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Your equation may have infinitely many complex solutions: sol = SolveValues[x^3 + x^2 - x == 20*Sin[x^2] - 5 && Abs[x] < 10, x]; Graphics[Point@ReIm[sol]] You can hardly expect `Solve` to give them all.

POSTED BY: Gianluca Gorni

Zhenyu Zeng

Posted 1 year ago

Hello. Thanks for your reply. May you tell me why they have infinitely complex solutions?

POSTED BY: Zhenyu Zeng

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

It's a consequence of Picard's theorem that there are infinitely many solutions in C. Also check MathWorld. It's the Great Picard theorem specifically that's of interest here.

POSTED BY: Daniel Lichtblau

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 1 year ago

... and in certain cases you will need to specify a region...

POSTED BY: Neil Singer

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 1 year ago

For transcendental equations you often need to choose a domain: Solve[x^3 + x^2 - x == 20Sin[x^2] - 5, x, Reals] Numericising the result: In[4]:= Solve[x^3 + x^2 - x == 20Sin[x^2] - 5, x, Reals] // N Out[4]= {{x -> -3.37217}, {x -> -3.19649}, {x -> -2.48921}, {x -> \ -1.70489}, {x -> -0.536109}, {x -> 0.496065}, {x -> 1.61444}} Regards Neil

For transcendental equations you often need to choose a domain:

Solve[x^3 + x^2 - x == 20*Sin[x^2] - 5, x, Reals]

Numericising the result:

In[4]:= Solve[x^3 + x^2 - x == 20*Sin[x^2] - 5, x, Reals] // N

Out[4]= {{x -> -3.37217}, {x -> -3.19649}, {x -> -2.48921}, {x -> \
-1.70489}, {x -> -0.536109}, {x -> 0.496065}, {x -> 1.61444}}

Regards

Neil

POSTED BY: Neil Singer