Apply Euler's method on the initial value problem $y'(t)=y(t)$ with $y(0)=1$ (in the interval $[0,1]$) and equidistant grid $I_h$, $h=\frac1n$. Give the approximation $y_h$ explicitly.
This question is a little bit related to: fixpoint iteration to solve $y'(t)=y(t), y(0)=1$
So I want to use Euler's method for the given IVP.
The method is stated as follows:
$y_{k+1}=y_k+hf(t_k, y_k)$
Here, we have $f(t_k,y_k)=y_k$ because of the differential equation $y'=y$.
So we get:
$y_1=y_0+\frac1n\cdot y_0=1+\frac1n\cdot 1=1+\frac1n$
$y_2=y_1+\frac1n\cdot y_1=1+\frac1n+\frac{1}{n}+\frac1{n^2}=1+\frac{2}{n}+\frac{1}{n^2}$
And so on. An explicit formula is given by:
$y_m=\sum_{k=0}^m \binom{m}{k}\frac{1}{n^k}$
I doubt, that this is correct. First of all this does not depend on $t$, and I am asked to show, that
$y_h(1)\to_{n\to\infty} y(1)=e$
Can you help me out? What have I done wrong? Is it correct to set $f(t_k, y_k)=y_k$?