An archived instance of discourse for discussion in a Fractal Introductory Colloquium.

# Iterating your personal logistic function

mark

Now that we've (almost) all created a Discourse account, updated our personal info, and uploaded at least one picture, let's try something a little more mathy. First, generate your own, personal logistic function on this web oracle that I set up for exactly that purpose. Then, reply to this post showing everybody

1. Exactly what your personal logistic function is,
2. The graph of your personal logistic function, and
3. The time series graph of the first few iterates starting from $x_0=0.1$.

A few comments are in order:

• You should enter the formula for you personal logistic function in LaTeX format. It's really not hard! If my function was $f(x)=x^2-1$, I'd enter $f(x)=x^2-1$. We'll practice in class.
• The graphs should be electronically produced logical tools include

All the graphs should look almost identical. If you examine the $y$-scale, however, you'll notice that their heights vary. You might mention the maximum height to help distinguish your graph.

The first two are, of course, very nice but you need an account to generate a PNG that you can upload here - unless, you're happy with a screen grab.

The time series plot is trickier but the tool on my iteration tutorial is designed to do exactly that.

Audrey

Sounds fun! According to the oracle, my personal logistic function is
$$f(x) = 3.832\,x(1-x).$$
The graph of my function looks like so:

I guess it's maximum value is $3.832/4 = 0.958$. The time series plot of the first 32 iterates from $x_0=0.1$ looks like:

I went with 32 iterates because I felt like a clear pattern had been established.

sjenkin1

## The Function

And the great Oracle dictates thus:
That the person known as Sam Jenkins has the function:
$f(x)=3.477x(x-1)$

## The Graph

Showing a graph such as this:

The Ymax $= 3477/4000=.86925$
The zeros of the graph are located at $x=0$, and $x=1$.

## The Iteration

The iteration ran to $n=40$
Once it settles into its pattern, the iterations approximately repeats every 4th N
Example:

31 0.4837844655412844
32 0.8683357452495927
33 0.39752116378825225
34 0.8327348524234404
35 0.4843027000260265
36 0.8683932490275537
37 0.3973738617245987
38 0.832629803957292
39 0.4845457268067225
40 0.8684195722351148

dbrown8

Hello, my beautiful and concise function is:

$$f(x)=2.809x(1-x)$$

$$The Iteration:$$

Wow, now that's'a nice'a time series!

Vanderbilt

Ok, so my function would be: $$f(x)=3.559x(1-x)$$

And then the graph would look like this:

(With Y Max = 3.559/4 = 0.88975)
(Courtesy of Wolfram Alpha)

Finally, the iteration would look like this:

(At 30 iterations)

tmorse

# Function

So, my personal logistic function is:
$$f(x)=4x(1-x)$$

# Graph

The graph that was spat out was:

The y max is $1$ and the zeros for this function are $(0,0)$ and $(1,0)$.

# Iteration

I went to 66 iterations because it shows kind of a pattern? It seems to change slightly over time.

katyagreb

My function is as follow:

## $$f(x)=3.68x−x^2$$

The graph that corresponds with this function is:

The maximum point is (0.5,0.92)

The time series plot of the first 35 iterates from $x_o=0.5$. It looks like the following:

I went with 35 iterates to show that certain patterns repeat themselves

Chief_Keith

$$f(x) = 3.649x(1-x).$$

My graph looks like this...

My maximum value is $0.91225 = 3.649/4.$ My time series plot is iterated 23 times from $x0 = 0.1.$
Now presenting... the WAVES OF DOOM!

I iterated 23 times to show an almost mirrored pattern.

eolberdi

The Oracle has given me the function of $f(x)= 2.609x(x-1)$
The Y Max (.5, .652)

0 0.1
1 0.23481000000000002
2 0.46877015451510007
3 0.649705433823331
4 0.5937778305655513
5 0.6293057195519385
6 0.6086276005950245
7 0.6214639158108171
8 0.6137581632597502
9 0.6184871404812228
10 0.615621720783381
11 0.6173718905351816
12 0.616307996765523
13 0.6169566217588124
14 0.6165618767672327
15 0.6168023750596617
16 0.6166559483157212
17 0.6167451357861584
18 0.6166908256621382
19 0.6167239023972066
20 0.6167037593593178

cbozarth

The function I was given is:
$$f(x)=3.173x(1-x)$$

The graph of this function is:

The iteration is:

afreundt

My personal logistic function, according to the Oracle is
$$f(x)=3.323*x(1-x)$$

The graph of my function looks like this:

Maximum value: $3.323/4=0.83075$

The time series of my function:

I chose 30 iterations from $x_0=0.1$. After Iteration 22, the series oscillates between the values of $0.8272$ and $0.4749$.

mark

Hi Dan - I reset the permissions so that you should be able to post more images in a single post now!

hcary

My functions is: $f(x)=2.929x(1-x)$
Which creates this graph:

With a y max of $2.929/4=.732$
The iteration:

With 50 iterates, which demonstrates that the graph converges to a number between .65 and .66
43 0.6612828129058163
44 0.6560604251310181
45 0.6609146259205149
46 0.6564078891898771
47 0.6605966199713604
48 0.6567073574399053
49 0.6603219732797869
50 0.6569655172442849

cjohns11

My personal logistic function is:
$$f(x)=2.978\,x(1-x).$$
The Graph for my function follows as such:

The maximum value: $2.978/4=.7445$

The Iteration for my function:

I chose 50 iterations from $x_o=0.1$.

By choosing 50 iterates, the graph oscillates between $.64$ and $.67$ as follows:

44 0.647481144393695
45 0.6797264512799152
46 0.6483058432713578
47 0.6790000122639799
48 0.6490818889250762
49 0.6783127301949141
50 0.6498132102057979

mark

@cjohns11 W00t! We're you able to use your own laptop?

The oracle gifted me this function:
$$f(x)=3.225*x*(1-x)$$
Which gave me this graph:

When I iterated this function, my $x_0$=0.1 and I iterated the function 40 times which made it look like this:

After about 15 iterations the graph begins to form a back and forth pattern, where every other iteration is within 0.001 of each other.

dcutchen

My personal logistic function is apparently: $$f(x)=3.375x(1-5)$$
The graph a the function appears like so:

The maximum is located at 0.844. This is the time series plot for the first 35 iterates at $x=0.1$:

There is a clear pattern that has been established.

cjohns11

Yes I was able to use my own laptop with the help of Firefox.

icrawfor

# Function

My Personal Logistic Function is:

$f(x)=3.141 x (1-x)$

(I love how close to Pi that number is)

# Graph

This function correlates with the following graph:

# Iteration

This function also yields the following iteration graph:

I ran the iteration 32 times and it seems to settle into a pattern after a while, alternating between .537 and .780

0 0.1
1 0.28269000000000005
2 0.6369205590099001
3 0.7263649207946281
4 0.6243017759922687
5 0.7367186241119122
6 0.6092418443084869
7 0.7477659952989842
8 0.5924303283523752
9 0.7584152886525112
10 0.5754988527131981
11 0.7673460588936701
12 0.5607504323380239
13 0.7736577781931043
14 0.5500249715844935
15 0.7773896544666458
16 0.543565690907502
17 0.779288478038437
18 0.5402455484958517
19 0.7801625091903079
20 0.5387096698832027
21 0.7805434045381127
22 0.5380388456524878
23 0.7807051181814998
24 0.5377537636444968
25 0.7807729861116517
26 0.5376340715779012
27 0.7808013283779704
28 0.5375840785148462
29 0.7808131397495184
30 0.5375632428887588
31 0.7808180577435393
32 0.5375545671362043

jhoneyc1

My personal logistic functions is as follows; $$3.936x(1-x)$$

$Graph$ My maximum value is 0.984 =3.936/4.

$Iteration$ $Graph$ $(50$ $Iterations)$