4 ms·

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 expo

by eliben 12y ago

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.

minimaxir 12y ago
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.
idlewords 12y ago
I think it means he can't math.
tdaltonc 12y ago
I think he's pointing either to this: http://en.wikipedia.org/wiki/Tetration http://en.wikipedia.org/wiki/Tetration or this: http://en.wikipedia.org/wiki/Double_exponential_function http://en.wikipedia.org/wiki/Double_exponential_function
- gipp 12y ago
  Neither of these are applicable to the situation he described, though.
sama 12y ago
combo of a n^2 and 2^n function--i think non-US english speakers call it a double-exponential function.
pzxc 12y ago
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.
- gipp 12y ago
  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.
- chimtim 12y ago
  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 http://en.wikipedia.org/wiki/Stirling's_approximation
d--b 12y ago
forget what he said, he's just a founder
colanderman 12y ago
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".
prajit 12y ago
"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.

by eliben 12y ago

minimaxir 12y ago

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.

idlewords 12y ago

I think it means he can't math.

tdaltonc 12y ago

I think he's pointing either to this: http://en.wikipedia.org/wiki/Tetration http://en.wikipedia.org/wiki/Tetration or this: http://en.wikipedia.org/wiki/Double_exponential_function http://en.wikipedia.org/wiki/Double_exponential_function

gipp 12y ago
Neither of these are applicable to the situation he described, though.

sama 12y ago

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

pzxc 12y ago

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.

gipp 12y ago
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.
chimtim 12y ago
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 http://en.wikipedia.org/wiki/Stirling's_approximation

d--b 12y ago

forget what he said, he's just a founder

colanderman 12y ago

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".

prajit 12y ago

"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.