Answer:
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
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:
[tex]f(x) = ln(1+x)[/tex]
knowing that the x point will be centered on:
[tex]x=0\\Pn(0,2)\\Error < 0.01[/tex]
By rewriting the equation we have to:
[tex][[Max(f^{(n+1)} (c))]/(n + 1)!] *0.2^{(n+1)} < 0.01[/tex]
So doing the derivatives related to the first function given in the exercise we have to:
[tex]f(x) = ln(1+x)[/tex]
Following this we have to:
[tex]Max|f^{(n+1)} (c)| < n![/tex]
Thus, error is;
[tex](n!/(n + 1)!) * 0.2^{(n + 1)} < 0.01[/tex]
[tex](1/(n + 1))* 0.2^{(n + 1)} < 0.01[/tex]
Let's try n = 1
[tex](1/(1 + 1)) *0.2^{(1 + 1)} = 0.02[/tex]
This is greater than 0.01 and so it will not work. Let's try n = 2
[tex](1/(2 + 1)) * 0.2^{(2 + 1)} = 0.00267[/tex]
This is less than 0.01. So,the least integer for n is 2.
See more about Taylor polynomial at brainly.com/question/23842376
