Question

\({\rm{f(x) = ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x,a = 1}}\)

Short answer

The Taylor polynomial is \({T_3}(x) = \frac{\pi }{4} + \frac{{(x - 1)}}{2} - \frac{{{{(x - 1)}^2}}}{4} + \frac{{{{(x - 1)}^3}}}{{12}}\).

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.

Sum of the Taylor polynomials

Solve the given equation,

\(\begin{array}{l}{T_n}(x){\rm{ centered at }}a{\rm{ is }}\sum\limits_{r = 0}^n {\frac{{{f^r}(a){{(x - a)}^r}}}{{r!}}} \\{T_3}(x) = \frac{{f(1) \cdot {{(x - 1)}^0}}}{{0!}} + \frac{{{f^\prime }(1) \cdot {{(x - 1)}^1}}}{{1!}} + \frac{{{f^{\prime \prime }}(1) \cdot {{(x - 1)}^2}}}{{2!}} + \frac{{{f^{\prime \prime \prime }}(1) \cdot {{(x - 1)}^3}}}{{3!}}\end{array}\)\(\begin{array}{l}f(x) = {\tan ^{ - 1}}x \Rightarrow f(1) = \frac{\pi }{4}\\{f^\prime }(x) = \frac{1}{{1 + {x^2}}} \Rightarrow {f^\prime }(1) = \frac{1}{2}\\{f^{\prime \prime }}(x) = - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} \Rightarrow {f^{\prime \prime }}(1) = - \frac{1}{2}\\{f^{\prime \prime \prime }}(x) = \frac{{6{x^2} - 2}}{{{{\left( {1 + {x^2}} \right)}^3}}} \Rightarrow {f^{\prime \prime \prime }}(1) = \frac{1}{2}\end{array}\)

On solving further,

Therefore

\(\begin{array}{l}{T_3}(x) = \frac{{f(1) \cdot {{(x - 1)}^0}}}{{0!}} + \frac{{{f^\prime }(1) \cdot {{(x - 1)}^1}}}{{1!}} + \frac{{{f^{\prime \prime }}(1) \cdot {{(x - 1)}^2}}}{{2!}} + \frac{{{f^{\prime \prime \prime }}(1) \cdot }}{{3!}}\\{T_3}(x) = \frac{\pi }{4} \cdot \frac{1}{1} + \frac{1}{2} \cdot \frac{{(x - 1)}}{1} - \frac{1}{2} \cdot \frac{{{{(x - 1)}^2}}}{2} + \frac{1}{2} \cdot \frac{{{{(x - 1)}^3}}}{6}\\{T_3}(x) = \frac{\pi }{4} + \frac{{(x - 1)}}{2} - \frac{{{{(x - 1)}^2}}}{4} + \frac{{{{(x - 1)}^3}}}{{12}}\end{array}\)

Therefore, the required Taylor polynomial is \({T_3}(x) = \frac{\pi }{4} + \frac{{(x - 1)}}{2} - \frac{{{{(x - 1)}^2}}}{4} + \frac{{{{(x - 1)}^3}}}{{12}}\).

Plotting the required graph

In the Graph below:

\({\rm{y = f(x)}}\) is the black curve