# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 6/refcontrol

    Polynomials
Mathematical mappings are usually given by a mathematical term, an expression which describes how to get (compute) from a given number its value. Here we consider polynomial functions, which are built in an easy manner. Their definition and basic properties work over any field.

## Definition

Let $K$ be a field.MDLD/field An expression of the form
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,,$
with $a_{i}\in K$ and $n\in \mathbb {N}$,
$K$
The numbers $a_{0},a_{1},\ldots ,a_{n}$ are called the coefficients of the polynomial. Two polynomials are equal if all their coefficients coincide. The polynomials with $a_{i}=0$ for all $i\geq 1$ are called constant polynomials, we write simply $a_{0}$ for them. In the zero polynomial, all coefficients equal $0$. Using the sum symbol, one can write a polynomial briefly as $\sum _{i=0}^{n}a_{i}X^{i}$.

## Definition

The degree of a nonzero polynomial
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,$
with $a_{n}\neq 0$
$n$
The zero polynomial does not have a degree. The coefficient $a_{n}$, where $n$ is the degree of the polynomial, is called the leading coefficient of the polynomial. The term $a_{n}X^{n}$ is called the leading term of the polynomial.
The set of all polynomials over a field $K$ is called polynomial ring over $K$, it is denoted by $K[X]$, where $X$ is the variable of the polynomial ring.
Two polynomials
    $P=\sum _{i=0}^{n}a_{i}X^{i}\,\,{\text{ and }}\,\,Q=\sum _{i=0}^{m}b_{i}X^{i}$
can be added by adding the components, i.e. the coefficients of the sum $P+Q$ are just the sums of the coefficients of the two polynomials. In case $n>m$, the "missing“ coefficients of $Q$ are to be interpreted as $0$. This addition is obviously associative and commutative, the zero polynomial is the neutral element and the negative polynomial $-P$ is obtained by taking from every coefficient of $P$ its negative.
One can also multiply two polynomials, one puts
    $X^{n}\cdot X^{m}:=X^{n+m}\,$
and extends this multiplication rule so that distributivity holds. This means one multiplies each summand with each summand and adds then everything together. One can describe this multiplication also explicitly by the rule:
    ${\left(\sum _{i=0}^{n}a_{i}X^{i}\right)}\cdot {\left(\sum _{j=0}^{m}b_{j}X^{j}\right)}=\sum _{k=0}^{n+m}c_{k}X^{k}{\text{ with }}c_{k}=\sum _{r=0}^{k}a_{r}b_{k-r}.$
For the degree the following rules hold.
    $\operatorname {deg} \,(P+Q)\leq \max\{\operatorname {deg} \,(P),\,\operatorname {deg} \,(Q)\}\,.$
    $\operatorname {deg} \,(P\cdot Q)=\operatorname {deg} \,(P)+\operatorname {deg} \,(Q)\,.$
If a polynomial $P\in K[X]$ and an element $a\in K$ are given, then one can insert $a$ into $P$ meaning that one replaces everywhere the variable $X$ by $a$. This yields a mapping
    $K\longrightarrow K,a\longmapsto P(a),$
which is called the corresponding polynomial function.
If $P$ and $Q$ are polynomials, then the composition $P\circ Q$ is described in the following way: one has to replace everywhere in $P$ the variable $X$ by $Q$ and then simplify this expression. The result is again a polynomial. The order of $P$ and $Q$ is important for this.
    Euclidean division
If we have a polynomial over the reals, we are interested in its zeros, its growth behavior, local maxima and minima. For these questions, the euclidean division for polynomials (long division) is important.

## TheoremTheorem 6.3 change

Let $K$ be a fieldMDLD/field and let $K[X]$ be the polynomial ringMDLD/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
    $P=TQ+R{\text{ and with }}\operatorname {deg} \,(R)<\operatorname {deg} \,(T){\text{ or }}R=0.$

### Proof

We prove the statement about the existence by induction over the degreeMDLD/degree (polynomial) of $P$. If the degree of $T$ is larger than the degree of $P$, then $Q=0$ and $R=P$ is a solution.
Suppose that $\operatorname {deg} \,(P)=0$. By the remark just made also $\operatorname {deg} \,(T)=0$ holds, so $T$ is a constant polynomial, and therefore (since $T\neq 0$ and $K$ is a field) $Q=P/T$ and $R=0$ is a solution.
So suppose now that $\operatorname {deg} \,(P)=n$ and that the statement for smaller degrees is already proven. We write $P=a_{n}X^{n}+\cdots +a_{1}X+a_{0}$ and $T=b_{k}X^{k}+\cdots +b_{1}X+b_{0}$ with $a_{n},b_{k}\neq 0,\,k\leq n$. Then setting $H={\frac {a_{n}}{b_{k}}}X^{n-k}$ we have the relation
    ${\begin{aligned}P'&:=P-TH\\&=0X^{n}+{\left(a_{n-1}-{\frac {a_{n}}{b_{k}}}b_{k-1}\right)}X^{n-1}+\cdots +{\left(a_{n-k}-{\frac {a_{n}}{b_{k}}}b_{0}\right)}X^{n-k}+a_{n-k-1}X^{n-k-1}+\cdots +a_{0}.\end{aligned}}$
The degree of this polynomial $P'$ is smaller than $n$ and we can apply the induction hypothesis to it. That means there exist $Q'$ and $R'$ such that
    $P'=TQ'+R'{\text{ and with }}\operatorname {deg} \,(R')<\operatorname {deg} \,(T){\text{ or }}R'=0.$
From this we get altogether
    $P=P'+TH=TQ'+TH+R'=T(Q'+H)+R'\,,$
so that $Q=Q'+H$ and $R=R'$ is a solution.
To prove uniqueness, let $P=TQ+R=TQ'+R'$, both fulfilling the stated conditions. Then $T(Q-Q')=R'-R$. Since the degree of the difference $R'-R$ is smaller than $\operatorname {deg} \,(T)$, this implies $R=R'$ and so $Q=Q'$.
$\Box$
The computation of the polynomials $Q$ and $R$ is also called long division. The polynomial $T$ is a factor of $P$ if and only if the division of $P$ by $T$ yields the remainder $0$. The proof of this theorem is constructive, meaning it can be used to do the computation effectively. For this, one has to be able to do the computing operations in the field $K$. We give an example.

## Example Create referencenumber

We want to apply the Euclidean divisionMDLD/Euclidean division (over $\mathbb {Q}$)
    $P=6X^{3}+X+1{\text{ divided by }}T=3X^{2}+2X-4.$
So we want to divide a polynomial of degree $3$ by a polynomial of degree $2$, hence the quotient and also the remainder have (at most) degree $1$. For the first step, we ask with which term we have to multiply $T$ to achieve that the product and $P$ have the same leading term. This is $2X$. The product is
    $2X{\left(3X^{2}+2X-4\right)}=6X^{3}+4X^{2}-8X\,.$
The difference between $P$ and this product is
    $6X^{3}+X+1-{\left(6X^{3}+4X^{2}-8X\right)}=-4X^{2}+9X+1\,.$
We continue the division by $T$ with this polynomial, which we call $P'$. In order to get coincidence with the leading coefficient we have to multiply $T$ with ${\frac {-4}{3}}$. This yields
    $-{\frac {4}{3}}T=-{\frac {4}{3}}{\left(3X^{2}+2X-4\right)}=-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\,.$
The difference between this and $P'$ is therefore
    $-4X^{2}+9X+1-{\left(-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\right)}={\frac {35}{3}}X-{\frac {13}{3}}\,.$
This is the remainder and altogether we get
    $6X^{3}+X+1={\left(3X^{2}+2X-4\right)}{\left(2X-{\frac {4}{3}}\right)}+{\frac {35}{3}}X-{\frac {13}{3}}\,.$

## LemmaLemma 6.5 change

Let $K$ be a fieldMDLD/field and let $K[X]$ be the polynomial ringMDLD/polynomial ring over $K$. Let $P\in K[X]$ be a polynomial and $a\in K$. Then $a$ is a zeroMDLD/zero (polynomial)
$P$
$P$
$X-a$

### Proof

If $P$ is a multiple of $X-a$, then we can write
    $P=(X-a)Q\,$
with another polynomial $Q$. Inserting $a$ yields
    $P(a)=(a-a)Q(a)=0\,.$
In general, there exists, due to Theorem 6.3 , a representation
    $P=(X-a)Q+R\,,$
where either $R=0$ or the degree of $R$ is $0$, so in both cases $R$ is a constant. Inserting $a$ yields
    $P(a)=R\,.$
So if $P(a)=0$ holds, then the remainder must be $R=0$, and this means $P=(X-a)Q$.
$\Box$

## CorollaryCorollary 6.6 change

Let $K$ be a fieldMDLD/field and let $K[X]$ be the polynomial ringMDLD/polynomial ring over $K$. Let $P\in K[X]$ be a polynomial ($\neq 0$) of degreeMDLD/degree (polynomial)
$d$
$P$
$d$

### Proof

We prove the statement by induction over $d$. For $d=0,1$ the statement holds. So suppose that $d\geq 2$ and that the statement is already proven for smaller degrees. Let $a$ be a zero of $P$ (if $P$ does not have a zero at all, we are done anyway). Hence, $P=Q(X-a)$ by Lemma 6.5
and the degree of $Q$ is $d-1$, so we can apply to $Q$ the induction hypothesis. The polynomial $Q$ has at most $d-1$ zeroes. For $b\in K$ we have $P(b)=Q(b)(b-a)$. This can be zero, due to Lemma 5.5 , only if one factor is $0$, so the zeroes of $P$ are $a$ or a zero of $Q$. Hence, there are at most $d$ zeroes of $P$.
$\Box$
    Fundamental theorem of algebra
The following fundamental theorem of algebra, for which we do not provide a proof, shows the importance of complex numbers.

## TheoremCreate referencenumber

Every nonconstant polynomialMDLD/polynomial (1K) $P\in \mathbb {C} [X]$ over the complex numbersMDLD/complex numbers has a
The fundamental theorem of algebra implies that for every polynomial $P\in \mathbb {C} [X]$ $P\neq 0$, there is a factorization into linear factors, i.e. one can write
    $P=c(X-z_{1})(X-z_{2})\cdot (X-z_{n})\,$
with uniquely determined complex numbers $c$ and $z_{1},\ldots ,z_{n}$ (where repetitions are allowed).
    Polynomial interpolation
The following theorem is called theorem about polynomial interpolation and describes the interpolation of given function values by a polynomial. If just one function value at one point is given, then this determines a constant polynomial, two values at two points determine a linear polynomial (the graph is a line), three values at three points determine a quadratic polynomial, etc.

## TheoremTheorem 6.8 change

Let $K$ be a field,MDLD/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
polynomialMDLD/polynomial (1K) $P\in K[X]$ of degree $\leq n-1$, such that $P{\left(a_{i}\right)}=b_{i}$
$i$

### Proof

We prove the existence and consider first the situation where $b_{j}=0$ for all $j\neq i$ for some fixed $i$. Then
    $(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$
is a polynomial of degree $n-1$ which at the points $a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}$ has value $0$. The polynomial
    ${\frac {b_{i}}{(a_{i}-a_{1})\cdots (a_{i}-a_{i-1})(a_{i}-a_{i+1})\cdots (a_{i}-a_{n})}}(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$
has at these points still a zero, but additionally at $a_{i}$ its value is $b_{i}$. We denote this polynomial by $P_{i}$. Then
    $P=P_{1}+P_{2}+\cdots +P_{n}\,$
is the polynomial looked for, because for the point $a_{i}$ we have
    $P_{j}(a_{i})=0\,$
for $j\neq i$ and $P_{i}(a_{i})=b_{i}$.
The uniqueness follows from Corollary 6.6 .
$\Box$

## Remark

If the data $a_{1},\ldots ,a_{n}$ and $b_{1},\ldots ,b_{n}$ are given, then one can find the interpolating polynomial $P$ of degree $\leq n-1$, which exists by Theorem 6.8 , in the following way: We write
    $P=c_{0}+c_{1}X+c_{2}X^{2}+\cdots +c_{n-2}X^{n-2}+c_{n-1}X^{n-1}\,$
with unknown coefficients $c_{0},\ldots ,c_{n-1}$, and determine then these coefficients. Each interpolation point $(a_{i},b_{i})$ yields a linear equation
    $c_{0}+c_{1}a_{i}+c_{2}a_{i}^{2}+\cdots +c_{n-2}a_{i}^{n-2}+c_{n-1}a_{i}^{n-1}=b_{i}\,$
over $K$. The resulting system of linear equations has exactly one solution $(c_{0},\ldots ,c_{n-1})$, which gives the polynomial.
We will deal with systems of linear equations later in more detail, it should however be known from school how to find the solutions.
    Rational functions
Next to the polynomial functions, the simplest functions are the rational functions.

## Definition

For polynomialsMDLD/polynomials (R) $P,Q\in \mathbb {R} [X]$, $Q\neq 0$, the functionMDLD/function (R)
    $D\longrightarrow \mathbb {R} ,z\longmapsto {\frac {P(z)}{Q(z)}},$
where $D$ is the complementMDLD/complement of the zeroesMDLD/zeroes
$Q$