Forward Euler and 1st-order linear ODE with exponential solution

Question

Let us consider the simple ODE $y'=y$ with $y(0)=1$ on the interval $[0,1]$. The solution is obviously $y(t)=\mathrm{e}^t$. Now, consider a Forward Euler approximation of the solution to the ODE. The sequence $y_k=(1+h)^k$ with $h=1/n$, $n$ being the number of time instants $t_k$, $k=0,1,\ldots,n$ in the interval $[0,1]$ is generated. It is true that $\lim_{n\to\infty}y_k=y(kh)$ for all $k=0,1,\ldots,n$? How can we show this using $y_k=(1+1/n)^k$?

Answer 1

• If we have a look at the last entry $y_n$, we have: $$ \lim_{n\to\infty}y_n=\lim_{n\to\infty}(1+1/n)^n=\exp(1) $$
• For intermediate locations, say $\gamma$ in $[0,1]$, we have to specify $k$ such that $k/n=\gamma$, that is $k=\gamma n$. We then have $$ \lim_{n\to\infty}y_{k(n)}=\lim_{n\to\infty}(1+1/n)^{k(n)}=\lim_{n\to\infty}(1+1/n)^{\gamma n}=\lim_{p\to\infty}(1+(\gamma/p))^p=\exp(\gamma) $$ and the exact solution is recovered.