Taylor Polynomials

Question

Let f(x)=1+x+x²+x³+x⁴+x⁵

a) Find T₃(x), the Taylor polynomial polynomial of f at x=0 with degree 3

b) Find the remainder R₃(x)=f(x)-T₃(x)

c) Find the maximum value of f(4)(x) on the interval lxl≤.1

d) Justify Taylor's inequality holds true for R₃(.1) using your result from the previous question

Answer 1

Given:

f(x) = 1 + x + x^2 + x^3 + x^4 + x^5

Taylor polynomial at x = 0 (also called the Maclaurin polynomial) of degree 3 just means:

keep terms up to x^3 and drop anything x^4 and higher.

a) Find T3(x), the Taylor polynomial of f at x = 0 with degree 3

Since f(x) is already a polynomial, we literally just cut it off after x^3:

T3(x) = 1 + x + x^2 + x^3

b) Find the remainder R3(x) = f(x) - T3(x)

Subtract:

R3(x) = (1 + x + x^2 + x^3 + x^4 + x^5) - (1 + x + x^2 + x^3)

Everything up to x^3 cancels, so:

R3(x) = x^4 + x^5

c) Find the maximum value of f(4)(x) on the interval |x| <= 1

First, take derivatives. Because f is a polynomial, this is straightforward.

f(x) = 1 + x + x^2 + x^3 + x^4 + x^5

f'(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4

f''(x) = 2 + 6x + 12x^2 + 20x^3

f'''(x) = 6 + 24x + 60x^2

f(4)(x) = 24 + 120x

Now we want the maximum value of f(4)(x) = 24 + 120x on |x| <= 1, meaning -1 <= x <= 1.

This is a line with positive slope (120), so it increases as x increases.

So the maximum happens at the right end, x = 1.

f(4)(1) = 24 + 120(1) = 144

So the maximum value is:

max f(4)(x) on [-1, 1] is 144

d) Justify Taylor's inequality for R3(0.1) using your result from part (c)

Taylor’s inequality (for degree 3) says:

|R3(x)| <= (M * |x|^4) / 4!

Where:

1. M is the maximum value of |f(4)(x)| on the interval between 0 and x
2. Here x = 0.1, and the interval between 0 and 0.1 is inside [-1, 1]
3. From part (c), we can use M = 144 (this is a safe upper bound)

Now plug into the bound:

4! = 24

|x|^4 = (0.1)^4 = 0.0001

So:

|R3(0.1)| <= (144 * 0.0001) / 24

Compute step by step:

144 / 24 = 6

6 * 0.0001 = 0.0006

So Taylor’s inequality gives:

|R3(0.1)| <= 0.0006

Now we can also compute the exact remainder from part (b) to check:

R3(x) = x^4 + x^5

R3(0.1) = (0.1)^4 + (0.1)^5

= 0.0001 + 0.00001

= 0.00011

So:

|R3(0.1)| = 0.00011

And this is indeed less than 0.0006, so the inequality holds.

Final check:

0.00011 <= 0.0006 (true)