Interesting Log Relationships (3)

The formula, log n – log (n – 1) = 1/n + 1/2n² + 1/3n³ + 1/4n⁴ +…., can be used to generate a simple infinite series expression for every number.

Basically if n is of the form t/(t – 1), then log n – log (n – 1) is of the form,

log t/(t – 1) – log (1/t – 1) = log t.

So, for example to find log 2 we set t = 2.

Therefore log (2/1) – log (1/1) =  log 2.

Thus with n in this case = 2/1,

log 2 = 1/1.2 + 1/2.2² + 1/3.2³ + 1/4.2⁴ +…  

So summing the first four terms of the series we obtain .68229…, which already approximates well to the correct answer .693147…

Likewise to find the corresponding expression for log 3 we set t = 3, with n thereby = 3/2.

Thus log 3 = 2/1.3 + 2²/2.3² + 2³ /3.3³ + 2⁴/4.3⁴ +…  

So again by summing the first four terms we obtain 1.03703… as against the correct answer, 1.09861…

This indicates that in this case the answer converges more slowly to the actual result and this will be increasingly the case as n becomes larger!.

Indeed, in principle we can use the same formula i.e..

log n – log (n – 1) = 1/n + 1/2n² + 1/3n³ + 1/4n⁴ +…,

to find an unlimited number of infinite expressions for the log of any number.

For example, 4 = 2² and 8 = 2³, so we equally express log 2 in terms of any number that is a power of 2 and by extension the log of any number in terms of a number that is a corresponding power of that number..

So log 2 = (log 4)/2 and (log 8)/3 respectively!

Thus these two alternative infinite series expressions for log 2 are,

 (3/1.4 + 3²/2.4² + 3³ /3.4³ + 3⁴/4.4⁴ +… )/2 and

(7/1.8 + 7²/2.8² + 7³ /3.8³ + 7⁴/4.8⁴ +… )/3 respectively.

However, a significant practical problem is that they converge ever more slowly to the actual result.

This problem can however be solved in an interesting manner!  

For example to find a quickly converging series for log 2, we can set

t = 2 raised to a small fractional power!

For example, for t = 2^1/10, n = t/(t – 1) = 2^1/10/(2^1/10 – 1)

Therefore

log 2 = 10{(2^1/10 – 1)/2^1/10 + (2^1/10 – 1)²/2. 2^2/10 + …….}

= 10(.066967… + .002242… + ….)

= .69209…

So, the result has already converged quite close to the actual result (i.e. .693147…)

However by raising here 2 to a sufficiently small power, we can approximate the log 2 as closely as we wish to the actual result through sole consideration of the first term.

So in general if a is the number whose log we wish to estimate, and 1/t is the power to which it is raised then the approximation = t(a^1/t – 1)/a^1/t

Therefore for example, when a = 2 and t = 1,000,000 the approximation for log 2 is given as,

1,000,000{(2^1/1,000,000 – 1)/2^1/1,000,000} 

= 1,000,000} (1.0000006931474… – 1)/1.0000006931474},

= 1,000,000(.0000006931469…) = .6931469…

So this answer is already correct to 6 significant figures with respect to the actual result.

Indeed as the denominator a/^1/t approximates ever more closely to 1 for very large t, the log for any number a can be approximated closely by the even more simple expression,

t(a^1/t – 1) where t is sufficiently small.

Again for example, log 145 = 4.97673374242…

So in this case setting t = 1,000,000,000,

a^1/t – 1 = 1.0000000049767337548… – 1 = .0000000049767337548…

Therefore log 145 ~ 1,000,000,000(.0000000049767337548…)

= 4.9767337548…

As we can see this approximation is already correct to 7 significant figures.

And we can continually improve this approximation by making t progressively larger.