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Taylor's Series Generated by a Real Function
#
Infinite Taylor approximation
#
Suppose $f: [a, a + r] \to \R$. And $f \in C^\infty[a, a + r]$.
Define $\widetilde f(x): [a, a + r] \to \R$ as:
$$
\widetilde f(x) := \sum _ {n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n}
$$
For some $\xi(a, x) \in (a, x)$, we have:
$$
f(x)=\sum _ {k=0}^{n-1} \frac{f^{(k)}(a)}{k !}(x-a)^{k}+\frac{f^{(n)}\left(\xi\right)}{n !}(x-a)^{n}
$$
Clearly
$$
\lim _ {n\to \infty}\frac{f^{(n)}\left(\xi\right)}{n !}(x-a)^{n} = 0 \iff f(x) = \widetilde f(x)
$$
A sufficient condition would be $\exists M > 0, \forall x \in (a, a+r):|f^{(n)}(x)| \le M^n$. Notice that
$$
\left |\frac{f^{(n)}(\xi(a, x))}{n!} (x-a)^n\right| \le (rM)^n/n! \to 0
$$
Another sufficient condition is $\forall n \in \N, \forall x \in [a, a+r]: f^{(n)}(x) \ge 0$. TODO (Bernstein 9.30 Apostol II).
Real Binomial Series
#
Generalized Binomial Coefficients
#
Suppose $a \in \C$, define
$$
\left(\begin{array}{l}
\alpha \\
k
\end{array}\right):=\frac{\alpha(\alpha-1)(\alpha-2) \cdots(\alpha-k+1)}{k !}
$$
We have the following similar to the integer cases:
$$
\left(\begin{array}{l}
\alpha \\
0
\end{array}\right)=1;\quad
\left(\begin{array}{c}
\alpha \\
k+1
\end{array}\right)=\left(\begin{array}{l}
\alpha \\
k
\end{array}\right) \frac{\alpha-k}{k+1}; \quad \left(\begin{array}{c}
\alpha \\
k-1
\end{array}\right)+\left(\begin{array}{l}
\alpha \\
k
\end{array}\right)=\left(\begin{array}{c}
\alpha+1 \\
k
\end{array}\right)
$$
Binomial Series
#
Consider the following Taylor approximation for $f: \R - \{-1\} \to \R$, known as the real binomial series.
$$
f(x) = (1+x)^{a} \sim \tilde f(x) = \sum _ {n=0}^{\infty}\left(\begin{array}{l}
a \\
n
\end{array}\right) x^{n}
$$
Suppose $a \in \N$, $f(x) = \tilde f(x)$. The sum is finite, and $r = \infty$.
Consider $g(x) = (1 - x)^{-c}$ on $x \in (-\infty, 1]$ for some $c > 0$. Notice that
$$
f^{(n)}(x)=c(c+1) \cdots(c+n-1)(1-x)^{-c-n} = \left(\begin{array}{l}
-c \\
n
\end{array}\right) (1-x)^{-c -n}
$$
By Bernstein's result, the following is true on $x\in(-1, 1)$.
$$
\frac{1}{(1-x)^{c}}=\sum _ {k=0}^{\infty}\left(\begin{array}{l}
-c \\
k
\end{array}\right)(-1)^{k} x^{k}
$$
Suppose $a < 0$, the following is true on $x \in (-1, 1)$.
$$
(1+x)^a=\sum _ {k=0}^{\infty}\left(\begin{array}{l}
a \\
k
\end{array}\right) x^{k}
$$
Suppose $a \in \R^+ - \N$, integration on both sides.
For $a = -1 / 2$. $\frac{1}{\sqrt{1-x}}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
2 n \
n
\end{array}\right) 4^{-n} x^{n}$.
Boundaries of Power Series
#
Abel's Limit Theorem in Real
#
Suppose $a _ n \in \R$. Suppose $f(x) = \sum _ {n=0}^ \infty a _ n x^n$ for $x \in (-1, 1)$. Suppose $\sum _ {n=0}^\infty a _ n$ converges, then
$$
\lim _ {x \rightarrow 1} f| _ {(-1, 1)}(x)=\sum _ {n=0}^{\infty} a _ {n} = f(1)
$$
That is $f(x)$ is left-continuous at $r$.
Abel's Limit Theorem in Complex
#
Suppose $a _ n \in \C$. Suppose $f(z) = \sum _ {n=0}^\infty a _ n z^n$ for $x\in B(0, r)$. Suppose $\sum _ {n=0}^\infty a _ n r^n$ converges, then
for any $M > 1$, for any Stolz sector $S=\{z \in B(0, 1) \mid |1-z| \leq M(1-|z|) \}$.
$$
\lim _ {z \to 1}f| _ S(z) = \sum _ {n=0}^\infty a _ n = f(1)
$$