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6

Let $x=u^2$, $dx = 2 u \,du $. Then the integral is $$2 \int_0^1 \frac{du}{\sqrt{1-u^2}} $$ Now let $u=\sin{t}$, $du = \cos{t} \, dt$. Then the integral is $$2 \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}} = 2 \frac{\pi}{2} = \pi$$


6

If to get from my home to the university I need to take a bus to Jerusalem, and then an internal bus; on the way back I first need to take an internal bus, and then a bus from Jerusalem. If you think about $AB$ as applying $B$ and then applying $A$, reversing it means first undoing $A$ and then undoing $B$. (Of course, as noted on this page, you need to ...


4

There is a combinatorial proof that is quite easy to follow (and remember, too). Assume that $G$ is a finite group and $p\mid |G|$. Let $X$ be the following set: $$ X = \{ (g_1,g_2,\ldots,g_p) : g_1 g_2\cdot\ldots\cdot g_p = e\}.$$ We have $|X| = |G|^{p-1}$, since for every element of $X$ we are free to take the first $(p-1)$ coordinates as we want, then ...


3

This is false in general. A counter example: Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ Then $AB=\begin{bmatrix}1&2\\0&2\end{bmatrix}$ and $(AB)^{-1}=\begin{bmatrix}1&-1\\0&\frac{1}{2}\end{bmatrix}\neq ...


3

If $A$ and $B$ are invertible then $$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$$ and it follows that (the definition of the inverse of $X$ is $XX^{-1} = I$) $$(AB)^{-1} = B^{-1}A^{-1}$$ The order of $A$ and $B$ have to be reversed in the equation for the inverse of $AB$ in order for the defining relation $(AB)(AB)^{-1} = I$ to be satisfied.


3

This is a way to explain the intuition behind de Rham cohomology: Cohomolgy comes up as a dual answer to homology. Homology identifies the shape of an object finding ‘holes’. More concretely, it looks for objects without boundary which are not the boundary of an object (and therefore the definition $H_k(M)=\text{ker}\partial_n/\text{im}\partial_{n+1}$). ...


3

I don't know of a totally elementary proof of this result, but here is some context for it. More generally, an $n$-manifold $M$ has a classifying map $f : M \to BO(n)$ for its tangent bundle. Knowing when $M$ is parallelizable is equivalent to knowing when $f$ is null-homotopic. There is a general machine for doing this involving lifting $f$ higher and ...


3

It's definitely late in the game to this question, but Martin kindly pointed me here, and I think there's something else to be said. The search for the elementary proof has two points of view I think are related to the soul of mathematics as it stood during the time when the preoccupation was great. The First Further Advancement of the subject ...


3

I think that many mathematicians (including me) create long proofs similarly to you, so this is a general way. Therefore I cannot add much to my answer to the linked question. Sometimes long proof are guided by a priori intuition, as Poincaré wrote, sometimes they are a posteriori glued together, for instance, as a classification of finite simple groups. ...


3

If you look at a graph of $y=x^3$, to my eyes, as I read it left to right, I can see the downward concavity peter down to $0$ before turning into upward concavity. But this is not what I see with $y=\frac{x^3}{|x|}=x|x|$. In that graph as I read it left to right, there is strong downward concavity until we reach $0$, and then the concavity jarringly ...


2

From a physical point of view, the connection between the heat equation and the normal distribution is clear, because the heat equation can be derived by considering Brownian motion. Brownian motion means adding up lots of independent, identically distributed numbers - each time a particle moves, we assume it moves a random amount, with each of these being ...


2

Using a self-similar solution which is $$ \partial_t T = \nu\partial_{xx}T $$ we can look at self similar transform via $$ \frac{\bar{T}}{t}\sim \nu\frac{\bar{T}}{x^2}\implies x\sim (\nu t)^{1/2} $$ using a transform of the form $$ \zeta = \dfrac{x}{(\nu t)^{1/2} } $$ then lets look for solutions of the form $$ T = \bar{T}f(\zeta) $$ thus we obtain $$ ...


2

Scroll to the bottom of Clifford Algebra: A Visual Introduction, right under Clifford the Big Red Algebra and you will find two links: The story continues with Geometric Algebra: Projective Geometry. The final chapter is Geometric Algebra: Conformal Geometry. If you like this intuitive graphical presentation of Clifford Algebra you might also be ...


2

I don't know how 'real' this is, but here's one way to visualise $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. There are two copies of $\mathbb{Z}$, call them $\mathbb{Z}\times\{0\}$ and $\mathbb{Z}\times\{1\}$. $$\mathbb{Z}\times\{0\}\qquad :\qquad \dots\qquad -3\qquad -2\qquad -1\qquad 0\qquad 1\qquad 2\qquad 3\qquad \dots$$ $$\mathbb{Z}\times\{1\}\qquad ...


2

Here's a simple geometric interpretation. Consider the figure below, with marked fenceposts (or copies of the letter C) at every integer point on a horizontal line. Its group of symmetries is $\Bbb Z\times \Bbb Z/2\Bbb Z$, generated by the horizontal translation and reflection across the horizontal.


2

You will find a lot of intuitive reasons why Fermat last theorem holds in the following book : "Modular Forms and Fermat’s Last Theorem", from Cornell, Silverman and Joseph, I quite liked it in my young years.


2

Functional dependence of $k$ given functions $F_i:\>\Omega\to{\mathbb R}$ $\>(1\leq i\leq k)$ with common domain $\Omega\subset{\mathbb R}^n$ means, intuitively, that there is a nontrivial function $$g:\quad{\mathbb R}^k\to{\mathbb R},\qquad y\mapsto g(y)$$ such that $$g\bigl(F_1(x),F_2(x),\ldots, F_k(x)\bigr)\equiv 0\qquad \forall x\in\Omega\ ...


2

Functional independence between two functions means that the level set of each of the functions intersects transversely the level set of the other function. In the plane, this means that a function $f$ functionally independent of another function $g$ cannot be written as $f=F(g)$ where $F$ is another function, because in this case, the level sets would be ...


2

Let's first see how we arrive at the correct negation using predicate logic and then justify: $$\neg \forall \epsilon > 0 \exists \delta > 0 \forall x, y \in I [ |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon] \\ \exists \epsilon > 0 \neg \exists \delta > 0 \forall x,y \in I [ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon] \\ ...


2

$\newcommand{\Z}{\mathbb{Z}}$$\renewcommand{\phi}{\varphi}$I am expanding on the comment of @mixedmath If $\gcd(e, \phi(n)) > 1$, then the function $\Z_{n} \to \Z_{n}$ given by $x \mapsto x^{e}$ is not injective. Thus there are two different messages $x \ne y$ that are mapped onto the same $x^{e} = y^{e}$, and there it goes any possibility of (uniquely) ...


1

Dot project is symmetric: $\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}$. Scalar projection is not. Dot project is sensitive to the magnitude of the second vector : $\vec{u} \cdot k\vec{v}=k(\vec{u} \cdot \vec{v})$. Scalar projection is not. If your distances are measured in metres then the dot product is measured in square metres but the scalar ...


1

How to order the set of ordered pairs of positive integers, from George Boolos & John Burgess & Richard Jeffrey, Computability and Logic (5th ed - 2007), page 8 : As a preliminary to enumerating them, we organize them into a rectangular array. We then traverse the array in Cantor’s zig-zag manner [indicated in a similar figure]. This ...


1

Firstly, the exponent $e$ is not private rather public and shared with everyone. The $e$ is chosen to be relatively prime to $\phi(n)$ because $Z{^*}_n$ is a multiplicative group i.e. the set of all +ve integers less than $n$ that are relatively prime to $n$. Since the multiplicative inverse $d$ of $e$ is unique and finding it is very hard without knowing ...


1

The derivative of $F$ is $DF=[\nabla F_1,\cdots,\nabla F_k]^T$ ; let $a\in\mathbb{R}^n$, $b=F(a)$ and $V=F^{-1}(b)$. $V$ is the intersection of the $k$ hypersurfaces $F_i(x)=b_i$. The normal vector in $a$ to such a hypersurface is $\nabla F_i(a)\in \mathbb{R}^n$. Here $rank(DF_a)=k$, that is $W=span(\nabla F_1(a),\cdots,\nabla F_k(a))$ has dimension $k$. ...


1

You could start with Flegg's From Geometry to Topology. It may not be extremely modern, but it is well-illustrated and prepares the ground for general topology. Another possibility is Richeson's Euler's Gem, which is indeed a gem of a book and gently leads into topology by way of Euler's polyhedral formula. A modern follow-up would be Introduction to ...


1

A function is also $\,f(z)=a\,z^{14}+b\,z^{15}+c\,z^{16}$ is just $z^{14}(a+b\,z+c\,z^2)$. And if you do a taylor expansion of $z^19+3z^6$ around $z_0=5$, you will find a linear term as well. Regarding the physics aspect with quantities where the minimum matters: Take any partition function of the form $z$ times something, average a couple of quadratic ...


1

Otherwise stated $\mathcal{A}_{\tau}$ is formed of events that can be defined by questions that can be answered (positively or negatively) only if you know whether $\tau$ as happened at time any time $n$. Notice that involving the knowledge of $X$ in the informal definition is not necessary IMO unless the filtration is the natural filtration generated by ...


1

Usually one requires certain properties (axioms) for an entropy: Nonnegativity; the bigger the uncertainty, the bigger the entropy; and additivity for independent observations/measurements. The last property implies that there should be a logarithmic dependence. For more details on the axiomatic formulation of entropy see ...


1

Here is (in my estimate) the approach outlined in Milnor-Stasheff. Let $M$ be a closed, orientable 3-fold, $w_i$ denote the $i$th Whitney class, and $o_i$ the $i$th obstruction class as it is defined in Steenrod's Topology of Fibre Bundles . As $M$ is orientable, $w_1=0$ and a straightforward calculation with Wu's formula gives $w_2$ and $w_3=0$ as well. ...


1

The integral equals $$ \mathrm{B}(1/2,1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt \pi \sqrt \pi}{1} = \pi. $$



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