Given the equations for the harmonic oscillator
$\frac{dy}{dz}=z, \frac{dz}{dt}= -y$
if the system is approximated by the symplectic Euler method, then it gives
$z_{n+1}= z_{n}-hy_{n}, \\ y_{n+1}= y_{n}+hz_{n+1}$
which shows that the circle $y^2_{n} + z^2_{n} = 1$ is mapped into an ellipse.
Deduce that to order $h^2$ the ellipse has the same enclosed area as the circle.
=> my attempt so far
in this case, area of ellipse = area of circle.
Truncation error from the symplectic euler method gives
$hT_{z}= h(z'+y) +O(h^2)$
$hT_{z}= O(h^2)$ since, $h(z'+y)$ goes to zero
$T_{z}= O(h)$
Similarly,
$hT_{y}= h(y'-z) +O(h^2)$
$hT_{y}= O(h^2)$ since, $h(y'-z)$ goes to zero
$T_{y}= O(h)$
Both have first order
did my answer satisfy the question?