Solve exp + trig function with solve()

Posted on 2008-06-20 06:24 By sur_aiya (2)

Categories:

How do I ...? with Maple

Hi,

want to solve this equation for t but getting 'Solutions may be lost' error.

eq3 := exp(-1/k*t)*k/(1+k^2)+(-cos(t)*k+sin(t))/(1+k^2);

solve(eq3=0, t);

Solutions maybe lost.

Is this too much for the symbolic engine? Does anyone know of a technique to solve this type of equation?

Appreciated.

solving

Comment on 2008-06-20 07:59 from Axel Vogt ( Maple Rating 4

955)

A better method is to use RootOf (and to multiply by 1+k^2), but it gives no 'answer'
even after applying 'allvalues'.

You may want to interprete it as: "the system can not provide a satisfactory answer
automatically in terms of its known functions", so it is quite likely that there is none
(even if there may be a solution [depending on conditions for k]).

Of course t=0 will do, but for others ...

 exp(-t)-cos(t)+sin(t); # k=1
 fsolve(%,t,complex,avoid={t=0.});

                 -6.5025973236926 - 6.8491704362806 I

symbolic in terms of k..

Comment on 2008-06-20 09:33 from sur_aiya (2)

Is there a way to get solution for t in terms of k (symbolic) for the range of k values for which there exists solution? obviously k=0 is a discontinuity so possibly a symbolic solution t in terms of k for k=0.25..3?

cheers for help.

cant tends towards can/

for k=1 the inverse is not a function

Comment on 2008-06-20 10:07 from Axel Vogt ( Maple Rating 4

955)

for k=1 the inverse is not a function, see above ... I think there is no name for it (as already said)

may be you can use Langrange inversion in t=Pi or Pi/2 to write down a series

Inverse function?

Comment on 2008-07-03 15:10 from edgar ( Maple Rating 2

181)

How about solving for the inverse function? solve for k in terms of t...
solve(eq3,k) yields an answer, involving not only exp, sin, cos, but also LambertW.

k = t*sin(t)/(LambertW(-t/(exp(cos(t)*t/sin(t))*sin(t)))*sin(t)+cos(t)*t)
---
G A Edgar

Non-Algebraic

Comment on 2008-06-20 10:30 from jpmay ( Maple Rating 2

132)

Even with k constant, this type of system is hard for solvers (Maple just returns a RootOf that allvalues doesn't expand as Axel points out and other computer algebra systems return similar things). Solvers can give symbolic answers to some exp-trig expressions but generally only if the answer can be found by replacing the exp and trig terms with symbols (i.e. if the solution is fundamentally algebraic and not analytic).

John