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.

a, b, c gives

a=1/2, b=1/2, c=1

Thus Rn =1/2n²+1/2n+1

When n=5, R5=1/2*5^2+1/2*5+1=16, which is also corresponds to the tabulated value for n=5 above.

Rn =1/2n²+1/2n+1 is the same with Rn=(n^2+n+2)/2

So far we have formed a conjecture that the maximum number of regions from n-chords is given by R= (n^2+n+2)/2.

Proof:

Let P(n) be the proposition that the maximum number of regions that can be separated from n-chords is given by R= (n^2+n+2)/2 for n0

Step1:    P(n) is true for n=1 as R= (1^2+1+2)/2=2 which is the maximum number of regions from 1 chord.

Step2:   Assume that P(n) is true for a k-chords circle i.e., that Rk= (k^2+k+2)/2. We consider the effect that adding an extra cut will have on the result.

Step3:   Looking at the tabulated value for n and R, you will found that adding an extra chord to a circle produces an extra (n+1) regions, so that we can say that

     

Rk+1=Rk+ the extra regions added by the extra chords.

=Rk+ (k+1)

              = (k^2+k+2)/2+ (k+1)

              = (k^2+3k+4)/2

Step4: (k^2+3k+4)/2 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 n0.

That is, by the principle of mathematical induction. P(n) is true.

Now we try to rewrite the formula in the form R=X+S, when X is an algebraic expression in n:

Compare these two tables, we can find out that

R2=R1+S1=2+2=4

R3=R2+S2=4+3=7

R4=R3+S3=7+4=11

R5=R4+S4=11+5=16

Thus R=X+ (n+1) X=R-(n+1) X= (n^2-n)/2

Therefore, R=X+S= (n^2-n)/2+ (n+1)

• Three-dimensional object

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)

= (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 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=12=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 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. 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:

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

Looking at the spreadsheet’s value of different dimensional object for n, form a difference array, you will found that

        

   Sequence    26    15     8      4      2   (Three-dimensional)

 

    differences  11     7      4      2    (Two-dimensional)

    differences     4      3      2      (One-dimensional)

     differences       1      1

As we found that all the dimensional objects separated into 2 parts when n=1, therefore when n=1, Q1=2. Use the difference array again to find out the results of the four-dimensional.

                                    

  Sequences   57   31   16     8    4     2  (Four dimensional)

 differences   26    15    8    4     2    (Three-dimensional)

  differences     11    7    4     2     (Two-dimensional)

      differences    4     3    2    (One-dimensional)

        differences     1     1

The results is shown below in a table:

①A conjecture for the relationship between the maximum number of parts(Q) and the number of cuts(n) in a four-dimensional object. As it is a four= dimensional object, so we might assume that

     Q=

Substituting all the values for

n gives:

n=12=a+b+c+d+e

n=24=16a+8b+4c+2d+e

n=38=81a+27b+9c+3d+e

n=416=256a+64b+16c+4d+e

n=531=625a+125b+25c+5d+e

n=657=1296a+216b+36c+6d+e

Solve these six equations for

a, b, c, d, e gives

a=1/24, b= -1/12, c=11/24, d=7/12, e=1

Thus Qn = .

When n=6, Q6==57 which is also corresponds to the tabulated value for n=6 above.

Proof:

Let W (n) be the proposition that the maximum number of parts that can be separated from n-cuts in a four-dimensional object is given by              

Step1:W(n) is true for n=1 as

 which is the maximum number of parts from 1 cut.

Step2:   Assume that W(n) is true for a k-cuts four-dimensional boject i.e., that . 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 four-dimensional object produces an extra (n^3+5n+6)/6 parts, so that we can say that

     

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:

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)

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