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# Use a Taylor polynomial to approximate the function with an error of less than 0.001.ln (1.75)

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Answered 2 years ago
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First, we will find the series for the function $g(x)= \ln x$ and then we will substitute $x$ with $1,75$ to get the series for our function $\ln 1.75$.

From the table of $\textit{Power Series for Elementary Functions}$ we see that

$g(x)=\ln x=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} (x-1)^{n}}{n}$

So, the series is

\begin{align*} \boldsymbol{\ln 1.75} &=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} (1.75-1)^{n}}{n} \boldsymbol{\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} 0.75^{n}}{n}} \end{align*}

According to $\textit{Theorem 8.15}$ the given inequality is

$\left|S-S_N \right| = \left|R_N \right| \leq a_{N+1 }=\dfrac{(-1)^{N} 0.75^{N+1}}{N+1}$

Hence, since the error is less than $0.001$ we have the new inequality

$\dfrac{(-1)^{N} 0.75^{N+1}}{N+1} <0.001$

It is correct for $N=14$.

Therefore, the required number for correct approximation is $\boldsymbol{14}$.

Therefore,

\begin{align*} \boldsymbol{\ln 1.75} &= \sum_{n=1}^{14} \dfrac{(-1)^{n-1} 0.75^{n}}{n} \\ &\approx \boldsymbol{0.559092} \end{align*}

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