- Level: International Baccalaureate
- Subject: Maths
- Word count: 2313
In this essay, I am going to investigate the maximum number of pieces obtained when n-dimensional object is cut, and then prove it is true.
Extracts from this document...
Introduction
How many pieces
Mathematics can be considered to be the study of patterns. A useful ability in maths can be forming a rule to describe a pattern. Of course any rule that we develop must be true in all relevant cases. In this essay, I am going to investigate the maximum number of pieces obtained when n-dimensional object is cut, and then prove it is true.
- One-dimensional object
Processing: Find the maximum number of segments that can be cut in n-cuts of a line segment. To begin the solution, consider the results for n=1, 2,3,4,5.
Let S represent the maximum number of segments in n-cuts of a line segment. The value for S is shown in the table.
n | 1 | 2 | 3 | 4 | 5 |
S | 2 | 3 | 4 | 5 | 6 |
Plotting the points related to the variables
n and S, suggests that the relationship
between them could be linear, and so we
might assume that y=kn+1.
Substituting the first value for n gives:
n=1, 2=k+1, k=1
Therefore, the rule which related the
maximum number of segments obtained
from n cuts is S=n+1.
- Two-dimensional object
Processing: Find the maximum number of regions that can be obtained when n chords are drawn. To begin the solution, consider the results for n=1, 2,3,4,5.
Let R represent the maximum number of pieces in n-cuts of a line segment. The value for R is shown in the table.
n | 1 | 2 | 3 | 4 | 5 |
R | 2 | 4 | 7 | 11 | 16 |
Middle
4
8
15
26
①Recursive rule:
P1=2=R1
P2=4=2+2= R1+P1
P3=8=4+4= R2+P2
P4=15=7+8= R3+P3
P5=26=11+15= R4+P4
Thus Pn= Pn-1+Rn-1
= Pn-2+ Rn-2+……+Rn-1
=Pn-(n-1) +Rn-(n-1) + Rn-2+……+ Rn-1
=P1+R1+ R2+ R3+……+ Rn-1
= 2+(1^2+1+2)/2+(2^2+2+2)/2 +……+ (n^2+n+2)/2
= 2+(n-1)+[1*2+2*3+3*4+……+n*(n-1)]/2
= n+1+ (n-1)*n*(n+1)]/ (3*2)[e]
= (n^3+5n+6)/6
Therefore, the recursive rule to generate the maximum number of parts is
(n^3+5n+6)/6
When n=5, P5= (5^3+5*5+6)/6=16, which corresponds to the tabulated value for n=5 above.
②A conjecture for the relationship between the maximum number of parts(P) and the number of cuts(n).
Use TI-84 to sketch[f] the
graph related to the variables n and P
suggests that the relationship between
them could be cubic, and so we
might assume that
P=an³+bn²+cn+d
Substituting all the values for
n gives:
n=1
2=a+b+c+d
n=24=8a+4b+2c+d
n=38=27a+9b+3c+d
n=415=64a+16b+4c+d
n=526=125a+25b+5c+d
Solve these five equations for
a, b, c, d gives
a=1/6, b=0, c=5/6, d=1
Thus Pn =1/6n³+5/6n+1
When n=5, P5=1/6*5³+5/6*5+1, which is also corresponds to the tabulated value for n=5 above.
Pn = (n^3+5n+6)/6 is the same with Pn =1/6n³+5/6n+1
So far we have formed a conjecture that the maximum number of parts from n-cuts is given by Pn = (n^3+5n+6)/6.
Proof:
Let T(n) be the proposition that the maximum number of parts that can be separated from n-cuts is given by P= (n^3+5n+6)/6 for n>0
Step1: T(n) is true for n=1 as P= (1^3+5+6)/6=2 which is the maximum number of parts from 1 cut.
Step2: Assume that T(n) is true for a k-cuts cuboid i.e., that Pk= (k^3+5k+6)/6. We consider the effect that adding an extra cut will have on the result.
Step3: Looking at the tabulated value for n and P, you will found that adding an extra cut to a cuboid produces an extra (n^2+n)/2 parts, so that we can say that
Pk+1=Pk+ the extra parts added by the extra cuts.
=Pk+ (k^2+k+2)/2
= (k^3+5k+6)/6 + (k^2+k+2)/2
= (k^3+3k^2+8k+12)/6
Step4: (k^3+3k^2+8k+12)/6 is the
assertion
Therefore, if the proposition is true for n=k, then it is true for n=k+1. As it is true for n=1, then it must be true for n=1+1=2. As it is true for n=2 then it must hold for n=2+1=3 and so on for all integers n>0.
That is, by the principle of mathematical induction. T (n) is true.
Now we try to rewrite the formula in the form P=Y+X+S where Y is an algebraic expression in n:
n | 1 | 2 | 3 | 4 | 5 |
S | 2 | 3 | 4 | 5 | 6 |
n | 1 | 2 | 3 | 4 | 5 |
R | 2 | 4 | 7 | 11 | 16 |
n | 1 | 2 | 3 | 4 | 5 |
P | 2 | 4 | 8 | 15 | 26 |
Compare these three tables, we can find out that
P2=R1+P1=2+2=4
P3=R2+P2=4+4=8
P4=R3+P3=7+8=15
P5=R4+P4=11+15=26
P=Y+X+S=Y+ (n^2-n)/2+ (n+1)
Y=P-(n^2-n)/2+ (n+1)
= (n^3+5n+6)/6-[(n^2-n)/2+ (n+1)]
= (n^3+5n+6)/6-(n^2+n+2)/2
= n^3-3n^2+2n
Therefore, P=Y+X+S= (n^3-3n^2+2n) + (n^2-n)/2+ (n+1)
- Four-dimensional object
n Object | One-dimensional | Two-dimensional | Three-dimensional | |
1 | 2 | 2 | 2 | |
2 | 3 | 4 | 4 | |
3 | 4 | 7 | 8 | |
4 | 5 | 11 | 15 | |
5 | 6 | 16 | 26 | |
Conclusion
Qk+1=Qk+ the extra parts added by the extra cuts.
=Qk+ (n^3+5n+6)/6
= 
+ (n^3+5n+6)/6
= (n^4+2n^3+11n^2+34n+48)/24
Step4: (n^4+2n^3+11n^2+34n+48)/24 is theassertion
Therefore, if the proposition is true for n=k, then it is true for n=k+1. As it is true for n=1, then it must be true for n=1+1=2. As it is true for n=2 then it must hold for n=2+1=3 and so on for all integers n>0.
That is, by the principle of mathematical induction. W (n) is true.
Now we try to rewrite the formula in the form Q=Z+Y+X+S where Z is an algebraic expression in n:
n | 1 | 2 | 3 | 4 | 5 |
S | 2 | 3 | 4 | 5 | 6 |
n | 1 | 2 | 3 | 4 | 5 |
R | 2 | 4 | 7 | 11 | 16 |
n | 1 | 2 | 3 | 4 | 5 |
P | 2 | 4 | 8 | 15 | 26 |
n | 1 | 2 | 3 | 4 | 5 | 6 |
Q | 2 | 4 | 8 | 16 | 31 | 57 |
Compare these four tables, we can find out that
Q2=P1+Q1=2+2=4
Q3=P2+Q2=4+4=8
Q4=P3+Q3=7+8=15
Q5=P4+Q4=11+15=26
Q=Z+Y+X+S=Z+ (n^3-3n^2+2n) + (n^2-n)/2+ (n+1)
Z=Q-(n^3-3n^2+2n) + (n^2-n)/2+ (n+1)
=
－[ (n^3-3n^2+2n)+(n^2-n)/2+ (n+1) (n^3+5n+6)/6-(n^2+n+2)/2]
=(n^4－26n^3+71n^2－46n)/24
=Therefore, Q=Z+Y+X+S=(n^4－26n^3+71n^2－46n)/24 +(n^3-3n^2+2n)+(n^2-n)/2+ (n+1)
http://tieba.baidu.com/f?kz=280407483
[a]Why can this formula be used?
[b]Inappropriate notation for multiplication.
[c]Wrong notation for “to the power of “
[d]最好用规范的分数线的形式表示。使用公式编辑器或MATH TYPE。
[e]这两步怎么来的？
[f]How to？说明一下。
[g]说明一下为什么可以这么做。
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