33:25, the idea that the functions on the R_inf sheet is mapped towards function g. But I wonder whether functions like g_0, g_1 are saddle points as well? Strogatz said that these sheets are not invariant, but I can't see how a R_n sheet is strictly mapped to a R_(n-1) sheet. The operation T can't really change the value of R_n to R_(n-1), so I suppose here we only rely on intuition that the degree of cycle is degraded, which symbolically change value of R_n to R_(n-1) in terms of its orbital nature. The bits remain is then to check the existence of a superstable 2^(i-1) cycle in the resultant functional. (In contrast, "T( gi ) = g_i-1 " can be proved algebraically without the intuition he stressed here.)
1. g_0, g_1, ... are not fixed points. So i guess they aren't saddle points? 2. T maps every function with a superstable parameter (R=n) representing all kinds of functions (sin, cosine, logistic, etc) with R = n to points for the same functions representing R=n-1 is what I understood
Professor Strogatz, this is another deep lecture on Renormalization, however the calculations gives a solid understanding of period doubling. The calculations is not meaningless.
17:10 Right, for g1. At least, as I see it, for a cycle with 2 values, it is a little bit as if we started with xa, then f(xa) = xb, then f(xb) = xa, ... in short, we have the sequence: xa, xb, xa, xb, xa, xb, ... Now, using f(f(x)) instead of f(x), the iteration leads to the sequence xa, f(f( xa) == xa, ... losing all occurences of xb, so we "degraded" from a cycle of 2 values to a cycle of one. NOW, the problem that I have is, with the same "spirit", a cycle of 3 values would "generate" the sequence, say, xa, f(xa)= xb, f(xb) = xc, f(xc)= xa, xb, xc, xa, xb, xc, ... nice, but then using f(f(x)) simply skips one step of each iteration will lead to the sequence: xa, xc, xb, xa, xc, xb, ... which is STILL a cycle of 3 values. It is NOT degraded to a cycle of 2 values. So, intuitively, T gi = g|i-1 would work only for the pre-chaos part, and won't work in the chaos intermixed with windows having cycles other than 2^n cycles. Right?
period 3 is born in a tangent bifurcation, while period doubling is born in a pitchfork bifurcation. As far as I can see, "T( gi ) = g_i-1 " only applies to the latter one. i.e. the set of periods that are connected with period doubling/pitchfork bifurcation.
I can’t think of any single lecture I’ve ever had through my bachelors or masters which I was able to completely digest within the 1-1.5 hours that the lecture was delivered in. And I think it was the case with most of my colleagues too. So, you’re not alone. Besides, everyone learns at their own pace; I don’t think it’s fair to judge yourselves if you take a little longer. As long as you were able to get it and maybe even develop some own thoughts and questions of your own :)
This is the best thing I’ve learnt in a long time! Thanks for uploading this! :)
33:25, the idea that the functions on the R_inf sheet is mapped towards
function g. But I wonder whether functions like g_0, g_1 are saddle
points as well?
Strogatz said that these sheets are not invariant, but I can't see how a R_n sheet is strictly mapped to a R_(n-1) sheet. The operation T can't really change the value of R_n to R_(n-1), so I suppose here we only rely on intuition that the degree of cycle is degraded, which symbolically change value of R_n to R_(n-1) in terms of its orbital nature. The bits remain is then to check the existence of a superstable 2^(i-1) cycle in the resultant functional.
(In contrast, "T( gi ) = g_i-1 " can be proved algebraically without the intuition he stressed here.)
1. g_0, g_1, ... are not fixed points. So i guess they aren't saddle points?
2. T maps every function with a superstable parameter (R=n) representing all kinds of functions (sin, cosine, logistic, etc) with R = n to points for the same functions representing R=n-1 is what I understood
Professor Strogatz, this is another deep lecture on Renormalization, however the calculations gives a solid understanding of period doubling. The calculations is not meaningless.
44:35
I can't understand why if the eigenvalue is -1 we have flip bifurcation???
Think of what happens to the slope of the logistic map at the fixed point when it reaches r=3 (the first flip bifurcation)…
17:10 Right, for g1. At least, as I see it, for a cycle with 2 values, it is a little bit as if we started with xa, then f(xa) = xb, then f(xb) = xa, ... in short, we have the sequence: xa, xb, xa, xb, xa, xb, ... Now, using f(f(x)) instead of f(x), the iteration leads to the sequence xa, f(f( xa) == xa, ... losing all occurences of xb, so we "degraded" from a cycle of 2 values to a cycle of one.
NOW, the problem that I have is, with the same "spirit", a cycle of 3 values would "generate" the sequence, say, xa, f(xa)= xb, f(xb) = xc, f(xc)= xa, xb, xc, xa, xb, xc, ... nice, but then using f(f(x)) simply skips one step of each iteration will lead to the sequence: xa, xc, xb, xa, xc, xb, ... which is STILL a cycle of 3 values. It is NOT degraded to a cycle of 2 values. So, intuitively, T gi = g|i-1 would work only for the pre-chaos part, and won't work in the chaos intermixed with windows having cycles other than 2^n cycles. Right?
period 3 is born in a tangent bifurcation, while period doubling is born in a pitchfork bifurcation. As far as I can see, "T( gi ) = g_i-1 " only applies to the latter one. i.e. the set of periods that are connected with period doubling/pitchfork bifurcation.
Is it normal that I need to spend more than 2 hours to finish these 1-hour recordings?
I can’t think of any single lecture I’ve ever had through my bachelors or masters which I was able to completely digest within the 1-1.5 hours that the lecture was delivered in. And I think it was the case with most of my colleagues too. So, you’re not alone.
Besides, everyone learns at their own pace; I don’t think it’s fair to judge yourselves if you take a little longer. As long as you were able to get it and maybe even develop some own thoughts and questions of your own :)
Take however long you need. This stuff is hard. Besides, didn't the tortoise win the race?
v stressed rn, shifting to the third board with 10 min left in the lecture