Question

(a) Approximate f by a Taylor polynomial with degree n at the number a.

(b) Use Taylor's Formula to estimate the accuracy of the approximation \(f(x) \approx {T_n}(x)\) when x lies in the given interval.

(c) Check your result in part (b) by graphing \(\left| {{{\rm{R}}_{\rm{n}}}{\rm{(x)}}} \right|\) \({\rm{f(x) = }}\sqrt {\rm{x}} {\rm{,}}\) \({\rm{a = 4,}}\) \({\rm{n = 2,}}\) \(4 \le x \le 4.2\)

Short answer

a)Taylor polynomial\({T_2}(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{{64}}{(x - 4)^2}\).

b) Estimated accuracy is\(1.5625 \times {10^{ - 5}}\).

c) HINT: All you have to do is graph\(\left| {{{\rm{R}}_{\rm{2}}}{\rm{(x)}}} \right|{\rm{ and g(x) = 1}}{\rm{.5626 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\).

Step-by-step solution

Concept Introduction

Taylor polynomials are function approximations that improve in quality as n rises. Taylor's theorem calculates the inaccuracy introduced by such approximations in terms of numbers. If a function's Taylor series is convergent, the sum of its Taylor polynomials is the limit of the infinite sequence of Taylor polynomials.

Approximate \({\rm{f}}\)by a Taylor polynomial with degree at the number \({\rm{a}}\).

a)

Find the values of the first two derivatives of \(f\) at\({\rm{x = 4}}\).

\(\begin{array}{c}f(x) = \sqrt x ,\\f(4) = 2\\{f^\prime }(x) = \frac{1}{2}{x^{ - 1/2}},\\{f^\prime }(4) = \frac{1}{4}\\{f^{\prime \prime }}(x) = - \frac{1}{4}{x^{ - 3/2}},\\{f^{\prime \prime }}(4) = - \frac{1}{{32}}\end{array}\)

\({T_2}(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{{64}}{(x - 4)^2}\)

To find, the Taylor polynomial of degree 2, use the values of the derivatives (at\(x = 4\)) determined in the previous step\({T_2}\).

Use Taylor’s Formula to estimate the accuracy

b)

Simplify the given values,

\(f(x) = \sqrt x ,\)\({\rm{ a = 4,}}\) \({\rm{n = 2,}}\) \(4 \le x \le 4.2\)

The \((n + 1) = 3rd\) derivative is written as \({f^{\prime \prime \prime }}(x) = \frac{3}{{8{x^{5/2}}}}\)

The 4th derivative will be negative; we know \({f^{\prime \prime \prime }}\) that the interval will be shrinking. As a result, the maximum of \({f^{\prime \prime \prime }}\) will be at \({\rm{x = 4,}}\) and we may use this as \(M\).

\(\begin{array}{l}\left| {{f^{\prime \prime \prime }}(4)} \right| \le M\\\left|{\frac{3}{{8{{(4)}^{5/2}}}}} \right| \le M\end{array}\)\(0.01171875 \le M\)

Apply the number from Taylor's inequality now. Because the series is centred at 4, we should insert into x the far end of the interval, 4.2, rather than x=4, because x=4 would result in 0 and the inaccuracy increases as you move further from the centre.

\(\begin{array}{l}\left| {{R_n}(x)} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\\left| {{R_n}(x)} \right| \le \frac{{0.01171875}}{{(2 + 1)!}}|4.2 - 4{|^{2 + 1}}\\\\ \le 0.000015625 = 1.5625 \times {10^{ - 5}}\end{array}\)

Hence, estimated accuracy is\(1.5625 \times {10^{ - 5}}\).

Check your result in part (b) by graphing

c)

Let's look \({R_2}(x)\) for:

\(\begin{array}{l}{R_2}(x) = f(x) - {T_2}(x)\\ = \sqrt x - \left( {2 + \frac{1}{4}(x - 4) - \frac{1}{{64}}{{(x - 4)}^2}} \right)\\ = - 2 + \sqrt x - \frac{1}{4}(x - 4) + \frac{1}{{64}}{(x - 4)^2}\end{array}\)

Let's look at a graph \(\left| {{R_2}(x)} \right|\) and \(g(x) = 1.5626 \cdot {10^{ - 5}}\) see if we can show the inequality: