Essay on How Many Pieces

How Many Pieces?

If a line segment were to be cut into smaller segments with n number of cuts, each individual cut splits the segment into two, and any additional cut splits a newly cut segment it splits into two, which only increases the amount of segments by one; the amount of segments (S) created would be n+1. n=1 ————————————————|—————————————————
S=2
n=2
————————————|———————|—————————————
S=3 n=3 ————|————————————|————————————|————
S=4
n=20
—————|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—|—————
S=21

A line segment is a one-dimensional object, and so the rule n+1 would not necessarily apply to a object of different dimensions, such as a circle, a two-dimensional object.

Number of cuts (n) | Number of separated areas within the circle (R) | 0 | 1 | 1 | 2 | 2 | 4 | 3 | 7 | 4 | 11 | 5 | 16 |

The sequence seems to be moving in a parabolic fashion.

The sequence adds n to the f(n-1), so f(n-1)= f(n-2)+n-2 and the like,
The first term f(1)=2, so R=n+1+kn(n-1) because the third term of the equation must equal zero at f(1), so the k can then be solved for with the basic formula the n outside of the parentheses is to create a parabolic function: f(2)=R(2)=4=n+1+kn(n-1) with n=an integer ≥0
So:
4=(2)+1+k(2)(n-1)
1=k(2)(1)
so k=1/2
R=n+1+(1/2)n(n-1)

R=n+1+(1/2)n(n-1)=(1/2)n(n+1)+1

Which holds true for all values found through the recursive formula.

The arithmetic sequence

Rn-1= Rn-2+n-1
Rn can be expanded to Rn= n+(n-1)+Rn-2 which becomes, through the same process;
Because R(0)=1, there must be a 1+the sequence:
R=1+n+(n-1)+(n-2)+Rn-3

and so the sum of the arithmetic sequence is given by the formula:
Sn=(n/2)(R1+Rn)
So:
R=1+(n/2)(1+Rn)
and Rn=R1+ (n− 1)d so:(d=n) R=1+(n/2)(1+(1+n(n-1))
R=1+(n/2)(2+n(n-1))=n(1+n(n-1))=
1+(1/2)n(n+1)

So:
R=0.5n(n+1)+1
R=(0.5(n-1)n)+(n+1)

If using a three dimensional object, such as a cuboid: Number of Cuts (n) | Number of Regions (P) | 0 | 1 | 1 | 2 | 2 | 4 | 3 | 8 | 4 | 15 | 5 | 26 | n=1n=2n=3 n=4n=5

The sequence of the points seem to show a function that is too flat at close to the origin to be a parabolic function, perhaps it is a cubic function:
If the relationship is a cubic function, then the standard equation would equal:
P(n)=an3+bn2+cn+1
(The +1 is t make when n=0, P(n)=1, because, when there are no cuts, there is still one region.)

P(1)= a+b+c+1=2 a+b+c=1 P(2)=8a+4b+2c+1=4
8a+4b+2c=3

P(3)=27a+9b+3c+1=8
27a+9b+3c=7

8a+4b+2c=3
-2(a+b+c=1)
6a+2b=1 b=0.5-3a 27a+9b+3c=7
-3(a+ b+ c=1)
24a+6b=4
12a+3b=2

12a+3(0.5-3a)=2
3a=0.5
a=1/6

6(1/6)+2b=1 b=0 So: c=5/6 The Equation would be:
P(n)=(1/6)n3+(5/6)n+1
P(n)=(1/6)(n3-3n2+8n-6)+0.5(n2+3n+2)+(n+1)
Graphing this on the same plane as the points, and the equation proves true, as it contains every y-value for every n whole number equal to or greater than zero:

If the domain is extended, the equation still proves true.

So if P(5)=26, both from the reclusive and the conjectured equation, and

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