Question

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

(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 \[\left| {{R_n}(x)} \right|\]

\[f(x) = {e^{{x^2}}},\;\;\;a = 0,\;\;\;n = 3,\;\;\;0 \le x \le 0.1\]

Short answer

a) Taylor polynomial with degree n at the number a is \[f(x) = {T_3} - 1 + 0 + {x^2}\] .

b) Estimated accuracy of the approximation \[{\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x)}}\] when x lies in the given interval is 0.00006

c) From the graph, the error is less than 0.0000502 on the interval.

Step-by-step solution

Concept of polynomials

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: \[f(x) = \sqrt x ,\;\;\;a = 4,\;\;\;n = 2,\;\;\;4 \le x \le 4.2\]

Finding Taylor polynomial with degree n at the number a.

a)

Let us consider the given equation.

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

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

\[f(x) - {e^{{x^2}}} \to f(0) - 1{\rm{ }}\]

\[{f^\prime }(x) - 2x{e^{{x^2}}} \to {f^\prime }(0) - 0{\rm{ }}\]

\[{f^{\prime \prime }}(x) - 2{e^{{x^2}}} + 4{x^2}{e^{{x^2}}} \to {f^{\prime \prime }}(0) - 2{\rm{ }}\]

\[{f^{\prime \prime \prime }}(x) - 4x{e^{{x^2}}} + \left( {8x{e^{{x^2}}} + 8{x^3}{e^{{x^2}}}} \right)\]

\[ \to {f^{\prime \prime }}(0) - 0\]

Taylor polynomial with degree \[n - 3\] at the number \[a - 0\] is given by:

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

Therefore, Taylor polynomial is \[f(x) = {T_3} - 1 + 0 + {x^2} + 0\].

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

Part b)

Evaluating further,

\[f(x) - {e^{{x^2}}},\;\;\;a - 0,\;\;\;n - 3,\;\;\;0 \le x \le 0.1\]

The\[(n + 1) - 4\]th derivative is\[{f^{(4)}}(x) - 4{e^{{x^2}}}\left( {4{x^4} + 12{x^2} + 3} \right)\].

\[\left| {{f^{(4)}}(0.1)} \right| \le M\]

\[\left| {4{e^{{{(0.1)}^2}}}\left( {4{{(0.1)}^4} + 12{{(0.1)}^2} + 3} \right)} \right| \le M\]

\[12.607 \le M\]

Because the error is greatest there:

\[\begin{array}{l}\left| {{R_n}(x)} \right|\\ \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\ \le \frac{{12.607}}{{(3 + 1)!}}|0.1 - 0{|^{3 + 1}}\\ \le 0.0000525\end{array}\]

The book says \[0.00006\] which I guess they got because they wanted to use one significant digit and rounded up to maintain the inequality.

Checking result in part b)

Part c)

On solving further,

\[f(x) - {e^{{x^2}}}\]

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

\[y - \left| {f(x) - {T_3}(x)} \right| - \left| {{e^{{x^2}}} - \left( {1 + {x^2}} \right)} \right|\]

Get a good view of the interval \[0 \le x \le 0.1\]and stretch the y-axis.

Hence, for \[0 \le x \le 0.1\] the difference is at most about \[0.0000502\] close to the number calculated in part b).