I found an example of FindRoot
usage that I couldn't find documented.
It looks like:
FindRoot[f[x] == 0.4, {x, 0, 1, 0, 1}]
The documentation shows this case FindRoot[lhs==rhs, {x, xstart, xmin, xmax}]
searches for a solution, stopping the search if x ever gets outside the range. But the example I found has an extra argument. What does it mean?
Thanks.
True
in that position and got the error FindRoot::bbound: Search region bound True for variable number 1 is not a number or Infinity. >> so my guess is search region bound.FindRoot[Cos[x] == 0.4, {x, 0, 1, 0, 3}]
finds the solution $x \approx 1.59$. Another odd thingFindRoot[Cos[x] == 0.4, {x, 0, 1, 0, 0.5}]
FindRoot::streg: The starting point {1.} is not in the search region {{0.},{0.5}}. >> $\endgroup$ – ssch Jan 7 '13 at 15:41FindRoot[x^2 - 2 == 0, {x, 1.1, 1.9, 1.2, 2}, Method -> "Secant"]
and move around the 1.1 or the 1.9. $\endgroup$ – b.gates.you.know.what Jan 7 '13 at 15:43