The Number e and the Exponential Function
Disclaimer: these notes are not mathematically rigorous. Instead, they present quick, and, I hope, plausible, derivations of the properties of and the natural logarithm.
Consider the following series: where runs through the positive integers. What happens as gets very large?
It’s easy to find out with a calculator using the function x^y. The first three terms are 2, 2.25, 2.37. You can use your calculator to confirm that for = 10, 100, 1000, 10,000, 100,000, 1,000,000 the values of are (rounding off) 2.59, 2.70, 2.717, 2.718, 2.71827, 2.718280. These calculations strongly suggest that as goes up to infinity, goes to a definite limit. It can be proved mathematically that does go to a limit, and this limiting value is called The value of e is 2.7182818283… .
To try to get a bit more insight into for large let us expand it using the binomial theorem. Recall that the binomial theorem gives all the terms in , as follows:
To use this result to find we obviously need to put giving:
We are particularly interested in what happens to this series when gets very large, because that’s when we are approaching In that limit, tends to 1, and so does . So, for large enough we can ignore the -dependence of these early terms in the series altogether!
When we do that, the series becomes just:
And, the larger we take the more accurately the terms in the binomial series can be simplified in this way, so as goes to infinity this simple series represents the limiting value of . Therefore, must be just the sum of this infinite series.
(Notice that we can see immediately from this series that is less than 3, because 1/3! is less than 1/22, and 1/4! is less than 1/23, and so on, so the whole series adds up to less than 1 + 1 + + 1/22 + 1/23 + 1/24 + … = 3.)
The Exponential Function ex
Taking our definition of as the infinite limit of it is clear that is the infinite limit of . Let us write this another way: put so Therefore, is the infinite limit of The strategy at this point is to expand this using the binomial theorem, as above, and get a power series for
(Footnote: there is one tricky technical point. The binomial expansion is only simple if the exponent is a whole number, and for general values of won’t be. But remember we are only interested in the limit of very large so if is a rational number where and are integers, for ny multiple of will be an integer, and pretty clearly the function is continuous in so we don’t need to worry. If is an irrational number, we can approximate it arbitrarily well by a sequence of rational numbers to get the same result.)
So, we need to do the binomial expansion of where is an integerto make this clear, let us write
Notice that this has exactly the same form as the binomial expansion of in the paragraph above, except that now a power of appears in each term. Again, we are only interested in the limiting value as goes to infinity, and in this limit goes to 1, as does Thus, as we take to infinity, the dependence of each term disappears, leaving
so when we differentiate we just get back. This means is the solution to the equation and also the equation etc. More generally, replacing by in the series above gives
and now differentiating the series term by term we see etc., so the function is the solution to differential equations of the form or of the form and so on.
Instead of differentiating term by term, we could have written
where we have used in the limit
The Natural Logarithm
We define the natural logarithm function as the inverse of the exponential function, by which we mean
Notice that we’ve switched and from the paragraph above! Differentiating the exponential function in this switched notation,
That is to say,
Therefore, can be written as an integral,
You can check that this satisfies the differential equation by taking the upper limit of the integral to be then subtracting the second from the first, dividing by and taking very small. But why have I taken the lower limit of the integral to be 1? In solving the differential equation in this way, I could have set the lower limit to be any constant and it would still be a solutionbut it would not be the inverse function to unless the integral has lower limit 1, since that gives for the value that We need this to be true to be consistent with since
Exercise: show from the integral form of that for small is approximately equal to Check with your calculator to see how accurate this is for