Thorben
08-17-2008, 01:49 PM
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
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