Really Maxima can't solve this?

Question

I want to get inverse function of this:
-0.5*( (%e^(-4*x))*(%e^(2*x)+1

I tried:

ex1: -0.5*( (%e^(-4*x))*(%e^(2*x)+1) );
ex2: ratsimp(ex1);
ex2;
solve([y=ex2],[x]);

But Maxima return empty array. Really Maxima can't solve this (Wolfram alpha can) or I do something wrong?

Maxima input and output

I need some powerful CAS (computer algebra system) for my product... I thought, that Maxima is strong enough, but it says the opposite — Данил Черкашин, May 20 '20 at 10:44
Does this answer your question? [How can I solve exponential equation in Maxima CAS](https://stackoverflow.com/questions/50260551/how-can-i-solve-exponential-equation-in-maxima-cas) — jacob, May 20 '20 at 14:20
As suggested by jacob, I also recommend trying the add-on package `to_poly_solve`. — Robert Dodier, May 20 '20 at 15:57
FYI - `Derive for Windows` solves it as [four complex](https://i.imgur.com/ciqHNlE.png) expressions. — John Alexiou, May 20 '20 at 16:35

JohanC · Answer 1 · 2020-05-20T16:21:40.917

2

You could try to solve it with Python's sympy. As sympy is not very fond of floating point numbers, I changed * 0.5 to / 2.

from sympy import symbols, exp, solve, Eq

x, y = symbols('x y')
expr1 = -((exp(-4*x))*(exp(2*x)+1))/2

res = solve(Eq(y, expr1), x)

Result:

[log(-sqrt(-sqrt(1 - 8*y)/y - 1/y)/2),
 log(sqrt(-sqrt(1 - 8*y)/y - 1/y)/2),
 log(-sqrt(sqrt(1 - 8*y)/y - 1/y)/2),
 log(sqrt(sqrt(1 - 8*y)/y - 1/y)/2)]

To check, each of the results can be substituted back into expr1:

for r in res:
    print(expr1.subs(x, r))
    print(expr1.subs(x, r).simplify())

This outputs 4 times y as the simplified expression:

-8*(1 - sqrt(1 - 8*y)/(4*y) - 1/(4*y))/(-sqrt(1 - 8*y)/y - 1/y)**2
y
-8*(1 - sqrt(1 - 8*y)/(4*y) - 1/(4*y))/(-sqrt(1 - 8*y)/y - 1/y)**2
y
-8*(1 + sqrt(1 - 8*y)/(4*y) - 1/(4*y))/(sqrt(1 - 8*y)/y - 1/y)**2
y
-8*(1 + sqrt(1 - 8*y)/(4*y) - 1/(4*y))/(sqrt(1 - 8*y)/y - 1/y)**2
y

edited May 20 '20 at 16:21

answered May 20 '20 at 09:32

JohanC

71,591
8
33
66

Thanks! But you want to say, that sympy is better than Maxima? – Данил Черкашин May 20 '20 at 10:45
Are there any similar systems as sumpy written in python but only in javascript? – Данил Черкашин May 20 '20 at 10:49
1

Interestingly, I tried OP's problem w/ `to_poly_solve` in Maxima and I get a somewhat different result, involving `log(sqrt(...))` but not `log(sqrt(sqrt(...)))`. I will try to look at whether these are actually the same later today. – Robert Dodier May 20 '20 at 15:58

Really Maxima can't solve this?

1 Answers1