Hello,
I was reading chapter 17.3.1, Modified Midpoint Method, of Edition 3 and do not understand eq. 17.3.4:

Why is the estimate (17.3.4) for y(x+H) _fourth_-order accurate? I wrote

y_n - y(x+H) = alpha_1*h^2 + alpha_2*h^4 + O(h^6)
y_(n/2) + y(x+H) = alpha_1*4*h^2 + alpha_2*16*h^4 + O(h^6)

which can be combined to

4y_n - y_(n/2) - 3y(x+H) = -alpha_2*12*h^4 + o(h^6)

<==>

( 4y_n - y_(n/2) )/3 = y(x+H) + o(h^4)

So why should this be fourth order accurate? And where do we gain TWO order of magnitude which the book points out is the main advantage of using the midpoint method in the first place?

Thanks for your help,

Thorben

Dear Thorben,

The text is not completely clear because it doesn't distinguish between local and global error. In terms of local error (i.e. in one step), the midpoint method is 2nd order accurate (error of order h^3). An extrapolation combining steps of size h and h/2 eliminates this error, and since the error series goes up in powers of h^2, this makes the method 4th order.

Eqn. 17.3.3 is the equation for the global error, where one takes n steps to cover an interval [a,b]. Since n =(b-a)/h, the global error typically has one lower power of h than the local error. Hence the error series has only even powers of h.

Gaining two orders at a time by extrapolation follows because the series is a series in h^2, not h.

Hope this helps.
Saul Teukolsky