How many terms of the Maclaurin series of $f (x) = \ln(1 + x)$ are needed to compute $\ln(1.2)$ with an error of at most $0.0001$?
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$\begingroup$The Maclaurin series of $\ln(1+x)$ is $$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}\ \,\ \text{ for }-1<x\le 1$$
Notice that, for $|x|<1$, this series converges absolutely. In particular, since $\left\vert\dfrac{x^n}{n}\right\vert\le \vert x^n\vert$, it holds $$\left\vert\sum_{n=k+1}^\infty(-1)^{n+1}\frac{x^{n}}{n}\right\vert\le\sum_{n=k+1}^\infty \vert x\vert^n=\frac{\vert x\vert^{k+1}}{1-\vert x\vert}$$
So, the error $\varepsilon_k$ at the $k$-th step is smaller than $\dfrac{|x|^{k+1}}{1-\vert x\vert}$. in your case, since $x=\dfrac15$, you'd get $$\varepsilon_k\le \frac1{4\cdot5^k}$$
Another way, that works for the series of $\ln(1+x)$ under the hypothesis $x\in [0,1]$, is considering the fact that, for those values of $x$, the series we're considering satisfies the hypothesis of Leibniz criterion. Therefore, $$\left\vert\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}-\sum_{n=1}^k(-1)^{n+1}\frac{x^n}{n}\right\vert\le \left\vert \frac{x^{k+1}}{k+1}\right\vert$$
In your case, ($x=\frac15$) it holds $\varepsilon_k\le \dfrac1{(k+1)5^{k+1}}$
Now, you can use these bounds to decide when to stop your approximation.
$\endgroup$ 3 $\begingroup$Let $f(x)$=$ln(1+x)$; $f(0)=0$
$f'(x)=(1+x)^{-1}$; $f'(0)=1$
$f''(x)=-1(1+x)^{-2}$; $f''(0)=-1$
$f'''(x)=2(1+x)^{-3}$; $f'''(0)=2$
$f^{1v}(x)=-6(1+x)^{-4}$; $f^{1v}(0)=-6$
$f^v(x)=24(1+x)^{-5}$; $f^v(0)=24$
Hence $ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$+...
$ln(1.2)=ln(1+0.2)$. $x$=0.2
Therefore $ln(1.2)=0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}-\frac{0.2^4}{4}+\frac{0.2^5}{5}$+...=0.18233
Exact value of $ln(1.2)$=0.18232
Error=0.18233-0.18232=0.00001, Hence error is accepted (0.00001<0.0001).
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