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t == -1/2*(2/45*I*sqrt(4613)*sqrt(3) - 82/5)^(1/3)*(I*sqrt(3) + 1) + 1/3*(10*I*sqrt(3) - 10)/(2/45*I*sqrt(4613)*sqrt(3) - 82/5)^(1/3)

asked 2020-06-22 22:01:28 +0200

rgta8478jo96 gravatar image

t == -1/2(2/45Isqrt(4613)sqrt(3) - 82/5)^(1/3)(Isqrt(3) + 1) + 1/3(10Isqrt(3) - 10)/(2/45Isqrt(4613)sqrt(3) - 82/5)^(1/3)

This is an example of solution(s) I am getting when solving equations. I am not mathematician and I do not know what these numbers means. Can you please tell me if there is way to 'ask' program to deliver results in a way that mortals can understand it. Please help. Thank you

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answered 2020-06-23 00:19:50 +0200

tmonteil gravatar image

You can use the show function to display the result in LaTeX :

https://sagecell.sagemath.org/?z=eJwr...

If you have multiple such formulas in a jupyter notebook, you can also add:

%display latex

in some cell, so that every formula will be turned into LaTeX once that cell is evaluated.

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answered 2020-06-23 17:42:00 +0200

Emmanuel Charpentier gravatar image

You can also ask for a numerical apriximation via the funcion/method vumerical_approx (with handy abbreviations n and N) ; in this case, this yelds a complex value with very small imaginary part, which Sage has trouble proving null but probably is:

Let's call Sol your solution. Then :

sage: Sol.rhs().imag_part().log().n()
-36.0436533891172
sage: Sol.rhs().imag_part().log().n(digits=30)
-70.0078652365544762511404442673
sage: Sol.rhs().imag_part().log().n(digits=60)
-infinity

Trying to obtain your solution by other means (e. g. as a root of a polynomial in an exact ring) may enable Sage to prove that thos solution is real.

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Asked: 2020-06-22 22:01:28 +0200

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Last updated: Jun 23 '20