2.3 The Base of the Natural Logarithm

As discussed in the previous chapter, a logarithm is an exponent of a base number that gives a desired value after the exponentiation is performed. So before proceeding to find actual values of the natural logarithm, an immediate question arises. If \(\ln x\) as defined above is a logarithm, what is the base of this logarithm? We give this special base number the symbol “\(e\)” recognizing that it will be used to describe the natural exponential operation. For any real number \(x>0\) the number \(e\) and the natural logarithm \(\ln x\) should satisfy \[ \ln x = \log_e x = p, ~~~~~{\rm if}~~~~~ x = e^p. \] This tells us that \(\ln e\) = 1, another property of any logarithm and its base. So geometrically we can see that, equivalently, if we perform our integration above between the limits of 1 to \(e\), we should arrive at a value of 1: \[ \ln e = \int_1^e \frac{1}{u}\;du \equiv 1. \] So we could, in principle, vary the value of the upper limit of the integration until a value of 1 is achieved for the integral, to an acceptable accuracy. This indeed can be done using iterative numerical methods and would provide a value for the base number \(e\). However, let’s take a slightly different approach. Let’s consider that \(\ln x\) and \(e^x\) are inverse functions. That is, given a value \(x\), if we take \(e^x\) and then take its natural logarithm, we get \(x\) back again. So, in general,
\[ \ln e^x = x. \] We now perform another calculus operation and take the derivative with respect to \(x\) on both sides of the above equation. Since our logarithm is defined as an integral of the function \(1/x\), then the derivative of \(\ln x\) will be \(1/x\). This in itself is an interesting result: \[ \frac{d}{dx} \ln x = \frac{1}{x}. \] But go back to our previous relationship, \(\ln e^x = x,\) and let’s take the derivative of both sides: \[ \frac{d}{dx} \ln e^x = \frac{d}{dx} x. \] Through the chain rule for derivatives, the left-hand side of this equation is simplified by defining the function \(v = e^x\) and then \[ \frac{d}{dx}\;\ln v = \frac{1}{v}\frac{dv}{dx} = \frac{1}{e^x} \;\frac{d}{dx}e^x. \] But since the derivative \(\frac{d}{dx}x\) is just equal to 1, then re-equating the two sides we get \[ \frac{1}{e^x} \;\frac{d}{dx}e^x = 1 \] or, \[ \frac{d}{dx}e^x = e^x. \] The derivative of the function \(e^x~~\) is \(~~e^x~!~~~~\)

This remarkable result gives us all we need to easily compute \(e\) (and \(e^x\)) to any desired accuracy. To do so, we use another result from calculus, namely a Taylor Series expansion. If the values of the derivatives of a function \(f(x)\) are known at a point \(x=a\), then it can be shown that \[ f(x) = f(a) + f'(a) (x-a) + \frac{1}{2!}f''(a)\; (x-a)^2 + \frac{1}{3!}f'''(a) \;(x-a)^3 + ... \] where \(f'(x)\), \(f''(x)\), etc., are the first, second, etc., derivatives of \(f(x)\), while \[ n! = n\cdot(n-1)\cdot(n-2)\cdot \ldots \cdot 3\cdot 2\cdot 1 \] is the factorial of \(n\). For our case we take \(a = 0\) and \(f(x) =e^x\). Then, since each derivative will be \(e^x\) as well, we find that \(f(0) = f'(0) = f''(0) =... = e^0 = 1\). We thus arrive at a straightforward expression for the exponential function in terms of an infinite series containing terms of \(x\) raised to integer powers: \[ e^x = 1 + x + \frac{1}{2!}x^2 +\frac{1}{3!}x^3+\frac{1}{4!}x^4 +\frac{1}{5!}x^5+ ... \] And, for \(x=1\), we have a way to compute the value of our natural base, \[ e = 1 + \frac{1}{1!} + \frac{1}{2!} +\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+ ... \]

Let’s compute \(e\) using a computer program. We’ll look at the results when using a variable number of terms, nTerms, in the above infinite series, up to 12 terms, and look for when the series converges to within 6 decimal places:

eEst  = cumsum( 1/factorial( seq(0,11) ) )
eErr  = eEst - exp(1)
nTerms eEst eErr
1 1.000000 -1.7182818
2 2.000000 -0.7182818
3 2.500000 -0.2182818
4 2.666667 -0.0516152
5 2.708333 -0.0099485
6 2.716667 -0.0016152
7 2.718056 -0.0002263
8 2.718254 -0.0000279
9 2.718279 -0.0000031
10 2.718281 -0.0000003
11 2.718282 0.0000000
12 2.718282 0.0000000

We see that the series converges to within 6 decimal places after about 11 terms or so, with a final value of about \(e\) = eCalc = 2.718282.

We have just computed a fairly accurate approximation of the base number for the natural logarithm.6 As a check, if we integrate \(1/x\) from 1 to \(e\) we should get 1, since \(\ln e = 1\) if it is the base number. We can do that with our computer programming language which has an integration function built in:

fx = function(x){ 1/x }

integrate(fx, 1, eCalc)
## 1 with absolute error < 1.1e-14

The base number, \(e\), for the natural logarithm is called Euler’s number, so-named after the Swiss mathematician Leonhard Euler who implemented the use of this symbol.7 It is also sometimes called Napier’s constant as John Napier actually arrived at its value during his pioneering studies into logarithms about 100 years earlier. Napier’s original “base” number was actually \(1/e\) in his calculations of logarithms. His colleague Henry Briggs, in 1615, convinced Napier to revise his system of logarithms to use Base 10, for which \(\log 10 = 1\) and thus simplifies the use of logarithms in everyday computations.8 The properties of the number \(e\), only a few of which we have mentioned, make it a special number indeed, and it is regarded as one of the most important numbers in all of mathematics.

  1. In actuality, \(e\) is an irrational number, as is \(\pi\), and hence its decimal representation will never repeat.↩︎

  2. Euler was indeed the first to use this symbol, but it isn’t clear if he meant for the “e” to stand for his name, or if it stood for “exponential,” or if it was just the next vowel to use after “a,” which was the other symbol he used in his work at the time. He is credited for using the symbols “f(x)” to define a function, for instance, as well as “i” for \(\sqrt{-1}\), among many other standard notations.↩︎

  3. See The Development of Mathematics, Dover 1992, (second edition, McGraw-Hill, 1945), pp. 161-162.↩︎