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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?

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

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