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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.

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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.image10.pngimage08.pngimage09.pngimage00.pngimage01.png

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

image16.png

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.

image14.pngimage12.pngimage13.pngimage11.png

image15.png

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

...read more.

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
=P
n-(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.image04.png

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 Pimage23.png

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=1image17.png2=a+b+c+d

n=2image17.png4=8a+4b+2c+d

n=3image17.png8=27a+9b+3c+d

n=4image17.png15=64a+16b+4c+d

n=5image17.png26=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.image05.png

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 theimage24.pngassertion

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

image17.pngP=Y+X+S=Y+ (n^2-n)/2+ (n+1)

image17.pngY=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

...read more.

Conclusion

Qk+1=Qk+ the extra parts added by the extra cuts.

              =Qk+ (n^3+5n+6)/6

              = image36.png + (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 theimage24.pngassertion

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

image17.pngQ=Z+Y+X+S=Z+ (n^3-3n^2+2n) + (n^2-n)/2+ (n+1)

image17.pngZ=Q-(n^3-3n^2+2n) + (n^2-n)/2+ (n+1)

        =image37.png-[        (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。

...read more.

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