What does ‘a power b’ mean?


How do you define a^b?

My daughter has just finished her 10th and has taken up humanities group in +2. However, I wanted her to learn Maths at home and have started teaching her Trigonometry, Complex Numbers and Calculus. While introducing functions, I noticed that she has already seen all kinds of algebraic functions (composed of integral powers of x) and trigonometric functions. But e^x and subsequently \ln(x) were alien to her.

In this connection, I needed to touch upon a^b.

It is easy to define the above when b is a positive integer. a could be any real number. We need not worry about whether it is an integer or a rational number etc. It could be any real number.

a^2 = a . a

a^3 = a . a . a

a^b = a . a ... a (b times)

If b is a negative integer, then there exists m, a positive integer such that -b = m. Then,

a^{-2} = (\frac{1}{a})^2 = \frac{1}{a} . \frac{1}{a}

a^{-3} = (\frac{1}{a})^3 = \frac{1}{a} . \frac{1}{a} . \frac{1}{a}

a^b = (\frac{1}{a})^{m} = \frac{1}{a} . \frac{1}{a} ... \frac{1}{a} (m times)

This is fairly easy to explain.

What if b is a rational fraction between 0 and 1? That is, b = 1/m where m is a positive integer.

a^{(1/2)} = \sqrt{a}

a^{(1/3)} = \sqrt[3]{a}

a^b = a^{(1/m)} = \sqrt[m]{a}

All that we need to know is, given any real number a, it is possible for us to find another real number c, whose mth power is a. That is, c^m = a.

Extending this to b being any rational number of the form b = m/n, where m, n are integers, we can well define

a^b = a^{(m/n)} = (\sqrt[n]{a})^m

Now let us take the next jump. What if b is not rational? Then what does a^b mean? How do you visualise it? What is the meaning of a^{\pi} or a^{\sqrt{2}}? Note that you have to do this without resorting to logarithms because we have not talked about them yet.

One way of defining this is through limits. Given any irrational number, we can find a rational number close to that to any degree of accuracy. a^b is well defined for b being a rational number. By selecting rational numbers close enough to the given irrational number, you can calculate the value of a^b. Thus you can extend this to all the real numbers and define a^b where both a, b are real.

Once you understand this, then you can understand a function like e^x. Then you define \ln(x) from this.

Actually, you can extend this to the complex numbers as well. I will only show this for a simple case, when a=b=i. That is, what is the meaning of i^i where i has the usual meaning, that is, i^2 = -1.

Suppose, y = i^i.

Then, \ln y = i \ln i

Here, use e^{ix} = \cos x + i \sin x.

For x = \pi/2, this would be, e^{i \pi/2} = i

Alternately, i \frac{\pi}{2} = \ln i

(Note that, e^{i \pi/2} is one of the numbers that will result in i. All of

x = \frac{\pi}{2} + 2n \pi

where n is an integer will satisfy this. We will take the simplest solution.)

Substituting the above in the previous equation,

\ln y = i \ln i = i . i \frac{\pi}{2}

Or, \ln y = - \frac{\pi}{2}

Or, y = i^i = e^{-\pi/2} (being one answer).

So it is possible to make sense of a complex number to the power of another complex number as well.

3 thoughts on “What does ‘a power b’ mean?

  1. When b is real, and a is complex, the case is straightforward – it follows the pattern of repeated multiplications where b is integer, or repeated multiplications of numerator and denominator when b is rational, or an approximation when b is irrational.
    When b is complex, or purely imaginary as i ^ i, , isn’t it somewhat circular to use exp(ix) = cos(x) + i sin(x) ?
    Here a is e, and b is ix. Is it a leap of faith to say a ^ b works the same way for imaginary /complex b also?

  2. /// However, I wanted her to learn Maths at home and have started teaching her Trigonometry, Complex Numbers and Calculus. ///

    Why this kolaveri 🙂

  3. Nagarajan: I am simply demonstrating how a^b works if b is complex. I am not doing any leap of faith thing here. e^(ix) can be used anywhere you want – it is not circuitous. It is quite logical. Just check google on i^i and you will see so many links:-)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s