Solve a transcendental equation

Question

I have the following complicated equation $e^{a t} = 2 \cos(bt)$, needs to be solved for t. Any possibility of getting analytic solution? If not, can we solve it in Mathematica?

Answer 1

You won't find a full analytic solution. If $t$ is small you can use the first terms of the Taylor series to get an approximation, solving $1+at=1-\frac {(bt)^2}2$. You can solve it numerically with any root finder. You should have an infinite set of solutions where $t$ has the opposite sign from $a$. Once $t$ gets a little large $e^{at}$ will be small and the solutions will be close to odd multiples of $\pi$.

Answer 2

Hint:

$$ 1 = 2 (e^{iz} - e^{-i\bar z}) = 2 (e^{i(b+ia)} - e^{-i(b-ia)}) $$

Answer 3

As already said, only numerical methods could solve this transcendental equation.

As Ross Millikan answered, the case where the solution $t$ would be small is quite interesting. But, instead of using Taylor expansion for an approximate solution, we can also consider Padé approximants using degree $1$ for numerator and degree $n$ for denominator. The first solutions would then be $$t_{(1)}=\frac{2 a}{3 a^2+2 b^2}$$ $$t_{(2)}=\frac{3 \left(3 a^2+2 b^2\right)}{a \left(13 a^2+12 b^2\right)}$$ $$t_{(3)}=\frac{4 a \left(13 a^2+12 b^2\right)}{75 a^4+96 a^2 b^2+22 b^4}$$ $$t_{(4)}=\frac{5 \left(75 a^4+96 a^2 b^2+22 b^4\right)}{a \left(541 a^4+880 a^2 b^2+340 b^4\right)}$$ $$t_{(5)}=\frac{6 a \left(541 a^4+880 a^2 b^2+340 b^4\right)}{4683 a^6+9240 a^4 b^2+5160 a^2 b^4+602 b^6}$$

For illustration purposes, considering only the positive solution for $a=4$ and $b=3$, we should get, as approximations, $$\left( \begin{array}{ccc} n & t_{(n)} & t_{(n)} \approx \\ 1 & \frac{4}{33} & 0.121212 \\ 2 & \frac{99}{632} & 0.156646 \\ 3 & \frac{2528}{17403} & 0.145262 \\ 4 & \frac{87015}{585512} & 0.148614 \\ 5 & \frac{1171024}{7932791} & 0.147618 \end{array} \right)$$ while the "exact" solution would be $0.147845$.