The discussion centers on the evaluation of the expression ##\sqrt[a]{-1}## as the variable ##a## varies over real numbers. It is established that for even integers (e.g., ##a = 2##), the result is complex, while for odd integers (e.g., ##a = 3##), the result is real and equals -1. The conversation also clarifies that there is no standard definition for the "π-th root" of a number, as the index of a radical must be a positive integer of 2 or larger. The logarithmic representation of negative bases is discussed, emphasizing that the natural logarithm is only defined for positive real numbers.
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As far as I know, there is no such thing as the "π-th root" of a number. The index of a radical is a positive integer that is 2 or larger. You can however raise a number to an arbitrary power. For example, ##2^{1/\pi} = (e^{\ln 2})^{1/\pi} = e^{\frac{\ln 2}{\pi}}##, but see the link that @SlowThinker posted.Mr Davis 97 said:How does the value of ##\displaystyle \sqrt[a]{-1}## vary as ##a## varies as any real number? When is this value complex and when is it real? For example, we know that when a = 2 it is complex, but when a = 3 it is real. What about when ##a = \pi##, for example?
if ##x^{1/a} = e^{\frac{\ln x}{a}}##, and we know that ##(-1)^{1/3} = -1##, does that mean that ##(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1##?Mark44 said:As far as I know, there is no such thing as the "π-th root" of a number. The index of a radical is a positive integer that is 2 or larger. You can however raise a number to an arbitrary power. For example, ##2^{1/\pi} = (e^{\ln 2})^{1/\pi} = e^{\frac{\ln 2}{\pi}}##, but see the link that @SlowThinker posted.
Not if ln means the usual natural logarithm function whose domain is positive real numbers.Mr Davis 97 said:if ##x^{1/a} = e^{\frac{\ln x}{a}}##, and we know that ##(-1)^{1/3} = -1##, does that mean that ##(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1##?
##(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}=-1## this is correct but you don't consider the other results,Mr Davis 97 said:if ##x^{1/a} = e^{\frac{\ln x}{a}}##, and we know that ##(-1)^{1/3} = -1##, does that mean that ##(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1##?
No. ##\sqrt 4## is generally accepted to mean the principal square root of 4, a positive number.MAGNIBORO said:##(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}=-1## this is correct but you don't consider the other results,
##\sqrt4=2## is correct but not complete, because ##\sqrt4=\pm 2##
MAGNIBORO said:in the same way are other results for ##(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}##
look the post https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
Mr Davis 97 said:if ##x^{1/a} = e^{\frac{\ln x}{a}}##