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Topic: Metcalfe's Law: Bitcoin Price and Adoption Analysis for the Future (Read 14808 times)

newbie
Activity: 8
Merit: 0
what about 2018
legendary
Activity: 1937
Merit: 1001
Hi, can you please post update for 2015 Smiley ?


I doubt it. It will look totally ridiculous.
full member
Activity: 167
Merit: 101
Hi, can you please post update for 2015 Smiley ?
sr. member
Activity: 311
Merit: 250
Quote
Any model for price should take this into account.

That is already priced in to the real price.  We don't need to adjust anything.


It is not in your model that is supposed to give an estimation of the price.

You should think first. Read it again.
sr. member
Activity: 364
Merit: 250
Quote
Any model for price should take this into account.

That is already priced in to the real price.  We don't need to adjust anything.
hero member
Activity: 544
Merit: 500
 Relating (Unique Addresses ^2) to user adoption would become flawed with the increased use of HD wallets. Or at least the prediction will probably be increasingly skewed  from this point onwards. ?
sr. member
Activity: 311
Merit: 250
What would be the purpose of dividing adoption-squared by number of coins?  I don't see how that is useful.

This is basic. Imagine that the number of coins doubles. Then the price is divided by two.

Therefore the objective price is inversely proportional to the number of coins in existence. Any model for price should take this into account. Indeed, one could only consider the number of coins that are not lost, but the estimates on these differ.

This will give a small correction to your formula, that may be irrelevant now but not in 1 or 2 years.

sr. member
Activity: 364
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What would be the purpose of dividing adoption-squared by number of coins?  I don't see how that is useful.
sr. member
Activity: 311
Merit: 250
Sounds like you'd rather use market cap than coin price.  I use coin price because that is what everyone in this subforum is interested in.

If I have time I will update with most recent data tonight.


No, this is not what I mean. I mean coin price, but it has to take account of the creation of new coins.

Read again:

Quote
The formula will be:

Price = C x (Unique number of addresses used)^2 / (Number of bitcoins created) x C

where C is the appropriate constant (different from the one you used but approximatively equal to the one you use multiplied by 13.000.000).

C/C = 1.  So you are just proposing dividing by number of coins.  Why?  Price is already per coin.  As number of coins increases, it is reflected in the price.


No my friend, it is C/(Number of coins created) that is not constant equal to one since "number of coins created changes over time (C stays the same).

The constant C has to be found by historical interpolation (in the same way you found your multipicative constant).

Sorry, there was a typo in the formula (just multiply once by C)

Price = C x (Unique number of addresses used)^2 / (Number of bitcoins created)
legendary
Activity: 1470
Merit: 1007
Interesting thread. Thanks, for the thorough analysis, raystonn.

I guess you know the posts by Peter R., who is also tracking price/mcap as a function of network size under the metcalfe assumption (MA).

Here's one funny idea I had on this topic when I looked at it myself a while ago (it's not fleshed out yet, just did some imprecise calculatios "by hand" on this, so be forgiving please...)

When modeling price/mcap as a function of network size under MA, the empirically determined parameter is the exact coefficient that determines the ratio of network size to price.

Here's the idea I had... perhaps there need to be /two/ (largely unrelated) parameter sets: one for "bubble times", and one for "regular market time". It seems that price is strongly related to network size both in normal times and in bubble times, but with different parameters determining the exact price at the time.
sr. member
Activity: 364
Merit: 250
Sounds like you'd rather use market cap than coin price.  I use coin price because that is what everyone in this subforum is interested in.

If I have time I will update with most recent data tonight.


No, this is not what I mean. I mean coin price, but it has to take account of the creation of new coins.

Read again:

Quote
The formula will be:

Price = C x (Unique number of addresses used)^2 / (Number of bitcoins created) x C

where C is the appropriate constant (different from the one you used but approximatively equal to the one you use multiplied by 13.000.000).

C/C = 1.  So you are just proposing dividing by number of coins.  Why?  Price is already per coin.  As number of coins increases, it is reflected in the price.
newbie
Activity: 9
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Thanks for the chart Smiley and Hope Bitcoin prices will be higher...
sr. member
Activity: 311
Merit: 250
Sounds like you'd rather use market cap than coin price.  I use coin price because that is what everyone in this subforum is interested in.

If I have time I will update with most recent data tonight.


No, this is not what I mean. I mean coin price, but it has to take account of the creation of new coins.

Read again:

Quote
The formula will be:

Price = C x (Unique number of addresses used)^2 / (Number of bitcoins created) x C

where C is the appropriate constant (different from the one you used but approximatively equal to the one you use multiplied by 13.000.000).
sr. member
Activity: 364
Merit: 250
Sounds like you'd rather use market cap than coin price.  I use coin price because that is what everyone in this subforum is interested in.

If I have time I will update with most recent data tonight.
sr. member
Activity: 311
Merit: 250

Hi,

May I suggest that you improve the model dividing by the number of bitcoins created and adjusting the multiplicative coefficient?

If you think about it, this will give a better estimate. It gives a little correction to your graphs.

For example, if the number of bitcoins is twice, then the price will drop by half...thus we should keep track of the creation of bitcoins overtime.

The formula will be:

Price = C x (Unique number of addresses used)^2 / (Number of bitcoins created) x C

where C is the appropriate constant (different from the one you used but approximatively equal to the one you use multiplied by 13.000.000).
legendary
Activity: 1008
Merit: 1000
Minor suggestion: in the zoomed in graph, keep the axes labeled (I would find that helpful, anyway)
legendary
Activity: 1652
Merit: 1265
2017 looks promissing Wink

ya, why not 2027 2037 or 2047?

Good question. It all depends on how much money you need to be set for life (buy a nice house, etc.)
If I can buy a nice house for 2-5btc I will, thus making 2017 promissing Smiley
Ofcourse I will be holding on to btc all the way to 2047 and higher long after the final capitulation of the USD Tongue
member
Activity: 112
Merit: 10
2017 looks promissing Wink

ya, why not 2027 2037 or 2047?
legendary
Activity: 1652
Merit: 1265
2017 looks promissing Wink
sr. member
Activity: 364
Merit: 250
We have now hit an all time high for Bitcoin adoption, and it looks to continue its upward trend.  As you can see, Bitcoin's price is heavily correlated with the Metcalfe's Law value of Unique Addresses ^2.  I see no signs of this correlation changing, nor of adoption slowing.  This implies a continuation of the exponential rise in Bitcoin's price.  Sometimes adoption leads price, and sometimes price leads adoption.  You can see this if you analyze the long-term trend in the top chart.  At this point it looks like adoption is going to lead price.



Here is the chart zoomed into the right side of the chart for better analysis:


Edit: Updated 9/8/2014 with latest data.


I have updated the first post of the thread with the latest data.


An all-time-high for adoption should come with an all-time-high for price soon.  Deviations from correlation are generally due to speculation and short term technical market movements.  Price and adoption do eventually converge.  In the past when adoption rose without price, you can see that price eventually rejoined adoption and began the next bubble.
member
Activity: 112
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the more businesses "accept" bitcoin (immediately dump it on exchanges via market sells), the more it will crash.

Of course you ignore the fact that the graph proves you completely wrong, don't you?  Adoption and price are strongly correlated.  Please leave your trolling outside my thread.


sr. member
Activity: 364
Merit: 250
the more businesses "accept" bitcoin (immediately dump it on exchanges via market sells), the more it will crash.

Of course you ignore the fact that the graph proves you completely wrong, don't you?  Adoption and price are strongly correlated.  Please leave your trolling outside my thread.
member
Activity: 112
Merit: 10
the more businesses "accept" bitcoin (immediately dump it on exchanges via market sells), the more it will crash.
sr. member
Activity: 364
Merit: 250
We have now hit an all time high for Bitcoin adoption, and it looks to continue its upward trend.  As you can see, Bitcoin's price is heavily correlated with the Metcalfe's Law value of Unique Addresses ^2.  I see no signs of this correlation changing, nor of adoption slowing.  This implies a continuation of the exponential rise in Bitcoin's price.  Sometimes adoption leads price, and sometimes price leads adoption.  You can see this if you analyze the long-term trend in the top chart.  At this point it looks like adoption is going to lead price.



Here is the chart zoomed into the right side of the chart for better analysis:


Edit: Updated 9/8/2014 with latest data.


I have updated the first post of the thread with the latest data.
hero member
Activity: 870
Merit: 585
That's why I'm surprised. On the chart we can see a curve that fluctuates around a straight line. Correct? The vertical scale is log. Correct? Therefore the straight line is an exponential line. Because no other line can be straight in the log scale. Correct? Therefore rate of adoption (squared) fluctuates around an exponential rate. Correct? Exponential line squared is still an exponential line. Correct? Therefore rate of adoption grows at an exponential rate. Correct?

Right now adoption is increasing at a linear rate.  This means we gain more users at some average constant rate of new addresses per day.  The Metcalfe value is this value squared.  So, to approximate, N(y) = y^2, where N is the number of addresses and y the year number.  N increases exponentially as time marches on in a linear fashion.

As we approach the center of the S curve, adoption will begin increasing at an exponential rate.  The Metcalfe value is this exponential rate, squared.  So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number.  This is simplified to N(y) = y^4.  N increases much faster here.

Now the S-curve. It is basically an exponential curve that at it's last third slows down to horizontal line. Until that stage it's rate of growth is constant.
I will stop you here.  What you describe is not an S curve.  This is an S curve:


That's a linear chart.  OP is log chart.
Please don't get them mixed up.
newbie
Activity: 1
Merit: 0
The correlation between unique addresses and market capitalization seems to be better and makes more sense for me.

I calculated it with data from blockchain.info, starting 08/17/2010 and ending today.
The correlation coefficient for unique addresses and market price is 0.9278 and for unique addresses and market capitalization it is 0.9355.
legendary
Activity: 2884
Merit: 1115
Leading Crypto Sports Betting & Casino Platform
So far Bitcoin's adoption looks to continue its upward trend.  The rate of adoption has been on a consistent daily rise again after hitting a local dip on 5/3/2014.  As you can see, Bitcoin's price is heavily correlated with the Metcalfe's Law value of Unique Addresses ^2.  I see no signs of this correlation changing, nor of adoption slowing.  This implies a continuation of the exponential rise in Bitcoin's price.

First post updated with new data.  Adoption increases.  The price dump is thus not based on fundamental data, but more on speculation.


Thanks for the update
Anyways in my opinion its just the market selling some coins for fiat more than a price dump metcalfe still applies here more unique users will correlate with exponential increases in price over some duration.
sr. member
Activity: 364
Merit: 250
So far Bitcoin's adoption looks to continue its upward trend.  The rate of adoption has been on a consistent daily rise again after hitting a local dip on 5/3/2014.  As you can see, Bitcoin's price is heavily correlated with the Metcalfe's Law value of Unique Addresses ^2.  I see no signs of this correlation changing, nor of adoption slowing.  This implies a continuation of the exponential rise in Bitcoin's price.



Here is the chart zoomed into the right side of the chart for better analysis:


Edit: Updated 8/13/2014 with latest data.  We continue to have higher lows on the Metcalfe data.


First post updated with new data.  Adoption increases.  The price dump is thus not based on fundamental data, but more on speculation.
legendary
Activity: 1596
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Sine secretum non libertas
The math of the logistic (S-Curve) model is simple. f(x) = 1 / (1 + e-x).

It's the sort of math that can emerge out of a lot of complexity.  The CDF of a gaussian, for example, is a sigmoid a la style du logistique, but the gaussian arises as described by the central limit theorem, as the asymptotic distribution of a mean of (functionally) random processes.  And that concept of randomness may conceal a chaotic process - arguably always does, given axiomatic physical determism.  Or it might just be, for example, a lotka-volterra system in which the quasiperiodic factors are on incomensurate time scales, and hence not evident above the noise of observation.

Given that the system described (the living economy) is an agent system physically, I don't think you can have a satisfactory understanding of the domain of applicability of a simple model, unless you can show how it is caused by the agent interactions.  When I say satisfactory, I mean a model such that you know when it must apply, and when it may or must fail to apply.

Make no mistake, I like the model.  I think it is useful - until it is no longer useful; and, I think it is even more useful to have a principled understanding of why and (hopefully, even) when it will cease to be useful.  Meanwhile, it can serve as a pretext or stimulus to hypothesize theories which explain why the model is presently applicable, an important creative process, but one which should include in its scope hypotheses which imply evanescent application.

TL;DR: I just don't want anyone to be mislead into thinking that because a model is elegantly simple, and a good fit, that therefore it is reliable.  It is often a precondition of, or indicator of reliability.  It is rarely an assurance of reliability.  When the model is the result of a sound theory, then reliability is indicated more strongly, and its conditions begin to be understood.
legendary
Activity: 1806
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Learning the troll avoidance button :)
I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.


Funny how human psychology makes most humans buy when the price is rising , but no one dares buying when the price is low after a bubble. Even when it's pretty obvious a new bubble will come.

I mean right now we are sitting on 600s but people will only really start buying once we past 1000 or so. And then once we peaked at about 5000 or so, people will stop buying until we are once again past the previous ATH rather than just buying during the downtrend.

You know markets people want to wait for a strong momentum movement before they start piling onto the ride
Otherwise looking back you just spent 2 to 3 months going sideways well unless you started at 420 to 600 then that was a nice 30% return in a short time period Smiley
sr. member
Activity: 364
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Any chance of getting some updated data? Smiley

Here is the latest:

sr. member
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I'm not an expert at all but what I can say is that the adoption is still at the beginning and it's far away from the midpoint.
hero member
Activity: 686
Merit: 501
Stephen Reed
I have a thread on the logistic model as applied to bitcoin prices here . . .

https://bitcointalksearch.org/topic/stephen-reeds-million-dollar-logistic-model-366214

and a shared spreadsheet with graphs here . . .

https://docs.google.com/spreadsheet/ccc?key=0ArD8rjI3DD1WdFIzNDFMeEhVSzhwcEVXZDVzdVpGU2c

The math of the logistic (S-Curve) model is simple. f(x) = 1 / (1 + e-x). On a linear graph you get the familiar S-curve, but that is not very useful yet for bitcoin because prices have increased on average 10x every year so a log graph is best right now.

If bitcoin prices were to exactly follow the logistic function then the following properties would hold . . .

1. At the beginning starting from zero adoption, the growth is approximately exponential, decreasing rapidly near the midpoint.

2. At the midpoint of adoption, the growth is linear.

3. At the ending with full adoption, the growth is zero, decreasing rapidly after the midpoint.

Note that currently prices are substantially lower than my model projects. I do not believe that we are yet near the midpoint of adoption - so I expect either to revise the model lower or to witness a surge in prices sometime in the next few months.


legendary
Activity: 1596
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Sine secretum non libertas
i think metcalfe's law is wrong. it assumes all nodes are equally valuable, which is pretty controversial. here is a good rundown of why:

http://spectrum.ieee.org/computing/networks/metcalfes-law-is-wrong

perhaps growth is not quadratic, but rather n log(n), where network size = n.

It is probably ~n*log(n) in the early phase, and ~n*log(n) -> n in the mature phase, but in the hockey-stick phase, it is closer to n^2, while the highest value links are being aggregated to the network.
legendary
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Leading Crypto Sports Betting & Casino Platform
Any chance of getting some updated data? Smiley

That's a good question hard to believe that a six month old chart is already getting outdated  Grin
member
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Any chance of getting some updated data? Smiley
newbie
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Could we make a parallel to the internet itself using domain names instead of wallets? Not sure about this but just asking as this is the speculation forum Smiley

domain names 2002-2014, not far from an s-curve
http://www.registrarstats.com/GeneratedFiles/ChartImages/TldHistoryCom_7192014.png?32ad38f3-de49-4ea5-8946-10d708ec470c
It's getting flatter, have you tried to buy a domain name lately? Anything made with actual words or short or remotely nice sounding is already taken.

not sure how we value the internet, perhaps market cap tech/web companies 2006-2014 (sorry this graph suck no time to find a better one)
http://www.statista.com/statistics/216657/market-capitalization-of-us-tech-and-internet-companies

http://www.caseyresearch.com/sites/default/files/resize/AMZNSharePricing-490x355.jpg

Where we are on the adoption curve (avail as live charts on coinometrics)
http://www.genibits.com/wp-content/uploads/2014/04/Fitch_Bitcoin_Transaction-Volumes.jpg
sr. member
Activity: 364
Merit: 250
Which is following which? Higher prices = more attention = more addresses used?

It depends on where you look in the graph.  Sometimes prices leads adoption.  Sometimes adoption leads price.  In the center of the graph you can clearly see adoption heading up without price.  Eventually price follows up and rejoins adoption.  So there can be periods of divergence due to speculation.  But historically they always converge again.
hero member
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Merit: 608
Which is following which? Higher prices = more attention = more addresses used?
legendary
Activity: 1596
Merit: 1030
Sine secretum non libertas
logistic adoption occurs repeatedly, as btc breaks into widening circles of transactors.  expect the adoption curve to be fractal.  there have been periods in the past when superexponential fit the price better than exponential.  i expect that the latent network was adding a markov blanket in those times.
sr. member
Activity: 453
Merit: 254
i think metcalfe's law is wrong. it assumes all nodes are equally valuable, which is pretty controversial. here is a good rundown of why:

http://spectrum.ieee.org/computing/networks/metcalfes-law-is-wrong

perhaps growth is not quadratic, but rather n log(n), where network size = n.

The nodes joining the network first are the nodes receiving the most benefit, others will follow.
The nodes joining first are not, usually, the largest.
Often , the traffic increase a lot when some subset of nodes is able to form closed loops where there is a positive feedback loop.


sr. member
Activity: 448
Merit: 250
i think metcalfe's law is wrong. it assumes all nodes are equally valuable, which is pretty controversial. here is a good rundown of why:

http://spectrum.ieee.org/computing/networks/metcalfes-law-is-wrong

perhaps growth is not quadratic, but rather n log(n), where network size = n.
full member
Activity: 336
Merit: 100
I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?

I think it's unique addresses used in transactions PER DAY.

Thanks, that makes sense Smiley.
legendary
Activity: 1008
Merit: 1000
I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?

I think it's unique addresses used in transactions PER DAY.
full member
Activity: 336
Merit: 100
I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?
legendary
Activity: 1610
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i believe we will go super exponential this year and hit $10,000. 

just around the corner...
sr. member
Activity: 364
Merit: 250
I have updated the graph with the latest data.  We continue to have higher lows on the Metcalfe data.  I have also added a zoomed chart to the first post.  We can clearly see we're working ourselves into a tighter and tighter range on adoption.  We should see a large price move upward when we break above the Metcalfe high of June 1st.
hero member
Activity: 798
Merit: 1000
Who's there?
I think you are just playing at definitions and missing the larger point.  As rate of adoption increases beyond linear, rate of price appreciation will increase beyond the current exponential.  Until we pass the 50% point on the S curve, the rate of price appreciation should therefore accelerate, not diminish.

If you don’t believe me or don’t get it, I don’t have time to try to convince you, sorry. (c)-you know who  Wink
sr. member
Activity: 364
Merit: 250
N(y) = y^2, where N is the number of addresses and y the year number.  N increases exponentially as time marches on in a linear fashion. As we approach the center of the S curve, adoption will begin increasing at an exponential rate.  The Metcalfe value is this exponential rate, squared.  So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number.  This is simplified to N(y) = y^4.  N increases much faster here.
OK. I think I get the root of our misunderstanding. You seem to think that y^2 or y^4 is an exponent. It is not. If you don't trust me, ask somebody whom you trust. Smiley

Quote
I will stop you here.  What you describe is not an S curve.  This is an S curve:
rogerwilco already posted reference to the the S-curve aka logistic function: http://en.wikipedia.org/wiki/Logistic_function. In particular, this article says: "the logistic curve shows early exponential growth for negative argument, which slows to linear growth of slope 1/4 for an argument near zero, then approaches one with an exponentially decaying gap.". Which means exactly what I've said: exponential curve that slows down to horizontal line. Again, sorry for bothering you with calculus 101.

I think you are just playing at definitions and missing the larger point.  As rate of adoption increases beyond linear, rate of price appreciation will increase beyond the current exponential.  Until we pass the 50% point on the S curve, the rate of price appreciation should therefore accelerate, not diminish.
sr. member
Activity: 364
Merit: 250
I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.


Funny how human psychology makes most humans buy when the price is rising , but no one dares buying when the price is low after a bubble. Even when it's pretty obvious a new bubble will come.

I mean right now we are sitting on 600s but people will only really start buying once we past 1000 or so. And then once we peaked at about 5000 or so, people will stop buying until we are once again past the previous ATH rather than just buying during the downtrend.

Without buyers near the top, we'd have no one to sell to.  I plan to keep no more than 90% of my liquid wealth in Bitcoin.  I will need to sell at regular price intervals to keep that ratio.
legendary
Activity: 1106
Merit: 1005
I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.


Funny how human psychology makes most humans buy when the price is rising , but no one dares buying when the price is low after a bubble. Even when it's pretty obvious a new bubble will come.

I mean right now we are sitting on 600s but people will only really start buying once we past 1000 or so. And then once we peaked at about 5000 or so, people will stop buying until we are once again past the previous ATH rather than just buying during the downtrend.
hero member
Activity: 798
Merit: 1000
Who's there?
N(y) = y^2, where N is the number of addresses and y the year number.  N increases exponentially as time marches on in a linear fashion. As we approach the center of the S curve, adoption will begin increasing at an exponential rate.  The Metcalfe value is this exponential rate, squared.  So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number.  This is simplified to N(y) = y^4.  N increases much faster here.
OK. I think I get the root of our misunderstanding. You seem to think that y^2 or y^4 is an exponent. It is not. If you don't trust me, ask somebody whom you trust. Smiley

Quote
I will stop you here.  What you describe is not an S curve.  This is an S curve:
rogerwilco already posted reference to the the S-curve aka logistic function: http://en.wikipedia.org/wiki/Logistic_function. In particular, this article says: "the logistic curve shows early exponential growth for negative argument, which slows to linear growth of slope 1/4 for an argument near zero, then approaches one with an exponentially decaying gap.". Which means exactly what I've said: exponential curve that slows down to horizontal line. Again, sorry for bothering you with calculus 101.
legendary
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Leading Crypto Sports Betting & Casino Platform
Thanks for the chart I was wondering what the chart looks like normalized for Price Volatility
I assume that as the market price moves upward or downward that the price affects the users and unique transactions that occur in the network
That said more data is always a good thing
sr. member
Activity: 364
Merit: 250
That's why I'm surprised. On the chart we can see a curve that fluctuates around a straight line. Correct? The vertical scale is log. Correct? Therefore the straight line is an exponential line. Because no other line can be straight in the log scale. Correct? Therefore rate of adoption (squared) fluctuates around an exponential rate. Correct? Exponential line squared is still an exponential line. Correct? Therefore rate of adoption grows at an exponential rate. Correct?

Right now adoption is increasing at a linear rate.  This means we gain more users at some average constant rate of new addresses per day.  The Metcalfe value is this value squared.  So, to approximate, N(y) = y^2, where N is the number of addresses and y the year number.  N increases exponentially as time marches on in a linear fashion.

As we approach the center of the S curve, adoption will begin increasing at an exponential rate.  The Metcalfe value is this exponential rate, squared.  So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number.  This is simplified to N(y) = y^4.  N increases much faster here.

Now the S-curve. It is basically an exponential curve that at it's last third slows down to horizontal line. Until that stage it's rate of growth is constant.
I will stop you here.  What you describe is not an S curve.  This is an S curve:

hero member
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Who's there?
I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.
Sorry, one more correction: Super-exponential curve is a curve that looks exponential on log scale. Something like y=exp(exp(x)). While exponential curve, squared, is still just an exponential: y = (exp(x))^2 = exp(2x). In log scale it will still looks like straight line, just more steep, than non-squared one.

P.S. Sorry for nitpicking, nothing personal (and thanks for the OP), I just wanted to clarify some math details.
legendary
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the price will keep increasing with new merchants especially the big ones!
hero member
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Who's there?
I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No.  The rate of adoption is currently linear.  The Metcalfe value is exponential.  That is the number of unique addresses used per day, squared.  Vertical is an expression.  The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.

I *am* the OP.  The Metcalfe value displayed on the chart is the rate of adoption, *squared*.  That is the definition of the value from Metcalfe's Law.

That's why I'm surprised. On the chart we can see a curve that fluctuates around a straight line. Correct? The vertical scale is log. Correct? Therefore the straight line is an exponential line. Because no other line can be straight in the log scale. Correct? Therefore rate of adoption (squared) fluctuates around an exponential rate. Correct? Exponential line squared is still an exponential line. Correct? Therefore rate of adoption grows at an exponential rate. Correct?

Now the S-curve. It is basically an exponential curve that at it's last third slows down to horizontal line. Until that stage it's rate of growth is constant. Say, it grows 10x each year. It was growing this way at the beginning, it grows that way now, it will keep growing this way till it starts slowing down. The rate of growth in relative number, percentage-wise will never increase. That's why I said that S-curve has no vertical stage. The area that you are referring to as "vertical stage" is the area somewhere around 50% where the growth in absolute numbers, will reach the maximum. But the growth rate in relative numbers, in %, will be decreasing by then and will be lower than it is now.

Edit: typo
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

Actually, the S-curve is the colloquial term for the Logistic function, which can be intuitively understood as a system initially taking on exponential growth but eventually running into limiting factors.

The fastest rate of expansion ("going vertical") is halfway between the lower and upper limits. This inflection point represents a paradigm shift, because whereas before the growth could seemingly expand anywhere in the system, now it's become the norm so there are fewer places left to expand into. As an example, in the outside chance Bitcoin becomes the world reserve currency, this would be the point at which people would no longer value a bitcoin in dollars, but rather value a dollar in terms of bitcoins.
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No.  The rate of adoption is currently linear.  The Metcalfe value is exponential.  That is the number of unique addresses used per day, squared.  Vertical is an expression.  The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.

I *am* the OP.  The Metcalfe value displayed on the chart is the rate of adoption, *squared*.  That is the definition of the value from Metcalfe's Law.
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No.  The rate of adoption is currently linear.  The Metcalfe value is exponential.  That is the number of unique addresses used per day, squared.  Vertical is an expression.  The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.
sr. member
Activity: 364
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No.  The rate of adoption is currently linear.  The Metcalfe value is exponential.  That is the number of unique addresses used per day, squared.  Vertical is an expression.  The middle of an S curve looks nearly vertical.
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.
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can you explain metcalfe?
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Just speaking as a consumer, if I can buy bitcoin without fees and spend it at merchants that provide a better discount than my credit cards' 1% reward points

This and this.  If these two things happen, then the late adopter mass consumers will eventually get won over.  If not, then it's doubtful, sadly.  Today bitcoin still remains a very inconvenient and costly way to pay for shit.
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Sounds good, I'm hoping services like Circle (if they ever send me that invite I requested) will help quite a bit with mainstream adoption. Just speaking as a consumer, if I can buy bitcoin without fees and spend it at merchants that provide a better discount than my credit cards' 1% reward points, then I'll ditch the cards and switch to bitcoin and gift cards bought with bitcoin. I'm sure there are a ton of people out there who would also switch if they start seeing these discounts pop up.
sr. member
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I've been analyzing further.  When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential.  Metcalfe will go super-exponential at that point, along with price.
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We have now hit an all time high for Bitcoin adoption, and it looks to continue its upward trend.  As you can see, Bitcoin's price is heavily correlated with the Metcalfe's Law value of Unique Addresses ^2.  I see no signs of this correlation changing, nor of adoption slowing.  This implies a continuation of the exponential rise in Bitcoin's price.  Sometimes adoption leads price, and sometimes price leads adoption.  You can see this if you analyze the long-term trend in the top chart.  At this point it looks like adoption is going to lead price.



Here is the chart zoomed into the right side of the chart for better analysis:


Edit: Updated 9/8/2014 with latest data.
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