Show that natural exponential function is well-defined

Question

Thomas' Calculus p. 440 (Early Transcendental version, 14th edition), author defines the natural exponential function this way:

For every real number $x$, we define the natural exponential function to be $e^{x} = \exp(x)$.

Where $\exp$ denotes the inverse function of $\ln$, and $e$ is a number such that $\ln(e) = 1$.

The natural logarithm is defined to be:

$$ \ln x=\int_1^x \frac{1}{t}dt, x>0. $$

The author gives a very readable proof of the validity of the definition of natural exponential function (the equality $e^{x} = \exp(x)$ to be specific), but only for all x that are rational. So, is there any proof of the validity for irrational x also?

The reason I believe the proof for the irrational part is necessary is because several important results are deduced based on the equality. Only accept that it's true and move on is fine, but it's not satisfactory.

Answer 1

Having defined $\ln(x)$, $\exp(x)$, and $e$ as you described, there are two ways one might define the function $e^x$:

• Define $e^x = \exp(x)$.
• Define $e^x$ for rational values of $x$ so that it satisfies all the usual laws of exponents, and then extend continuously to the full set of real numbers.

I understand the question to be why these two definitions are equivalent for irrational values of $x$, and the reason is that there is there is only one way to extend the second definition to a continuous function, and since $\exp(x)$ agrees with the second definition for rational values and is itself continuous, it must be equal to the unique continuous extension of the second definition to the reals.

Answer 2

You can prove first that $\ln(x^p)=p\ln(x)\tag{$\star$}\label{1},$ then $\exp(x)=\exp(x\ln(e))=\exp(\ln(e^x))=e^x$ because $\exp(\cdot)$ is the inverse function of $\ln(\cdot)$ and $\ln(e)=1.$

However, $\eqref{1}$ requires the power rule.

We can start by proving the power rule for non-negative integers, then negative integers, then rational numbers and finally irrational numbers. Doing so, we would have proved the power rule for all real numbers.

For non-negative integer powers, we only need to realise that $(x+h)^n=x^n+nhx^{n-1}+h^2f(x,h)$ where $n$ is a non-negative integer and we don't care what that $f(x,h)$ is for our purposes. You can think about why there's only one $x^n$ by setting $h=0.$ For $hx^{n-1},$ we realise that we are multiplying $n$ $(x+h)$ brackets together so we need to have $n$ variables multiplied together per term, so if we have $n-1$ $x$'s, we should have a single $h.$ As for why we have $n$ of those, it's better to count the number of ways of choosing one $h$ out of the $n$ brackets. It's then easy to see there are $n$ ways for $n$ brackets. Now, we can differentiate using the limit definition. \begin{eqnarray} \frac{dx^n}{dx}&=&\lim_{h\to0}\frac{(x+h)^n-x^n}{h}\\ &=&\lim_{h\to0}\frac{x^n+nhx^{n-1}+h^2f(x,h)-x^n}{h}\\ &=&\lim_{h\to0}(nx^{n-1}+hf(x,h))\\ &=&nx^{n-1} \end{eqnarray} Now for negative powers, we only need to consider the special case $x^{-1}$ and the rest follows by the chain rule. \begin{eqnarray} \frac{dx^{-1}}{dx}&=&\lim_{h\to0}\frac{\frac1{x+h}-\frac1x}h\\ &=&\lim_{h\to0}\frac{-1}{x(x+h)}\\ &=&-x^{-2} \end{eqnarray} Now consider the general case where we take $n$ as a non-negative integer. \begin{eqnarray} \frac{dx^{-n}}{dx}&=&\frac{d(x^n)^{-1}}{dx}\\ &=&-(x^n)^{-2}(nx^{n-1})\\ &=&-nx^{-n-1} \end{eqnarray} So it's the same rule. Now for rational powers, we can agree that for any rational power $r,$ there exist some integers $n,m$ such that $$(x^r)^m=x^{rm}=x^n$$ We can use the chain rule to find the derivative of the rational power. \begin{eqnarray} m(x^r)^{m-1}\frac{dx^r}{dx}&=&nx^{n-1}\\ mx^{rm-r}\frac{dx^r}{dx}&=&rmx^{rm-1}\\ \frac{dx^r}{dx}&=&rx^{r-1} \end{eqnarray} Again, the same rule. Now for irrational powers $q,$ we can use the fact that an irrational number can be expressed as the sum of an infinite number of rational numbers $r_k,$ an infinite series; $q=\sum_{k=0}^\infty r_k.$ We can use the product rule to show \begin{eqnarray} \frac{dx^q}{dx}&=&\frac{dx^{\sum_{k=0}^\infty r_k}}{dx}\\ &=&\frac{d(x^{r_0}\cdot x^{r_1}\cdot x^{r_2}\cdots)}{dx}\\ &=&(x^{r_0}\cdot x^{r_1}\cdot x^{r_2}\cdots)\sum_{k=0}^\infty x^{-r_k}\frac{dx^{r_k}}{dx}\\ &=&x^q\sum_{k=0}^\infty r_kx^{-1}\\ &=&qx^{q-1} \end{eqnarray} Again, the same rule. So, for all real powers $p,$ $$\frac{dx^p}{dx}=px^{p-1}.$$

Now, we can derive $\ln(x^p)=p\ln(x)$ using the integral definition and the substitution $t=u^p.$ \begin{eqnarray} \ln(x^p)&=&\int_1^{x^p}\frac{dt}t\\ &=&\int_1^x\frac{pu^{p-1}du}{u^p}\\ &=&p\int_1^x\frac{du}u\\ &=&p\ln(x) \end{eqnarray} The final step is showing $\exp(x)=e^x$ as we want which I did in the beginning of the answer.