0

The expression $\displaystyle (-10)^\frac{1}{3}$ returns errror when calculated with scientific calculator returns an imaginary number. But it's power is $\dfrac{1}{\text{odd number}}$.

V.G
  • 4,196
Arpit Bhalla
  • 123
  • My scientific calculator shows it's value. Your's calculator must be programmed for evaluating powers of positive numbers only. – V.G Sep 02 '20 at 05:15
  • Read the first section of this page : https://wiki.analytica.com/Exponentiation_of_negative_numbers – Sarvesh Ravichandran Iyer Sep 02 '20 at 05:15
  • 1
    Every nonzero complex number has three cube roots. Looks like your calculator's favourite cube root of $-10$ isn't the same as your favourite one. – Angina Seng Sep 02 '20 at 05:17
  • Also https://math.stackexchange.com/questions/1602160/why-do-i-get-an-imaginary-result-for-the-cube-root-of-a-negative-number – Chris Culter Sep 02 '20 at 06:01

2 Answers2

2

Most calculators utilize the $log$ function to calculate powers, using$$x^y=e^{y\ln x}$$ given that $e^x$ and $\ln x$ are calculated quickly thanks to the math (embedded) processor.

A negative $x$ is either treated directly (using $i^2=-1$) or it's made positive first, and the negative is dealt with after.

In your case, probably that $(-10)^{\frac13}$ is calculated as $e^{\frac13 \ln(-10)}$, and $\ln(-10)$ would be calculated as $\ln(10i^2)$

Déjà vu
  • 934
2

It might be giving one of the complex roots of the number for some reason. For example $x^3 = -8$ has three solutions $-2, 1-i\sqrt{3}, 1+i\sqrt{3}$.

Q the Platypus
  • 4,626