Lesson 18 - Taylor Series

Overview of Taylor/Maclaurin Series

Consider a function $f$ that has a power series representation at $x = a$ . Then the series has the form

\sum_{n = 0}^{\infty} C_{n} (x - a)^{n} = C_{0} + C_{1} (x - a) + C_{2} (x - a)^{2} + \dots

What should the coefficients be? If the series above is a representation for $f$ at $x = a$ , we certainly want the series to equal $f (a)$ at $x = a$ . Evaluating the series at $x = a$ , we see that

\begin{matrix} \sum_{n = 0}^{\infty} C_{n} (x - a)^{n} = C_{0} + C_{1} (a - a) + C_{2} (a - a)^{2} + \dots \\ = c_{0} \end{matrix}

Thus, the series equals $f (a)$ if the coefficient $C_{0} = f (a)$ . In addition, we would like the first derivative of the power series to equal $f^{'} (a)$ at $x = a$ . Differentiating the equation above term-by-term, we see that

\frac{d}{d x} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}) = C_{1} + 2 C_{2} (x - a) + 3 C_{3} (x - a)^{2} + \dots

Therefore, at $x = a$ , the derivative is

\begin{array}{l} \frac{d}{d x} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}) = C_{1} + 2 C_{2} (a - a) + 3 C_{3} (a - a)^{2} + \dots \\ = C_{1} . \end{array}

Therefore, the derivative of the series equals $f^{'} (a)$ if the coefficient $C_{1} = f^{'} (a)$ . Continuing in this way, we look for coefficients $C_{n}$ such that all the derivatives of the power series above will agree with all the corresponding derivatives of $f$ at $x = a$ . The second and third derivatives of the equation are given by

\begin{array}{l} \frac{d^{2}}{d x^{2}} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}) = 2 C_{2} + 3 \times_{2} C_{3} (x - a) + 4 \times_{3} C_{4} (x - a)^{2} + \dots \\ a n d \\ \frac{d^{3}}{d x^{3}} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}) = 3 \times 2 C_{3} + 4 \times 3 \times 2 C_{4} (x - a) + 5 \times 4 \times 3 C_{5} (x - a)^{2} + \dots \end{array}

Therefore, at $x = a$ , the second and third derivatives.

\frac{d^{2}}{d x^{2}} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{n}) = 2 C_{2} + 3 \times 2 C_{3} (a - a) + 4 \times 3 C_{4} (a - a)^{2} + \dots

and

\frac{d^{3}}{d x^{3}} (\sum_{n = 0}^{\infty} C_{n} (x - a)^{m} n) = 3 \times 2 C_{3} + 4 \times 3 \times 2 C_{4} (a - a) + 5 \times 4 \times 3 C_{5 (a - a)^{2}} + \dots

= 3 \times 2 C_{3}

equal $f^{″} (a)$ and $f^{‴} (a)$ , respectively, if $C_{2} = \frac{f^{″} (a)}{2}$ and $C_{3} = \frac{f^{‴} (a)}{3 \times 2}$ . More generally, we see that if $f$ we has a power series representation at $x = a$ , then the coefficients should be given by $C_{n} = \frac{f^{(n)} (a)}{n!}$ . That is, the series should be

$\sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} (x - a)^{n} = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2!} (x - a)^{2} + \frac{f^{‴} (a)}{3!} (x - a)^{3} + \dots$

This power series for $f$ is known as the Taylor series for $f$ at $a$ . If $a = 0$ , then this series is known as the Maclaurin series for f.

Theorem: Uniqueness of Taylor Series

If a function $f$ has a power series at $a$ that converges to $f$ on some open interval containing $a$ , then that power series is the Taylor series for $f$ at $a$ .

Taylor Polynomials

The $n$ th partial sum of the Taylor series for a function $f$ at $a$ is known as the $n$ th Taylor polynomial. Here are the first 4 partial sums of the Taylor series.

\begin{array}{l} p_{0} (x) = f (a) \\ p_{1} (x) = f (a) = f^{'} (a) (x - a) \\ p_{2} (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2!} (x - a)^{2} \\ p_{3} (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2!} (x - a)^{2} + \frac{f^{‴} (a)}{3!} (x - a)^{3} + \dots \end{array}

If $f$ has $n$ derivatives at $x = a$ , then the $n$ th Taylor polynomial for $f$ at $a$ is

p_{n} (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2!} (x - a)^{2} + \frac{f^{‴} (a)}{3!} (x - a)^{3} + \dots + \frac{f^{n} (a)}{n!}

Example: Finding Taylor Polynomials

Find the Taylor polynomials $p_{0}, p_{1}, p_{2}$ and $p_{3}$ for $f (x) = \ln x$ at $x = 1$ .

Solution

To find these Taylor polynomials, we need to evaluate $f$ and its first three derivatives at $x = 1$ .

\begin{array}{l} f (x) = \ln x & f (1) = 0 \\ f^{'} (x) = \frac{1}{x} & f^{'} (1) = 1 \\ f^{″} (x) = - \frac{1}{x^{2}} & f^{″} (1) = - 1 \\ f^{‴} (x) = \frac{2}{x^{2}} & f^{‴} (1) = 2 \end{array}

Therefore,

\begin{array}{l} p_{0} (x) = f (1) = 0, \\ p_{1} (x) = f (1) + f^{'} (1) (x - 1) = x - 1 \\ p_{2} (x) = f (1) + f^{'} (1) (x - 1) + \frac{f^{″} (1)}{2} (x - 1)^{2} + \frac{f^{‴} (1)}{3!} (x - 1)^{3} \\ = (x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} \end{array}

Example: Try For Yourself

Find the Taylor polynomials $p_{0}, p_{1}, p_{2}$ and $p_{3}$ for $f (x) = \frac{1}{x^{2}}$ at $x = 1$ .

Next Lesson: Lesson 19 - Taylor Polynomial, Power Series and Shifting Index

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