Difference between revisions of "Taylor polynomial"

Revision as of 17:53, 9 March 2022

The degree- $n$ Taylor polynomial of a function $f(x)$ about $x = a$ is the unique polynomial of degree $n$ whose value and first $n$ derivatives match the value and first $n$ derivatives of $f(x)$ at $x = a$ .

The formula for a degree- $n$ Taylor polynomial of $f(x)$ about $x = a$ is $\sum_{k=0}^{n} \frac{f^{(k)}(a)(x-a)^k}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots + \frac{f^{(n)}(a)(x-a)^n}{n!}.$ In the formula above, $f^{(k)}$ denotes the order- $k$ derivative of $f$ .

Taylor polynomials are often used to approximate non-polynomial functions that cannot be calculated exactly, such as trigonometric functions, exponential functions, and logarithms.

Derivation of the formula

We want the Taylor polynomial to have $k$ -th derivative $f^{(k)}(a)$ at $x = a$ . The Power Rule for derivatives gives that the derivative of $(x-a)^j$ is $j(x-a)^{j-1}$ for all positive integers $j$ , and $0$ for $j = 0$ (because when $j = 0$ the function is a constant $1$ ). Here the Chain Rule is used implicitly with the fact that $x - a$ has derivative $1$ for all $x$ .

For $m < k$ , the degree- $m$ term in $x - a$ has $k$ th derivative $0$ , because after $k$ differentiations the degree of the term will have reached $0$ and then at least one more differentiation ensures that the term is eliminated.

For $m > k$ , the degree- $m$ term in $x - a$ has $k$ th derivative $0$ at $x = a$ , because the $k$ differentiations leave a term with a positive power of $(x - a)$ , which is zero at $x = a$ .

The degree- $k$ term undergoes $k$ differentiations, leaving a constant term and accumulating all of the factors $j$ for $k \geq j \geq 1$ . As such, its $k$ th derivative is $k!$ times its original coefficient for all $x$ , so the coefficient of $(x-a)^k$ should be defined as $\frac{f^{(k)}(a)}{k!}$ .

Special cases

Maclaurin polynomial

A Maclaurin polynomial is a Taylor series with $a = 0$ . Setting $a = 0$ simplifies the appearance of the polynomial somewhat, since every instance of $(x-a)$ in the formula is replaced with $x$ .

For some functions, like $e^x$ and $\sin x$ , Maclaurin polynomials are generally effective across the domain (although using a different $a$ -value might allow greater accuracy for the same choice of degree). However, for functions like $\ln x$ , Maclaurin polynomials cannot be defined because the function and its derivatives are undefined at $x = 0$ . For other functions, Maclaurin polynomials can be defined, but do not in general approximate the function well (see Taylor series), so a value of $a$ closer to the $x$ -value of the desired approximation must be chosen.

Tangent-line approximation

A tangent-line approximation is a first-degree Taylor polynomial, given by $f(a) + f'(a)(x - a)$ . The name "tangent-line approximation" comes from the fact that the graph is a line tangent to the graph of $f(x)$ at $x = a$ . Tangent-line approximations are used in Euler's method and Newton's method.

Error bound

Taylor series

The Taylor series of an infinitely differentiable function $f(x)$ is the infinite series $\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.$ The partial sums of the Taylor series are the Taylor polynomials of $f(x)$ about $x = a$ of each degree.

The Taylor series is the Maclaurin series is the Taylor series chosen with $a = 0$ . The partial sums of the Maclaurin series are the Maclaurin polynomials of $f(x)$ of each degree.

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 ==Derivation of the formula==
+We want the Taylor polynomial to have <math>k</math>-th derivative <math>f^{(k)}(a)</math> at <math>x = a</math>. The [[Derivative/Formulas|Power Rule]] for derivatives gives that the derivative of <math>(x-a)^j</math> is <math>j(x-a)^{j-1}</math> for all positive integers <math>j</math>, and <math>0</math> for <math>j = 0</math> (because when <math>j = 0</math> the function is a constant <math>1</math>). Here the [[Chain Rule]] is used implicitly with the fact that <math>x - a</math> has derivative <math>1</math> for all <math>x</math>.
+For <math>m < k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math>, because after <math>k</math> differentiations the degree of the term will have reached <math>0</math> and then at least one more differentiation ensures that the term is eliminated.
+For <math>m > k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math> at <math>x = a</math>, because the <math>k</math> differentiations leave a term with a positive power of <math>(x - a)</math>, which is zero at <math>x = a</math>.
+The degree-<math>k</math> term undergoes <math>k</math> differentiations, leaving a constant term and accumulating all of the factors <math>j</math> for <math>k \geq j \geq 1</math>. As such, its <math>k</math>th derivative is <math>k!</math> times its original coefficient for all <math>x</math>, so the coefficient of <math>(x-a)^k</math> should be defined as <math>\frac{f^{(k)}(a)}{k!}</math>.
 ==Special cases==

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