Numerical Analysis Question

Phyxius117 · #1 $Old$ November 1st, 2009, 05:16 PM

How would you solve this problem?

Show that Max |G(x)| = h^(2)/4
xj<=x<=x(j+1)

Where G(x) = (x-jh)(x-(j+1)h)

j = iteration number
h = step size

At first I realize that

xj = j * h
x(j+1) = (j+1) * h

but I still can't reach the final answer h^(2)/4

Help Please!!

CaptainBlack · #2 $Old$ November 2nd, 2009, 12:22 AM

Quote:

Originally Posted by Phyxius117 $View Post$

How would you solve this problem?

Show that Max |G(x)| = h^(2)/4
xj<=x<=x(j+1)

Where G(x) = (x-jh)(x-(j+1)h)

j = iteration number
h = step size

At first I realize that

xj = j * h
x(j+1) = (j+1) * h

but I still can't reach the final answer h^(2)/4

Help Please!!

Can you please clarrify your notation.

CB