REMINDER The notation \(\approx\) means “approximately equal to.” The accuracy of the Linear Approximation is discussed at the end of this section.

Linear Approximation of Δf

If \(f\) is differentiable at \(x = a\) and \(\Delta x\) is small, then

where \(\Delta f = f(a + \Delta x) - f(a)\).

GRAPHICAL INSIGHT

The Linear Approximation is sometimes called the tangent line approximation. Why? Observe in Figure 8.33 that \(\Delta f\) is the vertical change in the graph from \(x = a\) to \(x = a + \Delta x\). For a straight line, the vertical change is equal to the slope times the horizontal change \(\Delta x\), and since the tangent line has slope \(f'(a)\), its vertical change is \(f'(a)\Delta x\). So the Linear Approximation approximates \(\Delta f\) by the vertical change in the tangent line. When \(\Delta x\) is small, the two quantities are nearly equal.

Figure 8.33: Graphical meaning of the Linear Approximation \(\Delta f \approx f'(a)\Delta x\).

EXAMPLE 1

Use the Linear Approximation to estimate \(\frac{1}{10.2} - \frac{1}{10}\). How accurate is your estimate?

Solution We apply the Linear Approximation to \(f(x)=\frac{1}{x}\) with \(a = 10\) and \(\Delta x = 0.2\):

We have \(f'(x) = -x^{-2}\)and\( f'(10) = -0.01\), so \(\Delta f\) is approximated by

In other words,

A calculator gives the value \(\frac{1}{10.2} - \frac{1}{10} \approx -0.00196\) and thus our error is less than \(10^{-4}\):

Differential Notation

The Linear Approximation to \(y = f(x)\) is often written using the “differentials” \(dx\) and \(dy\). In this notation, \(dx\) is used instead of \(\Delta x\) to represent the change in \(x\), and \(dy\) is the corresponding vertical change in the tangent line:

Let \(\Delta y = f(a + dx) - f(a)\). Then the Linear Approximation says

This is simply another way of writing \(\Delta f \approx f'(a)\Delta x\).

EXAMPLE 2 Differential Notation

Approximately how much larger is \(\sqrt[3]{8.1}\) than \(\sqrt[3]{8} = 2\)?

Solution We are interested in \(\sqrt[3]{8.1} - \sqrt[3]{8}\), so we apply the Linear Approximation to \(f(x) = x^{\frac{1}{3}}\) with \(a = 8\) and change \(\Delta x = dx = 0.1\).

Step 1. Write out \(\Delta y\).

Step 2. Compute \(dy\).

Therefore, \(dy = f'(8)dx =\frac{1}{12}(0.1)\approx 0.0083\).

Step 3. Use the Linear Approximation.

Thus \(\sqrt[3]{8.1}\) is larger than \(\sqrt[3]{8}\) by approximately 0.0083, and \(\sqrt[3]{8.1}\approx2.0083\).

EXAMPLE 3 Thermal Expansion

A thin metal cable has length \(L = 12\text{ cm}\) when the temperature is \(T = 21^\circ\text{C}\). Estimate the change in length when \(T\) rises to 24°C, assuming that

where \(k = 1.7 \times10^{-5}(^\circ\text{C})^{-1}\) (\(k\) is called the coefficient of thermal expansion).

Solution How does the Linear Approximation apply here? We will use the differential \(dL\) to estimate the actual change in length \(\Delta L\) when \(T\) increases from 21° to 24°—that is, when \(dT = 3^\circ\). By Eq. (2), the differential \(dL\) is

By Eq. (4), since \(L = 12\),

The Linear Approximation \(\Delta L \approx dL\) tells us that the change in length is approximately

EXAMPLE 4 Effect of an Inexact Measurement

The Bonzo Pizza Company claims that its pizzas are circular with diameter 50 cm (Figure 8.35).

Figure 8.35: The border of the actual pizza lies between the dashed circles.

(a) What is the area of the pizza?

(b) Estimate the quantity of pizza lost or gained if the diameter is off by at most 1.2 cm.

Solution First, we need a formula for the area \(A\) of a circle in terms of its diameter \(D\). Since the radius is \(r = \frac{D}{2}\), the area is

(a) If \(D = 50\text{ cm}\), then the pizza has area \(A(50) = \left(\frac{\pi}{4}\right)(50)^2\approx1963.5\text{ cm}^2\).

(b) If the actual diameter is equal to \(50 + \Delta D\), then the loss or gain in pizza area is \(\Delta A = A(50 + \Delta D) - A(50)\). Observe that \(A'(D) = \frac{\pi}{2}D\) and \(A'(50) = 25\pi \approx 78.5\text{ cm}\), so the Linear Approximation yields

Because \(\Delta D\) is at most \(\pm 1.2 \text{ cm}\), the loss or gain in pizza is no more than around

This is a loss or gain of approximately \(4.8\%\).

Approximating f(x) by Its Linearization

If \(f\) is differentiable at \(x = a\) and \(x\) is close to \(a\), then

CONCEPTUAL INSIGHT

Keep in mind that the linearization and the Linear Approximation are two ways of saying the same thing. Indeed, when we apply the linearization with \(x = a + \Delta x\) and re-arrange, we obtain the Linear Approximation:

EXAMPLE 5

Compute the linearization of \(f(x) = \sqrt{x}e^{x-1}\) centered at \(a = 1\).

Solution By the Product Rule:

Therefore, the linearization at \(a = 1\) is

EXAMPLE 6

Estimate \(\tan(\frac{\pi}{4}+0.02)\) and compute the percentage error.

Solution We find the linearization of \(f(x)=\tan x\) at \(a=\frac{\pi}{4}\):

At \(x=\frac{\pi}{4}+0.02\), the linearization yields the estimate

A calculator gives \(\tan\left(\frac{\pi}{4}+0.02\right)\approx1.0408\), so

Error Bound:

where \(K\) is the max of \(|f''|\) on the interval \([a, a + \Delta x]\).

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 7 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 8.42 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 8.42: Maclaurin polynomials for \(f(x)=e^x\).

EXAMPLE 8 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 8.43): \[ T_{4}(x) =2+\frac{1}{4}\left( x-3\right) -\frac{1}{64}\left( x-3\right) ^{2}+\frac{1}{512}\left( x-3\right) ^{3}-\frac{5}{16{,}384}\left( x-3\right) ^{4}\]

Figure 8.43: Graph of \(f(x)=\sqrt{x+1}\) and \(T_4(x)\) centered at \(x=3\).

EXAMPLE 9 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 10 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 8.44 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 11 How far is the horizon?

Valerie is at the beach, looking out over the ocean (Figure 8.45). 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 8.45: View from the beach

Figure 8.46: Valerie can see a distance \(d = R\theta\), the length of arc \(AH\).

Solution Let \(R\) be the radius of the earth. Figure 8.46 shows that Valerie can see a distance \(d = R\theta\), the length of the circular arc \(AH\) in Figure 8.46. 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 12 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 8.47 that \(\ln x\) and \(T_3(x)\) are indistinguishable near \(x=1.2\).

Figure 8.47: \(\ln x\) and \(T_3(x)\) are indistinguishable near \(x=1.2\).

EXAMPLE 13 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 \[ \left\vert \int_a^b f(x)\, dx \ \right\vert \le\int_a^b \big| \ f(x) \ \big|\, dx \]

which is valid for all integrable functions.