Taylor and Maclaurin Series Examples 1

Taylor and Maclaurin Series Examples 1

Recall from the Taylor and Maclaurin Series page that if a power series $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges to $f(x)$ on the interval $(c - R, c + R)$ where $c$ is the center of convergence and $R$ is a radius of convergence, then $a_n = \frac{f^{(n)}(c)}{n!}$ for $n = 0, 1, 2, ...$.

A Taylor series for a function $f$ is a series in the form:

(1)
\begin{align} \quad \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x - c)^n = f(c) + f'(c)(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!} (x - c)^3 + ... \end{align}

We call a Taylor series with the center of convergence $c = 0$ a Maclaurin series.

Furthermore, we said that a function was analytic at $c$ if the Taylor series at $x = c$ converges to $f(x)$ on an open interval containing $c$. Some common analytic functions and their Maclaurin series are given below:

• $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$, $\forall x \in \mathbb{R}$.
• $\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n + 1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$, $\forall x \in \mathbb{R}$.
• $\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - ...$, $\forall x \in \mathbb{R}$.

We will now look at some examples regarding Taylor and Maclaurin series.

Example 1

Reprove that if $f(x) = \sum_{n=0}^{\infty} a_n(x - c)^n$ on the interval $(c - R, c + R)$ then $a_n = \frac{f^{(n)}(c)}{n!}$ for each $n = 0, 1, 2, ...$.

Suppose that $f(x) = \sum_{n=0}^{\infty} a_n(x - c)^n$ on $(c - R, c + R)$. Then we have that:

(2)
\begin{align} \quad f(x) = a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3 + ... \end{align}

Plugging in $x = c$ and we have that $a_0 = \frac{f^{(0)}(c)}{0!} = f(c)$.

If we differentiate the equation from above, we have that:

(3)
\begin{align} \quad f'(x) = a_1 + 2a_2(x - c) + 3a_3(x - c)^2 + ... \end{align}

Plugging in $x = c$ once again and we have that $a_1 = \frac{f^{(1)}(c)}{1!} = f'(c)$.

If we differentiate the equative above yet again, then we have that:

(4)
\begin{align} \quad f''(x) = 2a_2 + 3\cdot 2 a_3 (x - c) + ... \end{align}

Plugging in $x = c$ and we see that $2a_2 = f^{(2)}(c)$ and so $a_2 = \frac{f^{(2)}(c)}{2!}$. We can repeat this process inductive and have that for $n = 0, 1, 2, ...$:

(5)
\begin{align} \quad a_n = \frac{f^{(n)}(c)}{n!} \end{align}

Example 2

Prove that the function $f(x) = e^x$ is analytic.

We note that the derivatives of $f(x) = e^x$ are $f^{(n)}(x) = e^x$ for all $n = 0, 1, 2, ...$. Therefore, the Taylor series of $e^x$ about $x = c$ is given by:

(6)
\begin{align} \quad \sum_{n=0}^{\infty} \frac{e^c}{n!} (x - c)^n \end{align}

Using the ratio test, we have that:

(7)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \frac{e^c}{(n+1)!} \frac{n!}{e^c} = \lim_{n \to \infty} \frac{1}{n+1} = 0 \end{align}

Therefore the radius of convergence for this power series is $\infty$.

Now suppose that $g(x)$ is the sum of this series. Then we have that:

(8)
\begin{align} \quad g(x) = \sum_{n=0}^{\infty} \frac{e^c}{n!} (x - c)^n \\ \quad g(x) = e^c + e^c(x - c) + \frac{e^c}{2!}(x - c)^2 + \frac{e^c}{3!}(x - c)^3 + ... \end{align}

Plugging in $x=c$ gives us that $g(c) = e^c$. Now differentiate the equation above to get that:

(9)
\begin{align} \quad g'(x) = e^c + e^c(x - c) + \frac{e^c}{2!}(x - c)^2 + ... \end{align}

Plugging in $x=c$ gives us that $g'(c) = e^c$. Therefore, $g(x) = g'(x)$ which implies that $g(x) = Ce^x$. Plugging in $x = c$ gives us that $g(c) = Ce^c$. But $g(c) = e^c$, and so $C = 1$. Therefore $g(x) = e^x$, that is:

(10)
\begin{align} \quad e^x = \sum_{n=0}^{\infty} \frac{e^c}{n!} (x - c)^n \end{align}

So, $f(x) = e^x$ is analytic for all $x \in \mathbb{R}$, that is, the power series $\sum_{n=0}^{\infty} \frac{e^c}{n!} (x - c)^n$ converges to $e^x$ for all $x \in \mathbb{R}$.