The Integral Remainder Examples 1

Recall from The Integral Remainder page that if $f$ is $n + 1$ times differentiable on some interval containing the center of convergence $c$ and $x$, and if $P_n(x) = f(c) + \frac{f^{(1)}(c)}{1!}(x - c) + \frac{f^{(2)}(c)}{2!}(x - c)^2 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$ is the $n^{\mathrm{th}}$ order Taylor polynomial of $f$ at $x = c$. Then for $f(x) = P_n(x) + E_n(x)$ the error term $E_n(x)$ of $P_n(x)$ from $f(x)$ can be computed with the following formula:

(1)

\begin{align} \quad E_n(x) = \frac{1}{n!} \int_c^x (x - t)^n f^{(n+1)}(t) \: dt \end{align}

Before we look at some examples, we must first acknowledge the following inequality which the reader should attempt to verify:

(2)

\begin{align} \quad \biggr \rvert \int_a^b f(x) \: dx \biggr \rvert ≤ \int_a^b \mid f(x) \mid \: dx \end{align}

We will now look at some examples of applying this form of the error/remainder formula.

Example 1

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

Since we are dealing with the Maclaurin series for $e^x$ we have that $c = 0$ and so the integral remainder for our error is:

(3)

\begin{align} \quad E_n(x) = \frac{1}{n!} \int_0^x (x - t)^n f^{(n+1)}(t) \: dt \end{align}

Now we note that since $f(x) = e^x$ we have that $f^{(n+1)}(x) = e^x$ for $n ≥ 0$ and thus we obtain the integral:

(4)

\begin{align} \quad E_n(x) = \frac{1}{n!} \int_0^x (x - t)^n e^t \: dt \\ \quad E_n(x) = \frac{1}{n!} \left [ \frac{(x - t)^{n+1}e^t}{(n+1)} \right ]_{t=0}^{t=x} \\ \quad E_n(x) = \frac{1}{n!} \left [ 0 - \frac{x^{n+1}}{(n+1)} \right] \\ \quad E_n(x) = -\frac{x^{n+1}}{(n+1)!} \end{align}

Taking the absolute value of $E_n(x)$ and we get that:

(5)

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

Note that $\lim_{n \to \infty} \frac{\mid x \mid^{n+1}}{(n+1)!} = 0$ so $\lim_{n \to \infty} P_n(x) = f(x)$ and 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 first note that since $f(x) = \sin x$ we have that $f^{(n+1)}(x) = \pm \sin x$ or $f^{(n+1)}(x) = \pm \cos x$. We already know that $-1 ≤ \pm \sin x ≤ 1$ and $-1 ≤ \pm \cos x ≤ 1$ for all $x \in \mathbb{R}$ and so $\mid f^{(n+1)}(x) \mid ≤ 1$ for all $x \in \mathbb{R}$. Now since we're dealing with a Maclaurin series, our center of convergence $c = 0$, and thus:

(6)

\begin{align} \quad E_n(x) = \frac{1}{n!} \int_0^x (x - t)^n f^{(n+1)}(t) \: dt \end{align}

Now let's take the absolute value of $E_n(x)$ and using the inequality we mentioned at the top of this page along with the fact that $\mid f^{(n+1)}(x) \mid ≤ 1$. For $x > 0$ we have that:

(7)

\begin{align} \quad \quad \mid E_n(x) \mid = \biggr \rvert \frac{1}{n!} \int_0^x (x - t)^n f^{(n+1)}(t) \: dt \biggr \rvert ≤ \frac{1}{n!} \int \mid (x - t)^n f^{(n+1)}(t) \mid \: dt = \frac{1}{n!} \int_0^x (x - t)^n \mid f^{(n+1)}(t) \mid \: dt = \frac{1}{n!} \int_0^x (x - t)^n \: dt = \frac{x^{n+1}}{(n+1)!} \end{align}

Now for $x < 0$ we have that:

(8)

\begin{align} \quad \mid E_n(x) \mid = \biggr \rvert - \frac{1}{n!} \int_x^0 (x - t)^n f^{(n+1)}(t) \: dt \biggr \rvert ≤ \frac{1}{n!} \int_x^0 (t - x)^n \: dt = \frac{(-x)^{n+1}}{(n + 1)!} \end{align}

So whether $x > 0$ or $x < 0$ we have that $\mid E_n(x) \mid ≤ \frac{\mid x \mid^{n+1}}{(n + 1)!}$ and so $\lim_{n \to \infty} \frac{\mid x \mid^{n+1}}{(n + 1)!} = 0$ and by the Squeeze Theorem we have that $\lim_{n \to \infty} E_n(x) = 0$ and so $\lim_{n \to \infty} P_n(x) = f(x)$, so $f(x) = \sin x$ is analytic for all $x \in \mathbb{R}$.