Question

Let ſbe a function having derivatives of all orders for all real numbers. Selected values off and its first four derivatives are shown in the table above. a. Write a second-degree Taylor polynomial for fabout x = 0 and use it to approximate f (0.2). a b. Let g be a function such that g(x) = f (x?). Write the fifth-degree Taylor polynomial for g', the derivative of g, about x=0. 2 c. Write the third-degree Taylor polynomial for f about x =1. = d. It is known that (s("(x) < 300 for 0​

Answer 1

Hi there!

a.

Recall the equation for a Taylor expansion:

$P_n(x) = f(c) + \frac{f^{'}(x)}{1!}(x - c)^1 + ... +\frac{f^{n}(x)}{n!}(x - c)^n$

**Where the numerator of the coefficient is the derivative evaluated at the point, and c = where the polynomial is centered.

We can plug in the given values to solve.

$P_5(0.2) = f(0) + \frac{f^{'}(0)}{1!}(0.2 - 0)^1 + \frac{f^{''}(0)}{2!}(0.2 - 0)^2 + \frac{f^{'''}(0)}{3!}(0.2 - 0)^3 + \frac{f^{4}(0)}{4!}(0.2 - 0)^4 \\\\P_5(0.2) = 4 + \frac{5}{1}(0.2) + \frac{-1}{2}(0.2)^2 + \frac{-15}{12}(0.2)^3 + \frac{23}{4!}(0.2)^4\\\\P_5(0.2) = 4 + 1 -0.02 -0.01 + 0.0015 = \boxed{4.9715}$

b.

At x = 0, f(x³) = f(x) because 0³ = 0, so we can simply take the derivative of the polynomial to find g'(x).

Differentiate the following.
$P_5(x) = f(0) + \frac{5}{1!}(x- 0)^1 + \frac{-1}{2!}(x - 0)^2 + \frac{-15}{2* 3!}(x- 0)^3 + \frac{23}{4!}(x - 0)^4$
Simplify:
$P_5(x) = 4 + 5x+ \frac{-1}{2}x^2 + \frac{-15}{12}x^3 + \frac{23}{24}x^4\\\\\boxed{\frac{d}{dx}P_5(x) = 5 -x - \frac{15}{4}x^2 + \frac{23}{6}x^3}$

c.

The third degree will include n = 0, 1, and 2. Also, c = 1 in this instance, so using the format above:
$P_n(x) = f(1) + \frac{f^{'}(1)}{1!}(x - 1)^1 + +\frac{f^{''}(1)}{2!}(x - 1)^2\\\\= 8 + \frac{3}{1!}(x - 1)^1 +\frac{-2}{2!}(x - 1)^2\\\\\boxed{= 8 + 3(x - 1) -(x - 1)^2}$

d.

Using the equation for the Lagrange error bound:
$| R_n| \leq \frac{f^{n + 1}(z) |(x - c)|^{n+ 1}}{(n + 1)!}$

$f^{n+1} (z)$ is the maximum value of the fourth derivative (since we are doing a third-degree polynomial), which is given to us in the problem.

We also know that:
n = 3

c = 1

x = 1.1

We can plug in the values:

$|R_n| \leq \frac{(300) |(1.1 - 1)|^{3+ 1}}{(3 + 1)!} = \frac{300(0.1^4)}{4!} = \boxed{0.00125}$