Topics From “Precalculus”: Part 2 (Explaining Calculus #7)

Because of the directions I’d like to be able to go later on in this series, we need to make an aside about more topics that aren’t specifically calculus related. If you’ve made it this far, great for you! You’re now getting into the realm of calculus. This is a fun ride, and it is sometimes good to slow down and appreciate what we’ve done.

We now know how to compute slopes of curved shapes. Just as the slope of a line tells us a lot of information, so too the derivative of a graph carries important meaning. The discovery of calculus by such individuals as Newton and Leibniz, along with those precursors who came close like Pascal and Fermat, was one of the key components in the growth of science. Every single field of science you could imagine involves calculus in either simple or advanced forms as an indispensible component without which that science could not be done in its modern form. We are, to borrow the phrase Newton himself used at the end of one of his greatest works, playing on the shore, finding some nice looking shells, while the great ocean of calculus lies undiscovered beyond us.

Because of the importance of the methods of calculus, surely we want to bring calculus into everything we possibly can. This means, of course, expanding our list of functions beyond those discussed in the first “Topics from Precalculus” post in this series (see the link at the end for that article). In particular, we will lay out some basic trigonometry along with a review of exponential functions and their cousin, the logarithmic functions.

Exponential Function Review

Exponential functions were discussed in the first installment of precalculus topics, so I’ll make this one quick. An exponential function is one that looks like f(x) = b^x for a constant b. For positive whole numbers, exponentials are defined by repeated multiplication, so b^2 = b*b, b^3 = b*b*b, and so on. For negative whole numbers, exponentials are defined by division, so b^{-1} = 1/b, b^{-2} = 1/b^2, and so on. In light of the equation b^x b^y = b^{x+y} that is true for all exponentials, we can say b^0 = 1. If we want to do exponentials with fractional exponents, like b^{x/y}, we define b^{x/y} as the number that satisfies the equation (b^{x/y})^y = b^x. If you want to define exponentials for exponents that are irrational, say like b^{\sqrt{2}}, that is basically a matter of “smoothing out” the values you’ve already defined. This is too detailed to be worth talking about here.

A New and Important Number

While talking about exponentials, there is an important number that comes up a lot in calculus that we should go ahead and define. This number is often associated with the word natural because, in the world of calculus, using this number turns out to make things easier than they otherwise would be. This number is abbreviated e, and is usually defined by the rather seemingly-esoteric equation

e = \lim\limits_{n \to \infty} \bigg( 1 + \dfrac{1}{n} \bigg)^n.

You don’t really have to know all that much about e yet – we will bring up important facts about this very special number as we need them. The curious reader could do some research on e – there is a lot of very interesting material there. The key property of e for the purposes of calculus will emerge in the post after this one.

Logarithmic Functions

We have by now a pretty good intuition for what inverse (or perhaps reverse) process is. In mathematics, the two most obvious examples are addition/subtraction and multiplication/division. If you add two, you can undo that by subtracting two. If you multiply by 2, you undo that by dividing by 2. In the same way, exponential functions also have “reverse functions”. In mathematical lingo, we normally call these “inverse functions,” but the word reverse serves the same purpose if it helps you understand better. The inverse function to the exponential function has the name of logarithm.

For every exponential function b^x, there is an logarithm function \log_b(x). To explain the key property of logarithms, it may help to use a simpler example first – addition and subtraction. I mentioned before that adding two and subtracting two are inverses. This means that the “inverse function” of f(x) = x+2 must be g(x) = x - 2. The way you can tell that these are “inverses” is that when you ‘execute’ one of these functions followed by the other, you wind up where you started. In math-terms,

f(g(x)) = f(x-2) = (x-2)+2 = x \text{ and } g(f(x)) = g(x+2) = (x+2)-2 = x.

Notice that we initially plug in x, and that is also exactly what we get out. This is what we mean by “inverses.” Then, in this way, the logarithm is the special function \log_b(x) for which b^{\log_b(x)} = x and \log_b(b^x) = x. Because of the extremely close relationship between logarithms and exponential functions, all the rules of exponents have counterpart rules for logarithms. Let me show how this work:

Fact: \log_b(xy) = \log_b(x) + \log_b(y).

Proof: First, remember that b^x = b^y only if x = y. So, to prove the identity we want, we can instead prove that b^{\log_b(xy)} = b^{\log_b(x) + \log_b(y)}. Using the foundational property of logarithms, we know immediately that b^{\log_b(xy)} = xy. The same fundamental property, along with the rule b^{x+y} = b^x b^y for exponentials, leads us to conclude that b^{\log_b(x) + \log_b(y)} = b^{\log_b(x)} b^{\log_b(y)} = xy. Combining everything we’ve done shows that the original equation must be true.

By using the same essential strategy, we can prove that log_b(a^x) = x \log_b(a) for any numbers a,x. This is the equivalent of the rule (b^a)^x = b^{ax}. For every rule of exponents, you can find a similar rule for logarithms. That is all we need to know about logarithms in general for now.

The last thing we need is to briefly comment on a special logarithm. The logarithm associated to the special number e from earlier is called the natural logarithm. We could write it as \log_e(x), but it comes up so frequently that very often mathematicians just use \ln(x) or \log(x) as shorthand. Thus, if you see a logarithm without a base, the implied base is the number e.

Trigonometric Functions

Here, we have to delve a bit into the geometry of triangles and circles. Recall the Pythagorean theorem – that for a right triangle with sides a \leq b < c the equation a^2 + b^2 = c^2 is always true. Trigonometry arises from this equation. In particular, if we divide both sides by c^2, then we conclude that

1 = \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} = \bigg(\dfrac{a}{c}\bigg)^2 + \bigg(\dfrac{b}{c}\bigg)^2.

In geometry class, the ratios between these sides are called sine and cosine. The ratio between b and a is called the tangent. Instead of using ratios, I’ll define the sine and cosine in a slightly different way.

Set up the unit circle – that is, a circle with center at the origin point (0,0) and radius equal to 1. Pick out the point (1,0) on the circle as the “starting point.” Now, imagine rotating around the circle, counterclockwise, along x units of the perimeter. Then we define the functions \sin{x}, \cos{x} so that the point you land on has the coordinates (\cos{x}, \sin{x}). This is actually the same definition as the fractions used before, with the added twist that side lengths that go left or down are given a negative sign and those that go right or up have positive sign.

For example, rotating a length of \dfrac{\pi}{6}, which is 30 degrees, the resulting triangle has a base of \dfrac{\sqrt{3}}{2} and a height of \dfrac{1}{2}. This means that \sin{\dfrac{\pi}{6}} = \dfrac{1}{2} and \cos{\dfrac{\pi}{6}} = \dfrac{\sqrt{3}}{2}. If you had instead rotated 30 degrees the other direction, we would find that \sin{\dfrac{-\pi}{6}} = \dfrac{-1}{2} and \cos{\dfrac{-\pi}{6}} = \dfrac{\sqrt{3}}{2}.

We won’t have much occasion for actually calculating values of these functions, so we won’t do too much of that. There is really only one fact that will be helpful to point out (so we can use it to calculate a derivative later). We want to be able to calculate \sin{(x+y)} and \cos{(x+y)}. Here is how we do that

Fact: For any values of x,y, we have

\sin{(x+y)} = \sin{(x)}\cos{(y)} + \sin{(y)} \cos{(x)}


\cos{(x+y)} = \cos{(x)}\cos{(y)} - \sin{(x)}\sin{(y)}.

The proof of this fact comes from geometry, it isn’t terribly complicate but goes too far astray, so I won’t prove it here. The only other thing we must talk about in passing are the other trigonometric functions, all of which are built out of sine and cosine. These are the tangent, cotangent, secant, and cosecant. They are defined, in that order, below:

\tan{x} = \dfrac{\sin{x}}{\cos{x}}, \hspace{0.3in} \cot{x} = \dfrac{\cos{x}}{\sin{x}}, \hspace{0.3in} \sec{x} = \dfrac{1}{\cos{x}}, \hspace{0.3in} \csc{x} = \dfrac{1}{\sin{x}}.

Along with the identity \sin^2{x} + \cos^2{x} = 1 (which is just the Pythagorean theorem), this is all we need from the realm of “pre-calculus” topics in order to move forward.


First “Precalculus” Post: in a new tab)

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