Apex CALCULUS v4.0

Solution

# 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 determine the Taylor Polynomial pattern by finding the first few coefficients.

The formula for Taylor Polynomial is:

Let us expand out the first few terms so that way we can obtain the first few coefficients.

Each \(\frac{f^{(n)}(c)}{n!}\) term is the coefficient in front of each polynomial. Let’s find the first few coefficients.

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

2. \(\frac{f^{(1)}(c)}{1!}(x-c)^1\). First find the first derivative of function f. \(f'(x) = \frac{-1}{x^2}\). Therefore, the coefficient of the first degree polynomial (x???c) is: \(\frac{-1/{x^2}}{1!} = -\frac{1}{x^2}\). If we plug in for c=1, the coefficient here is \(f'(1)=\frac{???1}{1!}=???1\)

3. \(\frac{f^{2}(c)}{2!}(x-c)^2\). The second derivative of function f is: \(f''(x) = \frac{2}{x^3}\). Therefore, the coefficient of the second degree polynomial (x???c) is: \(\frac{2/x^3}{2!} = \frac{1}{x^3}\). Plugging in for c=1, the coefficient is \(f''(1)=\frac{1}{1^3}=1\)

4. \(\frac{f^{3}(c)}{3!}(x-c)^3\). The third derivative of function f is: \(f'''(x) = \frac{-6}{x^4}\). Therefore, the coefficient of the third degree polynomial (x???c) is: \(\frac{-6/x^4}{3!} = -\frac{1}{x^4}\). Plugging in for c=1, the coefficient is \(f'''(1)=\frac{-1}{1^4}=-1\)

As we can see, the formula for the Taylor Polynomial is starting to look like:

\[P(x) = 1 - (x-c) + (x-c)^2 - (x-c)^3 + ... \]

Plugging in c=1, the formula will appear as: \[P(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + ... \]

Which can be generalized for the nth term as: \[P^n(x) = \sum _{n=0}^{\infty} (-1)^n(x-1)^n\]