3.15 Taylor Series

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We find the Taylor Series for a function.

1 Taylor Series

In this section, we will find a power series expansion centered at $x = a$ for a given infinitely differentiable function, $f(x)$ . Such a series is called a Taylor Series. In the case that $a = 0$ , we call the series a Maclaurin Series.

In other words, for a given function $f(x)$ and a given center, $x = a$ , we wish to create a power series so that $f(x) = \sum _{n=0}^\infty c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots ,$ where the series converges in an interval of the form $(a-r, a+r)$ for some $r >0$ . This is an interval centered at $x = a$ of radius $r$ . The key to creating the Taylor Series is to find the coefficients, $c_n$ . To find $c_0$ , let $x = a$ in the equation above. We get: $f(a) = \sum _{n=0}^\infty c_n (a-a)^n = c_0 + c_1(a-a) + c_2(a-a)^2 + \cdots = c_0.$ Next, to find $c_1$ , we differentiate both sides of the equation above, and then let $x = a$ . Differentiating gives $f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \cdots ,$ and substituting in $x = a$ , this simplifies to $f'(a) = c_1 + 2c_2(a-a) + 3c_3(a-a)^2 + \cdots = c_1.$

Next, to find $c_2$ , we differentiate both sides of the equation above, and then let $x = a$ . Differentiating gives $f''(x) = 2c_2 + 3\cdot 2 c_3(x-a) + 4\cdot 3 c_4(x-a)^2 + \cdots ,$ and substituting in $x = a$ , this simplifies to $f''(a) = 2c_2 + 3\cdot 2 c_3(a-a) + 4\cdot 3 c_4(a-a)^2 + \cdots = 2c_2,$ so that $c_2 = \frac {f''(a)}{2}$ .

Continuing in this fashion, we can find $c_3$ : $f'''(x) = 3 \cdot 2 \cdot 1 c_3 + 4\cdot 3 \cdot 2 c_4 (x-a) + 5 \cdot 4 \cdot 3 c_5 (x-a)^2 + \cdots ,$ and letting $x = a$ gives $f'''(a) = 3 \cdot 2 \cdot 1 c_3 + 4\cdot 3 \cdot 2 c_4 (a-a) + 5 \cdot 4 \cdot 3 c_5 (a-a)^2 + \cdots = 6c_3,$ so that $c_3 = \frac {f'''(a)}{6}$ . The pattern continues, so that $c_4 = \frac {f^{(4)}(a)}{4\cdot 3 \cdot 2 \cdot 1} = \frac {f^{(4)}(a)}{24}$ and $c_5 = \frac {f^{(5)}(a)}{5\cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac {f^{(5)}(a)}{120}$ . In general, we have $c_n = \frac {f^{(n)}(a)}{n\cdot (n-1) \cdot (n-2) \cdot \dots \cdot 1} = \frac {f^{(n)}(a)}{n!}.$ Thus, the Taylor Series representation of $f(x)$ centered at $x = a$ is given by

$f(x) = \sum _{n=0}^\infty c_n (x-a)^n = \sum _{n=0}^\infty \frac {f^{(n)}(a)}{n!} (x-a)^n.$

We present this results in the following theorem.

Taylor Series Suppose the function $f(x)$ has derivatives of all orders at $x = a$ . Then the Taylor Series representation for $f(x)$ centered at $x = a$ is given by $f(x) = \sum _{n=0}^\infty \frac {f^{(n)}(a)}{n!} (x-a)^n.$

In the case that $a = 0$ , the series has the form $f(x) = \sum _{n=0}^\infty \frac {f^{(n)}(0)}{n!} x^n,$ and is called the Maclaurin Series for $f(x)$ .

example 1 Find the first four terms of the Taylor Series $f(x) = \sqrt x$ centered at $x = 4$ .
Since we are asked to find the first four terms, we must find the coefficients $c _0, c_1, c_2$ and $c_3$ and the answer will have the form $\sqrt x = c_0 + c_1(x-4) + c_2(x-4)^2 + c_3(x-4)^3 + \cdots .$ To find the coefficients, we create the following table:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), \; a = 4} & \bf {c_n} \\[1em] \hline 0 & x^{1/2} & 4^{1/2} = 2 & 2 \\[1em] \hline 1 & \frac 12 x^{-1/2} & \frac 12 \cdot 4^{-1/2} = \frac 14 & \frac 14 \\[1em] \hline 2 & -\frac 14 x^{-3/2} & -\frac 14 \cdot 4^{-3/2} = -\frac 14 \cdot \frac 18 = -\frac {1}{32} & -\frac {1}{64} \\[1em] \hline 3 & \frac 38 x^{-5/2} & \frac 38 \cdot 4^{-5/2} = \frac 38 \cdot \frac {1}{32} = \frac {3}{256} & \frac {1}{512}\\[1em] \hline \end {array}$

We can now use the coefficients in the right-hand column of the table to write the final answer. The first four terms of the Taylor Series for the function $f(x) = \sqrt x$ centered at $x = 4$ are \begin{align*} \sqrt x &= c_0 + c_1(x-4) + c_2(x-4)^2 + c_3(x-4)^3 + \cdots \\ &= 2 + \frac 14(x-4) -\frac{1}{64}(x-4)^2 + \frac{1}{512}(x-4)^3 + \cdots \end{align*}

(problem 1) Find the first four terms of the Taylor Series $f(x) = \sqrt x$ centered at $x = 1$ .

The final answer has the form:

a) $\sqrt x = c_0 + c_1(x-1) + c_2(x-1) + c_3(x-1) + \cdots$ b) $\sqrt x = c_0 + c_1(x- \sqrt 1) + c_2(x-\sqrt 2) + c_3(x-\sqrt 3) + \cdots$ c) $\sqrt x = c_0 + c_1(\sqrt x-1) + c_2(\sqrt x -1)^2 + c_3(\sqrt x-1)^3 + \cdots$ d) $\sqrt x = c_0 + c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + \cdots$

Fill in the following table to find the coefficients:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a),\; a = 1} & \bf {c_n} \\[1em] \hline 0 & f(x) = x^{1/2} & f(1) = 1 & c_0 = \answer {1} \\[1em] \hline 1 & f'(x) = \answer {\frac 12 x^{-1/2}} & f'(1) = \answer {1/2} & c_1 = \answer {\frac 12} \\[1em] \hline 2 & f''(x) =\answer {-\frac 14 x^{-3/2}} & f''(1) = \answer {-1/4} & c_2 = \answer {-\frac {1}{8}} \\[1em] \hline 3 & f'''(x) =\answer {\frac 38 x^{-5/2}} & f'''(1) = \answer {\frac 38 } & c_3 = \answer {\frac {1}{16}}\\[1em] \hline \end {array}$

The first four terms of the Taylor Series for the function $f(x) = \sqrt x$ centered at $x = 1$ are $\sqrt x = \answer {1 + \frac 12(x-1) -\frac {1}{8}(x-1)^2 + \frac {1}{16}(x-1)^3} + \cdots$

In the last example and problem, we computed only the first four terms of the Taylor Series, yielding a polynomial of degree 3. This is a special polynomial.

Taylor Polynomial Give a function $f(x)$ and its first $n$ derivatives at $x = a$ , $f^{(k)}(a), k= 0, 1, 2, \ldots , n$ , the polynomial $T_n (x) = \sum _{k=0}^n\frac {f^{(k)}(a)}{k!} (x-a)^k.$ is called the Taylor Polynomial of degree $n$ for $f(x)$ centered at $x=a$ .

The Taylor Polynomial of degree 1 for $f(x)$ centered at $x = a$ , $T_1(x) = f(a) + f'(a)(x-a)$ is also known as the Linearization of $f(x)$ at $x = a$ which is used in making linear approximations.

example 2 Find the Taylor Polynomial of degree 3, $T_3(x)$ , for $f(x) = \sqrt [3] x$ centered at $x = -1$ .
Since we are asked to find $T_3(x)$ , we must find the coefficients $c _0, c_1, c_2$ and $c_3$ and the answer will have the form $T_3(x) = c_0 + c_1(x+1) + c_2(x+1)^2 + c_3(x+1)^3.$ To find the coefficients, we create the following table:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), \; a = -1} & \bf {c_n} \\[1em] \hline 0 & x^{1/3} & (-1)^{1/3} = -1 & -1 \\[1em] \hline 1 & \frac 13 x^{-2/3} & \frac 13 \cdot (-1)^{-2/3} = \frac 13 & \frac 13 \\[1em] \hline 2 & -\frac 29 x^{-5/3} & -\frac 29 \cdot (-1)^{-5/3} = -\frac 29 \cdot (-1) = \frac 29 & \frac {1}{9} \\[1em] \hline 3 & \frac {10}{27} x^{-8/3} & \frac {10}{27} \cdot (-1)^{-8/3} = \frac {10}{27} \cdot (1) = \frac {10}{27} & \frac {5}{81}\\[1em] \hline \end {array}$

We can now use the coefficients in the right-hand column of the table to write the final answer. The Taylor Polynomial of degree 3 for $f(x) = \sqrt [3] x$ centered at $x = -1$ is \begin{align*} T_3(x) &= c_0 + c_1(x+1) + c_2(x+1)^2 + c_3(x+1)^3 \\ &= -1 + \frac 13(x+1) + \frac{1}{9}(x+1)^2 + \frac{5}{81}(x+1)^3 \end{align*}

(problem 2a) Find the Taylor Polynomial of degree 3, $T_3(x)$ , for $f(x) = \sqrt [3] x$ centered at $x = 8$ .

The final answer has the form:

a) $\sqrt [3] x = c_0 + c_1(x-8) + c_2(x-8) + c_3(x-8)$ b) $\sqrt [3] x = c_0 + c_1(x- \sqrt [3] 1) + c_2(x-\sqrt [3] 2) + c_3(x-\sqrt [3] 3)$ c) $\sqrt [3] x = c_0 + c_1(x-8) + c_2(x-8)^2 + c_3(x-8)^3$ d) $\sqrt [3] x = c_0 + c_1(\sqrt [3]x-8) + c_2(\sqrt [3] x -8)^2 + c_3(\sqrt [3] x-8)^3$

Fill in the following table to find the coefficients:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), a = 8} & \bf {c_n} \\[1em] \hline 0 & f(x) = x^{1/3} & f(8) = 2 & c_0 = \answer {2} \\[1em] \hline 1 & f'(x) = \answer {\frac 13 x^{-2/3}} & f'(8) = \answer {1/12} & c_1 = \answer {1/12} \\[1em] \hline 2 & f''(x) =\answer {-\frac 29 x^{-5/3}} & f''(8) = \answer {-1/144} & c_2 = \answer {-\frac {1}{288}} \\[1em] \hline 3 & f'''(x) =\answer {\frac {10}{27} x^{-8/3}} & f'''(8) = \answer {5/3456 } & c_3 = \answer {\frac {5}{20736}}\\[1em] \hline \end {array}$

The first four terms of the Taylor Series for the function $f(x) = \sqrt [3] x$ centered at $x = 8$ are $T_3(x) = \answer {2 + \frac {1}{12}(x-8) -\frac {1}{288}(x-8)^2 + \frac {5}{20736}(x-8)^3}$

(problem 2b) Find the Taylor Polynomial of degree 3, $T_3(x)$ , for $f(x) = x\sqrt x$ centered at $x = 4$ .

The function $f(x) = x\sqrt x$ can be written as a power function, $x^k$ , for an appropriate value of $k$ . Write $x\sqrt x$ as a power function:

$x\sqrt x = x*x^{1/2}$ ; add exponents

$x\sqrt x = \answer {x^{3/2}}$

The final answer has the form:

a) $x\sqrt x = c_0 + c_1(x-4) + c_2(x-4) + c_3(x-4)$ b) $x\sqrt x = c_0 + c_1(x- 1\sqrt 1) + c_2(x-2\sqrt 2) + c_3(x-3\sqrt 3)$ c) $x\sqrt x = c_0 + c_1(x\sqrt x -4) + c_2(x\sqrt x -4)^2 + c_3(x\sqrt x-4)^3$ d) $x\sqrt x = c_0 + c_1(x-4) + c_2(x-4)^2 + c_3(x-4)^3$

Fill in the following table to find the coefficients:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), a = 4} & \bf {c_n} \\[1em] \hline 0 & f(x) = x^{3/2} & f(4) = 8 & c_0 = \answer {8} \\[1em] \hline 1 & f'(x) = \answer {\frac 32 x^{1/2}} & f'(4) = \answer {3} & c_1 = \answer {3} \\[1em] \hline 2 & f''(x) =\answer {\frac 34 x^{-1/2}} & f''(4) = \answer {3/8} & c_2 = \answer {3/16} \\[1em] \hline 3 & f'''(x) =\answer {-\frac 38 x^{-3/2}} & f'''(4) = \answer {-3/64} & c_3 = \answer {-\frac {1}{128}}\\[1em] \hline \end {array}$

The first four terms of the Taylor Series for the function $f(x) = x\sqrt x$ centered at $x = 4$ are $T_3(x) = \answer {8 + 3(x-4) +\frac {3}{16}(x-4)^2 - \frac {1}{128}(x-4)^3} + \cdots$

example 3 Find the Taylor Polynomial of degree 4, $T_4(x)$ , for $f(x) = \ln (x)$ centered at $x = 1$ .
Since we are asked to find the forth degree Taylor Polynomial, we must find the coefficients $c _0, c_1, c_2, c_3$ and $c_4$ and the answer will have the form $\ln (x) = c_0 + c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + c_4(x-1)^4.$ To find the coefficients, we create the following table:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), \; a = 1} & \bf {c_n} \\[1em] \hline 0 & \ln (x) & \ln (1) = 0 & 0 \\[1em] \hline 1 & \frac {1}{x} & 1 & 1 \\[1em] \hline 2 & -\frac {1}{x^2} & -1 & -1/2 \\[1em] \hline 3 & \frac {2}{x^3} & 2 & \frac 13\\[1em] \hline 4 & -\frac {6}{x^4} & -6 & -\frac 14\\[1em] \hline \end {array}$

We can now use the coefficients in the right-hand column of the table to write the final answer. The first five terms of the Taylor Series for the function $f(x) = \ln (x)$ centered at $x = 1$ are \begin{align*} T_4(x) &= c_0 + c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + c_4(x-1)^4\\ &= 0 + 1(x-1) -\frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4\\ &= (x-1) -\frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4 \end{align*}

(problem 3) Find the Taylor Polynomial of degree 4, $T_4(x)$ , for $f(x) = \ln (2x)$ centered at $x = 1$ .

The final answer has the form:

a) $\ln (2x) = c_0 + c_1(x-1) + c_2(x-1) + c_3(x-1) + c_4(x-1)$ b) $\ln (2x) = c_0 + c_1(x-1) + c_2(x-1)^2 + c_3(x-1)^3 + c_4(x-1)^4$ c) $\ln (2x) = c_0 + c_1(\ln (2x) -1) + c_2(\ln (2x) -1)^2 + c_3(\ln (2x)-1)^3+ c_4(\ln (2x)-1)^4$ d) $\ln (2x) = c_0 + c_1(x- \ln (1)) + c_2(x-\ln (2)) + c_3(x-\ln (3))+ c_4(x-\ln (4))$

Fill in the following table to find the coefficients:

$\begin {array}{|c|c|c|c|} \hline \bf {n} & \bf {f^{(n)}(x)} & \bf {f^{(n)}(a), a = 1} & \bf {c_n} \\[1em] \hline 0 & f(x) = \ln (2x) & f(1) = \ln (2) & c_0 = \answer {\ln (2)} \\[1em] \hline 1 & f'(x) = \answer {\frac {1}{x}} & f'(1) = \answer {1} & c_1 = \answer {1} \\[1em] \hline 2 & f''(x) =\answer {-\frac {1}{x^2}} & f''(1) = \answer {-1} & c_2 = \answer {-1/2} \\[1em] \hline 3 & f'''(x) =\answer {\frac {2}{x^3}} & f'''(1) = \answer {2} & c_3 = \answer {1/3}\\[1em] \hline 4 & f^{(4)}(x) =\answer {-\frac {6}{x^4}} & f^{(4)}(1) = \answer {-6} & c_4 = \answer {-1/4}\\[1em] \hline \end {array}$

The first five terms of the Taylor Series for the function $f(x) = \ln (2x)$ centered at $x = 1$ are $T_4(x) = \answer {\ln (2) + (x-1) -\frac 12(x-1)^2 + \frac 13(x-1)^3-\frac 14(x-1)^4}$

Taylor Polynomials can be used to approximate function values, and the following theorem gives us an expression for the error in such an approximation.

Taylor’s Remainder Theorem If the function $f(x)$ is $n+1$ times differentiable at $x = a$ then the difference between $f(x)$ and $T_n(x)$ is given by $f(x) - T_n(x) = \frac {f^{(n+1)}(c)}{(n+1)!}(x-a)^n$ where $c$ is number between $a$ and $x$ .

2 Video Lessons

Here is a detailed, lecture style video on Taylor Series:

Here is a detailed, lecture style video on Maclaurin Series:

2024-09-27 13:57:27