Respuesta :
The first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:
- The first order Maclaurin polynomial is f(x)=x
- The second order Maclaurin polynomial is also f(x)=x
- The third order Maclaurin polynomial is [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
- The fourth order Maclaurin polynomial is also [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
- You can see the graph on the attached picture.
So let's start by finding the first order maclaurin polynomial:
f(x)=f(0)+f'(0)x
so let's find each part of the function:
f(0)=arctan(0)
f(0)=0
now, let's find the first derivative of f(x)
f(x)=arctan(x)
This is a usual derivative so there is a rule we can use here:
[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]
so now we can find f'(0)
[tex]f'(0)=\frac{1}{(0)^{2}+1}[/tex]
f'(0)=1
So we can now complete the first order Maclaurin Polynomial:
f(x)=0+1x
which simplifies to:
f(x)=x
Now let's find the second order polynomial, for which we will need to get the second derivative of the function:
[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}[/tex]
so:
[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]
we can rewrite this derivative as:
[tex]f'(x)=(x^{2}+1)^{-1}[/tex]
and use the chain rule to get:
[tex]f''(x)=-1(x^{2}+1)^{-2}(2x)[/tex]
which simplifies to:
[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]
now, we can find f''(0):
[tex]f''(0)=-\frac{2(0)}{((0)^{2}+1)^{2}}[/tex]
which yields:
f''(0)=0
so now we can complete the second order Maclaurin polynomial:
[tex]f(x)=0+1x+\frac{0}{2!}x^{2}[/tex]
which simplifies to:
f(x)=x
Now let's find the third order polynomial, for which we will need to get the third derivative of the function:
[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}[/tex]
so:
[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]
In this case we can use the quotient rule to solve this:
Quotient rule: Whenever you have a function in the form , then it's derivative is:
[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]
in this case:
p=2x
p'=2
[tex]q=(x^{2}+1)^{2}[/tex]
[tex]q'=2(x^{2}+1)(2x)[/tex]
[tex]q'=4x(x^{2}+1)[/tex]
So when using the quotient rule we get:
[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]
[tex]f'''(x)=\frac{(2)(x^{2}+1)^{2}-(2x)(4x)(x^{2}+1)}{((x^{2}+1)^{2})^{2}}[/tex]
which simplifies to:
[tex]f'''(x)=\frac{-2x^{2}-2+8x^{2}}{(x^{2}+1)^{3}}[/tex]
[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]
now, we can find f'''(0):
[tex]f'''(0)=\frac{6(0)^{2}-2}{((0)^{2}+1)^{3}}[/tex]
which yields:
f'''(0)=-2
so now we can complete the third order Maclaurin polynomial:
[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}[/tex]
which simplifies to:
[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
Now let's find the fourth order polynomial, for which we will need to get the fourth derivative of the function:
[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}[/tex]
so:
[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]
In this case we can use the quotient rule to solve this:
[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]
in this case:
[tex]p=6x^{2}-2[/tex]
p'=12x
[tex]q=(x^{2}+1)^{3}[/tex]
[tex]q'=3(x^{2}+1)^{2}(2x)[/tex]
[tex]q'=6x(x^{2}+1)^{2}[/tex]
So when using the quotient rule we get:
[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]
[tex]f^{4}(x)=\frac{(12x)(x^{2}+1)^{3}-(6x^{2}-2)(6x)(x^{2}+1)^{2}}{((x^{2}+1)^{3})^{2}}[/tex]
which simplifies to:
[tex]f^{4}(x)=\frac{12x^{3}+12x-6x^{3}+12x}{(x^{2}+1)^{4}}[/tex]
[tex]f^{4}(x)=\frac{6x^{3}+24x}{(x^{2}+1)^{4}}[/tex]
now, we can find f^{4}(0):
[tex]f^{4}(x)=\frac{6(0)^{3}+24(0)}{((0)^{2}+1)^{4}}[/tex]
which yields:
[tex]f^{4}(0)=0[/tex]
so now we can complete the fourth order Maclaurin polynomial:
[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}+\frac{0}{4!}x^{4}[/tex]
which simplifies to:
[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
you can find the graph of the four polynomials in the attached picture.
So the first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:
- The first order Maclaurin polynomial is f(x)=x
- The second order Maclaurin polynomial is also f(x)=x
- The third order Maclaurin polynomial is [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
- The fourth order Maclaurin polynomial is also [tex]f(x)=x-\frac{1}{3}x^{3}[/tex]
You can find further information on the following link:
https://brainly.com/question/17440012?referrer=searchResults
