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Taylor series
Approximating a function by polynomials built from its derivatives.
A Taylor series rebuilds a function from its derivatives at one point: f(x₀) + f′(x₀)(x − x₀) + f″(x₀)(x − x₀)²/2! + …. The dashed curve is the polynomial; watch how it hugs the function near the centre and drifts away from it.
Introduction
In Activity, we investigated several Taylor polynomials centered at \(0\) for the function \(f(x) = \frac{1}{1-x}\) and found that \[\begin{aligned}\end{aligned}\]. Moreover, we saw that it is natural to extend the degree \(n\) Taylor polynomial to an infinite Taylor series (which will be formally defined in Definition). For \(f(x) = \frac{1}{1-x}\), its corresponding Taylor series is \[\begin{aligned}\end{aligned}\]. Because \(T(x)\) is an infinite geometric series with \(a = 1\) and \(r = x\), whenever \(|r| = |x| \lt 1\), the series converges and by Equation its sum is \[\begin{aligned}\end{aligned}\]. Remarkably, this shows that the original function, \(f(x) = \frac{1}{1-x}\), is equal to its Taylor series for all values of \(x\) that satisfy \(|x| \lt 1\). So, not only do Taylor polynomials provide increasingly better approximations to a function as the degree \(n\) increases, but it appears that if we let \(n\) increase without bound we arrive at a new representation of the original function through its Taylor series. We note that this new representation may only be valid for a limited set of \(x\)-values, such as \(|x| \lt 1\) in the case of \(f(x) = \frac{1}{1-x}\).
In Preview Activity, we consider an infinite geometric series and explore the function to which the series is related and that function's Taylor polynomials.
Exploration
Exploration
Taylor series and the Ratio Test
From our work in Activity and Preview Activity, we have seen two examples where the Taylor polynomials of a given function lead to a corresponding infinite Taylor series that is actually equal to the original function itself. This suggests that one reason Taylor series are important is that they give rise to polynomial-like representations of non-polynomial functions.
Next we formally define the Taylor series of a function and introduce a tool for determining where its Taylor series converges.
In the special case where \(a=0\) in Definition, the Taylor series is also sometimes called the Maclaurin series for \(f\).
From Activity, we know the degree \(n\) Taylor polynomial centered at \(0\) for the function \(f(x) = \frac{1}{1-x}\) is \[\begin{aligned}\end{aligned}\], and thus the Maclaurin series for \(f(x) = \frac{1}{1-x}\) is \[\begin{aligned}\end{aligned}\]. As we noted earlier in Equation, because this particular series is geometric (with common ratio \(r = x\)), we are able to easily find a shortcut formula for its value and understand the values of \(x\) for which the infinite sum makes sense. In particular, for \(|x| \lt 1\), \[\begin{aligned}\end{aligned}\], which shows that the Taylor series is equal to the original function \(f(x)\) for this set of \(x\) values.
It turns out that while most Taylor series are not geometric series, we can compare any Taylor series to a geometric series in order to determine an interval of \(x\)-values for which the Taylor series converges. To see how to approach this issue, we consider the following example with a particular function with a known Taylor series and determine the \(x\)-values for which its Taylor series should converge.
We next introduce a more formal way to test how geometric a Taylor series is in an effort to understand where the Taylor series is guaranteed to converge.
Condensed: the full section is in Boelkins, Active Calculus.
Taylor series of several important functions
So far, we have established that the Taylor series of two important functions, \(\frac{1}{1-x}\) and \(\ln(1+x)\), converge on an open interval of \(x\)-values and on that interval converge to the respective functions. For these functions,
if \(|x| \lt 1\), then \[\begin{aligned}\end{aligned}\]
if \(|x| \lt 1\), thenIt turns out that this representation is also valid when \(x = 1\) because the series \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\) converges, and does so to the exact value \(\ln(2)\). This means that while the open interval of convergence is \(-1 \lt x \lt 1\) for this series, the complete interval of convergence is actually \(-1 \lt x \le 1\). \[\begin{aligned}\end{aligned}\].
Following Activity, we noted some of the higher degree Taylor polynomials centered at \(a = 0\) for \(\sin(x)\), \(\cos(x)\), and \(e^x\). For the sine function, we observed that \[\begin{aligned}\end{aligned}\]. In the next example, we determine the Taylor series of \(f(x) = \sin(x)\) and investigate the \(x\)-values for which the series converges.
In Example, we found that the Taylor series for the sine function appears to converge to the sine function, and to do so for every real number \(x\). This result can be proven formally using a famous theorem called the Lagrange Error Bound, a theorem that quantifies how accurate a Taylor polynomial approximation is on an interval, which we will study in Section. To summarize, we now know the remarkable result that the sine function is equal to its Taylor series for every value of \(x\), so \[\begin{aligned}\end{aligned}\]
Nearly identical reasoning shows a similar result for the cosine function: that its Taylor series centered at \(a = 0\) converges for every real number \(x\) and converges to the cosine function itself. Thus, for any real number \(x\), \[\begin{aligned}\end{aligned}\]
In the next activity, we conduct a similar investigation for \(f(x) = e^x\).
A function such as \(f(x) = \sin(x)\) that is infinitely differentiable and has a power series that converges on some interval to the function itself is said to be analytic. Analytic functions are amazing and among the nicest functions in all of mathematics: they are completely determined by what happens at a single point.
Condensed: the full section is in Boelkins, Active Calculus.
Summary
Provided that \(f\) is a function for which every derivative of \(f\) exists at \(x = a\), the Taylor series centered at \(a\) of \(f\) is the series \(T_f(x)\) defined by \[\begin{aligned}T_f(x) =\mathstrut \amp \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k \\ =\mathstrut \amp f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots\end{aligned}\].
We know that an infinite geometric series converges whenever its common ratio, \(r\), satisfies \(|r| \lt 1\). The Ratio Test enables us to consider a Taylor series and determine an open interval of \(x\)-values on which the series converges, essentially by comparing the Taylor series to a geometric series. By letting \(r_n(x)\) be the ratio of consecutive terms in the series and taking the limit \[\begin{aligned}\end{aligned}\], we can say that for all \(x\) for which \(|r(x)| \lt 1\), the Taylor series converges to a finite value.
We have found the Taylor series for five important functions, determined an open interval of \(x\)-values on which each converges, and seen the striking result that where each Taylor series converges, it converges to original function \(f(x)\) that we used to generate the Taylor series. These results are summarized in Table.
Taylor series
In mathematical analysis, the Taylor series or Taylor expansion of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor series are named after Brook Taylor, who introduced them in 1715. A Taylor series is also called a Maclaurin series when 0 is the point where the derivatives are considered, after Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century.
The partial sum formed by the first n + 1 terms of a Taylor series is a polynomial of degree n that is called the nth Taylor polynomial of the function. Taylor polynomials are approximations of a function, which become generally more accurate as n increases. Taylor's theorem gives quantitative estimates on the error introduced by the use of such approximations. If the Taylor series of a function is convergent, its sum is the limit of the infinite sequence of the Taylor polynomials. A function may differ from the sum of its Taylor series, even if its Taylor series is convergent. A function is analytic at a point x if it is equal to the sum of its Taylor series in some open interval (or open disk in the complex plane) containing x. This implies that the function is analytic at every point of the interval (or disk).
Definition
The Taylor series of a real or complex-valued function f(x), that is infinitely differentiable at a real or complex number a, is the power series \[f(a) + \frac {f'(a)}{1!}(x-a) + \frac{f''(a)}{2!} (x-a)^2+ \cdots = \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} (x-a)^{n}.\] Here, n! denotes the factorial of n. The function f(a) denotes the nth derivative of f evaluated at the point a. The derivative of order zero of f is defined to be f itself and (x − a) and 0! are both defined to be 1. This series can be written by using sigma notation, as in the right side formula. The corresponding Taylor polynomial of degree n is \[T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.\] With a = 0, the Maclaurin series takes the form: \[f(0)+\frac {f'(0)}{1!} x+ \frac{f''(0)}{2!} x^2+ \cdots = \sum_{n=0} ^ {\infty} \frac {f^{(n)}(0)}{n!} x^{n}.\]
Basic properties
Taylor series inherit the basic properties of power series. Taylor series can also be combined algebraically. Sums, differences, products, and scalar multiples of Taylor series are obtained by the corresponding operations on power series. In particular, the Taylor series of \(f(x)g(x)\) around a point \(x=a\) is the Cauchy product of the Taylor series of \(f(x)\) and \(g(x)\) about \(x=a\). Compositions of functions having Taylor series likewise have Taylor series, obtained by substituting one convergent power series into another when the substitution is valid.
A Taylor series may be differentiated and integrated term by term. Thus \[\frac{d}{dx}\sum_{n=0}^{\infty} c_n (x-a)^n = \sum_{n=1}^{\infty} n c_n (x-a)^{n-1},\] and \[\int \sum_{n=0}^{\infty} c_n (x-a)^n\,dx = C+\sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}.\] The differentiated and integrated series have the same radius of convergence as the original power series, although the convergence behavior at the boundary may be different.
These properties sometimes allow the Taylor series of functions, such as the arctangent, to be computed in terms of simpler series, such as the geometric series.
Calculation of Taylor series
Several methods can be used to calculate Taylor series. One may apply the definition directly, although this often requires first identifying a general formula for the derivatives or coefficients. In many cases, Taylor series can also be obtained from known expansions by algebraic manipulations of power series, such as substitution, multiplication, division, addition, or subtraction, as well as termwise differentiation and integration of known Taylor series. In some cases, they may also be derived by repeated integration by parts. In practice, Taylor series are often computed with the aid of computer algebra systems.
A number of standard Maclaurin series are frequently used as starting points for calculating other Taylor series. Some fundamental examples are listed below; a more comprehensive listing appears later in the article.
Approximation error and Taylor's theorem
The error incurred in approximating a function by its degree n Taylor polynomial is called the remainder and is denoted by the function Rn(x). Taylor's theorem can be used to obtain a bound on the size of the remainder.
In particular, Taylor's theorem writes a function, where the hypotheses of the theorem are satisfied, in the form \[f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x).\] The behavior of the remainder as n tends to infinity determines whether the Taylor series represents the original function, which are questions of convergence and analyticity.
Generalization in finite differences
One form of the Gregory-Newton interpolation formula can be written as \[f(x)=\sum_{k=0}^\infty\frac{\Delta^k [f](a)}{k!} \,(x-a)_k\] which interpolates a polynomial \(f\) in terms of its finite differences evaluated at a single point \(a\), and where \((x-a)_k\) is the falling factorial. For a polynomial, this series terminates and gives the polynomial exactly; more generally, a function admits a Gregory-Newton development under suitable analytic hypotheses, classically formulated by Niels Erik Nørlund in terms of holomorphy in a half-plane together with an exponential type growth condition.
One generalization of the Taylor series that does converge to the value of the function itself for any bounded continuous function on (0, ∞), and this can be done by using the calculus of finite differences. Specifically, the following theorem, due to Einar Hille, that for any t > 0,
\[\lim_{h\to 0^+}\sum_{n=0}^\infty \frac{t^n}{n!}\frac{\Delta_h^nf(a)}{h^n} = f(a+t).\]
Here Δ
h is the nth finite difference operator with step size h. The series is precisely the Taylor series, except that divided differences appear in place of differentiation. When the function f is analytic at a, the terms in the series converge to the terms of the Taylor series, and in this sense generalizes the usual Taylor series.
In general, for any infinite sequence ai, the following power series identity holds: \[\sum_{n=0}^\infty\frac{u^n}{n!}\Delta^na_i = e^{-u}\sum_{j=0}^\infty\frac{u^j}{j!}a_{i+j}.\] So in particular, \[f(a+t) = \lim_{h\to 0^+} e^{-t/h}\sum_{j=0}^\infty f(a+jh) \frac{(t/h)^j}{j!}.\]
The series on the right is the expected value of f(a + X), where X is a Poisson-distributed random variable that takes the value jh with probability e·(t/h)/j!. Hence, \[f(a+t) = \lim_{h\to 0^+} \int_{-\infty}^\infty f(a+x)dP_{t/h,h}(x).\]
The law of large numbers implies that the identity holds.
Convergence and analyticity
A Taylor series is formed from the values of all derivatives of a function at a single point, but this does not by itself imply that the series converges to the function. In general, a Taylor series may fail to converge, or it may converge to a function different from the original one.
For example, the function \[f(x) = \begin{cases} e^{-1/x^2} & \text{if } x \neq 0, \\[3mu] 0 & \text{if } x = 0 \end{cases}\] is infinitely differentiable at \(x=0\), and all of its derivatives at \(0\) are equal to zero. Its Taylor series at \(0\) is therefore the zero series, even though the function itself is not identically zero. This gives a standard example of a non-analytic smooth function.
More generally, the Taylor series of a function represents the function at a point \(x\) precisely when the remainder terms in Taylor's theorem tend to zero at that point. If \[f(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x),\] then the Taylor series converges to \(f(x)\) exactly when \[\lim_{n\to\infty} R_n(x)=0.\]
A function is called analytic at a point if it is equal to the sum of its Taylor series in some open interval around that point, or, in the complex case, in some open disk. Equivalently, a function is analytic in a region if it is locally given by a convergent power series. Thus, if \[f(x)=\sum_{n=0}^{\infty}a_n(x-b)^n\] near \(b\), then differentiating the series term by term and setting \(x=b\) gives \[a_n=\frac{f^{(n)}(b)}{n!}.\] Thus the power series expansion of an analytic function is its Taylor series.
In real analysis, infinite differentiability does not imply analyticity, as the example above shows. Borel's lemma implies that every power series is the Taylor series of some smooth function. In complex analysis, however, every holomorphic function is analytic. A function whose Taylor series converges to the function throughout the whole complex plane is called an entire function. Polynomials, the exponential function, and the sine and cosine functions are entire functions.
Radius of convergence
For any power series
\[\sum_{n=0}^\infty c_n(x-a)^n,\]
there is a number \(R\), called the radius of convergence, such that the series converges absolutely for |x-a|
Thus, when a Taylor series converges, it does so in an open interval centered at \(a\) in the real case, or in an open disk centered at \(a\) in the complex case. The behavior at the boundary points may vary: the series may converge at some, all, or none of them.
For a complex analytic function, the radius of convergence of the Taylor series at \(a\) is the distance from \(a\) to the nearest point where the function cannot be continued holomorphically. In many common examples this is the distance to the nearest singularity in the complex plane.
This explains the different radii of convergence for familiar Taylor series. The series for \(e^x\), \(\sin x\), and \(\cos x\) have infinite radius of convergence because these functions are entire. By contrast, the Taylor series for \(\log(1+x)\) at \(x=0\) has radius of convergence \(1\), because the nearest singularity is at \(x=-1\).
Complex singularities can determine the radius of convergence even for functions that are smooth on the real line. For example, \[\frac{1}{1+x^2}\] is smooth for every real \(x\), but its Taylor series at \(0\) has radius of convergence \(1\), because the corresponding complex function has singularities at \(x=i\) and \(x=-i\).
The radius of convergence should not be confused with the quality of approximation by a low-degree Taylor polynomial. A Taylor polynomial may approximate a function accurately near the center even if the full Taylor series has a small radius of convergence. Conversely, near the boundary of the interval or disk of convergence, the Taylor series may converge slowly. Outside the radius of convergence, the Taylor series does not represent the function.
Generalizations near singularities
A Taylor series cannot be centered at a point where the function is not analytic. Some singularities, namely poles, can be accounted for by a Laurent series. If \(f\) has a pole of order \(k\) at \(z=a\), then near \(a\) it has a Laurent series of the form \[\sum_{n=-k}^{\infty} a_n (z-a)^n .\] A meromorphic function is a function which is analytic except at isolated poles; near each pole it has a Laurent series with only finitely many negative-power terms.
More generally, a function analytic in an annulus
\[r<|z-a| Other types of singularities, namely branch points, can occur for algebraic functions. If \(f(z)\) is an algebraic function of a complex variable \(z\) and \(z=a\) is a branch point, then \(f(z)\) need not have a Taylor series based at \(z=a\). However, after a change of variables
\[z-a=t^e,\]
where \(e\) is a positive integer called the ramification index, a branch of the function becomes analytic as a function of \(t\). The resulting expansion in fractional powers of \(z-a\) is known as a Puiseux series.
Taylor series in multiple variables
The Taylor series may also be generalized to functions of more than one variable with \[\begin{aligned} T(x_1,\ldots,x_d) &= \sum_{n_1=0}^\infty \cdots \sum_{n_d = 0}^\infty \frac{(x_1-a_1)^{n_1}\cdots (x_d-a_d)^{n_d}}{n_1!\cdots n_d!}\,\left(\frac{\partial^{n_1 + \cdots + n_d}f}{\partial x_1^{n_1}\cdots \partial x_d^{n_d}}\right)(a_1,\ldots,a_d) \\ &= f(a_1, \ldots,a_d) + \sum_{j=1}^d \frac{\partial f(a_1, \ldots,a_d)}{\partial x_j} (x_j - a_j) + \frac{1}{2!} \sum_{j=1}^d \sum_{k=1}^d \frac{\partial^2 f(a_1, \ldots,a_d)}{\partial x_j \partial x_k} (x_j - a_j)(x_k - a_k) \\ & \qquad \qquad + \frac{1}{3!} \sum_{j=1}^d\sum_{k=1}^d\sum_{l=1}^d \frac{\partial^3 f(a_1, \ldots,a_d)}{\partial x_j \partial x_k \partial x_l} (x_j - a_j)(x_k - a_k)(x_l - a_l) + \cdots, \\ &= \sum_{|\alpha| \geq 0}\frac{(\mathbf{x}-\mathbf{a})^\alpha}{\alpha !} \left({\mathrm{\partial}^{\alpha}}f\right)(\mathbf{a}). \end{aligned}\] The last expression is the multivariate Taylor series in terms of multi-index notation with a full analogy to the single variable case.
For example, for a function f(x, y) that depends on two variables, x and y, the Taylor series to second order about the point (a, b) is \[f(a,b) +(x-a) f_x(a,b) +(y-b) f_y(a,b) + \frac{1}{2!}\Big( (x-a)^2 f_{xx}(a,b) + 2(x-a)(y-b) f_{xy}(a,b) +(y-b)^2 f_{yy}(a,b) \Big)\] where the subscripts denote the respective partial derivatives.
Second-order Taylor series in several variables
A second-order Taylor series expansion of a scalar-valued function of more than one variable can be written compactly as \[T(\mathbf{x}) = f(\mathbf{a}) + (\mathbf{x} - \mathbf{a})^\mathsf{T} D f(\mathbf{a}) + \frac{1}{2!} (\mathbf{x} - \mathbf{a})^\mathsf{T} \left \{D^2 f(\mathbf{a}) \right \} (\mathbf{x} - \mathbf{a}) + \cdots,\] where D f(a) is the gradient of f evaluated at x = a and D f(a) is the Hessian matrix.
Example
In order to compute a second-order Taylor series expansion around the point (a, b) = (0, 0) of the function \[f(x,y)=e^x\ln(1+y),\] one first computes all the necessary partial derivatives: \[\begin{aligned} f_x &= e^x\ln(1+y), & f_y &= \frac{e^x}{1+y}, \\ f_{xx} &= e^x\ln(1+y), & f_{yy} &= -\frac{e^x}{(1+y)^2}, \\ f_{xy} &= f_{yx} = \frac{e^x}{1+y}. \end{aligned}\]
Evaluating these derivatives at the origin gives the Taylor coefficients \[\begin{aligned} f_x(0,0) &= 0, & f_y(0,0) &= 1, \\ f_{xx}(0,0) &= 0, & f_{yy}(0,0) &= -1, \\ f_{xy}(0,0) &= 1. \end{aligned}\]
Substituting these values in to the general formula \[\begin{aligned} T(x,y) &= f(a,b) +(x-a) f_x(a,b) +(y-b) f_y(a,b) \\ &\qquad {}+\frac{1}{2!}\left( (x-a)^2f_{xx}(a,b) + 2(x-a)(y-b)f_{xy}(a,b) +(y-b)^2 f_{yy}(a,b) \right)+ \cdots \end{aligned}\] produces \[\begin{aligned} T(x,y) &= 0 + 0(x-0) + 1(y-0) + \frac{1}{2}\big( 0(x-0)^2 + 2(x-0)(y-0) + (-1)(y-0)^2 \big) + \cdots \\ &= y + xy - \tfrac12 y^2 + \cdots \end{aligned}\]
Since ln(1 + y) is analytic in |y| < 1, we have \[e^x\ln(1+y)= y + xy - \tfrac12 y^2 + \cdots, \qquad |y| < 1.\]
Applications
Taylor polynomials are used to approximate functions near a point. Keeping only the first nonzero terms often gives a simpler model of a more complicated expression. For example, the small-angle approximation \[\sin x \approx x\] comes from the first term of the Taylor series for sine, and higher-order approximations are obtained by retaining more terms. This approximation is widely used: for example, in Gaussian optics where the behavior of light rays making small angles with an axis is studied by replacing the sine function with its linear approximation.
Such approximations are used throughout mathematics, physics, and engineering. In perturbation theory, a complicated quantity is often expanded in powers of a small parameter, and the first few terms are used as an approximate solution. Taylor expansions also occur in the analysis of the simple pendulum and in numerical methods for approximating functions.
History
The ancient Greek philosopher Zeno of Elea considered the problem of summing an infinite series to achieve a finite result, but rejected it as an impossibility; the result was Zeno's paradox. Later, Aristotle proposed a philosophical resolution of the paradox, but the mathematical content was apparently unresolved until taken up by Archimedes, as it had been prior to Aristotle by the Presocratic Atomist Democritus. It was through Archimedes's method of exhaustion that an infinite number of progressive subdivisions could be performed to achieve a finite result. Liu Hui independently employed a similar method a few centuries later.
In the 14th century, the earliest examples of specific Taylor series (but not the general method) were given by the Indian mathematician Madhava of Sangamagrama. Though no record of his work survives, writings of his followers in the Kerala school of astronomy and mathematics suggest that he found the Taylor series for the trigonometric functions of sine, cosine, and arctangent; see Madhava series. During the following two centuries, his followers developed further series expansions and rational approximations.
In late 1670, James Gregory was shown in a letter from John Collins several Maclaurin series (sin x, cos x, arcsin x, and x cot x) derived by Isaac Newton, and told that Newton had developed a general method for expanding functions in series. Newton had in fact used a cumbersome method involving long division of series and term-by-term integration, but Gregory did not know it and set out to discover a general method for himself. In early 1671 Gregory discovered something like the general Maclaurin series and sent a letter to Collins including series for arctan x, tan x, sec x, ln sec x (the integral of tan), ln tan 1/2(1/2π + x) (the integral of sec, the inverse Gudermannian function), arcsec(√2 e), and 2 arctan e − 1/2π (the Gudermannian function). However, thinking that he had merely redeveloped a method by Newton, Gregory never described how he obtained these series, and it can only be inferred that he understood the general method by examining scratch work he had scribbled on the back of another letter from 1671.
In 1691-1692, Newton wrote down an explicit statement of the Taylor and Maclaurin series in an unpublished version of his work De Quadratura Curvarum. It was the earliest explicit formulation of the general Taylor series. However, this work by Newton was never completed and the relevant sections were omitted from the portions published in 1704 under the title Tractatus de Quadratura Curvarum.
It was not until 1715 that a general method for constructing these series for all functions for which they exist was finally published by Brook Taylor, after whom the series are now named.
The Maclaurin series was named after Colin Maclaurin, a Scottish mathematician, who published a special case of the Taylor result in the mid-18th century.
Ejemplo resuelto · taylor series of e^x
Paso a paso
- f(x) = e^{x}
Taylor series about x = 0: f(x) = Σ fᵏ(x₀)/k! · (x − x₀)ᵏ.
- f^{(0)}(0) = e^{x}\big|_{x=0} = 1
Derivative 0 at the centre.
- f^{(1)}(0) = e^{x}\big|_{x=0} = 1
Derivative 1 at the centre.
- f^{(2)}(0) = e^{x}\big|_{x=0} = 1
Derivative 2 at the centre.
- f^{(3)}(0) = e^{x}\big|_{x=0} = 1
Derivative 3 at the centre.
- f^{(4)}(0) = e^{x}\big|_{x=0} = 1
Derivative 4 at the centre.
- f^{(5)}(0) = e^{x}\big|_{x=0} = 1
Derivative 5 at the centre.
- \frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1
Assemble the terms up to degree 5.
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The examples we have considered so far in this section have all been for Taylor polynomials and series centered at 0, but Taylor polynomials and series can be centered at any value of \(a\).
Let \(f(x) = \cos(x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(a = \frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(a = \frac{\pi}{2}\). Why is the result not surprising?
Let \(f(x) = \frac{1}{1+x} = (1+x)^{-1}\). Find the Taylor polynomials up through order four of \(f\) centered at \(a = 1\). Then find the Taylor series for \(f(x)\) centered at \(a = 1\).
Revelar la respuesta
Let \(f(x) = \cos(x)\).
We know that \(f'(x) = -\sin(x)\), \(f''(x) = -cos(x)\), \(f'''(x) = \sin(x)\), and \(f^{(4)}(x) = cos(x)\), so \(f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1\), \(f''\left(\frac{\pi}{2}\right) = -cos\left(\frac{\pi}{2}\right) = 0\), \(f'''\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\), and \(f^{(4)}(x) = cos\left(\frac{\pi}{2}\right) = 0\). We also note that \(f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\)
It follows that the degree four Taylor polynomial of \(f\) centered at \(x = \frac{\pi}{2}\) is \[\begin{aligned}\end{aligned}\], from which we can also easily determine \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
From the pattern in \(T_4(x)\), we find that the Taylor series for \(f(x)\) centered at \(x = \frac{\pi}{2}\) is \[\begin{aligned}\end{aligned}\]
This result is not surprising, because \(f(x) = \cos(x) = -\sin\left( x - \frac{\pi}{2} \right)\). So, if we stated with the Taylor series for \(g(x) = \sin(x)\) centered at \(a = 0\), we'd have \[\begin{aligned}\end{aligned}\] from which it follows that \[\begin{aligned}\end{aligned}\]. Noting that \(-1 \cdot (-1)^k = -1^{k+1}\), we have established that \[\begin{aligned}\end{aligned}\] which matches our earlier result.
With \(f(x) = \frac{1}{1+x} = (1+x)^{-1}\), we observe that \(f'(x) = -(1+x)^{-2}\), \(f''(x) = (-1)(-2)(1+x)^{-3}\), \(f'''(x) = (-1)(-2)(-3)(1+x)^{-4}\), and \(f^{(4)}(x) = (-1)(-2)(-3)(-4)(1+x)^{-5}\). Evaluating each of these functions at \(a = 1\), we find \(f(1) = \frac{1}{2}\), \(f'(1) = (-1) \cdot \frac{1}{2^2}\), \(f''(1) = \frac{2!}{2^3}\), \(f'''(1) = (-1) \cdot \frac{3!}{2^4}\), and \(f^{(4)}(1) = \frac{4!}{2^5}\).
From these derivative values and the fact that \(c_k = \frac{f^{(k)}(1)}{k!}\), it follows that \[\begin{aligned}\end{aligned}\], from which we can also easily find \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
Finally, from the patterns in our preceding work, we see the Taylor series for \(f(x)\) centered at \(a = 1\) is \[\begin{aligned}\end{aligned}\].
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As we will see in more detail in the next section, we can use known Taylor series to obtain other Taylor series, and we preview that idea in this exercise.
Calculate the first four derivatives of \(\sin(x^2)\) and hence find the fourth order Taylor polynomial for \(\sin(x^2)\) centered at \(a=0\).
Part (a) demonstrates the direct approach to finding Taylor polynomials and series. Next we utilize a known Taylor series to make the process simpler. Recall that the Taylor series centered at 0 for \(f(x) = \sin(x)\) is \[\begin{aligned}\end{aligned}\].
Substitute \(x^2\) for \(x\) in the Taylor series \(T(x)\) in Equation. Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should result in the Taylor series for \(\sin(x^2)\) centered at 0.
For what interval of \(x\)-values should we expect the Taylor series for \(\sin(x^2)\) to converge?
Revelar la respuesta
Let \(f(x) = \sin(x^2)\). Then
\(\sin(x^2)\) \(f(0) = 0\) \(f'(x) = 2x\cos(x^2)\) \(f'(0) = 0\) \(f''(x) = -4x^2\sin(x^2) + 2\cos(x^2)\) \(f''(0) = 0\) \(f'''(x) = -8x^3\cos(x^2) - 12x\sin(x^2)\) \(f'''(0) = 0\) \(f^{(4)}(x) = 16x^4\sin(x^2) - 48x^2\cos(x^2) - 12\sin(x^2)\) \(f^{(4)}(0) = 0\) so the fourth order Taylor polynomial for \(f\) centered at \(a=0\) is \[\begin{aligned}\end{aligned}\].
Substituting \(x^2\) for \(x\) in the Taylor series for \(\sin(x)\), we find that since \[\begin{aligned}\end{aligned}\], we have \[\begin{aligned}\end{aligned}\] which should be the Taylor series for \(\sin(x^2)\) since \(\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots\) for every value of \(x\).
We expect the interval of convergence of the series for \(\sin(x^2)\) to be the same as the series for \(\sin(x)\), since \(\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots\) for every value of \(x\).
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The examples we have considered so far in this section have all been for Taylor polynomials and series centered at 0, but Taylor polynomials and series can be centered at any value of \(a\).
Let \(f(x) = \cos(x)\). Find the Taylor polynomials up through order four of \(f\) centered at \(a = \frac{\pi}{2}\). Then find the Taylor series for \(f(x)\) centered at \(a = \frac{\pi}{2}\). Why is the result not surprising?
Let \(f(x) = \frac{1}{1+x} = (1+x)^{-1}\). Find the Taylor polynomials up through order four of \(f\) centered at \(a = 1\). Then find the Taylor series for \(f(x)\) centered at \(a = 1\).
Revelar la respuesta
Let \(f(x) = \cos(x)\).
We know that \(f'(x) = -\sin(x)\), \(f''(x) = -cos(x)\), \(f'''(x) = \sin(x)\), and \(f^{(4)}(x) = cos(x)\), so \(f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1\), \(f''\left(\frac{\pi}{2}\right) = -cos\left(\frac{\pi}{2}\right) = 0\), \(f'''\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\), and \(f^{(4)}(x) = cos\left(\frac{\pi}{2}\right) = 0\). We also note that \(f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\)
It follows that the degree four Taylor polynomial of \(f\) centered at \(x = \frac{\pi}{2}\) is \[\begin{aligned}\end{aligned}\], from which we can also easily determine \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
From the pattern in \(T_4(x)\), we find that the Taylor series for \(f(x)\) centered at \(x = \frac{\pi}{2}\) is \[\begin{aligned}\end{aligned}\]
This result is not surprising, because \(f(x) = \cos(x) = -\sin\left( x - \frac{\pi}{2} \right)\). So, if we stated with the Taylor series for \(g(x) = \sin(x)\) centered at \(a = 0\), we'd have \[\begin{aligned}\end{aligned}\] from which it follows that \[\begin{aligned}\end{aligned}\]. Noting that \(-1 \cdot (-1)^k = -1^{k+1}\), we have established that \[\begin{aligned}\end{aligned}\] which matches our earlier result.
With \(f(x) = \frac{1}{1+x} = (1+x)^{-1}\), we observe that \(f'(x) = -(1+x)^{-2}\), \(f''(x) = (-1)(-2)(1+x)^{-3}\), \(f'''(x) = (-1)(-2)(-3)(1+x)^{-4}\), and \(f^{(4)}(x) = (-1)(-2)(-3)(-4)(1+x)^{-5}\). Evaluating each of these functions at \(a = 1\), we find \(f(1) = \frac{1}{2}\), \(f'(1) = (-1) \cdot \frac{1}{2^2}\), \(f''(1) = \frac{2!}{2^3}\), \(f'''(1) = (-1) \cdot \frac{3!}{2^4}\), and \(f^{(4)}(1) = \frac{4!}{2^5}\).
From these derivative values and the fact that \(c_k = \frac{f^{(k)}(1)}{k!}\), it follows that \[\begin{aligned}\end{aligned}\], from which we can also easily find \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
Finally, from the patterns in our preceding work, we see the Taylor series for \(f(x)\) centered at \(a = 1\) is \[\begin{aligned}\end{aligned}\].
-
As we will see in more detail in the next section, we can use known Taylor series to obtain other Taylor series, and we preview that idea in this exercise.
Calculate the first four derivatives of \(\sin(x^2)\) and hence find the fourth order Taylor polynomial for \(\sin(x^2)\) centered at \(a=0\).
Part (a) demonstrates the direct approach to finding Taylor polynomials and series. Next we utilize a known Taylor series to make the process simpler. Recall that the Taylor series centered at 0 for \(f(x) = \sin(x)\) is \[\begin{aligned}\end{aligned}\].
Substitute \(x^2\) for \(x\) in the Taylor series \(T(x)\) in Equation. Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should result in the Taylor series for \(\sin(x^2)\) centered at 0.
For what interval of \(x\)-values should we expect the Taylor series for \(\sin(x^2)\) to converge?
Revelar la respuesta
Let \(f(x) = \sin(x^2)\). Then
\(\sin(x^2)\) \(f(0) = 0\) \(f'(x) = 2x\cos(x^2)\) \(f'(0) = 0\) \(f''(x) = -4x^2\sin(x^2) + 2\cos(x^2)\) \(f''(0) = 0\) \(f'''(x) = -8x^3\cos(x^2) - 12x\sin(x^2)\) \(f'''(0) = 0\) \(f^{(4)}(x) = 16x^4\sin(x^2) - 48x^2\cos(x^2) - 12\sin(x^2)\) \(f^{(4)}(0) = 0\) so the fourth order Taylor polynomial for \(f\) centered at \(a=0\) is \[\begin{aligned}\end{aligned}\].
Substituting \(x^2\) for \(x\) in the Taylor series for \(\sin(x)\), we find that since \[\begin{aligned}\end{aligned}\], we have \[\begin{aligned}\end{aligned}\] which should be the Taylor series for \(\sin(x^2)\) since \(\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots\) for every value of \(x\).
We expect the interval of convergence of the series for \(\sin(x^2)\) to be the same as the series for \(\sin(x)\), since \(\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots\) for every value of \(x\).
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Cómo: Taylor series
- What is the Taylor series centered at a of a function f?
- What does it mean for an infinite Taylor series to converge to a finite sum, and how can we determine for which x-values the series converges?
- What are the Taylor series centered at a = 0 for \frac{1}{1-x}, \ln(1+x), \sin(x), \cos(x), and e^x, and for which x-values do these series converge?
Preguntas que la gente hace
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
Partes de esta página están adaptadas de Boelkins, Active Calculus (CC BY-SA 4.0), Wikipedia (CC BY-SA 4.0). Condensados y re-explicados aquí; los errores son nuestros.
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