- #1
FranzDiCoccio
- 342
- 41
Hi,
I know that, at least formally, the index of a radical should be positive and integer. That is if I introduce
[tex]\sqrt[x]{2}[/tex]
I need to assume [itex]x\in \mathbb N[/itex] and [itex]x>0[/itex].
However, my calculator has no problem in calculating the radical for any [itex]x\neq 0[/itex], say [itex]x=-\pi[/itex].
The result it gives is based on the assumption
[tex]\sqrt[x]{2} = 2^{\frac{1}{x}}[/tex]
and on the fact that the exponent of the exponential function can be any number. Therefore
[tex]\sqrt[-\pi]{2} = 2^{-\frac{1}{\pi}}\approx 0.802[/tex]
It seems to me that the above "assumption" that a radical can be replaced by an exponential holds true in any case, provided that the radicand is positive.
I wonder whether I'm overlooking some strange case where the above assumption fails.
I do not see any, so I guess that the bounds [itex]x\in \mathbb N[/itex] and [itex]x>0[/itex]. are just formal.
I think that the idea is that it is not really worth bothering with weird indexes in radicals.
Those really interested in [itex]\sqrt[x]{2}[/itex] for any [itex]x[/itex], should just stop using radicals and work with exponentials only.
Is that it, or is there more to it?
Thanks a lot for your input
Franz
I know that, at least formally, the index of a radical should be positive and integer. That is if I introduce
[tex]\sqrt[x]{2}[/tex]
I need to assume [itex]x\in \mathbb N[/itex] and [itex]x>0[/itex].
However, my calculator has no problem in calculating the radical for any [itex]x\neq 0[/itex], say [itex]x=-\pi[/itex].
The result it gives is based on the assumption
[tex]\sqrt[x]{2} = 2^{\frac{1}{x}}[/tex]
and on the fact that the exponent of the exponential function can be any number. Therefore
[tex]\sqrt[-\pi]{2} = 2^{-\frac{1}{\pi}}\approx 0.802[/tex]
It seems to me that the above "assumption" that a radical can be replaced by an exponential holds true in any case, provided that the radicand is positive.
I wonder whether I'm overlooking some strange case where the above assumption fails.
I do not see any, so I guess that the bounds [itex]x\in \mathbb N[/itex] and [itex]x>0[/itex]. are just formal.
I think that the idea is that it is not really worth bothering with weird indexes in radicals.
Those really interested in [itex]\sqrt[x]{2}[/itex] for any [itex]x[/itex], should just stop using radicals and work with exponentials only.
Is that it, or is there more to it?
Thanks a lot for your input
Franz