# X^2=X*X, so X^1/2=?

If X[sup]2[/sup] is just X times X, what does X[sup]1/2[/sup] mean? I know it’s the square root, but what does it actually mean? Anything? Is there a way to figure out the nth root of something without using derivatives? That is, what operations does a calculator do?

A calculator can easily do square roots without derivatives or logarithms. The algorithm is trivial.

As for other exponents (e.g. 5.17 raised to the 3.14 power) it uses logarithms. In the above example, it takes the natural logarithm of 5.17 (1.643), multiplies it by 3.14 (5.159). Then, raises e to that power: e^5.159= 173.92. The algorithm for logarithm and anti-logarithm are not as simple as square roots.

If you square X^n, you get X^2n. This could be used to justify the question you ask.

There is a process a bit like long division that you can use to find the square root of a number. While it’s not difficult, it is sort of hard to explain here. I’ll try if you want.

Logarithmic tables could presumably be used to find roots, remember that e^x has a very simple expansion series.

Derivatives can easily be used using hte power theorem; you could probably get a decent answer using McLaurin series and once you have done it once you have the formula any time you want it… so you only need to use derivatives once.

If you want some justification why X[sup]1/2[/sup] should be the square root of X, here you go:
X[sup]n[/sup] where n is a positive integer is defined to be X times itself n times. Under this definition, X[sup]1/2[/sup] is not defined. We wish to extend the defintion to allow for non-integral powers, but we would like to still have X[sup]n[/sup] still be the product of n copies of X if n is an integer. We notice that:
X[sup]n[/sup]X[sup]m[/sup] = X[sup]n+m[/sup]
If we allow non-integral powers and set n and m in the above equality to 1/2, we get:
X[sup]1/2[/sup]X[sup]1/2[/sup] = X[sup]1/2+1/2[/sup] = X[sup]1[/sup] = X
So, X[sup]1/2[/sup] times itself is X.