Skip to main content

Section 9.4 Taylor Series (PS4)

Subsection 9.4.1 Activities

Activity 9.4.1.

The following tasks will help us find a mechanism to produce a power series given information about its derivatives.
(a)
Find the 2nd derivative of \(x^2\text{.}\)
  1. \(\displaystyle 2x\)
  2. \(\displaystyle 2\)
  3. \(\displaystyle 4x\)
  4. \(\displaystyle 4\)
(b)
Find the 3rd derivative of \(x^3\text{.}\)
  1. \(\displaystyle 2\)
  2. \(\displaystyle 3x\)
  3. \(\displaystyle 6\)
  4. \(\displaystyle 12x\)
(c)
Find the 4th derivative of \(x^4\text{.}\)
  1. \(\displaystyle 18\)
  2. \(\displaystyle 24\)
  3. \(\displaystyle 32\)
  4. \(\displaystyle 64\)
(d)
Based on these results, which of the following should always equal the \(n\)th derivative of \(x^n\) with respect to \(x\text{?}\)
  1. \(\displaystyle n\)
  2. \(\displaystyle n^2\)
  3. \(\displaystyle n!\)
  4. \(\displaystyle n^n\)

Activity 9.4.2.

Let’s use derivatives to rediscover the sequence \(a_n\) which gives a power series representation for \(e^x\text{.}\)
(a)
Let’s say that
\begin{equation*} e^x=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4\dots\text{.} \end{equation*}
What must \(a_0\) be to satisfy \(e^0=1\text{?}\)
(b)
Then,
\begin{equation*} \frac{d}{dx}[e^x]=e^x=a_1+2a_2x+3a_3x^2+4a_4x^3\dots\text{.} \end{equation*}
What must \(a_1\) be to also satisfy \(e^0=1\text{?}\)
(c)
Then,
\begin{equation*} \frac{d^2}{dx^2}[e^x]=e^x=2a_2+6a_3x+12a_4x^2+\dots\text{.} \end{equation*}
What must \(a_2\) be to also satisfy \(e^0=1\text{?}\)
(d)
Then,
\begin{equation*} \frac{d^3}{dx^3}[e^x]=e^x=6a_3+24a_4x+\dots\text{.} \end{equation*}
What must \(a_3\) be to also satisfy \(e^0=1\text{?}\)
(e)
So this \(6a_3\) term was obtained from the fact that the \(3\)rd derivative of \(x^3\) is \(3!=6\text{.}\)
So finally, we may skip ahead to the \(n\)th derivative:
\begin{equation*} \frac{d^n}{dx^n}[e^x]=e^x=n!\cdot a_n+(n+1)!\cdot a_{n+1}\cdot x+\dots\text{.} \end{equation*}
What must \(a_n\) be to also satisfy \(e^0=1\text{?}\)
(f)
This reveals the power series we previously found for \(e^x\text{:}\)
\begin{equation*} e^x=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty \frac{1}{n!}x^n\text{.} \end{equation*}
So in general, if \(f(x)=a_0+a_1x+a_2x^2+\dots\text{,}\) then
\begin{equation*} \frac{d^n}{dx^n}[f(x)]=f^{(n)}(x)=n!\cdot a_n+(n+1)!\cdot a_{n+1}\cdot x+\dots\text{.} \end{equation*}
What must \(a_n\) be to produce the correct value for \(f^{(n)}(0)\text{?}\)

Definition 9.4.4.

The Taylor series generated by \(f(x)\) and centered at \(x=c\) is given by
\begin{align*} f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n\\ &=f(c)+f^\prime(c)(x-c)+\frac{f^{\prime\prime}(c)}{2!}(x-c)^2+\frac{f^{(3)}(c)}{3!}(x-c)^3+\ldots \end{align*}
with an interval of convergence determinable by series convergence rules.
When \(c=0\text{,}\)
\begin{align*} f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\\ &=f(0)+f^\prime(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\ldots \end{align*}
is called the Maclaurin series generated by \(f\text{.}\)

Activity 9.4.5.

Observe that \(f(x)=\sin(x)\) is a function such that:
\begin{equation*} \begin{array}{c|c|c|c|c|c|c|c} f(0) & f'(0) & f''(0) & f^{(3)}(0) & f^{(4)}(0) & f^{(5)}(0) & f^{(6)}(0) & f^{(7)}(0) \\ \hline \sin(0) & \cos(0) & -\sin(0) & -\cos(0) & \sin(0) & \cos(0) & -\sin(0) & -\cos(0) \\ \hline 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 \end{array} \end{equation*}
(a)
Given the zeros appearing for every even derivative above, which of these is a valid simplification of the Maclarin series \(\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\) for \(\sin(x)\text{?}\)
  1. \(\displaystyle \sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n\)
  2. \(\displaystyle \sum_{2n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\)
  3. \(\displaystyle \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\)
  4. \(\displaystyle \sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}\)
(b)
Now consider the following consolidated chart:
\begin{equation*} \begin{array}{c|c|c|c|c|c|c|c} f^{(1)}(0) & f^{(3)}(0) &f^{(5)}(0) & f^{(7)}(0) \\ \hline \cos(0) & -\cos(0) & \cos(0) & -\cos(0) \\ \hline 1 & -1 & 1 & -1 \end{array} \end{equation*}
Which formula yields these alternating \(1\)s and \(-1\)s appearing for \(f^{(2n+1)}(0)\text{?}\)
  1. \(\displaystyle f^{(2n+1)}(0)=(-1)^n\)
  2. \(\displaystyle f^{(2n+1)}(0)=(-1)^{n+1}\)
  3. \(\displaystyle f^{(2n+1)}(0)=(-1)^{2n}\)
  4. \(\displaystyle f^{(2n+1)}(0)=(-1)^{2n+1}\)

Definition 9.4.7.

For a function \(f(x)\) with a Taylor series centered at \(x=c\text{,}\)
\begin{align*} f(x) \amp\approx T_k(x)\\ \amp = \sum_{n=0}^k \frac{f^{(n)}(c)}{n!}(x-c)^n \\ \amp = f(c)+f^\prime(c)(x-c)+\frac{f^{\prime\prime}(c)}{2!}(x-c)^2+ \ldots+\frac{f^{(k)}(c)}{k!}(x-c)^k \end{align*}
where \(T_k(x)\) is called the \(k^{th}\) degree Taylor polynomial generated by \(f\) and centered at \(x=c\text{.}\)
The \(k^{th}\) degree Taylor polynomial can be seen as the “best” polynomial of degree \(k\) or less for approximating \(f(x)\) for values close to \(x=c\text{.}\) Note that the \(1^{st}\) degree Taylor polynomial is also known as the linearization of \(f\text{.}\)

Activity 9.4.8.

Let \(f(x)\) be a function such that:
\begin{equation*} \begin{array}{c|c|c|c|c|c|c} f(4) & f'(4) & f''(4) & f'''(4) & f^{(4)}(4) & f^{(5)}(4) & f^{(6)}(4) \\ \hline 0 & 1 & 2 & 3 & 4 & 5 & 6 \end{array} \end{equation*}
(a)
Find a Taylor polynomial for \(f(x)\) centered at \(x=4\) of degree \(3\text{.}\)
(b)
Using the table above, find a general closed form for \(f^{(n)}(4)\text{.}\)
(c)
Use (b) to find a Taylor series for \(f(x)\) centered at \(x=4\text{.}\)

Activity 9.4.9.

Let \(f(x)\) be a function such that:
\begin{equation*} \begin{array}{c|c|c|c|c|c|c} f(-2) & f'(-2) & f''(-2) & f'''(-2) & f^{(4)}(-2) & f^{(5)}(-2) & f^{(6)}(-2) \\ \hline 0 & 2 & -16 & 54 & -128 & 250 & -432 \end{array} \end{equation*}
(a)
Find a Taylor polynomial for \(f(x)\) centered at \(x=-2\) of degree \(3\text{.}\)
(b)
Using the table above, find a general closed form for \(f^{(n)}(-2)\text{.}\)
(c)
Use (b) to find a Taylor series for \(f(x)\) centered at \(x=-2\text{.}\)

Remark 9.4.10.

You might have seen \(\sqrt{-1}\) written as \(i\text{,}\) and know that \(z\) is a complex number if \(z=a+bi\) for some real numbers \(a\) and \(b\text{.}\) Note that \(i^2=-1\text{,}\) \(i^3=(i^2)i=-i\text{,}\) \(i^4=(i^2)^2=1\text{,}\) \(i^5=(i^4)i=i\text{,}\) and so on. This gives rise to the following notion.

Definition 9.4.11. Euler’s Identity.

For any real number \(\theta\text{,}\)
\begin{align*} e^{i\theta} & = 1+\displaystyle\frac{i\theta}{1!}+\displaystyle\frac{(i\theta)^2}{2!}+\displaystyle\frac{(i\theta)^3}{3!}+\displaystyle\frac{(i\theta)^4}{4!}+\displaystyle\frac{(i\theta)^5}{5!}+\displaystyle\frac{(i\theta)^6}{6!}+\displaystyle\frac{(i\theta)^7}{7!}+\displaystyle\frac{(i\theta)^8}{8!}+\ldots\\ & = 1+i\theta-\displaystyle\frac{\theta^2}{2!}-\displaystyle\frac{i\theta^3}{3!}+\displaystyle\frac{\theta^4}{4!}+\displaystyle\frac{i\theta^5}{5!}-\displaystyle\frac{\theta^6}{6!}-\displaystyle\frac{i\theta^7}{7!}+\displaystyle\frac{\theta^8}{8!}+\ldots\\ & = \left(1-\displaystyle\frac{\theta^2}{2!}+\displaystyle\frac{\theta^4}{4!}-\displaystyle\frac{\theta^6}{6!}+\ldots\right)+i\left(\theta-\displaystyle\frac{\theta^3}{3!}+\displaystyle\frac{\theta^5}{5!}-\displaystyle\frac{\theta^7}{7!}+\ldots\right)\\ & = \cos(\theta)+i\sin(\theta). \end{align*}

Activity 9.4.12.

Use Euler’s identity to evaluate \(e^{i\pi}\text{.}\)

Subsection 9.4.2 Videos

Figure 191. Video: Determine a Taylor or Maclaurin series for a function

Subsection 9.4.3 Exercises