# 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:

 $\exp(x)=\sum_{k=0}^\infty {x^k\over k!} =1+x+{x^2\over2}+{x^3\over6}+{x^4\over24}+\cdots$

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: $e^x=\exp(x)$:

$e=\exp(1)=\sum_{n=0}^\infty{1\over n!}=1+1+{1\over2}+{1\over6}+\dots= 2.7182818284590452353602874713526\dots$

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