n = 5, R = 16
To find a recursive rule to the maximum amount of regions, we have to view the R cases.
R1 = 2
R2 = 4 → R1 + 2
R3 = 7 → R2 + 3
R4 = 11 → R3 + 4
R5 = 16 → R4 + 5
So the recursive formula for generating the maximum amount of regions is Rn=Rn-1+ n.
Relationship between maximum number of regions (R) and number of chords (n) using technology:
1. 2.
3.
Conjecture for maximum number of regions (R) and number of chords (n): R = (½)n 2 + (½)n + 1
n = 1→ R = 2, n = 2→ R = 4, n = 3→ R = 7, n = 4→ R = 11, n = 5→ R = 16
n
Ʃ (½)n 2 + (½)n + 1 = (2 + 4 + 7 + 11 + 16 + …..+n)
n=1
-
Check n = 1: (½)(1) 2 + (½)(1) + 1 = 2 ✔
k
Ʃ (½)k 2 + (½)k + 1 = (2 + 4 + 7 + 11 + 16 + …..+k)
k=1
k+1
Ʃ (½)(k +1) 2 + (½)(k + 1) + 1 = (2 + 4 + 7 + 11 + 16 + …+k + (k + 1))
k+1=1
Assuming the n = k is correct for all values of n, then the n = k + 1 should also work for
(2 + 4 + 7 + 11 + 16 + …+k + (k + 1))
(2 + 4 + 7 + 11 + 16 + …+k + (k + 1)) = (2 + 4 + 7 + 11 + 16 + …+k) + (k + 1)
Substitute n = k case to get (½)k 2 + (½)k + 1 + (k + 1)
Split (k + 1) into (k + ½ + ½) → (½)k 2 + (½)k + 1 + (k + ½ + ½)
Rearrange (½)k 2 + k + ½ + (½)k + ½ + 1
Factor out ½ → ½(k 2 + 2k + 1) + ½ (k + 1) + 1
½(k + 1)2 + ½ (k + 1) + 1 ✔
(2 + 4 + 7 + 11 + 16 + …+k + (k + 1)) = ½(k + 1)2 + ½ (k + 1) + 1
-
Therefore, by verifying the n = 1 case, and assuming the n = k and n = k + 1 case is true, we can assume that (2 + 4 + 7 + 11 + 16 + …..+k) = (½)k 2 + (½)k + 1
Rewrite formula in the form R = X + S
Find pattern for Rn.
R1 = 2 → 1 + 1 → R0 + S0
R2 = 4 → 2 + 2 → R1 + S1
R3 = 7 → 4 + 3 → R2 + S2
R4 = 11 → 7 + 4 → R3 + S3
R5 = 16 → 11 + 5 → R4 + S4
Rn = Rn - 1 + Sn – 1
R = X + S → X = R – S
Plug R and S: X = (½)n 2 + (½)n + 1 – (n + 1)→ = (½)n 2 – ½n
R = (½)n 2 – ½n + (n + 1 ) = (½)n 2 + (½)n + 1
For a finite, three-dimensional cuboid, the maximum number of parts (P) that are obtained with n cuts are
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}]}]
Code on Wolfram Alpha Mathematica:
Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}], Orange, Opacity [1], Polygon [{{0, 15, -15}, {0, 15, 15}, {0, -15, 15}, {0, -15, -15}}]}]
Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}], Orange, Opacity [1], Polygon [{{0, 15, -15}, {0, 15, 15}, {0, -15, 15}, {0, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 0}, {-15,15, 0}, {15, 15, 0}, {15,-15, 0}}]}]
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,3,-15},{-15,4,15},{15,4,15},{15,4,-15}}], Orange, Opacity [1], Polygon [{{-1, 15, -15}, {-1, 15, 15}, {-1, -15, 15}, {-1, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 4}, {-15,15, 4}, {15, 15, -9}, {15,-15, -9}}], Yellow, Polygon [{{-6, -15, -15}, {-6, 15, -15}, {15, 15, 15}, {15, -15, 15}}]}]
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,3,-15},{-15,4,15},{15,4,15},{15,4,-15}}], Orange, Opacity [1], Polygon [{{-1, 15, -15}, {-1, 15, 15}, {-1, -15, 15}, {-1, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 4}, {-15,15, 4}, {15, 15, -9}, {15,-15, -9}}], Yellow, Polygon [{{-6, -15, -15}, {-6, 15, -15}, {15, 15, 15}, {15, -15, 15}}], Red, Polygon [{{-15, -15, -15}, {-15, 15, -15}, {15, 15, 1}, {15, -15, 1}}]}]
Recursive:
P5 = 26→R4+P4 → 11 + 15
P4 = 15→R3+P3 → 7 + 8
P3 = 8→R2+P2 → 4 + 4
P2 = 4→R1+P1 → 2 + 2
P1 = 2→R1 → 2
Pn-1+Rn-1
Relationship between maximum number of part (P) and number of (n) cuts using technology:
1. 2.
3.
Conjecture for P = (1/6)n 3 + (5/6)n + 1
n
Ʃ (1/6)n 3 + (5/6)n + 1 = (2 + 4 + 8 + 15 + 26 + …..+n)
n=1
n
Ʃ (1/6)n 3 + (5/6)n + 1 = (2 + 4 + 8 + 15 + 26 + …..+n)
n=1
Check n = 1: (1/6)(1) 3 + (5/6)(1) + 1 = 2 ✔
Assume n = k
k
Ʃ (1/6)k 3 + (5/6)k + 1 = (2 + 4 + 8 + 15 + 26 + …..+k)
k=1
Assume n = k + 1
k+1
Ʃ (1/6) (k +1) 3 + (5/6)(k + 1) + 1 = (2 + 4 + 8 + 15 + 26 + …+k + (k + 1))
k+1=1
Assuming the n = k is correct for all values of n, then the n = k + 1 should also work for
(2 + 4 + 8 + 15 + 26 + …+k + (k + 1))
(2 + 4 + 8 + 15 + 26 + …+k + (k + 1)) = (2 + 4 + 8 + 15 + 26 + …+k) + (k + 1)
Substitute n = k case to get (1/6)k 3 + (5/6)k + 1 + (k + 1)
Rewrite formula in form of P = Y + X + S → Y = P – X + S.
S + X was found earlier to be (½)n 2 + (½)n + 1
Therefore, after plugging in the equations for P and (X + S), respectively,
Y = (1/6)n 3 + (5/6)n + 1 – (½)n 2 – (½)n – 1 → (1/6)n 3 – (½)n 2 + (1/3n)
Plug Y, (X + S), respectively: P = (1/6)n 3 – (½)n 2 + (1/3)n + (½)n 2 + (½)n + 1
Simplify P = (1/6)n 3 + (5/6)n + 1
For a finite four-dimensional object, a recursive formula can be determined by looking at the previous results:
By looking at the previous recursive formulae Rn = Rn-1+ n, Rn = Rn -1 + Sn – 1 and Pn = Pn-1 + Rn-1, it can be assumed that Qn = Qn-1+ Pn-1. It can also be assumed from the previous cases, that when n = 1, Q1 = 2. The numbers generated for the equation Qn = Qn-1+ Pn-1 are listed above for n = 1, 2, 3, 4, and 5.
1. 2.
3.
Conjecture by technology: Qn = (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1
Prove Conjecture:
n
Ʃ (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 = (2 + 4 + 8 + 16 + 31 + …..+n)
n=1
Check n = 1: (1/24)(1) 4 – (1/12)(1)3 + (11/24)(1)2 + (7/12)(1) + 1 = 2 ✔
Assume n = k
k
Ʃ (1/24)k 4 – (1/12)k3 + (11/24)k2 + (7/12)k + 1 = (2 + 4 + 8 + 16 + 31 + …..+k)
k=1
Assume n = k + 1
k
Ʃ (1/24)(k + 1)4 – (1/12)(k + 1)3 + (11/24)(k + 1)2 + (7/12)(k + 1) + 1 = (2 + 4 + 8 + 16 + 31 + …
k=1 +k +(k + 1))
Assuming the n = k is correct for all values of n, then the n = k + 1 should also work for
(2 + 4 + 8 + 16 + 31 + …+k + (k + 1))
(2 + 4 + 8 + 16 + 31 + …+k + (k + 1)) = (2 + 4 + 8 + 16 + 31 + …+k) + (k + 1)
Substitute n = k case to get (1/24)k 4 – (1/12)k3 + (11/24)k2 + (7/12)k + 1 + (k + 1)
In the form, Q = Z + Y + X + S, (Y + X + S) was found earlier to be (1/6)n 3 + (5/6)n + 1, so we need to solve for Z by plugging in (Y + X + S).
Q = Z + (1/6)n 3 + (5/6)n + 1
Plug in (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 for Q
(1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 = Z + (1/6)n 3 + (5/6)n + 1
Solve for Z: Z = (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 – ((1/6)n 3 + (5/6)n + 1)
Z = (1/24)n 4– (1/4)n 3 + (11/24)n2 – (1/4)n
Q = (1/24)n 4– (1/4)n 3 + (11/24)n2 – (1/4)n + (1/6)n 3 + (5/6)n + 1
Q = (1/24)n 4 – (1/12)n 3 + (11/24)n2 + (7/12)n + 1
By using the first dimension, second dimension and third dimension, I was able to find a rule using the recursive formulae from the previous dimensions to find a pattern for the fourth dimension. The equation Qn = Qn-1+ Pn-1 can be used to find the n amount of cuts for the fourth dimension, and assuming this pattern is accurate, the n amount of cuts can be found for the fifth, sixth, seventh…infinite dimensions. For example, using G as the maximum amount of cuts for the fifth dimension, the pattern for the segments resulted from n amount of cuts can be found by doing Gn = Gn-1+ Qn-1, assuming that G1 is 2 because of the previous patterns. The resulting numbers can be found by using a regression test, for example, the equation for the fourth dimension, Qn = (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1, was found by using a quartic regression. This use of finding a pattern and using technology to find the equation for the pattern and verifying the pattern can help find the maximum number of parts made by n number of cuts for an object with infinite dimensions.
