4.2 The derivative of a function (Page 4/4)

Analysis of functions of a single Page 4 / 4

f^{'} (z) = \sum_{n = 0}^{\infty} n a_{n} z^{n - 1} = \sum_{n = 1}^{\infty} n a_{n} z^{n - 1} .

The proof will use part (3) of [link] . Fix an $z$ with $| z | < r .$ Choose $r^{'}$ so that $| z | < r^{'} < r,$ and write $α$ for $r^{'} - | z |,$ i.e., $| z | + α = r^{'} .$ Note first that the infinite series $\sum_{n = 0}^{\infty} | a_{n} | {r^{'}}^{n}$ converges to a positive number we will call $M .$ Also, from the Cauchy-Hadamard Formula, we know that the power series function $\sum n a_{n} w^{n}$ has the same radius of convergence as does $f,$ and hence the infinite series $\sum n a_{n} z^{n - 1}$ converges to a number we will denote by $L .$ We define a function $θ$ by $θ (h) = f (z + h) - f (z) - L h$ from which it follows immediately that

f (z + h) - f (z) = L h + θ (h),

which establishes [link] . To complete the proof that $f$ is differentiable at $z,$ it will suffice to establish [link] , i.e., to show that

lim_{h \to 0} \frac{θ (h)}{h} = 0 .

That is, given $ϵ > 0$ we must show that there exists a $δ > 0$ such that if $0 < | h | < δ$ then

| θ (h) / h | = | \frac{f (z + h) - f (z)}{h} - L | < ϵ .

Assuming, without loss of generality, that $| h | < α,$ we have that

\begin{matrix} | \frac{f (z + h) - f (z)}{h} - L | & = & | \frac{\sum_{n = 0}^{\infty} a_{n} {(z + h)}^{n} - \sum_{n = 0}^{\infty} a_{n} z^{n}}{h} - L | \\ = & | \frac{\sum_{n = 0}^{\infty} a_{n} (\sum_{k = 0}^{n} (\binom{n}{k}) z^{n - k} h^{k}) - \sum_{n = 0}^{\infty} a_{n} z^{n}}{h} - L | \\ = & | \frac{\sum_{n = 0}^{\infty} a_{n} ((\sum_{k = 0}^{n} (\binom{n}{k}) z^{n - k} h^{k}) - z^{n})}{h} - L | \\ = & | \frac{\sum_{n = 1}^{\infty} a_{n} (\sum_{k = 1}^{n} (\binom{n}{k}) z^{n - k} h^{k})}{h} - L | \\ = & | \sum_{n = 1}^{\infty} a_{n} (\sum_{k = 1}^{n} (\binom{n}{k}) z^{n - k} h^{k - 1}) - \sum_{n = 1}^{\infty} n a_{n} z^{n - 1} | \\ = & | \sum_{n = 1}^{\infty} a_{n} (\sum_{k = 1}^{n} (\binom{n}{k}) z^{n - k} h^{k - 1}) - \sum_{n = 1}^{\infty} (\binom{n}{1}) a_{n} z^{n - 1} | \\ = & | \sum_{n = 2}^{\infty} a_{n} (\sum_{k = 2}^{n} (\binom{n}{k}) z^{n - k} h^{k - 1}) | \\ \leq & \sum_{n = 2}^{\infty} \sum_{k = 2}^{n} | a_{n} | (\binom{n}{k}) {| z |}^{n - k} {| h |}^{k - 1} \\ \leq & | h | \sum_{n = 2}^{\infty} | a_{n} | \sum_{k = 2}^{n} (\binom{n}{k}) {| z |}^{n - k} {| h |}^{k - 2} \\ \leq & | h | \sum_{n = 2}^{\infty} | a_{n} | \sum_{k = 2}^{n} (\binom{n}{k}) {| z |}^{n - k} {| α |}^{k - 2} \\ \leq & | h | \frac{1}{α^{2}} \sum_{n = 0}^{\infty} | a_{n} | \sum_{k = 0}^{n} (\binom{n}{k}) {| z |}^{n - k} α^{k} \\ = & | h | \frac{1}{α^{2}} \sum_{n = 0}^{\infty} | a_{n} {| (| z | + α)}^{n} \\ = & | h | \frac{1}{α^{2}} \sum_{n = 0}^{\infty} | a_{n} | {r^{'}}^{n} \\ = & | h | \frac{M}{α^{2}}, \end{matrix}

so that if $δ = ϵ / \frac{M}{α^{2}},$ then $| θ (h) / h | < ϵ,$ whenever $| h | < δ,$ as desired.

REMARK [link] shows that indeed power series functions are differentiable, and in fact their derivatives can be computed, just like polynomials, by differentiating term by term.This is certainly a result we would have hoped was true, but the proof is not trivial.

The next theorem, the Chain Rule, is another nontrivial one. It deals with the differentiability of the composition of two differentiable functions.Again, the result is what we would have wanted, the composition of two differentiable functions is itself differentiable, but the argument required to prove it is tricky.

Chain rule

Let $f : S \to C$ be a function, and assume that $f$ is differentiable at a point $c .$ Suppose $g : T \to C$ is a function, that $T \subseteq C,$ that the number $f (c) \in T,$ and that $g$ is differentiable at $f (c) .$ Then the composition $g \circ f$ is differentiable at $c$ and

{(g \circ f)}^{'} (c) = g^{'} (f (c)) f^{'} (c) .

Using part (3) of [link] , write

g (f (c) + k) - g (f (c)) = L_{g} k + θ_{g} (k)

and

f (c + h) - f (c) = L_{f} h + θ_{f} (h) .

We know from that theorem that $L_{g} = g^{'} (f (c))$ and $L_{f} = f^{'} (c) .$ And, we also know that

lim_{k \to 0} \frac{θ_{g} (k)}{k} = 0 and lim_{h \to 0} \frac{θ_{f} (h)}{h} = 0 .

Define a function $k (h) = f (c + h) - f (c) .$ Then, by [link] , we have that ${lim}_{h \to 0} k (h) = 0 .$ We will show that $g \circ f$ is differentiable at $c$ by showing that there exists a number $L$ and a function $θ$ satisfying the two conditions of part (3) of [link] . Thus, we have that

\begin{matrix} g \circ f (c + h) - g \circ f (c) & = g (f (c + h)) - g (f (c)) \\ = & g (f (c) + k (h)) - g (f (c)) \\ = & L_{g} k (h) + θ_{g} (k (h)) \\ = & L_{g} (f (c + h) - f (c)) + θ_{g} (k (h)) \\ = & L_{g} (L_{f} h + θ_{f} (h)) + θ_{g} (k (h)) \\ = & L_{g} L_{f} h + L_{g} θ_{f} (h) + θ_{g} (k (h)) . \end{matrix}

We define $L = L_{g} l_{f} = g^{'} (f (c)) f^{'} (c),$ and we define the function $θ$ by

θ (h) = L_{g} θ_{f} (h) + θ_{g} (k (h)) .

By our definitions, we have established [link]

g \circ f (c + h) - g \circ f (c) = L h + θ (h),

so that it remains to verify [link] .

We must show that, given $ϵ > 0,$ there exists a $δ > 0$ such that if $0 < | h | < δ$ then $| θ (h) / h | < ϵ .$ First, choose an $ϵ^{'} > 0$ so that

| L_{g} | ϵ^{'} + | L_{f} | ϵ^{'} + {ϵ^{'}}^{2} < ϵ (4.3) .

Next, using part (b) of [link] , choose a $δ^{'} > 0$ such that if $| k | < δ^{'}$ then $| θ_{g} (k) | < ϵ^{'} | k | .$ Finally, choose $δ > 0$ so that if $0 < | h | < δ,$ then the following two inequalities hold. $| k (h) | < δ^{'}$ and $| θ_{f} (h) | < ϵ^{'} | h | .$ The first can be satisfied because $f$ is continuous at $c,$ and the second is a consequence of part (b) of [link] . Then:if $0 < | h | < δ,$

\begin{matrix} | θ (h) | & = | L_{g} θ_{f} (h) + θ_{g} (k (h)) | \\ \leq & | L_{g} | | θ_{f} (h) | + | θ_{g} (k (h)) | \\ < & | L_{g} | ϵ^{'} | h | + ϵ^{'} | k (h) | \\ = & | L_{g} | ϵ^{'} | h | + ϵ^{'} | f (c + h) - f (c) | \\ = & | L_{g} | ϵ^{'} | h | + ϵ^{'} | L_{f} h + θ_{f} (h) | \\ \leq & | L_{g} | ϵ^{'} | h | + ϵ^{'} | L_{f} | | h | + ϵ^{'} | θ_{f} (h) | \\ < & | L_{g} | ϵ^{'} | h | + ϵ^{'} | L_{f} | | h | + ϵ^{'} ϵ^{'} | h | \\ = & (| L_{g} | ϵ^{'} + | L_{f} | ϵ^{'} + {ϵ^{'}}^{2}) | h |, \end{matrix}

whence

| θ (h) / h | < (| L_{g} | ϵ^{'} + | L_{f} | ϵ^{'} + {ϵ^{'}}^{2}) < ϵ,

as desired.

Derive the familiar formulas for the derivatives of the elementary transcendental functions:
${exp}^{'} = exp, {sin}^{'} = cos,, s i n h^{'} = c o s h, {cosh}^{'} = sinh and {cos}^{'} = - sin .$
Define a function $f$ as follows. $f (z) = {cos}^{2} (z) + {sin}^{2} (z) .$ Use part (a) and the Chain Rule to show that $f^{'} (z) = 0$ for all $z \in C .$ Does this imply that ${cos}^{2} (z) + {sin}^{2} (z) = 1$ for all complex numbers $z ?$
Suppose $f$ is expandable in a Taylor series around the point $c :$ $f (z) = \sum_{n = 0}^{\infty} a_{n} {(z - c)}^{n}$ for all $z \in B_{r} (c) .$ Prove that $f$ is differentiable at each point of the open disk $B_{r} (c),$ and show that
$f^{'} (z) = \sum_{n = 1}^{\infty} n a_{n} {(z - c)}^{n - 1} .$
HINT: Use [link] and the chain rule.

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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