Reminder

\(k\)-factorial is the number \(k! = k(k-1)(k-2)\cdots (2)(1)\). Thus, \[ \begin{align*} 1!=1,\quad 2!=(2)1 = 2\\ 3!=(3)(2)1 =6 \end{align*} \]

By convention, we define \(0! = 1\).

THEOREM 1

The polynomial \(T_n(x)\) centered at \(a\) agrees with \(f(x)\) to order \(n\) at \(x = a\), and it is the only polynomial of degree at most \(n\) with this property.

EXAMPLE 1 Maclaurin Polynomials for \(e^x\)

Plot the third and fourth Maclaurin polynomials for \(f(x)=e^x\). Compare with the linear approximation.

Solution All higher derivatives coincide with \(f(x)\) itself: \(f^{(k)}(x) = e^x\). Therefore, \[ f(0) = f'(0) = f''(0) = f'''(0) = f^{(4)}(0) = e^0 = 1 \]

The third Maclaurin polynomial (the case \(a = 0\)) is \[ T_3(x) = f(0) + f'(0)x + \frac12f''(0)x^2 + \frac1{3!}f'''(0)x^3 = 1 + x + \frac12x^2+\frac16x^3\\\]

We obtain \(T_4(x)\) by adding the term of degree 4 to \(T_3(x)\): \[ T_4(x) = T_3(x) + \frac1{4!}f^{(4)}(0)x^4 = 1 + x + \frac12x^2+\frac16x^3 + \frac1{24}x^4 \]

Figure 8.35 shows that \(T_3\) and \(T_4\) approximate \(f(x) = e^x\) much more closely than the linear approximation \(T_1(x)\) on an interval around \(a=0\). Higher-degree Taylor polynomials would provide even better approximations on larger intervals.

Figure 8.35: Maclaurin polynomials for \(f(x)=e^x\).

EXAMPLE 2 Computing Taylor Polynomials

Compute the Taylor polynomial \(T_4(x)\) centered at \(a=3\) for \(f(x)=\sqrt{x+1}\).

Solution First evaluate the derivatives up to degree 4 at \(a=3\): \[ \begin{eqnarray*} f(x) &=& (x+1)^{1/2}, \quad\quad\quad\quad& f(3)&=&2 \\[3pt] f^{\prime }(x) &=&\frac{1}{2}(x+1)^{-1/2}, & f^{\prime }(3)&=&\frac{1}{4} \\[3pt] f''(x) &=& -\frac{1}{4}(x+1)^{-3/2}, & f''(3)&=&-\frac{1}{32} \\[3pt] f'''(x) &=&\frac{3}{8}(x+1)^{-5/2}, & f'''(3)&=& \frac{3}{256} \\[3pt] f^{(4)}(x) &=& -\frac{15}{16}(x+1)^{-7/2}, & f^{(4)}(3)&=& -\frac{15}{2048} \end{eqnarray*} \]

Then compute the coefficients \(\frac{f^{(j)}(3)}{j!}\): \[ \begin{alignat*}{3} &\textrm{Constant term}&& = f(3) = 2 \\ &\textrm{Coefficient of $(x-3)$}&& = f'(3)=\frac14\\ &\textrm{Coefficient of $(x-3)^2$}&& = \frac{f''(3)}{2!} = -\frac{1}{32}\cdot\frac1{2!}=-\frac1{64}\\[3pt] &\textrm{Coefficient of $(x-3)^3$}&& = \frac{f'''(3)}{3!} = \frac{3}{256}\cdot\frac1{3!} = \frac1{512}\\[3pt] &\textrm{Coefficient of $(x-3)^4$}&& = \frac{f^{(4)}(3)}{4!} =- \frac{15}{2048}\cdot\frac1{4!} = -\frac{5}{16{,}384} \end{alignat*} \]

The Taylor polynomial \(T_4(x)\) centered at \(a=3\) is (see Figure 8.36): \[ T_{4}(x) =2+\frac{1}{4}\left( x-3\right) -\frac{1}{64}\left( x-3\right) ^{2}+\frac{1}{512}\left( x-3\right) ^{3}-\frac{5}{16{,}384}\left( x-3\right) ^{4}\]

Figure 8.36: Graph of \(f(x)=\sqrt{x+1}\) and \(T_4(x)\) centered at \(x=3\).

EXAMPLE 3 Finding a General Formula for \(T_n\)

Find the Taylor polynomials \(T_n(x)\) of \(f(x) = \ln x\) centered at \(a=1\).

Solution For \(f(x) = \ln x\), the constant term of \(T_n(x)\) at \(a = 1\) is zero because \(f(1)=\ln 1 =0\). Next, we compute the derivatives: \[ f'(x)=x^{-1},\qquad f''(x) = -x^{-2}, \qquad f'''(x)=2x^{-3},\qquad f^{(4)}(x)=-3\cdot 2x^{-4} \]

Similarly, \(f^{(5)}(x) = 4\cdot 3\cdot 2x^{-5}\). The general pattern is that \(f^{(k)}(x)\) is a multiple of \(x^{-k}\), with a coefficient \(\pm (k-1)!\) that alternates in sign: \[ f^{(k)}(x) = (-1)^{k-1}(k-1)!\,x^{-k}\tag{1}\]

The coefficient of \((x-1)^k\) in \(T_n(x)\) is \[ \frac{f^{(k)}(1)}{k!} = \frac{(-1)^{k-1}(k-1)!}{k!} = \frac{(-1)^{k-1}}{k}\qquad (\textrm{for \(k\ge 1\)}) \]

Thus, the coefficients for \(k\ge 1\) form a sequence \(1, -\frac12, \frac13, -\frac14,\ldots,\) and \[ T_n(x) = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3-\cdots + (-1)^{n-1}\frac1n(x-1)^n \]

EXAMPLE 4 Cosine

Find the Maclaurin polynomials of \(f(x)=\cos x\).

Solution The derivatives form a repeating pattern of period 4: \[ \begin{alignat*}{3} f(x) &= \cos x,\qquad f^{\prime }(x) &= -\sin x,&\qquad f^{\prime \prime }&(x) = -\cos x,\qquad f^{\prime \prime \prime }(x)=\sin x,\\ f^{(4)}(x) &= \cos x,\qquad f^{(5)}(x)\, &= -\sin x,&\qquad\cdots & \end{alignat*} \]

In general, \(f^{(j+4)}(x)=f^{(j)}(x)\). The derivatives at \(x=0\) also form a pattern:

Therefore, the coefficients of the odd powers \(x^{2k+1}\) are zero, and the coefficients of the even powers \(x^{2k}\) alternate in sign with value \((-1)^k/(2k)!\): \[\begin {eqnarray} T_0(x) &=& T_1(x)= 1,\qquad T_2(x) = T_3(x) = 1 -\frac{1}{2!}x^2\notag\\ T_4(x) &=& T_5(x) = 1-\frac{x^2}{2} + \frac{x^4}{4!}\notag\\ T_{2n}(x) &=& T_{2n+1}(x) = 1-\frac{1}{2}x^{2}+\frac{1}{4!}x^{4}-\frac{1}{6!}x^{6}+\cdots +(-1)^{n}\frac{1}{(2n)!}x^{2n} \notag\end {eqnarray} \]

Scottish mathematician Colin Maclaurin (1698-1746) was a professor in Edinburgh. Newton was so impressed by his work that he once offered to pay part of Maclaurin's salary.

Figure 8.37 shows that as \(n\) increases, \(T_n(x)\) approximates \(f(x) = \cos x\) well over larger and larger intervals, but outside this interval, the approximation fails.

EXAMPLE 5 How far is the horizon?

Valerie is at the beach, looking out over the ocean (Figure 8.38). How far can she see? Use Maclaurin polynomials to estimate the distance \(d\), assuming that Valerie's eye level is \(h = 1.7\) m above ground. What if she looks out from a window where her eye level is \(20\) m?

Figure 8.38: View from the beach

Figure 8.39: Valerie can see a distance \(d = R\theta\), the length of arc \(AH\).

Solution Let \(R\) be the radius of the earth. Figure 8.39 shows that Valerie can see a distance \(d = R\theta\), the length of the circular arc \(AH\) in Figure 8.39. We have \[ \cos\theta = \frac{R}{R+h} \]

Our key observation is that \(\theta\) is close to zero (both \(\theta\) and \(h\) are much smaller than shown in the figure), so we lose very little accuracy if we replace \(\cos\theta\) by its second Maclaurin polynomial \(T_2(\theta) = 1-\frac12\theta^2\), as computed in Example 4: \[ 1-\frac12\theta^2 \approx \frac{R}{R+h}\quad\Rightarrow\quad \theta^2 \approx 2 - \frac{2R}{R+h}\quad\Rightarrow\quad \theta \approx \sqrt{\frac{2h}{R+h}} \]

Furthermore, \(h\) is very small relative to \(R\), so we may replace \(R+h\) by \(R\) to obtain \[ d=R\theta \approx R\sqrt{\frac{2h}{R}}= \sqrt{2Rh} \]

The earth's radius is approximately \(R \approx 6.37 \times 10^6\) m, so \[ d=\sqrt{2Rh} \approx \sqrt{2(6.37 \times 10^6)h} \approx 3569\sqrt{h} \mbox{m} \]

In particular, we see that \(d\) is proportional to \(\sqrt{h}\).

If Valerie's eye level is \(h=1.7 \mbox{m}\), then \(d\approx3569\sqrt{1.7}\approx4653 \mbox{m}\), or roughly 4.7 km. If \(h=20 \mbox{m}\), then \(d\approx 3569\sqrt{20}\approx 15.96 \mbox{m}\), or nearly 16 km.

A proof of Theorem 2 is presented at the end of this section.

THEOREM 2 Error Bound

Assume that \(f^{(n + 1)}(x)\) exists and is continuous. Let \(K\) be a number such that \(|f^{(n+1)}(u)|\le K\) for all \(u\) between \(a\) and \(x\). Then \[ \boxed{\bbox[#FAF8ED,5pt]{|f(x) - T_n(x)| \le K\frac{|x-a|^{n+1}}{(n+1)!} }} \]

where \(T_n(x)\) is the \(n\)th Taylor polynomial centered at \(x=a\).

EXAMPLE 6 Using the Error Bound

Apply the error bound to \[|\ln 1.2 - T_3(1.2)|\] where \(T_3(x)\) is the third Taylor polynomial for \(f(x) = \ln x\) at \(a=1\). Check your result with a calculator.

Solution

Step 1. Find a value of \(K\).

To use the error bound with \(n=3\), we must find a value of \(K\) such that \(|f^{(4)}(u)|\le K\) for all \(u\) between \(a=1\) and \(x=1.2\). As we computed in Example 3, \(f^{(4)}(x) = - 6x^{-4}\). The absolute value \(|f^{(4)}(x)|\) is decreasing for \(x>0\), so its maximum value on \([1,1.2]\) is \(|f^{(4)}(1)| = 6\). Therefore, we may take \(K=6\).

Step 2. Apply the error bound. \[ |\ln 1.2 - T_3(1.2)| \le K\frac{\left| x-a\right|^{n+1}}{(n+1)!} = 6 \frac{\left| 1.2-1\right|^{4}}{4!} \approx 0.0004 \]

Step 3. Check the result.

Recall from Example 3 that \[ T_3(x) = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3 \]

The following values from a calculator confirm that the error is at most \(0.0004\): \[ |\ln 1.2 - T_3(1.2)| \approx |0.182667 - 0.182322| \approx 0.00035<0.0004 \]

Observe in Figure 8.40 that \(\ln x\) and \(T_3(x)\) are indistinguishable near \(x=1.2\).

Figure 8.40: \(\ln x\) and \(T_3(x)\) are indistinguishable near \(x=1.2\).

EXAMPLE 7 Approximating with a Given Accuracy

Let \(T_n(x)\) be the \(n\)th Maclaurin polynomial for \(f(x) = \cos x\). Find a value of \(n\) such that \[ |\cos 0.2 - T_{n}(0.2)| < 10^{-5} \]

Solution

Step 1. Find a value of \(K\).

Since \(|f^{(n)}(x)|\) is \(|\cos x|\) or \(|\sin x|\), depending on whether \(n\) is even or odd, we have \(|f^{(n)}(u)|\le 1\) for all \(u\). Thus, we may apply the error bound with \(K=1\).

Step 2. Find a value of \(n\).

The error bound gives us \[ |\cos 0.2 - T_{n}(0.2)|\le K\frac{\left| 0.2-0\right| ^{n+1}}{(n+1)!}=\frac{\left| 0.2\right| ^{n+1}}{(n+1)!} \]

To make the error less than \(10^{-5}\), we must choose \(n\) so that \[ \frac{| 0.2| ^{n+1}}{(n+1)!}<10^{-5} \]

It's not possible to solve this inequality for \(n\), but we can find a suitable \(n\) by checking several values:

We see that the error is less than \(10^{-5}\) for \(n=4\).

Taylor's Theorem

Assume that \(f^{(n+1)}(x)\) exists and is continuous. Then \[ \boxed{\bbox[#FAF8ED,5pt]{R_{n}(x)=\frac{1}{n!}\int_{a}^{x}(x-u)^{n}f^{(n+1)}(u)\, du}}\tag{2} \]

Proof

Set \[ I_{n}(x) =\frac{1}{n!}\int_{a}^{x}(x-u)^{n}f^{(n+1)}(u)\, du \]

Our goal is to show that \(R_{n}(x)=I_{n}(x)\). For \(n=0\), \(R_0(x) = f(x)-f(a)\) and the desired result is just a restatement of the Fundamental Theorem of Calculus: \[ I_{0}(x)=\int_{a}^{x}f^{\prime }(u)\, du=f(x)-f(a)=R_{0}(x) \]

To prove the formula for \(n>0\), we apply Integration by Parts to \(I_n(x)\) with \[ h(u)=\frac{1}{n!}(x-u)^n,\qquad g(u)=f^{(n)}(u) \]

Then \(g'(u) = f^{(n+1)}(u)\), and so \[\begin {eqnarray} I_n(x) &=&\int_{a}^{x}h(u)\,g^{\prime }(u)\, du = h(u)g(u)\bigg|_{a}^{x}-\int_{a}^{x}h^{\prime }(u)g(u)\,du \notag\\ &=& \frac{1}{n!}(x-u)^{n}f^{(n)}(u)\bigg|_{a}^{x}-\frac{1}{n!} \int_{a}^{x}(-n)(x-u)^{n-1}f^{(n)}(u)\, du \notag\\ &=&-\frac{1}{n!}(x-a)^{n}f^{(n)}(a) + I_{n-1}(x)\notag \end {eqnarray} \]

This can be rewritten as \[ I_{n-1}(x) =\frac{f^{(n)}(a)}{n!}(x-a)^{n}+I_{n}(x) \]

Now apply this relation \(n\) times, noting that \(I_0(x) = f(x) - f(a)\): \[ \begin {eqnarray} f(x) &=&f(a)+I_{0}(x) \notag\\ &=&f(a)+\frac{f^{^{\prime }}(a)}{1!}(x-a) +I_{1}(x) \notag\\ &=&f(a)+\frac{f^{^{\prime }}(a)}{1!}(x-a) +\frac{f^{\prime \prime }(a)}{2!}(x-a)^{2}+I_{2}(x) \notag\\ &\vdots & \notag\\ &=&f(a)+\frac{f^{^{\prime }}(a)}{1!}(x-a) +\cdots + \frac{f^{(n)}(a)}{n!}(x-a)^{n}+I_{n}(x)\notag \end {eqnarray} \]

This shows that \(f(x)=T_{n}(x)+I_{n}(x)\) and hence \(I_{n}(x)=R_{n}(x)\), as desired.

In Eq.(3), we use the inequality \[ \left\vert \int_a^b f(x)\, dx\right\vert \le\int_a^b |f(x)|\, dx \]

which is valid for all integrable functions.