Re: Calculation Of Exponentiated Values--How Is It Done?

From: Timothy Little (tim-via-n.i.net_at_little-possums.net)
Date: 02/05/05 

Date: 5 Feb 2005 21:57:37 GMT

Travis wrote:
> I understand 5^4 = 5 * 5 * 5 * 5 = 625. That's not the problem. That's
> easy.
> But how is 5 ^ 4.68723 calculated??? How do you get a value for that?

In practice, usually by e^(4.68723*ln(5)) with some approximation
method for e^x and ln(x).

Mathematically, x^(-n) is defined to be 1/(x^n). That extends the
definition to the integers.

For rationals, x^(n/m) is defined as the m'th root of x^n (for n, m
integers and x > 0). So your example would be the (postive real)
100000th root of 5^468723. Not very useful in practice, but that's
where the definition comes from.

For irrational y, you can choose any sequence of rationals (q_n) that
converges to y, and set x^y to be the limit of the sequence (x^{q_n}).

> Does anyone know how the above can be calculated on paper?

Messily. Very messily. Easier if you have good tables of logarithms.

> And as far as the calculator's computation, does anyone have any
> idea how that works, or know of any informative links? I assume the
> calculator operates on some sort of iterative process.

One of the easier methods to grasp is a Taylor series:
 e^x = 1 + x^2/2 + x^3/6 + ... + x^n/n! + ...

When |x| < 1 this converges rather quickly, but it does technically
work for any x. If x > 1 then you can repeatedly apply
 e^x = (e^(x/2))^2 to make it converge faster.

You can also use
 ln(x+1) = x - x^2/2 + x^3/3 - ... - (-1)^n x^n / n + ...

This is only valid for |x| < 1, and preferably |x| should be small.
If |x| > 1, you can use a similar trick to find a converging series:
 ln(x) = 2 ln(sqrt(x)).

These are far from the fastest methods, but they work and are
relatively easy to remember.

- Tim

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