Let $\eta(x;h)$ be the approximate solution furnished by Euler's method for the initial-value problem $y'=y, y(0)=1$. I proved that:
$i) \eta(x;h)=(1+h)^{x/h}$;
$ii) \eta(x;h)$ has the expansion $\eta(x;h)=\sum_{i=0}^{\infty}\tau_i(x)h^i$ with $\tau_0(x)=e^x$, which converges for $\mid{h}\mid<1$; the $\tau_i(x)$ here are analytic functions independent of h.
Now, I want:
$1)$ To determine $\tau_i(x), i=1,2,3$;
$2)$ To show that $\tau_i(x), i\ge1$ are the solutions of the initial value problems $\tau_i^\prime(x)=\tau_i(x)-\sum_{k=1}^{i}\frac{\tau^{k+1}_{i-k}(x)}{(k+1)!}, \tau_i(0)=0$.
In my calculations, I found $\tau_i(x)=\frac{1}{i!}\times\frac{\partial \eta}{\partial h}\eta(x;h)\Big|_{h=0}$, so how to write $\tau_i(x), i=1,2,3$ from this?
Could someone help me to solve this problem?
You could start with $$ \ln(η(x;h))=\frac xh\ln(1+h)=x(1-\frac h2+\frac{h^2}3-\frac{h^3}4\pm...) $$ so that $$ η(x;h)=e^x\cdot e^{-xh/2}\cdot e^{xh^2/3}\cdots $$ Another, probably less useful, identity is the binomial series $$ η(x;h)=\sum_{k=0}^\infty\binom{x/h}{k} h^k=\sum_{k=0}^\infty\frac{x(x-h)(x-2h)...(x-(k-1)h)}{k!} $$