Question

Explain the reason on how the Taylor polynomials converges to \(f(x)\).

Short answer

The value of \({T_n}(x) \to f(x)\) as \(n \to \infty \).

Step-by-step solution

Given data

The Taylor polynomials are, \({T_0}(x) = 1,{T_1}(x) = - x + 2,{T_2}(x) = {x^2} - 3x + 3\) and \({T_3}(x) = - {x^3} + 4{x^2} - 6x + 4\).

Concept used of the \(n\) th-degree of the Taylor polynomial

The\(n\)th-degree of the Taylor polynomial of\(f(x)\)at\(a\)is\({T_n}(x) = \sum\limits_{i = 0}^\infty {\frac{{{f^{(i)}}(a)}}{{i!}}} {(x - a)^i}\),

\({T_n}(x) = f(a) + \frac{{{f^\prime }(a)}}{{1!}}(x - a) + \frac{{{f^{\prime \prime }}(a)}}{{2!}}{(x - a)^2} + \frac{{{f^{\prime \prime \prime }}(a)}}{{3!}}{(x - a)^3} + \cdots {\rm{ }}\)

Draw the graph

The graph is,

Analyse the graph

The given function is \(f(x) = \frac{1}{x}\) and centered at \(a = 1\).

The corresponding Taylor series is \({T_n}(x) = 1 - (x - 1) + \left( {{x^2} - 2x + 1} \right) - \left( {{x^3} + 4{x^2} - 6x + 4} \right)\) and \(f\) is the sum of its Taylor series.

The Taylor polynomials are,

\({T_0}(x) = 1,{T_1}(x) = - x + 2,{T_2}(x) = {x^2} - 3x + 3\)and \({T_3}(x) = - {x^3} + 4{x^2}\)

\( - 6x + 4\)

When \(n\) increases, the Taylor series \({T_n}(x)\) becomes a good approximation to \(f(x)\) on a larger interval.

That is, the value of \({T_n}(x) \to f(x)\) as \(n \to \infty \).