• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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
=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. 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=2 4=8a+4b+2c+d

n=3 8=27a+9b+3c+d

n=4 15=64a+16b+4c+d

n=5 26=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 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. 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。

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

-ï¿½ï¿½(tm)\E7Cï¿½ï¿½ï¿½ï¿½=ï¿½ï¿½rï¿½ï¿½&Fï¿½oï¿½ï¿½-eï¿½`ï¿½3/4ï¿½ï¿½ï¿½'Yï¿½ï¿½Jï¿½X1/4ï¿½g:ï¿½5'ï¿½ï¿½ï¿½Mvï¿½'Mï¿½ï¿½(tm)\$6 ï¿½dï¿½ï¿½ï¿½0I(ï¿½3/4yï¿½u1/20ï¿½ï¿½ï¿½ï¿½1/2ï¿½m.hm{ï¿½ï¿½ï¿½Íxï¿½ï¿½dï¿½ï¿½ï¿½ï¿½ï¿½<ï¿½Nï¿½ï¿½ï¿½Vï¿½ï¿½ï¿½ï¿½-ï¿½_\ï¿½ï¿½ï¿½(tm)ï¿½ï¿½ï¿½;Wï¿½x"nvEï¿½(r)"2ï¿½ï¿½ï¿½vï¿½ï¿½ï¿½ï¿½\51ï¿½"ï¿½ï¿½ï¿½sÉ±dï¿½Sï¿½mxï¿½n>?ï¿½ï¿½1/4ï¿½(ï¿½7Cï¿½ï¿½g{pï¿½@:uÅ¶@ï¿½ï¿½ï¿½Tï¿½(ï¿½ ï¿½ï¿½Nï¿½ï¿½ï¿½ï¿½ï¿½vW'y(ï¿½ï¿½ï¿½nï¿½ï¿½4"Êï¿½hNï¿½Z0ï¿½vzeQï¿½-ï¿½zTï¿½ï¿½Ô¢0*3/4ï¿½ï¿½ï¿½Ee ï¿½ï¿½lï¿½=ï¿½ï¿½ï¿½b\$jRï¿½'ï¿½Fï¿½Cï¿½ï¿½"ï¿½1/2Õ¯ï¿½pvï¿½-Í´ï¿½Xï¿½ï¿½"ï¿½Uï¿½"vFT8ï¿½ï¿½ï¿½ï¿½fE[l,ï¿½Kï¿½xï¿½`ï¿½\ï¿½tï¿½ï¿½7Mï¿½(r)ï¿½ï¿½3w...Ùºï¿½{gï¿½wï¿½ï¿½+ï¿½ï¿½Ô´ Mï¿½ï¿½ï¿½Eï¿½ï¿½ï¿½qï¿½ -ï¿½ï¿½qõï¿½ï¿½ï¿½ï¿½gï¿½ ;wvï¿½~^ï¿½DsdN1>ï¿½"ï¿½;ï¿½..."ï¿½ohï¿½Yï¿½Ê-ï¿½>Lp"1/2tï¿½ï¿½Nb,*ï¿½^3/4ï¿½=ï¿½-ï¿½ï¿½ï¿½ï¿½Rï¿½ ï¿½ï¿½_ï¿½"~ï¿½- ï¿½ -wõï¿½-ï¿½ï¿½>ï¿½Uï¿½ ï¿½ï¿½^ï¿½~ï¿½ï¿½ï¿½^yPï¿½ï¿½Zï¿½%ï¿½ï¿½ï¿½ï¿½"A ï¿½hï¿½S[ï¿½ï¿½ï¿½ï¿½ï¿½3/4;ï¿½^/wï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½h>zï¿½ï¿½ï¿½aï¿½ï¿½)ï¿½ï¿½yï¿½A`pï¿½]Qï¿½ï¿½ ï¿½Âï¿½h!\ï¿½ï¿½m/ï¿½~ï¿½ï¿½ï¿½a{?lï¯³1/2 [email protected]ï¿½xzï¿½_ï¿½1/2ï¿½=imï¿½ï¿½rï¿½S/ï¿½dï¿½ï¿½'...1/2ï¿½91/2Ý¶"r"&ï¿½y`ï¿½&ï¿½'/'-ï¿½2ï¿½ï¿½(:ï¿½%ï¿½ï¿½oï¿½:ï¿½Û² ï¿½2jï¿½ï¿½-ï¿½H;ï¿½8ï¿½ï¿½ï¿½8ï¿½'ï¿½ï¿½k6ï¿½ï¿½vï¿½(c)Eï¿½;t-ï¿½ ï¿½ï¿½Í¨7Cï¿½:ï¿½0+2 "|j:ï¿½!Iï¿½ï¿½rï¿½(#ï¿½(c)/ï¿½ï¿½ï¿½ï¿½ï¿½ Mï¿½-ï¿½JÈ*Q×´ï¿½-1/4ï¿½5,ï¿½ ï¿½Æ­;4ï¿½#,1/4ï¿½ï¿½~<ï¿½ j|dï¿½wï¿½-ï¿½o/Oï¿½r1ï¿½×ï¿½Xï¿½`ï¿½&aï¿½ï¿½Þï¿½tq(tm)ï¿½ï¿½w'ï¿½-ï¿½=cï¿½ï¿½&ï¿½ï¿½"ï¿½ï¿½o[` ï¿½oï¿½ï¿½vï¿½ï¿½ï¿½Wï¿½ï¿½ï¿½ ï¿½6ï¿½ï¿½ï¿½ï¿½wï¿½Wk:"?mï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½dÏ¿ï¿½Ôï¿½ ï¿½ï¿½^ß ~ï¿½q I/ï¿½Oï¿½ï¿½;ï¿½OI' CHoï¿½Opï¿½) ï¿½ï¿½ ï¿½"ï¿½fï¿½.%ï¿½,( 1_...*ï¿½RfqéµÕ·%ï¿½yPï¿½3/4Zï¿½g((ï¿½ ï¿½# ï¿½ï¿½ï¿½ï¿½p ï¿½ ï¿½ ï¿½Bï¿½ß¢ P_ "ï¿½xï¿½ï¿½ 4c 0w \rï¿½ ï¿½ï¿½ï¿½zï¿½ï¿½Bï¿½uï¿½P_ï¿½/ï¿½P</ï¿½ï¿½ï¿½1/4!Hê¢zB'<ï¿½ï¿½ï¿½YEï¿½ï¿½[email protected]ï¿½ï¿½ï¿½ ï¿½qï¿½ï¿½3.Iï¿½ï¿½`"(c)ï¿½bTï¿½3/4ï¿½qvj"D;Tï¿½J,ï¿½ï¿½@@Qï¿½ ï¿½!ï¿½3/4 ï¿½Lï¿½Wï¿½|ï¿½Cï¿½|...ï¿½ï¿½Zï¿½ pï¿½#ï¿½ï¿½ï¿½j[ï¿½oï¿½@1/2EAï¿½5ï¿½Jï¿½PWï¿½ï¿½YZï¿½jAhï¿½'ï¿½ï¿½lz(tm)ï¿½ï¿½Lï¿½ï¿½ï¿½=Aï¿½^ï¿½ï¿½ï¿½dvC(c)0ï¿½Z`Úï¿½cDDï¿½ï¿½kB'*|K4ï¿½ï¿½*OB3/4ï¿½Kï¿½'ï¿½ï¿½ï¿½Mï¿½ï¿½(r)Xï¿½ -(tm)ÂGï¿½Sï¿½ï¿½5ï¿½1/2O_'kvï¿½-)ï¿½ï¿½vï¿½Nï¿½...ï¿½ï¿½=ï¿½F*18ï¿½3/4lï¿½k-ï¿½ï¿½;ï¿½ï¿½ï¿½ï¿½-ï¿½Lï¿½ï¿½"oï¿½ï¿½pï¿½ï¿½;&vï¿½Z?ï¿½ï¿½ï¿½~>ï¿½F!ï¿½ï¿½ï¿½ï¿½{ï¿½^ï¿½ï¿½ï¿½ï¿½\$ &ï¿½-ï¿½ 3/4ï¿½"ï¿½vvï¿½ï¿½|ï¿½F>ï¿½M~#jï¿½[xï¿½ï¿½ï¿½ï¿½"ï¿½6ï¿½<ï¿½Þb Dï¿½ï¿½"ï¿½ï¿½ï¿½;jï¿½Ï³ï¿½ï¿½ï¿½iï¿½"ï¿½ ï¿½yï¿½mï¿½/ε{

2. ## El copo de nieve de Koch

el numero de fase: Con la implementaci�n de este modelo se obtuvieron los siguientes resultados: Lo que concuerda con los datos de la Tabla 1. N�mero de fase 0 1 2 3 1 A continuaci�n se muestran la gr�fica de =longitud del per�metro, para =0, 1 , 2 y 3.

1. ## How many pieces? In this study, the maximum number of parts obtained by n ...

> Two Dimensional Circle n = 1, R = 2 n = 2, R = 4 n = 3, R = 7 n = 4, R = 11 n = 5, R = 16 n 1 2 3 4 5 R 2 4 7 11 16 To find a recursive

2. ## This essay will examine theoretical and experimental probability in relation to the Korean card ...

However, as stated above, some of the hands evenly appeared but there were also hands that were extremely higher or lower than the theoretical probability. Hence, it could be concluded that if the number of trials were increased, then the data will be evenly spread.

1. ## Statistics project. Comparing and analyzing the correlation of the number of novels read per ...

from 61-80 mark and most read a little of amount of 1 book per week. Their average mark is 63.7 and their median is 61-80. Their modal value is also 61-80. This mark with correlation to number of books read shows that number of books read has very little to do with modal mark earned.

2. ## Systems of Linear Equations. Investigate Systems of linear equations where the system constants ...

The solution to the second example was the same as the first example and the original system of equations. This seems to indicate that all 2x2 systems that follow this pattern will have the same solutions, but in order to prove that we need to solve a 2 by 2

1. ## Maths Project. Statistical Analysis of GCSE results at my secondary school summer 2010 ...

37 Ru 11 46 m 46 36 Sa 10 46 m 46 35 Sa 7 40 f 40 34 Sc 10 46 f 46 33 Se 9 0 m 0 32 Sh 12 46 f 46 31 Si 10 34 f 34 30 So 10 34 f 34 29 St

2. ## In this portfolio, I am required to investigate the number of regions obtained by ...

i.e ? 2, 4, 7, 11, 16 The differences are - 2, 3, 4, and 5 And the difference is equal to the number of chords drawn. Therefore, Number of regions = Number of regions obtained + number of chords drawn obtained in the previous sketch We can also consider • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 