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Comment by eliben | Hacker News Reader

I'm having a hard time understanding what he means by "hyperexponential". If N grows exponentially and there's some function of N that's O(N^2), it's still exponential.

He can't mean the hyperexponential probability distribution, that's something different.

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I think he was using that word not as a technical term, but as a way of saying "even faster/sharper than exponential". I.e., hyper- (the prefix meaning above) + exponential.

Certainly O(N^2) is not at the top of the food chain -- for example, it is beaten by O(2^N) or O(N!) or O(N^N). All of these could, colloquially, be referred to as "hyper"exponential.

Imagine a hockey stick graph (exponential). Now imagine that hockey stick graphed on a logarithmic scale -- the logarithms more or less cancel out the exponentiation if done right, making the graph appear to be linear even though the data is exponential. Now, imagine data that, even when graphed on a logarithmic scale, still looks like a hockey stick. That is hyperexponential.

And I think I just proved Sam's point -- that it is hard to visualize/grok.

O(N^2) is not exponential at all, it is polynomial. O(2^N) is exponential.

And the grandparent was pointing out -- correctly -- that that description doesn't really apply here. (2^N)^2 = 2^(2N) = still just 'exponential'. Plotted on a log scale it would still be a straight line.

O(N^2) is not exponential but just polynomial. O(2^N) is still exponential. O(N!) and O(N^N) are faster than exponential but not that much; O(N^N) grows slower than O(N!)[1].

Hyper-exponential is most likely double exponential O(a^b^N). There is an exponential growth of apps on these platforms and another exponential growth of users on these apps.

[1] http://en.wikipedia.org/wiki/Stirling's_approximation

"There’s the famous observation that the value of a network grows as a function of the square of the number of nodes"

The value of a network is measured as O(V^2), where V is the number of nodes (users) in the network.

"and also many of these services/products double their userbase every N months."

V = 2^N (the number of users doubles every N months), so the overall function for the value of a network is O(V^2) = O((2^N)^2) = O(2^(2N)) = O(4^n). O(4^n) > O(2^n), but it is not the same as O(2^(N^2)), which was what Sam was trying to convey.

Many laypeople just use "exponential" to mean "really fast" (or even worse, "much more than"). It's been used that way so much that it's been watered down to mean just "fast". So the author probably just said "hyperexponential" to get back the original lay-meaning of "really fast".

I think he's pointing either to this:

http://en.wikipedia.org/wiki/Tetration

or this:

http://en.wikipedia.org/wiki/Double_exponential_function

Neither of these are applicable to the situation he described, though.

Hyperexponential I would assume is an order of magnitude higher than exponential. Although if I remember my video game terminology correctly, the proper hierarchy is exponential, then superexponential, then hyperexponential.

combo of a n^2 and 2^n function--i think non-US english speakers call it a double-exponential function.

I think it means he can't math.

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