We define the function
by requiring that
and ![{\displaystyle \exp(0)=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8a538ff408ca744879fb353be27dbfae4d511b8)
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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e75975c1d100242d680d2a0f37a2b9f5b1a3a9ea)
The second defining equation now tells us that
for all
The result is a particularly simple Taylor series:
|
Let us check that a well-behaved function satisfies the equation
![{\displaystyle f(a)\,f(b)=f(a+b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd372fbadb69e59de38cd115c3c25d37b8d55d48)
if and only if
![{\displaystyle f^{(i+k)}(0)=f^{(i)}(0)\,f^{(k)}(0).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ba92d83fa157c630e8e578889fdd9ccc8d209fe)
We will do this by expanding the
's in powers of
and
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},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98e96b75904060f24717b8cb94adf9f475bb15d4)
and using the binomial expansion
![{\displaystyle (a+b)^{i}=\sum _{l=0}^{i}{\frac {i!}{(i-l)!\,l!}}\,a^{i-l}\,b^{l},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb5569f698136e72cad3bad6cc1182c06cb9edc0)
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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/386c50e45a79a3ceee28c72f9d6a7792b4b6f897)
Voilà.
The function
obviously satisfies
and hence
So does the function
Moreover,
implies
We gather from this
- that the functions satisfying
form a one-parameter family, the parameter being the real number
and
- that the one-parameter family of functions
satisfies
, the parameter being the real number ![{\displaystyle u.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edd5636410da69bac33da075162221527401793c)
But
also defines a one-parameter family of functions that satisfies
, the parameter being the positive number
Conclusion: for every real number
there is a positive number
(and vice versa) such that
One of the most important numbers is
defined as the number
for which
that is:
:
![{\displaystyle e=\exp(1)=\sum _{n=0}^{\infty }{1 \over n!}=1+1+{1 \over 2}+{1 \over 6}+\dots =2.7182818284590452353602874713526\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5674af5a5af45bd487985227202a94c3eb90241)
The natural logarithm
is defined as the inverse of
so
Show that
![{\displaystyle {d\ln f(x) \over dx}={1 \over f(x)}{df \over dx}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee26b8364b8ba7513c85fd9fbe8ec585fc7dcb83)
Hint: differentiate