23. Taylor Series
Recall: The Maclaurin series for \(f(x)\) is: (assuming it converges) \[\begin{aligned} f(x) &=\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!}x^n \\ &=f(0)+f'(0)x+\dfrac{f''(0)}{2}x^2 +\dfrac{f^{(3)}(0)}{3!}x^3+\cdots+\dfrac{f^{(n)}(0)}{n!}x^n+\cdots \end{aligned}\]
b. Standard Maclaurin Series
There are several series which are so important that you should commit them to memory. These are the Maclaurin series (Taylor series about \(x=0\)) for \(e^x\), \(\sin x\), \(\cos x\) and \(\dfrac{1}{1-x}\). A few other series are nearly as important. They are the Maclaurin series for \(\ln(1+x)\) and \(\arctan x\). Here are the series and their intervals of convergence:
| \(\begin{aligned} e^x &=\sum_{n=0}^\infty \dfrac{x^n}{n!} \\ &=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3+\cdots \end{aligned}\) | \[-\infty < x < \infty\] |
| \(\begin{aligned} \sin x &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \\ &=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\cdots \end{aligned}\) | \[-\infty < x < \infty\] |
| \(\begin{aligned} \cos x &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n} \\ &=1-\dfrac{1}{2}x^2+\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\cdots \end{aligned}\) | \[-\infty < x < \infty\] |
| \(\begin{aligned} \dfrac{1}{1-x} &=\sum_{n=0}^\infty x^n \\ &=1+x+x^2+x^3+\cdots \end{aligned}\) | \[-1 \lt x \lt 1 \] |
| \(\begin{aligned} \ln(1+x) &=\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}x^n \\ &=x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-\dfrac{1}{4}x^4+\cdots \end{aligned}\) | \[-1 \lt x \le 1 \] |
| \(\begin{aligned} \arctan x &=\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}x^{2n+1} \\ &=x-\dfrac{1}{3}x^3+\dfrac{1}{5}x^5-\dfrac{1}{7}x^7+\cdots \end{aligned}\) | \[-1 \le x \le 1 \] |
The series for \(\dfrac{1}{1-x}\) is just the geometric series. The series for \(\ln(1+x)\) and \(\arctan x\) and their intervals of convergence were derived in the previous chapter by manipulating the series for \(\dfrac{1}{1-x}\). The series for \(e^x\), \(\sin x\) and \(\cos x\) are derived in the following exercises:
Derive the Maclaurin series for \(e^x\) and find its radius and interval of convergence.
\(\begin{aligned}\displaystyle
e^x
&=\sum_{n=0}^\infty \dfrac{x^n}{n!} \\
&=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3+\cdots
\end{aligned}\)
\(R=\infty\)
\((-\infty,\infty)\)
The function, its derivatives and their values at \(x=0\) are: \[\begin{aligned} f(x)&=e^x&f(0)&=1 \\ f'(x)&=e^x&f'(0)&=1 \\ \vdots& &\vdots& \\ f^{(n)}(x)&=e^x\qquad& f^{(n)}(0)&=1 \end{aligned}\] So the Taylor series is \[\begin{aligned} f(x) &=\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!}x^n \\ &=f(0)+f'(0)x+\dfrac{f''(0)}{2}x^2+\cdots \\ e^x &=\sum_{n=0}^\infty \dfrac{1}{n!}x^n \\ &=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots \end{aligned}\] To find the radius of convergence, we use the ratio test: \[\begin{aligned} \rho=\lim_{n\rightarrow\infty}\dfrac{|a_{n+1}|}{|a_n|} &=\lim_{n\rightarrow\infty}\dfrac{|x|^{n+1}}{(n+1)!}\dfrac{n!}{|x|^n} \\ &=\lim_{n\rightarrow\infty}\dfrac{|x|}{n+1}=0 \lt 1 \end{aligned}\] So the series is convergent for all \(x\), the radius of convergence is \(R=\infty\) and the interval of convergence is \((-\infty,\infty)\).
Derive the Maclaurin series for \(\sin x\) and find its radius and interval of convergence.
\(\begin{aligned}\displaystyle
\sin x
&=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \\
&=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\cdots
\end{aligned}\)
\(R=\infty\)
\((-\infty,\infty)\)
The function and \(4\) derivatives and their values at \(x=0\) are:
\[\begin{aligned} f(x)&=\sin x \\ f'(x)&=\cos x \\ f''(x)&=-\sin x \\ f'''(x)&=-\cos x \\ f^{(4)}(x)&=\sin x \end{aligned}\]
\[\begin{aligned} f(0)&=0 \\ f'(0)&=1 \\ f''(0)&=0 \\ f'''(0)&=-1 \\ f^{(4)}(0)&=0 \end{aligned}\]
Thereafter, the derivatives repeat cyclically every \(4\) derivatives. Notice the even derivatives are all \(0\) and the odd derivatives alternate between \(\pm 1\). Thus the Maclaurin series is: \[\begin{aligned} f(x) &=f(0)+f'(0)x+\dfrac{f''(0)}{2}x^2 \\ &\quad+\dfrac{f'''(0)}{3!}x^3+\dfrac{f^{(4)}(0)}{4!}x^4+\cdots \\ \sin x &=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\cdots \\ &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \end{aligned}\] To find the radius of convergence, we use the ratio test: \[\begin{aligned} \rho=\lim_{n\rightarrow\infty}\dfrac{|a_{n+1}|}{|a_n|} &=\lim_{n\rightarrow\infty}\dfrac{|x|^{2n+3}}{(2n+3)!}\dfrac{(2n+1)!}{|x|^{2n+1}} \\ &=\lim_{n\rightarrow\infty}\dfrac{|x|^2}{(2n+3)(2n+2)}=0 \lt 1 \end{aligned}\] So the series is convergent for all \(x\), the radius of convergence is \(R=\infty\) and the interval of convergence is \((-\infty,\infty)\).
Derive the Maclaurin series for \(\cos x\) and find its radius and interval of convergence.
\(\begin{aligned}\displaystyle
\cos x
&=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n} \\
&=1-\dfrac{1}{2}x^2+\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\cdots
\end{aligned}\)
\(R=\infty\)
\((-\infty,\infty)\)
The function and \(4\) derivatives and their values at \(x=0\) are:
\[\begin{aligned} f(x)&=\cos x \\ f'(x)&=-\sin x \\ f''(x)&=-\cos x \\ f'''(x)&=\sin x \\ f^{(4)}(x)&=\cos x \end{aligned}\]
\[\begin{aligned} f(0)&=1 \\ f'(0)&=0 \\ f''(0)&=-1 \\ f'''(0)&=0 \\ f^{(4)}(0)&=1 \end{aligned}\]
Thereafter, the derivatives repeat cyclically every \(4\) derivatives. Notice the odd derivatives are all \(0\) and the even derivatives alternate between \(\pm 1\). Thus the Maclaurin series is: \[\begin{aligned} f(x) &=f(0)+f'(0)x+\dfrac{f''(0)}{2}x^2 \\ &\quad+\dfrac{f'''(0)}{3!}x^3+\dfrac{f^{(4)}(0)}{4!}x^4+\cdots \\ \cos x &=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\cdots \\ &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n} \end{aligned}\] To find the radius of convergence, we use the ratio test: \[\begin{aligned} \rho=\lim_{n\rightarrow\infty}\dfrac{|a_{n+1}|}{|a_n|} &=\lim_{n\rightarrow\infty}\dfrac{|x|^{2n+2}}{(2n+2)!}\dfrac{(2n)!}{|x|^{2n}} \\ &=\lim_{n\rightarrow\infty}\dfrac{|x|^2}{(2n+2)(2n+1)}=0 \lt 1 \end{aligned}\] So the series is convergent for all \(x\), the radius of convergence is \(R=\infty\) and the interval of convergence is \((-\infty,\infty)\).
On subsequent pages you will be given this However, on an exam you may be expected to remember them.
Notice that \(\sin x\) and \(\arctan x\) are odd functions and their series
have only odd powers and odd factorials.
Similarly, \(\cos x\) is an even function and its series has only even powers
and even factorials.
Heading
Placeholder text: Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum