Trapezoid method, fixed-point iteration

Hi,

**problem:**

The trapezoid method for solving ordinary differential equations is,

$\displaystyle y(x_{i+1}) = y(x_i) + h \frac{f(y(x_{i+1})) + f(y(x_i))}{2}$.

Since this method is implicit, we must solve for $\displaystyle y(x_{i+1})$ in every step. If we use fixed-point iteration $\displaystyle x=g(x)$, we will have convergence if $\displaystyle |g'(x)|<1$ for $\displaystyle x \in[y(x_i), y(x_{i+1})]$.

(a) Give the expression for $\displaystyle g$.

(b) Give a condition on $\displaystyle h$ such that the fixed-point iterations converges at each step.

**attempt:**

(a)

$\displaystyle g(x) = y(x_i) + h \frac{f(x) + f(y(x_i))}{2}$

For each iteration $\displaystyle i$, we have to run fixed-point iteration enough times to make sure we get the wanted accuracy.

We take a guess at $\displaystyle x=y_{i+1}$, calculate $\displaystyle g(x)$ and see how big $\displaystyle |x-g(x)|$ is. If it's close enough to $\displaystyle 0$, we stop the fixed-point iteration.

(b)

I'm not really sure what to make of this question...

Fixed-point methods converge if $\displaystyle |g'(x)|<1$.

$\displaystyle

\begin{aligned}

&|g'(x)| = \frac{h}{2}\Big[f'(x) + f'(y(x_i))\Big] < 1&\\

&\Rightarrow h < \frac{2}{\Big[f'(x) + f'(y(x_i))\Big]} &

\end{aligned}

$

I do not see much sense in my last expression.

Thanks!