This Quantum World/Appendix/Exponential function

The exponential function

We define the function $\exp (x)$ by requiring that

${\exp }^{\prime }(x)=\exp (x)$  and  $\exp (0)=1.$

The value of this function is everywhere equal to its slope. Differentiating the first defining equation repeatedly we find that

${\exp }^{(n)}(x)={\exp }^{(n-1)}(x)=\cdots =\exp (x).$

The second defining equation now tells us that ${\exp }^{(k)}(0)=1$ for all $k.$ The result is a particularly simple Taylor series:

$\exp (x)=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots$

Let us check that a well-behaved function satisfies the equation

$f(a)f(b)=f(a+b)$

if and only if

$f^{(i+k)}(0)=f^{(i)}(0)f^{(k)}(0).$

We will do this by expanding the $f$'s in powers of $a$ and $b$ and compare coefficents. We have

$f(a)f(b)=\sum _{i=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(i)}(0)f^{(k)}(0)}{i!k!}{a}^i{b}^k,$

and using the binomial expansion

$(a+b{)}^i=\sum _{l=0}^i\frac{i!}{(i-l)!l!}{a}^{i-l}{b}^l,$

we also have that

$f(a+b)=\sum _{i=0}^{\mathrm{\infty }}\frac{f^{(i)}(0)}{i!}(a+b{)}^i=\sum _{i=0}^{\mathrm{\infty }}\sum _{l=0}^i\frac{f^{(i)}(0)}{(i-l)!l!}{a}^{i-l}{b}^l=\sum _{i=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(i+k)}(0)}{i!k!}{a}^i{b}^k.$

Voilà.

The function $\exp (x)$ obviously satisfies $f^{(i+k)}(0)=f^{(i)}(0)f^{(k)}(0)$ and hence $f(a)f(b)=f(a+b).$

So does the function $f(x)=\exp (ux).$

Moreover, $f^{(i+k)}(0)=f^{(i)}(0)f^{(k)}(0)$ implies $f^{(n)}(0)=[f^{\prime }(0){]}^n.$

We gather from this

• that the functions satisfying $f(a)f(b)=f(a+b)$ form a one-parameter family, the parameter being the real number $f^{\prime }(0),$ and

• that the one-parameter family of functions $\exp (ux)$ satisfies $f(a)f(b)=f(a+b)$, the parameter being the real number $u.$

But $f(x)=v^x$ also defines a one-parameter family of functions that satisfies $f(a)f(b)=f(a+b)$, the parameter being the positive number $v.$

Conclusion: for every real number $u$ there is a positive number $v$ (and vice versa) such that $v^x=\exp (ux).$

One of the most important numbers is $e,$ defined as the number $v$ for which $u=1,$ that is: $e^x=\exp (x)$:

$e=\exp (1)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=1+1+\frac{1}{2}+\frac{1}{6}+\dots =2.7182818284590452353602874713526\dots$

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The natural logarithm $\ln (x)$ is defined as the inverse of $\exp (x),$ so $\exp [\ln (x)]=\ln [\exp (x)]=x.$ Show that

$\frac{d\ln f(x)}{dx}=\frac{1}{f(x)}\frac{df}{dx}.$

Hint: differentiate $\exp \{\ln [f(x)]\}.$