not 2, one can take h(x) = 0).
A hyperelliptic function is an element of the function field of such a curve, or of the Jacobian variety on the curve; these two concepts are identical for elliptic functions, but different for hyperelliptic functions.
== Genus ==
The degree of the polynomial determines...
One day he 'taught' something in class and then at the end, he said-ah, I confused this class with another class. This is not related to your syllabus.
@TedShifrin I should have said it correctly. Instead of saying loop at $\sqrt 2$, I meant to say loop at $(\sqrt 2,\sqrt 2)$.
Instead of rational no., I should have said: a point ( , ) in R^2 with both entries rational.
I think uncountability of the group is because: I take a straight line through $\sqrt 2$ and consider circles $\gamma_a$ through $(\sqrt 2, \sqrt 2)$ and point a on the line. So it can be seen that $[\gamma_a]\ne [\gamma_b]$.
This is because between $\gamma_a$ and $\gamma_b$, there is rational no. hence both are not hpy. equivalent.
So we have {$[\gamma_a]: a$ is on the line and on complement of $Q^2$} uncountable.
how do you know those circles are paths in R^2 \ Q^2? i don't really care about the underlying problem, it just seems that there's no immediately clear reason why such a circle would have to avoid Q^2, or what you'd need to worry about in order to ensure that
I should replace circles by loops: So from base point $(\sqrt 2, \sqrt 2)$ to $a$, we have a path constructed as follows: there are only countable elements of $Q^2$ so we avoid all them to join to a.
Then we come back to basepoint using another path.
Concatenation of these two paths is a loop at the basepoint.
I should consider this loop instead of circle.
But I see this requires lot more details. I'll think about this.
does anyone have any insight into why this ought to be true?
$A$ is an integer matrix
and $p$ is a positive integer prime
then $\text{tr}(A^{p^{n-1}) $ is equal to $\text{tr}(A^{p^n})$ mod $p^{n}$ for $n \geq 1$ a natural number
or at least does anyone see some straightforward way to prove this
sorry $\text{tr}(A^{p^{n-1}})$ is equal to $\text{tr}(A^{p^{n}})$ mod $p^n$
so $n=1$ is basically just a consequence of fermats little theorem
it does not look like there is going to be necessarily some straightforward way to be very careful with congurences and extend the $n=1$ approach for higher $n$ though
(this result is true and now basically considered 'classical')
If a divisor $D$ on a compact riemann surface of genus $g$ has degree $2g-2$, then $D$ is a canonical divisor? I know that if $D$ is a canonical divisor then its degree is $2g-2$.
if $g=1$, i.e. the Riemann surface is a torus, a divisor is canonical iff it is principal. do you know if every degree $0$ divisor on a torus is principal?
I'm reading Evans Semigroup theory (7.4.1). We have $\mathbf{u}'(t)=A\mathbf{u}(t)$ and $\mathbf{u}(0)=u$. He said $A:D(A)\to X$ a linear operator.
Where $D(A)$ is a subspace of $X$ ($X$ is a Banach space.)
I don't understand few things 1) operator means same space to same space, 2) If domain is not full space then $A\mathbf{u}(t)$ may not be defined.
He said we'll write $\mathbf{u}(t)=S(t)u$ to display explicitly the dependence of $\mathbf{u}(t)$ on the initial value $u\in X$.
How to read this equation? I am asking this because I want to explain myself the meaning of $S(t)S(s)u$.
I don't understand why $S(t+s)u=S(t)S(s)u$.
By definition, $\mathbf{u}(t+s)=S(t+s)u$. Similarly, $\mathbf{u}(s)=S(s)u$ so now $S(t)S(s)u=S(t)\mathbf{u}(s)$. Here $\mathbf{u}(s)$ is our initial function. So we get another solution of the ODE $\mathbf{v}(t)=S(t)\mathbf{u}(s)$.
@PNDas Yes, now that's a good one! My proof only shows, that if f is linear, it satisfies the conditions, but the point is, I have to show: Any function that satisfies the conditions, can't be anything apart from being a linear function, right?
@PNDas I said, it just like that. Now, of course for non-zero c it's valid, so I asserted it. Btw, your point is good, but don't you think, you are interpreting in a different way?
@Franklin try proving $\forall n \in \mathbb N \exists k$ such that $n < k$, and $\forall q \in \mathbb Q \exists m,n \in \mathbb Q$ where $m,n \neq 0$ and such that $mq+p = 1$
koro: i will be idle very shortly but i think the idea would be to investigate conditions on the map A: I x [0, 2pi] to X that allow you to meaningfully define a map from D to X given by g(r e^{it}) = A(r,t) and check that maybe those conditions are satisfied in this case
Leslie, I tried to do it like this now. To show continuity of F, I used the definition of coontinuity: take any open set U in X and show that $F^{-1}(U)$ is open.
Suppose that $f: \mathbb R\to \mathbb R$ is a function that satisfies Cauchy’s functional equation that is, $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb R$.
It can be shown that there exists some constant $c$ such that $f(x)=cx$ for all $x\in \mathbb Q$.
Now, one theorem says that if $f$ is bounded...
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Suppose $$f(x)=\frac{1}{t}\int_0^t(f(x+y)-f(y))dy$$ for all $x\in\mathbb{R}$ and all $t>0$. Then show that there exists a constant $c$ such that $f(x)=cx$ for all $x$.
My approach: It is given that $f:\mathbb{R}\to\mathbb{R}$ is continuo...
Let $F\subset\Bbb{R} $ intersect every uncountable $\mathcal{F}_{\sigma}$ set.
$B\subset \Bbb{R}$ is said to have the property of Baire if $B=U\triangle M$ where $U$ is open and $M$ is meager.
Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the ...
@Franklin Either my question has a simple answer, which you could just give, or it does not, in which case, I have no interest in attempting to read through the two or three questions and images posted above which might give me the context I requested. Nevermind.
I will only remark that when I came in ages ago and looked at the posted question, I noted that the necessary hypothesis of continuity was missing ... and I moved on.
@shintuku $f'$? The only hypothesis I see given is that $f$ is continuous. $f'$ is not a priori guaranteed to exist. However, if $f$ is differentiable, then this makes life a bit easier.
the quoted picture includes proof $f$ is differentiable
i haven't checked it personally, just layed out the structure of the proof for franklin
@Franklin once you understand that you can go from $f'$ constant to there is $c$ such that $f(x) = xc$, you'll understand that you need to prove $f$ is differentiable to get $f'$ constant
I am very confused about what is actually being asked.
My understanding of the question is that @Franklin is trying to show that if $f(x+y) = f(x) + f(y)$ for all $x,y\in\mathbb{R}$ and $f$ is continuous, then $f(x) = cx$ for some constant $c$. Is this not the case?
@shintuku Yes, that is why I pinged @Franklin, and not you. I am trying to get him to clarify the question. But most of my input here has been ignored, so maybe it is time for me to bow out.
Last try: @Franklin as I understand it, you are trying to show that if $f$ is a continuous function on $\mathbb{R}$ and $f(x+y) = f(x) + f(y)$ for all $x,y\in \mathbb{R}$, then $f(x) = cx$ for some constant $c$. Is this correct?
@TedShifrin Bah! Kids these days! GET OFFA MAH LAWN!
@Franklin Okay, that seems to be a "yes". So, can you show that if a function $f$ has the property that $f(x+y) = f(x) + f(y)$ for any two rational numbers $x$ and $y$, then there is a constant so that $f(x) = cx$ for any rational number $x$?
The Cauchy functional equation asks about functions $f \colon \mathbb R \to \mathbb R$ such that
$$f(x+y)=f(x)+f(y).$$
It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for s...
However, if by "i agree" you mean "yes, I can prove this fact", then apply continuity. Suppose that $x \in \mathbb{R}$. Choose a sequence $(q_j) \subseteq \mathbb{Q}$ with $q_j \to x$. What can you say about $\lim_{j\to\infty} f(q_j)$? What does this have to do with continuity?
What I think is: he needs to try to prove it by himself. Instead of looking at others answers. He should take some time. It'll automatically get cleared up.
@TedShifrin I do this myself, what can I say about others. When I study for semester exams, if a problem takes too much time just look for solutions online. (This gives their pov, and different ways to the problems). Otherwise, I take my time.
I grew up in the dark ages (or even earlier, if you listen to @leslie). We had no internet to pilfer from. And searching the library for non-obvious journal articles or books was beyond an art form. So I grew up figuring things out for myself (or with my colleagues).
@TedShifrin A good friend of mine studied under VS Varadarajan. He describes his phd program as being one day after another of being screamed at in an Indian accent: "YOU ARE NOT THINKING! WHY AREN'T YOU THINKING!?"
I will admit that it has changed the framework of research. It's much easier to find out what's known about certain questions. Years ago a graduate student who was auditing my undergraduate differential geometry class asked me a cool question to which I didn't know the answer. I asked a colleague. He didn't know. I emailed a knowledgeable expert. He didn't know. I then googled (duh) and found a paper by Arnold on exactly that topic.
i agree with ted, although the existence of a resource like MSE also allows conversations, when they do occur, to be more focused. sometimes instead of working all of the context up from first principles in a long back and forth, one can just link to a question/answer, and say "where in this do you get stuck." which puts a little bit of work on the asker to engage on their own the way they might with a textbook or clay tablet like ted used to use.
My one Ph.D. student would tell you that I was actually surprisingly encouraging and gentle. Perhaps the thousands of undergrads will say something else.
That one Ph.D. student has taught for her career at a 2-year (which then merged with 4-year) college. Even as a grad student, she was a phenomenal teacher.
@TedShifrin The thing is: Probably you'll admit, life grew much faster and faster. Now, everyone is drowned with exams and time-frames. Everyone, including me, prefers things to work out with some hours at max.
@XanderHenderson Btw, I proved it. Yes, the part to show, the derivative is constant, just needs some manipulations. That's it... but as I gather, this functional equation is a very standard one
We had plenty of depression, too. I am not minimizing stress, but each generation thinks it's uniquely plagued. Apparently, chat doesn't want to post ...
I will admit that Covid and imminent world war are things that I didn't deal with as a student, although I almost got drafted to go to the Vietnam war and would have fled the country had I been.
@Franklin You haven't answered my question, which has nothing to do with the derivative.
@Franklin When I was an undergraduate, I was told that a 12 credit hour load was considered a "full load", and that a "full load" should be thought of as the equivalent of a full time job, i.e. 40 hours per week. So every hour of lecture was supposed to correspond to at least two hours of work outside of class (homework, office hours, study groups, etc).
I typically took at least 15 hours per semester. My worst semester as an undergrad was 23 hours.
I was always drowning in exams, papers, and problem sets.
That is the life of an undergrad. But you have to do that work if you are going to learn.
I wish I could get my students to understand that they aren't going to just passively absorb material in lecture, and that they have to actually work outside of class. :/
However, if by "i agree" you mean "yes, I can prove this fact", then apply continuity. Suppose that $x \in \mathbb{R}$. Choose a sequence $(q_j) \subseteq \mathbb{Q}$ with $q_j \to x$. What can you say about $\lim_{j\to\infty} f(q_j)$? What does this have to do with continuity?
ted successfully avoided the draft by reminding his local draft board that he had taught most of their parents calculus and had not been of draftable age since before the US even existed
Can anyone tell me the learning road map of functional analysis? Real analysis (including measure theory) --linear algebra-->complex analysis-->metric space---> general topology--->normed space---->topological vector space.
Let $f: R \to R$ be a continuous function.Suppose $$ f(x) = \frac{1}{t}\int_{0}^{t} (f(x+y)-f(y))dy$$
for all $x \in R $ and $t>0$. Then show that there exists a constant $c$ such that $f(x)=cx$ for all $x$
At first I want to ask how shall I integrate the function give above? And also how am I s...
The second part is to prove that: assuming that every map $S^1\to X$ can be extended to $D^2\to X$, we must have $\pi_1(X,x_0)=0$ for every $x_0\in X$.
@Franklin Please stop linking to other things, and do some work for yourself. I asked you if you were trying to prove that if $f(x)+f(y) = f(x+y)$ and $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}$. You said "Yes". I asked if you could prove the result for $x \in \mathbb{Q}$. You said "Yes". So, given that the result is true in $\mathbb{Q}$, how do you extend to $\mathbb{R}$?
Or is there some other place in this conversation where we have miscommunicated?
I don't see how the linked question is relevant to the approach that I have been trying to guide you down, given the answers you have provided to my previous questions.
So I take [f] in $\pi_1$ at $x_0$. Here, we have $f: I\to X$. We want to change the domain to $S^1$. I wrongly thought earlier that $I\cong S^1$ considering compactness. But then I realized that it is not so and recalled that $S^1\cong I/$~, where ~ identifies the end points.
With that and using property of quotient maps, there exists $\tilde f: S^1\to X$ such that $\tilde f\circ q=f$, where $q$ is the quotient map $I\to S^1$.
sourav: i don't know that there is a single road map that would work for everybody, but all of those are helpful. i think the most central prerequisites would be real/complex analysis and linear algebra. not so much general topology (you can learn as you go). "normed space and topological vector space" is so close to functional analysis that it's maybe indistinguishable from it (FA contains a lot of the key examples)
I want your advice on my solution of this problem.
problem
Show that for a space $X$, the following three are equivalent:
(a) Every map $S^1 \rightarrow X$ is homotopic to a constant map, with image a point.
(b) Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$.
(c) $\pi_1(X...
@XanderHenderson The thing is: A set of rationals converge to a real , is possible, can be seen more clearly, if he has done a fair amount of real analysis course. But I have not done it so much. True, I know this as a fact up until now. But yes, this actually makes tge argument lot simpler
Maybe, as the real analysis course progresses in my college, I will grow more easy going with these
I don't know what definition of the real numbers you're using in your course, but I would take it as almost axiomatic that every real number is the limit of a sequence of rational numbers.
@Franklin I suggested the same thing earlier. I asked you to find f on integers, then extend it rationals and then use continuity. There is a general result: If f and g are continuous real valued functions on R and if they agree on a dense set (Q is dense in R and so is R-Q for example) then they are equal.