I need to solve a system similar to the following (Except it is quite large. Solving this ought to do the job):
$$ \tan[2f(t)] = 1+ t^2\ $$
and
$ f(t) $ is $ k $, such that$$ \tan[2kt]-(1+k^2) = 0\ $$
I have used RootSearch
for both equations as I also know the range in which $k$ and $T$ lie. The sample of my code is below.
f[T_] := k /. Extract[Flatten[Quiet[RootSearch[Tan[2*k*T] - (1 + k^2) == 0, {k, 0, 1}],
$MinPrecision::precset]], -1];
Extract[Flatten[Quiet[RootSearch[Tan[2*f[T]] - (1 + T^2) == 0, {T, 0, 1}],
$MinPrecision::precset]], -1]
and I end up getting the following error
ReplaceAll::reps: {k,0,1} is neither a list of replacement rules nor a valid dispatch table,
and so cannot be used for replacing. >>
Please suggest a solution. It would be preferable to use something as powerful as RootSearch
because my other equations seem to be unsolvable via FindRoot
or NSolve
.
FindAllCrossings2D[]
on the system{Tan[2 y t] == 1 + y^2, Tan[2 y] == 1 + t^2}
; you already know the ranges ofy
andt
anyway, so... $\endgroup$ – J. M.'s ennui♦ Aug 26 '12 at 11:41T=0
for instance. $\endgroup$ – b.gates.you.know.what Aug 26 '12 at 11:47