Back to Riemann, Part 3

I’ve already talked about “i” as a placeholder that represents rotation around the origin, and how f(x) = x^z works when z = a + bi across the complex plane. “a” represents changes in magnitude of the vector line drawn from the origin to the point x^(a+bi), and “bi” contributes to the angle of that vector. As “b” varies, the angle varies (the angle is referred to as the “argument” of the function).

So, how does e^i tie into all this? In the last entry, I introduced the natural logarithm ln(), and said that it is the log of base e. And I talked about real exponents from minus to plus infinity. Before we go to the next step, for imaginary exponents of base e, let’s look at “e” a bit more.

Jacob Bernoulli (1655-1705) was a Swiss mathematician. In the early 1680’s he was studying a problem on compound interest. If you have simple interest, then you spend whatever money you make on your investment, and your investment always makes you the same amount each time. That is, $100 at 5% annually is $5. After 10 years, you still have your initial $100 investment, plus the extra $50 in interest which you’d spent on coffee at the end of each year. However, if you put the interest you make back into savings, then you get something like this for those same 10 years:

100 ———— 5
105 ———— 5.25
110.25 ——— 5.5125
115.7625 ——- 5.788125
121.550625 —- 6.07753125
127.6281563 — 6.381407813
134.0095641 — 6.700478203
140.7100423 — 7.035502113
147.7455444 — 7.387277219
155.1328216 — 7.75664108

As can be seen here, you’re making interest on your interest, plus the principle, and the result is that your total net worth goes up. The question, though, is what happens when you calculate the interest monthly rather than annually, or even if you calculate it daily. Bernoulli used the formula:

Which is read as taking the limit of (1 + 1/n) ^ n as “n” goes to infinity. In his example, the account pays 100% interest annually. If the payments are made every 6 months, at 5%, then my formula becomes $100 * (1 + 0.05/2)^2, or $105.06 for the year. Monthly, it’s $100 * (1 + 0.05/12)^12 = $105.10. Granted, it’s not much of an extra gain, but it does show that the more often you calculate interest during the compounding period, the more money you make. So, compounding daily would give you $105.13 in the first year. Now, what Bernoulli did was to use that 100% interest value, and he noticed that the formula tended to reach a limit as “n” went to infinity, specifically, that limit was 2.7182818. (As an approximation, calculating with 100% on one dollar daily for one year, $1 * (1 + 1/365)^365) = 2.707.)

The more general rule is that compounding = e^Rt, where R is the rate as a decimal value (5% => R = 0.05), and t is the number of years you’re going to hold the account.

If the principal is $100, R = 0.05, and we want to save the money for 10 years, the total with compounded interest would be,
y = 100 * e^(0.05 * 10) = $164.87

According to the wiki entry, the first known reference to this constant was in correspondence from Gottfried Leibniz to Christiaan Huygens in 1690 and 1691, where it was given the letter “b”. Leonhard Euler assigned it to the symbol epsilon (e) in 1731 as the base for the natural logarithm, ln().

One of the starting points in teaching calculus is the idea of measuring the area of a curve by summing the areas of rectangles under the curve that all have the same width (I talked about this in the Riemann Prose series, and it’s the basis of what Bernoulli did above in finding e). The larger the number of boxes, the smaller their widths and the closer the sum total gets to the actual area under that curve. When the number of boxes approaches infinity, their widths approach 0, as does their individual areas. Doing this is called taking the derivative of that equation, f(x). So, the derivative of f(x) -> f'(x) = d/dx f(x).

If f(x) = x^2, then the derivative is 2 * x. The derivative of 2 * x is 2. If you drop a rock from the top of a tall building and measure how long it takes to reach the bottom, and you figure that gravity is an acceleration constant of g feet per second per second, then if the rock covers 400 feet in 5 seconds, and:

x = 1/2 * g * t^2
velocity at time t = g * t
acceleration = g

g = 2 * x / t^2 = 2 * 400 / 5^2 = 32 feet per sec^2
velocity = 32 * 5 = 160 feet per second.

The general rule of taking the derivative of an exponential function in the form of x ^ b is to multiply the function by “b”, and subtract 1 from the exponent.

That is:
d/dx of the function f(x) = x^2 is: 2 * x
d/dx of the function f(x) = x^3 is: 3 * x^2

The reason to go through all this is that e^x is a special case.
d/dx e^x = e^x

And, d/dx of ln(x) = 1/x

To put it graphically, when you draw a line through 2 points on a curve (x1,y1) and (x2,y2), the difference between the two points (x2 – x1, y2 – y1) is the slope of that line running through those points. And, as those points get closer together, the line more closely approximates the tangent line to that curve. Therefore, taking the derivative f'(x) of a function f(x) yields the line tangent to that curve at point x. And, the slope of the curve e^x at point x is the line defined by e^x.

(Graph of e^x for 0.1 <= x <= 3.0, and the tangent line for the derivative at an arbitrary point x ~ 2.3.)

Ok, so we now know one way to calculate e, why e is useful, and how to take the derivatives of both e^x and ln(x).

Let’s get back on topic for e^bi.
Next week.

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