Question

How do you determine the third and fourth Taylor Polynomials of ${{x}^{3}}+9x-1$ at x = -1 ?

Answer 1

To determine the third and fourth degree Taylor Polynomials of the given equation we should initially find the derivatives of degree 1, 2, 3 and 4 for this problem for the polynomial ${{x}^{3}}+9x-1$ and then apply the values obtained in the formula of Taylor series expansion and then get the third and fourth Taylor Polynomials at x $\Rightarrow$ -1

Complete step-by-step solution:
Firstly note that as we have a polynomial of degree 3 then its Taylor Series will be exact for$O({{x}^{3}})$.
Let us define the function, f(x), by:
$f\left( x \right)\Rightarrow {{x}^{3}}+9x-1$
Let us find all the derivatives:

& {{f}^{\left( 1 \right)}}\left( x \right)\Rightarrow 3{{x}^{2}}+9 \\\ & {{f}^{\left( 2 \right)}}\left( x \right)\Rightarrow 6x \\\ & {{f}^{\left( 3 \right)}}\left( x \right)\Rightarrow 6 \\\ & {{f}^{\left( 4 \right)}}\left( x \right)\Rightarrow 0 \\\ \end{aligned}$$ Now let us find the values of the above at a = −1 $$\begin{aligned} & {{f}^{\left( 0 \right)}}\left( -1 \right)\Rightarrow -11 \\\ & {{f}^{\left( 1 \right)}}\left( -1 \right)\Rightarrow 12 \\\ & {{f}^{\left( 2 \right)}}\left( -1 \right)\Rightarrow -6 \\\ & {{f}^{\left( 3 \right)}}\left( -1 \right)\Rightarrow 6 \\\ & {{f}^{\left( 4 \right)}}\left( -1 \right)\Rightarrow 0 \\\ \end{aligned}$$ The Taylor Series about x $$\Rightarrow $$ a is given by the formula: $$f\left( x \right)\Rightarrow f\left( a \right)+{{f}^{\left( 1 \right)}}\left( x \right)\left( x-a \right)+\dfrac{{{f}^{\left( 2 \right)}}\left( x \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{{{f}^{\left( 3 \right)}}\left( x \right)}{3!}{{\left( x-a \right)}^{3}}+\dfrac{{{f}^{\left( 4 \right)}}\left( x \right)}{4!}{{\left( x-a \right)}^{4}}+......$$ So, if we want to write the truncated Taylor Series, we can just truncate the series as required. Thus the 3rd order Taylor Series about a $$\Rightarrow $$ −1 is: $$\begin{aligned} & {{T}_{3}}\left( x \right)\Rightarrow \left( -11 \right)+\left( 12 \right)\left( x+1 \right)+\dfrac{-6}{2!}{{\left( x+1 \right)}^{2}}~~~~~ \\\ & \Rightarrow -11+12\left( x+1 \right)-\dfrac{6}{2}{{\left( x+1 \right)}^{2~}}~~~~ \\\ & \Rightarrow -11+12\left( x+1 \right)-3{{\left( x+1 \right)}^{2}} \\\ \end{aligned}$$ And similarly, truncating at the next term we have: $$\begin{aligned} & {{T}_{4}}\left( x \right)\Rightarrow {{T}_{3}}\left( x \right)+\dfrac{6}{3!}{{\left( x+1 \right)}^{3}}~~~~~ \\\ & \Rightarrow {{T}_{3}}\left( x \right)+\dfrac{6}{6}{{\left( x+1 \right)}^{3}}~~~~~ \\\ & \Rightarrow -11+12\left( x+1 \right)-3{{\left( x+1 \right)}^{2}}+{{\left( x+1 \right)}^{3}} \\\ \end{aligned}$$ And as initially, as the starting function is a polynomial of degree 3 then this truncated polynomial $${{T}_{4}}\left( x \right)$$ is in fact exact as all higher derivatives (and therefore terms) are zero. So, from this we get the following $${{T}_{3}}\left( x \right)\Rightarrow -11+12\left( x+1 \right)-3{{\left( x+1 \right)}^{2}}$$ $${{T}_{4}}\left( x \right)\Rightarrow -11+12\left( x+1 \right)-3{{\left( x+1 \right)}^{2}}+{{\left( x+1 \right)}^{3}}$$ **Note:** It is very important here to remember the formula of Taylor series expansion and the method of obtaining the derivative of simple polynomials. We have to be thorough in simplifying the equations obtained in the process of determining the Taylor polynomials. Students may go wrong in performing successive differentiation as the constant terms may confuse us during differentiation.