Skip to 0 minutes and 14 secondsWe have arrived at the last section of course. I hope you've learned a lot about linear relations, quadratic relations, inverse relations, and power law relations too. We should have a good familiarity now with these really essential aspects of Pre-Calculus, from the geometrical and algebraic point of view and with lots of examples to support your intuition and understanding. It's important, however, to realise that when it comes to relationships between quantities in the real world, what we've gone through in this course is just part of the story. The story goes further once we introduce calculus. Calculus is an important 17th century development, which really empowers us to think in novel ways beyond the algebraic set up so far.

Skip to 1 minute and 7 secondsWe've already had some glimpses of that when we were talking about the log function. I remind you that the log function was intimately connected with the inverse relation, y equals 1/x. In fact, if we looked at the area underneath this 1/x function, from 1 to a certain point, that cumulative area defines for us this log function. So it's a function that's defined not algebraically, but in terms of an area operation from an elementary function. It's certainly a step up because the notion of area involves calculus ideas. Very closely related to the log function is its inverse function. That's what you get when you reflect the graph in the line y equals x.

Skip to 1 minute and 59 secondsThat's the famous exponential function called y equals e to the x. Now, you may have seen the exponential function defined in other ways. In fact, there are quite a few different ways of introducing the exponential function. But this is one of them. You can think of it as being the inverse of the graph of y equals ln x, which is obtained from 1/x by looking at areas. Once we have a log and the exponential function under our belts, then we can look at a lot of other kinds of relations. And the story, then, becomes richer and more involved.

Skip to 2 minutes and 41 secondsJust to give you one example that we'll have a look at in these activities is that-- this question that was asked by Arthur Cayley, which actually comes from graph theory. He asked, if you have four points labelled 1, 2, 3, 4, and you want to create a tree on that four points-- that is you want to join the points with line segments, so that there's no loops, a graph without any cycles-- then the number of trees on n labelled vertices is n to the n minus 2? So in case n is 4, you would get 4 squared, or 16. So this is really going beyond this kind of relation.

Skip to 3 minutes and 25 secondsThere's a power here, but the power itself depends on n, so it's really combining this kind of idea with the exponential kind of idea as well. So we've had an overview of a lot of important Pre-Calculus notions, and you've strengthened your geometrical understanding, algebraic manipulation power, and also connected with quite a lot of actual real life applications so that you have some appreciation of the connection between mathematics and the world in which we actually live. So for those of you who are going to go on to do calculus, well, we've already had a taste of that, the way we introduce the log function here in terms of areas under 1/x.

Skip to 4 minutes and 8 secondsBut you'll be able to go on and learn about new functions and applications really building on that solid foundation that you've learned about in this course.

# Beyond power and polynomial relations

In this video we discuss other kinds of relations going beyond the elementary polynomial and power laws, culminating with a formula of Cayley on the number of labelled trees on \(\normalsize{n}\) vertices.

## The logarithm function \(\normalsize y=\ln x\) and its inverse

We’ve already seen that the (natural) logarithm function \(\normalsize y=\ln x\) could be defined, at least approximately, as an area under the hyperbolic function \(\normalsize y=1/x\). And we’ve seen that purely on geometric grounds, using only the basic scaling associated with the hyperbola, that

\[\Large \ln (ab)=\ln a + \ln b. \tag{1}\label{1}\]The inverse function to the logarithm is \(\normalsize y=e^x\), called the *exponential function*, which is defined by the condition that

It follows that for any non-zero positive value \(\normalsize x\) we have the relation

\[\Large e^{ln x}=x . \tag{3}\label{3}\]Geometrically the meaning is that the graph of the exponential function and the logarithm function are reflections of each other in the line \(\normalsize y=x\).

An important property of the exponential function is that

\[\Large e^0=1. \tag{4}\label{4}\]

Q1(E): Explain why this formula holds.

An even more important property is that

\[\Large e^{(x+y)}=e^x e^y \tag{5}\label{5}\]for all \(\normalsize x\) and \(\normalsize y\). So the following is a good exercise, requiring a bit of conceptual thinking.

Q2(M): Using equations \((\ref{1})\), \((\ref{2})\) and \((\ref{3})\), prove that \((\ref{5})\) holds.

The laws \((\ref{4})\) and \((\ref{5})\) show that this function \(\normalsize y=e^x\) behaves like the more familiar exponential functions \(\normalsize y=2^x\) or \(\normalsize y=3^x\).

The diagram provides some evidence that there is indeed a remarkable approximate value, sometime known as *Euler’s number*:

In actual life, the exponential and logarithm functions, although they cannot in general by calculated precisely like the polynomial functions, are very useful for a wide variety of applications that go beyond the ones we have looked at in this course. In this final section of the course, we give some indications of such possible directions.

## Answers

A1.Recall from the definition that \(\normalsize \ln x\) is the area under the hyperbola \(\normalsize y=1/t\) from \(\normalsize t=1\) to \(\normalsize t=x\). So \(\normalsize \ln 1 = 0\) which is to say \(\normalsize e^0 = 1\).\[\Large \ln(ab) = \ln a + \ln b.\]

A2.We use equations \(\normalsize (\ref{1})\) and \(\normalsize (\ref{2})\) in combination. Starting withExponentiate both sides

\[\Large ab = e^{\ln a + \ln b}.\]Now let \(\normalsize a = e^x\) and \(\normalsize b = e^y\), so that

\[\Large e^x e^y = e^{\ln(e^x) + \ln(e^y)} = e^{x+y}.\]

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