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I have a set of arithmetic functions from $D\subset\mathbb C$ to $\mathbb C$ (addition, division, trigonometric functions, ...). Each of those functions can also be restricted from $E\subset \mathbb R$ to $\mathbb R$.

By using integer constant, I can define many different reals using my set of functions, for example $\pi=4\tan^{-1}(1)$.

  1. Can I define some reals using the complex that I can't define using only the real definition ? For example, if I have only exponential function, I can define $\Re(e^{i.a})=\cos(a)$ But can I define $\cos(a)$ without trigonometric function and without complex ? I don't think so. Are they other examples where I do not restrict any use of usual functions ?

  2. Are they some sets of functions "complete" (and which ones)? I mean that using such a set of functions, if I restrict my functions to the reals, I can define the same reals that If I can use the more general complex notations. (Regarding the previous question, the set $\{\Re,\exp\}$ would not be complete for example).

  3. Is the set $\{+,-,\times,\div,\exp,\ln,\cos,\sin,\tan^{-1},\Re\}$ complete ?

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    http://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions2012-08-18
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    And how do you get $\sqrt 2$ with your functions? You need to add at least $exp^{-1}$ so that you can use $\sqrt 2 = 2^{\frac{1}{2}}$2012-08-18
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    I add $\ln$ to the list2012-08-18
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    I'm not sure I understand #2 and #3. What do you mean by "define"? For any real $y$ there is a real $x$ such that $\ln x = y$, so would you say that $\{\ln\}$ is complete?2012-08-18
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    My only constants are integers, so I can define any fraction using the division, then I can define any power (like $\sqrt{2}$) using $\exp$ and $\ln$, and so one... But I think some algebric numbers can not be define (written) this way.2012-08-18
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    Ok, now I take $\pi$ and wherever there is a digit 7, I put a 9. How do you express that number?2012-08-19
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    I'm pretty sure I can't in this scope.2012-08-19

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I'm not sure if this is what you have in mind, but are you familiar with the casus irreducibilis? If a cubic with rational coefficients is irreducible over the rationals and has three real roots, then those roots can be expressed in terms of arithmetic operations, square roots, and cube roots, but one must allow square roots of negative numbers. More details at Wikipedia.

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    That's exactly what I had in mind, thank you :)2012-08-19