Ask Uncle Colin: A logarithmic coincidence?

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I noticed that $2^{\frac{1}{1,000,000}} = 1.000 000 693 147 2$ or so, pretty much exactly $\left(1 + \frac{1}{1,000,000} \ln(2)\right)$. Is that a coincidence?

Nice Interesting Numbers; Jarring Acronym

Dear NINJA,

The easiest way to see that it's not a coincidence is to check out $3^{\frac{1}{1,000,000}} $, which is about $\left(1 + \frac{1}{1,000,000} \ln(3)\right)$ -- but that leaves unanswered the question "why?".

My first thought was to use the binomial theorem -- to check that $\left(1 + \frac{1}{1,000,000} \ln(2)\right)^{1,000,000} \approx 2$, or in general, $\left(1 + \frac 1n \ln(k)\right)^n \approx k$, you'd expand it to get $1 + \ln(k) + \frac{1}{2}(n-1) (\ln(k))^2 + ...$, which actually isn't all that helpful1.

Instead, the trick is to use logarithms straight off, and rewrite the left-hand side as $e^{\ln\left(2^{\frac{1}{1,000,000}}\right) } = e^{\frac {1}{1,000,000} \ln(2)}$. For small $x$, $e^x \approx 1 + x$, so $e^{\frac {1}{1,000,000} \ln(2)} \approx 1 + \frac{1}{1,000,000} \ln(2)$. This, clearly, works if you replace 2 or 1,000,000 with whatever numbers you like.

-- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. That said, it does give an interesting identity involving powers of a natural logarithm. []

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