Hot answers tagged

14

Intuitive interpretation: A curve has an infinity of points. So many that the amount counteracts the abscence of dimensions, like an $0\times\infty$ undeterminacy. More precisely, a curve still has no dimension transversally, but a finite (or infinite) dimension longitudinally. You can think of it as the limit of a necklace of pearls (points of finite ...


10

Euclid did say that that "A point is that which has no part". It is nothing more that a whisper of an indication that "You are here." You can't really see a point since there is nothing there. "...if point has no dimensions, i.e. in other words there is nothing, then how is it possible to draw any curve?" It isn't possible. Pick a ridiculously small ...


5

A curve is completely determined by two facts: Knowledge of all of the points lying on the curve Knowledge that the curve is drawn on the Euclidean plane When it's said that a curve is made out of points, one really means to include in the latter fact too, or something similar (e.g. a topology or a metric on the collection of points). There are more ...


5

You are confusing theory with parctise. A point does not exist in the real universe. Anything you draw will have some dimensions.


4

It is hard to know what will give you an intuition without knowing what you want and what background you already have. If you are asking roughly what kind of geometric or topological information Cech cohomology reveals about a space, then answer is the same as for all cohomology theories: it reveals the global connectivity of spaces with relatively simple ...


4

Note that $\lfloor x \rfloor = x - \{x\}$, where $\{x \}$ (the fractional part of $x$) is periodic with period $1$, with a jump discontinuity at integer points. Let's try to write $\{x\}=f\left(\cot \pi x\right)$, since $\cot \pi x$ has the same property. We need $f(y)\rightarrow 0$ as $y\rightarrow \infty$, $f(y)\rightarrow 1$ as $y\rightarrow -\infty$, ...


4

$$\frac{1}{(1-x)^{a+1}}=\left(1+x+x^2+x^3+\ldots\right)^{a+1} $$ hence the coefficient of $x^n$ in this power series is the number of elements of the set: $$ E_n = \{(n_1,n_2,\ldots,n_{a+1}):\, n_i\in\mathbb{N},\, n_1+n_2+\ldots+n_{a+1}=n \} $$ that by stars and bars ig given by $\binom{n+a}{a}$. It follows that: $$ \frac{1}{(1-x)^{a+1}}=\sum_{n\geq ...


4

You are confusing reality and the idealized world of maths. In an ideal world, a point has no dimensions. You could never actually draw a point. In reality, you can. You take a pencil, make a point on a piece of paper and say "that is the point". But the fact that you can see the point already means that is has some dimensions (the ink occupies a non-empty ...


4

If you have some physics background, then $L^2$ norm can often be interpreted as the "energy" of the wave functions. Physical interpretation of L1 Norm and L2 Norm In quantum physics, the $L^2$ norm represents the probability of detecting a particular pure state amount many mixed states. In statistic, minimizing the $L^2$ norm of the difference between 2 ...


4

I'm not sure how intuitive is this, but the norm $||f||_{L^2([a,b])}$ of an integrable function defined on $[a,b]$ is the square root of the "area under the graph" of $|f|^2$ on the interval $[a,b]$. For example, if $f \equiv C$ is a constant function, then the area under the graph of $f^2 = C^2$ is the area of a rectangle given by $(b - a)C^2$ and so ...


4

There are several good answers here, one accepted. Nevertheless I'm surprised not to see the $L^2$ norm described as the infinite dimensional analogue of Euclidean distance. In the plane, the length of the vector $(x,y)$ - that is, the distance between $(x,y)$ and the origin - is $\sqrt{x^2 + y^2}$. In $n$-space it's the square root of the sum of the ...


3

A point in an n-dimensional Euclidean space is that point which consists of n coordinates which give a distinct location of that point. A function in Euclidean space essentially provides a rule for defining a set coordinates in space for which that function is true. You are correct in that the physical realization of a collection of 0-dimensional points ...


3

If $p = 2$ then $-1$ is a square mod $p$, so we may now assume that $p$ is odd. Suppose that $n^2 \equiv -1 \bmod p$ for some $n \in \mathbb{N}$. Then, using that $\mathbb{Z}/p\mathbb{Z}$ is a field, $$ 1 \equiv n^{p-1} \equiv (n^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p - 1}{2}} \bmod p $$ which implies that $\frac{p-1}{2}$ is even, thus, that $p \equiv 1 ...


3

For $p>2$ prime, $-1$ is a quadratic residue modulo $p$ if and only if $p \equiv 1 \bmod 4$. So you can determine whether $-1$ is a quadratic residue modulo $p$ as quickly as you can find the remainder of $p$ upon division by $4$.


3

Sufficiently constructive mathematics has semantics in a topos. In a topos there is on the one hand the subobject classifier $\Omega$, and on the other hand the coproduct $2 = 1 + 1$ of two copies of the terminal object. I believe the inclusion $1 \to 1 + 1$ of, say, the left copy of $1$ is a monomorphism in a topos, and hence it is classified by a map $2 ...


3

Yes, there is. In elliptic case, we have that $A:=(a_{ij})$, the matrix corresponding to second derivatives, is positive definite. Psitive definiteness implies that $$v^t.A.v$$ is positive for all $v\neq 0$ and zero for $v=0$. If one draws all the $v$'s which satisfy $ v^t.A.v =c$ for a constant $c$, it will be an oval-like shape, an ellipsis (in n-dimension ...


3

A differential is a $1$-form. At each point a $1$-form gives a linear functional in a tangent space, and the level sets of a linear functional are parallel affines subspaces. Now $df$ in $\mathbb R^3$ gives a sort of a ruling in each tangent. For a given $p$, the level sets of $df_p$ are what we can see if we zoom in on the level sets of $f$ in a ...


3

OK let's see if this helps you. Suppose you have two functions $f,g:[a,b]\to \mathbb{R}$. If someone asks you what is distance between $f(x)$ and $g(x)$ it is easy you would say $|f(x)-g(x)|$. But if I ask what is the distance between $f$ and $g$, this question is kind of absurd. But I can ask what is the distance between $f$ and $g$ on average? Then it is ...


2

Just try and calculate. The easiest example of a cusp is $\text{Spec }k[t^2,t^3]\to\text{Spec }k[x,y]$, where $x\mapsto t^2$ and $y\mapsto t^3$. The completion of the plane at the origin is $\text{Spec }k[[x,y]]$, formal power series in two variables. The fiber of the cusp over the completion is $k[[x,y]]\otimes_{k[x,y]}k[t^2,t^3]=k[[t^2,t^3]]$, formal power ...


2

Preliminary remarks: 1) I have tried, in the following lines, to stick to the kind of answer desired by @Lui, i.e., an intuitive understanding. Of course, a rigorous demonstration is needed, but, for that, one can find explanations a little everywhere, in particular in the extraordinary book "Concrete Mathematics" : ...


2

The way I like to think about this is that if we have two vector spaces $V$ and $W$, their tensor product $V \otimes W$ intuitively is the idea "for each vector $v\in V$, attach to it the entire vector space $W$." Notice however that since we have $V\otimes W \cong W \otimes V$ we can rephrase this as the idea "for every $w \in W$ attach to it the entire ...


2

1) There is no line field on $S^2$. More or less this follows because $\chi(M)$ is nonzero. 2) The condition $\omega \wedge d\omega = 0$ is known as being involutive. The Frobenius theorem says that involutive distributions are integrable: there is always a chart $U$ such that, in this chart, the distribution is spanned by the first $k$ tangent vectors ...


2

As at least one other answer has mentioned, it takes an infinitude of points to make up a curve. (Or to make up any other interval, like the rather simple "curve" of the closed interval [0,1], for example.) At least one other answer has also mentioned the notion of "measure," and that's probably the more appropriate way to think about the small-ness of a ...


2

Consider function $f(z)=z^n$. As you go around a small contour around $z=0$ which is $n$-fold zero of $f$, $f(z)$ "goes around" $0$ in complex plane $n$ times. The same will be true for any holomorphic function with $n$-fold zero at some point (meaning that it is zero there together with its first $n-1$ derivatives). This multiplicity of zeros is therefore ...


1

Suppose $F$ is homeomorphic to a compact subspace of $\Bbb R^\infty$. (There is some nice classification of which spaces this is true of, but I forget it. Maybe every compact Hausdorff space? At the very least it is true of any compact manifold and of any finite CW complex.) Consider the space "$B(F)$" (terrible notation, sorry; this is meant to evoke ...


1

In Natural Deduction, it is correct given $A$, to derive $C \to A$. See: Jan von Plato, Elements of Logical Reasoning (2013), page 22: Consider as another case $A ⊃ (B ⊃ A)$. Verbally, if we assume $A$, then $A$ follows under any other assumption $B$: $A$ --- hypothesis: goal $B ⊃ A$ $B ⊃ A$ --- 1,$⊃$I $A ⊃ (B ⊃ A)$ ...


1

If we're talking of the field $\;\Bbb F_{p^2}\;$ , then the multiplicative group of this, $\;\left(\Bbb F_{p^2}\right)^*:=\Bbb F_{p^2}\setminus\{0\}\;$ is a cyclic group (as is any finite subgroup of the multiplicative group of any field), and in this case it is a group of order $\;p^2-1\;$ , so you're looking for the number of generators this group has, and ...


1

Start with $$\oint_{\gamma}f'(z)\textrm{d}z=\oint_{\gamma}\textrm{d} f(z) = 0.$$ This shows that Cauchy's theorem holds for all functions that are the derivative of some other function. This includes $z^m$ for all $m\in\mathbb{Z}\setminus\{-1\}$. So it holds for all polynomials and therefore also for all functions that can be approximated by polynomials on ...


1

I have grappled with this question myself, and here is what I came up with. Holomorphic functions are highly constrained. A priori, all we know is that they must satisfy the Cauchy–Riemann equations, but by doing a little math we see that that infinitesimal constraint propagates outward into local and even global constraints. This propagation is, to be ...


1

An attempt at intuitive reasoning (given high dimensional results are often counter-intuitive): Consider a unit $n$-ball centred on the origin, inscribed in an $n$-cube of sidelength $2$. The vertices of the $n$-cube lie at distance $\sqrt{n}$ from the origin, which for large $n$ makes the radius $1$ look negligible. For large $n$ we could generously ...



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