@Xander I will keep that in mind. It'll be tricky to not do that for this exam since I've got my classes and they're currently my highest priority, but I guess that's why I'm taking the test twice just in case
@MatheinBoulomenos the point of it is that there's a lot of variance for us schools so there should be some way to measure if everyone has a certain base of knowledge. Plus i think grad school here pretty much always includes classes that you would get at an earlier level in a european school
@Daminark we have some international students, yeah. I don't personally know American math students, but why not? I would recommend you do a masters, first though.
@XanderHenderson i think part of the weirdness is that "European bachelors" means more than "US bachelors" bc there's waaaay more variation for US schools
which means that a lot of students have to get through a year of calculus, then a year of linear algebra and differential equations, before they are considered "ready" for upper division math
@EricSilva Indeed. At my current institution, most math majors can get away with only a couple of proofs based courses, and neither analysis nor algebra is required
yeah at the school i mentioned their first year phd courses were a dummit and foote course and a baby rudin course and since they mostly just accepted their own students it wouldve been most of the students first analysis and algebra courses
@Secret To backtrack a bit, my advisor would say that a fractal is "an object [whatever that means] to which we can associate an appropriate zeta function [whatever that means], and that this zeta function can be analytically continued to a meromorphic[?] function on a domain where it has poles [or perhaps some kind of natural boundary of singularities, or something like that] that are strictly complex [and therefore not real]"
That's a strange characteristic, as superficially I don't see how zeta functions (assuming your advisor is referring to something similar to riemann zeta) are related to how a geometry behaves. But I guess I would had to be in your field to understand better...
Has anyone tried to make a connection between the Mandelbrot set and the non-trivial zeros the zeta function?
Looking at the Mandelbrot set, it appears that all points are to the left of the line 0.5 + t*I, and to the right of the point -2. So the set is bounded on the left by the trivial zeros...
Background
Imagine a computer that has finite memory but produces an infinite output. Our aim is to find if it using a deterministic algorithm or a non-deterministic(in layman's term having an element of randomness in it's) algorithm. The below algorithm I have proposed seems to do exactly that....
In mathematics, particularly linear algebra and numerical analysis, the Gram–Schmidt process is a method for orthonormalising a set of vectors in an inner product space, most commonly the Euclidean space Rn equipped with the standard inner product. The Gram–Schmidt process takes a finite, linearly independent set S = {v1, ..., vk} for k ≤ n and generates an orthogonal set S′ = {u1, ..., uk} that spans the same k-dimensional subspace of Rn as S.
The method is named after Jørgen Pedersen Gram and Erhard Schmidt, but Pierre-Simon Laplace had been familiar with it before Gram and Schmidt. In the theory...
@MoreAnonymous Anything can be considered fractal, depending on what you mean by fractal; my advisor's answer, I suspect, would be "What is the associated zeta function?"
@MatheinBoulomenos My advisor seems to find interesting connections all over the place; he started in QM and wrote a couple of books on Feynman's operational calculus
the theory of fractal strings is related to the wave equation, which requires a Laplacian, which quickly gets you into the analysis of unbounded operators, which is basically where QM lives ;)
@MatheinBoulomenos It was easy to do using Newton's method. It gives you after three iterative 1.4857... while the exact value is 1.47..... Thank you for your help!
Mona: Actually I recall the algorithm I use for gran schmidt in my PhD, is basically reading the section in wikipedia which about implementing the numerical stable version of gran schmidt, and then just manually translate that text into python code
(plus coverting all instance of addition into kahan summation to increase the numerical stablity)
The resulting code works very fast and done 360 deg of 3x3 rotation matrix multiplications in roughly 2 seconds
@MoreAnonymous If youve ever taken a basic calculus class youd know that sequences are defined to be real valued functions on the domain of the natural numbers. There is nothing that special about that fact. The fact that you dont know that tells me you need to first brush up on whatever branch of math is termed as "the study of sequences". At least the elementary stuff.
@TheGreatDuck yea ... I've always paid little heed to definitions and always used an intuitive feel instead for the subject. But you are correct this I should pay more attention to definitions so as to avoid miscommunication
How do I go about showing that Z/nZ is not a group under multiplication? I am able to show that 1) it is closed under multiplication, 2) has an identity $\bar {1}$, 3) associative
I am planning to hide my profile on one of my SE site. I am not asking to hide my network profile. While that would be my main goal, but I know that is not possible.
What I want is no one can click on my username to see my profile page. That way I have effectively disabled my network profile. I ...
@MonaJalal Note the y axis, which is times *10^4. Your fhat function gives a very huge value between (-3,3). Thus your sin t is actually there, just the scale is too large to see it. Try restricting the y axis range with ylim
quick question, im trying to explain to some one who is new to u substitution with integrals, and am trying to show how it cancels out a trig function.
they understood it but i am wondering, have i actually written my integrals syntactically incorrect here: https://i.imgur.com/ZSZy0ZX.png
I refer to the the fact that i have integrals of x but i have du at the end, is that still valid since i showed what dx equals? I'm under the impression i would need dx technically but couldn't figure out a neater way to show it.
with a standard for problem Ax = b and x >= 0 and A mxn, I am struggling to find an example that has an x in the feasible region with only m positive components that isn't a basic feasible solution.
I think the sign is throwing me off cuz http://prntscr.com/i70eci or maybe I'm messing myself up because the numerator is that bar version and then regular z is z^3+8i OMG
throws a baseball at the problem I know taking the complex conjugate could be a bad idea
In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by LCM(a, b), is the smallest positive integer that is divisible by both a and b. Since division of integers by zero is undefined, this definition has meaning only if a and b are both different from zero. However, some authors define lcm(a,0) as 0 for all a, which is the result of taking the lcm to be the least upper bound in the lattice of divisibility.
The LCM is the "lowest common denominator" (LCD) that can be used before fractions can be added...
In mathematics, the affinely extended real number system is obtained from the real number system ℝ by adding two elements: + ∞ and – ∞ (read as positive infinity and negative infinity respectively). These new elements are not real numbers. It is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. The affinely extended real number system is denoted
R
¯
{\displaystyle {\overline {\mathbb...
because if both fixed something, and had different number of elements being fixed
one needs to use this argument
in the case of both of them fixing same number of elements , those arguments wont work
@anon anyways thanks anon :D i hope we can talk more tomorrow =P I need to sleep its 8 am and i have lecture in 6 hours >< hope i can get 5 hours sleep
if we take the limit of Arg(-1) is that -pi? Arg(z) since z = x+iy then Arg(x+iy) but then there's only x = -1 and y =0 because -1 is the real part and 0 is the imaginary part so Arg(-1+0i) = Arg(-1)
Hi; To solve "The perpendicular dropped from the vertex of the right angle upon the hypotenuse divides it into two segments of 9 and 16 feet respectively. Find the lengths of the perpendicular, and the two legs of the triangle.", what I did was set up the equations ${ a }^{ 2 }+{ b }^{ 2 }={ 5 }^{ 2 }\\ { 9 }^{ 2 }+p^{ 2 }=a^{ 2 }\\ { 16 }^{ 2 }+p^{ 2 }={ b }^{ 2 }$. But after I plugged in, I got a negative value for p. Does anyone know why?
Ok! The order of $f(a)$ must divide the order of a, which is 4, right? @LeakyNun
So, we have to findhow many permutations of S4 have order that divides 4, or not? @LeakyNun
We have 1 identity (order 1), 6 transpositions (order 2), 3 products of two disjoint transpositions (order 2), 6 4-cycles (order 4). So in total we have 1 + 6 + 3 + 6 = 16 elements of S4 that have order that divides 4, right? @LeakyNun
would it be right if I say at some point the sequence must go past the element with the biggest distance to the the limit and after this point apply the definition?
Maybe simple question, if $f(x)$ is monic and $p_i(x)$ is monic, $i = 1,\dots,n$, then from $P(x)Q(x)f(x) = \Pi_{i=1}^n p_i(x)$, why must $P(x)Q(x) = 1$?
I'm struggling with the following calculus question.
Let there be two functions $f,g : [a, \infty) \to \mathbb R$ such that:
$g$ is monotonic, differentiable and has a limit at zero
$f$ is continues such that $\int_a^b (f) < M \in \mathbb R$.
Prove that integral $\int_a^{\infty} f(x) * g(x)$...
Basically I don't understand why he can justify the existence of the limit by Cauchy (i.e. $(4)$) but then needs an additional proof that the limit exists (i.e. $(2),(3)$)... Why is it not enough to just use Cauchy?
let $Q \subset \mathbb R^3$ be the subset where all coordinates are positive. let $f: Q \to Q$ be continous. how do i show that there exists a $v\in S^2 \cap Q : f(v)=\lambda v$ ?
@Vrouvrou Yes. Draw a picture. To prove: show that $f(t)-f'(s)(t-s)$ is increasing for $t\in[s,\infty)$ and decreasing for $t\in(-\infty,s]$. (Remember that [$f$ is convex iff $f'$ is increasing.)