Applications of Series — Question 7

*Question 7 Derive one forward Euler step for $y'=f(t,y)$ from Taylor’s theorem. Identify the discarded term. Then apply one step with $h=0.1$ to $y'=y$, $y(0)=1$, and bound its local error.

*Question 7 – Solution

Taylor’s theorem about $t_n$ gives $$y(t_n+h)=y(t_n)+h y'(t_n)+\frac{h^2}{2}y''(\xi_n)$$ for some $\xi_n$ between $t_n$ and $t_n+h$.

Since $y'(t_n)=f(t_n,y(t_n))$, dropping the quadratic remainder produces Euler’s update $$\boxed{y_{n+1}=y_n+h f(t_n,y_n)}.$$ The one-step local truncation error is $h^2y''(\xi_n)/2$, hence is order $h^2$.

For $y'=y$, $y_0=1$, and $h=0.1$, $$y_1=1+0.1(1)=\boxed{1.1}.$$ The exact value is $e^{0.1}$. On $[0,0.1]$, $y''(t)=e^t\le e^{0.1}$, so $$|e^{0.1}-1.1|\le\frac{e^{0.1}(0.1)^2}{2}<0.00553.$$