Methods of Approximating ln(a+b)

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

ln(2+3)\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):

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

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

f(x)=ln(3)+x3x22(3)2)+....f(x) =\ln(\sqrt{3}) + \frac{x}{\sqrt{3}}-\frac{x^2}{2 \cdot (\sqrt{3})^2)}+....

and put x=2x=2, and ln(3)=ln32\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 (3)\sqrt(3) we use.

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

Method 2

Another approach requires a little bit more work.

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


x=ln[3(23+1)]x=\ln[\sqrt{3} \cdot (\frac{2}{\sqrt{3}}+1)]

x=ln3+ln(23+1)x=\ln \sqrt{3}+\ln(\frac{2}{\sqrt{3}}+1)

xln32=ln(23+1)x-\frac{\ln3}{2} = \ln(\frac{2}{\sqrt{3}}+1)

Now for any real aa

a[xln32]=aln(23+1)a[x-\frac{\ln 3}{2}] = a\ln(\frac{2}{\sqrt{3}}+1)

=ln[(23+1)a]= \ln[(\frac{2}{\sqrt{3}}+1)^a]

Now (1+x)b1+bx(1+x)^b \approx 1+bx by taking the first two terms of the binomial expansion, so:

a[xln32]ln(1+2a3)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|a| to minimise the error term, so we chose

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


3n[xln32]ln(n+2n)\frac{\sqrt{3}}{n}[x-\frac{\ln 3}{2}] \approx \ln(\frac{n+2}{n})

So xn3[lnn+2lnn]+ln32x \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 nn appropriately to approximate the original expression. Again, we must estimate the value of 3\sqrt{3} as it appears in our approximating expression.

Do comment on any other methods of approximating this value!

Note by Jihoon Kang
4 years, 10 months ago

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1 vote

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Be careful with trying to apply the Taylor series to a large xx 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 - 4 years, 10 months ago

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Interesting question! If you don't want to approximate 3\sqrt3 you can use the identity ln(r+3)=12[ln(2r)+ln(r2+32r+3)]\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. 12(ln4+12(ln72+ln(9756+3)))\frac12 (\ln 4+\frac12 (\ln\frac72+\ln (\frac {97}{56}+\sqrt3)))

Note it isn't a coincidence that 97563\frac {97}{56} \approx \sqrt3 and so after simplification the expression becomes 4ln4+2ln7+ln38\frac {4\ln4+2\ln7+\ln3}{8}

If you repeat it once more, 8ln4+4ln74ln2+2ln972ln28+ln12161.31695789684\frac{8\ln4+4\ln7-4\ln2+2\ln{97}-2\ln{28}+\ln{12}}{16} \approx 1.31695789684

In comparison, ln(2+3)1.31695789692\ln(2+\sqrt3) \approx 1.31695789692

Shaun Leong - 4 years, 10 months ago

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Thank you very much! I didn't come across this so this was very interesting

Jihoon Kang - 4 years, 10 months ago

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