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) + (f^(')(x))/(1!)(x - c)^1 + ... +(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) + (f^(')(0))/(1!)(0.2 - 0)^1 + (f^('')(0))/(2!)(0.2 - 0)^2 + (f^(''')(0))/(3!)(0.2 - 0)^3 + (f^(4)(0))/(4!)(0.2 - 0)^4 \\\\P_5(0.2) = 4 + (5)/(1)(0.2) + (-1)/(2)(0.2)^2 + (-15)/(12)(0.2)^3 + (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) + (5)/(1!)(x- 0)^1 + (-1)/(2!)(x - 0)^2 + (-15)/(2* 3!)(x- 0)^3 + (23)/(4!)(x - 0)^4`
Simplify:

`P_5(x) = 4 + 5x+ (-1)/(2)x^2 + (-15)/(12)x^3 + (23)/(24)x^4\\\\\boxed{(d)/(dx)P_5(x) = 5 -x - (15)/(4)x^2 + (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) + (f^(')(1))/(1!)(x - 1)^1 + +(f^('')(1))/(2!)(x - 1)^2\\\\= 8 + (3)/(1!)(x - 1)^1 +(-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 (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 ((300) |(1.1 - 1)|^(3+ 1))/((3 + 1)!) = (300(0.1^4))/(4!) = \boxed{0.00125}`