Question

(a) Approximate f by a Taylor polynomial with degree natthe number a.

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

(c) Check your result in part (b) by graphing \[\left| {{R_n}(x)} \right|\]

Short answer

Part a) Taylor polynomial with degree n at the number a is \[\ln 3 + \left( {\frac{2}{3}} \right)(x - 1) - \left( {\frac{2}{9}} \right){(x - 1)^2} + \left( {\frac{8}{{81}}} \right){(x - 1)^3}\].

Part b) Estimated accuracy of the approximation \[{\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x)}}\] when x lies in the given interval is \0.016.

Part c) From the graph, the error is less than 0.004231 on the interval.

Step-by-step solution

Concept of polynomials

Polynomial is created of two terms, namely Poly meaning “many” and nominal meaning “terms".
A polynomial is defined as an expression that's composed of variables, constants, and exponents, that are combined using mathematical operations like addition, subtraction, multiplication, and division.
A polynomial comprises constants and variables, but we cannot perform division operations by a variable in polynomials.

Given parameters

An equation:\[f(x) = \ln (1 + 2x),\;\;\;a = 1,\;\;\;n = 3,\;\;\;0.5 \le x \le 1\].

Finding Taylor polynomial with degree n at the number a.

Part a)

Let us consider the given equation.

Taylor polynomial with degree n at the number a is given by:

\[f(x) \approx {T_n} = \sum\limits_{r = 0}^n {\frac{{{f^r}(a){{(x - a)}^r}}}{{r!}}} \]

\[\begin{array}{l}f(x) = \ln (1 + 2x)\\ \Rightarrow f(1) = \ln 3\\{f^\prime }(x) = \frac{2}{{(1 + 2x)}}\\ \Rightarrow {f^\prime }(1) = \frac{2}{3}\\{f^{\prime \prime }}(x) = - \frac{4}{{{{(1 + 2x)}^2}}}\\ \Rightarrow {f^{\prime \prime }}(1) = - \frac{4}{9}\\{f^{\prime \prime \prime }}(x) = \frac{{16}}{{{{(1 + 2x)}^3}}}\\ \Rightarrow {f^{\prime \prime \prime }}(1) = \frac{{16}}{{27}}\end{array}\]

Taylor polynomial with degree \[n = 3\]at the number \[a = 1\]is given by

\[\begin{array}{l}f(x) \approx {T_3} = \sum\limits_{r - 0}^3 {\frac{{{f^r}(1){{(x - 1)}^r}}}{{r!}}} \\f(x) \approx {T_3} = f(1) + {f^\prime }(1)(x - 1) + \frac{{{f^{\prime \prime }}(1){{(x - 1)}^2}}}{2} + \frac{{{f^{\prime \prime \prime }}(1){{(x - 1)}^3}}}{6}\\f(x) \approx {T_3} = \ln 3 + \left( {\frac{2}{3}} \right)(x - 1) + \left( { - \frac{4}{9}} \right)\frac{{{{(x - 1)}^2}}}{2} + \left( {\frac{{16}}{{27}}} \right)\frac{{{{(x - 1)}^3}}}{6}\\f(x) \approx {T_3} = \ln 3 + \left( {\frac{2}{3}} \right)(x - 1) - \left( {\frac{2}{9}} \right){(x - 1)^2} + \left( {\frac{8}{{81}}} \right){(x - 1)^3}\end{array}\]

Therefore Taylor polynomial is \[\ln 3 + \left( {\frac{2}{3}} \right)(x - 1) - \left( {\frac{2}{9}} \right){(x - 1)^2} + \left( {\frac{8}{{81}}} \right){(x - 1)^3}\].

Estimated accuracy of the approximation \[{\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x)}}\].

Part b)

Evaluating further,

\[f(x) = \ln (1 + 2x),\;\;\;a = 1,\;\;\;n = 3,\;\;\;0.5 \le x \le 1.5\]

The \[\left( {n + 1} \right) = 4\]th derivative is \[{f^{(4)}}(x) = - \frac{{96}}{{{{(2x + 1)}^4}}}\].

The\[5\]th derivative will be positive, so\[{f^{(4)}}(x)\]is increasing, but negative. So the maximum of\[\left| {{f^{(4)}}(x)} \right|\]will be at the left interval endpoint. If we let\[x = 0.5\]we can use this as\[M.\]

\[|{f^{(4)}}(0.5)|\, \le M\]

\[\left| { - \frac{{96}}{{{{(2(0.5) + 1)}^4}}}} \right| \le M\]

\[6 \le M\]

Step 2

Apply the number from Taylor's inequality now. The series is centred at\[1\], which is the middle of the supplied interval, so we may plug into either end of it. Because the inaccuracy is highest at the farthest point from the centre, we utilise that number.

\[\left| {{R_n}(x)} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\]

\[ \le \frac{6}{{(3 + 1)!}}|1.5 - 1{|^{3 + 1}}\]

\[ \le 0.016\]

Checking the graph for \[\left| {{R_n}(x)} \right| = \left| {{T_n}(x) - f(x)} \right|\], the error is closer to \[ \le 0.0042\], but Taylor's inequality is an estimate, and sometimes the difference between it and the actual can be somewhat large like for this problem.

Hence, estimated accuracy of the given interval is 0.016.

Checking result in part b)

Part c)

On solving further,

\[f(x) = \ln (1 + 2x)\]

Graph the absolute value of the difference between\[f(x)\]and\[{T_2}\]

\[\begin{array}{l}y = \left| {f(x) - {T_3}(x)} \right|\\ = \left| {\ln (1 + 2x) - \left( {\ln (3) + \frac{2}{3}(x - 1) - \frac{2}{9}{{(x - 1)}^2} + \frac{8}{{81}}{{(x - 1)}^3}} \right)} \right|\end{array}\]

Get a good view of the interval \[0.5 \le x \le 1.5\] and stretch the \[y\]-axis.

For \[0.5 \le x \le 1.5\] the difference is at most about 0.004231, less than the number calculated in part b).

Therefore, error is less than 0.004231 on the interval.