Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
en.wikipedia.org/wiki/… –  xavierm02 Aug 18 '12 at 14:58
    
And how do you get $\sqrt 2$ with your functions? You need to add at least $exp^{-1}$ so that you can use $\sqrt 2 = 2^{\frac{1}{2}}$ –  xavierm02 Aug 18 '12 at 15:02
    
I add $\ln$ to the list –  Xoff Aug 18 '12 at 15:07
    
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? –  Antonio Vargas Aug 18 '12 at 16:55
    
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. –  Xoff Aug 18 '12 at 17:58

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
That's exactly what I had in mind, thank you :) –  Xoff Aug 19 '12 at 7:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.