# 5.12. Analytical Functions

The limit functions of convergent power series are quite easy to evaluate approximatively as we only need addition and multiplication to do this. On this fact alone it would be rewarding to see as many functions as possible as limit functions. But their domains are always intervals, so a lot of even elementary functions, e.g. the reciprocal function, are excluded from the outset.

For a single value however it is sufficient if a function coincides only locally with a limit function. The following definition thus will increase the number of easy to evaluate functions considerably..

Definition:  Let $A\subset ℝ$ be a non-empty subset of $ℝ$.

We call a function $f:A\to ℝ$  analytical if every $a\in A$ allows a convergent power series $\left(\sum _{i=0}^{n}{a}_{i}{\left(\mathrm{X}-a\right)}^{i}\right)$ with r as radius of convergence and an s with $0 such that

 [5.12.1]

We use the symbol ${\mathcal{C}}^{\ast }\left(A\right)$ to denote the set of all analytical functions on A.

Consider:

• The task to verify a function f as analytical, i.e. to provide a representation for f according to [5.12.1], is often phrased as: 'Expand f into a power series for each point of A.' A difficult task to cope with on the basis of our recent techniques. When we study calculus this situation will improve.

• If $a\in A$ is isolated, i.e. if

$A\cap \right]a-s,a+s\left[=\left\{a\right\}$

for a suitable $s>0$, every function  f is expendable for a infact by any power series where ${a}_{0}=f\left(a\right)$. In this case the property [5.12.1] is no longer special to  f.

• If  $f:A\to ℝ$ is analytical and $B\subset A$ is non-empty, then the restriction $f|B:B\to ℝ$ is analytical as well.

Example:

 The reciprocal function $\frac{1}{\mathrm{X}}$ is analytical. [5.12.2] Let $a\ne 0$ be arbitrary, then for every x with $|x-a|<|a|$, i.e. with $|\frac{x}{a}-1|<1$, the following calculation holds (note the limit of the geometric series [5.9.4]): $\begin{array}{ll}\frac{1}{x}\hfill & =\frac{1}{a-a+x}\hfill \\ \hfill & =\frac{1}{a}\cdot \frac{1}{1-\left(1-\frac{x}{a}\right)}\hfill \\ \hfill & =\frac{1}{a}\cdot \sum _{i=0}^{\infty }{\left(1-\frac{x}{a}\right)}^{i}\hfill \\ \hfill & =\frac{1}{a}\cdot \sum _{i=0}^{\infty }\frac{1}{{a}^{i}}{\left(a-x\right)}^{i}\hfill \\ \hfill & =\sum _{i=0}^{\infty }\frac{{\left(-1\right)}^{i}}{{a}^{i+1}}{\left(x-a\right)}^{i}\hfill \end{array}$
 Every limit function $\sum _{i=0}^{\infty }{a}_{i}{\left(\mathrm{X}-a\right)}^{i}$ of a convergent power series is analytical due to [5.11.20]. [5.12.3]
 So all polynomials as well as sin, cos and exp are analytical and (with [5.12.8]) also tan, cot and every quotient of two polynomials. [5.12.4]

Locally analytical functions mimic limit functions. We may thus expect that they also share their properties. And indeed from the calculation rules for convergent power series we can show that analyticity is compatible with basic arithmetics.

Proposition:  If  f und g are analytic on A, i.e.  $f,g\in {\mathcal{C}}^{\ast }\left(A\right)$, then

 $f+g\in {\mathcal{C}}^{\ast }\left(A\right)$ [5.12.5]
 $f-g\in {\mathcal{C}}^{\ast }\left(A\right)$ [5.12.6]
 $f\cdot g\in {\mathcal{C}}^{\ast }\left(A\right)$ [5.12.7]
 [5.12.8]

Proof:  All the proofs are nearly identical. So we only show

 1. ► For any $a\in A$ there are numbers ${s}_{1},{s}_{2}>0$ and suitable power series with radii of convergence ${r}_{i}\ge {s}_{i}$ such that With  $s≔\mathrm{min}\left\{{s}_{1},{s}_{2}\right\}$ we conclude from [5.11.16]: Additionally we note with 4. ► $g\left(a\right)\ne 0$ implies ${b}_{0}\ne 0$ so that the condition in [5.11.19] is satisfied.

Consider:

• The function space ${\mathcal{C}}^{\ast }\left(A\right)$ is hence underlaid with an algebraic structure: As the constant function $\mathbf{0}≔0|A$ is analytic [5.12.5] and [5.12.6] can be combined to:

$\left({\mathcal{C}}^{\ast }\left(A\right),+\right)$ is an abelian group  i The addition + is associative and commutative. 0 is the neutral element, i.e.  $f+\mathbf{0}=f$  for all  f. Each f has a unique inverse, $-f$ in this case, such that  $f+\left(-f\right)=\mathbf{0}$.
.

And with [5.12.7] we even get ( $\mathbf{1}≔1|A$ is analytical as well):

$\left({\mathcal{C}}^{\ast }\left(A\right),+,\cdot \right)$ is a commutative ring with identity element  i The axioms for an abelian group are satisfied. The multiplication · is associative and commutative. · is distributive with respect to +. 1 is the neutral element for the multiplication, i.e. $\mathbf{1}·f=f$ for all f.
.

There is an interesting property of analytical functions related to [5.12.8]: If  $f\left(a\right)$ is non-zero this remains true in a complete neighbourhood of a.

Proposition:  If  $f:A\to ℝ$ is analytical with  $f\left(a\right)\ne 0$ there is an $n\in {ℕ}^{\ast }$ such that

 [5.12.9]

Proof:  Let

be an expansion of  f for a. If  $f\left(a\right)\ne 0$, i.e. ${a}_{0}\ne 0$ we follow an argument in the proof of [5.11.19] and find an $n\in {ℕ}^{\ast }$, without restriction $\frac{1}{n}, such that

.

Another important result, the identity theorem for convergent power series [5.11.14] is also transferable to analytical functions:

Proposition:  Let $f:A\to ℝ$ be an analytical function and let $\left({x}_{n}\right)$ be a sequence in A with $a\ne {x}_{n}\to a\in A$. If all the sequence members ${x}_{n}$ are zeros of  f there is an $s>0$ such that:

 [5.12.10]

Proof:  We expand  f into a power series for a i.e. we find a suitable s such that:

.

As ${x}_{n}\to a$ we may assume that $\left({x}_{n}\right)$ is a sequence in $\right]a-s,a+s\left[$. According to [5.11.14] we thus have ${a}_{k}=0$ for all the coefficients so that  $f\left(x\right)=0$ for all $x\in A\cap \right]a-s,a+s\left[$.

For analytical functions on an interval
 i We understand this as a common notation for open and closed intervals. In the open case we also allow the values $\infty$ for the right and $-\infty$ for the left boundery. Thus $ℝ$ and e.g. ${ℝ}^{>0}$ are regarded as intervals as well.
this result can be enhanced considerably:

Proposition (Identity Theorem for Analytical Functions):  Let $f:I\to ℝ$ be an analytical function on an interval, $a\in I$ and $\left({x}_{n}\right)$ a sequence in I with $a\ne {x}_{n}\to a$. If all sequence members ${x}_{n}$ are zeros of  f then:

 [5.12.11]

Proof:  It is sufficient to show for each $b\in I$ that:

1. $a

2. $a>b\text{ }⇒\text{ }f|\left[b,a\right]=0$

We prove 1. as an example, 2. is verified the same way. As I is an interval we have $\left[a,b\right]\subset I$. Furthermore we know that  $f|\left[a,b\right]$ is analytical.

According to [5.12.10] the bounded set

$C≔\left\{x\in \left[a,b\right]|f|\left[a,x\right]=0\right\}\subset \left[a,b\right]$

is non-empty as we find an $\epsilon >0$, without restriction $\epsilon , such that $a+\epsilon \in C$. Thus there is the supremum  $c≔\mathrm{sup}C\in \right]a,b\right]$.

We show that $f\left(x\right)=0$ for all $x\in \left[a,c\left[$: Any  $x is no upper bound for C, so there is an  $y>x$ within C. Now  $f|\left[a,y\right]=0$ implies  $f\left(x\right)=0$.

As $a we may choose a sequence $\left({x}_{n}\right)$ in $\left[a,c\left[$ such that ${x}_{n}\to c$. Employing [5.12.10] again we get an $\epsilon >0$ such that

Hence we have  $f\left(c\right)=0$ and subsequently $c=b$, because otherwise we would find an x with $c such that $f|\left[a,x\right]=0$, but that would be a member $x\in C$ exceeding the supremum of C.

Consider:

• [5.12.11] is only valid for intervals. As an example the function  $f:{ℝ}^{\ne 0}\to ℝ$ where

is analytical but not the zero function.

• The premise in [5.12.11] is certainly true if there is a subinterval J such that $f|J=0$. The following version of [5.12.11] is more pleasing to the eye:

If  $f\in {\mathcal{C}}^{\ast }\left(I\right)$ and if $J\subset I$ is a subinterval of I then:

 $f|J=0\text{ }⇒\text{ }f=0$ [5.12.12]
• In addition the ring $\left({\mathcal{C}}^{\ast }\left(I\right),+,\cdot \right)$ is an integral domain  i $\text{\hspace{0.28em}}f\cdot g=0\text{ }⇒\text{ }f=0\text{\hspace{0.28em}}\vee \text{\hspace{0.28em}}g=0$
, i.e. there are no zero divisors.

To see this, take  $f,g\in {\mathcal{C}}^{\ast }\left(I\right)$ such that  $f\cdot g=0$. If  $f\left(x\right)\ne 0$ for a single $x\in I$ we have $f\left(x\right)\ne 0$ for all x in a subinterval $J\subset I$ due to [5.12.10]. Thus g equals zero on this subinterval:  $g|J=0$. Consequently:  $g=0$.

Employing the usual difference trick [5.12.12] allows to compare two analytical functions: Two such functions already coincide if they do so only on a subinterval.

Proposition:  If  $f,g\in {\mathcal{C}}^{\ast }\left(I\right)$ and $J\subset I$. Then

 $f|J=g|J\text{ }⇒\text{ }f=g$ [5.12.13]

Proof:  $f-g$ is analytical on I and equal to 0 on J. Hence  $f-g=0$, i.e. $f=g$.

As a disadvantage the identity theorem [5.12.13] narrows the plasticity of analytical functions. There will be no analytical function on $ℝ$ e.g. equal to 0 if $x<-\frac{1}{2}$ and equal to 1 if $x>\frac{1}{2}$ (There are no analytical stairs!).

It is in 9.12 where we construct certain high class functions (so-called ${\mathcal{C}}^{\infty }$-hats) which can be used to produce stairs.