This Quantum World/Appendix/Exponential function

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The exponential function[edit]

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)]\}.