\chapter{\padic power series}
	\section{Elementary functions}
		In many of the following reasonings we'll use the next proposition, which will give us the possibility to manipulate formal power series, knowing their behaviour in some neighbourhood of $0$.
		\begin{prop}
			\label{prop:formal-series}
			Let $f(X_1, \dots, X_n) \in \C\llbracket X_1, \dots, X_n \rrbracket$ be a power series and  let $\epsilon > 0$ such that $f$ is absolutely convergent on $[-\epsilon, \epsilon]^n$ and $f(x_1, \dots, x_n) = 0$ for every $x_i \in [-\epsilon, \epsilon]$. Then $f \equiv 0$, i.e. all terms of $f$ vanishes.
		\end{prop}
		\begin{proof}
			We prove the proposition by induction on $n$.
			\begin{itemize}
				\item $n=1$: let
				\[
					f(X) = \sum_{i=0}^{+\infty} a_iX^i.
				\]
				Obviously $f(0) = a_0 = 0$ so we can write $f(X) = X \cdot(a_1 + a_2X + \dots) =: X \cdot f_1(X)$. Now $f_1(X) \in \C\llbracket X \rrbracket$ vanishes for every $x \in [-\epsilon, \epsilon] \setminus \{0\}$. It is well known from complex analysis that a formal power series in $\C$ defines a holomorphic function where it converges (so, in particular, it's continuous). We then obtain that $f_1$ is continuous so $f_1(0) = 0$, i.e. $a_1 = 0$. Then we can write $f(X) = X^2 \cdot (a_2 + a_3X + \dots) =: X^2 \cdot f_2(X)$, where $f_2(x) = 0$ for every $x \in [-\epsilon, \epsilon] \setminus \{0\}$. Iterating this process we obtain $a_n = 0$ for each $n \in \N$ so $f \equiv 0$.
				\item $n>1$: let's assume that the thesis holds for every $i < n$ and let's prove it for $n$. For brevity, let $Y = (X_1, \dots, X_{n-1})$. Since $f$ is absolutely convergent we can write
				\[
					f(Y, X_n) = \sum_{i=0}^{+\infty} g_i(Y)X_n^i, \qquad g_i(Y) \in \C\llbracket Y \rrbracket.
				\]
				For $x_n = 0$ we have $f(y, 0) = g_0(y) = 0$ for $y \in [-\epsilon, \epsilon]^{n-1}$. Then, by induction, $g_0 \equiv 0$ so
				\[
					f(Y, X_n) = X_n \cdot (g_1(Y) + g_2(Y)X_n + \dots) =: X_n \cdot f_1(Y, X_n)
				\]
				and by hypothesis $f_1(y, x_n) = 0$ for every $y \in [-\epsilon, \epsilon]^{n-1}, x_n \in [-\epsilon, \epsilon] \setminus \{0\}$. Clearly, fixed $y \in [-\epsilon, \epsilon]^{n-1}$, the function $X \mapsto f_1(y, X)$ is a continuous function so $f_1(y, 0) = \lim_{x \to 0} f_1(y, x) = 0$. We have obtained $0 \equiv f_1(Y, 0) = g_1(Y)$ so, by inductive hypothesis, $g_1(Y) \equiv 0$ and we can write
				\[
					f(Y, X_n) = X_n^2 \cdot (g_2(Y) + g_3(Y)X_n + \dots) =: X_n^2 \cdot f_2(Y, X_n)
				\]
				where $f_2(y, x_n) = 0$ if $y \in [-\epsilon, \epsilon]^{n-1}, x_n \in [-\epsilon, \epsilon] \setminus \{0\}$. Iterating this process we obtain $g_n(Y) \equiv 0$ for every $n \in \N$, i.e. $f \equiv 0$.
			\end{itemize}
		\end{proof}
		Another important lemma about \padic power series is the Dwork's lemma. It expresses an important phenomenon in \padic analysis: if we know $F(X^p)/(F(X)^p)$ then we also know something about $F$. This ratio represents how far off $F$ is from commuting with the $p$-power map, which is a very important map also in different contexts (e.g. Frobenius morphism for characteristic $p$ fields).
		\begin{lemma}[Dwork's lemma]
			\label{lemma:dwork}
			Let $F(X) \in 1 + X\Qp\ser{X}$. Then $F(X) \in 1 + X\Zp\ser{X}$ if and only if $\tfrac{F(X^p)}{F(X)^p} \in 1 + pX\Zp\ser{X}$.
		\end{lemma}
		\begin{proof}
			If $F(X) \in 1 + X\Zp\ser{X}$ then, since $(a + b)^p \equiv a^p + b^p \mod p$ and $a^p \equiv a \mod p$ if $a \in \Zp$, we have
			\[
			F(X)^p = F(X^p) + pG(X) \qquad \exists G(X) \in X\Zp\ser{X}.
			\]
			Then
			\[
			\frac{F(X^p)}{F(X)^p} = 1 - p\cdot\frac{G(X)}{F(X)^p} \in 1 + pX\Zp\ser{X},
			\]
			because $F(X)^p \in 1 + X\Zp\ser{X}$ so it can be inverted.\newline
			For the other implication let
			$F(X) = \sum a_iX^i$; by hypothesis we know that $\exists G(X) = \sum b_iX^i$ such that $G(X) \in 1 + pX\Zp\ser{X}$ and
			\[
			F(X^p) = F(X)^p\cdot G(X)
			\]
			We'll prove by induction that $a_i \in \Zp$. By assumption $F(X) \in 1 + X\Qp\ser{X}$ so $a_0 = 1$. Let's now suppose that $a_i \in \Zp$ for every $i < n$. Looking at the coefficients of $X^n$ on both sides of the above equation we obtain
			\begin{gather*}
				\text{coefficient of $X^n$ in } \left(\sum_{i=0}^n a_iX^i\right)^p\cdot\left(1 + \sum_{i=1}^n b_iX^i\right) =
				\begin{cases}
					a_{n/p}, & \text{if $p$ divides $n$;}\\
					0, & \text{otherwise;}
				\end{cases}.
			\end{gather*}
			Expanding the expression for the coefficient of $X^n$ on the left and subtracting $a_{n/p}$ (recall that $a_{n/p}^p \equiv a_{n/p} \mod p$) we notice that the resulting expression consists of $pa_n$ added to some terms in $p\Zp$ so we can conclude that $pa_n \in p\Zp$, i.e. $a_n \in \Zp$. (To see why this is true it can be convenient to recall the formula $(x_1 + \dots + x_n)^m = \sum_{i_1 + \dots + i_n = m} \binom{m}{i_1, \dots, i_n} x_1^{i_1}\dots x_n^{i_n}$).
		\end{proof}

		We'll also prove here a technical lemma, which we will use to study the \padic logarithm.
		\begin{lemma}
			\label{exercise:7-p.74}
			Let $a$ be a primitive $m$-th root of $1$ in $\Qpa$. Then
			\begin{enumerate}[label=(\roman*)]
				\item if $m = p^n$ for some $n \in \N$ then $\pabs{a-1} = p^{-1/\phi(p^n)}$;
				\item otherwise, $\pabs{a-1} = 1$.
			\end{enumerate}
		\end{lemma}
		\begin{proof}
			Let $\Phi_n(X) \in \Zp[X]$ be the $n$-th cyclotomic polynomial. \newline
			\textit{(i)} To prove this case we'll do an induction on $n$. The case $n=0$ is trivial, since $a = 1$. If $n=1$ then $a$ is a primitive $p$-th root of $1$, i.e. $a^p = 1, a \neq 1$, so $\Phi_p(a) = 0$. By Eisenstein criterion (\cref{prop:eisenstein}) it is easy to prove that $\Phi_p(X)$ is irreducible over $\Qp$. We consider then
			\[
				f(X) := \Phi_p(X + 1) = \frac{(X + 1)^p - 1}{X} =  X^{p-1} + \binom{p}{p-1} X^{p-2} + \dots + \binom{p}{2}X + p.
			\]
			Clearly $f(X) \in \Qp[X]$ is irreducible and $f(a - 1) = 0$ so, recalling how we extended $\pabs{\ }$ to $\Qpa$, we have
			\[
				\pabs{a - 1} = \pabs{p}^{1/(p-1)} = p^{-1/(p-1)}
			\]
			which concludes the proof of the case $n=1$. Now, suppose we know the thesis holds for $m=p^i, i < n$ and let's prove it also holds for $m=p^n$. If we consider the extension $K = \Qp(a)$ with the usual notation ($A$ is the maximal subring and $M$ is its maximal ideal), it is easy to note that $\pabs{a} = 1$ and
			\[
				a^{p^n} = 1 \implies a^{p^n} \equiv 1 \mod M
			\]
			and since $A/M$ is a finite field of characteristic $p$ we obtain
			\[
				a \equiv 1 \mod M
			\]
			which means exactly $a = 1 + b$ for some $b \in K, \pabs{b} < 1$. We recall the easy facts
			\[
				\deg \Phi_{p^n}(X) = \phi(p^n) = p^n - p^{n-1}, \qquad \Phi_{p^n}(X) = \Phi_p \left(X^{p^{n-1}}\right)
			\]
			which imply $\Phi_{p^n}(1) = \Phi_p(1) = p$. Since $a$ is a primitive $p^n$-th root of $1$, every other primitive $p^n$-th root of $1$ is $a^j$, with $p \nmid j$, so
			\[
				\Phi_{p^n}(X) = \prod_{1 \leq j < p^n, p \nmid j} (X - a^j).
			\]
			Evaluating at $X = 1$ we obtain
			\[
				\pabs{p} = \prod_{1 \leq j < p^n, p \nmid j} \pabs{1 - a^j}.
			\]
			Using the fact $a \equiv 1 \mod M$ we can see that
			\[
				\frac{1 - a^j}{1 - a} = 1 + a + \dots + a^{j-1} \equiv j \mod M
			\]
			and if $p \nmid j$ we obtain
			\[
				\pabs{\frac{1 - a^j}{1 - a}} = 1
			\]
			so $\pabs{1 - a^j} = \pabs{1 - a}$, which implies $\pabs{1 - a} = \pabs{p}^{1/\phi(p^n)}$.\newline
			\textit{(ii)} First of all let's consider the basic case $p \nmid m$. Then, $a -1$ is a root of the polynomial
			\[
				f(X) = \Phi_m(X+1) = X^{m-1} + \dots + m = g_1(X) \dotsm g_r(X)
			\]
			where $g_i(X) \in \Qp[X]$ is an irreducible factor of $f$, and we can assume that every $g_i(X)$ is monic and has coefficients in $\Zp$. By hypothesis $\pabs{m} = 1$ and if $b_i$ is the constant term of $g_i(X)$ we have
			\[
				\pabs{b_1b_2\dotsm b_r} = \pabs{m} = 1, b_i \in \Zp  \implies \pabs{b_i} = 1 \quad \forall \, i=1,\dots,r.
			\]
			Since $f(a-1)=0$ there is at least one $g_i(X)$ such that $g_i(a - 1)=0$ so $\lambda_{\Qp}(a - 1) = g_i(X)$. Then
			\[
				\pabs{a-1}^{\deg g_i(X)} = \pabs{b_i} = 1 \implies \pabs{a-1} = 1.
			\]
			Now let $m = p^nq$ with $p \nmid q$ (clearly $q \in \N_{>1}$) and suppose the thesis holds for every $m \in \N$ such that $m$ is not a power of $p$ and $p^n \nmid m$. Then, if $a$ is a primitive $m$-th root of $1$, $a^p$ is a primitive $(p^{n-1}q)$-th root of $1$ and, by inductive hypothesis, we know
			\[
				\pabs{a - 1}\cdot\pabs{a^{p-1} + a^{p-2} + \dots + a + 1} = \pabs{a^p - 1} = 1.
			\]
			Since $\pabs{a} = 1$ we have
			\begin{gather*}
				\pabs{a-1} \leq 1, \qquad \pabs{a^{p-1} + \dots + 1} \leq 1 \\
				\implies \pabs{a-1} = 1
			\end{gather*}
			which proves the statement.
		\end{proof}
		We recall that in an ultrametric space, like $(\Cp, \pabs{\ })$, a sequence is Cauchy if and only if the difference between adjacent terms tends to $0$, and if the space is also complete, an infinite series converges if and only if its general term tends to $0$ (see \cref{lemma:cauchy-sequence-ultrametric} and \cref{prop:summable_families}). Now we are ready to define analytic functions on $\Cp$ and prove some of their basic properties.
		\begin{defn}
			A function $f$ is an \emph{analytic function} if
			\[
				f(X) = \sum_{n=0}^{+\infty} a_nX^n, \qquad a_n \in \Cp.
			\]
			We can define $f(x)$ for every $x \in \Cp$ such that the series converges, i.e. $\pabs{a_nx^n} \to 0$ as $n \to +\infty$.
		\end{defn}
		Like in complex analysis, given an analytic function $f$, we can define its \emph{radius of convergence}. Surprisingly we have the exact same formula as in classic analysis.
		\begin{prop}
			Let $f(X) = \sum_{n=0}^{+\infty} a_nX^n$ be an analytic function. We can define its radius of convergence as
			\[
				r := \frac{1}{\limsup \pabs{a_n}^{1/n}},
			\]
			with the usual meaning: $f$ converges if $\pabs{x} < r$ and diverges if $\pabs{x} > r$.
		\end{prop}
		\begin{proof}
			We recall the definition of $\limsup$: $1/r$ is the least real number such that for any $C > 1/r$ there are only finitely many $\pabs{a_n}^{1/n} > C$. \newline
			Let's first consider the case $\pabs{x} < r$: we can write $\pabs{x} = (1 - \epsilon)r$ for some $\epsilon > 0$. We have
			\[
				\pabs{a_nx^n} = \left(r\pabs{a_n}^{1/n}\right)^n\cdot (1 - \epsilon)^n
			\]
			and, if $n$ is big enough, by definition of $r$ we have
			\[
				\pabs{a_n}^{1/n} \leq \frac{1}{r - \frac{1}{2}\epsilon r}.
			\]
			Then
			\[
				\lim_{n \to +\infty} \pabs{a_nx^n} \leq \lim_{n \to +\infty} \left( \frac{(1 - \epsilon)r}{(1 - \frac{1}{2}\epsilon)r} \right)^n = \lim_{n \to +\infty} \left( \frac{1 - \epsilon}{1 - \frac{1}{2}\epsilon} \right)^n = 0,
			\]
			which gives us the desired convergence. \newline
			Let's now prove that if $\pabs{x} > r$ (and $r < +\infty$) the series diverges. Let's choose an element $\pabs{x} > r$ and an $\epsilon > 0$ such that $\pabs{x} \geq (1 + \epsilon)r$. By definition of $\limsup$ we can find a subsequence $(a_{n_k})_k$ such that $\pabs{a_{n_k}}^{1/n_k} \geq 1/(r + \tfrac{1}{2}\epsilon r)$. Then
			\[
				\lim_{k \to +\infty} \pabs{a_{n_k}x^{n_k}} \geq \lim_{k \to +\infty} \left( \frac{1 + \epsilon}{1 +\frac{1}{2} \epsilon} \right) ^ {n_k} = +\infty,
			\]
			which implies that $f$ cannot converge. \newline
			Finally, if $r = +\infty$, i.e. $\lim_{n \to +\infty} \pabs{a_n}^{1/n} = 0$, chosen an element $x \in \Cp^\times$ ($x = 0$ is trivial) we have that, if $n$ is big enough, $\pabs{a_n} \leq 1/(2^n\pabs{x}^n)$ so
			\[
				\lim_{n \to +\infty} \pabs{a_nx^n} \leq \lim_{n \to +\infty} 2^{-n} = 0
			\]
			and $f$ converges everywhere.
		\end{proof}
		This proposition tells us nothing about the case $\pabs{x} = r$. In classical analysis there isn't a simple answer: for example the well known function
		\[
			\log(1 + X) = \sum_{n=1}^{+\infty} (-1)^{n+1} \frac{X^n}{n}
		\]
		has radius of convergence $r = 1$ on $\C$. When $\abs{x} = 1$ this series can diverge (for example if $x = -1$ we obtain the divergent series $- \sum 1/n$) or converge (if $x = 1$ we obtain $\sum (-1)^{n+1}/n$, which converges by Leibniz criterion). This happens because, over $\R$, there are conditionally convergent series that aren't absolutely convergent. In \padic analysis this cannot happen because convergence only depends on $\pabs{x}$: a given analytic function behaves exactly in the same way for every $\pabs{x} = r$. We will study more deeply this formal series in $\Cp$ when we'll talk about \padic logarithm.\newline
		Let's prove two basic facts about analytic functions. For brevity we'll adopt the notation $D_a(r) := B_{\leq r}(a)$ and $D_a(r^-) := B_{<r}(a)$, where we consider the balls in $\Cp$. We'll also omit the subscript $a$ if $a=0$, for example $D(r) = D_0(r) = B_{\leq r}(0)$.
		\begin{prop}
			\label{prop:convergence-power-series-Zp}
			Every $f(X) \in \Zp \llbracket X \rrbracket$ converges in $D(1^-)$.
		\end{prop}
		\begin{proof}
			Let $f(X) = \sum_{n=0}^{+\infty} a_nX^n \in \Zp \llbracket X \rrbracket$ and let $x \in D(1^-)$. Then
			\begin{gather*}
				\pabs{x} < 1, \pabs{a_n} \leq 1 \,\forall\, n\in\N \\
				\implies \lim_{n \to +\infty} \pabs{a_nx^n} \leq \lim_{n \to +\infty} \pabs{x}^n = 0. \qedhere
			\end{gather*}
		\end{proof}
		\begin{prop}
			\label{prop:continuity-analitic-function}
			Every $f(X) = \sum_{n=0}^{+\infty} a_nX^n \in \Cp \llbracket X \rrbracket$ which converges in a disc $D = D(r)$, or $D(r^-)$, is continuous on $D$.
		\end{prop}
		\begin{proof}
			Let's first prove continuity in $0$. Let $x \in D$ such that $\pabs{x} < \delta < r$ ($\delta > 0$ will be chosen later); then, by continuity of absolute value, we have
			\[
				\pabs{f(x) - f(0)} = \pabs{\sum_{n=1}^{+\infty} a_nx^n} \leq \max_{n \in \N^{\times}} \,\pabs{a_nx^n} \leq \max_{ n \in \N^{\times}} \, \left( \pabs{a_n}\cdot\delta^n\right).
			\]
			Clearly, since $f$ converges on $D$, we must have $1/r' > \limsup \pabs{a_n}^{1/n}$ where $\delta < r' < r$ so, for a large enough $N$, $\pabs{a_n} < r'^{-n}$ if $n > N$. Let's introduce
			\[
				C(\delta) := \max_{1 \leq n \leq N}\,\left(\pabs{a_n} \cdot \delta^n \right);
			\]
			it's obvious that $C(\delta) \to 0^+$ as $\delta \to 0^+$. Instead, if $n > N$, we have
			\[
				\pabs{a_n}\cdot\delta^n \leq \left( \frac{\delta}{r'}\right)^n \leq \left( \frac{\delta}{r'}\right)^N,
			\]
			since $\delta/r' < 1$.
		  	Then
			\[
				\pabs{f(x) - f(0)} \leq \max\left\{C(\delta), \left(\frac{\delta}{r'}\right)^N\right\}
			\]
			and we can make the right member as small as we want by choosing smaller $\delta$. This proves continuity in $0$. \newline
			Let's now prove continuity in $0 \neq x \in D$ and consider $y \in D$ such that $\pabs{x - y} < \delta$, where $\delta < \pabs{x}$ will be chosen later, as before. Then, by the isosceles triangle principle, $\pabs{x} = \pabs{y}$. We have
			\begin{gather*}
				\pabs{f(x) - f(y)} = \pabs{\sum_{n=1}^{+\infty} (a_nx^n - a_ny^n)} \leq \max_{n \in \N^{\times}}\,\left(\pabs{a_n}\cdot\pabs{x^n - y^n}\right)  \leq \\
				\leq \max_{n \in \N^{\times}}\,\left(\pabs{a_n}\cdot\pabs{(x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})} \right)
			\end{gather*}
			but $\pabs{x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1}} \leq \max_{1 \leq i \leq n}\, \pabs{x^{n-i}y^{i-1}} = \pabs{x}^{n-1}$ hence
			\[
				\pabs{f(x) - f(y)} \leq \max_{n \in \N^{\times}}\,\left(\pabs{x - y}\cdot\pabs{a_n}\pabs{x}^{n-1} \right) < \frac{\delta}{\pabs{x}} \cdot \max_{n \in \N^{\times}}\,\left(\pabs{a_n}\cdot\pabs{x}^n \right).
			\]
			We know that $\lim_{n \to +\infty} \pabs{a_n}\pabs{x}^n = 0$ so as $\delta \to 0^+$ we have $\pabs{f(x) - f(y)} \to 0$, which proves the statement.
		\end{proof}
		\begin{defn}
			The (partial) function $\log_p(1 + X)\colon \Cp \to \Cp$ defined by
			\[
				\log_p(1 + x) := \sum_{n=1}^{+\infty} (-1)^{n+1} \frac{x^n}{n}
			\]
			is the \emph{\padic logarithm}.
		\end{defn}
		\begin{prop}
			The function $\log_p(1 + X)$ converges on $D(1^-)$ and diverges elsewhere.
		\end{prop}
		\begin{proof}
			It's immediate to verify that the series converges if $\pabs{x} < 1$ and diverges if $\pabs{x} \geq 1$. In-fact $\pabs{a_n} = p^{\ord n}$ so $\lim_{n \to +\infty} \pabs{a_n}^{1/n} = \lim_{n \to +\infty} p^{(\ord n)/n} = 1$ and we obtain the desired radius of convergence. Lastly, if $\pabs{x} = 1$, we have $\pabs{a_nx^n} = p^{\ord n} \geq 1$ so the series diverges.
		\end{proof}
		From now on, unless otherwise specified, we'll use $\log_p$ meaning the \padic logarithm we have just defined. Let's now prove the basic property of logarithms, which also holds in \padic environment.
		\begin{prop}
			\label{prop:padic-logarithm-product-sum}
			The logarithm of a product is the sum of the logarithms. More precisely, if $x, y \in D(1^-)$ then $\log_p \left[ (1 + x)(1+y)\right] = \log_p(1 + x) + \log_p(1 + y)$.
		\end{prop}
		\begin{proof}
			First of all let's observe that $x, y \in D(1^-) \implies x + y + xy \in D(1^-)$, so we can compute the logarithms. By definition
			\[
				\log_p\left[(1 + x)(1 + y)\right] = \sum_{n=1}^{+\infty} (-1)^{n+1} \frac{(x + y + xy)^n}{n}.
			\]
			If we work in $\R$ with the usual metric we already know that $\log\left[ (1+x)(1+y)\right] = \log(1 + x) + \log(1+y)$ and, using the Taylor expansion of $\log$, we have
			\[
				\sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{x^n}{n} + \sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{y^n}{n} = \sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{(x + y + xy)^n}{n}
			\]
			for every $x, y \in \left[-\tfrac{1}{2}, \tfrac{1}{2}\right]$. Thanks to \cref{prop:formal-series} we infer that this relation also holds in the ring of formal power series in two variables $\Q\llbracket X,Y \rrbracket$. Then, using the fact that if a series converges in $\Cp$ its terms can be rearranged in any order without changing the sum, we can write
			\begin{gather*}
				\log_p\left[(1 + x)(1 + y)\right] = \sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{(x + y + xy)^n}{n} =\\
				= \sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{x^n}{n} + \sum_{n = 1}^{+\infty} (-1)^{n+1} \frac{y^n}{n} = \log_p(1 + x) + \log_p(1 + y)
			\end{gather*}
			which concludes the proof.
		\end{proof}
		\begin{corollary}
			\label{corollary:log-root-of-1}
			If $1 + x \in \Cp$ is a root of $1$ and $\pabs{x} < 1$, then $\log_p(1 + x) = 0$. In particular if $1+x$ is a $p^m$-th root of $1$ then $\log_p(1 + x) = 0$.
		\end{corollary}
		\begin{proof}
			Let's first observe that we can actually compute the logarithm of $1 + x$ since by hypothesis $\pabs{x} < 1$ (if $1+x$ is a $p^m$-th root of $1$ then automatically $\pabs{x} < 1$ by \cref{exercise:7-p.74}). Now we have
			\[
				k\cdot\log_p(1 + x) = \log_p\left[(1 + x)^k\right] = \log_p(1) = 0,
			\]
			which concludes the proof.
		\end{proof}
		We have obtained a function, defined on a particular disc of $\Cp$, using the Taylor expansion of the classical $\log(1 + X)$. Now we would like to define the exponential function, beginning from the classical $\exp(x) = \sum_{n=0}^{+\infty} x^n/n!$, and study its relation with the logarithm.
		\begin{defn}
			The (partial) function $\exp_p(X)\colon \Cp \to \Cp$ defined by
			\[
				\exp_p(x) := \sum_{n=0}^{+\infty} \frac{x^n}{n!}
			\]
			is the \emph{\padic exponential}.
		\end{defn}
 		Looking at this series we immediately see that, unlike in the classical case where the $n!$ in the denominator makes sure the series converges for every $x \in \C$, there can be some problems. In-fact if $n!$ is divisible by a high power of $p$, its reciprocal will have a big absolute value. More precisely, we can compute exactly $\pabs{1/n!} = p^{\ord(n!)}$.
		\begin{lemma}
			\label{exercise:14-p.7}
			Given $n \in \N$,
			\[
				\emph{ord}_p (n!) = \frac{n-S_n}{p-1}
			\]
			where $S_n$ is the sum of digits in $n$ to base $p$.
		\end{lemma}
		\begin{proof}
			Let's write $n$ in base $p$:
			\[
				n = a_0 + a_1p + \dots + a_rp^r, \qquad a_i \in \{0, \dots, p-1\},\, a_r \neq 0.
			\]
			Then $S_n = a_0 + a_1 + \dots + a_n$. By definition, $\ord(n!)$ is the maximum $t$ such that $p^t \mid (n!)$. We can use this little formula to compute it:
			\[
				\ord(n!) = \sum_{k=1}^{+\infty} \left[\frac{n}{p^k} \right] = \sum_{k=1}^r \left[\frac{n}{p^k} \right]
			\]
			where $[x]$ is the integer part of $x \in \R$, i.e. the only integer such that $[x] \leq x < [x]+1$. Using the representation of $n$ in base $p$ we have that $\left[n / p^k \right] = 0$ if $k > r$ and, otherwise,
			\[
				\left[\frac{n}{p^k} \right] = \frac{n - a_0 - \dots - a_{k-1}p^{k-1}}{p^k}
			\]
			so if we add them together we obtain
			\[
				\sum_{k=1}^r \left[\frac{n}{p^k} \right] = \sum_{k=1}^r \frac{n - \sum_{j=0}^{k-1} a_jp^j}{p^k}.
			\]
			With a little bit of computation, recalling that $1 + p + \dots + p^{k-1} = \frac{p^k - 1}{p-1}$, we obtain the desired formula.
		\end{proof}
		\begin{prop}
			The function $\exp_p(X)$ converges on $D(r_p^-)$ and diverges elsewhere, where $r_p := p^{-1/(p-1)}$.
		\end{prop}
		\begin{proof}
			Using \cref{exercise:14-p.7} we obtain
			\[
			\pabs{1/n!} = p^{\frac{n - S_n}{p - 1} }
			\]
			and, recalling the formula for the radius of convergence $r = 1/(\limsup \pabs{a_n}^{1/n})$, we can write
			\[
			\ord r = -\ord \left( \limsup p^{-(\ord a_n)/n} \right) = -\ord \left( p^{-\liminf (\ord a_n)/n} \right) = \liminf \left( \frac{\ord a_n}{n} \right)
			\]
			so, in our case where $a_n = 1/n!$, we obtain
			\[
			\ord r = \liminf \left(-\frac{n - S_n}{n(p-1)} \right).
			\]
			We can use the easy upper bound $S_n \leq (p-1) \cdot \ord n$ to prove
			\[
				\lim_{n \to +\infty} \frac{S_n - n}{n(p-1)} = -\frac{1}{p-1}
			\]
			so the exponential series $\sum_{n=0}^{+\infty} x^n/n$ converges if $\pabs{x} < p^{-1/(p-1)} = r_p$ and diverges if $\pabs{x} > p^{-1/(p-1)} = r_p$. If $\pabs{x} = r_p$, i.e. $\ord x = 1/(p-1)$, we have
			\[
				\ord (a_nx^n) = -\frac{n-S_n}{p-1} + \frac{n}{p-1} = \frac{S_n}{p-1}
			\]
			and, choosing $n = p^m$ so $S_n = 1$, we have $\pabs{a_{p^m}x^{p^m}} = p^{-1/(p-1)} > 0$; we have found a subsequence of $(\pabs{a_nx^n})_n$ which does not converge to zero so we conclude that if $\pabs{x} =r_p$ the exponential series diverges.
		\end{proof}
		We immediately note that $D(r_p^-) \subsetneq D(1^-)$, i.e. $\exp_p$ converges in a smaller disc than $\log_p$. We now prove that, like in the classical case, the \padic exponential transforms sums into products.
		\begin{prop}
			If $x, y \in D(r_p^-)$ then $\exp_p(x + y) = \exp_p(x) \cdot \exp_p(y)$.
		\end{prop}
		\begin{proof}
			Let's first observe that $x, y \in D(r_p^-) \implies x + y \in D(r_p^-)$ so we can compute $\exp_p(x + y)$. The rest of the proof is completely analogue to the proof of \cref{prop:padic-logarithm-product-sum}, using the fact that $\exp(x + y) = \exp(x) \cdot \exp(y)$ if $x, y \in \R$ (which then can be translated to a relation between power series by \cref{prop:formal-series}).
		\end{proof}
		Finally we have all the tools we need to prove the relation between \padic exponential and logarithm.
		\begin{prop}
			\label{prop:exp-and-log-inverse}
			The functions $\log_p$, defined by $x \mapsto \log_p(1 + (x-1))$, and $\exp_p$ give mutually inverse isomorphisms between the multiplicative group $(D_1(r_p^-), \cdot)$ and the additive group $(D(r_p^-), +)$.
		\end{prop}
		\begin{proof}
			First of all let's observe that $\exp_p\colon D(r_p^-) \to D_1(r_p^-)$ and $\log_p\colon D_1(r_p^-) \to D(r_p^-)$ so that the proposition actually makes sense. To prove that $\exp_p(x) \in 1 + D(r_p^-) \subset 1 + D(1^-)$ let's note that
			\[
				x \in D(r_p^-) \implies \ord\left(\frac{x^n}{n!}\right) = n\cdot\ord(x) - \ord(n!) > \frac{n}{p-1} - \frac{n - S_n}{p-1} = \frac{S_n}{p-1} \geq \frac{1}{p-1}
			\]
			so we have
			\begin{gather*}
				\ord(\exp_p(x) - 1) = \ord\left( \sum_{n=1}^{+\infty} \frac{x^n}{n!} \right) \geq \min_{n \geq 1} \left\{\ord\left(\frac{x^n}{n!}\right) \right\} > \frac{1}{p-1} \\
				\implies \exp_p(x) \in 1 + D(r_p^-).
			\end{gather*}
			Instead, to prove that $\log_p(1 + x) \in D(r_p^-)$ if $x \in D(r_p^-)$ let's observe that
			\[
				\ord\left(\frac{x^n}{n}\right) - \frac{1}{p-1} > \frac{n}{p-1} - \ord(n) - \frac{1}{p-1} = \frac{n-1}{p-1} - \ord(n) =: f(n).
			\]
			We claim that $f$ has its minima at $n=1$ and $n=p$, where it's zero. To see why this is true let's first observe that we can just consider the case where $n = p^k$ for $k \in \N$ since if $n' = p^km$ with $p \nmid m$ we have $f(n') \geq f(n)$. It is then an easy calculation to verify that $f(p^{k+1}) \geq f(p^k)$ for $k \in \N$. Thus we have
			\[
				\ord(\log_p(1 + x)) = \ord\left(\sum_{n=1}^{+\infty} (-1)^{n+1}\frac{x^n}{n}\right) \geq \min_{n > 0} \left\{ \ord\left(\frac{x^n}{n}\right) \right\} > \frac{1}{p-1}
			\]
			which means precisely $\log_p(1 + x) \in D(r_p^-)$. \newline
			We have already proved in some previous propositions that $\exp_p\colon (D(r_p^-),+) \to (D_1(r_p^-), \cdot)$ and $\log_p\colon (D_1(r_p^-), \cdot) \to (D(r_p^-), +)$ are group morphisms so now we have only to prove that they are mutually inverse.\newline
			To see that $\log_p \circ \exp_p\colon D(r_p^-) \to D(r_p^-)$ is the identity function we compute
			\[
				\log_p(\exp_p(x)) = \sum_{n=1}^{+\infty} (-1)^{n+1}\frac{(\exp_p(x) - 1)^n}{n} = \sum_{n=1}^{+\infty} (-1)^{n+1}\frac{\left(\sum_{m=1}^{+\infty} \frac{x^m}{m!}\right)^n}{n}.
			\]
			Since if $x \in \R$ we have $\log(\exp(x)) = x$ we infer, by \cref{prop:formal-series}, that the following formal identity  holds in $\Q\llbracket X\rrbracket$:
			\[
				\sum_{n=1}^{+\infty} (-1)^{n+1}\frac{\left(\sum_{m=1}^{+\infty} \frac{X^m}{m!}\right)^n}{n} = X
			\]
			which implies $\log_p(\exp_p(x)) = x$ for $x \in D(r_p^-)$. The same exact reasoning can be also used to prove $\exp_p(\log_p(1 + x)) = 1 + x$ for $x \in D(r_p^-)$.
		\end{proof}
		This proposition implies, in particular, that $\log_p$ is injective on $D_1(r_p^-)$. It is easy to see that this is the biggest disc where this is true: in-fact if $\zeta \in \Cp$ is a primitive $p$-th root of $1$ then, by \cref{exercise:7-p.74}, $\pabs{\zeta - 1} = p^{-1/(p-1)} = r_p$ and $\log_p(\zeta) = 0 = \log_p(1)$.
		\begin{defn}
			The (partial) functions $\sin_p\colon \Cp \to \Cp$ and $\cos_p\colon \Cp \to \Cp$ defined by
			\begin{gather*}
				\sin_p(x) := \sum_{n=0}^{+\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\\
				\cos_p(x) := \sum_{n=0}^{+\infty} (-1)^n \frac{x^{2n}}{(2n)!}
			\end{gather*}
			are the \padic sine and the \padic cosine.
		\end{defn}
		It's easy to prove that $\sin_p$ and $\cos_p$ are defined on $D(r_p^-)$.

		Another important function in classical analysis is the binomial expansion
		\[
			B_a(x) = \sum_{n=0}^{+\infty} \binom{a}{k} x^n
		\]
		where $x, a \in \C$ and we used the generalized binomial coefficient defined by:
		\begin{gather*}
			\binom{a}{k} :=
			\begin{cases*}
				1, & \text{if $k = 0$} \\
				\frac{a(a-1)\dots (a - k + 1)}{k!}, & \text{otherwise} \\
			\end{cases*}.
		\end{gather*}
	 	This is exactly the MacLaurin series of $f(x) = (1 + x)^a$. Using ratio test it can be proved that for any $a \in \C$ this series converges if $\abs{x} < 1$ and, unless $a \in \N$, diverges if $\abs{x} > 1$. Its behaviour when $\abs{x} = 1$ is a little more complicated and depends on the value of $a$. We'll now try to define an analogue function in the \padic environment.
		\begin{defn}
			Fixed $a \in \Cp$, the (partial) function $B_{a, p}(X)\colon \Cp \to \Cp$ defined by
			\[
				B_{a,p}(X) := \sum_{n=0}^{+\infty} \binom{a}{n}X^n = 1 + \sum_{n=1}^{+\infty} \frac{a(a-1)\dots (a - n + 1)}{n!}X^n
			\]
			is the \padic binomial expansion
		\end{defn}
		We'll now try to study where it converges (it will be more complicated than the previous functions, since this is actually the first series with coefficient in $\Cp$, and not in $\Q$).
		\begin{prop}
			\label{prop:convergence-binomial}
			If $\pabs{a} > 1$ then the region of convergence of $B_{a, p}(X)$ is $D((r_p/\pabs{a})^-)$.  Instead, if $\pabs{a} \leq 1$ the binomial expansion surely converges on $D(r_p^-)$ (although the region of convergence can be bigger). Finally, if $a \in \Zp$ then $B_{a, p}(X) \in \Zp\llbracket X \rrbracket$ so it surely converges on $D(1^-)$.
		\end{prop}
		\begin{proof}
			Let's suppose $\pabs{a} > 1$. Then, by the isosceles triangle principle, if $i \in \Z$ then $\pabs{a - i} = \pabs{a}$ and we obtain that the $n$-th term of the series has norm $\pabs{ax}^n/\pabs{n!}$. Thus, with a little computation, we obtain that the radius of convergence is $r = p^{-1/(p-1)}/\pabs{a} = r_p/\pabs{a}$. Similarly to the exponential case it's easy to prove that the region of convergence if $D((r_p/\pabs{a})^-)$.\newline
			If $\pabs{a} \leq 1$ it is more difficult to find the exact region of convergence; anyway if $i \in \Z$ we have $\pabs{a - i} \leq \max\{\pabs{a}, \pabs{i}\} \leq 1$ so $\pabs{\binom{a}{n}x^n} \leq \pabs{x^n/n!}$. Then $B_{a,p}(X)$ surely converges on $D(r_p^-)$.\newline
			To prove that if $a \in \Zp$ then $B_{a,p}(X) \in \Zp\llbracket X \rrbracket$ we just need to show that $\binom{a}{n} \in \Zp$ for every $n \in \N$ (we already know $\binom{a}{n} \in \Qp$). Let's fix $n$ and choose $a_0 \in \Z$ such that $a_0 > n$ and $\ord(a - a_0) > N$, where $N$ will be chosen later (to choose $a_0$ we can just truncate the \padic expansion of $a \in \Zp$ at some index greater than $N$). Now $\binom{a_0}{n} \in \Z \subset \Zp$ and it suffices to show that $\pabs{\binom{a_0}{n} - \binom{a}{n}} \leq 1$ for a suitable $N$ (then we can conclude using the ultrametric inequality). This easily follows from the continuity of the polynomial $X(X-1)\dots(X - n + 1)$ (special case of \cref{prop:continuity-analitic-function}). Then $B_{a, p}(X) \in \Zp\llbracket X \rrbracket$ if $a \in \Zp$ so, by \cref{prop:convergence-power-series-Zp}, it converges in $D(1^-)$.
		\end{proof}
		We can now prove the main property of the binomial expansion, and justify the shorthand $B_{a,p}(X) = (1 + X)^a$, at least for $a \in \Q$.
		\begin{prop}
			If $a \in \Q^\times$ and $x \in \Cp$ is in the region of convergence of $B_{a,p}(X)$, then $\left[B_{a, p}(x)\right]^{1/a} = 1 + x$.
		\end{prop}
		\begin{proof}
			Let's first consider $a = 1/m$ with $m \in \Z^\times$. The idea behind the proof is the usual one: if $x \in \R$ and $\abs{x} < 1$ we have $B_{1/m}(x) = (1 + x)^{1/m}$ so $B_{1/m}(x)^m = 1 + x$ which, by \cref{prop:formal-series}, gives us the formal identity between the two power series in $\Q\llbracket X \rrbracket$ (observe that $m < 0$ doesn't create problems since $B_{1/m}(X)$ is an invertible element of $\Q\llbracket X \rrbracket$) that we then translate into an equality between \padic analytic functions (observe the trivial fact $a \in \Q \implies B_{a, p}(X) \in \Q\llbracket X \rrbracket$). We must pay attention only to the last step, i.e. we can substitute only $x$ in the region of convergence of $B_{1/m, p}(X)$ so, for example, if $p \mid m$ we can use $x \in D((r_p\pabs{m})^-)$ and if $p \nmid m$ we can choose $x \in D(r_p^-)$. We have proved that $B_{1/m, p}(x)^m = 1 + x$ for every $x$ where $B_{1/m, p}(X)$ converges. \newline
			Now let $a = n/m$ with $n, m \in \Z^\times$. It is easy to prove, using the same technique as before, that $B_{n/m, p}(X) = B_{1/m, p}(X) ^ n$. Then we can write
			\[
				B_{n/m, p}(X)^{m/n} = B_{1/m, p}(X)^m = 1 + X,
			\]
			which proves the thesis.
		\end{proof}
		We can use the \padic binomial expansion to study an interesting example of how the same convergent series in $(\Q, \abs{\ })$ and in $(\Qp, \pabs{\ })$ can have different sums.
		\begin{example}
			Let's consider the following power series:
			\begin{gather*}
				B_{1/2}\left(\frac{7}{9}\right) = \sum_{n=0}^{+\infty} \binom{1/2}{n} \left(\frac{7}{9}\right)^n \qquad \in \Q\llbracket X \rrbracket, \\
				B_{1/2, 7}\left(\frac{7}{9}\right) = \sum_{n=0}^{+\infty} \binom{1/2}{n} \left(\frac{7}{9}\right)^n \qquad \in \Q_7 \llbracket X \rrbracket.
			\end{gather*}
			They are exactly the same power series but they converge to different numbers, both of which are of course square roots of $\tfrac{16}{9}$ (clearly its square roots are the same both in $\Q$ and in $\Q_7$). In the first case, working in $(\Q, 	\abs{\ })$, we have
			\[
				B_{1/2}\left(\frac{7}{9}\right) = \left(1 + \frac{7}{9}\right)^{1/2} = \frac{4}{3} > 0.
			\]
			Instead, in the second case, we have
			\[
				B_{1/2, 7}\left(\frac{7}{9}\right) = \left(1 + \frac{7}{9}\right)^{1/2} = -\frac{4}{3} < 0.
			\]
			In-fact, $\textrm{ord}_7\left(\tfrac{7}{9}\right) = 1$ so for $n \geq 1$ we have
			\[
				\abs{\frac{1/2(1/2 - 1)\dots(1/2 - n + 1)}{n!}\cdot \left(\frac{7}{9}\right)^n}_7 \leq \frac{7^{-n}}{\abs{n!}_7} = 7^{\frac{-5n - S_n}{6}} < 1
			\]
			so it must be $B_{1/2, 7}\left(\tfrac{7}{9}\right) \equiv 1 \mod 7$. Now it's easy to see that $-\tfrac{4}{3} \equiv 1 \mod 7$ and $\tfrac{4}{3} \equiv -1 \mod 7$. We conclude that necessarily $B_{1/2, 7}\left(\tfrac{7}{9}\right) = -\tfrac{4}{3}$.
		\end{example}
	 	This example also warns us about the danger of using the notation $B_{a, p}(X) = (1 + X)^a$, which comes certainly handy sometimes but we have to remember that it can yield different results than the ones we would expect on $\R$.
	 \section{The Iwasawa logarithm and Artin-Hasse exponential}
	 	\begin{defn}
	 		Let $X \subseteq \Cp$ be a set with no isolated points. A function $f\colon X \to \Cp$ is \emph{differentiable} at $a \in X$ if
	 		\[
	 			\exists \lim_{X \ni x \to a} \frac{f(x) - f(a)}{x - a} =: f'(a) \in \Cp.
	 		\]
	 		Equivalently, $f$ is differentiable at $a \in X$ if
	 		\[
	 			f(x) = f(a) + (x - a)f'(a) + (x - a)\phi(x), \qquad \lim_{X \ni x \to a} \phi(x) = 0.
	 		\]
	 	\end{defn}
 		We also introduce a stronger notion of differentiability for \padic functions, which will give us some analogue theorems to the classical case.
 		\begin{defn}
 			Let $X \subseteq \Cp$ be a set with no isolated points. A function $f\colon X \to \Cp$ is \emph{strictly differentiable} at $a \in X$ (and we write $f \in S^1(a)$) if the difference quotients
 			\[
 				\Phi f(x, y) := \frac{f(x) - f(y)}{x - y}
 			\]
 			tends to $\Cp \ni \ell = f'(a)$ as $X \times X \setminus \Delta_X \ni (x, y) \to (a, a)$.
 			Here we used the notation $\Delta_X = \{(x, x) : x \in X \} \subset X \times X$.
 			We say $f \in S^1(X)$ if $f \in S^1(a)$ for every $a \in X$.
 		\end{defn}
 		In the classical case this definition is not very useful: in-fact if $I \subset \R$ is an open interval and $f \in \mathcal{C}^1(I, \R)$ then $f$ is strictly differentiable at every point of $I$. In the next example we'll see this is not the case in \padic analysis.
 		\begin{example}
 			Let's consider the sequence of disjoint open balls $(B_n)_{n \geq 1}$ defined by
 			\[
 				B_n := \{x \in \Zp: \pabs{x - p^n} < \pabs{p^{2n}} \} \subseteq \{x \in \Zp: \pabs{x} = \pabs{p^n}\}
 			\]
 			and let $f\colon \Zp \to \Cp$ defined by
 			\begin{gather*}
 				f(x) :=
 				\begin{cases}
	 				p^{2n}, & \text{if $x \in B_n$;} \\
	 				0, & \text{otherwise};\\
 				\end{cases}.
 			\end{gather*}
 			The function $f$ is constant on each open ball $B_n$, hence $f$ is locally constant outside the origin. Then $f$ is differentiable at every $\Zp \ni x \neq 0$ and $f'(x) = 0$. At the origin
 			\[
 				\lim_{\Zp \ni x \to 0} \frac{f(x) - f(0)}{x} = \lim_{\Zp \ni x \to 0} \frac{f(x)}{x} = 0
 			\]
 			so $f'(0) = 0$ (to see why it is true, let $x = up^n, u \in \Zp^\times$; then $f(x) = p^{2n}$ and so $\tfrac{f(x)}{x} = u^{-1}p^n$). Then $f'\colon \Zp \to \Cp$ is identically $0$ so it is obviously continuous (i.e. $f \in \mathcal{C}^1$) but $f$ is not strictly differentiable at $0$. In-fact, let's consider $\Phi f(x, y)$ where $x = x_n = p^n$ and $y = y_n = p^n - p^{2n}$:
 			\[
 				\Phi f(x_n, y_n) = \frac{f(x_n) - f(y_n)}{x_n - y_n} = \frac{p^{2n} - 0}{p^{2n}} = 1
 			\]
 			so, if we consider this particular path $(x_n, y_n) \to (0,0)$ as $n \to +\infty$ we obtain
 			\[
 				0 = f'(0) \neq \lim_{n \to +\infty} \Phi f(x_n, y_n) = 1,
 			\]
 			which implies $f$ is not strictly differentiable at $0$ (we have used that $\pabs{y_n} = \pabs{p^n}$ and $y_n \notin B_n$).
 		\end{example}
 		We'll now prove a proposition which we're very familiar with in classical analysis.
 		\begin{prop}
 			If $f\colon X \to \Cp$ is strictly differentiable at $a \in X$ and $f'(a) \neq 0$, then there is a neighbourhood $V$ of $a \in X$ in which $f$ is injective.
 		\end{prop}
 		\begin{proof}
 			Since $f \in S^1(a)$ and $\pabs{f'(a)} > 0$ we can find a neighbourhood $V$ of $a$ such that
 			\[
 				\pabs{\Phi f(x, y) - f'(a)} < \pabs{f'(a)} \qquad \text{for $(x, y) \in V \times V \setminus \Delta_V$}.
 			\]
 			Then, by the isosceles triangle principle, we must have $\pabs{\Phi f(x, y)} = \pabs{f'(a)}$ which means exactly
 			\[
 				\pabs{f(x) - f(y)} = \pabs{f'(a)}\pabs{x - y} \qquad \text{for $(x, y) \in V \times V$}. \qedhere
 			\]
 		\end{proof}
 		Let's now focus on analytic functions; they are, like in the classical case, everywhere strictly differentiable any number of times (i.e. they're in $\bigcap_{k > 0} S^k$). We'll only prove it for $k=1$.
 		\begin{thm}
 			\label{thm:derivative-power-series}
 			Let $f(X) = \sum_{n \geq 0} a_nX^n$ be an analytic function which converges on $D = D(r^-)$. Then $f \in S^1(D)$ and $f'$ is given by
 			\[
 				f'(X) = \sum_{n=1}^{+\infty} na_nX^{n-1}.
 			\]
 		\end{thm}
 		\begin{proof}
 			It's immediate to note that the radius of convergence of $f'$ is greater or equal to $r$, the radius of convergence of $f$, since $\pabs{n} \leq 1$ for every $n \in \N$. First of all let's fix $x \in D$ and prove that
 			\[
 				\lim_{h \to 0}\, \pabs{\frac{f(x+h) - f(x)}{h} - f'(x)} = 0.
 			\]
 			We can re-write this limit as
 			\[
 				\lim_{h \to 0}\,\pabs{\sum_{n = 2}^{+\infty} a_n \cdot \left( \frac{(x+h)^n - x^n}{h} - nx^{n-1} \right) } = 0;
 			\]
 			and using the binomial theorem on $(x + h)^n$ we can then write
 			\[
 				\lim_{h \to 0}\,\pabs{\sum_{n = 2}^{+\infty} a_n \cdot \left( \sum_{i=0}^{n-2} \binom{n}{i} x^ih^{n-1-i} \right) } = 0.
 			\]
 			Let's now distinguish two cases: $x = 0$ and $x \neq 0$. \newline
 			If $x = 0$ then we must prove $\lim_{h \to 0}\,\pabs{\sum_{n = 2}^{+\infty} a_n \cdot h^{n-1} } = 0$,	which easily follows from
 			\[
 				\lim_{h \to 0}\,\pabs{\sum_{n = 2}^{+\infty} a_n \cdot h^{n-1} }  \leq \lim_{h \to 0}\,\left( \pabs{h} \cdot \max_{n \geq 2} \left\{\pabs{a_nh^{n-2}} \right\} \right)= 0,
 			\]
 			where we considered $0 < \pabs{h} < r$ and exploited the fact that $\lim_{n \to +\infty} \pabs{a_nh^{n-2}} = 0$ so the maximum in the limit above is bounded.\newline
 			Now, assuming $x \neq 0$ and $0 < \pabs{h} < \pabs{x}$, it's easy to see that
 			\[
 				\pabs{\sum_{i=0}^{n-2} \binom{n}{i} x^ih^{n-1-i}} \leq \pabs{h}^{n-1} \cdot \max_{0 \leq i \leq n-2}\left\{\pabs{x^ih^{-i}}\right\} \leq \pabs{h}^{n-1}\cdot \left(\frac{\pabs{x}}{\pabs{h}}\right)^{n-2} = \pabs{h}\cdot \pabs{x}^{n-2}.
 			\]
 			Then we have
 			\[
 				\pabs{\sum_{n = 2}^{+\infty} a_n \cdot \left( \sum_{i=0}^{n-2} \binom{n}{i} x^ih^{n-1-i} \right) } \leq \pabs{h}\cdot\max_{n \geq 2}\left\{\pabs{a_nx^{n-2}}\right\},
 			\]
 			and since $\lim_{n\to +\infty} \pabs{a_nx^{n-2}} = 0$ the maximum above is bounded so
 			\[
 				\lim_{h \to 0}\,\pabs{\sum_{n = 2}^{+\infty} a_n \cdot \left( \sum_{i=0}^{n-2} \binom{n}{i} x^ih^{n-1-i} \right) } \leq \lim_{h \to 0} \left( \pabs{h}\cdot\max_{n \geq 2}\left\{\pabs{a_nx^{n-2}}\right\}\right) = 0.
 			\]
 			We have proved that $f$ is differentiable everywhere and $f'$ is its derivative. \newline
 			We won't prove here that $f$ is actually strictly differentiable: a proof of this statement for a particular case (where $r \geq 1$, i.e. $\lim_{n\to+\infty}\pabs{a_n}=0$) can be found at \cite[239]{robert:padic-analysis}.
 		\end{proof}
 		\begin{example}
 			We can now prove one well known result of classical analysis: the derivative of $e^x$ is $e^x$. More precisely, if $x \in D(r_p^-)$ then $\frac{\mathrm{d}}{\mathrm{d}x}\exp_p(x) = \exp_p(x)$. It easily follows applying \cref{thm:derivative-power-series}:
 			\[
 				\frac{\mathrm{d}}{\mathrm{d}x}\exp_p(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left( \sum_{n=0}^{+\infty} \frac{x^n}{n!} \right) = \sum_{n=1}^{+\infty} \frac{x^{n-1}}{(n-1)!} = \exp_p(x).
 			\]
 		\end{example}
 		\begin{defn}
 			Let $f\colon \Cp \to \Cp$ be a (partial) function. If for every $x$ in its domain there exists a neighbourhood where $f$ is a power series, we say that $f$ is \emph{locally analytic}.
 		\end{defn}
 		We present now two \padic locally analytic functions: the \emph{Iwasawa logarithm} and the \emph{Artin-Hasse} exponential.
 		\begin{prop}
 			There exists a unique function $\Log_p\colon \Cp^\times \to \Cp$ such that:
 			\begin{enumerate}[label=(\arabic*)]
 				\item $\Log_p$ agrees with $\log_p$ in $D_1(1^-)$, i.e.,
 				\[
 					\Log_p(x) = \sum_{n=1}^{+\infty} (-1)^{n+1}\frac{(x-1)^n}{n} \qquad \text{for $\pabs{x-1} < 1$};
 				\]
 				\item $\Log_p(xy) = \Log_p(x) + \Log_p(y)$ for all $x, y \in \Cp^\times$;
 				\item $\Log_p(p) = 0$.
 			\end{enumerate}
 		\end{prop}
 		\begin{proof}
 			We recall from \cref{prop:structure-Cp} that any non-zero $x \in \Cp$ can be written as $x = p^r\omega(x_1)\langle x_1\rangle$, where $p^r$ is a root of the equation $X^b - p^a = 0$ where $r=\tfrac{a}{b}=\ord(x)$, $\omega(x_1)$ is a root of $1$ and $\pabs{\langle x_1 \rangle - 1} < 1$. If such an extension of the logarithm exists, then, by \textit{(2)} and \textit{(3)}, it must be
 			\[
 				\Log_p(x) = \Log_p(p^r) + \Log_p(\omega(x_1)) + \Log_p(\langle x_1 \rangle) = 0 + 0 + \Log_p(\langle x_1 \rangle) = \log_p(\langle x_1 \rangle),
 			\]
 			since $\langle x_1 \rangle \in D_1(1^-)$. Then there is at most one extension of the logarithm and it is the one defined by
 			\[
 				\Log_p(x) := \log_p(\langle x_1 \rangle).
 			\]
 			First of all we have to show that this is well defined: in-fact we could have chosen another root of $X^b - p^a= 0$ and we would have obtained a different factorization of the same element. Let's suppose that
 			\[
 				x = p^r\cdot\omega(x_1)\cdot\langle x_1\rangle = \frac{p^r}{\zeta}\cdot\omega\left(x_1\overline{\zeta}\right)\cdot\left\langle x_1\overline{\zeta}\right\rangle,
 			\]
 			where $\zeta \in \Cp$ is a $b$-th root of unity and $\overline{\zeta} = \zeta + M \in A/M$ (we recall that $A = D(1), M = D(1^-)$ in $\Cp$). We have to prove then that $\log_p(\langle x_1\rangle) = \log_p(\left\langle x_1\overline{\zeta}\right\rangle)$. Let's first recall how the Teichm{\"u}ller representatives are defined: if $\overline{\Fp}$ is the algebraic closure of $\Fp$ then $\omega\colon \overline{\Fp} \to \Zpu$ is a section of the projection $\pi\colon \Zpu \twoheadrightarrow \Zpu/p\Zpu = \overline{\Fp}$ such that $\omega(x)^{p^f - 1} = 1$ if $x \in \F_{p^f}^\times$ and $\omega(0)=0$ (it is immediate that since $\omega$ can be defined on every finite field of characteristic $p$, see the proof of \cref{prop:structure-finite-extension}, it can be extended to $\overline{\Fp}$). It's easy to see that $\omega\colon \overline{\Fp}^\times \to \left(\Zpu\right)^\times$ is a group morphism, i.e. $\omega(xy) = \omega(x) \cdot \omega(y)$: in-fact if $x, y \in \F_{p^f}$ then $\omega(xy)$ is defined as the only element of $\Zpu$ such that $\omega(xy)^{p^f} = \omega(xy)$ and $\pi(\omega(xy))=xy$ and it's clear that $\omega(x)\cdot\omega(y)$ satisfies both these conditions.
 			We also recall from \cref{prop:structure-Cp} that we can find a big enough $f \in \N$ such that
 			\begin{gather*}
	 			\F_{p^f} \ni x_1 = \frac{x}{p^r} + M,\\
	 			\F_{p^f} \ni x_1\overline{\zeta} = \frac{x\zeta}{p^r} + M = \left(\frac{x}{p^r} + M\right) \cdot (\zeta + M).
 			\end{gather*}
 			Explained all the notations, we can finally write
 			\[
 				\omega\left(x_1\overline{\zeta}\right) = \omega(x_1)\cdot\omega\left(\overline{\zeta}\right) \implies \left\langle x_1\overline{\zeta}\right\rangle = \frac{x\zeta}{p^r\cdot\omega\left(x_1\overline{\zeta}\right)} = \langle x_1\rangle \cdot \frac{\zeta}{\omega\left(\overline{\zeta}\right)}.
 			\]
 			Now, since $\zeta^b = 1$, we have $\overline{\zeta}^b = 1$ so $\omega\left(\overline{\zeta}\right)^b = 1$, since $\omega$ is a group morphism and $\omega(1) = 1$. Finally,
 			\[
 				\left(\frac{\zeta}{\omega\left(\overline{\zeta}\right)}\right)^b = \frac{\zeta^b}{\omega\left(\overline{\zeta}\right)^b} = 1
 			\]
 			so $\xi := \tfrac{\zeta}{\omega\left( \overline{\zeta}\right)}$ is a root of $1$. Let's prove that $\pabs{\xi - 1} < 1$; let's suppose by contradiction that $\xi = 1 + \Delta$ with $\pabs{\Delta} \geq 1$ and let's write $\langle x_1 \rangle = 1 + \delta$ with $\pabs{\delta} < 1$. We know by hypothesis that $\langle x_1\overline{\zeta} \rangle \in D_1(1^-)$ so we must have
 			\[
 				\pabs{(1 + \delta) \cdot (1 + \Delta) - 1} = \pabs{\delta + \Delta \cdot(1 + \delta)} < 1
 			\]
 			but since $\pabs{\delta} < 1$ and $\pabs{1 + \delta} = 1$, we have $\pabs{\Delta \cdot (1 + \delta)} = \pabs{\Delta} \geq 1$ so
 			\[
 				\pabs{\delta + \Delta\cdot(1 + \delta)} = \max\left\{\pabs{\delta}, \pabs{\Delta \cdot (1 + \delta)}\right\} = \pabs{\Delta} \geq 1,
 			\]
 			which is absurd (here we have used several times the isosceles triangle principle). We have proved that $\xi$ is a root of $1$ with $\pabs{\xi - 1} < 1$ so we can compute $\log_p(\xi) = 0$. Then
 			\[
 				\log_p\left(\left\langle x_1\overline{\zeta}\right\rangle\right) = \log_p(\langle x_1 \rangle) + \log_p(\xi) = \log_p(\langle x_1 \rangle)
 			\]
 			and the function $\Log_p$ is well defined. \newline
 			Properties \textit{(1)} and \textit{(3)} are now obvious from the definition: if $x \in D_1(1^-)$ then we can choose $x = \langle x \rangle$  so $\Log_p(x) = \log_p(x)$ and $p = p^1 \cdot 1 \cdot 1$ so $\Log_p(p) = \log_p(1) = 0$. To prove \textit{(2)} let $x = p^r\omega(x_1)\langle x_1 \rangle$, $y = p^s \omega(y_1) \langle y_1 \rangle$ and $z = xy = p^{r+s}\omega(z_1)\langle z_1 \rangle$. Now $p^{r+s}$ isn't necessarily the same fractional power as $p^rp^s$ (it can differ by a root of unit), but we can choose to use exactly $p^rp^s$, since the value of $\Log_p$ doesn't depend on the choice of the fractional power. In this case we'll have $z_1 = \tfrac{z}{p^rp^s} + M = x_1y_1$ so $\omega(z_1) = \omega(x_1)\cdot\omega(y_1)$ and $\langle z_1 \rangle = \langle x_1 \rangle \cdot \langle y_1 \rangle$. Then $\Log_p(xy) = \Log_p(x) + \Log_p(y)$.
 		\end{proof}
 		\begin{prop}
 			$\Log_p$ is locally analytic on $\Cp^\times$ with derivative $\Cp^\times \ni x \mapsto \tfrac{1}{x}$.
 		\end{prop}
 		\begin{proof}
 			Let's fix a point $x_0 \in \Cp^\times$ and let $r := \pabs{x_0}$. For every $x \in D_{x_0}(r^-)$ (the largest disc about $x_0$ which doesn't contain $0$) we have $\pabs{\tfrac{x}{x_0} - 1} < 1$ and so
 			\[
 				\Log_p(x) = \Log_p\left(x_0 \cdot \left(1 + \frac{x}{x_0} - 1\right)\right) = \Log_p(x_0) + \sum_{n=1}^{+\infty} (-1)^{n+1}\cdot\frac{(x - x_0)^n}{n\cdot x_0^n}.
 			\]
 			We have just proved that, in a neighbourhood of $x_0$, $\Log_p$ can be represented by a convergent power series in $x - x_0$. Since this reasoning can be done for any $x_0 \in \Cp^\times$ we can conclude that $\Log_p$ is locally analytic.\newline
 			Let's consider $x \in D_{x_0}(r^-)$ as above: using the locally analyticity of $\Log_p$ and \cref{thm:derivative-power-series} we obtain:
 			\begin{gather*}
				\frac{\mathrm{d}}{\mathrm{d}x}\Log_p(x) = \sum_{n=1}^{+\infty} (-1)^{n+1}\cdot\frac{(x-x_0)^{n-1}}{x_0^n} = x_0^{-1} \cdot \sum_{n=0}^{+\infty} \left(1 - \frac{x}{x_0}\right)^n = \frac{x_0^{-1}}{(x/x_0)} = \frac{1}{x}. \qedhere
 			\end{gather*}
 		\end{proof}
 		We have found a locally analytic function defined on $\Cp^\times$ which extends $\log_p$ and has the same basic properties.

 		It is now natural to try to build a homomorphism $f\colon \Cp \to \Cp^\times$ extending the exponential, which is only defined in $D(r_p^-)$. If there exists such an extension then, fixed $x \in \Cp^\times$ and $n \in \N$ such that $p^nx \in D(r_p^-)$, then
 		\[
 			f(x)^{p^n} = f(p^nx) = \exp_p(p^nx)
 		\]
 		so $f(x)$ must be a $p^n$-th root of $\exp_p(p^nx)$. As stated in the next proposition, this extension can actually be done in a coherent way.
 		\begin{prop}
 			There exists a continuous homomorphism $\mathrm{Exp}\colon \Cp \to D_1(1^-)$ extending $\exp_p$.
 		\end{prop}
 		\begin{proof}
 			The idea behind the proof exploits the fact that, since $(D_1(1^-), \cdot)$ is a divisible group, there is an extension property for homomorphisms defined over subgroups. For the whole proof see \cite[259]{robert:padic-analysis}.
 		\end{proof}
 		Unlike the Iwasawa logarithm, the extensions $\mathrm{Exp}$ of the exponential are not defined in a canonic way so they're not very useful. Anyway it is easy to prove that, chosen such an extension $\mathrm{Exp}$, $\log_p \circ\, \mathrm{Exp} = \mathrm{id}_{\Cp}$. In-fact:
 		\[
 			p^n\cdot \left(\log_p\circ\,\mathrm{Exp}(x)\right) = \log_p\left(\mathrm{Exp}(x)^{p^n}\right) = \log_p\left(\mathrm{Exp}\left(p^nx\right)\right) = \log_p\left(\exp_p\left(p^nx\right)\right) = p^nx.
 		\]

 		We'll now describe a slightly different exponential function which converges in $D(1^-)$: the Artin-Hasse exponential. Before defining it we'll need to study some basic properties of the well known M{\"o}bius function.
 		\begin{defn}
 			Let $\mu\colon \N^\times \to \N$ be defined by
 			\[
 				\mu(n) :=
 				\begin{cases}
	 				0, & \text{if $n$ is divisible by a perfect square greater than 1;}\\
	 				(-1)^k, & \text{if $n$ is a product of $k$ distinct prime factors;}
 				\end{cases}.
 			\]
 			This is the \emph{M{\"o}bius function}.
 		\end{defn}
 		\begin{prop}
	 		\label{prop:mobius-function}
	 		Let $n \in \N^\times$, then
	 		\[
	 			\sum_{d \mid n} \mu(d) =
	 			\begin{cases}
		 			1, & \text{if $n=1$;} \\
		 			0, & \text{otherwise;}
	 			\end{cases}.
	 		\]
	 		In particular, if $p$ is a prime,
	 		\[
	 			\sum_{\mathclap{d \mid n,\,p \nmid d}} \mu(d) =
	 			\begin{cases}
		 			1, & \text{if $n$ is a power of $p$;} \\
		 			0, & \text{otherwise;}
	 			\end{cases}.
	 		\]
 		\end{prop}
 		\begin{proof}
 			The case $n=1$ is trivial ($\mu(1) = 1$). Let $n = p_1^{a_1}\dots \,p_s^{a_s}$ with $s \geq 1$ and $p_i$ prime for every $i=1,\dots,s$. Then, by an easy combinatoric argument, we have
 			\[
 				\sum_{d \mid n} \mu(d) = \sum_{\epsilon_i = 0 \lor 1} \mu(p_1^{\epsilon_1}\dots\,p_s^{\epsilon_s}) = \sum_{\epsilon_i = 0 \lor 1} (-1)^{\sum \epsilon_i} = (1 - 1)^s = 0.
 			\]
 			The second statement is just a particular case of the first one applied to $n \cdot p^{-\ord(n)}$ in place of $n$.
 		\end{proof}
 		\begin{prop}
 			\label{prop:formal-identity-exp}
 			In $\Q\llbracket X \rrbracket$ the following holds:
 			\[
 				\exp(X) = \prod_{n=1}^{+\infty} B_{-\mu(n)/n}(-X^n).
 			\]
 		\end{prop}
 		\begin{proof}
 			First of all let's observe that the infinite product of series actually makes sense: in-fact $B_{-\mu(n)/n,\,p}(-X^n) = 1 + \tfrac{\mu(n)}{n}X^n + o(X^n)$ so the $n$-th factor has no power of $X$ less than the $n$-th, so only a finite number of series is involved to determine the coefficient of any power of $X$.
 			To prove that the identity holds we'll use \cref{prop:formal-series}; let $x \in \R$ with $\abs{x} < 1$, then we know that
 			\[
 				B_{-\mu(n)/n}(-x^n) = (1 - x^n)^{-\frac{\mu(n)}{n}}.
 			\]
 			Taking the (classical) $\log$ of the right side we obtain
 			\[
 				\log\left(\prod_{n=1}^{+\infty} (1 - x^n)^{-\frac{\mu(n)}{n}}\right) = -\sum_{n=1}^{+\infty}\frac{\mu(n)}{n} \cdot \log(1 - x^n) = \sum_{n=1}^{+\infty} \frac{\mu(n)}{n}\cdot\sum_{m=1}^{+\infty} \frac{x^{nm}}{m} = \sum_{j=1}^{+\infty}\left(\frac{x^j}{j}\cdot \sum_{n \mid j} \mu(n)\right)
 			\]
 			where in the last step we set $j = nm$ and we rearranged the terms of the series since it is absolutely convergent. To see why this is true, let's consider
 			\[
 				\sum_{n=1}^{+\infty} \abs{\frac{\mu(n)}{n}}\cdot\abs{\log(1 - x^n)} \leq \sum_{n=1}^{+\infty} \frac{\abs{\log(1 - x^n)}}{n}.
 			\]
 			Since $\abs{\log(1 - x^n)} \sim -\abs{x}^n$ as $n \to +\infty$, we can just study the convergence of the series
 			\[
 				\sum_{n=1}^{+\infty} \frac{\abs{x}^n}{n},
 			\]
 			which converges since it is dominated by the convergent geometric series $\sum_{n=1}^{+\infty} \abs{x}^n$ (we're using $\abs{x} < 1$). Now that we have justified why we can rearrange terms, using \cref{prop:mobius-function}, we obtain
 			\begin{gather*}
 				\log\left(\prod_{n=1}^{+\infty} (1 - x^n)^{-\frac{\mu(n)}{n}}\right) = \sum_{j=1}^{+\infty}\left(\frac{x^j}{j}\cdot \sum_{n \mid j} \mu(n)\right) = x = \log(\exp(x)) \\
 				\implies \exp(x) = \prod_{n=1}^{+\infty} (1 - x^n)^{-\frac{\mu(n)}{n}}
 			\end{gather*}
 			which, translated back to formal power series, concludes the proof.
 		\end{proof}
 		We have just proved that
 		\[
 			\exp_p(X) = \prod_{n=1}^{+\infty} B_{-\mu(n)/n,\,p}(-X^n)
 		\]
 		(recall that $B_{-\mu(n)/n,\,p}(X) = B_{-\mu(n)/n}(X)$ and $\exp_p(X) = \exp(X)$, as elements of $\Q \llbracket X \rrbracket$).
 		With this new expression of $\exp_p(X)$ we can understand where convergence ``problems'' arise. In-fact if $p \mid n$ and $n$ is square-free (so $\mu(n) \neq 0$) then $\pabs{\mu(n)/n} = \pabs{n}^{-1} \geq p$ so $B_{-\mu(n)/n,\,p}(-X^n)$ converges only if $\pabs{x}^n \in D((r_p\pabs{n})^-)$ (see \cref{prop:convergence-binomial}). If $n=p$ we have convergence precisely when
 		\[
 			\pabs{x} < \left(p^{-1/(p-1)}\cdot p^{-1}\right)^{\frac{1}{p}} = p^{-1/(p-1)} = r_p.
 		\]
 		Instead, if $p \nmid n$, we have no problems, since $-\tfrac{\mu(n)}{n} \in \Zp$ and, by \cref{prop:convergence-binomial}, we have $B_{-\mu(n)/n,\,p}(-X^n) \in \Zp\llbracket X \rrbracket$ so $x \in D(1^-)$ guarantees convergence.	This motivates the following definition.
 		\begin{defn}
 			\label{defn:artin-hasse}
 			The (partial) function $\E_p(X)\colon \Cp \to \Cp$  defined by
 			\[
 				\E_p(X) := \prod_{\substack{n=1 \\ p\nmid n}}^{+\infty} B_{-\mu(n)/n,\,p}(-X^n) = \prod_{\substack{n=1 \\ p\nmid n}}^{+\infty} (1 - X^n)^{-\frac{\mu(n)}{n}}
 			\]
 			is called the \emph{Artin-Hasse exponential}.
 		\end{defn}
 	 	We observe again that  $B_{-\mu(n)/n,\,p}(-X^n) \in 1 + X^n\Q\llbracket X \rrbracket$ so the infinite product makes sense.
 	 	\begin{prop}
 	 		\label{prop:artin-hasse-formula}
 	 		In $\Q\llbracket X \rrbracket$ the following holds:
 	 		\[
 	 			\E_p(X) = \exp_p\left(\sum_{i=0
 	 			}^{+\infty} \frac{X^{p^i}}{p^i}\right).
 	 		\]
 	 	\end{prop}
  		\begin{proof}
  			As usual let's consider $x \in \R$ with $\abs{x} < 1$: then
  			\[
  				\E_p(x) := \prod_{\substack{n=1 \\ p\nmid n}}^{+\infty} (1 - x^n)^{-\frac{\mu(n)}{n}}
  			\]
  			and taking logarithm we obtain
  			\[
  				\log\left(\E_p(x)\right) = -\sum_{\substack{n=1 \\ p \nmid n}} \frac{\mu(n)}{n} \left( \sum_{m=1}^{+\infty} \frac{x^{mn}}{m}\right) = \sum_{j=1}^{+\infty} \left(\frac{x^j}{j} \cdot \sum_{\mathclap{n \mid j,\,p \nmid n}} \mu(n) \right),
  			\]
  			where in the last step we set $j=nm$ and we rearranged the terms of the series (we can do it, the proof is analogue to the one given in \cref{prop:formal-identity-exp}). Using \cref{prop:mobius-function} we obtain the following relation (in $\R$):
  			\[
  				\log\left(\E_p(x)\right) = \sum_{m=0}^{+\infty} \frac{x^{p^m}}{p^m} \implies \E_p(x) = \exp\left(\sum_{m=0}^{+\infty} \frac{x^{p^m}}{p^m}\right).
  			\]
  			We can conclude immediately applying \cref{prop:formal-series} (recall that  $\exp_p$ and $\exp$ are exactly the same formal series in $\Q\llbracket X\rrbracket$).
  		\end{proof}
  		At this point it is very easy to prove that $\E_p(X)$ converges on $D(1^-)$ (much better than the smaller disc of convergence of $\exp_p(X)$).
  		\begin{prop}
  			\label{prop:artin-hasse-convergence}
  			The Artin-Hasse exponential $\E_p(X)$ converges on $D(1^-)$.
  		\end{prop}
  		\begin{proof}
  			We recall the definition of $\E_p$:
  			\[
  				\E_p(X) := \prod_{\substack{n=1 \\ p \nmid n}}^{+\infty} B_{-\mu(n)/n,\,p}(-X^n).
  			\]
  			Now if $p \nmid n$ we have already proved that $B_{-\mu(n)/n,\,p}(-X^n) \in 1 + X^n(\Zp\cap\Q)\llbracket X\rrbracket$ so the whole series $\E_p(X)$ has coefficients in $\Zp\cap\Q$. Then we conclude using \cref{prop:convergence-power-series-Zp}.
  		\end{proof}
  		We could have proved directly that $\exp_p\left(\sum_{n=0}^{+\infty} \frac{x^{p^n}}{p^n}\right) \in \Zp\llbracket X\rrbracket$ using the Dwork's lemma. In-fact
  		we already know that $\E_p(X) \in 1 + X\Qp\ser{X}$ and we can compute
  		\begin{gather*}
  			\E_p(X^p) = \exp_p\left(\sum_{n=0}^{+\infty} \frac{X^{p^{n+1}}}{p^n}\right), \\
  			\E_p(X)^p = \exp_p\left(\sum_{n=0}^{+\infty} \frac{X^{p^n}}{p^{n-1}}\right) = \exp_p\left(pX + \sum_{n=0}^{+\infty} \frac{X^{p^{n+1}}}{p^n}\right),
  		\end{gather*}
  		where we used $\exp_p(Y)^p = \exp_p(pY)$ (this formal identity can be easily verified using \cref{prop:formal-series}). Since $\tfrac{\exp_p(X)}{\exp_p(Y)} = \exp_p(X - Y)$ (also easy to prove) we have
  		\[
  			\frac{\E_p(X^p)}{\E_p(X)^p} = \frac{\exp_p\left(\sum_{n=0}^{+\infty} \frac{X^{p^{n+1}}}{p^n}\right)}{\exp_p\left(pX + \sum_{n=0}^{+\infty} \frac{X^{p^{n+1}}}{p^n}\right)} = \exp_p(-pX) \in 1 + pX\Zp\ser{X}
  		\]
  		and we can conclude that $\E_p(X) \in 1 + X\Zp\ser{X}$ thanks to \cref{lemma:dwork}.


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\chapter{Integral Equations}

\Section{11}{1}{Definition of an integral equation.}

An integral equation is one which involves an unknown function under
the sign of integration; and the process of determining the unknown
function is called solving the equation*.

The introduction of integral equations into analysis is due to Laplace
(1782) who considered the equations

f x) = [e f (f) (t) dt, g x) = [ - 4> (t) dt

(where in each case (f) represents the unknown function), in connexion
with the solution of differential equations. The first integral
equation of which a solution was obtained, was Fourier's equation

f(.v) = l cos (xt) (f> (t) dt,

of which, in certain circumstances, a solution isf

2 f

(f)( x) = - I cos (ux)f(u) du,

TTJo

f x) being an even function of a;, since cos xt) is an even function.

Later, Abel+ was led to an integral equation in connexion with a
mechanical problem and obtained two solutions of it; after this,
Liouville investigated an integral equation which arose in the course
of his researches on differential equations and discovered an
important method for solving integral equations §, which will be
discussed in \hardsectionref{11}{4}.

In recent years, the subject of integral equations has become of some
importance in various branches of Mathematics; such equations (in
physical problems) frequently involve repeated integrals and the
investigation of them naturally presents greater difficulties than do
those elementary equations which will be treated in this chapter.

To render the analysis as easy as possible, we shall suppose
throughout that the constants a, h and the variables x, y, are real
and further that

* Except in the case of Fourier's integral \hardsectionref{9}{7}) we practically
always need continuom solutions of integral equations.

t If this value of (p be substituted in the equation we obtain a
result which is, effectively, that of \hardsectionref{9}{7}.

+ Solution de quelques problemes a Vaide d'integrales definies (1823).
See Oeuvres, i. pp. 11, 97.

§ The numerical computation of solutions of integral equations has
been investigated recently by Whittaker, Proc. Roijal Soc. xciv. (A),
(1918), pp. 367-383.

U-2

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O', y, i>' also that the given function* K (x, y), which occurs under
the integral sign in the majority of equations considered, is a real
function of a; and y and either (i) it is a continuous function of
both variables in the range (a x b, a i/ b), or (ii) it is a
continuous function of both variables in the range a y x b and K (x,
y) = when y > x; in the latter case K x, y) has its discontinuities
regularly distributed, and in either case it is

easily proved that, iff(y) is continuous when a y b, f(y) K x, y) dy
is a

J a

continuous function of x when a x b.

11 'll. An algebraical lemma.

The algebraical result which will now be obtained is of great
importance in Fredholm's theory of integral equations.

Let (a'i, 3/1, Zy), .V2, i/2, 22)) ( '35 3) 3) be three points at
unit distance from the origin. The greatest (numerical) value of the
volume of the parallelepiped, of which the lines joining the origin to
these points are conterminous edges, is +1, the edges then being
perpendicular. Therefore, if Xr + 7/ + Zr = l r = l, 2, 3), the upper
and lower bounds of the determinant

 2 2/2 22 I

 3 3 23 I

are ±1.

A lemma due to Hadamardt generalises this result.

Let

 21? 0 22, ... a n

A

 nl) )!2'  < n

n

where a,. is real and 2 a" r = l ( i = l, 2, ... n); let A j. be the
cofactor of a r in D and let A be the determinant whose elements are A
j., so that, by a well-known theorem |,

Since Z) is a continuous function of its elements, and is obviously
bounded, the ordinary theory of maxima and minima is applicable, and
if we consider variations in

" dD

air ( '=!) 2, ... n) only, D is stationary for such variations if 2 -
8aij. = 0, where Saj,...

r=l VOlir

n

are variations subject to the sole condition 2 a.ir8air=0; therefore
§

r = l

n

but 2 airAir=I), and so X'2a\ r = D; therefore Air = Da.ir-

r=l

* Bocher in his important work on integral equations (Camb. Math.
Tracts, No. 10), always considers the more general case in which A'
(x, y) has discontinuities regularly distributed, i.e. the
discontinuities are of the nature described in Chapter iv, example 10.
The reader will see from that example that the results of this chapter
can almost all be generalised in this way. To make this chapter more
simple we shall not consider such generalisations.

+ Bulletin des Sci. Math. (2), xvii. (1893), p. 240.

J Burnside and Panton, Theory of Equations, ii. p. 40.

§ By the ordinary theory of undetermined multipliers.

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Considering variations in the other elements of D, we see that D is
stationary for variations in all elements when Amr=Damj. (m=l, 2, ...
n; r=l, 2, ... n). Consequently A = D". D, and so I)' ' - D ~ . Hence
the maximum and minimum values of I) are +1.

Corollary. If a r be real and subject only to the condition | a I < j
since

2 a,,/( ii/) 2 1,

r=l

we easily see that the maximum value of | i9 | is (/i J/)" = ?i "i/".

\Section{11}{2}{Fredholms* equation and its tentative solution.}
An important
integral equation of a general type is

J a

where f(x) is a given continuous function, X is a parameter (in
general complex) and K (oo, ) is subject to the conditions f laid down
in \hardsectionref{11}{1}. K (x, ) is called the nucleusX of the equation.

This integral equation is known as Fredholms equation or the integral
equation of the second kind (see \hardsectionref{11}{3}). It was observed by Volterra
that an equation of this type could be regarded as a limiting form of
a system of linear equations. Fredholm's investigation involved the
tentative carrying out of a similar limiting process, and justifying
it by the reasoning given below in \hardsubsectionref{11}{2}{1}. Hilbert (Gottinger Nach.
1904, pp. 49-91) justified the limiting process directly.

We now proceed to write down the system of linear equations in
question, and shall then investigate Fredholm's method of justifying
the passage to the limit.

The integral equation is the limiting form (when 5-*-0) of the
equation

<i> x) =f x) + \ i K X, X,) (t> (.r,) 8, where x - Xq\ i = 8, XQ = a,
.r = b.

Since this equation is to be true when a x b, it is true when x takes
the values Xi, Xi, ... Xni and so

n

-\ 8 2 K x.p, x ) (.rg) + (f) (Xj,) =/ (Xj,) ip = l,2,... n).

q=l

* Fredholm's first paper On the subject appeared in the Ofversigt af
K. Vetenskaps-Akad. Forhandlingar (Stockholm), lvii. (1900), pp.
39-46. His researches are also given in Acta Blath. xxvii. (1903), pp.
365-390.

t The reader will observe that if K x, |) = |>x), the equation may be
written

<p (x) =f (x) +\ Tk (x, I) ( ) d|.

This is called an equation with variable upper limit.

1;. Another term is kernel; French noyau, German Kern.

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This system of equations for < (.rp), p=l, 2, ... 7i) has a unique
solution if the determinant formed by the coefficients of x ) does not
vanish. This determinant is

Z> (X) = 1 - X8 K (. 1,xi) -X8K .vi, x. ) ... -UK x, .v ) -\ 8K x.,
, A'l) 1 - XS K x.,, .r.,) ... - X S A' (.rg, . )

-\ 8K (.r, xi) - \ 8 K x, x.,) ... 1-X8K (x, x ) = 1 - X 2 8A' x,
.Vp) +- 2 S2 1 ''''' '  '" " '

p = l 2 !p, q=l I K X, Xp) A [Xq, Xg)

X3 )i

';>> <I, r=l

+ ...

 "(" 'p' ' P' V" p " '9/ ' (." p> " r)

K Xq,Xp) K Xg,.Vq) K Xg,Xr)

lK x,.,Xp) K Xr,Xq) K Xr, Xr)

on expanding* in powers of X.

Making S-3-O, ti-s-qo, and writing the summations a-s integrations, we
are thus led to consider the series

/6 x2 [ [

i)(X) = l-X| A' (li, 0 1 + 2] / j

 ( 2, 1) />:( 2, I2)

d idi;-...

Further, if Z) (.r, x,) is the cofactor of the term in I)n ) which
involves K x\, x J, the solution of the system of linear equations is

,, /( i)i) (.r, xi)+f x.2) Z) (av, .v.i) + ...+f xJD x, Xn)

Now it is easily seen that the appropriate limiting form to be
considered in association with Da Xf, x ) is D ); also that, if /n +
i',

Dn (AV, -iV) = U\ K x, x ) - XS 2

K(x,x ) K x,Xp)

A Xpi Xi,) A (A'p, Xp)

+,\ \ 8 2

 ( A'- V) Xf,.Vp) K x,Xg)

K Xp, .r ) K (Xp, .?7p) (Xp,; g)

A' (.Pg, X ) K (Xg, Xp) K Xg, Xg)

So that the limiting form for 8~ I) x ., x ) to be considered t is

D (.r,Xt,; X) = X A" x

>, Av)-X2 j

2! ja Ja A:( i,

'' I K x,x ) A'(.r, 1) I

K 2,X;) K -2, l) ( 2,6)

 Il' l2-----

Consequently we are led to consider the possibility of the equation
<P(x)=f x) + J \ [ d .x,; ) fii)d giving the solution of the integral
equation.

* The factorials appear because each determinant of s rows and columns
occurs s ! times as p, q, ... take all the values 1, 2, ... n, whereas
it appears only once in the original determinant for D (X).

t The law of formation of successive terms is obvious from those
written down.

] 1 -21] INTEGRAL EQUATIONS

Example 1. Shew that, in the case of the equation

  x) = x->r\ \ xii(l) y)dy, J

D ) = 1\, D (.r, y; X) = \ xy Zx

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% 215
%

we have

and a solution is

0( ) =

3-X'

we have

Example 2. Shew that, in the case of the equation

<j> x) = x + \ I (xy+y )<f>(i/)dy,

J

D x, y; X) = X xy + ?/2) +X xy -,xy - If + y\ and obtain a sohition
of the equation.

\Subsection{11}{2}{1}{Investigation of Fredholm's solution.}

So far the construction of the solution has been purely tentative; we
now start ah initio and verify that we actually do get a solution of
the equation; to do this we consider the two functions D ( ), D x,
y; ) arrived at in \hardsectionref{11}{2}.

We write the series, by which D ( -) was defined in \hardsectionref{11}{2}, in the
form 1 + 2 so that

 =i n\

rb fb rb J a -1 a J a

d,d 2  d n;

since K x, y) is continuous and therefore bounded, we have \ K x,y)\ <
M, where M is independent of x and y; since K x, y) is real, we may
employ Hadamard's lemma \hardsubsectionref{11}{1}{1}) and we see at once that

Write n 'M'' (b - a)" = n ! 6; then

lim(6.WM=lim f-" - f|(l+jYf

u oo M oo (71+1)5 (\ nj )

since ( 1 +

The series 2 bnX"- is therefore absolutely convergent for all values
of X,;

w = l

and so \hardsubsectionref{2}{3}{4}) the series 1+2 - ~ converges for all values of X and
there- fore \hardsubsectionref{5}{6}{4}) represents an integral function of X.

Now Tfrite the series for D (x, y;X) in the form 2 - - '- - .

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% 216
%

Then, by Hadamard's lemma (\hardsubsectionref{11}{1}{1}),

and hence ' < Cn, where c is independent of a; and y and 2 Cn> ' is n\
=o

absohitely convergent.

Therefore D x, y; ) is an integral function of \ and the series for D
x, y\ \ )- \ K (x, y) is a uniformly convergent \hardsubsectionref{3}{3}{4}) series of
continuous* functionsj|of x and y when a x b, a y b.

Now pick out the coefficient of A" (a;, y) in D(x, y;X); and we get

D x,y;X) \ D ) K x, y) + 2 (-) X +', where

Expanding in minors of the first column, we g t Q x, y) equal to the
integral of the sum of n determinants; writing | i, o, ... m-i, |,
,n,   n-\ in place of fi, o, ... in the ??ith of them, we see that
the integrals of all the determinants + are equal and so

Qn x,y) = -n\ K x, y) Pnd d i    d n-i,

J a J a J a

where

Pn=\ K X, I), K X,,), ... K(x, | \ 0

i di.a K A ... A-(inf -o

It follows at once that

D(x,y\ X) = \ D(X)K(x,y)+\ ( D (x, \ X)K ly)dl

. a

Now take the equation

<i> )=fi ) + \' K ly)<f> y)dy,

J a

multiply by D (x,; X) and integrate, and we get

l'f(BD(,; )d

J a

= I % (I) 0 -. ?; ) ? - r [' (' '; ) < ' ) > y* '

.'a . a J i(

the integrations in the repeated integral being in either order.

* It is easy to verify that every term (except possibly the first) of
the series for D (x, y; ) is a continuous function under either
hypothesis (i) or hypothesis (ii) of \hardsectionref{11}{1}.

h

t The order of integration is immaterial \hardsectionref{4}{3}).

t

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% 217
%

That is to say

J a

= r<j>( )D(w,; ) d -r D w,y; ) -\ D( ) K x,y)]<j>(y)dy

J a J a

= \ D ) \' K x,y)<i> y)dy

in virtue of the given equation.

Therefore if D X) 0 and if Fredholm's equation has a solution it can
be none other than

< X) =f x) +jy( ) d;

and, by actual substitution of this value of <f> x in the integral
equation, we see that it actually is a solution.

This is, therefore, the unique continuous solution of the equation if
i)(X)4=0.

Corollary. If we put /( ) = 0, the ' homogeneous ' equation

J a

has no continuous solution except (j) (.i')=0 unless D X) = 0.

Example 1. By expanding the determinant involved in \$ (.r, y) in
minors of its first row, shew that

D x,y; ) = \ D K)K x,y) + \ \ ' K x, )D,y; ) d .

J a

Example 2. By using the formulae

2)(X) = 1+ i ""fi, D x,y; ) = \ D X) K(x, y)+ i ( - )" " ' ' "/'' \
/i=i 'I 11=1 "

shew that f " i> (,; X) f l = - X .

Example 3. If K x, y) = 1 (y 0) i \ y) = (y > ' )-> shew that D ) = b-
a) X.

Example 4. Shew that, if K (.r, y)=fi (x) .f [y), and if

-6

fi )f2 x)dx = A,

then

D ) = A\ D x,y; ) = \ h x)f., (y),

and the solution of the corresponding integral equation is

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%

Example 5. Shew that, if

K x, y) =/i x) gi y) +f x) 2 y), then D (X) and D x, y; X) are
quadratic in X; and, more generally, if

II K x,y)= 2 f x)g,n ),

m = l

then i)(X) and D x, y, X) are polynomials of degree n in X.

\Subsection{11}{2}{2}{Volterra's reciprocal functions.}

Two functions K x, y), k x, y; X) are said to be reciprocal if they
are bounded in the ranges a- x, y - 1), any discontinuities they may
have are regularly distributed (\hardsectionref{11}{1}, footnote), and if

K x,y)+k x,y] ) = \ \ k x, \ \ ) K, y)d .

J a

We observe that, since the right-hand side is continuous*, the surn of
two reciprocal functions is continuous.

Also, a function K x, y) can only have one reciprocal if Z) (X) 4=;
for if there were two, their difference k x, y) would be a continuous
solution of the homogeneous equation

h x,y; X) = X f >(-,(.r, )K k,y)di,

(where x is to be regarded as a parameter), and by \hardsubsectionref{11}{2}{1} corollary,
the only continuous solution of this equation is zero.

By the use of reciprocal functions, Volterra has obtained an elegant
reciprocal relation between pairs of equations of Fredholm's type.

We first observe, from the relation

B x,y; ) = \ D X)K x,y) + \ \ D x, -X) K H y) d

proved in \hardsubsectionref{11}{2}{1}, that the value of A;(, y; X) is

-D x,y;X)l[\ D ) ], and from \hardsubsectionref{11}{2}{1} example 1, the equation

k x, y; ) + K x, y) = X f K x, ) k, y;X)d

J a M

is evidently true.

Then, if we take the integral equation

< > x)=f(x) + xl'K x, )<f>( )d,

J a

when a'> x h, we have, on multiplying the equation

J a * By example 10 at the end of Chapter iv.

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% 219
%

by k (x,; ) and integrating,

J a

= r k(x, : ) f( )d + xf' I' k(x, : ) K 1,)<f>(,)d,dl

J a J II J a

Reversing the order of integration* in the repeated integral and
making use of the relation defining reciprocal functions, we get

\ \ x, : ) 4>( )di

J a

= !'k(w, : ) f( )d +r K(x,,) + k w,,;X) < (|0 i,

J a J a

and so X f '(, X)/( )fZ = -X l K (x, ) <f> (,) d,

J a n

= -<P(x)+f x). Hence f(x) = ( ) + \ f ' (a;, f; ) /( ) d;

. a

similarly, from this equation we can derive the equation

4> x)=f(x) + \ f'K(x, )4>( )d,

J II

so that either of these equations with reciprocal nuclei may be
regarded as the solution of the other.

\Subsection{11}{2}{3}{Homogeneous integral equations.}

rb The equation < x) = X ) K x, ) </> ( ) d is called a homogeneous
integral

- n

equation. We have seen (\hardsubsectionref{11}{2}{1} corollary TODO) that the only continuous
solution of the homogeneous equation, when D (X) 4= 0, is ( x) = 0.

The roots of the equation D (X) = are therefore of considerable
importance in the theory of the integral equation. They are called the
characteristic numbers of the nucleus.

It will now be shewn that, when D (X) = 0, a solution which is not
identically zero can be obtained.

Let+ X = Xo be a root m times repeated of the equation D (X) = 0.

Since D (X) is an integral function, we may expand it into the
convergent series

D (X) = Cm (X - Xo)" + c,n, (X - Xor +1 + . . . (m > 0, c, + 0).

* The reader will have no diflSculty in extending the result of \hardsectionref{4}{3}
to the integral under consideration.

t French valeurs caracteristiques, German Eigenicerthe.

J It will be proved in \hardsubsectionref{11}{5}{1} that, if K (x, ij) = K y, x), the
equation D (X) = has at least one root.

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%

Similarly, since D x, y; X) is an integral function of \, there
exists a Taylor series of the form

by \hardsubsectionref{3}{3}{4} it is easily verified that the series defining g (cc, y), (n
= 1, 1 + 1, ...) converges absolutely and uniformly when a x b, a y -
b, and thence that the series for D (x, y; ) converges absolutely and
uniformly in the same domain of values of x and y.

But, by \hardsubsectionref{11}{2}{1} example 2,

L

i)(f.f;X)df=-X >

now the right-hand side has a zero of order m - 1 at A,o, while the
left-hand side has a zero of order at least I, and so we have m-\' I.

Substituting the series just given for D ( ) and D x,y\ X) in the
result of \hardsubsectionref{11}{2}{1} example 2, viz.

D x,y- ) = XD (X) K x, y) + \ \ ''K x, ) D I y; ) d

J a

dividing by (X, - XoY and making X -* X,o. we get

9i, y) = \ K x, I) gi (, y) d .

J a

Hence if y have any constant value, gi x, y) satisfies the homogeneous
integral equation, and any linear combination of such solutions,
obtained by giving y various values, is a solution.

Corollary. The equation

<l> G ) =/(. -) + Xo f ' iT (.r, ) < ( ) d

J a

has no solution or an infinite number. For, if ( x) is a solution, so
is x) + c gi (x, y),

y where Cy may be any function of y.

Example 1. Shew that solutions of

< x) = \ I cos'<(.r-|)( ( )c/

are </> (.r) = cos (?i - 2r) .r, and ( (.r) = sin (% - 2?-) .r; where
r assumes all positive integral values (zero included) not exceeding
hi.

Example 2. Shew that

  x) = \ y cos'' (.r-h I) ( ( ) o?|

has the same solutions as those given in example 1, and shew that the
corresponding values of X give all the roots of D (X) = 0.

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\Section{11}{3}{Integral equations of the first and second kinds. }
Fredholra's
equation is sometimes called an integral equation of the second kind;
while the equation

f x) = \ \ \ \ {x, ), )d

J a

is called the integral equation of the first kind.

In the case when K x, ) = Q ii > x, we may write the equations of the
first and second kinds in the respective forms

f x) = \ [ K x, ) )dl

J a

4> a )=f x) + xrK x, )cf>( )d .

J a

These are described as equations with variable upper limits.

\Subsection{11}{3}{1}{Volterra's equation.}

The equation of the first kind with variable upper limit is frequently
known as Volterra's equation. The problem of solving it has been
reduced by that writer to the solution of Fredholm's equation.

Assuming that K x, ) is a continuous function of both variables when
\$ X, we have

f x) = \ \ K x, )< )dl

J a

The right-hand side has a differential coefficient \hardsectionref{4}{2} example 1)
if -T- exists and is continuous, and so

ox

f (x) = \ K (x, x) <p X) + Xj -cf> ) dl

This is an equation of Fredholm's type. If we denote its solution by
x), we get on integrating ft-om a to x,

f x)-f a)=\ \ K x, )<l> )dl

J a

and so the solution of the Fredholm equation gives a solution of
Volterra's equation if /(a) = 0.

The solution of the equation of the first kind with constant upper
limit can frequently be obtained in the form of a series *.

\Section{11}{4}{The Liouville- Neumann method of successive substitutions "f TODO. }
A
method of solving the equation

cf> x)=f(x) + \ \ ' K(x, )cf ( )d,

J a

which is of historical importance, is due to Liouville.

* See example 7, p. 231; a solution valid under fewer restrictions is
given by Bocher. t Journal de Math. ii. (1837), iii. (1838). K.
Neumann's investigations were later (1870); see his Untersuchungen
ilber das logarithmische und Newton'sche Potential.

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%

It consists in continually substituting the value of 4) oo) given by
the right-hand side in the expression < ( ) which occurs on the
right-hand side.

This procedure gives the series S x)=f x) + \ \ ' K x, aA?) + S X- f K
x, 0 [* K I,)

J a m = 2 J a J a

J a

Since | K x, y) \ and \ f(x) \ are bounded, let their upper bounds be
M, M'. Then the modulus of the general term of the series does not
exceed

|\ [ ' l/' ir(6-a)'". The series for S x) therefore converges
uniformly when

\ \ \ < M-' h-a)-; and, by actual substitution, it satisfies the
integral equation.

If Kix, y) = when y>x\ we find by induction that the modukis of the
general term in the series for S x) does not exceed

X I * Jf'" M' x - a)>"l m l) \ X\'>' i/' M' (6 - ay>'jm !,

and so the series converges uniformly for all values of X; and we
infer that in this case Fredholm's solution is an integral function of
X.

It is obvious from the form of the solution that when | X. | < M~ (b -
a)~\ the reciprocal function k (x,; X) may be written in the form

k x, X)=.-K x, )- t X-- f' K(x, 0 f /l (I,, eO

m=2 J a J a

for with this definitipn of k x,; A,), we see that

S x)=f x)-\ \ ' k(x,; ) f( )d,

J a

so that k x, |;X) is a reciprocal function, and by \hardsubsectionref{11}{2}{2} there is
only one reciprocal function.

Write

K (X, I) = K, X, ), f K X, r) Kn (r, B r - J n+r (, ),

and then we have

-k x,;X)= S V K, +,(x, ),

m =

while f K (x, r ) Kn (r, e r = A .+ (*'. )>

J a

as may be seen at once on writing each side as an (m + n - l)-tuple
integral. The functions K, (x, f) are called iterated functions.

J a

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\Section{11}{5}{Symmetric nuclei.}

Let Ki x, y) = K (y, x); then the nucleus K x, y) is said to be
symmetric. The iterated functions of such a nucleus are also
symmetric, i.e. Kn, y) = Kn (y, x) for all values of n; for, if Kn
x, y) is symmetric, then

Kn+r (, y) = f K,, ) Kn I y) d = j K, (|, o:) K, y, ) d

J a 'a

= Kn(y, )K,i,x)d = Kn, y,x),

J a

and the required result follows by induction.

Also, none of the iterated functions are identically zero; for, if
possible, let Kp (x, y) = 0; let n be chosen so that 2"" <]) - 2",
and, since Kp (x, y) = 0, it follows that K n x, y) = 0. from the
recurrence formula.

But then = K n x, x) = I, \ i x, ) K -i (, ) d

J a

= f [Kn-.i . rr-d,

J a

and so K n-i i, ) - j continuing this argument, we find ultimately
that Ki (x, y) = 0, and the integral equation is trivial.

\Subsection{11}{5}{1}{Schmidt's* theorem that, if the nucleus is symmetric, the equation J) ( ) = has at least one root.}

To prove this theorem, let

Un = Kn (X, X) dx,

J a

SO that, when \ \ \ < M~ (b - a)~\ we have, by \hardsubsectionref{11}{2}{1} example 2 and
\hardsectionref{11}{4},

\ 1 dDiX)

D ) dx ri "

Now since I I fjLKn+i (x, |) +,i\ i (x, )]- d dx

J a J a

for all real values of fi, we have

tl'U . + 2/JLU.>n + U,n-, > 0, and so U n+i U.n-- U n-, Uon-1 > 0.

Therefore U.,, Ui, ... are all positive, and if- JZ /C, = i, it
follows, by in- duction from the inequality ?7o +2 C/2,1-2 C 2 i'.
that Unn+n/U.n > v -

X

Therefore when j X- v~\ the terms of S Un'. ' ~ do not tend to zero;

and so, by \hardsectionref{5}{4}, the function, - -7- - has a singularity inside or
on the

* The proof given is due to Kneser, Palermo Bendiconti, xxii. (1906),
p. 236.

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circle \ \ = v~ \ but since D( ) is an integral function, the only
possible

sinerularities of -r., . - j- are at zeros of D(X); therefore D (X)
has a zero ° 1) (X,) d\

inside or on the circle \ \ \ = v~ .

[Note. By \hardsubsectionref{11}{2}{1}, Z> (X) is either an integral function or else a
mere polynomial; in the latter case, it has a zero by \hardsubsectionref{6}{3}{1} example
1; the point of the theorem is that in the former case D (X) cannot
be such a function as e ', which has no zeros.]

\Section{11}{6}{Orthogonal functions.}

The real continuous functions (jj (x), (f)o (x), . . . are said to be
orthogonal and normal* for the range a, b) if

If Ave are given n real continuous linearly independent functions u-
x), Uo x), ...Un(x), we can form n linear combinations of them which
are orthogonal.

For suppose we can construct m - 1 orthogonal functions </>!, ...
(f>m-i such that (f>p is a linear combination of u-, ii, ... Up
(where p = 1, 2, ... ni - I); we shall now shew how to construct the
function, such that c i, <p.2,  (f>m are all normal and
orthogonal.

Let (f>m (x) = Ci, m (f>i ( ) + C2, m < 2 ( ) +  + Cm-i <pm-i (i )
+ Urn ),

so that i( is a function of Mj, lu, ... w, . Then, multiplying by and
integrating,

i< ni x) (t>p oc) dx = Cp,, + I Um ) <j>p (x) dx p < m). Hence I i4>m
i ) 4>p oc) dx =

J a

if Cp m == - Uin (jc) <pp ( ) dx;

J a

a function i, (x), orthogonal to (f) (x), (f>. (x), . . . <pm-i (x),
is therefore con- structed.

rb

Now choose a so that a- I [i(f>r,i (x)]- dx=l;

J a

and take <p,n (x) = a. i<, . (x).

Then j </) (x) 4>p (x) dx |~ '

We can thus obtain the functions (/)i, c/).., ... in order.

* They are said to be orthogonal if the first equation only is
satisfied; the systematic study of such functions is due to Murphy,
Camb. Phil. Trans, iv. (1833), pp. 353-408, and v. (1835), pp.
113-148, 31.5-394.

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The members of a finite set of orthogonal functions are linear! '
inde- pendent. For, if

a,(j), (x) + a.2(f>. (x)+ ... + an(f>n ( v) = 0,

we should get, on multiplying by (f>p (a;) and integrating, a =;
therefore all the coefficients a vanish and the relation is nugatory.

It is obvious that tt ~ cost/u;, tt ~ - sin mx form a set of normal
orthogonal functions foifthe range ( - tt, tt).

Example 1. From the functions 1, .r, x-, ... construct the following
set of functions which are orthogonal (but not normal) for the range (
- 1, 1) :

1, X, - i, x -'jx, x* - x- + - g, ....

Example 2. From the functions 1, x, x, ... construct a set of
functions

which are orthogonal (but not normal) for the range (a, h); where

[A similar investigation is given in >5 15-14.]

11 "61. The connexion of orthogonal functions with homogeneous
integral equations.

Consider the homogeneous equation

<l> x) = \,\% )K x, )dl

  a

where Xq is a real* characteristic number for K(x, ); we have already
seen how to construct solutions of the equation; let n linearly
independent solutions be taken and construct from them n orthogonal
functions (pi, < .\,, ... (/> .

Then, since the functions, are orthogonal, rr i <l>Uy)f K x, )cf>, (
)di\'dy = i f\ < p.n(y)fK(x, )<f>, )Ydi

J a |\ t = 1 J a J m = lJ a [\ J a J

and it is easily seen that the expression on the right may be written
in the form

rb

i If K(x. )4>, ( )d

H = 1 (. J a

>H = 1

on performing the integration with regard to y; and this is the same
as i r K x, y) cf>,n (y) dyi' K x, |),, ( ) rff

m. = \ J a J a

Therefore, if we write K for K x, y) and A for

i 4>Ay)\' K, )<l>.nX )dl

m=\ J a

* It will be seen immediately that the characteristic numbers of a
symmetric nucleus are all real.

W. M. A. 15

Therefore

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rb rb

we have A?dy = \ KAdy,

J a J a

rb rb rb

and so 1 A-dij = K'dy - (K - Kfdy.

la J a J a

a \ m = \ f o ) J a

and SO Xo~' <, ( ) [ [K x,y)Ydy.

w = 1 J a

Integrating, we get

n <: Xo'- I / [K x, y)Y-dydx.

J a J a

This formula gives an upper limit to the number, 7i, of orthogonal
functions corresponding to any characteristic number Xo-

These n orthogonal functions are called characteristic functions (or
auto- functions) corresponding to Xq.

Now let '"* (x), < <i' (x) be characteristic functions corresponding
to different characteristic numbers Xq, Xj.

Then </>' > (x) </><!' (x) = xJ K x, ) ( c) x) </)' ' ) d,

J a

and so

[ (f> ' >(x)(f) ' (x)dx = \,\ i K(x, )4> '> (x)cf)'' )d dx ...(1),

and similarly

[ (/)' ' (x) </) w (x) dx = \ of I K x, ) 0 " ( ) </)(!> (x) d dx

la J a J a

= X [ [ K(lx)cf> '' ix)(f> ' ( )dxd ...(2),

J a ' a

on interchanging x and .

We infer from (1) and (2) that if \ \ and if K x, ) = K, x),

I 4> ' (x)(f> ' (x)dx=0,

J a

and so the functions < "'' (x), '" (x) are mutually orthogonal.

If therefore the nucleus be symmetric and if, corresponding to each
characteristic number, we construct the complete system of orthogonal
functions, all the functions so obtained will be orthogonal.

Further, if the nucleus be symmetric all the characteristic numbers
are real; for if Xq, Xj be conjugate complex roots and if* Uq x) = v
(x) + iw (x) be

* V (x) and w (x) being real.

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a solution for the characteristic number X, then iii (x) = v (x) -
iiu (x) is a solution for the characteristic number \; replacing 0*''
(x), 'i* (x) in the equation

I ( 'o' (x) </>< ' (x) dx =

J a

by V (x) + iw (x), v x) - iw (x), (which is obviously permissible), we
get

f [\ v(x)Y+ \ w x)Y]dx = 0,

J a

which implies v (x) = w (x) = 0, so that the integral equation has no
solution except zero corresponding to the characteristic numbers Xq.
il this is contrary to \hardsubsectionref{11}{2}{3};
hence, if the nucleus be symmetric, the characteristic
numbers are real.

11 7. The development* of a symmetric nucleus.

Let < j x), (f)., (x), 3 (x), ... be a complete set of orthogonal
functions satisfying the homogeneous integral equation with symmetric
nucleus

cf>(x) = xfK(x, )<f>( )dl

J a

the corresponding characteristic numbers beingf Xj, X, X, -

XT J. 1, f' < n (* ) < i(v)  -r 1

Now supposel that the series - - - - is umiormly convergent

when a x b, a y %b. Then it luill he shewn that

K(x.y)=i M My). =i

For consider the symmetric nucleus

71 = 1 / n

If this nucleus is not identically zero, it will possess (\hardsubsectionref{11}{5}{1}) at
least one characteristic number /j,.

Let ylr(x) be any solution of the equation

 jr x)=fJ,f H(x.. )ylri )dl

' a

which does not vanish identically.

Multiply by cf)n (x) and integrate and we get

ry!r(x)cf>,, x)dx = f,f Hxix, )- I "t if l l ( ) cj U ) dxd;

J a a ' a [ n = l 7i )

* This investigation is due to Schmidt, the result to Hilbert.

t These numbers are not all different if there is more than one
orthogonal function to each characteristic number.

X The supposition is, of course, a matter for verification with any
particular equation.

15-2

>t\

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since the series converges uniformly, we may integrate term by term
and get

f ir (.v) 4> x) dx = \ \ ( ) (/> i )d - r < (I) f ( ) d

= 0.

Therefore t/ (x) is orthogonal to < i (x), < 2 ( ),  ', and so
taking the equation

J a [ n = l n )

rb

we have yjr (x) = y"- / K (x, )' ) d .

J a

Therefore /i, is a characteristic number of K (x, y), and so -v/r x)
must be a linear combination of the (finite number of) functions < n
(*') corresponding to this number; let

m

Multiply by (j>m (x) and integrate; then since -yfr (x) is orthogonal
to all the functions (j)n (x), we see that a, = 0, so, contrary to
hypothesis, yjr (x) = 0.

The contradiction implies that the nucleus H (x, y) must be
identically zero; that is to say, K (x, y) can be expanded in the
given series, if it is uniformly convergent.

Example. Shew that, if Xq be a characteristic number, the equation

4> x)=f .v) + \ J'' E x, ) )d

J a

certainly has no sokition . unless f(x) is orthogonal to all the
characteristic functions corresponding to .

11 "71. The solution of Fredhohns equation by a series. Retaining the
notation of \hardsectionref{11}{7}, consider the integral equation

   x) =f x) +\ f K (x, )< ) dl

-' a

where K x, ) is symmetric.

If we assume that ( ) can be expanded into a uniformly convergent

00

series 2 an4>n (1), we have

00 00 A

2 an(f)n x!)=f(x)+ 1 -an(f>n( ), ' w = l M=l n

SO that f(x) can be expanded in the series

00 A -A

S an -\ (f)n x).

Hence if the function f(x) can be expanded into the convergent series

< . " b \

2 bn< >n(ix), then the series 2 . " \ <f>n ), if it converges
uniformly in

n = l n-l n

the range a, b), is the solution of Fredhohns equation.

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X

To determine the coefficients 6, we observe that S bn (f)n ( )
converges uni-

n=l

formly by \hardsubsectionref{3}{3}{5}*; then, multiplying by (f)n(x) and integrating, we
get

J a

11 "8. Solution of AbeVs integral equation. This equation is of the
form

  -llw

d\$ 0<fi<l, a x b),

where/' (x) is continuous aud/(a)=0; we proceed to find a continuous
solution u (x)

Let (f)(x)= I u (I) o?|, and take the formula t

J a

7T n dx

sin/iTT ~ J z-xY~> x - y

multiply by u ( ) and integrate, and we get, on using Dirichlet's
formula \hardsubsectionref{4}{5}{1} corollary\

  f' f x)dx

Since the original expression has a continuous derivate, so has the
final one; therefore the continuous solution, if it exist, can be
none other than

"" ''V~dzj z-xy-> '

and it can be verified l)y substitution J that this function actually
is a solution.

11 "SI. Schlomilch's integral equation.

Let $f(x)$ have a continuous differential coefficient when - tt .r tt.
Then the equation

2 /"i" f x)=- (f) xsmd)de

  J n

has one solution with a continuoiis differential coefficient when - it
x it, namely

/ in < x)=f 0) + x ' f' xsm6)d6. Jo From \hardsectionref{4}{2} it follows that

f (.v) = - sin Oct)' x sin (9) dd j

(so that we have (0)=/(0), ' (0) = |7r/' (0)).

* Since the numbers X, are all real we may arrange them in two sets,
one negative the other positive, the members in each set being in
order of magnitude; then, when ' X ! > X, it is evident that X /(X -
X) is a monotonic sequence in the case of either set.

t This follows from \hardsubsectionref{6}{2}{4} example 1, by writing (z - x)l(x - f ) in
place of x.

X For the details we refer to Bocher's tract.

§ Zeitschrift filr Math, und Phys. ii. (1857). The reader will easily
see that this is reducible to a case of Volterra's equation with a
discontinuous nucleus.

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Write .r sin // for x\ and we have on multiplying by x and integrating

X j /' x sin \//-) d\ lr - - I < I sin ( ' (.i- sin 6 sin \//-) c? \
dyjr.

Change the order of integration in the repeated integral \hardsectionref{4}{3}) and
take a new variable x in iDlace of \ f defined by the equation sin; =
sin 6 sin yj/.

rr,. fi" .,, .s,, 2.V fi ( f (b' (x sin v) cos Y dv],

Then xj f'ixsn.i.)d = j \ j -~J -~ - \ de.

Changing the order of integration again \hardsubsectionref{4}{5}{1}),
$$
TODO
$$
and so x \ f x sin y\ r) d = x \ </>' (.r sin x) cos;>( o?;

Jo J

= 0Gt-)-( (O).

Since (f) (0) =/ (0), we must have

(j) (.r) =/ (0)+x ' f (x sin;/.) c a/;

y

and it can be verified by substitution that this function actually is
a solution.

REFERENCES.

H. Bateman, Report to the British Association*, 1910.

M. BocHER, Introditction to Integral Equations (Cambridge Math.
Tracts, No. 10, 1909).

H. B. Heywood et M. Fr chet, Liquation de Fredkolm. (Paris, 1912).

V. Volte ra, Lego7is sur les equations integrales et les equations
integro-differentielles (Paris, 1913).

T. Lalesco, Introduction a la the'orie des equations integrales
(Paris, 1912).

I. Fredholm, Acta Mathematica, xxvii. (1903), pp. 365-390.

D. HiLBERT, Orundzilge einer allgemeinen Theorie der linearen
Integralgleichungen (Leipzig, 1912).

E. Schmidt, Math. Ann. lxiii. (1907), pp. 433-476.

E. Goursat, Cours d' Analyse, iii. (Paris, 1915), Chs. xxx-xxxill.

MiSCELT.ANEOUS EXAMPLES.

1. Shew that if the time of descent of a particle down a smooth curve
to its lowest point is independent of the starting point (the particle
starting from rest) the curve is a cycloid. \addexamplecitation{Abel.}

* The reader will find a more complete bibliugrapby in this Report
than it is possible to give here.

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2. Shew that, if/( ) is continuous, the sohition of

(f) (x) =f x) + \ j COS (2.rs) <f) (s) ds

/(.r) + X I f s)cos 2jcs)ds

'" *( >= Hrjxv .

assuming the legitimacy of a certain change of order of integration.

3. Shew that the Weber-Hermite functions

satisfy 4> x) = \ I e '*- (s) ds

for the characteristic values of X. (A. Milne.)

4. Shew that even periodic solutions (with period Stt) of the
diflferential equation

- - + ( 2 + >l-2 cos2 x) cf) (.r) = satisfy the integral equation

<f> x) = \ j e < oo'=< '(f, s)ds. (Whittaker; see \hardsubsectionref{19}{2}{1}.)

5. Shew that the characteristic functions of the equation

are <f (x) = cos m.v, sin inx,

where \=m' and m is any integer.

6. Shew that <p (x) = / " - -f (|) d\$

has the discontinuous solution <f) (x) = Lr ~ . \addexamplecitation{Bocher.}

7. Shew that a solution of the integral equation with a symmetric
nucleus

f :v)=rK x, )<t>i\$)di

J a

is 0( )= 2 a \ \ (f) (x),

n=l

provided that this series converges uniformly, where X, 0 (x) are the
characteristic numbers and functions of K x, |) and 2 a 0 (x) is the
expansion off(x).

n=l

8. Shew that, if | A ] < 1, the characteristic functions of the
equation

'f' '' =LI[ i-2hcL i- Hh'' "

are 1, cos?n.r, sin mx, the corresponding characteristic numbers being
1, l/A'", Ijh'", where m takes all positive integral values.