For lamda=0.1 and lamda=0.3, the squirrel population declines at a rate of 1/10 and 1/3 times that of the previous year per year, respectively. This means that the squirrel population decreases from that of the previous year approaching extinction. For lamda=1, the squirrel population remains constant from year to year maintaining its population. For lamda=2 and lamda=5, the squirrel population grows at a rate of 2 and 5 times that of the previous year per year, respectively causing a squirrel population explosion. In otherwords, growth and decay are exponential. When x=0, there can be no growth or decay, since we have no squirrels.
Now look at the slope of each graph. Do you see a relation between the long term behavior of the system and the slope of its graph?
The slope of 1 (or lamda=1) is the dividing line separating growth from decay. Slopes greater than 1 cause population growth to eventual overpopulation; Slopes less than 1 cause population decay and eventual extinction.
For negative lamdas, we can not discuss squirrels. Instead, we will approach it from the purely mathematical point of view. The behavior of the function for values of lamda less than -1 is negatively exponential and approaches negative infinity. For lamda=-1, the function is merely a mirror of the behavior of lamda=1 about the x-axis. For values of lamda greater than -1 and less than zero, we see that the function appoaches zero.
This can be related to the positive lamdas by simply noting that the relations of going to infinity for lamda greater than 1 mirrors for lamda less than negative 1. Likewise, the behavior of lamda greater than zero and less than 1 are also mirrored about the x-axis, thus creating two regions. One region approaches infinity and hte other approaches zero. The regions are divided by the lines y=x and y=-x.
Draw the graph of the square root function, together with the two special lines y=x and y=-x. What is special about the points where the graph of the square root function intersects one of the two lines?
Taking the square root repeatedly shows us that f(x) approaches one as x increases from zero to one and approaches infinity for values of x greater than 1. At x=1, there is no change in the value of f(x) when repeated square roots are taken.
Note that the function y=sqrt(x) only intersects y=x and y=-x at x=1, the point where repeated square roots always yield y=1.
The graph of f(x)=cos(x) where x is given in radians is bounded by the lines y=1 and y=-1. The points of intersection between y=cos(x) and y=(plus or minus)x are at x=(plus or minus)0.99984774, the point where (plus or minus)x=cos(x).
The fixed points of y=x^2 are at x=0 and x=1. These points are the same points as the intersections of y=x^2 with lines y=x and y=-x.
For x=0.1, with iteration, we see that f(x) approaches 0. For x=-0.1, with iteration, once again, f(x) approaches 0. For x=0.9, with iteration, not surprisingly, f(x) approaches 0, but for x=1.1, with iteration, f(x) approaches positive infinity. Values of the point between one and negative one will converge to zero. All other values of the point will diverge to positive infinity.
The fixed point x=0 is attracting, since points near it converge to it. The fixed point x=1 is repelling, since points near it diverge away from it.
As lamda increases from 1 to 4, f(x) changesfrom a convergence to 0 to oscillations between two values of y and eventually chaos between 0 and 1. At x0=0, the function is the straight line y=0. At x0=1, the function is the straight line y=1.
Is it attracting for all x0?
No. The attracting/repelling/neutral characteristics of the function are controlled by lamda.
Is it attracting for all lamda? If not, what happens right after it stopped being attracting?
No. Right after it stops attracting, it repells and then becomes neutral approaching chaos.
This will be hand-sketched and turned into your mailbox.
f(0)=3.2*0*(1-0)=0
f(11/16)=3.2*11/16*(1-11/16)=11/16
Therefore, x=0 and x=11/16 are fixed points for f(x).
x=(21+sqrt(21))/32=.799455
f((21+sqrt(21))/32)=3.2*(21+sqrt(21))/32*(1-(21+sqrt(21))/32)=0.5130445
f(f(21+sqrt(21))/32)=(f(0.5130445)=3.2*0.5130445*(1-0.5130445)=.799455
x=(21-sqrt(21))/32=0.5130445
f((21-sqrt(21))/32)=3.2*(21-sqrt(21))/32*(1-(21-sqrt(21))/32)=.799455
f(f(21+sqrt(21))/32)=(f(0.5130445)=3.2*0.5130445*(1-0.5130445)=.5130445
So, x=(21+sqrt(21))/32) and x=(21-sqrt(21))/32 are period-2 points.
The graph with lamda=2 has no period-2 points.
The graph with lamda=3.1 has a period-2 point at x=.323. It is repelling.
The graph with lamda=3.5 has a period-2 point at x=.283. It is repelling.
The graph will be presented by hand in your box. Period-4 is the period of the attracting orbit. The order of the attracting orbit here is x=.38 -> x=.83 -> x=.51 -> x=.87 -> x=.38.
The period-2 points attract to the points where the slope of the second iterate is 0. Points of neutrality occur where the second iterate curve crosses the line y=x. Attracting points lie in the area in which the second iterate slope is between -1 and 1. All other points are repelling.
The points of neutrality of the second iteration will always be intersection points of all period-n iterations where n is not 1.
Compute the slope of the graph of the Logistic map at the fixed point 0 in terms of lamda. What is the slope for the other fixed point?
f(0)=lamda*0*(1-0)=0
f((lamda-1)/lamda)=lamda*(lamda-1)/lamda*(1-(lamda-1)/lamda)
=(lamda-1)((lamda-(lamda-1)/lamda1))
=(lamda-1)/lamda
Therefore, x=0 and x=(lamda-1)/lamda are fixed points for f(x).
f'(x)=lamda-2*lamda*x
slope = f'(x)
f'(0)=lamda-2*lamda(0)=lamda
f'((lamda-1)/lamda)=lamda-2*lamda((lamda-1)/lamda)
=lamda-2(lamda-1)
=1-lamda
Values of lamda between 0 and 2 gives attracting fixed points. At lamda=2 and lamda=0, the fixed point is neutral, and at values of lamda greater than 2, the points repel.
The exact lamda value at which bifurcation occurs is 2.0. We see this by solving for lamda, since we know that at slopes of -1 and 1 the fixed point is neutral.
-1 is less than or equal to (1- lamda) is less than or equal to 1.
-2 is less than or equal to (- lamda) is less than or equal to 0.
Then, 0 is less than or equal to (lamda) is less than or equal to 2. This indicates that the exact lamda value of bifurcation (where the graph switches from attracting to repelling) is 2.
There is only one fixed point (and it is an attracting fixed point).
The bifurcations always double, and therefore the period always doubles at a bifurcation.
At lamda=-0.75 a bifurcation occurs, and we see a shift from attracting (for values larger than lamda=-0.75) to repelling (for values smaller than lamda=-0.75).
The orbit diagrams are the same shape, but that of f(x)=x^2+lamda is a rotation of 180 degrees of the other orbit diagram.
What is the range of lamda values in the diagram for the Logistic family?
The lamda values are greater than or equal to 0.
What is the range of lamda values in the diagram for the Quadratic family?
The lamda values are less than or equal to 0.
For lamda=4, the graph of the Logistic map is special. The interval [0,1/2] is mapped entirely onto [0,1] twice. Notice that this is not true for smaller (positive) values of lamda. Find the value for lamda such that the graph of f(x)=x^2+lamda covers the interval [-2,2] exactly twice, as described above for the Logistic map.
This occurs at lamda=-2.0.