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
The binomial coefficients
are defined recursively as
follows:
1.
for all
integers

2.
for all integers
3.
for all integers
Plot the central binomial coefficient
of

as a function
for
Find upper bounds for, or determine, the rate of growth
of
as a function of
.
A.) To obtain the following function in Mathematica I plugged in the
following code section. The following graphs are also a great visual
representation.
REDDOT[n_, k_] :=
REDDOT[n, k] = ReleaseHold[REDDOT[n - 1, k] + REDDOT[n - 1, k - 1]];
REDDOT[n_, 0] = 1;
REDDOT[0, k_] = 0;
T = Table[{n, REDDOT[(2*n), n]}, {n, 0, 25}];
ListPlot[T]
ListLogPlot[T]
The upper bounds of C(2n,n) as a function of n can be calculated using
the Stirling’s Formula. Developed by James Stirling, this formula gives
an approximation for large factorials. The simple lower and upper
bounds are given by:
. More accurate bounds are given
by:
polynomial
. In Big-Oh notation, the upper bound
is a
. The upper bounds of this
graph give us all the information we need to know as we can calculate
any values needed.
I also used mathematics fit function in order to obtain another
possible option. The code follows this excerpt.
Fit[T, {0, n}, n]
= 0.+ 7.57432*10^11 n
B.)
WALK THE GRID
Imagine walking on the grid below, from the red dot (the “origin”) to
any other dot by walking UP one step or RIGHT one step at a time:

Show that the number of different paths from the origin to a dot
that is k steps UP and n-k steps RIGHT (or vice versa),
where

is C(n,k)
For the dots that are 15 steps away from the origin calculate how
many different paths there are to that dot.

Use Mathematica to expand
coefficients of the powers
and examine the
. Is this a coincidence?
Taking only one step in the upward or rightward direction when trying
to maneuver your way through this maze can be looked at as pascals
triangle. To show what I mean by this I have attached a picture to
show the process.
This image was taken from wikipedia. It explains perfectly how
walking through this grid would occur. You simply add the numbers of
the 2 adjacent honeycombs and it subsequently gives you the answer
to the following spot.
The above table taken from a google image shows us exactly what our
value would be at row (15,15) which is 40116600.
In expanding
we get the following equation
This equation shows that no matter what the values for a or b are they
will always add up to 15. There is also nothing special about 15 as you
can do this with any number in mathematica and recieve very similar
results.
BLOCKS
A block tower of 2 colors is built as a vertical stack of blocks, with a
bottom and a top. Below are several block towers of height 5:

How many block towers of height 5 are there and what are
they?

What does this have to do with the binomial expansion
of

?
What does it have to do with the number of walks on the grid,
above, to points that are 5 steps from the origin?

What connection is there between block towers of height n,
the binomial expansion of
and walks on the grid to
points that are n steps from the origin?
When first looking at this problem I didn’t know how to approach it but
after reevaluating it I realized after
expanding
That
there could be 32 different combinations according to the expansion.
This is derived by taking the fact that there is two possible states for
each block. A block can either be black or white. When there is one
block the formula gives you two states, which is correct. When there
are two blocks, the formula will give you four possible states.
EX.) 2^N= Number of combinations, Where N = number of blocks ;
2^(5)= 32
Dijkstra’s fusc function
We define the function fusc(n) and plot it as the following image will
show.
for values known as
the following code gave me the values for n<100
fusc[n_]:=If[EvenQ[n]==True,fusc[n/2],fusc[(n-1)/2]+fusc[(n+1)/2]];
fusc[0]:=0;
fusc[1]:=1;
t1=SessionTime[];
A=Table[{n,fusc[n]},{n,0,99}];
Grid[A]
For every value of n that is a multiple of 3, the table says true if fusc(n)
is even. The table shows that this is true for every multiple of three.
For proof by induction i have used a couple different fusc functions
below making sure to walk through every step in order to show that
through induction each is true. For the example I will be using fusc(6)
but to find fusc(6) you must first find fusc(3)
fusc[3] = fusc[(3-1)/2] + fusc[(3+1)/2] = fusc[1] + fusc[2] = 1 + 1 = 2
fusc[6] = fusc[6/2] = fusc[3] = 2
Since the two outputs have the same number it shows that the fusc
function is correct and that for every value of n that is a multple of 3
the fusc n will be even.
For proof that fusc[n] is not a multiple 2 when n is not a multiple of 3
please follow the next proof by induction.
fusc[4] = fusc[4/2] = fusc[2] = fusc[2/2] = fusc[1] = 1
fusc[7] = fusc[(7-1)/2] + fusc[(7+1)/2] = fusc[3] + fusc[4] = 2 + 1 = 3
The previous examples show how
Fibonacci numbers
The fibonacci numbers f(n) are determined by the following recursive
relation:
Use Mathematica to calculate and plot
The growth rate is growing exponentially, so f(n) will increase
proportionally to n so that means they will have the same growth rate.
The above code and graph were produced using mathmatica given to
find the growth rate. Via the graph 1.61803 is the b term in our a*b^n
Formula. Using the log function to find a slope that matches one of the
graphs we find out that the growth rate of the graph is approximately
1.58.
An example of the growth rate of Fn there is a chart below that
describes the situation a lot better and gives an example of how the
growth rate is increasing in our case.
Generating Primes
The graph below shows Computation of dp(n) or the number of distinct
primes generated as g(n):=a(n)-a(n-1), n>1 along with the graph of dp(n)
as a function of n:
As we can tell from the graph dp(n) is always increasing as n becomes
bigger. With this in mind we can also say that it is getting slower and
slower the further we go through the graph therefor it is increasing at
a slow rate. For example just in order to get a distinct 11 prime
numbers it would take a value of n to be close to 960 . The function is
not one-to-one because it has the same value for dp(n) for multiple
values of n. It also appears to be increasing exponentially as a lot of
the graphs in this project have been.