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 xlies in the given interval.

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

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

Short answer

Part a) Taylor polynomial with degree n at the number a is \[{T_3}(x) - (x - 1) + \frac{1}{2}{(x - 1)^2} - \frac{1}{6}{(x - 1)^3}\].

Part b) Estimated accuracy of the approximation \[{\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x)}}\] when x lies in the given interval is \[ \le 0.042\].

Part c) From the graph, the error is less than 0.0076 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: \[f(x) = x\ln x,\;a = 1,\;n = 3,\;0.5 \le x \le 1.5\].

Finding Taylor polynomial with degree n at the number a.

Part a)

Let us consider the given equation.

For n = 3, we will need to find up to the 3rd derivative. Evaluate each at a = 1.

\[\begin{array}{*{20}{c}}{f(x) = x\ln x}&{f(1) = 0}\\{{f^\prime }(x) = (1)\ln (x) + x \cdot \frac{1}{x} = \ln (x) + 1}&{{f^\prime }(1) = 1}\\{{f^{\prime \prime }}(x) = \frac{1}{x}}&{{f^{\prime \prime }}(1) = 1}\\{{f^{\prime \prime \prime }}(x) = - \frac{1}{{{x^2}}}}&{{f^{\prime \prime \prime }}(1) = - 1}\end{array}\]

Put everything into the Taylor series formula.

\[\begin{array}{l}f(x) = f(a) + \frac{{{f^\prime }(a)}}{{1!}}(x - a) + \frac{{{f^{\prime \prime }}(a)}}{{2!}}{(x - a)^2} + \frac{{{f^{\prime \prime \prime }}(a)}}{{3!}}{(x - a)^3} + \cdots \\{T_3}(x) = 0 + \frac{1}{1}(x - 1) + \frac{1}{{2!}}{(x - 1)^2} + \frac{{ - 1}}{{3!}}{(x - 1)^3}\\ = (x - 1) + \frac{1}{2}{(x - 1)^2} - \frac{1}{6}{(x - 1)^3}\end{array}\]

Therefore, Taylor polynomial is

\[{T_3}(x) = (x - 1) + \frac{1}{2}{(x - 1)^2} - \frac{1}{6}{(x - 1)^3}\].

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

Part b)

Evaluating further, f(x) = x,

In \[x,\;a = 1,\;n = 3,\;0.5 \le x \le 1.5\]

The\[(n + 1) = 4\]th derivative is\[{f^{(4)}}(x) = \frac{2}{{{x^3}}}\]

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

\[\]

\[\begin{array}{l}\left| {{f^{(4)}}(0.5)} \right| \le M\\\left| {\frac{2}{{{{(0.5)}^3}}}} \right| \le M\end{array}\]

\[16 \le M\]

Now use the number in Taylor’s inequality. The series is centred at 1 which is middle of the given interval to plug into x We can use the number farthest from the centre because the error is greatest there.

\[\begin{array}{l}\left| {{R_n}(x)} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\ \le \frac{{16}}{{(3 + 1)!}}|1.5 - 1{|^{3 + 1}}\\ \le 0.042\end{array}\]

Hence, estimated accuracy is 0.042.

Checking result in part b)

Part c)

On solving further,

\[f(x) = x\ln x\]

Graph the absolute value of the difference between\[f(x)\]and\[{T_3}\]

\[\begin{array}{l}y = \left| {f(x) - {T_3}(x)} \right|\\ = \left| {x\ln (x) - \left( {(x - 1) + \frac{1}{2}{{(x - 1)}^2} - \frac{1}{6}{{(x - 1)}^3}} \right)} \right|\end{array}\]

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 0.0076 less than the number calculated in part b.

Hence the error is less than 0.0076.