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:

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