Someone recently told me certain real numbers could only be expressed in closed form via an expression involving complex numbers.
Is this true? If so do these numbers have a name?
What is a simple example?
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2$\begingroup$ $i^i$, where $i$ is the imaginary unit, comes to mind. It is a real number, with value $e^{-\pi/2}$, but it doesn't fit your example because I just also gave a closed form with no imaginary units. Interesting. $\endgroup$– The CountCommented Feb 22, 2017 at 18:22
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$\begingroup$ @OmnipotentEntity: I would like you to ask to the person who told you this, to give at least one such example. Is it possible ? $\endgroup$– AdrenCommented Feb 22, 2017 at 18:28
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$\begingroup$ He didn't know of an example. He said he had read it somewhere. $\endgroup$– OmnipotentEntityCommented Feb 22, 2017 at 18:42
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$\begingroup$ Note that this needs a very specific definition of 'closed-form', but it's true under that definition; 'nested radicals of rational expressions' is a pretty good version. $\endgroup$– Steven StadnickiCommented Feb 22, 2017 at 19:05
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$\begingroup$ That seems to be that that doesn't make any sense. I could be wrong but if it's expressible in closed for in terms of non-real complex numbers and a nonreal complex number is expressible in terms of reals and square roots of negative reals, then all reals and indeed all complex numbers that are expressible in any closed form are expressible in closed form via reals. $\endgroup$– fleabloodCommented Feb 22, 2017 at 19:38
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2 Answers
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This phenomenon occurs prominently in the (historically interesting) casus irreducibilis of third degree polynomial equations with three real roots. Take the equation $$x^3-2x^2-6x+5=0$$ as an example.
answered Feb 22, 2017 at 18:57
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To answer my final part about a nice, minimal example:
$$\sqrt[3]{1+i \sqrt{7}}+\sqrt[3]{1-i \sqrt{7}}$$
Which $\approx$ 2.6016791318831542525
This value was adapted from one of the roots of the polynomial posted in the accepted answer, then back solved for the generating cubic to ensure it is a real casus irreducibilis. (Previous version using $\sqrt{5}$ did not have a rational cubic.)
The generating cubic is simply $x^3 - 6x - 2 = 0$.
As noted by a @Steven Stadnicki , this is only irreducible under the radicals. Using Euler's formula we can find a trigonometric representation of this number.
$$2 \sqrt{2} \cos{\left(\frac{\tan^{-1}{\left(\sqrt{7}\right)}}{3}\right)}$$
answered Feb 22, 2017 at 20:57