| Author |
Comment/Response |
yehuda
|
 01/05/13 5:38pm
I took few more minutes to look at your post
there is enough information there to look at
1. You mix exact numbers (integers) with machine precision numbers. One way to get more accurate results is to use rationals or determine the precision of the estimated numbers yourself
2. Using FindRoot is both fast and accurate
d1 = -((9.31323*10^-10 (-64. +
R)^7 (3. + (-1 + 2.32831*10^-10 (-64. + R)^8) (80. +
R)))/(-1 + 2.32831*10^-10 (-64. + R)^8)^2) + (-1 +
2.32831*10^-10 (-64. + R)^8 +
1.86265*10^-9 (-64. + R)^7 (80. + R))/(2 (-1 +
2.32831*10^-10 (-64. + R)^8));
d2=D[d1,R];
sol=FindRoot[d1==0,{R,70,64,80}]
returns
{R -> 77.6988}
so
{d1,d2}/.sol
returning
{0., -0.492557}
meets your requirements
yehuda
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