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\begin{document}

\noindent
\centerline{{\bf Elliptic Curves MT25: Solutions to Sheet 3}}
\bigskip

\begin{enumerate}[{\bf (1)}]


\item
Following the hint, there exist polynomials $p(x),q(x)\in R[x]$
such that $p(x) f(x) + q(x) f'(x) = D$,
and so: $p(a_0) f(a_0) + q(a_0) f'(a_0) = D$.
Since $|a_0|\leqslant 1$, and since $p(x),q(x),f(x),f'(x)\in R[x]$,
we must have \[|p(a_0)|,|f(a_0)|,|q(a_0)|,|f'(a_0)| \leqslant 1\]
and so $|D| \leqslant \hbox{max}( |p(a_0) f(a_0)|,|q(a_0) f'(a_0)| )
\leqslant 1$, which also implies $|D^2| = |D|^2 \leqslant |D|$.
Now, we are given that $|f(a_0)| < |D|^2$
and so $| p(a_0) f(a_0) | \leqslant | f(a_0) | < |D^2| \leqslant |D|$.
Hence $|D| \not= |p(a_0) f(a_0) |$, and
so $|q(a_0) f'(a_0)| = |D - p(a_0) f(a_0)| = \hbox{max}( |D|, |p(a_0) f(a_0)|)
= |D|$ [using the fact that, if $|u| \not= |v|$
then $|u\pm v| = \hbox{max}(|u|,|v|)$].
Since also $|q(a_0)| \leqslant 1$, this gives
that $|f'(a_0)| \geqslant |D|$. Finally,
$|f(a_0)| < |D|^2 \leqslant |f'(a_0)|^2$, and so $f(X)$
has a root $a\in R$ by Hensel's Lemma.
\medskip


\item
In projective form, the point is:
$(-64/25,59/125,1)$. The coordinate with largest $5$-adic value
is $59/125$, and on dividing all coordinates
through by this, we can also represent
the point as: $(-320/59,1,125/59)$, which is in $5$-adic standard
form (that is, $1$ is the maximum of the $5$-adic valuations
of the coordinates). Reducing mod~$5$ gives $(0,1,0)$ on $\widetilde \e$,
which is the point at infinity.
\medskip

%\item
%Let $\c$ be the curve $2 Y^2 = X^4 - 17$.
%Over~$\R$, the curve has the point
%$(4,\sqrt{239/2})$. 
%In $\Q_2$, let $x=11$. Then
%$11^4 - 17 = 2^5 \cdot 457$; but 457 is an integer congruent to~1 mod~8,
%and so, from lectures, there exists $\gamma \in \Z_2 \subset \Q_2$ such that
%$457 = \gamma^2$; then $(11, 4\gamma)$ is a point on the curve
%over~$\Q_2$. For the next few primes,
%we shall repeatedly use the result from lectures that, if $p\not= 2$
%and $|a|_p = 1$, then $a$ is a square in $\Q_p$ iff it
%is a square mod~$p$. Now, note that $-8,-34,-34,-8,10,-8$ are
%quadratic residues modulo~$p=3,5,7,11,13,17$, respectively,
%and so there exists $\gamma \in \Q_p$
%such that $\gamma^2 = -8,-34,-34,-8,10,-8$, respectively;
%this gives the $\Q_p$-rational points on the
%curve: $(1,\gamma), (0,\gamma/2), (0,\gamma/2),
%(1,\gamma), (4,\gamma/2), (1,\gamma)$, respectively.
%Hence $\c$ has $\Q_p$-rational points for all primes
%up to and including $p=17$ [in fact, it was only necessary to
%do here $p=2,3,5,7,11,17$ here].
%\par Let $p$ be any prime $p\geqslant 13$ and $p\not= 17$.
%Our curve~$\c$ (after multiplying both sides by~$2$) is
%birationally equivalent to $V^2 = 2 X^4 - 34$, where $V = 2Y$;
%note that $2 X^4 - 34$ has discriminant $-2^{11} 17^3$.
%Recall Theorem~1.15 from lectures, 
%that any curve $y^2 = Q(x)$, where $Q(x) = f_4 x^4 + \ldots + f_0$
%has nonzero discriminant over~$\F_p$, 
%has at least $p - 1 - 2\sqrt{p} > 4$
%affine points over~$\F_p$, and so at least~$5$ affine points.
%At most~$4$ of these can have $2 x^4 - 34 \equiv 0$~(mod~$p$),
%and so there exists
%$x_0 \in \Z$ 
%such that $2(x_0^4 - 17)$ is a nonzero quadratic residue mod~$p$.
%As above, there exists $\gamma \in \Q_p$ such that
%$\gamma^2 = 2(x_0^4 - 17)$, and so $(x_0,\gamma/2)$ is a
%$\Q_p$-rational point on our original curve~$\c$.
%Hence $\c$ has points in~$\R$ and every~$\Q_p$.
%\newpage
%
%\par {\it Alternative method for the above parts:}
%Note that for $p \notdiv 2\Delta$
%(that is: $p \not= 2,17$)
%$p \geqslant 7$, we have (from a theorem from lectures)
%the number of affine points in 
%${\widetilde \c} (\F_p) \geqslant p-1 - 2\sqrt{p} > 0$, and so the
%number of affine points is $ > 0$; these are non-singular, since
%$2\Delta \not= 0$ in~$\F_p$. Also, 
%by a theorem from lectures, any non-singular point on 
%${\widetilde \c} (\F_p)$ lifts to a point on $\c (\Q_p)$; 
%also, any affine non-singular point on ${\widetilde \c} (\F_p)$ 
%lifts to an affine point on $\c (\Q_p)$
%(since the point at infinity on $\c (\Q_p)$ maps to the point at 
%infinity on ${\widetilde \c} (\F_p)$ under the reduction map mod~$p$).
%This shows the existence of such $x,y \in \Q_p$ for
%all~$p$ except $p = 2,3,5$ and bad primes, so only still need
%to consider $p = 2,3,5,17$. Also, note that for $p = 3$ there
%is the non-singular affine point $(1,1)$ on ${\widetilde \c} (\F_3)$, 
%for $p = 5$ there
%is the non-singular affine point $(0,2)$ on ${\widetilde \c} (\F_5)$, 
%and for $p = 17$ there is the non-singular affine point 
%$(1,3)$ on ${\widetilde \c} (\F_{17})$, and as before these must 
%lift to affine points $(x,y) \in \c(\Q_p)$.
%Over~$\R$, the curve has the point
%$(4,\sqrt{239/2})$.
%In $\Q_2$, let $x=11$. Then
%$11^4 - 17 = 2^5 \cdot 457$; but 457 is an integer congruent to~1 mod~8,
%and so, from lectures, there exists $\gamma \in \Z_2 \subset \Q_2$ such that
%$457 = \gamma^2$; then $(11, 4\gamma)$ is a point on the curve
%over~$\Q_2$.
%This completes the alternative method of showing
%that $\c$ has points in~$\R$ and every~$\Q_p$.
%
%\par Now, imagine that $\c$ had a $\Q$-rational point; that is,
%imagine that there exist $X,Y\in \Q$ such that $2 Y^2 = X^4 - 17$.
%Let $X = t/r$, where $r,t\in \Z$ and $\hbox{gcd}(r,t) = 1$.
%Then $2(r^2 Y)^2 = t^4 - 17 r^4 \in \Z$. For any prime~$p$,
%this gives that $|2|_p |r^2 Y|_p^2 \leqslant 1$. When $p\not= 2$,
%we have $|2|_p = 1$ and so $|r^2 Y|_p^2 \leqslant 1$
%and so $|r^2 Y|_p \leqslant 1$. When $p=2$,
%we have $2^{-1} |r^2 Y|_2^2 \leqslant 1$; but this still
%gives $|r^2 Y|_2 \leqslant 1$ [since $|r^2 Y|_2 > 1$
%would mean $|r^2 Y|_2 \geqslant 2^1$ and so $2^{-1} |r^2 Y|_2^2 
%\geqslant 2^{-1} 2^2 > 1$]. Overall, we have shown that
%$|r^2 Y|_p \leqslant 1$ for all~$p$; since also $r^2 Y \in \Q$,
%this gives that $r^2 Y \in \Z$; let us say: $s = r^2 Y \in \Z$.
%Therefore $r,s,t\in\Z$ satisfy $2 s^2 = t^4 - 17 r^4$
%and $\hbox{gcd}(r,t) = 1$.
%%% \newpage 
%
%\par Let $q$ be any prime such that $q | s$. Note that
%$q\not= 17$ [since if $17 | s$ then our equation
%$2 s^2 = t^4 - 17 r^4$ would force $17 | t$, and so
%$17^2 | 2 s^2$ and $17^2 | t^4$, which would give $17^2 | 17 r^4$,
%forcing $17 | r$, which
%would contradict $\hbox{gcd}(r,t) = 1$]. 
%Note also that
%we must not then have $q|r$ [since if $q|s$ and $q|r$, then
%our equation $2 s^2 = t^4 - 17 r^4$ would force $q | t$, which
%would contradict $\hbox{gcd}(r,t) = 1$]. 
%Reducing our equation modulo~$q$ gives that $0 \equiv t^4 - 17 r^4$
%and so $(t^2/r^2)^2 \equiv 17$ mod~$q$ [note that division
%by $r^2$ is allowable mod~$q$ since $r$ is not divisible by~$q$].
%Hence $17$ is a nonzero quadratic residue mod~$q$.
%For $q\not= 2$, quadratic reciprocity then allows us to
%deduce that~$q$ is a quadratic residue mod~$17$
%[since $17 \equiv 1$ (mod~$4$)]. Furthermore, one can check
%directly that $2$ is a quadratic residue mod~$17$, since
%$2 \equiv 6^2$ mod~$17$. Overall, we have shown that
%every prime~$q$ dividing~$s$ must be a quadratic residue
%mod~$17$, 
%and we can take $s>0$ (if necessary, replacing~$s$ by~$-s$),
%so that~$s$ is the product of its prime factors.
%It follows that $s$ itself must be a quadratic residue mod~$17$,
%since it is a product of quadratic residues mod~$17$.
%Hence $s^2$ is a nonzero fourth power [also known as
%a {\it quartic residue}] mod~$17$.
%Note also that reducing our equation mod~$17$ gives
%$2 s^2 \equiv t^4$ (mod~$17$), so that $2 s^2$ is also
%a nonzero fourth power mod~$17$. Since $s, 2s^2$ are
%both nonzero fourth powers mod~$17$, it follows that
%$2 = (2s^2)/s^2$ is also a fourth power mod~$17$.
%On the other hand, one can compute directly that
%$0^4,1^4,2^4,3^4,4^4,5^4,6^4,7^4,8^4,9^4,10^4,11^4,12^4,13^4,14^4,
%15^4,16^4$ are congruent mod~$17$ to: 
%$0,1,16,13,1,13,4,4,16,16,4,4,13,1,13,16,1$, respectively,
%so that in fact $2$ is not a fourth power mod~$17$
%(alternatively: if $2 \equiv w^4$ (mod~17) then $-1 \equiv 2^4
%\equiv w^{16} \equiv 1$ (mod~17) by FLT, a contradiction,
%avoiding the need for an enumeration).
%This contradication show that our original curve $\c$
%has no $\Q$-rational points.
%
%\medskip
%
%
%
%\item
%\par \noindent {\bf (a).} Let $\e : Y^2 = X^3 + p^2$. Then
%the point $(0,p)$ on $\e$ reduces modulo~$p$ to the
%cusp~$(0,0)$ on $\widetilde \e : Y^2 = X^3$, which is another
%way of saying that $(0,0)$ on $\widetilde \e$ lifts to the
%point $(0,p)$ in~$\e (\Q_p)$.  
%\par \noindent {\bf (b).} Question~3 is an example of this.
%\par \noindent {\bf (c).} Let $\e : Y^2 = X^3 + X^2 + p^2$. Then 
%the point $(0,p)$ on $\e$ reduces modulo~$p$ to the 
%node~$(0,0)$ on $\widetilde \e : Y^2 = X^3 + X^2$, which is another
%way of saying that $(0,0)$ on $\widetilde \e$ lifts to the  
%point $(0,p)$ in~$\e (\Q_p)$.
%\par \noindent {\bf (d).} Let $\e : Y^2 = X^3 + X^2 + p$ and
%$\widetilde \e : Y^2 = X^3 + X^2$. Then the node~$(0,0)$ on
%$\widetilde \e$ does not lift to any point in~$\e(\Q_p)$ by the
%same argument as in Question~2.
%\medskip
%
%\item
%By definition of a homomorphism, we need to show
%\[ [m] F(X,Y) \stackrel{?}{=} F([m](X),[m]Y).\]
%This is true when $m = 0$, since $[0](T) := 0$.
%Now for $m \geq 1$ we can use the recursive definition to rewrite this as
%\[ F([m-1]F(X,Y),F(X,Y)) \stackrel{?}{=} F(F([m-1](X),X), F([m-1]Y,Y)).\]
%By induction we can assume
%$[m-1] F(X,Y) =  F([m-1](X),[m-1]Y)$.
%Starting from the LHS we compute
%\[ F([m-1]F(X,Y),F(X,Y))  = F(F([m-1](X),[m-1](Y)), F(X,Y))\] \[ = 
%F([m-1](X),F([m-1](Y),F(X,Y))) = F([m-1](X),F(F(X,Y),[m-1](Y))) \] \[ = F([m-1](X),F(X,F(Y,[m-1](Y))))
%= F(F([m-1](X),X),F(Y,[m-1](Y)))\]
%which by commutativity equals the RHS.
%Here we have used, in order, induction, associativity, commutativiity, associativity, associativity.
%
%\item {\bf (a)} We can regard $F$ as a formal group law over $K$, and then $\log_F$ is a homomorphism from $F$ to $\widehat{\mathbb{G}}_a$. In particular, we have an identity of formal power series (with coefficients in $K$):
%$\log_F(X)+\log_F(Y) = \log_F(F(X,Y))$. When $x,y \in \M$, both sides converge, since the $n$th term in $\log_F$ contributes something of size $\le \frac{|\varpi|^n}{|n|}$ and this goes to $0$ as $n \to \infty$. This also shows that we have $\log_F(x)+\log_F(y) = \log_F(F(x,y))$: the identity of formal power series gives us an identity between terms of degree $\le n-1$ for each $n$, and the terms of degree $\ge n$ contribute something of size $\le \frac{|\varpi|^n}{|n|}$. Note that you do have to be careful when composing power series --- see Example 8.14 in \url{https://kconrad.math.uconn.edu/blurbs/gradnumthy/infseriespadic.pdf}.
%
%\item
%In all of the following, we use
%the result from lectures that (when the coefficients
%of $\cal E$ are in $\Bbb Z$) $\etq$ is isomorphic to a subgroup
%of $\widetilde {\cal E}$ mod~$p$, where $p\not=2$ is a prime
%not dividing the discriminant. Note that this typically
%gives a much faster way of computing $\etq$ than the
%Nagell-Lutz result. N.B. (a),(b),(c) are about the level that
%could be asked as part of a 3-hour exam.
%\par\noindent {\bf (a).} There are the obvious points
%$P = (0,1)$ of order~$3$ and $Q = (-1,0)$ of order~$2$ in $\etq$,
%which generate a subgroup of $\etq$ of size~$6$ (namely,
%the~$6$ points $mP+nQ$ for $0\leqslant m \leqslant 2$ and
%$0\leqslant n \leqslant 1$). Further, $\Delta = 4A^3 + 27B^2 = 27$,
%so we can reduce modulo any prime except~$2$ (which must always
%be avoided) and~$3$. Over ${\Bbb F}_5$, there are only
%six points: ${\bf o}, (0,\pm 1), (2,\pm 3), (4,0)$. So, we conclude
%that $\etq$ has size at most~$6$, which means that it consists
%of precisely the $C_6 = C_2\times C_3$ group
%of points we have found already.
%\par Note that, if $P=(0,1)$ and $Q=(-1,0)$, then the complete
%list of torsion points is: ${\bf o}, P = (0,1), 2P = (0,-1),
%Q = (-1,0), P+Q = (2,-3), 2P+Q = (2,3)$.
%\par\noindent {\bf (b).} Here, the (birational over~$\Bbb Q$)
%transformation $(X,Y) \mapsto (X-1,Y)$ takes the given
%curve to: $Y^2 = X(X+1)(X-1) = X^3 - X$, which has discriminant $-4$, 
%and so we can reduce modulo the prime~$3$
%Over~${\Bbb F}_3$
%there are the points: ${\bf o}, (0,0), (1,0), (2,0)$ giving
%that $\etq$ has size at most~$4$. But, in fact, $\etq$
%contains ${\bf o}, (0,0), (-1,0), (1,0)$, which means
%the this $C_2\times C_2$ group gives all of $\etq$.
%[{\it Alternatively, even without using a birational transformation
%to the form $Y^2=X^3 + AX + B$, we could just work entirely
%with the given equation of the curve, noting that
%the original cubic $X(X-1)(X-2)$ has no repeated
%roots~mod~$3$, so that $Y^2 = X(X-1)(X-2)$ is an elliptic curve
%mod~$3$, and then noting that the only points are:
%${\bf o}, (0,0), (1,0), (2,0)$.}]
%\par\noindent {\bf (c).} The (birational over~$\Bbb Q$)
%transformation $(X,Y) \mapsto (3^2X, 3^3Y)$ takes the
%given curve to: $Y^2 = X^3 + 1$ which we have already seen
%in part~(a) to have a $C_2\times C_3$ group as its
%torsion group.
%\par Note that, if $P = (0,1/27)$ and $Q = (-1/9,0)$ then the
%complete list of torsion points is given by: ${\bf o}, P=(0,1/27),
%2P=(0,-1/27), Q = (-1/9,0), P+Q = (2/9,-1/9), 2P+Q = (2/9,1/9)$.

\item[{\bf (8)}]
Take $F(X,Y) = X + Y + t X Y^p$,
clearly non-commutative. One only has to check associativity.
$$ F(F(X,Y),Z) = F( X + Y + t X Y^p, Z )
= X + Y + t X Y^p + Z + t(X + Y + t X Y^p) Z^p
$$
$$
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
= X + Y + t X Y^p + Z + t X Z^p + t Y Z^p,$$
since the remaining term $t^2 X Y^p Z^p \in I = t^2 \F_p[t]$
and so $t^2 X Y^p Z^p = 0$ in $R = \F_p[t]/I$.
Also:
$$ F(X, F(Y,Z))
= X + F(Y,Z) + t X F(Y,Z)^p
= X + Y + Z + t Y Z^p + t X (Y + Z + t Y Z^p)^p
$$
$$
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
= X + Y + Z + t Y Z^p + t X (Y^p + Z^p + (t Y Z^p)^p),
$$
since all other terms in the binomial expansion
of $(Y + Z + t Y Z^p)^p$ are divisible by~$p$ and
so are equal to~$0$ in $R$, which has characteristic~$p$.
But $(t Y Z^p)^p = t^2 (t^{p-2} Y^p Z^{p^2})\in I$ 
and so $(t Y Z^p)^p = 0$ in~$R$, so that the above
are the same, giving associativity, as required.


\item[{\bf (9)}]
Hint: consider the order $4$ automorphism (of elliptic curves over $\mathbb{C}$) $(x,y) \mapsto (-x,iy)$. This induces an automorphism of the formal group associated to $Y^2 = X^3 + AX$. 

See the full solutions file for the full solution!

\item[{\bf (10)}] First we show convergence of the right hand side. For each $i$, set $g_i = G_i(x_1,\ldots,x_l)$. We have $|g_i| \le \max(|x_i|) < 1$. Choose $\rho < 1$ such that $\max(|x_i|) = \rho\delta$ with $\delta < 1$.  Since $\{|a_n|\rho^{\sum n_i}\}$ is bounded, say by $C$, we have \[\left|\sum_{n: \sum n_i \ge N} a_n g_1^{n_1}\cdots g_k^{n_k}\right| \le C \delta^N.\] It follows that the right hand side converges, and is the limit of the sequence \[\sum_{n: \sum n_i \le N-1} a_n g_1^{n_1}\cdots g_k^{n_k}\] as $N \to \infty$. 

Write $G_i = \sum b_{i,m} X_1^{m_1} \ldots X_l^{m_l}$ and for a positive integer $M$ consider the finite truncations $G_{i,M} = \sum_{\sum m_i \le M-1} b_{i,m} X_1^{m_1} \ldots X_l^{m_l}$ with values $g_{i,M}$ when we substitute the $x_i$. 

We have \[\left|a_n g_1^{n_1}\cdots g_k^{n_k}-a_n g_{1,M}^{n_1}\cdots g_{k,M}^{n_k}\right| =\left|a_n(g_1^{n_1}\cdots g_k^{n_k}- g_{1,M}^{n_1}\cdots g_{k,M}^{n_k})\right|   \le |a_n|\rho^M\delta^M,\] since all the terms which appear in $g_1^{n_1}\cdots g_k^{n_k}$, but not in $g_{1,M}^{n_1}\cdots g_{k,M}^{n_k}$, have valuation $\le \rho^M\delta^M$. We deduce that \[F(g_1,\ldots,g_k) - \sum_{n: \sum n_i \le N-1} a_n g_{1,N}^{n_1}\cdots g_{k,N}^{n_k}\] \[\le \max(C\delta^N,|a_n| \rho^N\delta^N : \sum n_i \le N-1) = C\delta^N.\]

On the other hand, if we consider the formal power series $F\circ G$ then the difference \[(F\circ G)
-  \sum_{n: \sum n_i \le N-1} a_n G_{1,N}^{n_1}\cdots G_{k,N}^{n_k}\] only contains terms of degree $\ge N$. From this and the condition on the $|a_n|$, it follows easily that $(F\circ G)(x_1,\ldots,x_l)$ converges and its value is the limit of \[\sum_{n: \sum n_i \le N-1} a_n g_{1,N}^{n_1}\cdots g_{k,N}^{n_k}\] as $N \to \infty$.

\end{enumerate}

\end{document}


