Question

How many nonzero terms of the Maclaurin series for
ln(1 + x) do you need to use to estimate ln 1.4 to within 0.0001?

Answer 1

The Maclaurin series for $f(x)=\ln(1+x)$ is

$\ln(1+x)=\displaystyle\sum_{n\ge1}\frac{(-1)^{n+1}x^n}n$

and converges for $-1<x\le1$. To approximate $\ln1.4$ to within 0.0001 is the same as finding $n$ such that

$|f(x)-P_n(x)|=|R_n(x)|\le10^{-4}$

where $P_n$ is the $n$th order Taylor approximation to $f(x)$ and $R_n$ is the remainder term.

By Taylor's theorem, the remainder term is bounded above by

$|R_n(x)|\le\dfrac{M|x|^{n+1}}{(n+1)!}$

where $M$ is the maximum value achieved by $f^{(n+1)}(x)$ in some interval centered at $x=0$. We have

$f(x)=\ln(1+x)\implies f'(x)=\dfrac1{1+x}\implies f''(x)=-\dfrac1{(1+x)^2}\implies\cdots\implies f^{(n+1)}(x)=\dfrac{(-1)^nn!}{(1+x)^{n+1}}$
$\implies\left|f^{(n+1)}(x)\right|=\left|\dfrac{(-1)^nn!}{(1+x)^{n+1}}\right|=\dfrac{n!}{|1+x|^{n+1}}$

Let's consider the interval $0\le x\le0.4$. The $n+1$th derivative is maximized at $x=0$, and this leaves us with

$|f^{(n+1)}(x)|\le\dfrac{n!}{1^{n+1}}=n!$

From this we have that

$|R_n(x)|=\left|\dfrac{f^{(n+1)}(x)x^{n+1}}{(n+1)!}\right|\le\dfrac{n!|x|^{n+1}}{(n+1)!}=\dfrac1{n+1}|x|^{n+1}$

So now we look for $n$ such that

$\dfrac1{n+1}0.4^{n+1}\le10^{-4}$

Using a calculator, we can deduce that this occurs for all integers $n\ge7$, which means we need at least 7 terms of the series to achieve the desired accuracy.

Let's check this result:

$f(0.4)=\ln1.4\approx0.336472$

$P_6(0.4)=\displaystyle\sum_{n=1}^{n=6}\frac{(-1)^{n+1}0.4^n}n\approx0.336299$

$P_7(0.4)=\displaystyle\sum_{n=1}^{n=7}\frac{(-1)^{n+1}0.4^n}n\approx0.336533$

Taking the absolute difference, you'll see that $|\ln1.4-P_7(0.4)|\le10^{-4}$, but this is not the case with $P_6$.

Answer 2

We need 7 terms at least to estimate ln 1.4 to within 0.0001.

What is the Maclaurin series?

A Maclaurin series is a Taylor series centred about zero.

Given function is:

f(x)=ln(1+x)

We know the expansion of ln(1+x) about zero(up to 7 terms) is:

$ln(1+x) =x-\frac{x^{2} }{2} +\frac{x^{3} }{3} -\frac{x^4}{4} +\frac{x^5}{5} -\frac{x^6}{6} +\frac{x^7}{7} -.....$

Where $-1\leq x\leq 1$

For ln 1.4 put x=0.4 in the above series

$ln(1+0.4) =0.4-\frac{0.4^{2} }{2} +\frac{0.4^{3} }{3} -\frac{0.4^4}{4} +\frac{0.4^5}{5} -\frac{0.4^6}{6} -\frac{0.4^7}{7} +....$

$ln(1.4)=0.3365333$

Therefore,  we need 7 terms at least to estimate ln 1.4 to within 0.0001.

To get more about the Taylor series visit:

https://brainly.com/question/24188700