Thread: Adams-Bashforth derivation

1. Adams-Bashforth derivation

When deriving the Adams-Bashforth 2 step predictor, one must calculate the integral of:

t-tj-1/tj-tj-1 f(tj,y(tj)) + t-tj/tj-1-tj f(tj-1,y(tj-1))

between the limits tj and tj+1, where f(t,y) is dy/dt...

How does one go about doing this integral, is there a clever substitution that must be made.. can any1 help???

2. Originally Posted by johnbarkwith
When deriving the Adams-Bashforth 2 step predictor, one must calculate the integral of:

t-tj-1/tj-tj-1 f(tj,y(tj)) + t-tj/tj-1-tj f(tj-1,y(tj-1))

between the limits tj and tj+1, where f(t,y) is dy/dt...

How does one go about doing this integral, is there a clever substitution that must be made.. can any1 help???
This is very difficult to interpret, put more brackets in to clarrify what is meant.

Having said that, to me it looks linear in $t$, in which case the integral is trivial
except for the fiddlyness of the other terms (the $t_j$ and $t_{j-1}$ are constants as
far as the integral is concerned)

RonL

3. sorry about the scrappy notation, dont know how to get any proper mathematical notation on here... The quotients involving the t's should be in brackets, and the whole quotient is multiplying the f(...)

If anyone could help with the integration i would be very grateful, Thanks...

4. been thinking about this more, and as u stated before the tj can just be treated as constants. does that mean that the f(...) can just be treated as a constant too as it is a function of constants, and therefore the whole integration becomes trivial (i.e. integrating t )??

or is my reasoning completely wrong?

5. Originally Posted by johnbarkwith
been thinking about this more, and as u stated before the tj can just be treated as constants. does that mean that the f(...) can just be treated as a constant too as it is a function of constants, and therefore the whole integration becomes trivial (i.e. integrating t )??

or is my reasoning completely wrong?
That's what I think, the integrand is just a linear function of t.

RonL