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?