Linear and Quadratic Approximations
Linear Approximations
Basic Problem 1a. Given a function \(f\), which is differentiable at \(x_0\), find a linear function \(L(x) = a x + b\) such that \(L\) has the value \(f(x_0)\) and derivative \(f'(x_0)\) at \(x_0\).
This problem has an immediate solution. Let \(L\) be the function
L(x) = f(x_0) + f'(x_0)(x-x_0).
It is clear that \(L(x_0)= f(x_0)\) and \(L'(x_0)= f'(x_0)\). Thus, \(L\) satisfies the conditions stated in Basic Problem 1a above. Furthermore, \(L\) is the unique linear function with such properties, for, if \(L_1(x) = ax + b\) is another such linear function, then
\(L_1(x_0) = ax_0 + b = f(x_0) \;\; \text{ and }\;\; L1′(x_0) = a = f'(x_0)\;\; \Longrightarrow\;\; b=f(x_0)- f'(x_0)x_0 \;\; \Longrightarrow \)
\(\Longrightarrow \;\; L_1(x) = a x + b = f'(x_0)x+ f(x_0)-f'(x_0)x_0 = f'(x_0)(x-x_0)+ f(x_0) = L(x). \)
This justifies the following
Definition 1
The function \(L\) in Equation 7.4 is called the linearization of \(\pmb{f}\) \textbf{at} \(\pmb{x_0}\) or the linearization approximation of \(\pmb{f}\) \textbf{at} \(\pmb{x_0}\).
Remark.
(a) The graph of the linearization \(L\) of \(f\) is, in fact, the tangent line to the graph of \(f\) at the point \(A(x_0, f(x_0))\), as \(y= f(x_0)+ f'(x_0)(x-x_0)\) is precisely the equation of the tangent line at \(A\) (see Figure7.43(a)).
Figure 7.43
(b) The label linear spproximation of \(f\) at \(x_0\) implies that \(L\) provides an actual approximation to the values of \(f\). This is indeed the case, as Figure 7.43(a) shows. The value \(L(x)\) can be used as an approximation to the actual value \(f(x)\). The difference between these two values is given by \(R(x)\). \(R(x)\) depends on both \(x_0\) and \(x\). Naturally, unless both values \(f(x)\) and \(L(x)\) are known, the value of \(R(x)\) cannot be computed. However, as we shall see in Theorem 1 below, it can be estimated.
Example 1
Given the function \(f(x) = \sqrt{x}\), (a) find the linearization of \(f\) at \(\displaystyle x_0= 1\); using part (a), find the linear approximation of (b) \(\sqrt{1.5}\); (c) \(\sqrt{0.5}\); (d) \(\sqrt{0.9}\).
Solution (a)
\(f\left (1\right )=1\;\; \text{ and }\;\; f’\left (1\right ) =\frac{1}{2}\frac{1}{\sqrt{1}}= \frac{1}{2} \;\;\Longrightarrow \;\; L(x) = 1+\frac{1}{2}\left (x-1\right ) \)
(b) In this example, the linear approximation yields an overestimate as the graph is concave downward, and hence it lies below the tangent line (see Figure 7.43). We have
\(\sqrt{1.5} \approx L(1.5) = 1+ \frac{1}{2}\left (1.5 -1\right ) = 1+.25 = 1.25. \)
A calculator yields the approximation \(\sqrt{1.4}\approx 1.22474\).
(c)
\(\sqrt{0.5} \approx L(0.5) = 1+ \frac{1}{2}\left (0.5-1\right ) = 1-.25 = .75. \)
A calculator yields the approximation \(\sqrt{0.5}\approx 0.70711\).
(d)
\(\sqrt{0.9} \approx L(0.9) = 1+ \frac{1}{2}\left (0.9-1\right ) = 1-.05 = .95. \)
A calculator yields the approximation \(\sqrt{0.9}\approx 0.94868\).
Remark. The above example shows that, as intuitively expected, the closer the value \(x\) to \(x_0\), the more accurate is the linear approximation \(L(x)\) to \(f(x)\).
How accurate is a linear approximation?
Basic Problem 1b
Given a function \(f\) differentiable at \(x_0\), how accurate is the linear approximation \(L(x) = f(x_0)+f'(x_0)(x-x_0)\) to \(f\) at the point \(x_0\)? That is, what is the size of the difference
\( f(x) – L(x) = f(x) – \left [ f(x_0) + f'(x_0)(x-x_0)\right ] = R(x)\).
Definition 2
The function \(R\) in Equation 7.5 is called the \textbf{remainder of} \(\pmb{f}\) \textbf{at} \(\pmb{x_0}\).
Theorem 1
Assume that \(f\) is twice differentiable at \(x_0\) in the open interval \(I=(a,b)\) and that \(f’\) is continuous on the closed interval \([a,b]\). Let \(x\) and \(x_0\) be two points on \([a,b]\), with \(x\neq x_0\), then, there exists a point \(c\) between \(x\) and \(x_0\) such that
\(f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f”(c)}{2}(x-x_0)^2\),
that is,
\( R(x) = \frac{f”(c)}{2}(x-x_0)^2\)
At times it is convenient to find an upper bound for the size of the remainder \(R(x)\). This can be obtained as follows:
Collary 1
In the notation of Theorem 1, if \(|f”(x)| \leq M\) for all \(x\in (a,b)\), then
\( |R(x)| \leq \frac{M}{2}|x-x_0|^2 \)
Proof
We fix \(x\) and \(x_0\) in \([a,b]\), with \(x\neq x_0\), and note that \(R(x_0)=R'(x_0)=0\).
Next, we introduce a new function
\(r(t)=R(t) – \frac{R(x)}{(x-x_0)^2}(t-x_0)^2,\; t\in [a,b], \)
and note that \(r(x_0)=r'(x_0)=0\) (explain why).
We now apply Rolle’s theorem to \(r\) on \([x,x_0]\). To see that \(r\) satisfies the three assumptions of Rolle’s theorem:
[(i)] \(r\) is continuous on this interval;
[(ii)] \(r\) is differentiable on the open interval \((x,x_0)\);
[(iii)] \(r(x) = r(x_0)\),
note that
\(r(t)= f(t) – \left [ f(x_0) + f'(x_0)(t-x_0)\right ] – \frac{R(x)}{(x-x_0)^2}(t-x_0)^2,\; t\in [a,b], \)
and since, by assumption, \(f\) is continuous and differentiable on \([a,b]\), (i) and (ii) follow. Furthermore, \(r(x) = R(x)-R(x)\) (explain why), that is, \(r(x)=r(x_0)=0\). Thus, there exists \(c_1\) between \(x\) and \(x_0\) such that \(r'(c_1)=0\).
Proceeding in a similar fashion, we can verify that
[(1)] Rolle’s theorem can be applied to \(r’\) to yield a value \(c\) between \(c_1\) and \(x_0\) such \(r”(c)=0\);
[(2)]
\(r”(c)=0 \;\; \Longleftrightarrow \;\; f”(c) -2\frac{R(x)}{(x-x_0)^2} =0 \;\; \Longleftrightarrow \;\; R(x) = \frac{f”(c)}{2}(x-x_0)^2 \)
(justify (1) and (2)) (see Exercise 11).
Example 1
(Continuation.) Using (7.7) or (7.8), determine how accurate are the linear approximations to the values (a) \(\sqrt{1.5}\); (b) \(\sqrt{0.5}\); (c) \(\sqrt{0.9}\) found in Example 1 in 7.8.1.
Solution
Since \(f(x) = \sqrt{x}\), then
\(f'(x)= \frac{1}{2}\frac{1}{\sqrt{x}}, \;\;\; f”(x)= -\frac{1}{4}\frac{1}{x\sqrt{x}}, \)
we have
(a)
\(\sqrt{1.5} – L(1.5) \underset{(7.8)}{=}\frac{1}{2}\left ( -\frac{1}{4}\frac{1}{c\sqrt{c}}\right )(1.5-1)^2 = – \frac{1}{8}\frac{1}{c\sqrt{c}}(0.5)^2, \; c\in (1,1.5). \)
Since
\(|f”(x)| = \left |-\frac{1}{4}\frac{1}{x\sqrt{x}}\right | \leq \frac{1}{4}, \;\; x\in [1,1.5] \Longrightarrow |R(1.5)| \leq \frac{1}{8} (0.5)^2 = 0.03125. \)
This can be compared with the values obtained above via a calculator and via a linear approximations.
\(\left |\sqrt{1.5} – L(1.5)\right | \approx \left |1.22474 – 1.25\right | = 0.02526. \)
This shows that a good estimate of the error can be obtained without actually knowing the value obtained by other means such as a computer/calculator.
(b)
\(\sqrt{0.5} – L(0.5) \underset{(7.7)}{=}\frac{1}{2}\left ( -\frac{1}{4}\frac{1}{c\sqrt{c}}\right )(0.5-1)^2 = – \frac{1}{8}\frac{1}{c\sqrt{c}}(0.5)^2, \; c\in (0.5,1). \)
Since
\(|f”(x)| = \left |-\frac{1}{4}\frac{1}{x\sqrt{x}}\right | \leq \frac{1}{4}\frac{1}{0.5\sqrt{0.5}}, \;\; x\in [0.5,1] \;\; \Longrightarrow \)
\(\Longrightarrow\;\;|R(0.5)| \leq \frac{1}{8}\frac{1}{0.5\sqrt{0.5}} (0.5)^2\leq \frac{1}{8}\frac{0.5}{ \sqrt{0.5}} \leq \frac{1}{8}\frac{0.5}{ \sqrt{0.49}} =
\frac{1}{8}\frac{0.5}{0.7} \approx 0.0892857. \)
(c)
\(\sqrt{0.9} – L(0.9) \underset{(7.7)}{=}\frac{1}{2}\left ( -\frac{1}{4}\frac{1}{c\sqrt{c}}\right )(0.9-1)^2 = – \frac{1}{8}\frac{1}{c\sqrt{c}}(0.1)^2, \; c\in (0.9,1). \)
Since
\(|f”(x)| = \left |-\frac{1}{4}\frac{1}{x\sqrt{x}}\right | \leq \frac{1}{4}\frac{1}{0.9\sqrt{0.9}}, \;\; x\in [0.9,1] \Longrightarrow \)
\(\Longrightarrow |R(0.9)| \leq \frac{1}{8}\frac{1}{0.9\sqrt{0.9}} (0.1)^2 \leq \frac{1}{8}\frac{1}{0.9\sqrt{0.81}} (0.1)^2 = \leq \frac{1}{8}\frac{(0.1)^2}{(0.9)^2} \approx 0.00154321. \)
Obervation. Rather than estimating the size of the remainder point by point as done in the above example, it is more convenient to find an estimate over an interval using (7.8), as the following two examples show.
Example 2
Given the function \(f(x) = \sin x\), (a) find the linearization of \(f\) at \(\displaystyle x_0= 0\);
\\
(b) estimate the size of \(R(x)\).
Solution
(a)
\(f\left (0\right )=\sin0 = 0\;\; \text{ and }\;\; f’\left (0\right ) =\cos 0 = 1 \;\;\Longrightarrow \;\; L(x) = 0+0\left (x-0\right )\Longleftrightarrow L(x) = x. \)
(b) Since \(f”(x) = -\sin x\), then \(|f”(x)| = |-\sin x| \leq 1\) for all \(x \in \mathbb{R}\), that is, we cantake \(M=1\) in (7.8),
thus
\(|R(x)| \leq \frac{1}{2} \cdot \frac{4}{9(7.5)^{2} \sqrt[3]{7.5}}(x-1)^{2} \leq \frac{1}{2} \cdot \frac{4}{9(7.5)^{2} \sqrt[3]{7.5}}(0.5)^{2} \approx 0.000505 \forall x \in[0.5,1.5]. \)
For example, if \(x\in [-.1, .1]\), then \(\displaystyle | R(x) | \leq \frac{1}{2}(.1)^2 = 0.005\), that is, the linear approximation is accurate at least for the first three decimal places with a margin of error of \(\pm 0.005\).
Example 3
Given the function \(\displaystyle f(x) = \frac{1}{\sqrt[3]{9-x}}\), (a) find the linearization of \(f\) at \(\displaystyle x_0= 1\); (b) find the linear appoximation of \(\displaystyle \frac{1}{\sqrt[3]{7.8}} \) using part(a); (c) estimate the size of the remainder \(R(x)\) using (7.8), for \(x\in [x_0-\Delta x, x_0+\Delta x]\), \(\Delta = 0.5.\).
Solution
(a) Since
\(f(x) = \frac{1}{(9-x)^{1/3}}, \;\; f'(x) = \frac{1}{3(9-x)^{4/3}}, \;\; f”(x) = \frac{4}{9(9-x)^{7/3}}, \;\; f(1) = \frac{1}{2}, \;\; f'(1) = \frac{1}{48}, \;\; \Longrightarrow \)
\(\Longrightarrow \;\; L(x) = \frac{1}{2}+ \frac{1}{48}(x-1). \)
(b)
\(\frac{1}{\sqrt[3]{7.8}} = f(1.2) \approx L(1.2) = \frac{1}{2}+ \frac{1}{48}(1.2-1)= \frac{1}{2}+ \frac{1}{240} = \frac{121}{240}\approx 0.5042. \)
(c) Note that as \(x \in [0.5, 1.5]\) increases, the denominator of \(f”\) decreases, thus, \(f”\) is increasing. Alternatively, we compute
\(f”'(x) = \frac{28}{27(9-x)^{10/3}} > 0 \;\;\forall\, x\in [0.5, 1.5] \;\;\Longrightarrow \;\; f” \; \text{ is increasing on } \; [0.5, 1.5] \;\; \Longrightarrow \)
\(\Longrightarrow \;\;|f”(x) | < f”(1.5) = \frac{4}{9(7.5)^2\sqrt[3]{7.5}} \;\;\forall\, x\in [0.5, 1.5]. \)
Thus
\( |R(x)| \leq \frac{1}{2}\cdot \frac{4}{9(7.5)^2\sqrt[3]{7.5}}\,(x-1)^2 \leq \frac{1}{2}\cdot \frac{4}{9(7.5)^2\sqrt[3]{7.5}}\,(0.5)^2 \approx 0.000505 \;\; \forall \, x\in [0.5, 1.5]. \)
This means that a linear approximation will yield the value of the first four decimal places with a margin of error of \(\pm 0.0005\).
Exercises
In problems 1-10, (a) find the linearization of \(f\) at the given point \(x_0\); (b) find the linear appoximation of \(f(x_1)\) for the given point \(x_1\) using part(a); (c) estimate the remainder \(R(x)\) using (7.8), for \(x\) on the interval \([x_0-\Delta x, x_0+\Delta x]\).
Q1
\(\displaystyle f(x) = \sqrt{x-1} \), (a) \(\displaystyle x_0= 2\), (b) \(\displaystyle x_1=1.8\), \(\displaystyle \Delta x = 0.5\).
Q2
\(\displaystyle f(x) = \ln (1+x)\), (a) \(\displaystyle x_0=0\), (b) \(\displaystyle x_1=0.2\), (c) \(\displaystyle \Delta x = 0.5\).
Q3
\(\displaystyle f(x) =\sin x \), (a) \(\displaystyle x_0=\pi\), (b) \(\displaystyle x_1=\frac{11\pi}{10}\), (c) \(\displaystyle \Delta x =\frac{\pi}{6} \).
Q4
\(\displaystyle f(x) =e^x \), (a) \(\displaystyle x_0=0\), (b) \(\displaystyle x_1=0.3\), (c) \(\displaystyle \Delta x =0.5 \).
Q5
\(\displaystyle f(x) =\arctan x \), (a) \(\displaystyle x_0=0\), (b) \(\displaystyle x_1=0.5\), (c) \(\displaystyle \Delta x =1 \). Hint: \(\displaystyle f”'(x) = \frac{6x^2-2}{\left (1+x^2\right )^3}\).
Q6
\(\displaystyle f(x) = \arcsin x\), (a) \(\displaystyle x_0=\frac{1}{2}\), \(b) (\displaystyle x_1=\frac{3}{5}\), (c) \(\displaystyle \Delta x = \frac{1}{4}\). Hint: \(\displaystyle f”'(x) = \frac{1 + 2 x^2}{\left (1-x^2\right )^{5/2}}\).
Q7
\(\displaystyle f(x) =\arccos x \), (a) \(\displaystyle x_0=0\), (b) \(\displaystyle x_1=0.2\), (c) \(\displaystyle \Delta x = 0.6\).
Q8
\(\displaystyle f(x) =\frac{1}{\sqrt[4]{4-x}} \), (a) \(\displaystyle x_0=3\), (b) \(\displaystyle x_1=2.8\), (c) \(\displaystyle \Delta x = 1\).
In problems 9 – 12, use an appropriate linearization to approximate the value of the expression.
Q9
\(\displaystyle \arctan \sqrt{1.2} \).
Q10
\(\displaystyle \sqrt[3]{28} \).
Q11
\(\displaystyle \cos\frac{3\pi}{5} \).
Q12
\(\displaystyle \ln (0.8) \).
Quadratic Approximations
Basic Problem 2. Given a function \(f\), which is twice differentiable at \(x_0\),
[(a)] find a quadratic function \(Q(x) = a x^2 + bx + c\) such that \(Q\) has the value \(f(x_0)\) and first and second derivatives \(f'(x_0)\) \(f”(x_0)\) at the point \(x_0 \);
[(b)] determine how accurate is the approximation. Namely, estimate the size of the remainder \(R(x) = f(x) – q(x)\).
This problem has an immediate solution. Let \(Q\) be the function
(7.9)
Q(x) = f(x_0) + f'(x_0)(x-x_0) +\frac{1}{2}f”(x_0)(x-x0)^2
Furthermore, assuming that \(f”’\) exists
(7.10)
R(x) =f(x) – \left (f(x_0) + f'(x_0)(x-x_0) +\frac{1}{2}f”(x_0)(x-x0)^2\right ) = \frac{1}{3\cdot2}f”'(c)(x-x0)^2,
for a \(c\) between \(x\) and \(x_0\).
First, it is clear that \(Q(x_0)= f(x_0)\), \(Q'(x_0)= f'(x_0)\) and \(Q”(x_0)= f”(x_0)\). Thus, \(Q\) satisfies the conditions stated in Basic Problem 3. Furthermore, \(Q\) is the \textit{unique} quadratic function with such properties, for, if \(Q_1(x) = ax^2 + bx+c\) is another such quadratic function, that is,
\(Q_1(x_0) = f(x_0), \;\;\; Q_1′(x_0) = f'(x_0), \;\;\; Q_1”(x_0) = f”(x_0), \)
then, rewriting \(Q_1\) in terms of \(x-x_0\)
\begin{align*}
Q_1(x)& = ax^2 + bx+c = a(x-x_0)^2 +2a x_0 x -a x_0^2 + b(x-x_0)+ bx_0+c =\\
&= a(x-x_0)^2 +2a x_0( x-x_0) +a x_0^2 + b(x-x_0)+ bx_0+c =\\
&= a(x-x_0)^2 +\left (2a x_0 +b\right )( x-x_0) + a x_0^2 + bx_0+c=\\
&= a_1(x-x_0)^2 +b_1( x-x_0) + c_1
\end{align*}
with
\(a_1=a, \;\;\; b_1 = 2a x_0 +b, \;\;\; c_1 = a x_0^2 + bx_0+c, \)
we have that
\(a_1= \frac{1}{2}f”(x_0),\;\;\; b_1 = f'(x_0), \;\;\; c_1 = f(x_0) \)
that is, \(Q_1(x) = Q(x)\).
We will not prove (7.10).
Analyzing the linear and quadratic approximations, it is natural to conjecture that polynomial approximations of higher degree also exist and their remainders can be estimated. This is the subject of Taylor series expansions, which is normally studied in a standard Calculus II course. Thus, we will stop here, but if you are curious, you can “Google” Taylor series and find videos in YouTube.