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This problem asks for Taylor polynomials for f(x) = ln(1 + x) centered at a = 0. Show your work in an organized way.

(a) Find the 4th, 5th, and 6th degree Taylor polynomials for f(x) = ln(1 + x) centered at a = 0.
(b) Find the nth degree Taylor polynomial for f(x) centered at a = 0, written in expanded form.
(c) Find the nth degree Taylor polynomial for f(x) centered at a = 0, written in sigma (summation) notation.
(d) Use the 7th degree Taylor polynomial to estimate ln(2).
(e) Compare your answer to the estimate for ln(2) given by your calculator. How accurate were you?
(f) Looking at the Taylor polynomials, explain why this estimate is less accurate than the estimate you found for sin(3°) in Problem 6 of Homework 10. (In that problem, you were asked to use a 7th degree Taylor polynomial centered at a = 0 for sin(x) to approximate sin(3°)).

Respuesta :

LammettHash

Compute up to the 6th order derivative:

[tex]f(x)=\ln(1+x)[/tex]

[tex]f'(x)=\dfrac1{1+x}[/tex]

[tex]f''(x)=-\dfrac1{(1+x)^2}[/tex]

[tex]f'''(x)=\dfrac2{(1+x)^3}[/tex]

You might start to see a pattern here: the [tex]k[/tex]-th order derivative is

[tex]f^{(k)}(x)=(-1)^{k-1}\dfrac{(k-1)!}{(1+x)^k}[/tex]

For [tex]x=0[/tex], we have

[tex]f^{(k)}(0)=(-1)^{k-1}\dfrac{(k-1)!}{1^k}=(-1)^{k-1}(k-1)![/tex]

The [tex]n[/tex]-th degree Taylor series for [tex]f(x)[/tex] centered at [tex]x=0[/tex] is

[tex]\displaystyle T_n(x)=f(0)+\sum_{k=1}^n\frac{f^{(k)}(0)}{k!}x^k[/tex]

[tex]\displaystyle T_n(x)=\ln(1+0)+\sum_{k=1}^n\frac{(-1)^{k-1}(k-1)!}{k!}x^k[/tex]

[tex]\displaystyle T_n(x)=-\sum_{k=1}^n\frac{(-1)^k}kx^k[/tex]

a. Plug in [tex]n=4,5,6[/tex] above:

[tex]T_4(x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4[/tex]

[tex]T_5(x)=T_4(x)+\dfrac{x^5}5[/tex]

[tex]T_6(x)=T_5(x)-\dfrac{x^6}6[/tex]

b. This would just be

[tex]T_n(x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\cdots+\dfrac{(-1)^{n-1}}nx^n[/tex]

c. See above part (a).

d. The 7th degree Taylor polynomial is

[tex]T_7(x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\dfrac{x^5}5-\dfrac{x^6}6+\dfrac{x^7}7[/tex]

Then

[tex]\ln2=f(1)\approx T_7(1)[/tex]

[tex]\ln2\approx1-\dfrac12+\dfrac13-\dfrac14+\dfrac15-\dfrac16+\dfrac17[/tex]

[tex]\ln2\approx\dfrac{319}{420}\approx0.7595[/tex]

e. The actual value is closer to [tex]\ln2\approx0.6931[/tex], giving an error of about 0.0664.

f. The short answer is that this series converges much slower than the Taylor series for [tex]\sin x[/tex].

ashishdwivedilVT

Taylor's series expansion about x=0 was written by using the expansion formula.

[tex]f(x) =ln(1+x)[/tex]

[tex]f(0)=0[/tex]

[tex]f'(x)=\frac{1}{1+x}[/tex]

[tex]f'(0)=1[/tex]

[tex]f''(x)=\frac{-1}{(1+x)^2}[/tex]

[tex]f''(0)=-1[/tex]

[tex]f'''(x)=\frac{2}{(1+x)^3}[/tex]

[tex]f'''(0)=2[/tex]

What is Taylor's series expansion of f(x) about c?

Taylor's series expansion of f(x) about c is given by

[tex]f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!} (x-c)^2+\frac{f'''(c)}{3!} (x-c)^3+......[/tex]

So, Taylor's series expansion about x=0 will be given by

[tex]f(x)=f(0)+f'(0)(x-0)+\frac{f''(0)}{2!} (x-0)^2+\frac{f'''(0)}{3!} (x-0)^3+......[/tex]

[tex]f(x)=1+x- \frac{x^2}{2!} +\frac{2}{3!} x^3+......[/tex]

Hence, Taylor's series expansion about x=0 was written by using the expansion formula.

To get more about Taylor's series visit:

https://brainly.com/question/24188700