Question

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

(b) Graph f and \({{\rm{T}}_{\rm{n}}}\)on a common screen.

(c) Use Taylor's Formula to estimate the accuracy of the approximation \({\rm{f(x)}} \approx {{\rm{T}}_{\rm{n}}}{\rm{(x)}}\) when x lies in the given interval.

(d) Check your result in part (c) by graphing \(\left| {{{\rm{R}}_{\rm{n}}}{\rm{(x)}}} \right|{\rm{.}}\)

Short answer

a. The required answer is \({\rm{1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x - 1) - }}\frac{{\rm{1}}}{{\rm{8}}}{{\rm{(x - 1)}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{(x - 1)}}^{\rm{3}}}\).

b. The graph is,

c. The accuracy of the approximation is \({\rm{0}}{\rm{.000006}}\).

d. The graph is,

Step-by-step solution

Definition of Concept

Taylor polynomial :A function's Taylor series is an infinite sum of terms stated in terms of the derivatives of the function at a single point. Near this point, the function and the sum of its Taylor series are equivalent for most typical functions.

Find up to the 3rd derivation for \({{\rm{T}}_{\rm{3}}}\)

Consider the given values and simplify,

We'll need to find up to the third derivative, as well as their values at a=1, for\({T_3}\)

\(\begin{array}{l}f(x) = \sqrt x \\f(1) = 1\\{f^\prime }(x) = \frac{1}{2}{x^{ - 1/2}}\\{f^\prime }(1) = \frac{1}{2}\\{f^{\prime \prime }}(x) = - \frac{1}{4}{x^{ - 3/2}}\\{f^\prime }(1) = - \frac{1}{4}\\{f^{\prime \prime \prime }}(x) = \frac{3}{8}{x^{ - 5/2}}\\{f^\prime }(1){\rm{ }} = \frac{3}{8}\end{array}\)

Put the series together.

\(\begin{array}{c}f(x) = f(a) + \frac{{{f^\prime }(a)}}{{1!}}(x - a) + \frac{{{f^{\prime \prime }}(a)}}{{2!}}{(x - a)^2} + \frac{{{f^{\prime \prime \prime }}(a)}}{{3!}}{(x - a)^3} + \cdots \\\sqrt x \approx {T_3}\\ \approx 1 + \frac{1}{2}(x - 1) - \frac{1}{{4 \cdot 2!}}{(x - 1)^2} + \frac{3}{{8 \cdot 3!}}{(x - 1)^3}\\ \approx 1 + \frac{1}{2}(x - 1) - \frac{1}{8}{(x - 1)^2} + \frac{1}{{16}}{(x - 1)^3}\end{array}\)

Therefore, the required solution is \({\rm{1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x - 1) - }}\frac{{\rm{1}}}{{\rm{8}}}{{\rm{(x - 1)}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{(x - 1)}}^{\rm{3}}}\).

The Graph f and \({{\rm{T}}_{\rm{n}}}\) on a common screen

The value of\({{\rm{T}}_{\rm{3}}}\)is,

\({{\rm{T}}_3}{\rm{ = 1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x - 1) - }}\frac{{\rm{1}}}{{\rm{8}}}{{\rm{(x - 1)}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{(x - 1)}}^{\rm{3}}}\)

The graph for the value,

Therefore, a viewing rectangle of \({\rm{[0,2}}{\rm{.2] \times [0,1}}{\rm{.6]}}\)is adequate.

Find the accuracy of the approximation that lies in the given interval

The value of\({{\rm{T}}_{\rm{3}}}\)is ,

\({{\rm{T}}_3}{\rm{ = 1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x - 1) - }}\frac{{\rm{1}}}{{\rm{8}}}{{\rm{(x - 1)}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{(x - 1)}}^{\rm{3}}}\)

The 4^th derivation is \({{\rm{f}}^{{\rm{(4)}}}}{\rm{(x) = - }}\frac{{{\rm{15}}}}{{{\rm{16}}}}{{\rm{x}}^{{\rm{ - 7/2}}}}\).Because the 5th derivative is positive, the 4th derivative is increasing. Because it is negative, the maximum absolute value of the fourth derivative will be at the left end of the interval. That can be applied to \({\rm{M}}\)of the Taylor inequality.

\(\begin{array}{l}\left| {{f^{(4)}}(0.9)} \right| \le M\\ - \frac{{15}}{{16}}{(0.9)^{ - 7/2}}\mid \le M\\1.35557169932{\rm{ }} \le M\end{array}\)

For the \({\rm{x}}\)value, choose one of the interval endpoints.

\(\begin{array}{c}\left| {{R_n}} \right| \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\ \le \frac{{1.35557169932}}{{(4)!}}|0.9 - 1{|^4}\\ \le 0.00000565\end{array}\)

Round it up to,

\(\left| {{R_n}} \right| \le 0.000006\)

Therefore, the required value is \({\rm{0}}{\rm{.000006}}\).

Check the result in part c by graphically

The value of\({{\rm{T}}_{\rm{3}}}\)is ,

\({{\rm{T}}_3}{\rm{ = 1 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x - 1) - }}\frac{{\rm{1}}}{{\rm{8}}}{{\rm{(x - 1)}}^{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{16}}}}{{\rm{(x - 1)}}^{\rm{3}}}\)

\(\left| {{R_n}} \right| \le 0.000006\)

Therefore, Over the interval, the graph \({\rm{y = }}\left| {{{\rm{R}}_{\rm{n}}}} \right|{\rm{ = }}\left| {{\rm{f(x) - }}{{\rm{T}}_{\rm{3}}}{\rm{(x)}}} \right|\) is less than 0.000006.