attempt 1: 10 attempts remaining. find the first four taylor polynomials for f(x) centered at x = a, and use…

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function $f(x)$ centered at $x = a$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n$, where $f^{(n)}(a)$ is the $n$-th derivative of $f(x)$ evaluated at $x = a$. The $n$-th Taylor polynomial $p_n(x)$ is the sum of the first $n+1$ terms of the Taylor series: $p_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$.

Step2: Find $f(a)$ for $n = 0$

First, find $f(x)=\ln(x + 1)$ and $a = 0$. Then $f(0)=\ln(0 + 1)=0$. So, $p_0(x)=f(0)=0$.

Step3: Find the first - derivative and $p_1(x)$

Differentiate $f(x)=\ln(x + 1)$ using the chain - rule. $f^{\prime}(x)=\frac{1}{x + 1}$. Evaluate at $x = 0$: $f^{\prime}(0)=1$. Then $p_1(x)=f(0)+f^{\prime}(0)(x - 0)=0 + 1\cdot x=x$.

Step4: Find the second - derivative and $p_2(x)$

Differentiate $f^{\prime}(x)=\frac{1}{x + 1}=(x + 1)^{-1}$ using the power - rule. $f^{\prime\prime}(x)=-(x + 1)^{-2}$. Evaluate at $x = 0$: $f^{\prime\prime}(0)=-1$. Then $p_2(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}=0 + x-\frac{1}{2}x^{2}=x-\frac{1}{2}x^{2}$.

Step5: Find the third - derivative and $p_3(x)$

Differentiate $f^{\prime\prime}(x)=-(x + 1)^{-2}$ using the power - rule. $f^{(3)}(x)=2(x + 1)^{-3}$. Evaluate at $x = 0$: $f^{(3)}(0)=2$. Then $p_3(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}=0 + x-\frac{1}{2}x^{2}+\frac{2}{6}x^{3}=x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}$.

Answer:

$p_0(x)=0$ $p_1(x)=x$ $p_2(x)=x-\frac{1}{2}x^{2}$ $p_3(x)=x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}$