Exercise 6.6.5

Question

\begin{enumerate}[label=(\alph*)] 
  \item Generate the Taylor coefficients for the exponential function $f(x)=e^{x}$, and then prove that the corresponding Taylor series converges uniformly to $e^{x}$ on any interval of the form $[-R, R]$.
  \item Verify the formula $f^{\prime}(x)=e^{x}$.
  \item Use a substitution to generate the series for $e^{-x}$, and then informally calculate $e^{x} \cdot e^{-x}$ by multiplying together the two series and collecting common powers of $x$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $f(x) = e^{x} = \sum_{n=0}^\infty x^n/n!$ from $f^{(n)}(0) = 1$
  \item Differentiating the series is valid by Theorem 6.5.7
    $$f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n}{n!}$$
  \item Let $a_n = (-1)^n/n!$, $b_n = 1/n!$, and $d_n = \sum^{n}_{k=0} a_k b_{n-k}$; the informal power series representation of $e^x \cdot e^{-x}$ becomes $\sum^\infty_{n=0} d_n x^n$. By plugging in $n=0$ we have $d_0 = 1$. For $n > 0$, note that

$$(n!)d_n = \sum^n_{k=0}\frac{n!}{k! (n-k)!} (-1)^k = \sum^n_{k=0} {n \choose k} (-1)^k = \sum^n_{k=0,\ k \text{ even}} {n \choose k} - \sum^n_{k=0,\ k \text{ odd}} {n \choose k} $$
  The first term is the total number of ways to choose an even number of elements from a set of size $n$, while the second term is the number of ways to choose an odd number of elements. For $n$ odd, these two terms must be equal, since for every unique subset with an even number of elements, we get a unique subset with an odd number of elements by taking the set complement. Hence for $n$ odd we must have $d_n = 0$.

It takes a bit more work for $n$ even. Call this set $N$ and divide it into disjoint subsets $A$ containing the first $n-1$ elements and $B$ containing the remaining element. Let $P \choose \text{even}$ be the number of ways to choose an even number of elements from $P$ and $P \choose \text{odd}$ be defined similarly. Then
  $${N \choose \text{odd}} = {A \choose \text{odd}}{B \choose \text{even}} + {A \choose \text{even}}{B \choose \text{odd}} = {A \choose \text{odd}} + {A \choose \text{even}}$$
  while
  $${N \choose \text{even}} = {A \choose \text{even}}{B \choose \text{even}} + {A \choose \text{odd}}{B \choose \text{odd}} = {A \choose \text{even}} + {A \choose \text{odd}} = {N \choose \text{odd}}$$
  and we once again have $d_n = 0$. (Incidentally, this trick does not work for $n = 0$ since it relies on being able to remove an element from $N$ to form $A$.)

Putting everything together we have $e^x \cdot e^{-x} = d_0 = 1$ as expected.
   \end{enumerate}

Exercise 6.6.5

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Exercise 6.6.5

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