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 \(\left| {{{\rm{R}}_{\rm{n}}}{\rm{(x)}}} \right|\)

\(f(x) = {x^{2/3}},\;\;\;a = 1,\;\;\;n = 3,\;\;\;0.8 \le x \le 1.2\)

Short answer

a) Taylor polynomialis \({T_3}(x) = 1 + \frac{2}{3}(x - 1) - \frac{1}{9}{(x - 1)^2} + \frac{4}{{81}}{(x - 1)^3}\).

b) Estimated accuracy is \(0.000097\).

c) From the graph, the error is less than \(0.0000532845\)on the interval.

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)

Let's look for the function's derivatives:

\(\begin{array}{*{20}{c}}{f(x)}&{ = {x^{2/3}}}&{f(1)}&{ = 1}\\{{f^\prime }(x)}&{ = \frac{2}{{3{x^{\frac{1}{3}}}}}}&{{f^\prime }(1)}&{ = \frac{2}{3}}\\{{f^{\prime \prime }}(x)}&{ = - \frac{2}{{9{x^{\frac{4}{3}}}}}}&{{f^{\prime \prime }}(1)}&{ = - \frac{2}{9}}\\{{f^{\prime \prime \prime }}(x)}&{ = \frac{8}{{27{x^{\frac{7}{3}}}}}}&{{f^{\prime \prime \prime }}(1)}&{ = \frac{8}{{27}}}\end{array}\)

We're able to locate\({T_3}(x)\):

\(\begin{array}{c}{T_3}(x) = \sum\limits_{n = 0}^3 {\frac{{{f^{(n)}}(1)}}{{n!}}} {(x - 1)^n}\\ = 1 + \frac{2}{3}(x - 1) - \frac{1}{9}{(x - 1)^2} + \frac{4}{{81}}{(x - 1)^3}\end{array}\)

Therefore, the required Taylor polynomialis \({T_3}(x) = 1 + \frac{2}{3}(x - 1) - \frac{1}{9}{(x - 1)^2} + \frac{4}{{81}}{(x - 1)^3}\).

Use Taylor’s Formula to estimate the accuracy

b)

Using Taylor’s Formula to calculate,

\(f(x) = {x^{2/3}},\;\;\;a = 1,\;\;\;n = 3,\;\;\;0.8 \le x \le 1.2\)

The \((n + 1) = 4th\)derivative is\({f^{(4)}}(x) = - \frac{{56}}{{81{x^{10/3}}}}\)

\(\left| {{f^{(4)}}(0.8)} \right| \le M\)

\(\left| { - \frac{{56}}{{81{{(0.8)}^{10/3}}}}} \right| \le M\)

\(1.45457589 \le M\)

\(\begin{array}{l}\left| {{R_n}(x)} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\ \le \frac{{1.45457589}}{{(3 + 1)!}}|1.2 - 1{|^{3 + 1}}\\ \le 0.000097\end{array}\)

Hence, estimated accuracy is \(0.000097\).

Step 4: Check your result in part (b) by graphing

c)

Calculate the absolute value of the difference between \(f(x)\)and \({T_3}\)

\(f(x) = {x^{2/3}}\)

\(\begin{array}{l}y = \left| {f(x) - {T_2}(x)} \right|\\ = \left| {{x^{2/3}} - \left( {1 + \frac{{2(x - 1)}}{3} - \frac{1}{9}{{(x - 1)}^2} + \frac{4}{{81}}{{(x - 1)}^3}} \right)} \right|\end{array}\)

Get a good view of the interval\(0.8 \le x \le 1.2\)and stretch the\(y\)-axis.

From\(0.8 \le x \le 1.2\)The difference is at most approximately\(0.0000532845\), which differs by a little more than\({\rm{0}}{\rm{.00004}}\) from the value computed in section (b), \(0.000097\)therefore I guess everything is fine.

Therefore, the error on the interval is less than \(0.0000532845\)on the graph.