tag:blogger.com,1999:blog-5304409.post112762724729988257..comments2024-03-17T16:17:20.145-07:00Comments on Amit's Thoughts: The Singularity is Near?Amithttp://www.blogger.com/profile/12159325271882018300noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-5304409.post-16724316612702101892012-11-06T09:04:51.957-08:002012-11-06T09:04:51.957-08:00Attempting to define "knee".
The knee p...Attempting to define "knee".<br /><br />The knee point in a current transformer is the essentially point at which the transformer begins to saturate. At that point the transformer output becomes nonlinear.<br /><br />In computer systems, the knee point is: "Computer systems often reach a point at which the relative cost to increase some tunable parameter is no longer worth the corresponding performance benefit."<br /> --<i>Finding a “Kneedle” in a Haystack: Detecting Knee Points in System Behavior</i> by Ville Satop, Jeannie Albrecht, David Irwin, and Barath Raghavan<br /><br />The knee point in an ecological system might be defined as the point at which a change in a specific variable the system is irreversible from a pragmatic human perspective.<br /><br />Yes, there is no knee point one can mathematically specify on an exponential curve. Yet there may be a point of 'no return' when applied to a real-world system expressing exponential behavior.<br /><br />See<br />http://en.wikipedia.org/wiki/Exponential_growth<br /><br />where the story of the Lily Pond is presented:<br />French children are told a story in which they imagine having a pond with water lily leaves floating on the surface. The lily population doubles in size every day and if left unchecked will smother the pond in 30 days, killing all the other living things in the water. Day after day the plant seems small and so it is decided to leave it to grow until it half-covers the pond, before cutting it back. They are then asked, on what day that will occur. This is revealed to be the 29th day, and then there will be just one day to save the pond. (From Meadows et al. 1972, p. 29 via Porritt 2005)Gordonnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-77496725277062700242010-11-03T22:50:09.938-07:002010-11-03T22:50:09.938-07:00A minor correction. Exponential functions f(x)=exp...A minor correction. Exponential functions f(x)=exp(x) are not "scale-free". This is evidenced by the fact that the "knee" "moves" when you rescale. Were exp(x) scale-free nothing would move when you rescale. Mathematically "scale-free" for real functions is defined as f(x)dx=f(x')dx' for x'=a*x. In English scaling the variable scales the function in an equal and inverse manner. For example the function f(x)=1/x is scale-free<br><br>x' = a*x<br>f(x) = 1/x<br><br>f(x)dx = f(x')dx'<br>dx/x = f(a*x)d(a*x)<br>dx/x = a*dx/(a*x)<br>dx/x = dx/x<br><br>Now try this with exp(x)<br><br>x' = a*x<br>f(x) = exp(x)<br><br>exp(x)dx = exp(a*x)d(a*x)<br>exp(x)dx = a*exp(a*x)dx<br><br>which holds only for the special case of a=1 (no scaling) but is false otherwise.<br><br>Furthermore, if you allow for a constant (actually a function of the function f) c(f) scaling factor in the definition of scale-free f(x)dx=c(f)*f(x')dx' then the function f(x)=1/x is even scale-free for power scaling of the variable<br><br>x' = a*x^p<br>f(x) = 1/x<br><br>c = c(f)<br>f(x)dx = c*f(x')dx'<br>dx/x = d(a*x^p)/(a*x^p)<br>dx/x = c*a*p*x^(p-1)dx/(a*x^p)<br>dx/x = c*p*dx/x<br>c -> 1/p<br>dx/x = dx/x<br><br>Thus again scale free if we allow a constant 1/p which if a function only of the function f. This is one reason why 1/x is used in Bayesian analysis as a prior distribution for an unknown scaling factor, because it is scale free an thus imposes no information about the scale. In this usage it is called Jeffrey's prior.<br><br>Where is the hidden scale in exp(x)? It is that which multiplies x. In other words the "scale" or "lifetime" or "half-life" or "time constant" etc of the exponential function which is typically written as<br><br>f(x) = exp(x/t) = exp(k*x)<br><br>which shows up all over the place in science for one. Here is the exponential probability distribution<br><br>p(x|tau) = tau*exp(x/tau)<br><br>which endows the exponential distribution with the unambiguous scale tau which is also the mean. The fact that when tau = 1 exp(x/tau) can be written as exp(x) hiding tau does not make exp(x/tau) scale-free. The scale simply is 1.<br><br>Hmm ... I hope that long wind helped somewhat.<br><br>Keith H Duggar<br>duggar youknowwhattodo alum.mit.eduKeith H Duggarnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1136455886117954422006-01-05T02:11:00.000-08:002006-01-05T02:11:00.000-08:00e^2x - e^x has a definite knee at x=-ln(2) irrespe...e^2x - e^x has a definite knee at x=-ln(2) irrespective of scaling.<BR/><BR/>The question is I suppose whether e^2x-e^x counts as an exponential expression. I don't see why notAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1134848023323186702005-12-17T11:33:00.000-08:002005-12-17T11:33:00.000-08:00A minor correction. Exponential functions f(x)=exp...A minor correction. Exponential functions f(x)=exp(x) are not "scale-free". This is evidenced by the fact that the "knee" "moves" when you rescale. Were exp(x) scale-free nothing would move when you rescale. Mathematically "scale-free" for real functions is defined as f(x)dx=f(x')dx' for x'=a*x. In English scaling the variable scales the function in an equal and inverse manner. For example the function f(x)=1/x is scale-free<BR/><BR/>x' = a*x<BR/>f(x) = 1/x<BR/><BR/>f(x)dx = f(x')dx'<BR/>dx/x = f(a*x)d(a*x)<BR/>dx/x = a*dx/(a*x)<BR/>dx/x = dx/x<BR/><BR/>Now try this with exp(x)<BR/><BR/>x' = a*x<BR/>f(x) = exp(x)<BR/><BR/>exp(x)dx = exp(a*x)d(a*x)<BR/>exp(x)dx = a*exp(a*x)dx<BR/><BR/>which holds only for the special case of a=1 (no scaling) but is false otherwise.<BR/><BR/>Furthermore, if you allow for a constant (actually a function of the function f) c(f) scaling factor in the definition of scale-free f(x)dx=c(f)*f(x')dx' then the function f(x)=1/x is even scale-free for power scaling of the variable<BR/><BR/>x' = a*x^p<BR/>f(x) = 1/x<BR/><BR/>c = c(f)<BR/>f(x)dx = c*f(x')dx'<BR/>dx/x = d(a*x^p)/(a*x^p)<BR/>dx/x = c*a*p*x^(p-1)dx/(a*x^p)<BR/>dx/x = c*p*dx/x<BR/>c -> 1/p<BR/>dx/x = dx/x<BR/><BR/>Thus again scale free if we allow a constant 1/p which if a function only of the function f. This is one reason why 1/x is used in Bayesian analysis as a prior distribution for an unknown scaling factor, because it is scale free an thus imposes no information about the scale. In this usage it is called Jeffrey's prior.<BR/><BR/>Where is the hidden scale in exp(x)? It is that which multiplies x. In other words the "scale" or "lifetime" or "half-life" or "time constant" etc of the exponential function which is typically written as<BR/><BR/>f(x) = exp(x/t) = exp(k*x)<BR/><BR/>which shows up all over the place in science for one. Here is the exponential probability distribution<BR/><BR/>p(x|tau) = tau*exp(x/tau)<BR/><BR/>which endows the exponential distribution with the unambiguous scale tau which is also the mean. The fact that when tau = 1 exp(x/tau) can be written as exp(x) hiding tau does not make exp(x/tau) scale-free. The scale simply is 1.<BR/><BR/>Hmm ... I hope that long wind helped somewhat.<BR/><BR/>Keith H Duggar<BR/>duggar youknowwhattodo alum.mit.eduAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1132327540861375072005-11-18T07:25:00.000-08:002005-11-18T07:25:00.000-08:00This had vaguely bugged me too, so it's great to s...This had vaguely bugged me too, so it's great to see formulated.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1127844968102187892005-09-27T11:16:00.000-07:002005-09-27T11:16:00.000-07:00I think the "knee" here is more of a psychological...I think the "knee" here is more of a psychological/perceptual one. I'm not a Kurzweil disciple, but I have to believe he understands the mathematics of exponentiation.<BR/><BR/>Anyway, to give a wishy-washy definition of what he might mean by the "knee": The knee is the point where the curve crosses the threshold where it begins to affect the daily life of most* people. <BR/><BR/>So, look at the internet through the 80s - no one on the street knew what it was, and then in the early 90s middle-aged engineers and nerds started paying for compuserve accounts, and now you can search google on your phone. The knee in this case happened sometime in between when the pop-science rags started doing "information superhighway" cover stories and the year 2000.<BR/><BR/>Computation is another easy example. Through most of this century, computing devices were used only by researchers, governments, and huge corporations. Then, some time in the 70s people started buying 4 function electronic calculators that cost $100 and had to be plugged in to the wall.<BR/><BR/>So, obviously, this sort of knee eludes formalism, which makes it a bit of a shaky point to predict the future from. But I think there is something there...<BR/><BR/>* Where by "most" I mean "most people in the developed world". My big complaint with Kurzweil is that he forgets that most of the world is still living in squalor...Markhttps://www.blogger.com/profile/00251649932665374955noreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1127801639892518492005-09-26T23:13:00.000-07:002005-09-26T23:13:00.000-07:00This had vaguely bugged me too, so it's great to s...This had vaguely bugged me too, so it's great to see formulated. A related point which should be even more obvious (but I can imagine missing it) is that exponentials don't have vertical asymptotes.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5304409.post-1127774324801463972005-09-26T15:38:00.000-07:002005-09-26T15:38:00.000-07:00Hi, I think when Kurzweil is talking about knees, ...Hi, I think when Kurzweil is talking about knees, he is talking about his specific graphs, in which he has arranged the graph scale to extend to cover the particular timeperiod he is interested in discussing.<BR/><BR/>So, on those particular graphs, there is a visible knee.<BR/><BR/>He has arranged the graphs to portray the particular period of time that he wants to discuss.<BR/><BR/>I don't think these "graphology" issues detract from his primary arguments and data which DO demonstrate exponential technological development. So don't let it get under your skin too much.Anonymousnoreply@blogger.com