Remainder for picard iteration for exponential function Consider the initial-value problem:
$$y'=y,\quad y(0)=1.$$
The solution is of course the exponential function
$$y=\exp x=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}.$$
We can obtain this by Picard iteration where $y_0=1$ and 
$$y_n(x)=1+\int\limits_0^x y_{n-1}(x)\mathrm{d}x.$$
Doing this on page 2, Schlag  (http://www.math.uchicago.edu/~schlag/bookweb.pdf) gives the solution:
$$y(x)=1+\int_0^xy(x)\mathrm{d}x=1+x+\int_0^x (x-t)y(t)\mathrm{d}t=\cdots =\sum_{n=0}^N\frac{x^n}{n!}+\int_0^x (x-t)^Ny(x)\mathrm{d}x.$$
Where do the remainder terms $\int_0^x (x-t)y(t)\mathrm{d}t,\dotsc, \int_0^x (x-t)^Ny(x)\mathrm{d}x$ come from? For instance 
$$y_2(x)=1+\int_0^x\left(1+\int_0^t y(s)\mathrm{d}s\right)\mathrm{d}t=1+x+\int_0^x\int_0^t y(s)\mathrm{d}s\mathrm{d}t.$$
How does the last term here become the remainder term given by Schlag?

You just 
use the Fubini theorem:
$$
\int_0^x \int_0^t y(s) ds dt =
\int_0^t\left( \int_t^x dt\right)y(s) ds  = 
\int_0^t(x-t)y(s) ds  
$$
