Bracket Investigations

Bracket Investigations (1) Power 2:

( a + b )2 = a 2 + 2ab + b2 , = a 2 + b 2 + 2ab, ≡ ∑ a 2 + 2∑ ab. ab

ab

( a + b + c )2 = a 2 + b2 + c 2 + 2ab + 2bc + 2ca ≡ ∑ a 2 + 2∑ ab. abc

abc

( a + b + c + d )2 = a 2 + b2 + c 2 + d 2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd ≡

∑a

2

abcd

+ 2 ∑ ab. abcd

The pattern seems to be:

∑

( a + b + c + d + )2 =

a2 + 2

abcd …

Power 3:

∑

ab

abcd …

( a + b )3 = a 3 + 3 a 2b + 3 ab2 + b3 , = a 3 + b3 + 3 a 2b + 3 ab 2 , ≡ ∑ a 3 + 3∑ a 2b. ab

ab

( a + b + c )3 = a 3 + b 3 + c 3 + 3 a 2b + 3a 2c + 3b 2 a + 3b 2c + 3c 2 a + 3c 2b + 6 abc ≡ ∑ a 3 + 3∑ a 2b + 6 abc. abc

abc

( a + b + c + d )3 = a 3 + b 3 + c 3 + d 3 + 3 a 2b + 3 a 2c + 3 a 2 d + 3b 2 a + 3b 2c + 3b 2 d + 3c 2 a + 3c 2b + 3c 2 d + 3 d 2 a + 3 d 2b + 3 d 2c + 6 abc + 6 abd + 6bcd + 6 acd ≡

∑a

abcd

3

+ 3 ∑ a 2b + 6 ∑ abc. abcd

abcd

The pattern seems to be:

( a + b + c + d + )3 =

∑a

abcd …

3

+3

∑ a b + 6 ∑ abc 2

abcd …

abcd …

Power 4:

( a + b )4 = a 4 + 4 a 3b + 6 a 2b2 + 4 ab3 + b 4 , = a 4 + b 4 + 4 a 3b + 4 ab3 + 6 a 2b 2 , ≡ ∑ a 4 + 4 ∑ a 3 b + 6 a 2b 2 . ab

ab

( a + b + c )4 = a 4 + b 4 + c 4 + 4 a 3 b + 4 a 3 c + 4b 3 a + 4b 3 c + 4 c 3 a + 4 c 3 b + 6 a 2 b 2 + 6 a 2 c 2 + 6b 2 c 2 + 12a 2bc + 12b 2 ac + 12c 2 ab, ≡ ∑ a 4 + 4 ∑ a 3b + 6∑ a 2b 2 + 12∑ a 2bc. abc

abc

abc

abc

( a + b + c + d )4 = a 4 + b 4 + c 4 + d 4 + 4 a 3b + 4 a 3 c + 4 a 3 d + 4b 3 a + 4b 3 c + 4b 3 d + 4 c 3 a + 4 c 3 b + 4c 3 d + 4 d 3 a + 4 d 3b + 4 d 3 c + 6 a 2b 2 + 6 a 2c 2 + 6 a 2 d 2 + 6b 2c 2 + 6b 2 d 2 + 6c 2 d 2 + 6c 2 d 2 + 12a 2bc + 12a 2bd + 12a 2cd + 12b 2 ac + 12b 2 ad + 12b 2cd + 12c 2 ab + 12c 2 ad + 12c 2bd + 12d 2 ab + 12d 2 ac + 12d 2bc + 24 abcd , ≡

∑a

abcd

4

+ 4 ∑ a 3b + 6 ∑ a 2b 2 + 12 ∑ a 2bc + 24 abcd . abcd

abcd

abcd

The pattern seems to be:

( a + b + c + d + )4 =

∑

abcd …

a4 + 4

∑ a b+6 ∑ a b 3

abcd …

2 2

abcd …

+ 12

∑

a 2bc + 24

abcd …

∑

abcd .

abcd …

Higher Powers: In a similar manner we can find the pattern for the 5th power:

( a + b + c + d + e + )5 =

∑

abcde…

a5 + 5

∑

abcde…

a 4b + 10

∑

abcde…

a3b2 + 20

∑

abcde…

a 3bc + 30

∑

abcde…

a 2b2c + 60

∑

a 2bcd + 120

abcde…

Questions: 1. How do we work out the letters that multiply one another in respective terms? 2. How do we work out the number of different type of terms in the expressions? 3. How do we work out the coefficients (numbers) in front of each term?

∑

abcde…

abcde.

Answers and observations: 1. The letters that multiply one another each have powers, in particular notice that a1 = a and a 0 = 1 . Noting this we observe that every single term in the expansion can have every letter; a, 4 b, c, d, e,…in it. For example in the expansion of ( a + b + c + d ) , namely:

( a + b + c + d )4 = a 4 + b 4 + c 4 + d 4 + 4 a 3 b + 4 a 3 c + 4 a 3 d + 4 b 3 a + 4b 3 c + 4 b 3 d + 4 c 3 a + 4 c 3 b + 4c3 d + 4 d 3 a + 4 d 3 b + 4 d 3 c + 6 a 2 b 2 + 6 a 2 c 2 + 6 a 2 d 2 + 6b 2 c 2 + 6b 2 d 2 + 6 c 2 d 2 + 6 c 2 d 2 + 12a 2 bc + 12a 2 bd + 12a 2 cd + 12b 2 ac + 12b 2 ad + 12b 2 cd + 12c 2 ab + 12c 2 ad + 12c 2 bd + 12d 2 ab + 12d 2 ac + 12d 2 bc + 24 abcd ,

which could be rewritten as:

( a + b + c + d )4 = a 4 b0 c 0 d 0 + a 0 b 4 c 0 d 0 + a 0 b0 c 4 d 0 + a 0 b0 c 0 d 4 + 4 a 3 b1c 0 d 0 + 4 a 3 b 0 c1 d 0 + 4 a 3 b 0 c 0 d 1 + 4 a1b3 c 0 d 0 + 4 a 0 b3 c1 d 0 + 4 a 0 b3 c 0 d 1 + 4 a1b 0 c3 d 0 + 4 a 0 b1c3 d 0 + 4 a 0 b 0 c3 d 1 + 4 a1b 0 c 0 d 3 + 4 a 0 b1c 0 d 3 + 4 a 0 b0 c1 d 3 + 6a 2 b2 c0 d 0 + 6a 2 b0 c 2 d 0 + 6a 2 b0 c0 d 2 + 6a 0 b2 c 2 d 0 + 6a 0 b2 c 0 d 2 + 6a 0 b0 c2 d 2 + 6 a 0 b 0 c 2 d 2 + 12a 2 b1c1 d 0 + 12a 2 b1c 0 d 1 + 12a 2 b0 c1 d 1 + 12a1b2 c1 d 0 + 12a1b2 c 0 d 1 + 12a 0 b 2 c1 d 1 + 12a1b1c 2 d 0 + 12a1b 0 c 2 d 1 + 12a 0 b1c 2 d 1 + 12a1b1c 0 d 2 + 12a1b0 c1 d 2 + 12a 0 b1c1 d 2 + 24 a1b1c1 d 1 .

Therefore, it is clear that within each term the powers must add up to the power on the original bracket before its expansion. The problem now becomes one of making sure that all of the possible permutations are accounted for within a particular type of term. Notice that the above expansion can be written in “Summation” form:

( a + b + c + d + )4 =

∑

a4 + 4

abcd …

and for terms of the type

∑

∑ a b+6 ∑ a b 3

abcd …

2 2

abcd …

+ 12

∑ a bc + 24 ∑ 2

abcd …

abcd .

abcd …

a 2bc we must make sure that all terms of this type are accounted for

abcd

by noting the valid values at the bottom of the summation sign. For example:

∑ a bc = a bc + a bd + a be + a cd + a ce + a de 2

abcde

2

2

2

2

2

2

+ b 2 ac + b 2 ad + b 2 ae + b 2cd + b 2ce + b 2 de + c 2 ab + c 2 ad + c 2 ae + c 2bd + c 2be + c 2 de + d 2 ab + d 2 ac + d 2 ae + d 2bc + d 2be + d 2ce + e2 ab + e2 ac + e2 ad + e2bc + e2bd + e2cd .

where all permutations for the letters a, b, c, d and e for this particular form have been found.

2. The first thing to notice here is that we only get the full number of terms possible that the particular power is capable of producing if the number of letters inside of the original bracket is the same as or more than the value of the power. For example, considering the original power of 4:

( a + b ) 4 = ∑ a 4 + 4 ∑ a 3 b + 6 a 2b 2 . ab

ab

( a + b + c )4 = ∑ a 4 + 4 ∑ a 3b + 6∑ a 2b2 + 12∑ a 2bc. abc

abc

abc

abc

( a + b + c + d )4 = ∑ a 4 + 4 ∑ a 3b + 6 ∑ a 2b2 + 12 ∑ a 2bc + 24 abcd . abcd

Four or more letters

( a + b + c + d + )4 =

abcd

abcd

abcd

Five different possible term types

∑

abcd …

a4 + 4

∑ a b+6 ∑ a b 3

abcd …

2 2

abcd …

+ 12

∑ a bc + 24 ∑ 2

abcd …

abcd .

abcd …

Only in the last two expansions are there as many letters inside the bracket as is the value of the power, 4. Therefore, only in these two expansions do we see the full number of different term types associated with the original power of 4 on the original bracket, that is, 5 different term types. We can compile a list of different term type possibilities for the original power: Power on original Bracket 1 2 3 4 5 6 7 8 9 10

Number of possible term types in expansion 1 2 3 5 7 11 15 22 30 42

What we can observe from these numbers is that the term type number is just the number of ways of making up the original power from additions, that is: The Partitions of the number that is the power.

For example: For the power of 4 on the original bracket we have 4 = 4+0 = 3 +1 = 2+2 = 2 +1+1 = 1+1+1+1

Since there are 5 unique ways of partitioning the number (power) 4 this means that there can only be a maximum of 5 different term types in an expansion having a power of 4 – as is seen above. 3. Before we can succinctly answer this we need to define the Factorial function: n ! = n ( n − 1)( n − 2 )( n − 3 ) × … × 3 × 2 × 1 .

For example:

5! = 5 ( 5 − 1)( 5 − 2 )( 5 − 3 )( 5 − 4 ) = 5 × 4 × 3 × 2 × 1 = 120 , and where 0! = 1 .

Considering the possible term types in the expansion of ( a + b + c + d +

∑a ,∑ 4

abcd …

a 3b ,

abcd …

∑

abcd …

a 2b 2 ,

∑

abcd …

a 2bc and

∑

)4 , namely:

abcd ,

abcd …

we find the following formula calculates their respective coefficients:

∑a

4

:

abcd …

∑

a 3b :

Coeff

∑

a 2b 2 :

Coeff

∑

a 2 bc :

Coeff

∑

abcd :

Coeff

abcd …

abcd …

abcd …

abcd …

4! = 1, 4! 4! 4 × 3 × 2 ×1 = = = 4, 3! × 1! 3 × 2 × 1 × 1 4! 4 × 3 × 2 ×1 = = = 6, 2! × 2! 2 × 1 × 2 × 1 4! 4 × 3 × 2 ×1 = = = 12 , 2! × 1! × 1! 2 × 1 × 1 × 1 4! 4 × 3 × 2 ×1 = = = 24 . 1! × 1! × 1! × 1! 1 × 1 × 1 × 1

Coeff =

Which is exactly what is observed above. Looking at these coefficients more generally we see the pattern behind them as: n! Coeff = p1 !× p2 !× p3 !×… × pn ! where n is the power on the original bracket, and the pi are the powers on the letters in the respective term type. Notice that the above term types in the expansion of ( a + b + c + d + could have been written as:

)4

4! =1, 4! × 0! × 0!× 0!× 0!× … 4! = = 4, 3! × 1! × 0!× 0!× 0!× … 4! = = 6, 2! × 2! × 0! × 0! × 0!× … 4! = = 12 , 2! × 1! × 1! × 0! × 0!× … 4! = = 24 , 1! × 1! × 1! × 1! × 0! × …

∑ a b c d e …:

Coeff =

∑

a 3b1c 0 d 0 e0 … :

Coeff

∑

a 2b 2 c 0 d 0 e 0 … :

Coeff

∑

a 2b1c1d 0 e0 … :

Coeff

∑

a1b1c1d 1e0 … :

Coeff

4 0 0

0 0

abcd …

abcd …

abcd …

abcd …

abcd …

which may help to see how the coefficient formula is applied. The general form of this is called the Multinomial Expansion, which is stated:

( x1 + x2 +

+ xm ) = n

 n! pm  p1 p2 x x … x ∑  m  1 2 x1 x2 … xm  p1 ! × p2 ! × … × pm ! 

Bracket Investigations (2) Binomial Theorem: Binomials are powers of brackets in which there are only two variables, for example: 3 ( a + b )2 , ( x + 3 )4 , ( x + y ) , etc.

We observe the following pattern:

( a + b )0 = 1, ( a + b )1 = a + b, ( a + b )2 = a 2 + 2ab + b2 , ( a + b )3 = a3 + 3 a 2b + 3 ab2 + b3 , ( a + b )4 = a 4 + 4 a3b + 6 a 2b2 + 4 ab3 + b 4 , ( a + b )5 = a 5 + 5a 4b + 10a3b2 + 10 a 2b3 + 5ab 4 + b 5 , and so on ... If we just record the coefficient values of the terms we have: 1 1 1 1 1

1 2

3 4

1 3

6

1 4

1

1 5 10 10 5 1 15 6 The last line relates to the coefficients of terms in ( a + b ) and as can be seen, these coefficients can be found by the addition of the two coefficients directly above when set out in the above “Triangular” arrangement. In fact this is known as Pascal’s Triangle and it can be used to find the coefficients of any integer power of ( a + b ) . 6

Moreover, if we set out a general expansion of ( a + b ) as: n

( a + b )n = C0n a nb 0 + C1n a n −1b1 + C2n a n −2b 2 +

+ Crn a n − r b r +

+ Cnn a 0b n ,

then

Crn =

n! ( n − r ) !× r !

. 5 For example, to calculate the coefficient of a b in the expansion of ( a + b ) we notice that 3 2

a 3 b 2 ≡ a 5− 2b 2 .

Hence

C25 =

5! 5! 5 × 4 × 3 × 2 ×1 = = = 10 , 2! ( 5 − 2 ) ! 2! × 3! 2 × 1 × 3 × 2 × 1

as observed above.

n! n! can be written as Crn = , where the pi are the powers of the p1 ! p2 ! ( n − r ) !× r ! a and b respectively. Thus, we have a direct link to the Multinomial Expansion as described in general above.

Notice that Crn =

Bracket Investigations (3) Many Brackets:

( x + b1 )( x + b2 ) = x 2 + ( b1 + b2 ) x + b1b2 , ( x + b1 )( x + b2 ) ( x + b3 ) = x3 + ( b1 + b2 + b3 ) x2 + ( b1b2 + b2 b3 + b1b3 ) x + b1b2 b3 , ( x + b1 )( x + b2 ) ( x + b3 ) ( x + b4 ) = x 4 + ( b1 + b2 + b3 + b4 ) x3 + ( b1b2 + b1b3 + b1b4 + b2 b3 + b2 b4 + b3 b4 ) x 2 + ( b1b2 b3 + b1b2 b4 + b1b3 b4 + b2 b3 b4 ) x + b1b2 b3 b4 , or, in “Summation” notation as in “Bracket Investigations (1)” above: 



 b1b2



( x + b1 )( x + b2 ) = x2 +  ∑ b1  x + b1b2 , 







 b1b2b3



 b1b2b3



( x + b1 )( x + b2 ) ( x + b3 ) = x3 +  ∑ b1  x 2 +  ∑ b1b2  x + b1b2b3 , 

     b1  x 3 +  ∑ b1b2  x 2 +  ∑ b1b2b3  x + b1b2b3b4 ,  b1b2b3b4   b1b2b3b4   b1b2b3b4 

( x + b1 )( x + b2 ) ( x + b3 ) ( x + b4 ) = x 4 +  ∑

which is a much more straightforward notation to remember and builds upon knowledge from “Bracket Investigations (1)” above. Generalising this further we find that:

( a1 x + b1 )( a2 x + b2 ) = a1a2 x 2 + ( b1a2 + a1b2 ) x + b1b2 , ( a1 x + b1 )( a2 x + b2 ) ( a3 x + b3 ) = a1a2 a3 x3 + ( b1a2 a3 + a1b2 a3 + a1a2b3 ) x 2 + ( b1b2 a3 + a1b2b3 + b1a2b3 ) x + b1b2b3 , ( a1 x + b1 )( a2 x + b2 ) ( a3 x + b3 ) ( a4 x + b4 ) = a1a2 a3 a4 x 4 + ( b1a2 a3 a4 + a1b2 a3 a4 + a1a2b3 a4 + a1a2 a3b4 ) x3 + (b1b2 a3 a4 + b1a2b3 a4 + b1a2 a3b4 + a1b2b3 a4 + a1b2 a3b4 + a1a2b3b4 ) x 2 + ( b1b2b3 a4 + b1b2 a3b4 + b1a2b3b4 + a1b2b3b4 ) x + b1b2b3b4 ,

To remember the layout of this more general form it is best to remember the layout of the more basic form above (with only the bi's ) and then to “complete” each term with the necessary ai's . Once again this builds upon knowledge and understanding of the “Bracket Investigations (1)” above.

When doing this type of example it is often best to label the coefficients of x in the following manner and to work them out individually:

( a1 x + b1 )( a2 x + b2 ) = P2 x 2 + P1 x + P0 , ( a1 x + b1 )( a2 x + b2 ) ( a3 x + b3 ) = P3 x3 + P2 x 2 + P1 x + P0 , ( a1 x + b1 )( a2 x + b2 ) ( a3 x + b3 ) ( a4 x + b4 ) = P4 x 4 + P3 x3 + P2 x 2 + P1 x + P0 . For example:-

Expand ( 2 x + 3 )( x − 4 )( 3 x − 1)( 5 x + 2 ) .

We have: P4 = a1a2 a3 a4 = ( 2 )(1)( 3 )( 5 ) = 30 P3 = b1a2 a3 a4 + a1b2 a3 a4 + a1a2b3 a4 + a1a2 a3b4 = ( 3 )(1)( 3 )( 5 ) + ( 2 )( −4 )( 3 )( 5 ) + ( 2 )(1)( −1)( 5 ) + ( 2 )(1)( 3 )( 2 ) = −73 P2 = b1b2 a3 a4 + b1a2b3 a4 + b1a2 a3b4 + a1b2b3 a4 + a1b2 a3b4 + a1a2b3b4 = ( 3 )( −4 )( 3 )( 5 ) + ( 3 )(1)( −1)( 5 ) + ( 3 )(1)( 3 )( 2 ) + ( 2 )( −4 )( −1)( 5) + ( 2 )( −4 )( 3 )( 2 ) + ( 2 )(1)( −1)( 2 ) = −189 P1 = b1b2b3 a4 + b1b2 a3b4 + b1a2b3b4 + a1b2b3b4 = ( 3 )( −4 )( −1)( 5 ) + ( 3 )( −4 )( 3 )( 2 ) + ( 3 )(1)( −1)( 2 ) + ( 2 )( −4 )( −1)( 2 ) = −2 P0 = b1b2b3b4 = ( 3 )( −4 )( −1)( 2 ) = 24

Therefore:

( 2 x + 3 )( x − 4 )( 3 x − 1)( 5 x + 2 ) = 30 x 4 − 73 x3 − 189 x 2 − 2 x + 24