proof of equivalence of formulas for exp

We present an elementary proof that:


There are of course other proofs, but this one has the advantage that it carries verbatim for the matrix exponentialMathworldPlanetmath and the operator exponentialPlanetmathPlanetmath.

At the outset, we observe that k=0zk/k! converges by the ratio testMathworldPlanetmath. For definiteness, the notation ez below will refer to exactly this series.


We expand the right-hand in the straightforward manner:

(1+zn)n =k=0n(nk)(zn)k

where π(k,n) denotes the coefficient


Let |z|M. Given ϵ>0, there is a N such that whenever nN, then k=n+1Mk/k!<ϵ/2, since the sum is the tail of the convergent seriesMathworldPlanetmath eM.

Since limnπ(k,n)=1 for k, there is also a N, with NN, so that whenever nN and 0kN, then |π(k,n)-1|<ϵ/(2eM). (Note that k is chosen only from a finite set.)

Now, when nN, we have

|k=0nπ(k,n)zkk!-k=0zkk!| =|k=0n(π(k,n)-1)zkk!-k=n+1zkk!|
(In the middle sum, we use the bound |π(k,n)-1|=1-π(k,n)1 for all k and n.)

In fact, we have proved uniform convergenceMathworldPlanetmath of limn(1+zn)n over |z|M. Exploiting this fact we can also show:

(1+zn+o(1n))n=(1+z+o(1)n)nk=0zkk!(pointwise, as n)

|z|<M. Given ϵ>0, for large enough n, we have

|(1+wn)n-ew|<ϵ/2uniformly for all |w|M.

Since o(1)0, for large enough n we can set w=z+o(1) above. Since the exponential is continuousMathworldPlanetmath11follows from uniform convergence on bounded subsets of either expression for ez, for large enough n we also have |ez+o(1)-ez|<ϵ/2. Thus

Title proof of equivalence of formulas for exp
Canonical name ProofOfEquivalenceOfFormulasForExp
Date of creation 2013-03-22 15:22:52
Last modified on 2013-03-22 15:22:52
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 11
Author stevecheng (10074)
Entry type Proof
Classification msc 30A99
Related topic ComplexExponentialFunction
Related topic ExponentialFunction
Related topic MatrixExponential