1
$\begingroup$

The Taylor Series expansion for $\frac{1}{1-x}$ is convergent for every real number $-1 < x < 1$.

\begin{equation*} \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^{n} . \end{equation*}

Since $0 \leq x^{2} < 1$ if, and only if $-1 < x < 1$, the Taylor Series expansion for $\frac{1}{1-x^{2}}$ is convergent for every real number $-1 < x < 1$.

\begin{equation*} \frac{1}{1 - x^{2}} = \sum_{n=0}^{\infty} x^{2n} . \end{equation*}

This is a special case of a theorem about the composition of certain functions. For example, the Taylor Series expansion for $e^{x}$ implies that

\begin{equation*} e^{x^{2}} = \sum_{n=0}^{\infty} \frac{1}{n!} \, x^{2n} \end{equation*}

and that

\begin{equation*} e^{(x-1)^2} = \sum_{n=0}^{\infty} \frac{1}{n!} \, \left((x - 1)^2\right)^{n} = \sum_{n=0}^{\infty} \frac{1}{n!} \, (x - 1)^{2n} . \end{equation*}

I am looking for the precise statement for such a theorem and a rigorous demonstration of it (or a citation for a rigorous demonstration of it).

$\endgroup$
  • $\begingroup$ What do you mean? You are just letting $x\to f(x)$, where $f(x)$ is a basic polynomial... $\endgroup$ – Simply Beautiful Art Dec 20 '16 at 21:02
  • $\begingroup$ @Simple Art From the definition of Taylor Series, computing derivatives at $a$ of a composition of functions would be quite messy, even for $f(x)=x^{2}$. There is a theorem that states that we can simply replace $x$ with $f(x)$. $\endgroup$ – user366533 Dec 20 '16 at 21:06
  • $\begingroup$ Well, it'd probably just be substitution. Nothing special about it as far as I can see. $\endgroup$ – Simply Beautiful Art Dec 20 '16 at 21:08
  • $\begingroup$ @Simple Art The same expansion is derived in two different ways. The radius of convergence is the same, too. These are remarkable statements. $\endgroup$ – user366533 Dec 20 '16 at 21:18
  • $\begingroup$ I do not think so. Why would you think substituting a function into a power series is anything special? $\endgroup$ – Simply Beautiful Art Dec 20 '16 at 21:19
0
$\begingroup$

You can substitute in another function but that might completely change the disk of convergence.

Let $f(z) = \sum_{j=0}^\infty a_j (z - z_0)^j$ be a Taylor series and $D$ its disk of convergence. If $g:S \to \mathbb{C}$, then $f \circ g$ is convergent in $g^{-1}[D]$ to $\sum_{j=0}^\infty a_j (g(z) - z_0)^j$.

Proof: Suppose $w = g(z) \in D$, then $f \circ g(z) = f(w) = \sum_{j=0}^\infty a_j (w - z_0)^j = \sum_{j=0}^\infty a_j (g(z) - z_0)^j$.

$\endgroup$
0
$\begingroup$

I don't think this theorem has a name, but see for instance Theorem 3.4 of Lang's Complex Analysis, which explains the convergence of formal power series under composition.

$\endgroup$
  • $\begingroup$ I am sure the explanation in this textbook is adaptable to the real analysis case. Do you know of a citation from the context of real analysis? $\endgroup$ – user366533 Dec 20 '16 at 21:20
  • $\begingroup$ Formal power series naturally live in the setting of complex analysis, and anything you want to know about the real case just comes from doing the complex case and restricting to the real line. $\endgroup$ – Daniel McLaury Dec 20 '16 at 21:22
0
$\begingroup$

Complementing the other answers with an example in the context of reals:

In answer to one of your comments, the radius of convergence is not always the same:

$\sum_{n=0}^{\infty }\left ( \frac{1}{2} \right )^{n}x^{n}=\sum_{n=0}^{\infty }\left ( \frac{x}{2} \right )^{n}=\frac{1}{1-\frac{x}{2}}$ has radius of convergence $R=2$ which is not the same as the radius of convergence of $\sum_{n=0}^{\infty} x^{n}$.

Since the radius of convergence is not the same, the result you are aiming at, which could be stated as follows:

"The limit of the composition is the same as the composition of the limit" is not true.

Taking the example above again to clarify why the statement is wrong:

The limit of $\sum_{n=0}^{\infty }\left ( \frac{x}{2} \right )^{n}$ is $\frac{1}{1-\frac{x}{2}}$ for $\left | x \right |< 2$. So if we were to pick $x=1.5$, it is not true that $\sum_{n=0}^{\infty} x^{n}=\frac{1}{1 - (1.5)}$ as the series diverges to $\infty $ for $\left | x \right |\geq 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy