This Quantum World/Appendix/Exponential function

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

${\displaystyle \exp '(x)=\exp(x)}$  and  ${\displaystyle \exp(0)=1.}$

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

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

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

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

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

${\displaystyle f(a)\,f(b)=f(a+b)}$

if and only if

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

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

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

and using the binomial expansion

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

we also have that

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

Voilà.

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

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

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

We gather from this

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

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

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

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

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

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

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

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

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