A node like that could split into a pair of branch points under a perturbation e.g. $x^2-y^2=a^2$. I presume that to be the generic case, and the node surviving said perturbation to be nongeneric.
But I don't know what the terminology for the node to 'survive' a given perturbation is.
Right, @Semiclassic. I'm just saying that to me saying "the singular point $(0,0)$ is stable under the perturbation" means that the point itself stays singular, not that it moves.
It's the characteristic polynomial in $x$ of the matrix $$\begin{pmatrix} a-y & 0 & b c & d \\ 0 & a+y & cd & b \\ bc & cd & 1 & 0 \\ d & b & 0 &-1 \end{pmatrix}$$
To a high-school student, linear algebra would be "simultaneous systems of equations." To a college student taking an intro course (not a full one), it'd probably mean "matrices." And to someone taking a serious course in linear algebra, it'd be stuff like vector spaces.
And there's stuff which I don't say a lot but which I know is there, e.g. null space. (An eigenvector, for instance, is nothing but an element in the null space of a particular matrix.)
btw if I'm looking a space $C$ of maps $X\to Y$, whats the topology called that makes all the evaluation maps $ev_x: C\to Y$ $f\mapsto f(x)$ continuous?
So in the context of sets what is the appropriate way to say inbetween? The way I've been doing it is B = { x|x ϵ 1-100 }. Is that acceptable or is there a proper way to do it?
@BalarkaSen On an unrelated note: Would you by any chance know of any resources you could point me to that discuss how to compute the period matrix of an algebraic variety $X$ defined as the zero set of a polynomial, which is a smooth manifold?
I don't really know what a period matrix is. But you can wait for @TedShifrin, who's the resident complex algebraic geometer/differential geometer here.
Also I misread your previous "differential geometry" as "differential topology". I don't really know much about the former :P
@BalarkaSen The period matrix of a smooth genus-g curve has matrix elements $\int_{\gamma_i}\omega_j$ where $\{\gamma_i\},\{\omega_j\}$ are bases of cycles and co-cycles respectively.
@Semiclassical Yes, I'm aware of the textbook definition so to speak, it's just that evaluating it by that means is not always possible, or at least not the easiest way.
The main tool I know is to find the Picard-Fuchs equation for the given elliptic curve and solve the resulting ODE.
That only works in cases where you've got a family of Riemann surfaces parametrized by a parameter, though, and even there it can be pretty hard to actually find the Picard-Fuchs equation.
@JoeSlater You can work backwards, using the formula for the sine of a sum.
The phase shift is physically meaningful, and perhaps it's better if you understand this in the context of RLC circuits, as in your problem; Young and Freedman have a good section it.
Ok thanks. I will try to read more about it. But I am not able to see how using the eulers formula or formula for the sine of sum is useful in this case.
In yet another attempt to find the solution to the quintic polynomial, I started looking backwards at the solutions to the quartic, cubic, quadratic, and linear polynomials to see if I could pick up any patterns.
$\left\{
\begin{aligned}
0&=ax+b \\
0&=ax^2+bx+c \\
0&=ax^3+bx^2+cx+d \\
0&=ax^4+...
@Semiclassical I actually don't have a particular one in mind; the question is motivated by reading I'm doing into the Kawazumi-Zhang invariant of a Riemann surface, which turns out for certain genera to be written as a Fourier series in the period matrix.
@KasmirKhaan While I may be able to do random integrals with complex analysis, that's just a side thing, and I can't do triple integrals (at least not very well)
@KasmirKhaan No, because the volume that you are integrating over is bounded by several surfaces, one of which is the plane $x+y+z=1$. On this plane, the equation holds, but your integration includes points in the interior (after all it is a volume integral) that do not satisfy, $x+y+z=1$. Does that make sense to you?
I was wondering on how to evaluate the following indefinite integral for all $n\in\mathbb R$.
$$\int\frac1{1+x^n}dx$$
It seems to be peculiar in that we have
$$\begin{align}
\int\frac1{1+x^{-1}}dx&=x-\ln(x+1)+c\\
\int\frac1{1+x^0}dx&=\frac12x+c\\
\int\frac1{1+x^{1/2}}dx&=2\sqrt x-2\ln(1+\sqrt ...
It's definitely worth knowing the details though; I always like to know what my tools are doing. Furthermore, it can be used to prove in certain cases that no closed-form exists, which Mathematica won't do for you.
I always feel like I've cheated if I've used a table, though if I'm doing an immense problem where the integral is the least of my worries, then not so much :)
@Socrates i don't understand. We try proof by contradiction and it fails either because the original statement is false or because the statement is not provable. How does one prove it's the latter and not just lack of ability of the prover to find the contradiction?
@Socrates what...? We can prove that contradiction shows independence in the case of the parallel postulate of geometry. Or do you claim otherwise? Surely there was a method used in that?
@TheGreatDuck but there is no(known) method for all conjectures, if there was, we wouldn't have conjectures.
Suppose the following: i try to prove Riemann Zeta by disproving the contradiction. Now I fail, but does it mean that Riemann Zeta is true(or indeed false)?
The fact that anyone tries and fails says absolutly nothing
If there where a method to show that it's only my lack of ability, then Riemann would be proven
@Socrates regarding your point, we don't have algorithms to find all possible solutions to differential equations with closed forms. We don't know how to reduce any given number into it's factors without brute force. I'd argue that we know how to solve some of them and reduce many of them by memory. Surely a method exists. Whether it applies to everything is obviously expected to not be the case.
However, I'd imagine that one was able to prove the parallel postulate was independent therefore, some kind of method exists for at least some statements and situations.
Hi all, I'm looking for a confirmation on an attempt to answer a question of mine. It concerns the estimation of mean value of a function of random variables. @Batman provided me with precious info, but it would be nice if anyone else could also take a look. Could you check Second attempt? Thank you very much.
@Socrates perhaps the latter. the first is the textbook example.
@SimplyBeautifulArt that's why i said ONLY the one's that exist. but you still have to find it using the algorithm. I'm merely saying nonexistent things give any result. Don't worry about the case of no closed form.
If we do that. Why does "think about it this way: can you prove to 100% that the universe didn't start 5 seconds ago?" not mean "Good day fellow, how are you"?
if $a^2+b^2=c^2$ and $(a,b,c)=1$ then one of a,b is odd and the other is even, and c is odd. Let $a\mapsto 2a$ then $a^2=\frac{c+b}{2}\frac{c-b}{2}$ and if a product of two coprime integers is a square then both are squares. Let $\frac{c+b}{2}=p^2$, $\frac{c-b}{2}=q^2$ then $c=p^2+q^2$, $b=p^2-q^2$, $2a=2pq$ which means all pythagorean triples are parametrized by Euclid's equation
No, @JamalS, I've never even thought about this. But what do you mean by a period matrix for a general algebraic variety? This is normally used for abelian varieties (so these are algebraic varieties that happen to be tori).
@TedShifrin Well, an example would be the Burnside curve, $y^2 = x^6-1$ with automorphism group $S_2 \times \mathbb Z_2$; I'd like to compute, for example $\Omega$ for that curve.
Oh, you're talking about curves, so then it's the Jacobian variety of the curve which is the abelian variety.
You can probably do that bare-hands. What's the genus?
If you haven't looked at it before, you might try Phillip Griffiths's beautiful little book An Introduction to Algebraic Curves. He does a lot of concrete stuff in that book.