What is the significance of 3^π in understanding irrational powers?

[Shing]

Hi guys,
I am wondering what the physical meaning of a irrational power?
like3^\pi?

3 times itself ... times?

 

[arildno]

An excellent question!

Basically, you must generalize the concept of "power" beyond that which gives meaning when the exponent is a natural number.

In essence, 3^{\pi} is, by definition:
3^{\pi}\equiv\sum_{n=0}^{\infty}\frac{(\pi\ln(3))^{n}}{n!}
where n! is the factorial of n (0!=1), and ln(3) is the natural logarithm of 3.

Note that the powers in this sum are natural numbers, defined in the usual manner.

 

[statdad]

You can also define 3^\pi to be

<br /> \lim_{n \to \infty} 3^{a_n}<br />

for any sequence {a_n} that converges to \pi. (It must be shown that the limit exists, and doesn't depend on the particular sequence, however), but this means you can take a sequence of rational numbers, so at each step you are dealing with 3 to a rational power.

 

[Shing]

An excellent question!

Basically, you must generalize the concept of "power" beyond that which gives meaning when the exponent is a natural number.

In essence, 3^{\pi} is, by definition:
3^{\pi}\equiv\sum_{n=0}^{\infty}\frac{(\pi\ln(3))^{n}}{n!}
where n! is the factorial of n (0!=1), and ln(3) is the natural logarithm of 3.

Note that the powers in this sum are natural numbers, defined in the usual manner.

thank you very much to you both!

however...correct me if I am in the wrong; is it Tayor's theorem? ( I haven't learned Tayor's theorem yet,)
But isn't it an approximation when x=\pi?

 

[Feldoh]

Well arildno's solution is a taylor series, however the answer he gave is exact and not an approximation.

 

[HallsofIvy]

I, unlike arildno, think this is NOT an "excellent question". Generally speaking it is not a good idea to ask for "physical meanings" for mathematics concepts. Physical meanings are added to mathematical concepts for specific applications. Drop the word "physical" and just ask for the "meaning", i.e. the definition of a^x for irrational x and it is an "excellent question"!

Here's what I have done in pre-calculus classes:

Define aⁿ, for n a positive integer ("counting number"), to be (a)(a)(a)\cdot\cdot\cdot(a)(a), multiplying a by itself n times. Then show, using counting techniques, that a^na^m= a^{m+n} and (a^n)^m= a^{mn}.

Now, in order that those nice formulas still be true if, say, n= 0, we must have a^0a^m= a^{m+0}= a^m. As long as a is not 0, we can divide through by a^m and get a^0= 1. So we require that a be non-zero and define a^0 to be 1.

In order that those formulas still be true for n negative, we use the fact that any negative integer can be written as "-n" where n is a positive integer and n+ (-n)= 0.
a^na^{-n}= a^{n+(-n)}= a^0= 1. Dividing through by a^n we have a^{-n}= 1/a^n. That is, in order that a^na^m= a^{m+n}
be true for all integers m and n, we define e^{-n} to be 1/e^n.

Any rational number can be written as a fraction, m/n for some integers m and n (and n non-zero). Now we look at (a^m)^n= a^{mn} with m= 1/n: (a^{1/n})^n= a^{n(1/n)}= a^1= a. In order that that formula still be true we must have a^{1/n} is a number such that its n th power is a. But there is no number whose square is -4 so if we want this to work for all n, we must require that a be positive. Then we define a^{1/n} to be a number whose nth power is a and, in order that this be a single-valued function we define it to be the positive such number: \sqrt[n]{a}.

Then, of course, requiring that a^{m/n}= a^{m(1/n)}= (a^m)^{1/n}, we define a^{m/n} to be \sqrt[n]{a^m}.

Given any real number x, we can show that there exist at least one sequence of rational numbers that converges to x and that if \{r_n\} is a sequence of rational number converging to x, then, for positive a, the sequence \{a^{r_n}\} converges and all such sequences converge to the same thing. We then define a^x to be the limit of that sequence. That is, essentially, defining a to an irrational power in such a way as to make a^x a continuous function for all x.