Hi I recently came across a problem which peaked my interest. The problem was to approximate (or determine the value of):

\(\ln (2+\sqrt{3})\)

without the use of a calculator. You are given a table of logs in base *e* of natural numbers.
I am interested to see what various methods of approach there are.

*Method 1*

For me, the first approach was to define a function \(f(x)\):

\(f(x) = \ln(x+\sqrt{3})\)

and then use Taylor's expansion about the point \(x=0\). Then we obtain:

\(f(x) =\ln(\sqrt{3}) + \frac{x}{\sqrt{3}}-\frac{x^2}{2 \cdot (\sqrt{3})^2)}+....\)

and put \(x=2\), and \(\ln(\sqrt{3}) = \frac{\ln 3}{2}\) which we can determine using the table. The degree of accuracy can be manipulated by changing the number of terms in the expansion we use, and how many decimal places of \(\sqrt(3)\) we use.

The only problem with this is the issue of approximating \(\sqrt{3}\) still remains.

*Method 2*

Another approach requires a little bit more work.

Suppose \(x =\ln(2+\sqrt{3})\), then we want to approximate \(x\)

Now:

\(x=\ln[\sqrt{3} \cdot (\frac{2}{\sqrt{3}}+1)]\)

\(x=\ln \sqrt{3}+\ln(\frac{2}{\sqrt{3}}+1)\)

\(x-\frac{\ln3}{2} = \ln(\frac{2}{\sqrt{3}}+1)\)

Now for any real \(a\)

\(a[x-\frac{\ln 3}{2}] = a\ln(\frac{2}{\sqrt{3}}+1)\)

\(= \ln[(\frac{2}{\sqrt{3}}+1)^a]\)

Now \((1+x)^b \approx 1+bx\) by taking the first two terms of the binomial expansion, so:

\(a[x-\frac{\ln 3}{2}] \approx \ln(1+\frac{2a}{\sqrt{3}})\)

Now without going into too much detail, we want to minimise \(|a|\) to minimise the error term, so we chose

\(a = \frac{\sqrt{3}}{n}\) for some large positive integer \(n\) (the higher the value of \(n\), the better the approximation)

Then:

\(\frac{\sqrt{3}}{n}[x-\frac{\ln 3}{2}] \approx \ln(\frac{n+2}{n})\)

So \(x \approx \frac{n}{\sqrt{3}} \cdot [\ln{n+2} - \ln{n}] + \frac{\ln{3}}{2}\)

Hence, depending on the accuracy of the approximation, we can chose a positive integer \(n\) appropriately to approximate the original expression. Again, we must estimate the value of \(\sqrt{3}\) as it appears in our approximating expression.

Do comment on any other methods of approximating this value!

## Comments

Sort by:

TopNewestBe careful with trying to apply the Taylor series to a large \(x \) value.

Do you know the radius of convergence? Do you know the rate of convergence? You might have to calculate many many terms in order to get an approximation. – Calvin Lin Staff · 5 months, 1 week ago

Log in to reply

Interesting question! If you don't want to approximate \(\sqrt3\) you can use the identity \[\ln(r+\sqrt3)=\frac12[\ln(2r)+\ln(\frac{r^2+3}{2r}+\sqrt3)]\]

If you repeat this you can get an approximation to 5d.p. \[\frac12 (\ln 4+\frac12 (\ln\frac72+\ln (\frac {97}{56}+\sqrt3)))\]

Note it isn't a coincidence that \(\frac {97}{56} \approx \sqrt3\) and so after simplification the expression becomes \[\frac {4\ln4+2\ln7+\ln3}{8}\]

If you repeat it once more, \[\frac{8\ln4+4\ln7-4\ln2+2\ln{97}-2\ln{28}+\ln{12}}{16} \approx 1.31695789684\]

In comparison, \[\ln(2+\sqrt3) \approx 1.31695789692\] – Shaun Leong · 5 months, 1 week ago

Log in to reply

– Jihoon Kang · 5 months, 1 week ago

Thank you very much! I didn't come across this so this was very interestingLog in to reply