The Operator $z\frac d{dz}$

For a differentiable function $f(z),$ let $Tf$ be the function given by
MATH
Iterations of this ``operator'' $T=z\frac d{dz}$ yield many interesting identities.

The Sum MATH

For MATH we have
MATH

MATH

MATH

MATH

MATH

MATH
On the other hand $T$ may be applied to the function $\frac 1{1-z}$ to obtain
MATH

MATH

MATH

MATH
Comparing this with the above series expansion, we conclude that
MATH

MATH

MATH

MATH

Observe that after $k$ iterations, the resulting function takes the form
MATH
for some polynomial $p_{k-1}(z)$ of degree $k.$ It is therefore desirous to find the recurrence relation satisfied by the polynomials $\{p_{k}\}.$ From the identity
MATH
(which, again, can be calculated by Maple,) we see that
MATH
holds for $k\geq 1.$

The Sum MATH

For MATH we have
MATH

MATH

MATH

MATH

MATH

MATH

On the other hand $T$ may be applied to the function $\exp (z)$ to obtain
MATH

MATH

MATH

MATH

MATH
Comparing this with the above series expansion, we conclude that
MATH

MATH

MATH

MATH

MATH

Observe that after $k$ iterations, the resulting function takes the form
MATH
for some polynomial $p_{k-1}(z)$ of degree $k.$ The recurrence relation satisfied by the polynomials $\{p_{k}\}$ may be deduced from the identity
MATH
we see that
MATH[Maple Worksheet]

Absolute Convergent

The series MATH is absolutely convergent if the series MATH converges. (In more formal language, the sequence $\{a_n\}$ is absolutely summable if the sequence MATH is summable.)

Convergence of MATH and MATH

Theorem. Every absolutely convergent series is convergent. Moreover, a series is absolutely convergent if and only if the series formed from its positive terms and the series formed from its negative terms both converge.

Proof. If MATH converges, then, by the Cauchy criterion,
MATH
Since
MATH
it follows that
MATH
which shows that MATH converges.

To prove the second part of the theorem, let
MATH

MATH
so that MATH is the series formed from the positive terms of MATH, and MATH is the series formed from the negative terms.

If MATH and MATH both converge, then
MATH
also converges, so MATH converges absolutely.

On the other hand, if MATH converges, then MATH also converges. Therefore
MATH
and
MATH
both converge.

Leibniz's Theorem.

Suppose that and that
MATH
and that
MATH
Then the series
MATH
converges.

Proof. We prove that

(1) MATH

(2) MATH

(3) $s_k\le s_l$ if $k$ is even and $l$ is odd.

This is so since:

(1) MATH

(2) MATH

(3) If $k$ is even and $l$ is odd, choose $n$ such that
MATH
then
MATH

Now, the sequence $\{s_{2n}\}$ converges, because it is nondecreasing and is bounded above (by $s_l$ for any odd $1$). Let
MATH

Similarly, let
MATH

If follows from (3) that $\alpha \le \beta ;$ since
MATH
it is actually the case that $\alpha =\beta $. This proves that MATH

Uniform Convergence

DEFINITION: Let $\{f_n\}$ be a sequence of functions defined on $A$, and let $f$ be a function which is also defined on $A.$ Then $f$ is called the uniform limit of $\{f_n\}$ on $A$ if for every $\epsilon >0$ there is some $N$ such that for all $x$ in $A,$
MATH

We also say that $\{f_n\}$ converges uniformly to $f$ on $\QTR{bf}{A} $, or that $f_n$ approaches $f$ uniformly on $A.$

THEOREM Suppose that $\{f_n\}$ is a sequence of functions which are integrable on $[a,b],$ and that $\{f_n\}$ converges uniformly on $[a,b]$ to a function $f$ which is integrable on $[a,b]$. Then
MATH

Proof. Let $\epsilon >0.$ There is some $N$ such that for all $n>N$ we have
MATH

MATH
Since this is true for any $\epsilon $ $>0$, it follows that MATH

Theorem. Suppose that $\{f_n\}$ is a sequence of functions which are continuous on $[a,b]$, and that $\{f_n\}$ converges uniformly on $[a,b]$ to $f$. Then $f$ is also continuous on $[a,b]$.

Proof. For each $x$ in $[a,b]$ we must prove that $f$ is continuous at $x.$ We will deal only with $x$ in $(a,b)$; the cases $x=a$ and $x=b$ require the usual simple modifications.

Let $\epsilon >0$. Since $\{f_n\}$ converges uniformly to $f$ on $[a,b]$, there is some $n$ such that
MATH
In particular, for all $h$ such that $x+h$ is in $[a,b]$, we have

(1) MATH

(2) MATH

Now $f_n$ is continuous, so there is some $\delta >0$ such that for $\left| h\right| $ $<\delta $ we have

(3) MATH

Thus, if MATH, then
MATH

MATH

MATH

MATH

This proves that $f$ is continuous at $x.$

Suppose that $\{f_n\}$ is a sequence of functions which are differentiable on $[a,b]$, with integrable derivatives $f^{\prime }$, and that $\{f_n\}$ converges (pointwise) to $f.$ Suppose, moreover, that $\{f_n^{\prime }\}$ converges uniformly on $[a,b]$ to some continuous function $g.$ Then $f$ is differentiable and
MATH

Proof. Applying the previous theorem to the interval $[a,x]$, we see that for each $x$ we have
MATH
Since $g$ is continuous, it follows that MATH for all $x$ in the interval $[a,b].$