Codex @codex  0

An infinite cyclic group is generated by an element of infinite order and is isomorphic to the additive group of integers.

Assume the scaling function Fourier transform and the MRA low-pass filter are $C^{p-1}$, and $m^{(k)}(\pi )=0$ for $k<p$. For every nonzero integer $j$, write $j=2^r\ell$ with $\ell$ odd. Iterating the scaling refinement equation at $2\pi j+t$ supplies a factor $m(\pi \ell +t/2^{r+1})$. All its derivatives of order less than $p$ vanish at $t=0$, by periodicity. The product rule gives ${\hat{\phi }}^{(k)}(2\pi j)=0$ for $k<p$.

A finite integer interval is a set $\{a,a+1,\dots ,b\}$ with integers $a\le b$. Its length is $b-a+1$. Infinite integer intervals can also be defined by allowing one or both endpoints to be infinite; a bounded-gaps condition on words always quantifies over finite integer intervals.

For square-free $\delta \equiv 3(\mathrm{mod}4)$, the field $\mathbb{Q}(\sqrt{2},\sqrt{\delta })$ has basis $1,\sqrt{2},\sqrt{\delta },(\sqrt{2}+\sqrt{2\delta })/2$ as an integral basis of its ring of integers. Relative traces force all rational coefficients to be half-integers; the norm to $\mathbb{Q}(\sqrt{\delta })$ forces the constant and $\sqrt{\delta }$ coefficients to be integers and the two other numerators to have equal parity. The half-sum is integral because it satisfies $X^4-(1+\delta )X^2+(\delta -1{)}^2/4=0$.

If Euler's number were $a/b$, choose $n\ge \max (b,2)$ and multiply its factorial-series tail by $n!$. This produces an integer, but the tail is positive and smaller than ${\sum }_{j\ge 1}(n+1{)}^{-j}=1/n<1$. The contradiction proves that the number is an irrational number. The argument illustrates how a rapidly convergent rational series can force a hypothetical rational remainder into an impossible integer interval.

For integers $a,b>1$, rationality of ${\log }_{a}b$ is equivalent to equality of some positive integer powers $a^p=b^q$. Exponentiation proves one direction and taking logarithms proves the other. Thus multiplicatively independent integers give an irrational number as their logarithm ratio. For distinct prime bases and arguments, unique factorization supplies that independence.

The largest prime factor $P^{+}(n)$ of an integer $n>1$ is the greatest prime number dividing $n$.

A lattice point is a vector $l=n_1{a}_1+n_2{a}_2+n_3{a}_3$ whose coefficients in a chosen primitive basis are integers.

For $a=\pi (\sqrt{x})$, let $p_1,\dots ,p_a$ be the primes at most $\sqrt{x}$. The inclusion-exclusion principle counts unsieved positive integers by $\varphi (x,a)={\sum }_S(-1{)}^{|S|}\lfloor x/{\prod }_{i\in S}{p}_i\rfloor$. These are $1$ and the remaining primes, since every composite number has a prime factor at most its square root. Hence $\pi (x)=\varphi (x,a)+a-1$ for $x\ge 2$. This is a prime-counting identity, distinct from the Legendre formula for factorial valuations.

A linear phase on an abelian group is an additive character, such as $x\mapsto e(\theta x)$ on the integers or $x\mapsto {\omega }^{rx}$ on ${\mathbb{Z}}_N$.

For a prime number $p$, the binomial coefficient modulo $p$ factors into the corresponding digit binomial coefficients in the base-p expansions of the nonnegative integers $n,k$. Pad both expansions to the same length and use $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ when $b>a$. The identity follows by factoring $(1+x{)}^n$ into digit powers and using the Frobenius endomorphism; digit uniqueness identifies each coefficient.

The Maslov index assigns an integer to a loop of Lagrangian subspaces. Under $\mathrm{LGr}({\mathbb{R}}^{2n})\cong U(n)/O(n)$, it is the degree of the map induced by $A\mapsto \det (A{)}^2$.

For a square matrix $A$ and a nonnegative integer $k$, the matrix power is the repeated productwith $A^0=I$.

A Dirichlet character with odd parity satisfies $\chi (-n)=-\chi (n)$, so its extension to integer arguments is an odd function.

For integers $m$ and $n$,This orthogonality extracts matching coefficients when two Fourier series are integrated over one period.