Evaluating 1st Derivative of a Simple Function
- From: monir <monirg@xxxxxxxxxxxx>
- Date: Mon, 06 Aug 2007 11:57:38 -0700
Hello;
Function: G = [sum m=1,9] gm.sin(m.th), coef gm, m=1,9 are known
r = 0.6 - 0.4 cos(th)
range th = 0.0 to pi
G=0.0 at th = 0.0 (or r=0.2)
G=0.0 at th = pi (or r=1.)
The above function can easily be constructed numerically and plotted
as G vs r. It displays a smooth function with finite 1st derivative
(dG/dr) everywhere including the endpoints r=1 (or th=pi) and r=0.2
(or th=0).
Now try to evaluate (dG/dr) analytically and you'll get (1/0) 1st
derivative at both ends !!!
(dG/dth) = [sum m=1,9] m.gm.cos(m.th)
(dr/dth) = 0.4 sin(th)
and (dG/dr) = { [sum m=1,9] m.gm.cos(m.th) } / [0.4 sin(th)]
= (1/0) at th=0
= (1/0) at th=pi
Analytically, there doesn't appear to be a limiting finite (dG/dr) at
either end, but the displayed function shows otherwise !!
If you concur, then How would you evaluate (dG/dr) at th=0 & at
th=pi ??
Your suggestion would be greatly appreciated.
Monir
.
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