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Taylor and Maclaurin Series
Describe the procedure for finding a Taylor polynomial of a given order for a function.
Overview of Taylor/Maclaurin Series
Consider a function \(f\) that has a power series representation at \(x=a.\) Then the series has the form
\[\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n}={c}_{0}+{c}_{1}(x-a)+{c}_{2}{(x-a)}^{2}+\text{\cdots }.\]What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series 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
\[\begin{array}{ll}\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n} & ={c}_{0}+{c}_{1}(a-a)+{c}_{2}{(a-a)}^{2}+\text{\cdots } \\ & ={c}_{0}.\end{array}\]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 term-by-term, we see that
\[\frac{d}{dx}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n})={c}_{1}+2{c}_{2}(x-a)+3{c}_{3}{(x-a)}^{2}+\text{\cdots }.\]Therefore, at \(x=a,\) the derivative is
\[\begin{array}{ll} \\ \\ \frac{d}{dx}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n}) & ={c}_{1}+2{c}_{2}(a-a)+3{c}_{3}{(a-a)}^{2}+\text{\cdots } \\ & ={c}_{1}.\end{array}\]Therefore, the derivative of the series equals \({f}^{'}(a)\) if the coefficient \({c}_{1}={f}^{'}(a).\) Continuing in this way, we look for coefficients cn such that all the derivatives of the power series will agree with all the corresponding derivatives of \(f\) at \(x=a.\) The second and third derivatives of are given by
\[\frac{{d}^{2}}{d{x}^{2}}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n})=2{c}_{2}+3\cdot 2{c}_{3}(x-a)+4\cdot 3{c}_{4}{(x-a)}^{2}+\text{\cdots }\]and
\[\frac{{d}^{3}}{d{x}^{3}}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n})=3\cdot 2{c}_{3}+4\cdot 3\cdot 2{c}_{4}(x-a)+5\cdot 4\cdot 3{c}_{5}{(x-a)}^{2}+\text{\cdots }.\]Therefore, at \(x=a,\) the second and third derivatives
\[\begin{array}{ll}\frac{{d}^{2}}{d{x}^{2}}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n}) & =2{c}_{2}+3\cdot 2{c}_{3}(a-a)+4\cdot 3{c}_{4}{(a-a)}^{2}+\text{\cdots } \\ & =2{c}_{2}\end{array}\]\[\begin{array}{ll}\frac{{d}^{3}}{d{x}^{3}}(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n}) & =3\cdot 2{c}_{3}+4\cdot 3\cdot 2{c}_{4}(a-a)+5\cdot 4\cdot 3{c}_{5}{(a-a)}^{2}+\text{\cdots } \\ & =3\cdot 2{c}_{3}\end{array}\]\[\sum _{n=0}^{\infty }\frac{{f}^{(n)}(a)}{n\text{!}}{(x-a)}^{n}=f(a)+{f}^{'}(a)(x-a)+\frac{{f}^{″}(a)}{2\text{!}}{(x-a)}^{2}+\frac{f\text{‴}(a)}{3\text{!}}{(x-a)}^{3}+\text{\cdots }.\]Condensed — the full section is in OpenStax Calculus Volume 2.
Taylor Polynomials
The nth partial sum of the Taylor series for a function \(f\) at \(a\) is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by
\[\begin{array}{l}{p}_{0}(x)=f(a), \\ {p}_{1}(x)=f(a)+{f}^{'}(a)(x-a), \\ {p}_{2}(x)=f(a)+{f}^{'}(a)(x-a)+\frac{{f}^{″}(a)}{2\text{!}}{(x-a)}^{2}, \\ {p}_{3}(x)=f(a)+{f}^{'}(a)(x-a)+\frac{{f}^{″}(a)}{2\text{!}}{(x-a)}^{2}+\frac{f\text{‴}(a)}{3\text{!}}{(x-a)}^{3},\end{array}\]respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of \(f\) at \(a,\) respectively. If \(a=0,\) then these polynomials are known as Maclaurin polynomials for \(f.\) We now provide a formal definition of Taylor and Maclaurin polynomials for a function \(f.\)
We now show how to use this definition to find several Taylor polynomials for \(f(x)=\text{ln}\ x\) at \(x=1.\)
Example
Try it.
Find the Taylor polynomials \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}\) for \(f(x)=\text{ln}\ x\) at \(x=1.\) Use a graphing utility to compare the graph of \(f\) with the graphs of \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}.\)
Solution
To find these Taylor polynomials, we need to evaluate \(f\) and its first three derivatives at \(x=1.\)
\[\begin{array}{llllllll}f(x) & = & \text{ln}\ x & & & f(1) & = & 0 \\ {f}^{'}(x) & = & \frac{1}{x} & & & {f}^{'}(1) & = & 1 \\ {f}^{″}(x) & = & -\frac{1}{{x}^{2}} & & & {f}^{″}(1) & = & -1 \\ f\text{‴}(x) & = & \frac{2}{{x}^{3}} & & & f\text{‴}(1) & = & 2\end{array}\]Therefore,
\[\begin{array}{lll}{p}_{0}(x) & = & f(1)=0, \\ {p}_{1}(x) & = & f(1)+{f}^{'}(1)(x-1)=x-1, \\ {p}_{2}(x) & = & f(1)+{f}^{'}(1)(x-1)+\frac{{f}^{″}(1)}{2}{(x-1)}^{2}=(x-1)-\frac{1}{2}{(x-1)}^{2}, \\ {p}_{3}(x) & = & f(1)+{f}^{'}(1)(x-1)+\frac{{f}^{″}(1)}{2}{(x-1)}^{2}+\frac{f\text{‴}(1)}{3\text{!}}{(x-1)}^{3} \\ & = & (x-1)-\frac{1}{2}{(x-1)}^{2}+\frac{1}{3}{(x-1)}^{3}.\end{array}\]The graphs of \(y=f(x)\) and the first three Taylor polynomials are shown in .
Condensed — the full section is in OpenStax Calculus Volume 2.
Taylor’s Theorem with Remainder
Recall that the nth Taylor polynomial for a function \(f\) at a is the nth partial sum of the Taylor series for \(f\) at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials \(\{{p}_{n}\}\) converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to \(f.\) To answer this question, we define the remainder \({R}_{n}(x)\) as
\[{R}_{n}(x)=f(x)-{p}_{n}(x).\]For the sequence of Taylor polynomials to converge to \(f,\) we need the remainder Rn to converge to zero. To determine if Rn converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.
Here we look for a bound on \(|{R}_{n}|.\) Consider the simplest case: \(n=0.\) Let p0 be the 0th Taylor polynomial at a for a function \(f.\) The remainder R0 satisfies
\[\begin{array}{ll}{R}_{0}(x) & =f(x)-{p}_{0}(x) \\ & =f(x)-f(a).\end{array}\]If \(f\) is differentiable on an interval I containing a and x, then by the Mean Value Theorem there exists a real number c between a and x such that \(f(x)-f(a)={f}^{'}(c)(x-a).\) Therefore,
\[{R}_{0}(x)={f}^{'}(c)(x-a).\]Using the Mean Value Theorem in a similar argument, we can show that if \(f\) is n times differentiable on an interval I containing a and x, then the nth remainder Rn satisfies
\[{R}_{n}(x)=\frac{{f}^{(n+1)}(c)}{(n+1)\text{!}}{(x-a)}^{n+1}\]for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder Rn. If we happen to know that \(|{f}^{(n+1)}(x)|\) is bounded by some real number M on this interval I, then
\[|{R}_{n}(x)|\le \frac{M}{(n+1)\text{!}}{|x-a|}^{n+1}\]for all x in the interval I.
Condensed — the full section is in OpenStax Calculus Volume 2.
Representing Functions with Taylor and Maclaurin Series
We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.
Example
Try it.
Find the Taylor series for \(f(x)=\frac{1}{x}\) at \(x=1.\) Determine the interval of convergence.
Solution
For \(f(x)=\frac{1}{x},\) the values of the function and its first four derivatives at \(x=1\) are
\[\begin{array}{llllllll}f(x) & = & \frac{1}{x} & & & f(1) & = & 1 \\ {f}^{'}(x) & = & -\frac{1}{{x}^{2}} & & & {f}^{'}(1) & = & -1 \\ {f}^{″}(x) & = & \frac{2}{{x}^{3}} & & & {f}^{″}(1) & = & 2\text{!} \\ f\text{‴}(x) & = & -\frac{3\cdot 2}{{x}^{4}} & & & f\text{‴}(1) & = & -3\text{!} \\ {f}^{(4)}(x) & = & \frac{4\cdot 3\cdot 2}{{x}^{5}} & & & {f}^{(4)}(1) & = & 4\text{!.}\end{array}\]That is, we have \({f}^{(n)}(1)={(-1)}^{n}n\text{!}\) for all \(n\ge 0.\) Therefore, the Taylor series for \(f\) at \(x=1\) is given by
\[\sum _{n=0}^{\infty }\frac{{f}^{(n)}(1)}{n\text{!}}{(x-1)}^{n}=\sum _{n=0}^{\infty }{(-1)}^{n}{(x-1)}^{n}.\]To find the interval of convergence, we use the ratio test. We find that
\[\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{|{(-1)}^{n+1}{(x-1)}^{n+1}|}{|{(-1)}^{n}{(x-1)}^{n}|}=|x-1|.\]Thus, the series converges if \(|x-1|<1.\) That is, the series converges for \(0 diverges by the divergence test. Similarly, at \(x=0,\) diverges. Therefore, the interval of convergence is \((0,2).\)
We know that the Taylor series found in this example converges on the interval \((0,2),\) but how do we know it actually converges to \(f?\) We consider this question in more generality in a moment, but for this example, we can answer this question by writing
\[f(x)=\frac{1}{x}=\frac{1}{1-(1-x)}.\]That is, \(f\) can be represented by the geometric series \(\sum _{n=0}^{\infty }{(1-x)}^{n}.\) Since this is a geometric series, it converges to \(\frac{1}{x}\) as long as \(|1-x|<1.\) Therefore, the Taylor series found in does converge to \(f(x)=\frac{1}{x}\) on \((0,2).\)
\[\underset{n\to \infty }{\text{lim}}{p}_{n}(x)=f(x).\]Since the remainder \({R}_{n}(x)=f(x)-{p}_{n}(x),\) the Taylor series converges to \(f\) if and only if
\[\underset{n\to \infty }{\text{lim}}{R}_{n}(x)=0.\]We now state this theorem formally.
With this theorem, we can prove that a Taylor series for \(f\) at a converges to \(f\) if we can prove that the remainder \({R}_{n}(x)\to 0.\) To prove that \({R}_{n}(x)\to 0,\) we typically use the bound
\[|{R}_{n}(x)|\le \frac{M}{(n+1)\text{!}}{|x-a|}^{n+1}\]Condensed — the full section is in OpenStax Calculus Volume 2.
Key Concepts
- Taylor polynomials are used to approximate functions near a value \(x=a.\) Maclaurin polynomials are Taylor polynomials at \(x=0.\)
- The nth degree Taylor polynomials for a function \(f\) are the partial sums of the Taylor series for \(f.\)
- If a function \(f\) has a power series representation at \(x=a,\) then it is given by its Taylor series at \(x=a.\)
- A Taylor series for \(f\) converges to \(f\) if and only if \(\underset{n\to \infty }{\text{lim}}{R}_{n}(x)=0\) where \({R}_{n}(x)=f(x)-{p}_{n}(x).\)
- The Taylor series for ex, \(\text{sin}\ x,\) and \(\text{cos}\ x\) converge to the respective functions for all real x.
Key Equations
| Taylor series for the function \(f\) at the point \(x=a\) | \(\sum _{n=0}^{\infty }\frac{{f}^{(n)}(a)}{n\text{!}}{(x-a)}^{n}=f(a)+{f}^{'}(a)(x-a)+\frac{{f}^{″}(a)}{2\text{!}}{(x-a)}^{2}+\text{\cdots }+\frac{{f}^{(n)}(a)}{n\text{!}}{(x-a)}^{n}+\text{\cdots }\) |
Taylor and Maclaurin Series
In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.
In the following exercises, verify that the given choice of n in the remainder estimate \(|{R}_{n}|\le \frac{M}{(n+1)\text{!}}{(x-a)}^{n+1},\) where M is the maximum value of \(|{f}^{(n+1)}(z)|\) on the interval between a and the indicated point, yields \(|{R}_{n}|\le \frac{1}{1000}.\) Find the value of the Taylor polynomial pn of \(f\) at the indicated point. If \(\left|Rn\right|\) is not less than \(1/1000\), determine what it is.
In the following exercises, find the smallest value of n such that the remainder estimate \(|{R}_{n}|\le \frac{M}{(n+1)\text{!}}{(x-a)}^{n+1},\) where M is the maximum value of \(|{f}^{(n+1)}(z)|\) on the interval between a and the indicated point, yields \(|{R}_{n}|\le \frac{1}{1000}\) on the indicated interval.
In the following exercises, the maximum of the right-hand side of the remainder estimate \(|{R}_{1}|\le \frac{\text{max}|{f}^{″}(z)|}{2}{R}^{2}\) on \([a-R,a+R]\) occurs at a or \(a\pm R.\) Estimate the maximum value of R such that \(\frac{\text{max}|{f}^{″}(z)|}{2}{R}^{2}\le 0.1\) on \([a-R,a+R]\) by plotting this maximum as a function of R.
In the following exercises, find the Taylor series of the given function centered at the indicated point.
In the following exercises, compute the Taylor series of each function around \(x=1.\)
[T] In the following exercises, identify the value of x such that the given series \(\sum _{n=0}^{\infty }{a}_{n}\) is the value of the Maclaurin series of \(f(x)\) at \(x.\) Approximate the value of \(f(x)\) using \({S}_{10}=\sum _{n=0}^{10}{a}_{n}.\)
Condensed — the full section is in OpenStax Calculus Volume 2.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Find the Taylor polynomials \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}\) for \(f(x)=\text{ln}\ x\) at \(x=1.\) Use a graphing utility to compare the graph of \(f\) with the graphs of \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}.\)
Odkrij odgovor
To find these Taylor polynomials, we need to evaluate \(f\) and its first three derivatives at \(x=1.\)
\[\begin{array}{llllllll}f(x) & = & \text{ln}\ x & & & f(1) & = & 0 \\ {f}^{'}(x) & = & \frac{1}{x} & & & {f}^{'}(1) & = & 1 \\ {f}^{″}(x) & = & -\frac{1}{{x}^{2}} & & & {f}^{″}(1) & = & -1 \\ f\text{‴}(x) & = & \frac{2}{{x}^{3}} & & & f\text{‴}(1) & = & 2\end{array}\]Therefore,
\[\begin{array}{lll}{p}_{0}(x) & = & f(1)=0, \\ {p}_{1}(x) & = & f(1)+{f}^{'}(1)(x-1)=x-1, \\ {p}_{2}(x) & = & f(1)+{f}^{'}(1)(x-1)+\frac{{f}^{″}(1)}{2}{(x-1)}^{2}=(x-1)-\frac{1}{2}{(x-1)}^{2}, \\ {p}_{3}(x) & = & f(1)+{f}^{'}(1)(x-1)+\frac{{f}^{″}(1)}{2}{(x-1)}^{2}+\frac{f\text{‴}(1)}{3\text{!}}{(x-1)}^{3} \\ & = & (x-1)-\frac{1}{2}{(x-1)}^{2}+\frac{1}{3}{(x-1)}^{3}.\end{array}\]The graphs of \(y=f(x)\) and the first three Taylor polynomials are shown in .
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Find the Taylor polynomials \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}\) for \(f(x)=\frac{1}{{x}^{2}}\) at \(x=1.\)
Odkrij odgovor
\({p}_{0}(x)=1;\ {p}_{1}(x)=1-2(x-1);\ {p}_{2}(x)=1-2(x-1)+3{(x-1)}^{2};\ {p}_{3}(x)=1-2(x-1)+3{(x-1)}^{2}-4{(x-1)}^{3}\)
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For each of the following functions, find formulas for the Maclaurin polynomials \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}.\) Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}\) with \(f.\)
- \(f(x)={e}^{x}\)
- \(f(x)=\text{sin}\ x\)
- \(f(x)=\text{cos}\ x\)
Odkrij odgovor
- Since \(f(x)={e}^{x},\) we know that \(f(x)={f}^{'}(x)={f}^{″}(x)=\text{\cdots }={f}^{(n)}(x)={e}^{x}\) for all positive integers n. Therefore,
\[f(0)={f}^{'}(0)={f}^{″}(0)=\text{\cdots }={f}^{(n)}(0)=1\]
for all positive integers n. Therefore, we have
\[\begin{array}{lll}{p}_{0}(x) & = & f(0)=1, \\ {p}_{1}(x) & = & f(0)+{f}^{'}(0)x=1+x, \\ {p}_{2}(x) & = & f(0)+{f}^{'}(0)x+\frac{{f}^{″}(0)}{2\text{!}}{x}^{2}=1+x+\frac{1}{2}{x}^{2}, \\ {p}_{3}(x) & = & f(0)+{f}^{'}(0)x+\frac{{f}^{″}(0)}{2}{x}^{2}+\frac{f\text{‴}(0)}{3\text{!}}{x}^{3} \\ & = & 1+x+\frac{1}{2}{x}^{2}+\frac{1}{3\text{!}}{x}^{3}, \\ {p}_{n}(x) & = & f(0)+{f}^{'}(0)x+\frac{{f}^{″}(0)}{2}{x}^{2}+\frac{f\text{‴}(0)}{3\text{!}}{x}^{3}+\text{\cdots }+\frac{{f}^{(n)}(0)}{n\text{!}}{x}^{n} \\ & = & 1+x+\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{3}}{3\text{!}}+\text{\cdots }+\frac{{x}^{n}}{n\text{!}} \\ & = & \sum _{k=0}^{n}\frac{{x}^{k}}{k\text{!}}.\end{array}\]
The function and the first three Maclaurin polynomials are shown in .
- For \(f(x)=\text{sin}\ x,\) the values of the function and its first four derivatives at \(x=0\) are given as follows:
\[\begin{array}{llllllll}f(x) & = & \text{sin}\ x & & & f(0) & = & 0 \\ {f}^{'}(x) & = & \text{cos}\ x & & & {f}^{'}(0) & = & 1 \\ {f}^{″}(x) & = & \text{-}\text{sin}\ x & & & {f}^{″}(0) & = & 0 \\ f\text{‴}(x) & = & \text{-}\text{cos}\ x & & & f\text{‴}(0) & = & -1 \\ {f}^{(4)}(x) & = & \text{sin}\ x & & & {f}^{(4)}(0) & = & 0.\end{array}\]
Since the fourth derivative is \(\text{sin}\ x,\) the pattern repeats. That is, \({f}^{(2m)}(0)=0\) and \({f}^{(2m+1)}(0)={(-1)}^{m}\) for \(m\ge 0.\) Thus, we have
\[\begin{array}{l}{p}_{0}(x)=0, \\ {p}_{1}(x)=0+x=x, \\ {p}_{2}(x)=0+x+0=x, \\ {p}_{3}(x)=0+x+0-\frac{1}{3\text{!}}{x}^{3}=x-\frac{{x}^{3}}{3\text{!}}, \\ {p}_{4}(x)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0=x-\frac{{x}^{3}}{3\text{!}}, \\ {p}_{5}(x)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0+\frac{1}{5\text{!}}{x}^{5}=x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}},\end{array}\]
and for \(m\ge 0,\)
\[\begin{array}{ll}{p}_{2m+1}(x) & ={p}_{2m+2}(x) \\ & =x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}-\text{\cdots }+{(-1)}^{m}\frac{{x}^{2m+1}}{(2m+1)\text{!}} \\ & =\sum _{k=0}^{m}{(-1)}^{k}\frac{{x}^{2k+1}}{(2k+1)\text{!}}.\end{array}\]
Graphs of the function and its Maclaurin polynomials are shown in .
- For \(f(x)=\text{cos}\ x,\) the values of the function and its first four derivatives at \(x=0\) are given as follows:
\[\begin{array}{llllllll}f(x) & = & \text{cos}\ x & & & f(0) & = & 1 \\ {f}^{'}(x) & = & \text{-}\text{sin}\ x & & & {f}^{'}(0) & = & 0 \\ {f}^{″}(x) & = & \text{-}\text{cos}\ x & & & {f}^{″}(0) & = & -1 \\ f\text{‴}(x) & = & \text{sin}\ x & & & f\text{‴}(0) & = & 0 \\ {f}^{(4)}(x) & = & \text{cos}\ x & & & {f}^{(4)}(0) & = & 1.\end{array}\]
Since the fourth derivative is \(\text{cos}\ x,\) the pattern repeats. In other words, \({f}^{(2m)}(0)={(-1)}^{m}\) and \({f}^{(2m+1)}=0\) for \(m\ge 0.\) Therefore,
\[\begin{array}{l}{p}_{0}(x)=1, \\ {p}_{1}(x)=1+0=1, \\ {p}_{2}(x)=1+0-\frac{1}{2\text{!}}{x}^{2}=1-\frac{{x}^{2}}{2\text{!}}, \\ {p}_{3}(x)=1+0-\frac{1}{2\text{!}}{x}^{2}+0=1-\frac{{x}^{2}}{2\text{!}}, \\ {p}_{4}(x)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}}, \\ {p}_{5}(x)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}+0=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}},\end{array}\]
and for \(n\ge 0,\)
\[\begin{array}{ll}{p}_{2m}(x) & ={p}_{2m+1}(x) \\ & =1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}}-\text{\cdots }+{(-1)}^{m}\frac{{x}^{2m}}{(2m)\text{!}} \\ & =\sum _{k=0}^{m}{(-1)}^{k}\frac{{x}^{2k}}{(2k)\text{!}}.\end{array}\]
Graphs of the function and the Maclaurin polynomials appear in .
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Find formulas for the Maclaurin polynomials \({p}_{0},{p}_{1},{p}_{2}\) and \({p}_{3}\) for \(f(x)=\frac{1}{1+x}.\) Find a formula for the nth Maclaurin polynomial. Write your answer using sigma notation.
Odkrij odgovor
\({p}_{0}(x)=1;\ {p}_{1}(x)=1-x;\ {p}_{2}(x)=1-x+{x}^{2};\ {p}_{3}(x)=1-x+{x}^{2}-{x}^{3};\ {p}_{n}(x)=1-x+{x}^{2}-{x}^{3}+\text{\cdots }+{(-1)}^{n}{x}^{n}=\sum _{k=0}^{n}{(-1)}^{k}{x}^{k}\)
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Consider the function \(f(x)=\sqrt[3]{x}.\)
- Find the first and second Taylor polynomials for \(f\) at \(x=8.\) Use a graphing utility to compare these polynomials with \(f\) near \(x=8.\)
- Use these two polynomials to estimate \(\sqrt[3]{11}.\)
- Use Taylor’s theorem to bound the error.
Odkrij odgovor
- For \(f(x)=\sqrt[3]{x},\) the values of the function and its first two derivatives at \(x=8\) are as follows:
\[\begin{array}{llllllll}f(x) & = & \sqrt[3]{x} & & & f(8) & = & 2 \\ {f}^{'}(x) & = & \frac{1}{3{x}^{2\text{/}3}} & & & {f}^{'}(8) & = & \frac{1}{12} \\ {f}^{″}(x) & = & \frac{-2}{9{x}^{5\text{/}3}} & & & {f}^{″}(8) & = & -\frac{1}{144}.\end{array}\]
Thus, the first and second Taylor polynomials at \(x=8\) are given by
\[\begin{array}{lll}{p}_{1}(x) & = & f(8)+{f}^{'}(8)(x-8) \\ & = & 2+\frac{1}{12}(x-8) \\ {p}_{2}(x) & = & f(8)+{f}^{'}(8)(x-8)+\frac{{f}^{″}(8)}{2\text{!}}{(x-8)}^{2} \\ & = & 2+\frac{1}{12}(x-8)-\frac{1}{288}{(x-8)}^{2}.\end{array}\]
The function and the Taylor polynomials are shown in .
- Using the first Taylor polynomial at \(x=8,\) we can estimate
\[\sqrt[3]{11}\approx {p}_{1}(11)=2+\frac{1}{12}(11-8)=2.25.\]
Using the second Taylor polynomial at \(x=8,\) we obtain
\[\sqrt[3]{11}\approx {p}_{2}(11)=2+\frac{1}{12}(11-8)-\frac{1}{288}{(11-8)}^{2}=2.21875.\] - By , there exists a c in the interval \((8,11)\) such that the remainder when approximating \(\sqrt[3]{11}\) by the first Taylor polynomial satisfies
\[{R}_{1}(11)=\frac{{f}^{″}(c)}{2\text{!}}{(11-8)}^{2}.\]
We do not know the exact value of c, so we find an upper bound on \({R}_{1}(11)\) by determining the maximum value of \({f}^{″}\) on the interval \((8,11).\) Since \({f}^{″}(x)=-\frac{2}{9{x}^{5\text{/}3}},\) the largest value for \(|{f}^{″}(x)|\) on that interval occurs at \(x=8.\) Using the fact that \({f}^{″}(8)=-\frac{1}{144},\) we obtain
\[|{R}_{1}(11)|\le \frac{1}{144\cdot 2\text{!}}{(11-8)}^{2}=0.03125.\]
Similarly, to estimate \({R}_{2}(11),\) we use the fact that
\[{R}_{2}(11)=\frac{f\text{‴}(c)}{3\text{!}}{(11-8)}^{3}.\]
Since \(f\text{‴}(x)=\frac{10}{27{x}^{8\text{/}3}},\) the maximum value of \(f\text{‴}\) on the interval \((8,11)\) is \(f\text{‴}(8)\approx 0.0014468.\) Therefore, we have
\[|{R}_{2}(11)|\le \frac{0.0011468}{3\text{!}}{(11-8)}^{3}\approx 0.0065104.\]
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Find the first and second Taylor polynomials for \(f(x)=\sqrt{x}\) at \(x=4.\) Use these polynomials to estimate \(\sqrt{6}.\) Use Taylor’s theorem to bound the error.
Odkrij odgovor
\({p}_{1}(x)=2+\frac{1}{4}(x-4);{p}_{2}(x)=2+\frac{1}{4}(x-4)-\frac{1}{64}{(x-4)}^{2};\ {p}_{1}(6)=2.5;{p}_{2}(6)=2.4375;\)
\(|{R}_{1}(6)|\le 0.0625;|{R}_{2}(6)|\le 0.015625\)
-
From b., the Maclaurin polynomials for \(\text{sin}\ x\) are given by
\[\begin{array}{ll}{p}_{2m+1}(x) & ={p}_{2m+2}(x) \\ & =x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}-\frac{{x}^{7}}{7\text{!}}+\text{\cdots }+{(-1)}^{m}\frac{{x}^{2m+1}}{(2m+1)\text{!}}\end{array}\]for \(m=0,1,2,\text{\ldots }.\)
- Use the fifth Maclaurin polynomial for \(\text{sin}\ x\) to approximate \(\text{sin}(\frac{\pi }{18})\) and bound the error.
- For what values of x does the fifth Maclaurin polynomial approximate \(\text{sin}\ x\) to within 0.0001?
Odkrij odgovor
- The fifth Maclaurin polynomial is
\[{p}_{5}(x)=x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}.\]
Using this polynomial, we can estimate as follows:
\[\begin{array}{ll}\text{sin}(\frac{\pi }{18}) & \approx {p}_{5}(\frac{\pi }{18}) \\ & =\frac{\pi }{18}-\frac{1}{3\text{!}}{(\frac{\pi }{18})}^{3}+\frac{1}{5\text{!}}{(\frac{\pi }{18})}^{5} \\ & \approx 0.173648.\end{array}\]
To estimate the error, use the fact that the sixth Maclaurin polynomial is \({p}_{6}(x)={p}_{5}(x)\) and calculate a bound on \({R}_{6}(\frac{\pi }{18}).\) By , the remainder is
\[{R}_{6}(\frac{\pi }{18})=\frac{{f}^{(7)}(c)}{7\text{!}}{(\frac{\pi }{18})}^{7}\]
for some c between 0 and \(\frac{\pi }{18}.\) Using the fact that \(|{f}^{(7)}(x)|\le 1\) for all x, we find that the magnitude of the error is at most
\[\frac{1}{7\text{!}}\cdot {(\frac{\pi }{18})}^{7}\le 9.8\ \times \ {10}^{-10}.\] - We need to find the values of x such that
\[\frac{1}{7\text{!}}{|x|}^{7}\le 0.0001.\]
Solving this inequality for x, we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as \(|x|<0.907.\)
-
Use the fourth Maclaurin polynomial for \(\text{cos}\ x\) to approximate \(\text{cos}(\frac{\pi }{12}).\)
Odkrij odgovor
0.96593
-
Find the Taylor series for \(f(x)=\frac{1}{x}\) at \(x=1.\) Determine the interval of convergence.
Odkrij odgovor
For \(f(x)=\frac{1}{x},\) the values of the function and its first four derivatives at \(x=1\) are
\[\begin{array}{llllllll}f(x) & = & \frac{1}{x} & & & f(1) & = & 1 \\ {f}^{'}(x) & = & -\frac{1}{{x}^{2}} & & & {f}^{'}(1) & = & -1 \\ {f}^{″}(x) & = & \frac{2}{{x}^{3}} & & & {f}^{″}(1) & = & 2\text{!} \\ f\text{‴}(x) & = & -\frac{3\cdot 2}{{x}^{4}} & & & f\text{‴}(1) & = & -3\text{!} \\ {f}^{(4)}(x) & = & \frac{4\cdot 3\cdot 2}{{x}^{5}} & & & {f}^{(4)}(1) & = & 4\text{!.}\end{array}\]That is, we have \({f}^{(n)}(1)={(-1)}^{n}n\text{!}\) for all \(n\ge 0.\) Therefore, the Taylor series for \(f\) at \(x=1\) is given by
\[\sum _{n=0}^{\infty }\frac{{f}^{(n)}(1)}{n\text{!}}{(x-1)}^{n}=\sum _{n=0}^{\infty }{(-1)}^{n}{(x-1)}^{n}.\]To find the interval of convergence, we use the ratio test. We find that
\[\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{|{(-1)}^{n+1}{(x-1)}^{n+1}|}{|{(-1)}^{n}{(x-1)}^{n}|}=|x-1|.\]Thus, the series converges if \(|x-1|<1.\) That is, the series converges for \(0
diverges by the divergence test. Similarly, at \(x=0,\)
\[\sum _{n=0}^{\infty }{(-1)}^{n}{(0-1)}^{n}=\sum _{n=0}^{\infty }{(-1)}^{2n}=\sum _{n=0}^{\infty }1\]diverges. Therefore, the interval of convergence is \((0,2).\)
-
Find the Taylor series for \(f(x)=\frac{1}{2x}\) at \(x=2\) and determine its interval of convergence.
Odkrij odgovor
\(\sum _{n=0}^{\infty }\frac{{\left(2-x\right)}^{n}}{{2}^{n+1}}.\) The interval of convergence is \((0,4).\)
-
For each of the following functions, find the Maclaurin series and its interval of convergence. Use to prove that the Maclaurin series for \(f\) converges to \(f\) on that interval.
- ex
- \(\text{sin}\ x\)
Odkrij odgovor
- Using the nth Maclaurin polynomial for ex found in a., we find that the Maclaurin series for ex is given by
\[\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}.\]
To determine the interval of convergence, we use the ratio test. Since
\[\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{{|x|}^{n+1}}{(n+1)\text{!}}\cdot \frac{n\text{!}}{{|x|}^{n}}=\frac{|x|}{n+1},\]
we have
\[\underset{n\to \infty }{\text{lim}}\frac{|{a}_{n+1}|}{|{a}_{n}|}=\underset{n\to \infty }{\text{lim}}\frac{|x|}{n+1}=0\]
for all x. Therefore, the series converges absolutely for all x, and thus, the interval of convergence is \((\text{-}\infty ,\infty ).\) To show that the series converges to ex for all x, we use the fact that \({f}^{(n)}(x)={e}^{x}\) for all \(n\ge 0\) and ex is an increasing function on \((\text{-}\infty ,\infty ).\) Therefore, for any real number b, the maximum value of ex for all \(|x|\le b\) is eb. Thus,
\[|{R}_{n}(x)|\le \frac{{e}^{b}}{(n+1)\text{!}}{|x|}^{n+1}.\]
Since we just showed that
\[\sum _{n=0}^{\infty }\frac{|x{|}^{n}}{n\text{!}}\]
converges for all x, by the divergence test, we know that
\[\underset{n\to \infty }{\text{lim}}\frac{{|x|}^{n+1}}{(n+1)\text{!}}=0\]
for any real number x. By combining this fact with the squeeze theorem, the result is \(\underset{n\to \infty }{\text{lim}}{R}_{n}(x)=0.\) - Using the nth Maclaurin polynomial for \(\text{sin}\ x\) found in b., we find that the Maclaurin series for \(\text{sin}\ x\) is given by
\[\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n+1}}{(2n+1)\text{!}}.\]
In order to apply the ratio test, consider
\[\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{{|x|}^{2n+3}}{(2n+3)\text{!}}\cdot \frac{(2n+1)\text{!}}{{|x|}^{2n+1}}=\frac{{|x|}^{2}}{(2n+3)(2n+2)}.\]
Since
\[\underset{n\to \infty }{\text{lim}}\frac{{|x|}^{2}}{(2n+3)(2n+2)}=0\]
for all x, we obtain the interval of convergence as \((\text{-}\infty ,\infty ).\) To show that the Maclaurin series converges to \(\text{sin}\ x,\) look at \({R}_{n}(x).\) For each x there exists a real number c between 0 and x such that
\[{R}_{n}(x)=\frac{{f}^{(n+1)}(c)}{(n+1)\text{!}}{x}^{n+1}.\]
Since \(|{f}^{(n+1)}(c)|\le 1\) for all integers n and all real numbers c, we have
\[|{R}_{n}(x)|\le \frac{{|x|}^{n+1}}{(n+1)\text{!}}\]
for all real numbers x. Using the same idea as in part a., the result is \(\underset{n\to \infty }{\text{lim}}{R}_{n}(x)=0\) for all x, and therefore, the Maclaurin series for \(\text{sin}\ x\) converges to \(\text{sin}\ x\) for all real x.
-
Find the Maclaurin series for \(f(x)=\text{cos}\ x.\) Use the ratio test to show that the interval of convergence is \((\text{-}\infty ,\infty ).\) Show that the Maclaurin series converges to \(\text{cos}\ x\) for all real numbers x.
Odkrij odgovor
\(\sum _{n=0}^{\infty }\frac{{(-1)}^{n}{x}^{2n}}{(2n)\text{!}}\)
By the ratio test, the interval of convergence is \((\text{-}\infty ,\infty ).\) Since \(|{R}_{n}(x)|\le \frac{{|x|}^{n+1}}{(n+1)\text{!}},\) the series converges to \(\text{cos}\ x\) for all real x.
-
\(f(x)=1+x+{x}^{2}\) at \(a=1\)
-
\(f(x)=1+x+{x}^{2}\) at \(a=-1\)
Odkrij odgovor
\(f(-1)=1;{f}^{'}(-1)=-1;{f}^{″}(-1)=2;f(x)=1-(x+1)+{(x+1)}^{2}\)
-
\(f(x)=\text{cos}(2x)\) at \(a=\pi\)
-
\(f(x)=\text{sin}(2x)\) at \(a=\frac{\pi }{2}\)
Odkrij odgovor
\({f}^{'}(x)=2\ \text{cos}(2x);{f}^{″}(x)=-4\ \text{sin}(2x);{p}_{2}(x)=-2(x-\frac{\pi }{2})\)
-
\(f(x)=\sqrt{x}\) at \(a=4\)
-
\(f(x)=\text{ln}\ x\) at \(a=1\)
Odkrij odgovor
\({f}^{'}(x)=\frac{1}{x};{f}^{″}(x)=-\frac{1}{{x}^{2}};{p}_{2}(x)=0+(x-1)-\frac{1}{2}{(x-1)}^{2}\)
-
\(f(x)=\frac{1}{x}\) at \(a=1\)
-
\(f(x)={e}^{x}\) at \(a=1\)
Odkrij odgovor
\({p}_{2}(x)=e+e(x-1)+\frac{e}{2}{(x-1)}^{2}\)
-
[T] \(\sqrt{10};a=9,n=3\)
-
[T] \({(28)}^{1\text{/}3};a=27,n=1\)
Odkrij odgovor
\(\frac{{d}^{2}}{d{x}^{2}}{x}^{1\text{/}3}=-\frac{2}{9{x}^{5\text{/}3}}\ge -0.00092\text{\ldots }\) when \(x\ge 28\) so the remainder estimate applies to the linear approximation \({x}^{1\text{/}3}\approx {p}_{1}(27)=3+\frac{x-27}{27},\) which gives \({(28)}^{1\text{/}3}\approx 3+\frac{1}{27}=3.\overset{\bar}{037},\) while \({(28)}^{1\text{/}3}\approx 3.03658.\)
-
[T] \(\text{sin}(6);a=2\pi ,n=5\)
-
[T] e2; \(a=0,n=9\)
Odkrij odgovor
Using the estimate \(\frac{{2}^{10}}{10\text{!}}<0.000283\) we can use the Taylor expansion of order 9 to estimate ex at \(x=2.\) as \({e}^{2}\approx {p}_{9}(2)=1+2+\frac{{2}^{2}}{2}+\frac{{2}^{3}}{6}+\text{\cdots }+\frac{{2}^{9}}{9\text{!}}=7.3887\text{\ldots }\) whereas \({e}^{2}\approx 7.3891.\)
-
[T] \(\text{cos}(\frac{\pi }{5});a=0,n=4\)
-
[T] \(\text{ln}\ (2);a=1,n=1000\)
Odkrij odgovor
Since \(\frac{{d}^{n}}{d{x}^{n}}(\text{ln}\ x)={(-1)}^{n-1}\frac{(n-1)\text{!}}{{x}^{n}},{R}_{1000}\approx \frac{1}{1001}.\) One has \({p}_{1000}(1)=\sum _{n=1}^{1000}\frac{{(-1)}^{n-1}}{n}\approx 0.6936\) whereas \(\text{ln}\ (2)\approx 0.6931\text{\cdots }.\)
-
Integrate the approximation \(\text{sin}\ t\approx t-\frac{{t}^{3}}{6}+\frac{{t}^{5}}{120}-\frac{{t}^{7}}{5040}\) evaluated at πt to approximate \({\int }_{0}^{1}\frac{\text{sin}\ \pi t}{\pi t}dt.\)
-
Integrate the approximation \({e}^{x}\approx 1+x+\frac{{x}^{2}}{2}+\text{\cdots }+\frac{{x}^{6}}{720}\) evaluated at −x2 to approximate \({\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx.\)
Odkrij odgovor
\({\int }_{0}^{1}(1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{6}+\frac{{x}^{8}}{24}-\frac{{x}^{10}}{120}+\frac{{x}^{12}}{720})\ dx\)
\(=1-\frac{{1}^{3}}{3}+\frac{{1}^{5}}{10}-\frac{{1}^{7}}{42}+\frac{{1}^{9}}{9\cdot 24}-\frac{{1}^{11}}{120\cdot 11}+\frac{{1}^{13}}{720\cdot 13}\approx 0.74683\) whereas \({\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx\approx 0.74682.\)
-
\(f(x)=\text{sin}\ x\) on \([\text{-}\pi ,\pi ],a=0\)
-
\(f(x)=\text{cos}\ x\) on \([-\frac{\pi }{2},\frac{\pi }{2}],a=0\)
Odkrij odgovor
Since \({f}^{(n+1)}(z)\) is \(\text{sin}\ z\) or \(\text{cos}\ z,\) we have \(M=1.\) Since \(|x-0|\le \frac{\pi }{2},\) we seek the smallest n such that \(\frac{{\pi }^{n+1}}{{2}^{n+1}(n+1)\text{!}}\le 0.001.\) The smallest such value is \(n=7.\) The remainder estimate is \({R}_{7}\le 0.00092.\)
-
\(f(x)={e}^{-2x}\) on \([-1,1],a=0\)
-
\(f(x)={e}^{\text{-}x}\) on \([-3,3],a=0\)
Odkrij odgovor
Since \({f}^{(n+1)}(z)=\text{\pm }{e}^{\text{-}z}\) one has \(M={e}^{3}.\) Since \(|x-0|\le 3,\) one seeks the smallest n such that \(\frac{{3}^{n+1}{e}^{3}}{(n+1)\text{!}}\le 0.001.\) The smallest such value is \(n=14.\) The remainder estimate is \({R}_{14}\le 0.000220.\)
-
[T] ex approximated by \(1+x,a=0\)
-
[T] \(\text{sin}\ x\) approximated by x, \(a=0\)
Odkrij odgovor
Since \(\text{sin}\ x\) is increasing for small x and since \(\text{si}{n}^{″}x=\text{-}\text{sin}\ x,\) the estimate applies whenever \({R}^{2}\text{sin}(R)\le 0.2,\) which applies up to \(R=0.596.\) -
[T] \(\text{ln}\ x\) approximated by \(x-1,a=1\)
-
[T] \(\text{cos}\ x\) approximated by \(1,a=0\)
Odkrij odgovor
Since the second derivative of \(\text{cos}\ x\) is \(\text{-}\text{cos}\ x\) and since \(\text{cos}\ x\) is decreasing away from \(x=0,\) the estimate applies when \({R}^{2}\text{cos}\ R\le 0.2\) or \(R\le 0.447.\) -
\({x}^{4}\) at \(a=-1\)
-
\(1+x+{x}^{2}+{x}^{3}\) at \(a=-1\)
Odkrij odgovor
\({(x+1)}^{3}-2{(x+1)}^{2}+2(x+1)\)
-
\(\text{sin}\ x\) at \(a=\pi\)
-
\(\text{cos}\ x\) at \(a=2\pi\)
Odkrij odgovor
Values of derivatives are the same as for \(x=0\) so \(\text{cos}\ x={\sum _{n=0}^{\infty }(-1)}^{n}\frac{{(x-2\pi )}^{2n}}{(2n)\text{!}}\)
Symbols used here
Add a_k for k = 1 up to n.
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
x belongs to A; every element of A is in B.
Instantaneous rate of change; slope of the graph.
Inequalities that allow equality; < and > exclude it.
Least upper bound, greatest lower bound.
2.71828…, the base whose exponential is its own derivative.
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
The value f(x) approaches as x approaches a.
Antiderivative (indefinite) or signed area from a to b (definite).
Prime notation for derivatives with respect to x (or t).
Constants of integration fixed by initial conditions.
How to: Taylor and Maclaurin Series
- Describe the procedure for finding a Taylor polynomial of a given order for a function.
- Explain the meaning and significance of Taylor’s theorem with remainder.
- Estimate the remainder for a Taylor series approximation of a given function.
- Since
- For
- For
- Find the first and second Taylor polynomials for
- Use these two polynomials to estimate
Questions people ask
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.
Poskusi sam.
Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Več v Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests