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25

No-one in their right mind would denote $\sin\left(x^2\right)$ as $\sin(x)^2$. Why? Because the (round) brackets would become redundant. Brackets are used to remove ambiguity in algebraic operations. If you exclude the exponent $\quad ^2 \quad $ from the brackets, you're implicitly saying that the $\quad ^2 \quad$ belongs outside the brackets, and, ...


18

It's perfectly fine to have equality signs. When you solve equations, what you really do is say Assume that the following is true $$4x-4=\frac{(2x)^2}{x}$$ then $$-4=0$$ is true. Contradiction, the original assumption is false.


15

I put a question mark above an equal sign, like this: $\stackrel{?}=$. (In MathJax, just type $\stackrel{?}=$.)


8

It denotes the maximum (or, in case of sequence or function spaces, the supremum) norm (on both sides of the inequality), $$\lVert x\rVert_\infty = \max \{ \lvert x_k\rvert : 1 \leqslant k \leqslant n\}$$ if $x\in \mathbb{R}^n$ or $x\in\mathbb{C}^n$. For $1 \leqslant p < \infty$, one has the norms $$\lVert x\rVert_p = \left(\sum_{k=1}^n \lvert ...


7

I like to put the $\iff$ (if and only if) symbol at the beginning of every new line, like this: $$\begin{align} 4x-4 &=\frac{(2x)^2}{x} \\ \iff-4 &= \frac{4x^2}{x} -4x \\ \iff-4 &= 4x -4x \\[0.2em] \iff-4 &= 0\end{align}$$ So $4x-4=\frac{(2x)^{2}}{x}$ if and only if $-4=0$, which is true.


5

If anything, $\sin^2(x)$ is the ambiguous notation. To some it might mean $\sin(\sin(x))$ (this is why $\sin^{-1}$ is sometimes used as arcsine), and to others it might mean $(\sin(x))^2$. I cannot think of a case where anyone would see $\sin(x)^2=\sin(x^2)$ (except when, say, $x=0$). However, we mathematicians avoid ambiguity; I typically use $(\sin(x))^2$ ...


5

The difference is that $\mapsto$ denotes the function itself. Thus you need not name the function. $a\mapsto b$ fully describes the action of the function. $\to$, on the other hand, describes only the domain and codomain. Thus one might say $x\mapsto x+1$ is equivalent to $f(x)=x+1$, in which case we would say $f:\mathbb{R}\to\mathbb{R}$.


5

The $\to$ arrow points from the domain of a function to its codomain or target set. The $\mapsto$ arrow (fittingly called \mapsto in TeX) shows what an individual element of the domain will be mapped to, i.e. it shows what the function does while $\to$ shows where it operates (so to say). So, the elaborate way to write a function definition is ...


3

Typically: $R[x]$ denotes the set of polynomials over $R$ When $R$ is a domain, $R(x)$ denotes the set of rational polynomials over $R$ $R[[x]]$ denotes the formal power series over $R$ $R((x))$ denotes the Laurent series over $R$ vuur asked an interesting question in the comments which I can speak to here. The answer is "If $R$ is a commutative ...


3

You are close. I think some people use $B^A$ to denote the set of all functions from $A$ to $B$. So you could write $$\left\{f \in B^A: \left|f^{-1}(\{k\})\right|=c_k\right\}$$ to denote the set of all functions such that $c_k$ elements of $A$ are sent to $k \in B$.


3

This is really a question about English writing. The question could be rewritten as Determine whether the first-order differential equation $$u dv+(v+uv-e^u)du = 0$$ is linear in $u$. Determine also whether it is linear in $v$.


3

One common usage of this symbol is average: The integral divided by the measure of the domain of integration.


3

If you intended to write $\frac{\partial^2 g}{\partial u\partial v}$, then $$ \frac{\partial^2 g}{\partial u\partial v}= \frac{\partial}{\partial u}\left(\frac{\partial g}{\partial v} \right)= \frac{\partial}{\partial v}\left(\frac{\partial g}{\partial u} \right) $$


2

I personally write Re and Im on my preprints, too. It seems to me that Re $z$ is much more elegant than $\Re z$.


2

Due to my nearly unreadable handwriting, I prefer Im and Re instead of the calligraphic symbols, as noone would be able to read them. I've never seen anyone use the calligraphic symbols in real life either.


2

Also used for the Cauchy Principal Value of an integral.


2

HINT $$ A:B :: C:D \equiv \implies \dfrac{A}{B} = \dfrac{C}{D} $$ The notation is called proportion. $A$ is to $B$ as $C$ is to $D$.


2

The domain of the function is the set of inputs that are allowed into the function (i.e., the inputs that, when input into the function, make sense). You are right that we cannot have $x = z$. But look at how the function is defined. It is $f(x,y,z)$. It takes as input a vector with $3$ components, and it outputs a number. So since the domain is the set ...


2

$E$ stands for expected value and $d(\cdot, \cdot)$ means distortion, however you define distortion. $X^n$ is a vector or $n$-tuple of random variables $X$. You could write this as $(X_1, X_2, \ldots, X_n)$ where the $X_i$ are random variables with a common distribution (but need not be independent). $f_n(X^n)$ is the result of compressing or encoding ...


2

For an expression $f(x)$, there is the notation $$[x^3](f(x))$$ See MathWorld. This notation is verbose but can be useful for expressions in several variables. For an expression in a single variable, I'd go for something simpler, like $$[f(x)]_3$$ but be explicit when you introduce or use non-standard notation.


2

As this is a question about notation allow me a few remarks: You won't wind up with a "four-element set". As $B$ has only two elements, $f[A]$ - i.e. the image of $A$ - will have at most two elements. This also means that your $\{5,5,6,5\}$ and $\{5,6,6,5\}$ are the same set, namely just $\{5,6\}$. The functions will differ, though. If you write your ...


2

In fact, your construction works for any small category $\mathscr A$ for which there is a functor $$ \widehat{\mathscr A} \to \mathsf{Cat},\, F \mapsto {\mathscr A}\big/{F} $$ where ${\mathscr A}\big/{F}$ is just another notation for the comma category $(\mathbf{yon}_{\mathscr A}\downarrow F)$. (This notation is justified by viewing $\mathscr A$ as a ...


2

Here is the answer that I wrote a few months ago, which should pertain to this question as well: Both categories are denoted by $\mathrm{Rel}$. In most cases, however, the author either explicitly tells which $\mathrm{Rel}$ he/she is using, or reproduces the definition of $\mathrm{Rel}$. (Note: this applies to any definition/term in math which can be ...


2

Your operative definition of $\longrightarrow$ and $\longmapsto$ are correct. In your example you are dealing with vectors in $\mathbb{R}^n$, which are elements of a set, so you should use $\longmapsto$. see also this wikipedia article: http://en.wikipedia.org/wiki/List_of_mathematical_symbols


2

Parsing the much-maligned notation yields $$\sum_{i,j} a_{ij} (x_i + y_j) \ne \sum_i a_{ij}x_i + \sum_j a_{ij}y_j$$ which is an evident inequality.


2

How about this : $$ \{ x : d(x,y) \leq d(u,y) \text{ for all } u \}. $$ EDIT $$ \{ x_i : d(x_i,y) \leq d(u,y) \text{ for all } u \text{ and } i \le n\}. $$ Could be a bit ropey.


1

If, as you said, you are using the notation described here, then your example above is wrong as this notation already denotes a set. It would simply have been $\arg \max d(x,y)$ or better $\arg \max_{x\in X}d(x,y)$ to make clear where your $x$ are from. As to your question, there's no way to write $\{x \mid \dots\}$ in such a way that you at the same time ...


1

I will assume $x=(x_i)$ to be a $1\times N$ (row) vector, $y=(y_i)$ an $N \times 1$ (column) vector, and $A=(a_{ij})$ an $N\times N$ matrix. Then the terms on the RHS correspond to the matrix multiplications $x_i a_{ij} =(x A)_j$ and $a_{ij}y_j = (A y)_j$. Note that these are the components of a row vector and a column vector respectively. What about the ...


1

Mathematically a norm is a total size or length of all vectors in a vector space or matrices. And by definition, $$\left \| x \right \|_n=\sqrt[n]{\sum _i\left | x_i \right |^{n}}, n\in \mathbb{R}$$ Now let us denote the highest entry in the vector $x$ by $x_j$ (assume it is the $j$-th entry). Thus we we wolud have, $${\sum _ix_i^{\infty }}\cong ...



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