In Example 8.7.14 we found the \(8^\text{ th }\) degree Maclaurin polynomial of \(\cos(x)\text{.}\) In doing so, we created the table shown in Table 8.8.3.

Notice how \(f^{(n)}(0)=0\) when \(n\) is odd, \(f^{(n)}(0)=1\) when \(n\) is divisible by \(4\text{,}\) and \(f^{(n)}(0)=-1\) when \(n\) is even but not divisible by 4. Thus the Maclaurin series of \(\cos(x)\) is \begin{equation*} 1-\frac{x^2}2+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!} - \cdots \end{equation*}

We can go further and write this as a summation. The coefficients alternate between positive and negative. Since we only need the terms where the power of \(x\) is even, we write the power series in terms of \(x^{2n}\text{:}\) \begin{equation*} \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!}. \end{equation*}

Table 8.8.5 shows the \(n^\text{ th }\) derivative of \(\ln(x)\) evaluated at \(x=1\) for \(n=0,\ldots,5\text{,}\) along with an expression for the \(n^\text{ th }\) term: \begin{equation*} f^{(n)}(1) = (-1)^{n+1}(n-1)! \text{ for \(n\geq 1\). } \end{equation*}

Remember that this is what distinguishes Taylor series from Taylor polynomials; we are very interested in finding a pattern for the \(n^\text{ th }\) term, not just finding a finite set of coefficients for a polynomial.

Since \(f(1) = \ln(1) = 0\text{,}\) we skip the first term and start the summation with \(n=1\text{,}\) giving the Taylor series for \(\ln(x)\text{,}\) centered at \(x=1\text{,}\) as \begin{equation*} \infser (-1)^{n+1}(n-1)!\frac{1}{n!}(x-1)^n = \infser (-1)^{n+1}\frac{(x-1)^n}{n}. \end{equation*}

Given a value \(x\text{,}\) the magnitude of the error term \(R_n(x)\) is bounded by \begin{equation*} \abs{R_n(x)} \leq \frac{\max\abs{\,f^{(n+1)}(z)}}{(n+1)!}\abs{x^{n+1}}. \end{equation*}

Since all derivatives of \(\cos(x)\) are \(\pm \sin(x)\) or \(\pm\cos(x)\text{,}\) whose magnitudes are bounded by \(1\text{,}\) we can state \begin{equation*} \abs{R_n(x)} \leq \frac{1}{(n+1)!}\abs{x^{n+1}} \end{equation*} which implies \begin{equation} -\frac{\abs{x^{n+1}}}{(n+1)!} \leq R_n(x) \leq\frac{\abs{x^{n+1}}}{(n+1)!}. \label{eq_coseqtaylor}\tag{8.8.1} \end{equation}

For any \(x\text{,}\) \(\lim\limits_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0\text{.}\) Applying the Squeeze Theorem to Equation (8.8.1), we conclude that \(\lim\limits_{n\to\infty} R_n(x) = 0\) for all \(x\text{,}\) and hence \begin{equation*} \cos(x) = \infser[0] (-1)^{n}\frac{x^{2n}}{(2n)!} \text{ for all \(x\) } . \end{equation*}

When \(k\) is a positive integer, the Maclaurin series is finite. For instance, when \(k=4\text{,}\) we have \begin{equation*} f(x) = (1+x)^4 = 1+4x+6x^2+4x^3+x^4. \end{equation*}

The coefficients of \(x\) when \(k\) is a positive integer are known as the binomial coefficients, giving the series we are developing its name.

When \(k=1/2\text{,}\) we have \(f(x) = \sqrt{1+x}\text{.}\) Knowing a series representation of this function would give a useful way of approximating \(\sqrt{1.3}\text{,}\) for instance.

To develop the Maclaurin series for \(f(x) = (1+x)^k\) for any value of \(k\neq0\text{,}\) we consider the derivatives of \(f\) evaluated at \(x=0\text{:}\)

Thus the Maclaurin series for \(f(x) = (1+x)^k\) is \begin{align*} (1+x)^k\amp =1+ k(x-c) + \frac{k(k-1)}{2!}(x-c)^2 + \frac{k(k-1)(k-2)}{3!}(x-c)^3 + \ldots\\ \dots + \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}(x-c)^n+\ldots \end{align*}

It is important to determine the interval of convergence of this series. With \begin{equation*} a_n = \frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n, \end{equation*} we apply the Ratio Test: \begin{align*} \lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}\amp =\lim_{n\to\infty} \abs{\frac{k(k-1)\cdots(k-(n-1))(k-n)}{(n+1)!}x^{n+1}}\Bigg/\abs{\frac{k(k-1)\cdots\big(k-(n-1)\big)}{n!}x^n}\\ \amp =\lim_{n\to\infty} \abs{\frac{k-n}{n}x}\\ \amp = \abs{x}. \end{align*}

The series converges absolutely when the limit of the Ratio Test is less than 1; therefore, we have absolute convergence when \(\abs{x}\lt 1\text{.}\)

While outside the scope of this text, the interval of convergence depends on the value of \(k\text{.}\) When \(k>0\text{,}\) the interval of convergence is \([-1,1]\text{.}\) When \(-1\lt k\lt 0\text{,}\) the interval of convergence is \([-1,1)\text{.}\) If \(k\leq -1\text{,}\) the interval of convergence is \((-1,1)\text{.}\)

Key Idea 8.8.9 informs us that \begin{equation*} e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \text{ and } \cos(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots. \end{equation*}

Applying Theorem 8.8.10, we find that \begin{align*} e^x\cos(x) \amp = \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right).\\ \end{align*} Distribute the right hand expression across the left: \begin{align*} \amp = 1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+x\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\frac{x^2}{2!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\\ \amp +\frac{x^3}{3!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right) + \frac{x^4}{4!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)+\cdots\\ \end{align*} Distribute again and collect like terms. \begin{align*} \amp = 1 + x -\frac{x^3}{3}-\frac{x^4}{6} - \frac{x^5}{30}+\frac{x^7}{630}+\cdots \end{align*}

While this process is a bit tedious, it is much faster than evaluating all the necessary derivatives of \(e^x\cos(x)\) and computing the Taylor series directly.

Because the series for \(e^x\) and \(\cos(x)\) both converge on \((-\infty,\infty)\text{,}\) so does the series expansion for \(e^x\cos(x)\text{.}\)

Given that \begin{equation*} \sin(x) = \infser[0] (-1)^n\frac{x^{2n+1}}{(2n+1)!} = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots, \end{equation*} we simply substitute \(x^2\) for \(x\) in the series, giving \begin{align*} \sin(x^2) \amp = \infser[0] (-1)^n\frac{(x^2)^{2n+1}}{(2n+1)!}\\ \amp \infser[0] (-1)^n\frac{(x^{4n+2}}{(2n+1)!}\\ \amp =x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!} -\frac{x^{14}}{7!}\cdots\text{.} \end{align*}

Since the Taylor series for \(\sin(x)\) has an infinite radius of convergence, so does the Taylor series for \(\sin(x^2)\text{.}\)

The Taylor expansion for \(\ln(x)\) given in Key Idea 8.8.9 is centered at \(x=1\text{,}\) so we will center the series for \(\ln(\sqrt{x})\) at \(x=1\) as well. With \begin{equation*} \ln(x) = \infser(-1)^{n+1}\frac{(x-1)^n}{n} = (x-1)- \frac{(x-1)^2}{2} +\frac{(x-1)^3}{3}-\cdots, \end{equation*} we substitute \(\sqrt{x}\) for \(x\) to obtain \begin{equation*} \ln(\sqrt{x}) = \infser(-1)^{n+1}\frac{(\sqrt{x}-1)^n}{n} = (\sqrt{x}-1)- \frac{(\sqrt{x}-1)^2}{2} +\frac{(\sqrt{x}-1)^3}{3}-\cdots. \end{equation*}

While this is not strictly a power series, it is a series that allows us to study the function \(\ln(\sqrt{x})\text{.}\) Since the interval of convergence of \(\ln(x)\) is \((0,2]\text{,}\) and the range of \(\sqrt{x}\) on \((0,4]\) is \((0,2]\text{,}\) the interval of convergence of this series expansion of \(\ln(\sqrt{x})\) is \((0,4]\text{.}\)

We learned, when studying Numerical Integration, that \(e^{-x^2}\) does not have an antiderivative expressible in terms of elementary functions. This means any definite integral of this function must have its value approximated, and not computed exactly.

We can quickly write out the Taylor series for \(e^{-x^2}\) using the Taylor series of \(e^x\text{:}\) \begin{align*} e^x \amp = \infser[0] \frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\\ \end{align*} and so \begin{align*} e^{-x^2} \amp = \infser[0] \frac{(-x^2)^n}{n!}\\ \amp = \infser[0] (-1)^n\frac{x^{2n}}{n!}\\ \amp = 1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots. \end{align*}

We use Theorem 8.6.7 to integrate: \begin{equation*} \int e^{-x^2}\ dx = C + x - \frac{x^3}{3}+\frac{x^5}{5\cdot2!}-\frac{x^7}{7\cdot3!}+\cdots +(-1)^n\frac{x^{2n+1}}{(2n+1)n!}+\cdots \end{equation*}

This is the antiderivative of \(e^{-x^2}\text{;}\) while we can write it out as a series, we cannot write it out in terms of elementary functions. We can evaluate the definite integral \(\ds \int_0^1e^{-x^2}\ dx\) using this antiderivative; substituting 1 and 0 for \(x\) and subtracting gives \begin{equation*} \int_0^1e^{-x^2}\ dx = 1-\frac{1}{3}+\frac{1}{5\cdot 2!}-\frac{1}{7\cdot3!} + \frac{1}{9\cdot4!}\cdots. \end{equation*}

Summing the 5 terms shown above give the approximation of \(0.74749\text{.}\) Since this is an alternating series, we can use the Alternating Series Approximation Theorem, (Theorem 8.5.5), to determine how accurate this approximation is. The next term of the series is \(1/(11\cdot5!) \approx 0.00075758\text{.}\) Thus we know our approximation is within \(0.00075758\) of the actual value of the integral. This is arguably much less work than using Simpson's Rule to approximate the value of the integral.

We found the first 5 terms of the power series solution to this differential equation in Example 8.6.10 in Section 8.6. These are: \begin{equation*} a_0=1, a_1 = 2, a_2 = \frac42=2, a_3=\frac{8}{2\cdot3}=\frac43, a_4=\frac{16}{2\cdot3\cdot4} = \frac23. \end{equation*}

We include the “unsimplified” expressions for the coefficients found in Example 8.6.10 as we are looking for a pattern. It can be shown that \(a_n = 2^n/n!\text{.}\) Thus the solution, written as a power series, is \begin{equation*} y = \infser[0] \frac{2^n}{n!}x^n = \infser[0] \frac{(2x)^n}{n!}. \end{equation*}

Using Key Idea 8.8.9 and Theorem 8.8.10, we recognize \(f(x) = e^{2x}\text{:}\) \begin{equation*} e^x = \infser[0] \frac{x^n}{n!} \qquad \Rightarrow \qquad e^{2x} = \infser[0] \frac{(2x)^n}{n!}. \end{equation*}