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].
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]
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
