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'm an Android programmer and am working on a graphing calculator. I have been looking for the limits on which roots can be done. I have a decent understanding of mathematics but can not seem to find these limits. Any help would be great, thanks.

share|improve this question
3  
I don't understand the question. –  Qiaochu Yuan Mar 2 '12 at 1:32
    
What are the limits on which numbers of y get a real value when x<0 and one gets the yth root of x. –  jersam515 Mar 2 '12 at 1:34
    
The word limit has a specific meaning in mathematics. Are you using the word limit to denote shortcomings? Are you asking what is the domain of the function $\sqrt \ : \mathbb{R} \to \mathbb{R}$? –  user2468 Mar 2 '12 at 1:36
    
A limit as in the limit on x+1>0 is that x>-1. Yes, the domain of the function where the result is real. –  jersam515 Mar 2 '12 at 1:38
    
Ok. Help me out here. As far as I understand $y \in \mathbb{R}$, $x < 0$ and you want to calculate the $y$th root of $x$. That is, compute $$r = x^{\frac{1}{y}}.$$ What is you question then? –  user2468 Mar 2 '12 at 1:41
show 3 more comments

2 Answers

up vote 2 down vote accepted

Let $x < 0$ and $y \in \mathbb{R}$. Note that $\text{Arg } x = \pi$.

By the definition of the complex logarithm we have $$x^{1/y} = e^{(\log x)/y} = e^{[\log|x| + i(\pi+2\pi\ell)]/y} = e^{(\log |x|)/y} e^{i\pi(1+2\ell)/y}$$ where $\ell$ is any integer. Thus $x^{1/y}$ has a real value if and only if $(1+2\ell)/y$ is an integer for some $\ell$. This happens exactly when $y = (1+2m)/n$ for some integers $m$ and $n$ with $n \neq 0$.

share|improve this answer
    
..which also means that $y$ has to be a rational number. –  user2468 Mar 2 '12 at 12:13
    
@J.D.: Not all rationals will work, like $y= 2$. –  Antonio Vargas Mar 2 '12 at 14:49
1  
Oh. Sorry. I meant $y$ at least has to be rational, in the fashion of necessary but not sufficient condition. –  user2468 Mar 2 '12 at 18:21
    
@J.d.: Ah, yeah, definitely. –  Antonio Vargas Mar 2 '12 at 21:18
add comment

Given a real number $x < 0$ and an integer $y \in \mathbb{Z}$. The $y$th root of $x$, given by: $$ r = x^{\frac{1}{y}} $$ is a real number if and only if $y$ is an odd number.

share|improve this answer
    
How's that work if y is a decimal, as per wikipedia decimal values are neither odd nor even? –  jersam515 Mar 2 '12 at 1:57
    
In that case, the answer still applies. If $x$ is a negative real number and $y$ is a real number that is not an integer, then $y$ is not odd, and $x^{\frac1y}$ is not a real number. –  Tanner Swett Mar 2 '12 at 2:01
    
I feel there is something fundamentally wrong with what I said. Write $x = k e^{i \pi}$. Then $$r = x^{\frac{1}{y}} = k^{\frac{1}{y}} e^{i \pi/y + 2\pi n}.$$ For $r \in \mathbb{R}$ we have: $$ \pi/y + 2\pi n \equiv \pi m $$ where both $m,n \in \mathbb{Z}.$ Which does not infer the condition on $y$ being odd. –  user2468 Mar 2 '12 at 2:20
1  
@J.D.: $x = ke^{i\pi+2\pi n} \Rightarrow r = k^{1/y} e^{i\pi \frac{1+2n}{y}} \Rightarrow y = (1+2n)/k$ where $n$ and $k$ are integers and $k \neq 0$ if you want $r$ to be real. –  Antonio Vargas Mar 2 '12 at 2:38
1  
In my above comment I didn't mean to use $k$ twice... instead take $y = (1+2n)/m$. –  Antonio Vargas Mar 2 '12 at 5:22
add comment

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.