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Let $f(x)=\ln(x+1)$ then

(a) find the fourth Taylor polynomial of $f$ at $x=1$ and

(b) use part (a) find the approximate the value of $\ln(2.2)$ correct 4 decimal

(c) Find an estimate for the error in part(b) using Taylor's theorem

Taylor series for the function

$$\ln(1+x)=\int_0^x\frac{dt}{t+1}=\int_0^x\sum_{n=0}^\infty (-1)^n t^ndt=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}$$ then how do we proceed?

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Taylor series for the function can be written as $$\ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}, x\in(-1,1].$$

(a) The fourth Taylor ploynomial at $x=1$ is $(-1)^{n+1}\frac{x^n}{n}|_{x=1}=-\frac{x^4}{4}$.

(b) We cannot approximate the value of $\ln(1+1.2)$ since $x\in(-1,1]$.

Hence, I guess you mean Maclaurin series of $\ln(1+x)$ at $x_{0} = 1$. Maclaurin series of $\ln(1+x)$ at 1 is $$\ln(1+x) = \ln(2) + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^nn}(x-1)^n, x\in\mathbf{R}$$ (a) The fourth Taylor ploynomial at $x=1$ is $\frac{(-1)^{n-1}}{2^nn}(x-1)^n.$

(b) (c) I didn't fully understand the meaning "correct 4 decimal".

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To come back to the usual formula, define $x=2y+1$. This makes $$\log(1+x)=\log(2+2y)=\log\Big(2(1+y)\Big)=\log(2)+\log(1+y)$$ Now, close to $y=0$ $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+O\left(y^5\right)$$ Now, replace $y$ by $\frac{x-1}2$ to get, around $x=1$ $$\log(1+x)=\log (2)+\frac{1}{2}(x-1)-\frac{1}{8} (x-1)^2+\frac{1}{24} (x-1)^3-\frac{1}{64} (x-1)^4+O\left((x-1)^5\right)$$

I am sure that you can take it from here.

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