using puthon

Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter r, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of r ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005). 3. Iterate the function 250 time steps (i.e. tmax=250) and plot the last 50 points (i.e. last 50 values of N) for each value ofr 4. To get the last k values of an array A, you can use A[-k:) 5. For each value of r, you will have to plot 50 values of N. Rather than looping through the values of N, it is easier to create an array the same length as N (e.g. Rplot) where all the values take on a single value of r. To do this, you can use something like Rplot=r*np.ones (50) or Rplot=np.full(50,r) (search the documentation to verify what these functions do). [ ] ##### 95: YOUR CODE HERE ########################################### r=(1.8,3.5.0.1) UHUR IU TK sure the behaviour shown in your cobweb plots makes sense by comparing your results to the time series plotted above (Le figure 1 of May 1974) 11 04: YOUR CODE HERE Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of r ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005). 3.Iterate the function 250 time steps (.e. tmax 250) and plot the last 50 points (.e. lasts values of N) for each value ofr 4. To get the last k values of an array A, you can use Ald] 5. For each value of you will have to plot 50 values of N. Rather than looping through the values of N. it is easier to create an array the same length as N (e.g. Rplot) where all the values take on a single value ofr. To do this, you can use something like Xplot np.ones (50) or plotep.full .) (search the documentation to verify what these functions do). 11 051 YOUR CODE WERE r-(1.8.3.5.0.1) Check to make sure the behaviour shown in your cobweb plots makes sense by comparing your results to the time series plotted above (l.e. figure 1 of May 1974) T1 041 YOUR CODE HERE I ! Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter r, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005) 3. Here the function 250 time steps (Lemax250) and plot the last 50 points (.e. last 50 values of N) for each value of 4. To get the last values of an array A you can use AH! 5. For each value ofr. you will have to plot 50 values of N. Rather than looping through the values of N, it is easier to create an array the same length as N (0.9. Rplot) where all the values take on a single value of. To do this, you can use something like Rplot np.ones (50) or plot op.full.(30.11 (search the documentation to verify what these functions do). 051 YOUR COOK HERE Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter r, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of r ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005). 3. Iterate the function 250 time steps (i.e. tmax=250) and plot the last 50 points (i.e. last 50 values of N) for each value ofr 4. To get the last k values of an array A, you can use A[-k:) 5. For each value of r, you will have to plot 50 values of N. Rather than looping through the values of N, it is easier to create an array the same length as N (e.g. Rplot) where all the values take on a single value of r. To do this, you can use something like Rplot=r*np.ones (50) or Rplot=np.full(50,r) (search the documentation to verify what these functions do). [ ] ##### 95: YOUR CODE HERE ########################################### r=(1.8,3.5.0.1) UHUR IU TK sure the behaviour shown in your cobweb plots makes sense by comparing your results to the time series plotted above (Le figure 1 of May 1974) 11 04: YOUR CODE HERE Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of r ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005). 3.Iterate the function 250 time steps (.e. tmax 250) and plot the last 50 points (.e. lasts values of N) for each value ofr 4. To get the last k values of an array A, you can use Ald] 5. For each value of you will have to plot 50 values of N. Rather than looping through the values of N. it is easier to create an array the same length as N (e.g. Rplot) where all the values take on a single value ofr. To do this, you can use something like Xplot np.ones (50) or plotep.full .) (search the documentation to verify what these functions do). 11 051 YOUR CODE WERE r-(1.8.3.5.0.1) Check to make sure the behaviour shown in your cobweb plots makes sense by comparing your results to the time series plotted above (l.e. figure 1 of May 1974) T1 041 YOUR CODE HERE I ! Bifurcation diagram for the logistic difference equation In the code cell below, write a script that summarizes the dynamics of the logistic equation as a function of the growth parameter r, in the form of a bifurcation diagram (like the one shown in class). Hints: 1. Copy and paste as needed from the scripts above 2. Use values of ranging from 1.8 to 3.5 in steps of 0.1 (once you are sure all is working ok, you can decrease the step size to 0.005) 3. Here the function 250 time steps (Lemax250) and plot the last 50 points (.e. last 50 values of N) for each value of 4. To get the last values of an array A you can use AH! 5. For each value ofr. you will have to plot 50 values of N. Rather than looping through the values of N, it is easier to create an array the same length as N (0.9. Rplot) where all the values take on a single value of. To do this, you can use something like Rplot np.ones (50) or plot op.full.(30.11 (search the documentation to verify what these functions do). 051 YOUR COOK HERE