Taylor's Theorem and The Lagrange Remainder Examples 1

Recall from the Taylor's Theorem and The Lagrange Remainder page that Taylor's Theorem says that if $f$ is $n + 1$ times differentiable on some interval containing the center of convergence $c$ and $x$, and let $P_n(x) = f(c) + \frac{f^{(1)}(c)}{1!}(x - c) + \frac{f^{(2)}(c)}{2!}(x - c)^2 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$ be the $n^{\mathrm{th}}$ order Taylor polynomial of $f$ at $x = c$. Then $f(x) = P_n(x) + E_n(x)$ where $E_n(x)$ is the error term of $P_n(x)$ from $f(x)$ and for $\xi$ between $c$ and $x$, the Lagrange Remainder form of the error $E_n$ is given by the formula $E_n(x) = \frac{f^{(n+1)}(\xi)}{(n + 1)!}(x - c)^{n+1}$.

We will now look at some examples of applying the Taylor's Theorem.

Example 1

Prove that $f(x) = e^x$ is analytic by showing that the Maclaurin series for $e^x$ represents $e^x$ for all $x \in \mathbb{R}$.

We note that since we are dealing with the Maclaurin series of $e^x$, then our center of convergence is $c = 0$ and so for $\xi$ between $0$ and $x$, the Lagrange Remainder form of the error is given by:

(1)

\begin{align} \quad E_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} \end{align}

Now we note that since $f(x) = e^x$ then $f^{(n+1)}(x) = e^x$ for any $n ≥ 0$ and so taking the absolute value of $E_n(x)$ and replacing $f^{(n+1)}(\xi)$ with $e^{\xi}$ we get that

(2)

\begin{align} \quad \mid E_n(x) \mid = \frac{e^{\xi} \mid x \mid^{n+1}}{(n+1)!} \end{align}

For any $x \in \mathbb{R}$ we thus have that:

(3)

\begin{align} \quad \lim_{n \to \infty} \mid E_n \mid = \lim_{n \to \infty} \frac{e^{\xi} \mid x \mid^{n+1}}{(n+1)!} = 0 \end{align}

By the Squeeze Theorem we have that then $\lim_{n \to \infty} \mid E_n(x) \mid = 0$ and so $\lim_{n \to \infty} E_n(x) = 0$ and thus $\lim_{n \to \infty} P_n(x) = f(x)$ so $f(x) = e^x$ is analytic for all $x \in \mathbb{R}$.

Example 2

Prove that $f(x) = \sin x$ is analytic by showing that the Maclaurin series for $\sin x$ represents $\sin x$ for all $x \in \mathbb{R}$.

We note that since we are dealing with the Maclaurin series of $\sin x$, then our center of convergence is $c = 0$ and so for $\xi$ between $0$ and $x$, the Lagrange Remainder form of the error is given by:

(4)

\begin{align} \quad E_n(x) = \frac{f^{(n+1)}(\xi)}{(n + 1)!}x^{n+1} \end{align}

Now since $f(x) = \sin x$, then $f^{(n+1)}(x) = \pm \sin x$ or $f^{(n+1)}(x) = \pm \cos x$. Note that $-1 ≤ \pm \sin x ≤ 1$ and $-1 ≤ \pm \cos x ≤ 1$ for all $x \in \mathbb{R}$ and so we have that $-1 ≤ f^{(n+1)}(x) ≤ 1$ or rather, $\mid f^{(n+1)}(x) \mid ≤ 1$. Since $\xi$ is between $0$ and $x$, we further have that:

(5)

\begin{align} \quad \mid f^{(n+1)}(\xi) \mid ≤ 1 \end{align}

Taking the absolute value of the Lagrange remainder $E_n(x)$ and we have that:

(6)

\begin{align} \quad \mid E_n(x) \mid = \biggr \rvert \frac{f^{(n+1)}(\xi)}{(n + 1)!}x^{n+1} \biggr \rvert = \frac{\mid f^{(n+1)}(\xi) \mid}{(n + 1)!} \mid x^{n+1} \mid ≤ \frac{\mid x \mid^{n+1}}{(n + 1)!} \end{align}

Now note that $\lim_{n \to \infty} \frac{\mid x \mid^{n+1}}{(n + 1)!} = 0$. By the Squeeze Theorem, we get that $\lim_{n \to \infty} \mid E_n(x) \mid = 0$ and so $\lim_{n \to \infty} E_n(x) = 0$ and thus $\lim_{n \to \infty} P_n(x) = f(x)$, so $f(x) = \sin x$ is analytic.

Example 3

Prove that $f(x) = \cos x$ is analytic by showing that the Maclaurin series for $\cos x$ represents $\cos x$ for all $x \in \mathbb{R}$.

We will not solve example 3 as it is analogous to that of example 2 given above.