Learning Outcomes
- Determine a Taylor or Maclaurin series for a function.
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 \(x^2\text{.}\)
- \(\displaystyle 2x\)
- \(\displaystyle 2\)
- \(\displaystyle 4x\)
- \(\displaystyle 4\)
(b)
Find the 3rd derivative of \(x^3\text{.}\)
- \(\displaystyle 2\)
- \(\displaystyle 3x\)
- \(\displaystyle 6\)
- \(\displaystyle 12x\)
(c)
Find the 4th derivative of \(x^4\text{.}\)
- \(\displaystyle 18\)
- \(\displaystyle 24\)
- \(\displaystyle 32\)
- \(\displaystyle 64\)
(d)
Based on these results, which of the following should always equal the \(n\)th derivative of \(x^n\) with respect to \(x\text{?}\)
- \(\displaystyle n\)
- \(\displaystyle n^2\)
- \(\displaystyle n!\)
- \(\displaystyle n^n\)
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 \(f^{(n)}(0)\text{?}\)
Fact 9.4.3.
If \(f(x)\) can be written as a power series, then there is a real number \(c\) such that
\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*}
on some interval centered at \(x=c\text{.}\)
In fact, the functions that can be represented as power series are exactly those functions which are infinitely differentiable on some open interval.
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 \(\sin(x)\text{?}\)
- \(\displaystyle \sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n\)
- \(\displaystyle \sum_{2n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\)
- \(\displaystyle \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\)
- \(\displaystyle \sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}\)
(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 \(f^{(2n+1)}(0)\text{?}\)
- \(\displaystyle f^{(2n+1)}(0)=(-1)^n\)
- \(\displaystyle f^{(2n+1)}(0)=(-1)^{n+1}\)
- \(\displaystyle f^{(2n+1)}(0)=(-1)^{2n}\)
- \(\displaystyle f^{(2n+1)}(0)=(-1)^{2n+1}\)
Fact 9.4.6.
The power series we’ve introduced for each of the following functions are in fact their Maclaurin series (Taylor series centered at \(0\)).
\begin{align*}
\frac{1}{1-x} = \sum_{n=0}^\infty \frac{n!}{n!}x^n\amp= 1+x+x^2+x^3+\dots\\
e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n\amp= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\\
\cos(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\amp =1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots\\
\sin(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\amp =x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\dots
\end{align*}
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 \(e^{i\pi}\text{.}\)
