New answers tagged

0

There is a notation for an element to equal a specific value using indicator functions. You could thus define the proportion equal to $x$ as follows $$ p_x=\frac{1}{N} \sum_{s \,\in\, S}I(s=x) $$ Here $N$ is the number of elements in $S$ and $I$ is the indicator function that equals $1$ when $s=x$ and zero otherwise. There should be no problem using your ...


2

Short answer: the solution provided is wrong. Linearity should not be characterized by "the value of power to which $x$ is raised". A precise definition depends on the context - elementary school, algebra I, linear algebra in college. Your argument about lines in the plane is the right one for grade 6. With any reasonable definition, the sum of linear ...


1

Yes, $\mathbb{R^n}$ with the $p$-norm is a special $L^p$ space with the counting measure on $\{1,...,n\}$: Let $x \in \mathbb{R}^n$. Then we have that $$\int |x|^p \mathrm{d}\mu=\sum_{i=1}^n |x(i)|^p$$ Because we can identify every element of $\mathbb{R}^n$ as a function from $\{1,...,n\}$ to $\mathbb{R}$.


2

You calculate the expected value of a random variable, so the domain of the expectation operator is the set of random variables. These are measurable functions on the probability space. The codomain is the set where the random variable takes its values. Your example concerns vector valued random variables, so the expectation will be a vector.


1

Let $e(t)= \exp(2i\pi t)$ and $b_y = e(\alpha xy)$ if $y\le n/x$ and $b_y=0$ otherwise then $$\sum_{y\ge 1} b_y \log y=\sum_{y\ge 1} b_y \int_1^y \frac1{\gamma} d\gamma =\int_1^\infty (\sum_{y\ge \gamma} b_y) \frac1{\gamma} d\gamma=\int_1^{n/x}(\sum_{y\ge \gamma,y\le n/x} e(\alpha x y)) \frac1{\gamma} d\gamma$$


1

$n!!$ is known as a "double factorial," or "semifactorial," and it typically represents the product of all positive integers less than or equal to $n$ with the same parity (odd or even) as $n$. For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ $7!! = 1 \cdot 3 \cdot 5 \cdot 7$


0

While there is no ''standard'' notation, as you describe, that is analogous to the sigma and pi notations, you may use Knuth's up-arrow notation instead. This notation uses $\uparrow$ to denote exponentiation, $\uparrow\uparrow$ do denote tetration, $\uparrow\uparrow\uparrow$ to denote pentation, and so forth. For instnace, $$ {\displaystyle 2\uparrow 4=2\...


1

A useful resource for "speaking" mathematics is The Handbook for Spoken Mathematics by Lawrence A. Chang. This book suggests that we should use a description of the font or script, plus the name of the letter (see pages 3–5). Hence $\mathcal{E}$ might be read as "calligraphic capital $E$". In most contexts, this would likely be overkill and require ...


1

A common error committed by novice students is to confuse the name of a function with the values taken by that function. When we write $$ f : \mathbb{R}\to \mathbb{R} : x \mapsto x^2, $$ then: $f$ is the function itself. At a very basic level, $f$ is a set; specifically, a subset of the Cartesian product $\mathbb{R}\times\mathbb{R}$. Even more ...


4

I think the most straightforward way to regard it is just as a function composition, which is formally defined as $(f \circ g)(x) = f(g(x))$ (taking note of well-definedness in the respective domains). So in your specific example, let $g : \mathbb{R} \to \mathbb{R} : x \mapsto x - 1$, then $f(x - 1) = (f \circ g)(x)$. And yes, to add to your additional ...


3

$$ \nabla f= \begin{pmatrix} x_1^{n-1} \\ x_2^{n-1} \\ \vdots \\ x_n^{n-1} \\ \end{pmatrix} $$ Here is the syntax used: $$ \nabla f= \begin{pmatrix} x_1^{n-1} \\ x_2^{n-1} \\ \vdots \\ x_n^{n-1} \\ \end{pmatrix} $$


0

With parametrization $$x=\cos t, y=\sin t$$ we get $$|z-1|=\sqrt {(1-\cos t}$$ Therefore the integral is $$2\int _0^{2\pi}\sin (t/2)dt =8$$


1

Set $z:=e^{i\theta}$ for $0\leqslant\theta\leqslant 2\pi$ then you integral around the contour you say is equivalent to \begin{align} \int_{|z|=1}|z-1||\,dz|=\int_0^{2\pi}|e^{i\theta}-1||ie^{i\theta}\,d\theta|&=\int^{2\pi}_0\sqrt{(1-\cos\theta)^2+\sin^2\theta}\,d\theta\\&=\int^{2\pi}_0\sqrt{2-2\cos\theta}\,d\theta \end{align} Using double angle ...


3

Parametrize with the usual $z=e^{it}$ $$=\int_0^{2\pi} |e^{it}-1||ie^{it}dt| = \int_0^{2\pi} \sqrt{(1-\cos t)^2+\sin^2 t} \hspace{4 pt}dt$$ From here simplify and use trig identities. (Hint: if you get $0$ you did something wrong, perhaps related to the simplification of the square root. The answer should be $8$).


2

No, there's no such thing as a point infinitely close to another point. I've never seen the notation $\int_{a^+}^{b^-}$. If I did see that notation I can't imagine what it would mean other than $\int_a^b$. The notation $a^{\pm}$ comes up in limits, not integrals! $\lim_{x\to a^+}f(x)$ means the right-hand limit of $f$ at $a$. The notation can be ...


0

Adding + above a number in limit means that we are approaching that value from positive side( right side) whereas - means approaching from left side.


1

It's probably because the notation could make it clear that the function is $f^2=f\cdot f$. That is, if $g=f^2=f\cdot f,$ then we'd write $g(x)=f^2(x).$ Writing $f(x)^2$ kind of hides the functional relationship. Although, as noted in the comments, writing $f^2(x)$ could be confused with composition. It's probably best to clarify what you mean when you write ...


3

Piecewise: $$ t(x)= \begin{cases} 0.09x &\text{if $x \le 750$}\\ 0.09(750)+0.11(x-750) &\text{otherwise} \end{cases} $$ Shorter: $$t(x) = 0.09x+0.02\max(x-750,0)$$


0

If $\Gamma$ is normal in $SL_2(\Bbb{Z})$ for a fixed $\delta\in SL_2(\Bbb{Z})$ $$ \delta\Gamma=\Gamma \delta = \bigcup_l P g_l \qquad (disjoint\ union) $$ where $P = \Gamma \cap \langle p \rangle=\langle p^N \rangle$ with $p(z)=z+N$. $$E_{\Gamma,\delta,k}(z)=\sum_l (Pg_l)'(z)^{k/2} \qquad is \ well-defined :\ \ (p^m g_l)'(z)=g_l'(z)$$ For all $\beta \in \...


3

This is the collection of sets $$\{n+p\mathbb Z\mid n\in\mathbb Z\}$$ which forms a group $(G,+_G)$ with the group operation “$+_G$” defined by $$(n+p\mathbb Z)+_G (m+p\mathbb Z)=(n+m)+p\mathbb Z$$ Recall that $$p\mathbb Z=\{\ldots, -2p,-p,0,p,2p,\ldots\}$$ and that $$n+p\mathbb Z=\{\ldots, n-2p,n-p,n,n+p,n+2p,\ldots\}$$ For a given positive integer $p$, ...


2

In addition to this answer, the notation $A^*$ comes to mind. I've seen this in a few contexts: In symbolic dynamics, an alphabet $A$ is a (finite) set. A word is a finite sequence in which each term comes from $A$. You will often see the notations $$ A^0 := \varnothing,\qquad A^n := \{ (a_1, a_2, \dotsc, a_n) : a_j \in A \}, \qquad A^* := \bigcup_{j=0}^{...


2

In more familiar terms, by the formula for the square of a sum, $$\frac1n\sum_{i=1}^n x_i^2-\left(\frac1n\sum_{i=1}^n x_i\right)^2=\frac1n\sum_{i=1}^n x_i^2-\frac1{n^2}\sum_{i=1}^n x_i^2-\frac2{n^2}\sum_{i=1}^n\sum_{j=i+1}^n x_ix_j.$$ The notation $S(x_1x_2)$ is questionable, because it doesn't clearly shows that the summation indexes do not cover the ...


1

Try reading $S(x_1)$ as $\sum\limits_i x_i$, then $S(x_1^2)$ as $\sum\limits_i x_i^2$ and $S(x_1 x_2)$ as $\sum\limits_{i,j} x_ix_j$


1

For reflexive, it must contain all the pairs $xRx$, so your last example is correct. It can contain any other pairs desired, so $\{(1,1),(2,2),(3,3),(2,3)\}$ is reflexive as well. For symmetric, note that it says $aRb \implies bRa$. If $aRb$ is not true it does not impose a requirement. Both your examples are symmetric, as is the empty relation. ...


0

After having spent some time searching, I cannot find an obvious dupe target for this question. I am, frankly, quite surprised by this. However, it is not an unreasonable question, and I think it deserves to be answered on MSE. I am going to answer a slightly more general question: What does it mean to "extend a function"? Suppose that $\Omega \...


3

For a set $A$, a usual way to denote the set of finite sequences of elements of $A$ is $$A^{< \mathbb N}.$$


2

As explained by @kccu, using $e_{i,j}$ would be a rather poor choice. This often denotes the edges between vertices $i$ and $j$. I would use a capital "blackboard bold" P : $\mathbb{P}$, and hence $\mathbb{P}_{i,j}$


1

I post my attempt here. It would be great if someone helps me verify it ^o^ It follows from $G$ is a closed linear subspace of the Hilbert space $H$ that $G$ is also a Hilbert space. By Riesz theorem, there is a unique $x_G \in G$ such that $g(x) = \langle x_G,x \rangle$ for all $x \in G$ and that $\|g\| = \|x_G\|$. We define $f:H \to \mathbb R$ by $f(x) = \...


2

It means that there is a unique linear extension $f: H \to \mathbb R$ of $g$ which is norm preserving, i.e. satisfies $\lVert f \rVert = \lVert g \rVert$.


3

I think you cluttered the words exactly in the wrong way. The statement is, that you have a map, that is linear and defined on the whole Hilbert space $H$ (not just $G$). But its norm stays the same or you can say is preserved,e.g. $\|g\| = \|f\|$ and $f\bigg|_G = g$.


2

Yes, that is exactly what this symbol means here, bidirectional assignment. In the flow of the statement of the theorem, one has to interpret the first use as declaring the map from $a$ to the remainder tuple, later comes the claim of the theorem that this map is bijective, justifying the double arrow. The only irritating part may be that the bidirectional ...


0

No, this is not possible. Whenever you use any symbol, it should be clearly stated what that symbol represents and it should refer to one object only. It might be fixed number, or it may be some variable, but it can't change meaning before you are done with whatever you are trying to do. If you want to write something similar to allow different values on ...


1

Yes, it is a quantifier. You can write the equivalence $$\not\exists a:p(a)\equiv\forall a:\lnot p(a).$$


1

I don't think that it's a standard notation. Nevertheless one can see in some papers and books the notation : $$\frac{\partial^{n+m}f(x,y)}{\partial^n x\:\partial^m y}\equiv f^{(n,m)}(x,y) $$ which is also used by WolframAlpha. From the context, this is consistent with : $$ f^{(1,0)}(x,y) \equiv \frac{\partial f(x,y)}{\partial x}$$


0

I suspect that notation means $f_x (x,t)$ which means the partial derivative of $ f(x,t)$ .


1

Your problem is thinking about "the" subspace of $\mathbb{R}^3$ that is (isomorphic to) $\mathbb{R}$. There are infinitely many. You can single out the three coordinate axes as special since you are representing vectors as ordered triples, but the $x$-axis is no more special than the other two axes and does not have another common name. Even calling the $x$-...


0

It is indeed reasonable to use arbitrary sets as the index sets for matrices, rather than specifically sets of the form $\{ 1, 2, \ldots, n \}$. Note that even infinite sets are reasonable here, giving you infinite dimensional matrices! This introduces some extra complications that you have to deal with; e.g. matrix multiplication would involve an infinite ...


2

Short Answer This is just repeated iterations of the Hockey-stick Identity. Quick Overview of the Hockey-stick Identity Here's my attempt at a more approachable description: Notice that for each of the colors, the sum of all of the numbers in the diagonal is the last number on the bottom that goes off in the opposite direction. For instance, if we look ...


0

As you say, $(i,j)\in A \times B$ refers to the elements, not to the indices, sets have no notion of indices. So if you want to be really strict and rigorous with your notation, you have to use ordered sets $A=\langle a_1, a_2, ... \rangle$ and $B=\langle b_1, b_2, ... \rangle$ (or some kind of similar construction). However, what does $c_{ij}$ means here? ...


3

Originally I skimmed through most of your question and mainly answered the question in bold. I have since reread the question and decided to give some more comments. Ordinal vs Cardinal indices You claim that the ordinal notation in indices of cardinals makes the computation unwieldy, however there is a very distinct reason for using ordinals and not ...


4

In algebraic geometry, a variety is defined to be the solution set of a system of polynomials over an algebraically closed field. More precisely, if $k$ is an algebraically closed field and $f_1,\dots,f_m$ are polynomials in $n$ variables with coefficients in $k$, then $$V(f_1,\dots,f_n) = \{(x_1,\dots,x_n) \in k^n \mid f_1(x_1,\dots x_n)=\cdots = f_m(x_1,\...


1

It is a specific type of linear matrix pencil.


3

I am not aware of any special name for an operator or matrix of the form $\lambda I - T$ (or $cI - A$) in either linear algebra or functional analysis. However, some possibilities are an inverse of the resolvent, a pseudoinverse of the resolvent, or a pre-resolvent. All three of these are meant to capture the the notion that $cI - A$ has a relation to the ...


2

It is the integration-by-parts formula, although demonstrated in a quite confusing way. Especially, it is unclear which function $\int$ is applied to. Using parentheses to emphasize the scope of $\int$ for each instance, we may instead write $$\int(fg) = \left(\int f\right) g - \int \left(\left( \int f \right) g'\right). $$ It is still an unconventional ...


2

He is writing an integral of a product, and the two factors are $\int f$ and $g'$. So he wrote the outside integral bigger. $$ \int\;{\textstyle \int f} \cdot g' $$


0

For me the braces $\{X_n\}$ would suggest that the order is irrelevant (and so we're not taking a limit, say), while brackets $(X_n)_{n \ge 1}$ is much more suggestive that the order matters. "Family" normally suggests a set (so no order). Just my 2 cts.


0

The answer is -1, please let me show You why. In binary there is one integer with one digit, two integers with two digits, four integers with three digits and so on, so there always are $2^(n-1)$ integers with n digits. So there must be $2^(0-1)$ = 1/2 integers with 0 digits, $2^(-1-1)$ = 1/4 integers with -1 digits and so on. 1/2 + 1/4 + 1/8 + ... add up ...


2

This is the indicator function. $$f(x)=1_{[-1,1]}(x)=\begin{cases}1&x\in[-1,1]\\0&x\not\in[-1,1]\end{cases}$$ To compute its Fourier transform is simple, you integrate this function over $(-\infty,\infty)$ as usual, but since the function is zero outside this specific range, the integral limits become $\int_{-1}^1$.


0

The line segment between $[0,B]$ taken $n$ times, i.e. for $n$ different real numbers. A subset of $\mathbb R^n$ where each real number is between $[0,B]$. So a function takes $n$ real numbers in (each between $0$ and $B$) and gives one value between $0$ and $B$ out.


3

The Cartesian product of $[0,B]$ with itself n times. That is, the set of ordered $n$-tuples $(x_1,x_2,\ldots,x_n)$ where $0\leq x_i \leq B$ for $i=1,\ldots, n$.


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