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Question

Use a Taylor polynomial to approximate the function with an error of less than 0.001.

ln (1.75)

Solution

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Answered 2 years ago
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First, we will find the series for the function g(x)=lnxg(x)= \ln x and then we will substitute xx with 1,751,75 to get the series for our function ln1.75\ln 1.75.

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

g(x)=lnx=n=1(1)n1(x1)nng(x)=\ln x=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} (x-1)^{n}}{n}

So, the series is

ln1.75=n=1(1)n1(1.751)nnn=1(1)n10.75nn\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 Theorem 8.15\textit{Theorem 8.15} the given inequality is

SSN=RNaN+1=(1)N0.75N+1N+1\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 0.0010.001 we have the new inequality

(1)N0.75N+1N+1<0.001\dfrac{(-1)^{N} 0.75^{N+1}}{N+1} <0.001

It is correct for N=14N=14.

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

Therefore,

ln1.75=n=114(1)n10.75nn0.559092\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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