# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Important theorems

Suppose that for every natural number $n$ a statement $A(n)$ is given. Suppose further that the following conditions are fulfilled.
* $A(0)$ is true.
* For all $n$ we have: if $A(n)$ holds, then also $A(n+1)$ holds.
$A(n)$
$n$
Every natural number $n\in \mathbb {N}$, $n\geq 2$, has a factorization into prime numbers. That means there exists a representation
    $n=p_{1}\cdot p_{2}\cdots p_{r}\,$
with prime numbers $p_{i}$.
There does not exist a rational number such that its square equals $2$. This means that the real number ${\sqrt {2}}$ is irrational.
There exist infinitely many prime numbers.
Let $a,b$ be elements of a field and let $n$ denote a natural number. Then
    $(a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}$
holds.
The complex numbers form a field.
Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P,T\in K[X]$ be polynomials with $T\neq 0$. Then there exist unique polynomials $Q,R\in K[X]$ such that
    $P=TQ+R{\text{ and with }}\operatorname {deg} \,(R)<\operatorname {deg} \,(T){\text{ or }}R=0.$
Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P\in K[X]$ be a polynomial and $a\in K$. Then $a$ is a zero of $P$ if and only if $P$ is a multiple of the linear polynomial $X-a$.
Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P\in K[X]$ be a polynomial ($\neq 0$) of degree $d$. Then $P$ has at most $d$ zeroes.
Every nonconstant polynomial $P\in \mathbb {C} [X]$ over the complex numbers has a zero.
Let $K$ be a field, and let $n$ different elements $a_{1},\ldots ,a_{n}\in K$ and $n$ elements $b_{1},\ldots ,b_{n}\in K$ are given. Then there exist a unique
polynomial $P\in K[X]$ of degree $\leq n-1$, such that $P{\left(a_{i}\right)}=b_{i}$ holds for all $i$.
A real sequence has at most one limit.
A convergent real sequence is bounded.
Let ${\left(x_{n}\right)}_{n\in \mathbb {N} },\,{\left(y_{n}\right)}_{n\in \mathbb {N} }$ and ${\left(z_{n}\right)}_{n\in \mathbb {N} }$ denote real sequences. Suppose that
    $x_{n}\leq y_{n}\leq z_{n}{\text{ for all }}n\in \mathbb {N}$
and that ${\left(x_{n}\right)}_{n\in \mathbb {N} }$ and ${\left(z_{n}\right)}_{n\in \mathbb {N} }$ converge to the same limit $a$. Then also ${\left(y_{n}\right)}_{n\in \mathbb {N} }$ converges to $a$.
Let ${\left(x_{n}\right)}_{n\in \mathbb {N} }$ and ${\left(y_{n}\right)}_{n\in \mathbb {N} }$ be
* The sequence ${\left(x_{n}+y_{n}\right)}_{n\in \mathbb {N} }$ is convergent, and holds.
    $\lim _{n\rightarrow \infty }{\left(x_{n}+y_{n}\right)}={\left(\lim _{n\rightarrow \infty }x_{n}\right)}+{\left(\lim _{n\rightarrow \infty }y_{n}\right)}\,$
* The sequence ${\left(x_{n}\cdot y_{n}\right)}_{n\in \mathbb {N} }$ is convergent, and holds.
    $\lim _{n\rightarrow \infty }{\left(x_{n}\cdot y_{n}\right)}={\left(\lim _{n\rightarrow \infty }x_{n}\right)}\cdot {\left(\lim _{n\rightarrow \infty }y_{n}\right)}\,$
* For $c\in \mathbb {R}$, we have
    $\lim _{n\rightarrow \infty }cx_{n}=c{\left(\lim _{n\rightarrow \infty }x_{n}\right)}\,.$
* Suppose that $\lim _{n\rightarrow \infty }x_{n}=x\neq 0$ and $x_{n}\neq 0$ for all $n\in \mathbb {N}$. Then $\left({\frac {1}{x_{n}}}\right)_{n\in \mathbb {N} }$ is also convergent, and holds.
    $\lim _{n\rightarrow \infty }{\frac {1}{x_{n}}}={\frac {1}{x}}\,$
* Suppose that $\lim _{n\rightarrow \infty }x_{n}=x\neq 0$ and that $x_{n}\neq 0$ for all $n\in \mathbb {N}$. Then $\left({\frac {y_{n}}{x_{n}}}\right)_{n\in \mathbb {N} }$ is also convergent, and holds.
    $\lim _{n\rightarrow \infty }{\frac {y_{n}}{x_{n}}}={\frac {\lim _{n\rightarrow \infty }y_{n}}{x}}\,$
Every nonempty subset of the real numbers, which is bounded from above, has a supremum in $\mathbb {R}$.
A bounded and monotone sequence in $\mathbb {R}$ converges.
Let
    $\sum _{k=0}^{\infty }a_{k}$
be a series of real numbers. Then the series is convergent if and only if the following Cauchy-criterion holds: For every $\epsilon >0$ there exists some $n_{0}$ such that for all
    $n\geq m\geq n_{0}\,$
the estimate
    $\vert {\sum _{k=m}^{n}a_{k}}\vert \leq \epsilon \,$
holds.
Let
    $\sum _{k=0}^{\infty }a_{k}$
denote a convergent series of real numbers. Then
    $\lim _{k\rightarrow \infty }a_{k}=0\,.$
Let ${\left(x_{k}\right)}_{k\in \mathbb {N} }$ be an decreasing null sequence of nonnegative real numbers. Then the series $\sum _{k=0}^{\infty }(-1)^{k}x_{k}$ converges.
An absolutely convergent series of real numbers converges.
Let $\sum _{k=0}^{\infty }b_{k}$ be a convergent series of real numbers and ${\left(a_{k}\right)}_{k\in \mathbb {N} }$ a sequence of real numbers fulfilling $\vert {a_{k}}\vert \leq b_{k}$ for all $k$. Then the series
    $\sum _{k=0}^{\infty }a_{k}$
is absolutely convergent.
For all real numbers $x$ with $\vert {x}\vert <1$, the geometric series $\sum _{k=0}^{\infty }x^{k}$ converges absolutely, and the sum equals
    $\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.$
Let
    $\sum _{k=0}^{\infty }a_{k}$
be a series of real numbers. Suppose there exists a real number $q$ with $0\leq q<1$, and a $k_{0}$ with
    $\vert {\frac {a_{k+1}}{a_{k}}}\vert \leq q\,$
for all $k\geq k_{0}$ (in particular $a_{k}\neq 0$ for $k\geq k_{0}$). Then the series $\sum _{k=0}^{\infty }a_{k}$ converges absolutely.