Hi all,
If I expand the function
$\displaystyle y(t_{n+1})$ about $\displaystyle y(t_n) $and get
$\displaystyle y(t_{n+1}) = y(t_n) + h y'(t_n) + h^2/2 y''(t_n) + ....$
what do I get if I expand the function
$\displaystyle y(t_{n+2}) =$ ?
Hi all,
If I expand the function
$\displaystyle y(t_{n+1})$ about $\displaystyle y(t_n) $and get
$\displaystyle y(t_{n+1}) = y(t_n) + h y'(t_n) + h^2/2 y''(t_n) + ....$
what do I get if I expand the function
$\displaystyle y(t_{n+2}) =$ ?
My question is this;
Calculate the truncation error for the following scheme
$\displaystyle y_{n+2} - y_{n+1} = \frac{h}{2} (3f_{n+2}-f_{n}) $
So I get
$\displaystyle y_{n+2}-y_{n+1}-\frac{h}{2}(3f_{n+2}-f_{n})=0 $
I then assume $\displaystyle y_{n}=y(t_{n}) $
I also know that $\displaystyle f(y_{n},t_{n})=y' $
This gives me
$\displaystyle y(t_{n+2})-y(t_{n+1})-\frac{h}{2}(3y'(t_{n+2})-y'(t_{n})\approx0 $
Using a taylor series expansion about $\displaystyle t_{n} $
This is where I get confused.
I have to get the equation all in terms of $\displaystyle t_{n} $
I know that $\displaystyle y(t_{n+1}) $
is
$\displaystyle y(t_{n+1}) = y(t_{n}) + hy'(t_{n+1} + ... $
and that
$\displaystyle y(t_{n+2}) = y(t_{n}) + 2hy'(t_{n+1}) + ... $
since
$\displaystyle y(t_{n+2}) - y(t_{n}) = 2h $
but how do I expand the
$\displaystyle y'(t_{n+2}) $
as a Taylor series or am I doing something wrong
Thank you for your reply