@shintuku Look, you are the one that is getting angry about perceived aggression. That aggression simply doesn't exist. No one cares enough about you to downvote your questions just to be an "asshole". You are likely to be much happier if you stop assuming that people who simply disagree with you are specifically targeting you, or out to get you.
I don't know why you choose to be angry and decide that others are "assholes". That's on you, not me.
Though perhaps "combative" would be a better description than "angry". I can't really assess your emotional state, though I can certainly describe the style of your interaction.
@mick Why? Who cares? What is the motivation behind this question? Why $2/3$ and not some other parameter? Why does it need to be meromorphic on $\mathbb{C}$? What are you planning on doing with this function? How much do you know about the general theory of Dirichlet series?
@SineoftheTime "deer-REEKH-lay" (with the "KH" being like the "ch" in the Scottish word "loch"). I assume that this is the French pronunciation, as it is the one I received from my advisor, who is French.
@SineoftheTime Most American institutions don't use high stakes examinations to assign marks to a course of study.
Typically, a semester long class will include at least two or three examinations (including a final exam), as well as other graded work (quizzes, homework, projects, presentations, etc).
@TedShifrin Like I said, I assumed that the pronunciation was French, as Michel is French, but maybe he uses the German pronunciation? He is actually usually pretty good about "correct" pronunciation.
@SineoftheTime This is usually not possible. It is sometimes possible to withdraw from an examination before it is marked (e.g. if you think you have done poorly on the SATs (a college placement exam thing), then you can walk out and not have them marked).
@shintuku I guess the system of examination is different. In my uni, I follow the lesson for approx 3 months, then there is a session with all exams. There are three sessions (winter, summer and the one in September) and all teacher must schedule at least one "appeal" (don't know if it's the right name)
@onepotatotwopotato Many people do. Others really like them (personally, I would much rather be examined orally---there are more opportunities for me to demonstrate what I know, and to correct mistakes).
@onepotatotwopotato Every oral exam I have ever taken (or given) includes some back-and-forth between the examinee and the examiner. If you mess up something major, the examiner will typically give you some kind of hint that you screwed the pooch.
Panic of course. But I'm not very used to oral exams so it's hard for me to handle it when I'm panicking on oral exams. But I'm used to written exams so I can handle panic quite well and actually, I usually don't panic because I know how to prepare.
@onepotatotwopotato You might try working on that, some. In the Real World™, written examinations are rare, while oral examinations are quite common (in the form of, say, presenting an idea to a research group, or a proposal to a development team, or whatever).
Oral exams are much more like how the Real World™ functions (though still artificial).
One skill I learned is to keep talking about what I'm thinking when I don't know the answer immediately so that examiners can know what I'm thinking and give a hint if they want.
@onepotatotwopotato Yup. Don't claim to know something you don't, but clearly explain your thinking, and the bits and pieces which are relevant. If you're lucky, you will eventually either remember or work out out the details.
And, like, real math is collaborative. We explain to each other what we are thinking, and poke holes in the arguments of others.
@TedShifrin That's next week---we are supposed to turn in a max 7000 word fairy tale rewrite. I'm going for a Hansel and Gretel kind of thing, but with modern tech.
You should write a poem on the poetry of logical ideas :P
> Pure mathematics is, in its way, the poetry of logical ideas. One seeks the most general ideas of operation which will bring together in simple, logical and unified form the largest possible circle of formal relationships. In this effort toward logical beauty spiritual formulas are discovered necessary for the deeper penetration into the laws of nature.
Write the generating function for Legendre's Polynomials. Starting from generating function prove that:
$$(2n+1)P_{n}(x)=P'_{n+1}(x)-P'_{n-1}(x),$$ and hence prove $\int_{-1}^{1}xP_n(x)P_{n-1}(x)dx=\frac{2n}{4n^2-1},n=1,2,3,\cdots$
(Here, $P_n(x)$ denotes Legendre Polynomial of degree $n$)
I tr...
> Joey went to the store and bought a pack of chips. A bottle of water costs $3.00, a pack of chips costs $1.00 and a pack of gum costs $2.00. How much did he spend in total? […] Performing the sum to reach $6 is incorrect, but in a specific way. It is not random noise. […] 74% of participants responded correctly with $1, and 24% with the mindless math answer of $6 (N = 196). > In contrast, consider the same problem with harder numbers: Joey went to the store and bought a pack of chips. A bottle of water costs $1.05, a pack of chips costs $0.75 and a pack of gum costs $1.70. How much did he…
> When faced with difficult tasks, people are often eager to start doing something. But this eagerness may preclude them from correctly representing the problems that they are so eager to start solving. When we ask people to “take their time”, we have to be more specific about when they should take their time. The execution of operations gives the illusion of progress, but if the problem is represented incorrectly, it remains just that, an illusion.
Hello, all. I'm just trying to understand one of the exercises in Arnold's *Ordinary Differential Equations*. Neither the exercise nor the provided answer seems to be making sense. If anyone can help, that will be much appreciated.
Exercise 4 in chapter 1, section 4 on Quasi-homogeneous Equations:
Choose the weights of the variables so that the differential equation of the phase curves of Newton's equation $\ddot{x} = Cx^k$ is quasi-homogeneous.
Provided answer:
The equation of the phase curves is $dy/dx = Cx^k/y$. Consequently $2\beta = (k + 1)\alpha$.
In this context, quasi-homogeneous means a DE of the form $dy/dx = F(x,y)$, where $F(e^{\alpha s} x, e^{\beta s} y) = e^{rs} F(x,y)$.
In other words, if you multiply both of the inputs to $F$ by some special constants, it's the same as multiplying the output by some other special constant. If this works, the factor $r$ is called the "degree" of the quasi-homogeneous function.
One thing I am struggling to understand: the problem statement doesn't mention $y$ at all. It appears that the use of $\ddot{x}$ in the problem statement implies there is some other independent variable, which for almost all the examples in the book would be called $t$.
But then the provided answer talks about $dy/dx$.
The mention of "phase curves" in the problem and answer also seems to imply, from the way that term is used through the rest of the book, that there should be at least 2 dependent variables $x$ and $y$ which are functions of an independent variable $t$. The "phase curves" should be all the values of $(x,y)$ which you get if you start from some point in phase space and then follow the evolution of the system with increasing $t$.
But again, the problem statement doesn't say anything about what $y$ is in this case.
Not sure if this is worthy of posting as a question on MO. I may just be misunderstanding the terminology used in the problem statement.
@XanderHenderson In the Western world, written exams are a relatively new innovation, first becoming popular in the 1700s, but they were used in China at least 1000 years earlier. en.wikipedia.org/wiki/Exam#History They were mostly adopted to save time, which was necessary as more men sought higher education. Of course, they can also help to reduce subjectivity & bias.
@XanderHenderson Nice. I was reminded a few days ago of the Indian mathematician-astronomer Aryabhata (476–550 CE). He gave many important results in trigonometry, but rarely gave proofs. Fortunately, several later writers did give proofs in their commentaries on his work.
Back in those days, people didn't have the modern attitude towards rigorous proof. And in the Indian tradition, it was standard to present knowledge in a poetic form. I guess it makes sense when paper is precious and you want students to memorise your teachings. It's much easier to memorise stuff when it's written as rhyming couplets. I assume the students would recite that stuff in a sing-song voice. :)
From knowing $\mathbb{N}$ and $\mathbb{Q}$ are both countable infinite, and the fact that the power set of any set yields a cardinality strictly greater than the set, can we conclude that there is a bijection between $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{Q})$?
I do not think so, since we don't know how much the power set "raises" the cardinality of a set, right?
The reason for the question is a passage in my lecture notes stating that $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{Q})$ do have the same cardinality because $\mathbb{Q}$ is countable infinite. I do not see the reasoning behind that argument.
But since $\mathbb{N}$ and $\mathbb{Q}$ are both countable infinite we have a bijection between them, so they have the same cardinality, and hence so must their power sets.
If $|A|=|B|$, then $|P(A)|=|P(B)|$
This is what I have done so far:
Assume $|A|=|B| \Rightarrow \exists$ a bijection $f:A \to B$
Define $F: P(A) \to P(B)$ where $ F(S)=\{f(a)| a\in S\}\subseteq B $
Claim: $F$ is injective
Suppse $ F(S)=F(S') \Rightarrow s=s'$. Let $$ a \in S \Rightarrow f(...
Let G be any group and Let N be any subgroup of G. If the product of any two right cosets of N in G is also a right coset of N in G, prove that N is normal.
@sunny Any map $f\colon A\to B$ induces a map $P(A)\to P(B)$ by $f(X)=\{f(x)\mid x\in X\}$ for any $X\subseteq A$. Check that if $f$ is a bijection then so is the induced map on powersets
Does there exist a pair of smooth manifolds $(M,N),$ $M \ne N$ related by an isometry $g,$ s.t. every smooth regular foliation of $M$ satisfies some differential equation on $N,$ and every smooth regular foliation of $N$ satisfies some differential equation on $M?$
from what I understand, smooth regular foliations of $M$ and $N$ resp. satisfy particular types of differential equations whose forms depend on the equipped metrics. Restrict a foliation on $N$ to the metric of $M$ and ask whether it satisfies some D.E. w.r.t. metric of $M.$ And vice versa.
@SouravGhosh Well, the condition that a differential equation is exact, is that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.$ But in here, $\frac{\partial M_1}{\partial y}= \frac{\partial M\mu(x)}{\partial y}=\mu(x)\frac{\partial M}{\partial y}.$
Similarly, $\frac{\partial N_1}{\partial x}=\frac{\partial N\mu(x)}{\partial x}=N\frac{\partial \mu(x)}{\partial x}+\mu(x)\frac{\partial N}{\partial x}$. But how to show that indeed, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ ?
one satisfying thing to consider might be if you have a forgetful functor $U\colon\mathcal{C}\rightarrow\mathcal{D}$, then a "free object functor" is a left adjoint $F\colon\mathcal{D}\rightarrow\mathcal{C}$. if this exists and $U$ is faithful (which forgetful functors usually are), then it follows that the counit $FU\Rightarrow\mathrm{id}$ of the adjunction is an epi (exercise), so every object is a "quotient" of a free object.
usually, though, one doesn't care much about free objects and cares about the more general projective objects, specifically in very nice categories that are called abelian categories (of which abelian groups and modules are examples), the property that every object be a quotient of a projective object is then called "having enough projectives" and very important in homological algebra
Hi! There is question: if $T\in\mathcal{D}'(0,1)$ and $\frac{dT}{dx}\in L^2(0,1)$ then show that $T\in L^2(0,1)$. How to do this? I know that $T'=0$ implies $T=T_c$.
The problem is can I write if $f\in L^2(0,1)$ then can I write $T_f=T_g'$ for some $g\in L^2(0,1)$?
May be I can do like this: $\frac{dT}{dx}=T_f$. Let $f_n\in C_c^{\infty}$ such that $f_n\to f$. I can consider $T_{f_n}\to T_f$. Now I can find $g_n\in C_c^{\infty}$ such that $T_{g_n}'=T_{f_n}$.
Okay got it. I didn't need to do all that. The interval is bounded.
I'm just inquiring due to the old people comment....of course a bbq without red meat is blasphemy......I wonder if my opinion will change when I reach your guys' age....
I usually brush it with a light glaze of olive oil then salt and pepper on one side so it isn't excessively salty and enjoy. Maybe not full sized peppercorns, but cracked peppers corns from the grinder...
$x_0$ is in the singular support iff for every $x\ne x_0$, there is a smooth function $f$ so that $T=T_f$ in a neighborhood of $x$. Or something like that.
I'm not looking in books.
That was wrong. That assumes the singular support is just $x_0$. $\Sigma$ is the sing. supp. iff for every $x\notin\Sigma$ blah blah blah?
Okay I consulted another book. Their definition is different. sing. support(T)={$x\in \Omega$ such that there is no open set $x\in U\subset \Omega$ such that $T|_U$ is smooth.}
This definition of support is correct. If we have distributions on U and V such that they match on U \cap V then we have a distribution on U\cup V. Now, so if T is zero on U and V then there is a distribution on U \cup V. Now to prove that this distribution is also zero, you use partitions of unity.
If $\varphi\in\mathcal{D}(U\cup V)$ then $T(\varphi)=T_1(\varphi\psi_1)+T_2(\varphi\psi_2)=0$
i dropped from favour with my grandmother when i did not attend JP II's mass a few miles away from her place (where i was staying at the time) because i was too lazy to get up.
I'm long past optimism. Somebody posted somewhere on FB wondering how MTG could have been elected to Congress. My answer was simple: Because the people in NW GA are just like her.