Respuesta :
Answer:
a) The 4th degree , 5th degree and sixth degree polynomials
[tex]f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}[/tex]
[tex]f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}[/tex]
[tex]f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}[/tex]
b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.
[tex]log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..[/tex]
Step-by-step explanation:
Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0
f(x) = log(1+x) ……(1)
using formula [tex]\frac{d}{dx} logx =\frac{1}{x}[/tex]
Differentiating Equation(1) with respective to 'x' we get
[tex]f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)[/tex]
[tex]f^{l} (x) = \frac{1}{1+x} (1)[/tex] ….(2)
At x= 0
[tex]f^{l} (0) = \frac{1}{1+0} (1)= 1[/tex]
using formula [tex]\frac{d}{dx} x^{n-1} =nx^{n-1}[/tex]
Again Differentiating Equation(2) with respective to 'x' we get
[tex]f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))[/tex]
[tex]f^{ll} (x) = \frac{-1}{(1+x)^2} (1)[/tex] ….(3)
At x=0
[tex]f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1[/tex]
Again Differentiating Equation(3) with respective to 'x' we get
[tex]f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))[/tex]
[tex]f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}[/tex] ….(4)
At x=0
[tex]f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)[/tex]
[tex]f^{lll} (0) = 2[/tex]
Again Differentiating Equation(4) with respective to 'x' we get
[tex]f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))[/tex]
[tex]f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}[/tex] ....(5)
[tex]f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}[/tex]
[tex]f^{lV} (0) = -6[/tex]
Again Differentiating Equation(5) with respective to 'x' we get
[tex]f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))[/tex]
[tex]f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}[/tex] .....(6)
At x=0
[tex]f^{V} (x) = 24[/tex]
Again Differentiating Equation(6) with respective to 'x' we get
[tex]f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))[/tex]
[tex]f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}[/tex]
and so on...
The nth term is
[tex]f^{n} (x) = = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}[/tex]
Step :-2
Taylors theorem expansion of f(x) is
[tex]f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)[/tex]
At x=a =0
[tex]f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)[/tex]
Substitute all values , we get
[tex]f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)[/tex]
On simplification we get
[tex]log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..[/tex]
