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Scilla [17]
3 years ago
14

This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,

5th, and 6th degree Taylor polynomials forf(x) = ln(1 +x) centeredata= 0.(b) Find the nth degree Taylor polynomial forf(x) centered at= 0,written in expanded form.
Mathematics
1 answer:
stich3 [128]3 years ago
4 0

Answer:

a) The 4th degree , 5th degree and sixth degree polynomials

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}

f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

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} +..

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 \frac{d}{dx} logx =\frac{1}{x}

Differentiating Equation(1) with respective to 'x' we get

f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)

f^{l} (x) = \frac{1}{1+x} (1)  ….(2)

At x= 0

f^{l} (0) = \frac{1}{1+0} (1)= 1

using formula \frac{d}{dx} x^{n-1}  =nx^{n-1}

Again Differentiating Equation(2) with respective to 'x' we get

f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))

f^{ll} (x) = \frac{-1}{(1+x)^2} (1)    ….(3)

At x=0

f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1

Again Differentiating Equation(3) with respective to 'x' we get

f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))

f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)=  \frac{(-1)^2 (2)!}{(1+x)^3} ….(4)

At x=0

f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)

f^{lll} (0) = 2

Again Differentiating Equation(4) with respective to 'x' we get

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))

f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4} ....(5)

f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}

f^{lV} (0) = -6

Again Differentiating Equation(5) with respective to 'x' we get

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))

f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5} .....(6)

At x=0

f^{V} (x) = 24

Again Differentiating Equation(6) with respective to 'x' we get

f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))

f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}

and so on...

The nth term is

f^{n} (x) =  = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}

Step :-2

Taylors theorem expansion of f(x) is

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)

At x=a =0

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)

Substitute  all values , we get

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)

On simplification we get

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} +..

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