Back to Riemann, Part 2

The idea of exponents is weird, if you think about it. Take the function:

f(x) = n^x

If “n” is positive and greater than 1, then f(x) goes to 0 as x approaches negative infinity, and it goes to infinity very quickly for x-> infinity. There are two special cases: f(x) = 1 for all “n” when x=0; and f(x) = “n” when x=1. But, what’s really happening here?

(n^x for x from -3 to 3, and n=2.)

By definition, n^x is the area of a square when x=2 (n*n), and the volume of a cube when x=3 (n*n*n). When x is a positive integer > 1, then f(x) is simply the result of multiplying “n” by itself x times. By convention, f(x) becomes (1/n)^|x| if x is a negative integer. That is, you’re taking the reciprocal of “n”, and f(x) is then the result of multiplying (1/n) by itself positive x times.

If x is a fraction, such as 1/2, we treat this as taking the root of “n”. In other words, for a square of area 10, what is the length of side “n” that gives us the answer 10 when f(x) = n*n? Answer: f(x) = sqrt(10) = approx. 3.162 (3.162 x 3.162 => 10). For f(x) = 10 = n^(1/3), n = 2.154 (n*n*n = 10).

What initially seems odd about this is that we can take any random fractional root and get a reasonable answer back. This is because 1/2 = 0.5, and 1/3 = 0.33333, and the line drawn by f(x) = 10^x is continuous over the range from 0.33333 to 0.5. Plugging in x = 0.5 works just as well as x = 0.52.

Actually, x=0.52 isn’t that bad, because exponents with the same base are additive, so that 10^0.52 is the same as 10^(0.5) * 10^(0.02) = 10^(1/2) * 10^(1/50), where we multiply the square root of 10 (3.162) by the 50th root of 10 (1.047). But, still. Something like 10^3.14159 looks bizarre to me.

Anyway, negative real numbers work the same way. x = -3.14159 would be like taking the reciprocal of 10 cubed times the reciprocal of the 0.14159th power of 10 (1 / 0.14159 = the 7.06th root, so 10^1/7.06 = 1.385).

The point is that f(x) = n^x is on a continuous curve over minus infinity to plus infinity, for any real value of x.

Now, assume that I have f(x) = 3 = 10^x, and I want to find x. How do I do that?

Yes, this is where logarithms come in.

y = logb(x) for b^y = x.

where b is the base value for the exponential function. That is, if f(x) = 10^2 = 100, then 2 = log10(100).

To answer the above question, for 10^x = 3, log10(3) = x = 0.477, which is just a bit less than one half (1/2.096) (which is reasonable, because 3 is a bit under the square root of 10, so 3 = 10^0.477 makes sense).

What makes logarithms so useful is that while f(x) = 10^x is a curve that gets big really fast, x itself is a straight line with a slope of 1. This means that if you’re working with really big numbers with weird curves, it makes a lot more sense to draw the curves of the logarithms of those numbers.

I’ll end with the introduction of the natural log, or ln(). Keeping in mind that log(1) of any base number = 0 (f(x) = n^0 = 1) , and that log(0) is undefined (because it would mean f(x) = 0^x, which applies to all values of x, or that x is negative infinity), what value of “k” would we need for the area under the curve y=1/x, from x=1 to k, to equal 1?

(I may have drawn the left side of the shaded area a little too far to the right of 1.)

There are several ways to derive this value for k, but the answer at the moment is 2.71828…, and is identified by the constant “e”. Then, with “e” as the base of f(x) = y = e^x, x = loge(y). log base e is the natural logarithm, and is shortened to ln(). And, naturally, ln(e) = 1.

Logs and e also covered in Riemann Prose part 7.

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