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Estimate ln(3) using Taylor Expansion up to 3rd order (without the use of a calculator).

$$f(x)=ln(x)$$ $$f'(x)=1/x$$ $$f''(x)=-1/x^2$$ $$f'''(x)=2/x^3$$

$$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''+...(a)$$

If I let a=2, I can solve for the first to third order, but I will then have to estimate f(2)=ln(2). Using the same method as above to estimate ln(2), i.e. a=1, I will have a rather inaccurate estimate of ln(2).

Is there any other ways to estimate ln(3) more accurately?

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A chance is given by solving $e^x=3$ with the Newton's method with starting point $x_0=1$.

We have: $$ x_{k+1} = x_k-\frac{\exp(x_k)-3}{\exp(x_k)}=x_k-1+3\exp(-x_k)$$ and the difference between $\log 3$ and: $$ x_2 = -1+ 3 e^{-1} + 3 e^{-3/e} $$ is yet less than $2\cdot 10^{-5}$. Another chance is given by considering that:

$$ \log\frac{1+x}{1-x}=2\operatorname{arctanh} x = \sum_{n\geq 0}\frac{2\,x^{2n+1}}{2n+1} $$ for any $x$ such that $|x|<1$, and $\log 3$ is the LHS of the previous identity in $x=\frac{1}{2}$, hence:

$$ \log 3 = \sum_{n\geq 0}\frac{1}{(2n+1)4^n} \tag{1}$$

that converges pretty fast. Also notice that $(1)$ is just: $$ \log 3 = \int_{0}^{2}\frac{dx}{1+x}=2\int_{0}^{1}\frac{dx}{1+2x}=\int_{0}^{1}\frac{dx}{1-\frac{x^2}{4}}$$ in disguise. Since: $$ 3 = \frac{6}{4}\cdot\frac{4}{2}=\frac{1+1/5}{1-1/5}\cdot\frac{1+1/3}{1-1/3}$$ we also have:

$$ \log 3 = \frac{2}{15}\sum_{n\geq 0}\frac{1}{2n+1}\left(\frac{5}{9^n}+\frac{3}{25^n}\right)\tag{2}$$

that converges even faster than $(1)$.

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    $\begingroup$ This is a nice one ! Cheers $\endgroup$ – Claude Leibovici Mar 7 '15 at 12:39
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Here I'm going to pick $a=1$ so that $\log(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^{n}$. Applying the ratio test shows that this series converges for $0<x<2$. If you do not know about convergence of series, it means that the above series I wrote only works for values of $x$ between $0$ and $2$ (inclusive); generically, any $x$ values outside a series' radius of converges produces $\pm \infty$ i.e. divergence.

Here is the trick: let $x = 3^{-1}$. Clearly $0 < \frac{1}{3}<2$ so the above series converges for this particular $x$ value, then apply rules of logs: $\log 3^{-1} = - \log{3}$. In other words, $\log(3) = - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{3}-1\right)^{n}$.

Of course, you can truncate this series at the third order if that is the approximation you are looking for.

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