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13

If you insist on uncountable unions, then the measure of the unit interval is the sum of uncountably many measures of the individual points of the interval. These measures are either 0 (in which case $\mu(I) = 0$, which is bad) or nonzero, in which case $\mu(I)$ would be $\infty$,. A possibly more interesting question is whether finitely-additive measures ...


12

Consider the area of the surface of the $n-$Ball with radius $1$ . It is given by: $$ A_{n}=2\frac{\pi^{n/2}}{\Gamma(n/2)} $$ Our intuition tells us that for $n=1$ the surface "area" (or to be mathematically more precise the, Hausdorff measure as @Michael Galuza pointed out correctly)should be 2, because it consist of two points. To make this ...


9

It's quite simple: Allowing only finite union yields too weak results [cf. Jordan measure] and allowing uncountable unions makes the system of measurable sets too big. If you for example assume that all singletons are measurable, then uncountable unions would imply that any set is measurable. This already doesn't work well with the Lebesgue measure, as the ...


6

Inasmuch as the OP requested an approach that is intuitive or geometric, let's proceed accordingly. The ensuing heuristic discussion serves only to supplement the more analytical/rigorous approaches, and can help solidify the understanding of the relationship. Suppose we have a right triangle formed from the three coordinate points $(0,0)$, $(1,0)$, and ...


5

Another reason is that in a summable family $(x_i)_{i\in I}$, it can be shown the set $\;\{i\in I\mid x_i\neq 0\}$ is at most countable. Thus the theory of summable families comes down to the theory of series.


4

Countable has this very peculiar property of infinity. It can be indexed, such that every point is only a finite distance away. Namely, it can be indexed using $\Bbb N$. And $\Bbb N$ has that great property that given any $k\in\Bbb N$, you only need to travel finitely many steps to meet with $k$. Why is that important? Because (unfortunately) mathematics ...


4

The field of probability can be made mathematically rigorous. Introductory textbooks on probability tend to use terminology like 'sample space', 'outcome', 'event', and outcomes are given names like 'Heads' or 'HTT' or 'King of Spades', all in an attempt to keep things informal and intuitive. These texts often refrain from defining such concepts precisely, ...


4

It means that these antiderivatives exist and are perfectly fine functions, but they can not be expressed by other elementary functions (functions that we are already well familiar with). If such anti derivative comes up very often, it is often given a name. Then you can study its properties and use it like any other function. Such functions are often called ...


3

Remember that a function is simply a rule for turning an input into an output. This means that: $$f(x)=\int_0^xe^{-t^2}dt$$ is a perfectly fine function. The fundamental theorem of calculus tells us that $f(x)$ is an antiderivative of $e^{-x^2}$. However, $f(x)$ cannot be written in closed-form. What that means, in this context, is that $f(x)$ cannot be ...


3

$$f\left( x \right) =\arctan { \left( x \right) +\arctan { \left( \frac { 1 }{ x } \right) } } $$ $$f^{ \prime }\left( x \right) =\frac { 1 }{ 1+{ x }^{ 2 } } -\frac { 1 }{ 1+{ x }^{ 2 } } =0$$ $$\\ f\left( x \right) =c\\ f=f\left( 1 \right) $$


3

Since the random variables are independent, \begin{align}\operatorname{E}[(X_i-\mu)(X_j-\mu)]&=\operatorname{E}[X_i-\mu] \cdot \operatorname{E}[X_j-\mu]\\ &= (\operatorname{E}[X_i] - \mu)(\operatorname{E}[X_j]-\mu) \\ &= (\mu-\mu)(\mu-\mu)\\ &=0 \cdot 0\\ & = 0.\end{align}


3

Might not be the answer you are hoping for, but intuitively you can see that if you have a volume in 3-space, then you can take that volume's boundary, which will be a closed 2-surface. A closed 2-surface has an empty boundary, ergo there is no curve that would serve as its boundary. Through the use of the usual integral theorems in 3-space, the identity ...


3

Take $\theta=\pi/2$, and write down one of the options. It is fairly easy to check in this case whether you got a clockwise or anti-clockwise rotation.


2

You already have a good explanation in terms of the topology on $P$ induced by the ordering relation. This topology also accounts for the terminology "dense open", which one often sees in connection with forcing. Nevertheless, your suspicion about the Stone space of the regular completion $B$ of a separative $P$ is also nearly correct. Given a subset $E$ ...


2

There's another way to interpret density topologically: given a poset $\mathbb{P}$, let $Fil(\mathbb{P})$ be the set of all maximal filters in $\mathbb{P}$. Then each condition $p\in\mathbb{P}$ yields a subset of $Fil(\mathbb{P})$: $U_p=\{G: p\in G\}$. The sets $U_p$ generate a topology on $Fil(\mathbb{P})$. For example, if $\mathbb{P}=2^{<\omega}$, the ...


2

Hint: Check the intuitive interpretation of curl and then that of divergence and combine them to obtain the interpretation of $\text{div}(\vec{\text{curl}} \vec{F})$


2

Shouldn't the rank be constant on a neighborhood of that point to make this conclusion? It is sufficient to check at that point if the rank is maximal, but otherwise the rank can "jump" up suddenly, meaning you would need more coordinates. It can't jump down suddenly, because the determinant function is smooth. It takes some time for a nonzero continuous ...


2

Is there an intuitive explanation to this ? Yes. There is an umbilical connection between $\bigg(\dfrac1n\bigg){\large!}$ or $\Gamma\bigg(\dfrac1n\bigg)$, and geometric shapes of the form $X^n+Y^n=R^n\iff x^n+y^n=1\iff y=\sqrt[n]{1-x^n}$, whose area is $\displaystyle\int_0^1\sqrt[n]{1-x^n}~dx$, which is nothing else than the beta function in disguise. ...


2

Let $$\tan ^{-1}(x)=\theta\iff x=\tan \theta$$ Where, $0<\theta<\frac{\pi}{2}$ Now, we know $$\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta$$ $$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{\tan\theta}$$ Setting $\tan\theta=x$ $$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{x}$$ $$\frac{\pi}{2}-\theta=\tan^{-1}\left(\frac{1}{x}\right)$$ ...


2

Let $f(x)=\arctan(x)+\arctan(\frac1x)$. Then $f'(x)=\frac{1}{1+x^2}+\frac{1}{1+1/x^2}\frac{-1}{x^2}=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$ so $f(x)$ is constant for $x>0$. Then note that $f(1)=\frac{\pi}{2}$ and thus $f(x)=\frac{\pi}{2}$. There is a discontinuity at $x=0$, so the derivative only makes sense for $x\neq 0$. When $x<0$, you can check that ...


2

$\lim_{x \to a} f(x)$, if it exists, is the standard part of $f(a+dx)$ for every nonzero infinitesimal $dx$. Similarly, $\lim_{x \to a^+} f(x)$, if it exists, is the standard part of $f(a+dx)$ for every positive infinitesimal $dx$. Note the quantifier: the limit only exists if the standard part is the same no matter what infinitesimal you choose. We don't ...


2

For your first question if we write the axiom like so $$g' \ast (g \ast s) = (g'\bullet g) \ast s$$ then $\ast$ is the action of $G$ on $S$, so the left operand to $\ast$ is an element of $G$, the right operand of $\ast$ is an element of $S$, and the result of the operation $\ast$ is an element of $S$. On the other hand $\bullet$ is the group operation, ...


1

A function $f$ analytic in a full disk $D_r$ around $a$ can be written as derivative of some other function $F$ in $D$:$$f(z)=F'(z)\quad(z\in D_r)\ .$$ If $f$ has an isolated singularity at $a$ one may still ask whether $f$ has a primitive $F$ in the punctured disk $\dot D_r:=D_r\setminus\{a\}$, in other words: whether the ODE $y'=f(z)$ has a solution in ...


1

Can anyone tell me why these integrals “cannot be found” ? For instance, the roots of a $($general$)$ quintic polynomial cannot be found either, i.e., they cannot be expressed in closed form. Basically, what the Abel-Ruffini theorem is for polynomials, Liouville's theorem and the Risch algorithm are for calculus.


1

Closed form means that the integral cannot be expressed in standard mathematical functions like trigonometric, exponential or polynomial functions etc. See the example of $\int {{2^2}^2}^x dx$. Also, you may want to study elliptic integrals.


1

A better understanding of $\mathcal P_T$ can be achieved by looking at a special case: Namely when $P_U$ and $P_V$ are diagonal, with all the $1$ entries coming first. For example, consider $m=3$, $n=4$, $r=2$, and $$A = \begin{pmatrix} 1 & 2 & \color{grey}{0} & \color{grey}{0}\\ 3 & 4 & \color{grey}{0} & \color{grey}{0}\\ ...


1

Letting $x=\tan u$, we have $$\begin{array}{lll} \arctan(x)+\arctan(1/x)&=&\arctan(\tan u)+\arctan(1/\tan u)\\ &=&\arctan(\tan u)+\arctan(\cot u)\\ &=&\color{blue}{\arctan(\tan u)+\arctan(\tan (\frac{\pi}{2}-u))}\\ &=&u+\frac{\pi}{2}-u\\ &=&\frac{\pi}{2} \end{array}$$ The line highlighted in blue should be similar to ...


1

$\arctan(a)+\arctan(b) =\arctan(\frac{a+b}{1-ab}) $. Therefore $\arctan(x)+\arctan(1/x) =\arctan(\frac{x+1/x}{1-1}) =\arctan(\frac{x+1/x}{0}) =\pi/2 $. If this bothers you, $\begin{array}\\ \arctan(x-c)+\arctan(1/x) &=\arctan(\frac{x-c+1/x}{1-(x-c)/x)})\\ &=\arctan(\frac{x-c+1/x}{c/x})\\ &=\arctan(\frac{x^2-cx+1}{c})\\ ...


1

The integral gives the area under the graph of a function $f(x)$. However, the integral can also give the average value of a function. Reviewing the definition of the summation definition of the integral, which cuts up the function into rectangles and then sums them, we see that there is a relationship between the summation and the average value formula. ...


1

Computing Nash equilibria is not an easy business in general (if it were, then the world would be a weird place...). You may want to note the following facts: Nash equilibria are may not exist in pure strategies. They're only guaranteed to exist in mixed strategies. In general, the problem of computing equilibria for bimatrix games is NP-hard (in fact, ...



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