Chapter 8

Exercise 8.8

8.

Find a formula for the \(n^{th}\) term of the Taylor series of \(f(x)\), centered at \(c\), by finding the coefficients of the first few powers of \(x\) and looking for a pattern.

\(f(x) = \frac{1}{x}\); \(c = 1\)

# Reference: https://www.math.ucla.edu/~anderson/rw1001/library/base/html/curve.html
fct_exp <- function(x){1/x}
curve(fct_exp, from = 0.01, to = 5, type = "l", col = "blue", ylab = "1/x")

Let’s find the first few coefficients.

Each \(\frac{f^{(n)}(c)}{n!}\) is the coefficient in front of each polynomial.

  1. \(f(c) = \frac{1}{1} = 1\) 1st coefficient = 1

  2. \(\frac{f^{(1)}(c)}{1!}(x-c)^1\)
    find the first derivative
    \(f'(x) = \frac{-1}{x^2}\) coefficient of first degree polynomial = \(\frac{-1/x^2}{1!} = -\frac{1}{x^2}\)
    plug \(c = 1\); \(f'(1) = -\frac{1}{1!} = -1\)

  3. \(\frac{f^2(c)}{2!}(x-c)^2\) find the second derivative
    \(f''(x) = \frac{2}{x^3}\) coefficient of second degree polynomial = \(\frac{2/x^3}{2!} = \frac{1}{x^3}\)
    plug \(c = 1\); \(f''(1) = \frac{1}{1^3} = 1\)

  4. \(\frac{f^3(c)}{3!}(x-c)^3\)
    find the third derivative
    \(f'''(x) = \frac{-6}{x^4}\) coefficient of third degree polynomial = \(\frac{-6/x^4}{3!} =-\frac{1}{x^4}\)
    plug \(c = 1\); \(f'''(1) = \frac{-1}{1^4} = -1\)

As it appears, the formula for the Taylor polynomial look like

\(P(x)=1−(x−c)+(x−c)^2−(x−c)^3+...\)

Formula after plugging c=1,

\(P(x)=1−(x−1)+(x−1)^2−(x−1)^3+...\)

Can be generalized for the \(n^{th}\) term as follows,

\(P^n(x) = \sum_{n=0}^{\infty}(-1)^n(x-1)^n\)