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Question: How do you find the second-degree Taylor polynomial \( {T_2}(x) \) for the function \( f(x) = \sqrt ...

How do you find the second-degree Taylor polynomial T2(x){T_2}(x) for the function f(x)=3+x2f(x) = \sqrt {3 + {x^2}} at the number x = 1

Explanation

Solution

Hint : In order to solve this question we need to make the Taylors equation and then we will have to put the value of the function at the given point on x and after the appropriate calculation we will set the second degree polynomial.

Complete step-by-step answer :
A complete Taylor polynomial for the function f centered around x=c is given by:
T(x)=∑n=0∞fn(c)n!T(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}(c)}}{{n!}}}
So the second degree Taylor polynomial will be the sum of the terms n=0 through n=2, or:
T2(x)=f(0)n0!(x−c)0+f(1)n1!(x−c)1+f(2)n2!(x−c)2{T_2}(x) = \dfrac{{{f^{(0)}}n}}{{0!}}{(x - c)^0} + \dfrac{{{f^{(1)}}n}}{{1!}}{(x - c)^1} + \dfrac{{{f^{(2)}}n}}{{2!}}{(x - c)^2}
Now putting the value of c = 1 in this equation:
T2(x)=f(1)+f′(1)(x−1)+f′′(1)2(x−1)2{T_2}(x) = f(1) + {f'}(1)(x - 1) + \dfrac{{f''(1)}}{2}{(x - 1)^2}
Now putting all the value of x in the function given;
f(1)=3+1=2f(1) = \sqrt {3 + 1} = 2
And putting the value of x and finding the derivatives;
f′(x)=12(3+x2)−12(2x)f'(x) = \dfrac{1}{2}{(3 + {x^2})^{ - \dfrac{1}{2}}}(2x)
On further solving;
f′(x)=x3+x2f'(x) = \dfrac{x}{{\sqrt {3 + {x^2}} }}
On putting the value of x=1 on this expression;
f′(1)=13+1f'(1) = \dfrac{1}{{\sqrt {3 + 1} }}
On further solving:
f′(1)=12f'(1) = \dfrac{1}{2}
Similarly we will finding second derivative of it;
f′′(x)=(3+x2)−12+x−12(3+x2)−32(2x)f''(x) = {(3 + {x^2})^{ - \dfrac{1}{2}}} + x\\{ - \dfrac{1}{2}{(3 + {x^2})^{ - \dfrac{3}{2}}}\\} (2x)
On further solving;
f′′(x)=(3+x2)−12−x2(3+x2)−32f''(x) = {(3 + {x^2})^{ - \dfrac{1}{2}}} - {x^2}{(3 + {x^2})^{ - \dfrac{3}{2}}}
To the last stage of calculation;
f′′(x)=3(3+x2)32f''(x) = \dfrac{3}{{{{(3 + {x^2})}^{\dfrac{3}{2}}}}}
Putting the value of x=1 in this expression;
f′′(1)=3(4)32f''(1) = \dfrac{3}{{{{(4)}^{\dfrac{3}{2}}}}}
On further solving;
f′′(1)=38f''(1) = \dfrac{3}{8}
Now putting all the values in the Taylor expression;
T2(x)=2+12(x−1)+316(x−1)2{T_2}(x) = 2 + \dfrac{1}{2}(x - 1) + \dfrac{3}{{16}}{(x - 1)^2}
So this is the Taylor second order expression.
So, the correct answer is “ T2(x)=2+12(x−1)+316(x−1)2{T_2}(x) = 2 + \dfrac{1}{2}(x - 1) + \dfrac{3}{{16}}{(x - 1)^2} ”.

Note : As we have the second degree like it we can find the degree term up to which we want of the Taylor expression. There are also some more of the named expressions which we can find like we are finding Taylor expressions.