# This Quantum World/Appendix/Exponential function

#### The exponential function

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

$\exp '(x)=\exp(x)$   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:

 $\exp(x)=\sum _{k=0}^{\infty }{x^{k} \over k!}=1+x+{x^{2} \over 2}+{x^{3} \over 6}+{x^{4} \over 24}+\cdots$ Let us check that a well-behaved function satisfies the equation

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

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

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

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

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

Voilà.

The function $\exp(x)$  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 $f^{(n)}(0)=[f'(0)]^{n}.$

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 $f'(0),$  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: $e^{x}=\exp(x)$ :

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

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

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

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