In a homework assignment I had to come up with an example of a function: $f: \mathbb{Q} \longrightarrow \mathbb{Q} \ \ $such that $ \forall x,y \in \mathbb{Q}$ our function is contracted with factor $\frac12$ and has no fixed points. i.e. $$ \left | {f(x)-f(y)} \right |\leq \frac12 \left | x-y \right | ,\\ \nexists \ \ x^* \ s.t. \ f(x^*)=x^* \ $$
I came up with $$ f(x) := \left\{\begin{array}{lr} \frac12 (x-1) + \frac32, & \text{for } x < 1\\ \frac12- x^{-2}, & \text{for } 1\leq x\leq 2\\
@Astyx Oh, I remember. But that wasn't the exact task, I was allowed to use a function which is recursive and I could give that function my lambda expression as parameter
Alternative approach: $\dfrac{d}{dx}\cos^2 x = -2\sin x \cos x=-\sin 2x$ and $\dfrac{d}{dx}\dfrac{1}{2}(1+\cos 2 x)=-\frac12 (2\sin 2x)=-\sin 2x$. So the first-derivatives of both of them agree everywhere.
I reviewed 20 close votes today and only voted "close" twice, many of the questions were crap, but still people are way to anxious with their close votes :/
Hey guys! I'm doing a constraint question for a naive bayes proof, and I'm having trouble understanding how to take the lagrangian of a function with another function as the integral constraint.
I know that all values of the gaussian continuous distribution must add up to 1, but I don't know exactly how to put that in terms of an integral from -inf to inf
Yeah - I do have a lagrangian multiplier, but the multiplier $\lambda$ is multiplying an integral from -inf to inf
Normalization of a probability distribution is just that the probability of measuring anything is 1. So that's $\int_{-\infty}^\infty p(x)\,dx$ where $p(x)$ is the probability distribution function.
Anyone got an idea of how I can work out the area element of a stereographic projection? Supposedly for $\psi : \Bbb{R}^{n-1} \to S^{n-1} \subsetneq \Bbb{R}^{n}$ defined by $\psi(x_1, \dots, x_{n-1}) = (|x|^2+1)^{-1} \left( 2x_1, \dots, 2x_{n-1}, |x|^2 - 1 \right)$, it's supposed to follow that $$ \left\langle \dfrac{\partial \psi}{\partial x_i}, \dfrac{\partial \psi}{\partial x_j} \right\rangle = \delta_{ij} \dfrac{4}{(|x|^2+1)^2}$$
The mathematica chat isn't very lively, and I'm hoping someone would know the answer here: how can I get an exact value in Mathematica? I just calculated an integral, which gave me numerical value, even while the answer is 7/6.
I hope I'm not rude for asking a Mathematica question, but I was trying to calculate something while doing maths:)
@Astyx Trivial, that's what I expect when talking to Math people, very unusual conversations, I love it, however, I'm still going to say that I meant how is life going with you ?
One of my coauthors and I, just before the days of Mathematica, wrote a paper (of which I'm very proud), and we each did a certain Chern class computation — reasonably involved — about 20 times before we could confirm we had it right.
I'd avoid considering $x\to y$ as equality, for two reasons. 1) You'll run into indeterminate limits at some point, where they 'seem to' be ill-defined at $x=y$ and require some care there. 2) Not all functions are continuous, and therefore $\lim_{x\to y}f(x)$ need not coincide with $f(y)$.
Kind of funny: It's apparently open if all embeddings $\Bbb C\to \Bbb C^3$ are all related by an automorphism of $\Bbb C^3$ where I mean automorphisms in the sense of ehh... affine algebraic varieties?
After you've done a few of the exercises, it's appropriate to prove and then use all the limit theorems. But make sure you don't use the product rule for limits when one of those limits don't exist. Students love to screw up $\lim\limits_{x\to 0} x\sin(1/x)$.
I think the language use is synonymous. I guess you are wanting to specify which sort of indeterminate form it is ($0/0$, $0\times\infty$, $\infty/\infty$), but conceptually these all come about because we have an undefined quantity if we plug in the limit. You might be right, though, since $1/x$ is undefined at $0$, but not a typical "indeterminate form."
I'm lost in the forest of fancy mathematical terms, happy to have my intuition light up for me what's left of affordable knowledge, made for average people, beginning from myself.
There is no room for poetry in this virtual room, math is too strict and well-constructed, that the most inspiring words, simply turn into one line of Logical deduction.
@Mahmoud: From my exam last time I taught this course. Give me the $\delta$-$\epsilon$ proof for $$\lim_{x\to 2}\frac{x^2+5x-2}{x+1}=4.$$ (You can email me your proof.)
@MathWanderer Sure, but what sort of properties should be observable on the graph? I mean, we can do this basically by some random bijection between suitable sets, but that would hardly be interesting
@TedShifrin Hmm, just stumbled upon something rather interesting
This MO answer talks about a lot of different applications of the determinant
and in particular: "14. In algebraic geometry, most projective curves can be seen as the zero set of some determinantal equality det(xA+yB+zC)=0. The theory was developed by Helton & Vinnikov. For instance, a hyperbolic polynomial in three variables can be written as det(xI_n+yH+zK) with H,KH Hermitian matrices; this was conjectured by P. Lax."
Probably the place to start is to write it using real variables. The Cartesian form isn't really helpful here, but the polar form looks relatively nice.
That means that the inequality can only depend on the argument of a given point. So the boundaries of the region will be rays emanating from the origin.
@user379685 For that, I think it's best to use the double-angle identity to write $\cos 4a$ in terms of $\sin 2a$. That'll give you a quadratic equation in $\sin 2a$.
