I am having issues completing this. Please tell me my mistake. CHEGG ONLY ALLOWS $METO$ POST ONE PICTURE $AT$ A TIME $SO$ I WILL SEND THE OTHER TWO SEPERATLY $AS$ CODE $1$ AND CODE $2$ Bifurcation Diagram for the Logistic MapBifurcation Diagram for the Logistic Map

Compute the bifurcation diagram for the logistic map. The logistic map is given by

$x_{n+1}=x_n(1-x_n),$

Here is a resonable outline:

Loop $1$ Start at $=2\mathrm{.}4$ and finish at $=4.$

Set $x=x_0$

Loop $2$ Iterate logistic map a fixed number of times $($transient$)$

Compute $x$

Loop $2($end$)$

Loop $3$ Iterate logistic map a fixed number of times $($data$)$

Compute and save $x$

Loop $3($end$)$

Loop $1($end$)$

Warning #$1$: The array data will be graded. To pass the assessment, the indices and values in your $x_{?}$data must match exactly the indices and values in my $x\_$data. That

another $n_{?}$data times. The values of $x$ you obtain from these second $n_{?}$data iterations are to be stored in $x_{-}$data and are then graded. If your values in $x\_$data do not exactly

match mine, the grader will not pass you.

Warning #$2$: You will need to write your iteration in the form $x={}^{**}{x}^{**}(1-x).$ Roundoff errors and the butterfly effect for chaotic systems will prevent you from passing the

assessment if you write the equation in a different form, such as $x={}^{**}(x-x^{{?}^{{?}^{?}}}2)$ or $x=x^{**}(1-x{)}^{**}.$ You can try and compare the different results!Bifurcation Diagram for the Logistic Map

Compute the bifurcation diagram for the logistic map. The logistic map is given by

$x_{n+1}=x_n(1-x_n),$

Here is a resonable outline:

Loop $1$ Start at $=2\mathrm{.}4$ and finish at $=4.$

Set $x=x_0$

Loop $2$ Iterate logistic map a fixed number of times $($transient$)$

Compute $x$

Loop $2($end$)$

Loop $3$ Iterate logistic map a fixed number of times $($data$)$

Compute and save $x$

Loop $3($end$)$

Loop $1($end$)$

Warning #$1$: The array data will be graded. To pass the assessment, the indices and values in your $x_{?}$data must match exactly the indices and values in my $x\_$data. That

another $n_{?}$data times. The values of $x$ you obtain from these second $n_{?}$data iterations are to be stored in $x_{-}$data and are then graded. If your values in $x\_$data do not exactly

match mine, the grader will not pass you.

Warning #$2$: You will need to write your iteration in the form $x={}^{**}{x}^{**}(1-x).$ Roundoff errors and the butterfly effect for chaotic systems will prevent you from passing the

assessment if you write the equation in a different form, such as $x={}^{**}(x-x^{{?}^{{?}^{?}}}2)$ or $x=x^{**}(1-x{)}^{**}.$ You can try and compare the different results!Bifurcation Diagram for the Logistic Map

Compute the bifurcation diagram for the logistic map. The logistic map is given by

$x_{n+1}=x_n(1-x_n),$

Here is a resonable outline:

Loop $1$ Start at $=2\mathrm{.}4$ and finish at $=4.$

Set $x=x_0$

Loop $2$ Iterate logistic map a fixed number of times $($transient$)$

Compute $x$

Loop $2($end$)$

Loop $3$ Iterate logistic map a fixed number of times $($data$)$

Compute and save $x$

Loop $3($end$)$

Loop $1($end$)$

Warning #$1$: The array data will be graded. To pass the assessment, the indices and values in your $x_{?}$data must match exactly the indices and values in my $x\_$data. That

another $n_{?}$data times. The values of $x$ you obtain from these second $n_{?}$data iterations are to be stored in $x_{-}$data and are then graded. If your values in $x\_$data do not exactly

match mine, the grader will not pass you.

Warning #$2$: You will need to write your iteration in the form $x={}^{**}{x}^{**}(1-x).$ Roundoff errors and the butterfly effect for chaotic systems will prevent you from passing the

assessment if you write the equation in a different form, such as $x={}^{**}(x-x^{{?}^{{?}^{?}}}2)$ or