Exponentials.

Recall that the number $ e$ is defined by the following sum:

$\displaystyle e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots \, . $


Exercise:Prove that $ e$ is irrational.



Definition:We define an object called $ e^x$ by

$\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \,. $


This is an example of an object called a power series. ``Power'' because it is made of powers of $ x$ and ``series'' for obvious reasons. For us, a power series is not much different from a polynomial - we just have infinetely many terms (pieces) at the same time. We still can multiply and add power series according to the same rules.


Exercise:Check that $ (1-x)(1+x+x^2+x^3+\dots)=1$, so we can write

$\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+x^4+\dots \;.$     (4)



Exercise:Prove that $ de^x=e^x$.


Consider now the ring of polynomials in two variables $ x$ and $ y$. These are objects of the type

$\displaystyle \sum_{k,l}a_{kl}x^ky^l, $

which again form a ring (that is we can multiply and add such expressions). Note that we assumed $ xy=yx$. It means that we write our variables in any order we like (we wrote $ y$ on the right to $ x$). Here is an example: $ 2-3x+4.2y^2x^3$.


Exercise:Work out the addition, the multiplication, the differentiations with repect to $ x$ and $ y$.


The main property of the exponential function is

Lemma 3   $ e^{x+y}=e^xe^y$.

Proof. Just multiply everything out and use the Newton binomial formula:
$\displaystyle (x+y)^n=x^n+{n\choose 1}x^{n-1}y+ {n\choose 2}x^{n-2}y^2+{n\choose 3} x^{n-3}y^3+\dots +{n\choose n} y^n,$     (5)

where the bynomial coefficients are

$\displaystyle {n \choose k}= \frac{n!}{k!(n-k)!}. $


Exercise:Work out the details.


$ \qedsymbol$

Our next goal is to compute $ e^d$. Namely, if $ p(x)$ is a polynomial, what would be $ e^dp(x)$?

Lemma 4   We have $ e^dp(x)=p(x+1)$ and, more general, for any number $ h$, we have

$\displaystyle e^{hd}p(x)=p(x+h). $

Proof. It is enough to consider the case $ p(x)=x^n$ (why?). We have
$\displaystyle { e^{hd}x^n=(1+hd+\frac{h^2d^2}{2!}+\frac{h^3d^3}{3!}+\frac{h^4d^4}{4!}+\dots)x^n=}$
    $\displaystyle x^n+nhx^{n-1}+\frac{n(n-1)}{2}h^2x^{n-2}+ \frac{n(n-1)(n-2)}{3!}h^3x^{n-3}+\dots +h^n=(x+h)^n,$  

by the binomial formula. (Note that we have only finetely many non-zero terms.) $ \qedsymbol$


Remark:In calculus, the formula $ e^{hd}p(x)=p(x+h)$ is called the Taylor expansion formula. It is an important formula, and it includes many others as special cases. For example, if $ p(x)$ is a polynomial then as we saw this is just the Newton binomial formula.

If $ p(x)=1/(1-x)$ then the Taylor expansion formula (where $ x$ is set to zero) becomes $ ($4$ )$ (with $ x$ replaced by $ h$) - just the formula for the geometric progression.

Note that in calculus, the Taylor expansion formula does not necessarily hold for all values of $ x$. But in algebraic formal sense it always true.



Exercise:a) For what values of $ x$ is the formula $ ($4$ )$ true?

b) Let $ f(x)=0$ for $ x\leqslant 0$ and $ f(x)=e^{-1/x}$ for $ x>0$. Use calculus to show that $ (e^{hd}f)(x)\vert _{x\to 0}=0$ for all $ h$.


Now let us consider two operation: $ e^x$ and $ e^{hd}$.

Lemma 5   $ e^{hd}e^x=qe^xe^{hd}$ , where $ q=e^h$. It means that for any polynomial (even power series) $ p(x)$ we have $ e^{hd}e^x p(x)=qe^xe^{hd}p(x)$..

Proof.

$\displaystyle e^{hd}e^x p(x)=e^{x+h}p(x+h)=e^he^xp(x+h)=qe^xe^{hd}p(x). $

$ \qedsymbol$


Exercise:Let $ A$ and $ B$ are two letters such that $ AB-BA=h$. Prove that $ e^Ae^B=qe^Be^A$ by multiplying the corresponding series and moving all $ A$'s to the right and all $ B$'s to the left.



Remark:Our notations are not completely random - $ h$ is the standard notation for the Plank constant, and $ q$ is traditionally related to the word quantum.