Assume that f(x)=ln(1+x) is the given function and that Pn represents the nth Taylor Polynomial centered at x=0. Find the least integer n for which Pn(0.2) approximates ln(1.2) to within 0.01.

In this exercise we have to use the knowledge of Taylor Polynomial to calculate the requested function, this way we will have;

the least integer for n is 2    

The function given in this exercise corresponds to:

knowing that the x point will be centered on:

By rewriting the equation we have to:

So doing the derivatives related to the first function given in the exercise we have to:

Following this we have to:

Thus, error is;

Let's try n = 1

This is greater than 0.01 and so it will not work. Let's try n = 2

This is less than 0.01. So,the least integer for n is 2.

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the least integer for n is 2

Step-by-step explanation:

We are given;

f(x) = ln(1+x)

centered at x=0

Pn(0.2)

Error < 0.01

We will use the format;

[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01

So;

f(x) = ln(1+x)

First derivative: f'(x) = 1/(x + 1) < 0! = 1

2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1

3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2

4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6

This follows that;

Max|f^(n+1) (c)| < n!

Thus, error is;

(n!/(n + 1)!) × 0.2^(n + 1) < 0.01

This gives;

(1/(n + 1)) × 0.2^(n + 1) < 0.01

Let's try n = 1

(1/(1 + 1)) × 0.2^(1 + 1) = 0.02

This is greater than 0.01 and so it will not work.

Let's try n = 2

(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267

This is less than 0.01.

So,the least integer for n is 2

the answer is 48m^2!

step-by-step explanation:

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