# Irrational Exponents

1. Jun 12, 2010

How can we find irrational exponents of a real or imaginary number?? say 2 raise to the power pi

2. Jun 12, 2010

### phyzguy

We would normally do this as follows:
$$2^\pi = e^{\pi log(2)} ~= 8.825...$$

3. Jun 12, 2010

But this is also a irrational exponent of an irrational number. How we calculate it. I heard we can find it using limits but how i don't know.

4. Jun 12, 2010

### arildno

Okay!

Essentially, what you need to do will always involve:

1. Set up a procedure of nested limiting process, so that you at each step only use rational numbers.

2. Make theoretical investigations to ascertain:
a) That the procedure made in 1 converges theoretically

b) Find a way to estimate the maximal error any finite step in your algorithm is going to make

c) Decide upon the level of accuracy you want.

These points can be said to hold for every algorithm you might choose from.

In order to decide which algorithm you should choose, it is rational to choose the most cost-effective one, i.e, the one that yields greatest accuracy for fewest steps.

Much of what we call computational/applied maths is concerned with developing/improving such algorithms.

In your <i>particular</i> case, involving nested limiting procedures, remember that an "outer" algorithm cannot hope to achieve a greater accuracy than the maximal error already present in the inner algorithm!

This will, essentially, set a bound on the number of finite steps it will be meaningful to compute in the outer algorithm, whereas the number of steps in the inner algorithm will be chosen to suit overall speed&accuracy requiremements.

One such way to do this, is to compute your exponential by the identities:
$$2^{\pi}=e^{\pi\ln(2)}$$ (*)
$$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}(**)$$

Here, your inner algorithm must compute $\pi\ln(2)[/tex] to a good level of accuracy. Since both [itex]\pi$ and $\ln(2)$ are irrational numbers, most likely, you must approximath separately, prior to computing an approximate product.

Then, you put that into (**), and cut off the infinite series at a suitable term.
Note that this process will always involve only rational numbers.

Last edited: Jun 12, 2010
5. Jun 12, 2010

### arildno

To make an example:

Rounded to the third decimal, we have:
$$\pi\ln(2)\approx{2}.178$$

whereas we have:
$$2^{\pi}\approx{8}.825$$

How many included terms of the series gives a solid enough approximation?

1. Term included: 1
2. Term incl. 3.178
3. Term incl. 5.549842
4.Term:.. 7.271799292
5.term: ...8,209405037494
6. Term:... 8,6178261002312
7. Term:... 8,7660829460048
8. term:... 8,812212120045898
9. Term:... 8.8247707562456
10. Term:....8,827809....

Note that in the 10th term, there has come an overshoot, due to error accumalation.

Thus, we should truncate our series after 9 terms or so, for this particular approximation.

6. Jun 12, 2010

### Anonymous217

I just wanted to thank you. I found this information really interesting!

7. Jun 12, 2010

### arildno

You're welcome!

8. Apr 18, 2011

### eczeno

not much to add to what is said above, except that we can have a little fun with irrational exponents:

there are many values of 2^pi; what has been discussed is how to find the one real value. if we allow ourselves to venture into the complex plain then there are countably infinite values distributed (densely i believe) on a circle centered at zero with radius 2^pi (the real value (to find this see above)). they are easy to find, first write

2 = 2*e^(i*k*2*pi) where k=any integer.

then,

2^pi = (2^pi)*e^(i*k*2*pi^2)

since n*pi does not equal m*pi^2 for any integers n and m, these numbers never repeat.

it should also be clear, that all these numbers, when raised to the (1/pi) power equal 2. this is not true if you use the principle value of the argument, however. i.e. :

(2^pi)*e^(i*2*pi^2) is really (2^pi)*e^(i*2*pi^2-i*6*pi) which is not 2 when raised to the (1/pi) power.