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Subsection 2.7.2 Back to that Limit

Recall that we are trying to choose \(a\) so that

\begin{align*} \lim_{h\to0} \frac{a^h-1}{h} &= C(a) = 1. \end{align*}

We can estimate the correct value of \(a\) by using our numerical estimate of \(C(10)\) above. The way to do this is to first rewrite \(C(a)\) in terms of logarithms.

\begin{align*} a&= 10^{\log_{10} a} & \text{ and so }&& a^h &= 10^{h\log_{10} a}. \end{align*}

Using this we rewrite \(C(a)\) as

\begin{align*} C(a) &= \lim_{h\to0} \frac{1}{h} \left( 10^{h\log_{10} a}-1 \right)\\ \end{align*}

Now set \(H = h\log_{10}(a)\text{,}\) and notice that as \(h\to 0\) we also have \(H \to 0\)

\begin{align*} &= \lim_{H \to 0} \frac{\log_{10} a}{H} \left(10^H-1\right)\\ &= \log_{10} a \cdot \lim_{H \to 0} \frac{10^H-1}{H}\\ &= \log_{10} a \cdot C(10). \end{align*}

Below is a sketch of \(C(a)\) against \(a\text{.}\)

Figure 2.7.3

Remember that we are trying to find an \(a\) with \(C(a)=1\text{.}\) We can do so by recognising that \(C(a)=C(10)\,(\log_{10}a)\) has the following properties.

  • When \(a=1\text{,}\) \(\log_{10}(a) = \log_{10} 1 =0\) so that \(C(a) = C(10) \log_{10}(a) = 0\text{.}\) Of course, we should have expected this, because when \(a=1\) we have \(a^x = 1^x = 1\) which is just the constant function and \(\diff{}{x} 1 = 0\text{.}\)
  • \(\log_{10}a\) increases as \(a\) increases, and hence \(C(a)=C(10)\ \log_{10}a\) increases as \(a\) increases.
  • \(\log_{10}a\) tends to \(+\infty\) as \(a\rightarrow\infty\text{,}\) and hence \(C(a)\) tends to \(+\infty\) as \(a\rightarrow\infty\text{.}\)

Hence the graph of \(C(a)\) passes through \((1,0)\text{,}\) is always increasing as \(a\) increases and goes off to \(+\infty\) as \(a\) goes off to \(+\infty\text{.}\) See Figure 2.7.3. Consequently  5 We are applying the Intermediate Value Theorem here, but we have neglected to verify the hypothesis that \(\log_{10} (a)\) is a continuous function. Please forgive us — we could do this if we really had to, but it would make a big mess without adding much understanding, if we were to do so here in the text. Better to just trust us on this. there is exactly one value of \(a\) for which \(C(a) = 1\text{.}\)

The value of \(a\) for which \(C(a)=1\) is given the name \(e\text{.}\) It is called Euler's constant  6 Unfortunately there is another Euler's constant, \(\gamma\text{,}\) which is more properly called the Euler–Mascheroni constant. Anyway like many mathematical discoveries, \(e\) was first found by someone else — Napier used the constant \(e\) in order to compute logarithms but only implicitly. Bernoulli was probably the first to approximate it when examining continuous compound interest. It first appeared explicitly in work of Leibniz, though he denoted it \(b\text{.}\) It was Euler, though, who established the notation we now use and who showed how important the constant is to mathematics.. In Example 2.7.1, we estimated \(C(10)\approx 2.3026\text{.}\) So if we assume \(C(a)=1\) then the above equation becomes

\begin{align*} 2.3026 \cdot \log_{10} a &\approx 1\\ \log_{10} a &\approx \frac{1}{2.3026} \approx 0.4343\\ a &\approx 10^{0.4343} \approx 2.7813 \end{align*}

This gives us the estimate \(a \approx 2.7813\) which is not too bad. In fact  7 Recall \(n\) factorial, written \(n!\) is the product \(n\times(n-1)\times(n-2)\times\cdots\times2\times1\text{.}\)

We will be able to explain this last formula once we develop Taylor polynomials later in the course.

To summarise

We plot \(e^x\) in the graph below

And just a reminder of some of its  8 The function \(e^x\) is of course the special case of the function \(a^x\) with \(a = e\text{.}\) So it inherits all the usual algebraic properties of \(a^x\text{.}\) properties…

  1. \(e^0=1\)
  2. \(e^{x+y}=e^xe^y\)
  3. \(e^{-x}=\tfrac{1}{e^x}\)
  4. \(\big(e^x\big)^y=e^{xy}\)
  5. \(\lim\limits_{x\rightarrow\infty}e^x=\infty\text{,}\) \(\lim\limits_{x\rightarrow-\infty}e^x=0\)

Now consider again the problem of differentiating \(a^x\text{.}\) We saw above that

\begin{align*} \diff{}{x} a^x &= C(a) \cdot a^x \qquad\text{ and }\qquad C(a) = C(10) \cdot \log_{10} a \\ \amp\text{ which gives } \diff{}{x} a^x = C(10)\cdot \log_{10} a \cdot a^x \end{align*}

We can eliminate the \(C(10)\) term with a little care. Since we know that \(\diff{}{x} e^x = e^x\text{,}\) we have \(C(e)=1\text{.}\) This allows us to express

\begin{align*} 1 = C(e) &= C(10) \cdot \log_{10} e & \text{ and so}\\ C(10) &= \frac{1}{\log_{10} e} \end{align*}

Putting things back together gives

\begin{align*} \diff{}{x} a^x &= \frac{\log_{10} a}{\log_{10} e} \cdot a^x\\ &= \log_e a \cdot a^x. \end{align*}

There is more than one way to get to this result. For example, let \(f(x) = a^x\text{,}\) then

\begin{align*} \log_e f(x) &= x \log_e a\\ f(x) &= e^{ x \log_e a} \end{align*}

So if we write \(g(x) = e^x\) then we are really attempting to differentiate the function

\begin{align*} \diff{f}{x} &= \diff{}{x} g(x \cdot \log_e a). \end{align*}

In order to compute this derivative we need to know how to differentiate

\begin{gather*} \diff{}{x} g( q x) \end{gather*}

where \(q\) is a constant. We'll hold off on learning this for the moment until we have introduced the chain rule (see Section 2.9 and in particular Corollary 2.9.9). Similarly we'd like to know how to differentiate logarithms — again this has to wait until we have learned the chain rule.

Notice that the derivatives

\begin{align*} \diff{}{x} x^n &= n x^{n-1} & \text{ and }&& \diff{}{x} e^x &= e^x \end{align*}

are either nearly unchanged or actually unchanged by differentiating. It turns out that some of the trigonometric functions also have this property of being “nearly unchanged” by differentiation. That brings us to the next section.

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