Real Analytic Functions

Definition: Real Analytic Function

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $O$ be an open subset of $\mathbb{R}$ which is contained in $\mathcal{D}$.

We say that $f$ is analytic on $O$ if there exists a real power series $\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$ which converges on $O$ such that

$$f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n\forall x\in O$$

Theorem: Differentiation of Real Analytic Functions

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $O$ be an open subset of $\mathbb{R}$ which is contained in $\mathcal{D}$.

If $f$ is analytic on $O$ with $f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$, then $f$ is differentiable on $O$ and its derivative is also analytic on $O$ with

$$f^{\prime }(x)=\sum _{n=1}^{\infty }n{c}_n(x-a{)}^{n-1}$$

PROOF

TODO

Theorem: Antiderivatives of Real Analytic Functions

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $O$ be an open subset of $\mathbb{R}$ which is contained in $\mathcal{D}$.

If $f$ is analytic on $O$ with $f(x)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$, then its antiderivatives are also analytic on $O$ with

$$\int f(x)\mathrm{dx}=\text{const}+\sum _{n=0}^{\infty }{c}_n\frac{(x-a{)}^{n+1}}{n+1}$$

PROOF

TODO

Theorem: Automatic Differentiation via Dual Numbers

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real analytic function and let $a+b\epsilon$ be a dual number.

If $a\in \mathcal{D}$, then

$$f(a+b\epsilon )=f(a)+b{f}^{\prime }(a)\epsilon$$

Tip: Automatic Differentiation

This theorem allows us to find the derivative of $f$ at any $a\in \mathcal{D}$ without the need to find an expression for $f^{\prime }$, so long as we have a closed-form expression for $f$. All we have to do is set $b=1$ and then use this expression to evaluate $f(a+b\epsilon )$. In the end, the value of the derivative $f^{\prime }(a)$ will be the coefficient before $\epsilon$.

PROOF

TODO