Linearizing an explicit differentiation scheme

[Pietair]

Homework Statement

Consider the following implicit scheme:
y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})]

By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error O(\Delta t^{3})

Homework Equations

The solution is:
y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})

And the notation: y'(x)=f(x,y(x))

The Attempt at a Solution

I think I have to take the Taylor expansion of f(y_{n+1})

I get:
f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})

Substituting in the main scheme:

y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})]f(y_{n+1})

This expression is, as far as I can see, not equal to the expression of the solution.

Any help would be appreciated!

 

[SammyS]

Homework Statement

Consider the following implicit scheme:
y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n+1})+f(y_{n})\right]

By linearization one can obtain an explicit scheme which is an approximation to this - with approximation error O(\Delta t^{3})

Homework Equations

The solution is:
y_{n+1}=y_{n}+\Delta t\left [1-\frac{1}{2}\Delta t f'(y_{n}) \right ]^{-1}f(y_{n})

And the notation: y'(x)=f(x,y(x))

The Attempt at a Solution

I think I have to take the Taylor expansion of f(y_{n+1})

I get:
f(y_{n+1})=f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})

Substituting in the main scheme:

y_{n+1}=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta tf'(y_{n})+O(\Delta t^{2})+f(y_{n})\right]f(y_{n+1})=y_{n}+\frac{\Delta t}{2}\left [f(y_{n})+\Delta ty'(y_{n})+O(\Delta t^{2})+f(y_{n})\right]f(y_{n+1})

This expression is, as far as I can see, not equal to the expression of the solution.

Any help would be appreciated!

You are missing the "\right" code with several of your "]" symbols. (Corrected in Quote.)

 

[hunt_mat]

This is what I did, it may help you it may not.
<br /> \begin{array}{rcl}<br /> f(y_{n+1}) &amp; = &amp; f\left( y_{n}+\frac{\delta t}{2}X\right) \\<br /> &amp; = &amp; f(y_{n}+\frac{X\delta t}{2}f&#039;(y_{n}) \\<br /> &amp; = &amp; f(y_{n})+ \frac{\delta t}{2}(f(y_{n+1})+f(y_{n}))f&#039;(y_{n})<br /> \end{array}<br />
where X=f(y_{n+1})+f(y_{n}) and we can solve for f(y_{n+1}), to obtain:
<br /> f(y_{n+1})=\left( 1-\frac{\delta t}{2}f&#039;(y_{n})\right)^{-1}\left( 2f(y_{n})+\frac{\delta t}{2}f&#039;(y_{n})\right)<br />
Substitute this into your equation.