<s>[INST] Showing that a sequence of Picard iterates converges I have a sequence of functions:
$$y_{n}(x) = 1 + \int \limits_0^x 1 + t^2 + y_{n-1}^2(t)\,\mathrm dt$$
With $y_0 = 1$.
I'm trying to show that this converges in a box $-1 \le x \le 1$ and $-10 \le y \le 10$. However when I try to show this converges in this box, the bound on the integral gives $|y-1| \le xM$ where $M$ is the maximum of $1 + t^2 + y^2$ in the box, however this maximum is $102$ so the iterates only converge in the box when $|y-1| \le 102$ but this isn't true?
Am I missing something obvious here?
Thanks
 [/INST] A theorem for finding an interval of convergence for a Picard iteration scheme is usually stated as:
Given the IVP
\begin{equation*}
y^{\prime} = f(x,y), \quad y(x_{0}) = y_{0},
\end{equation*}
suppose $f$ and $df/dy$ are continuous in the rectangle $R$ given by
\begin{equation*}
x_{0} \le x \le x_{0}+a, \quad \left| y-y_{0}\right| \le b
\end{equation*}
compute the values
\begin{equation*}
M = \max_{(x,y)\in R}\left|f(x,y)\right|, \quad \alpha = \min\left(a, \frac{b}{M}\right).
\end{equation*}
Then the Picard iterates converge to a unique solution on the interval $[x_{0},x_{0}+\alpha]$.
(Note, the interval given by this theorem may be smaller that the actual interval of convergence).  
In your case we have that,
\begin{alignat*}{2}
x_{0} &= 0 \\
a &= 1, \\
b &= 10,\\
M &= \max_{R}\left| 1 + x^{2} + y^{2}\right| \\
&= 1 + a^{2} + b^{2} \\
&= 102 &&\Rightarrow \\
\alpha &= \min\left(1, \frac{10}{102}\right)\\
&\le 1 \\
&= a &&\Rightarrow
\end{alignat*}
which falls short of the desired result.  
Alternatively, we can try looking at the iterates to see if we can learn more about their radius of convergence.
For $y_{n}(x) = 1 + \int_{0}^{x}1+t^{2}+y^{2}_{n-1}(t)dt$, assume the differential equation has the form
\begin{alignat*}{2}
y(x) = 1 + \int_{0}^{x}f(t,y)dt &&\Rightarrow \\
f(t,y) = 1 + t^{2}+y^{2}(t)
\end{alignat*}
and 
\begin{equation*}
y(0) = 1 = y_{0}
\end{equation*}
Now,
\begin{alignat*}{2}
y_{1}(x) &= 1 + \int_{0}^{x}f(t,y_{0})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+1 dt \\
&= 1 + 2x + \frac{1}{2}x^{2},
\end{alignat*}
\begin{alignat*}{2}
y_{2}(x) &= 1 + \int_{0}^{x}f(t,y_{1})dt \\
&= 1 + \int_{0}^{x}1+t^{2}+(1 + 2t + \frac{1}{2}t^{2})^{2} dt \\
&= 1 + 2x + 2x^{2}+2x^{3}+\frac{1}{2}x^{4}+\frac{1}{20}x^{5}
\end{alignat*}
and for $y_{3}$ I get something like:
\begin{equation*}
y_{3}(x) = 1 + 2x + 2x^{2} + \frac{7}{3}x^{3} + 3x^{4} + \frac{13}{5}x^{5}+\frac{202}{120}x^{6} + \frac{52}{70}x^{7}+\frac{22}{80}x^{8}+\cdots
\end{equation*}
So, it seems (loosely) for some constant $M$ large enough
\begin{alignat*}{2}
M\sum_{k=0}^{\infty}x^{k} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1+\sum_{k=1}^{\infty}\frac{x^{k}}{k} \Rightarrow\\
M\frac{1}{1-x} &\ge \lim_{k\to\infty}y_{k}(x) &&\ge 1-\log(1-x)
\end{alignat*}
both of which are valid for $\left|x\right| < 1$, but not for the closed interval.  
</s>