## Surprising Math

### 1. A pleasant surprise

When I worked on the diameter problem for d-polytopes with n facets. I was aiming to prove an upper  bound of the form $n^{\log d}$ but my proof only gave $d^{\log n},$

It was a pleasant surprise to note that $n^{\log d}=d^{\log n}$.

### 2. A bigger surprise

A few weeks ago James Lee gave a talk and proved a bound of the form $(\log n)^{\log \log \log n}.$  I was surprised to learn from him that $(\log n)^{\log \log \log n} = ( \log \log n)^{\log \log n} .$

(Update: I got it wrong at first, thanks guys)

This is an even more surprising special case of the formula above.

### 3.  Is it better to have the discount first?

Question: What is a better deal: A store that gives 12% student-discount after it adds a 12% value added tax to the price of the product? Or a store that first adds 12% tax on the entire sum and only then deducts 12% student discount?

Ohh, The way I asked this question the two alternatives are precisely the same. Let me ask it again: What is a better deal: A store that gives 12% student discount after adding a 12% value added tax to the price of the product? Or a store that first deducts the 12% student discount, and only then adds 12% tax on the new price?

Answer: The same, by a surprisingly not obvious special case of commutativity of multiplication.

(See a related comment on Dave Bacon blog.)

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### 4 Responses to Surprising Math

1. RandomWalker says:

Ok, so point (1) is explained by n^log d = e ^ ((log n) (log d)) = d ^ log n

But then, similarly for (2), how is
(log log log log n) * (log n) = (log log log n) * (log log n) ?

Am I missing something?