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 AM  Vol.12 No.2 , February 2021
Powers of Octonions
Abstract: As it is known, Binomial expansion, De Moivre’s formula, and Euler’s formula are suitable methods for computing the powers of a complex number, but to compute the powers of an octonion number in easy way, we need to derive suitable formulas from these methods. In this paper, we present a novel way to compute the powers of an octonion number using formulas derived from the binomial expansion.

1. Introduction

An octonion number a can be expressed as:

a = a 0 + a 1 i 1 + a 2 i 2 + a 3 i 3 + a 4 i 4 + a 5 i 5 + a 6 i 6 + a 7 i 7 (1)

where a 0 , , a 7 are real numbers and i 1 , , i 7 are imaginary units. Their multiplication is given in Table 1. Octonion algebra is an eight-dimensional, non-commutative, non-associative and normed division algebra.

Table 1. Imaginary units multiplication.

Octonions have been used in many fields of mathematics, and they have many applications, about octonions and their applications see [1] [2] [3], and [4] [5] to take a historical overview.

In the matrix representation, an octonion a can be represented by 8 × 8 real matrices. One form of these matrices is a matrix A (the left matrix representation) [6]:

A = [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 1 a 0 a 3 a 2 a 5 a 4 a 7 a 6 a 2 a 3 a 0 a 1 a 6 a 7 a 4 a 5 a 3 a 2 a 1 a 0 a 7 a 6 a 5 a 4 a 4 a 5 a 6 a 7 a 0 a 1 a 2 a 3 a 5 a 4 a 7 a 6 a 1 a 0 a 3 a 2 a 6 a 7 a 4 a 5 a 2 a 3 a 0 a 1 a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 0 ]

The problem of computing the nth power of an octonion is still interesting to many researchers. Some methods are used, such as binomial expansion, De Moivre’s formula, and Euler’s formula. To solve this problem, we use a new technique to construct formulas computing the powers of an octonion.

2. The Methodology

For a complex number b = b 0 + b 1 i where b 0 , b 1 are real numbers, and i is the imaginary unit satisfies i 2 = 1 ,

b n = ( b 0 + b 1 i ) n = j = 0 n ( n n j ) ( b 0 n j ) ( b 1 i ) j , n is a positive integer number (2)

If n is an even number then there will be n 2 + 1 real terms and n 2 imaginary

terms, to simplify, we define Im [ b ] = b 1 i , Re [ b ] = b 0 and, let b 0 , b 1 be nonzero real numbers, we can write (2) as:

Re [ b n ] = j = 0 n / 2 ( n n 2 j ) b 0 n 2 j ( b 1 2 ) j Im [ b n ] = ( b b 0 ) j = 0 n / 2 1 ( n n 2 j 1 ) b 0 n 2 j 1 ( b 1 2 ) j (3)

If n is an odd number then there will be n + 1 2 real terms and n + 1 2 imaginary

terms, we can write (2) as:

Re [ b n ] = j = 0 ( n 1 ) / 2 ( n n 2 j ) b 0 n 2 j ( b 1 2 ) j Im [ b n ] = ( b b 0 ) j = 0 ( n 1 ) / 2 ( n n 2 j 1 ) b 0 n 2 j 1 ( b 1 2 ) j (4)

To prove (3), let us write b in a matrix representation form:

B = [ b 0 b 1 b 1 b 0 ]

For n = 2 , b 2 can be computed from:

[ b 0 b 1 b 1 b 0 ] [ b 0 b 1 ] = [ b 0 2 b 1 2 2 b 0 b 1 ] ,

so

Re [ b 2 ] = b 0 2 b 1 2 = j = 0 1 ( 2 2 2 j ) b 0 2 2 j ( b 1 2 ) j Im [ b 2 ] = 2 b 0 b 1 = ( b b 0 ) j = 0 0 ( 2 1 2 j ) b 0 1 2 j ( b 1 2 ) j (5)

Assume that (3) is true for n = 2 k , where k is a positive integer, therefore:

Re [ b 2 k ] = j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j ( b 1 2 ) j Im [ b 2 k ] = ( b b 0 ) j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j 1 ( b 1 2 ) j (6)

We can compute b 2 k + 2 = b 2 b 2 k by using the matrix:

B 2 = [ b 0 2 b 1 2 2 b 0 b 1 2 b 0 b 1 b 0 2 b 1 2 ]

Multiplying the first row of B 2 by ( b 2 k ) T (the column matrix representing ( b 2 k ) ) gives Re [ b 2 k + 2 ] , and multiplying the second row of B 2 by ( b 2 k ) T gives Im [ b 2 k + 2 ] .

Re [ b 2 k + 2 ] = ( b 0 2 b 1 2 ) j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j ( b 1 2 ) j 2 b 0 b 1 j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j 1 ( b 1 2 ) j = j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j + 2 ( b 1 2 ) j + j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j ( b 1 2 ) j + 1 + 2 j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j ( b 1 2 ) j + 1 = b 0 2 k + 2 ( 2 k 2 k 2 ) b 0 2 k ( b 1 2 ) + ( 2 k 2 k 4 ) b 0 2 k 2 ( b 1 2 ) 2 +

+ ( 2 k 2 k 2 r + 2 ) b 0 2 k 2 r + 4 ( b 1 2 ) r 1 + ( 2 k k 2 r ) b 0 2 k 2 r + 2 ( b 1 2 ) r + + ( 2 k 2 ) b 0 4 ( b 1 2 ) k 1 + b 0 2 ( b 1 2 ) k b 0 2 k ( b 1 2 ) + ( 2 k 2 k 2 ) b 0 2 k 2 ( b 1 2 ) 2 + + ( 2 k 2 k 2 r + 2 ) b 0 2 k 2 r + 2 ( b 1 2 ) r + ( 2 k 2 k 2 r ) b 0 2 k 2 r ( b 1 2 ) r + 1 + + ( 2 k 2 ) b 0 2 ( b 1 2 ) k + ( b 1 2 ) k + 1 + 2 [ ( 2 k 2 k 1 ) b 0 2 k ( b 1 2 ) + ( 2 k 2 k 3 ) b 0 2 k 2 ( b 1 2 ) 2 + + ( 2 k 2 k 2 r + 1 ) b 0 2 k 2 r + 2 ( b 1 2 ) r + ( 2 k 2 k 2 r 1 ) b 0 2 k 2 r ( b 1 2 ) r + 1 +

+ ( 2 k 3 ) b 0 4 ( b 1 2 ) k 1 + ( 2 k 1 ) b 0 2 ( b 1 2 ) k ] = b 0 2 k + 2 [ ( 2 k 2 k 2 ) + 1 + 2 ( 2 k 2 k 1 ) ] b 0 2 k ( b 1 2 ) + [ ( 2 k 2 k 4 ) + ( 2 k 2 k 2 ) + 2 ( 2 k 2 k 3 ) ] b 0 2 k 2 ( b 1 2 ) 2 + + [ ( 2 k 2 k 2 r ) + ( 2 k k 2 r + 2 ) + 2 ( 2 k 2 k 2 r + 1 ) ] b 0 2 k 2 r + 2 ( b 1 2 ) r + + [ ( 2 k 2 ) + 1 + 2 ( 2 k 1 ) ] b 0 2 ( b 1 2 ) k + ( b 1 2 ) k + 1

= b 0 2 k + 2 ( 2 k + 2 2 k ) b 0 2 k ( b 1 2 ) + ( 2 k + 2 2 k 2 ) b 0 2 k 2 ( b 1 2 ) 2 + + ( 2 k + 2 2 k 2 r + 2 ) b 0 2 k 2 r + 2 ( b 1 2 ) r + + ( 2 k + 2 2 ) b 0 2 ( b 1 2 ) k + ( b 1 2 ) k + 1 = j = 0 k + 1 ( 2 k + 2 2 k + 2 2 j ) b 0 2 k + 2 2 j [ b 1 2 ] j

Im [ b 2 k + 2 ] = i [ 2 b 0 b 1 j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j ( b 2 ) j + b 1 ( b 0 2 b 1 2 ) j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j 1 ( b 1 2 ) j ] = b 1 i [ 2 j = 0 k ( 2 k 2 k 2 j ) b 0 2 k 2 j + 1 ( b 1 2 ) j + j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j + 1 ( b 1 2 ) j + j = 0 k 1 ( 2 k 2 k 2 j 1 ) b 0 2 k 2 j 1 ( b 1 2 ) j + 1 ] = b 1 i [ 2 b 0 2 k + 1 2 ( 2 k 2 k 2 ) b 0 2 k 1 ( b 1 2 ) + 2 ( 2 k 2 k 4 ) b 0 2 k 3 ( b 1 2 ) 2 +

+ 2 ( 2 k 2 k 2 r + 2 ) b 0 2 k 2 r + 3 ( b 1 2 ) r 1 + 2 ( 2 k 2 k 2 r ) b 0 2 k 2 r + 1 ( b 1 2 ) r + + 2 ( 2 k 2 ) b 0 3 ( b 1 2 ) k 1 + 2 b 0 ( b 1 2 ) k + ( 2 k 2 k 1 ) b 0 2 k + 1 ( 2 k 2 k 3 ) b 0 2 k 1 ( b 1 2 ) + ( 2 k 2 k 5 ) b 0 2 k 3 ( b 1 2 ) 2 + + ( 2 k 2 k 2 r + 1 ) b 0 2 k 2 r + 3 ( b 1 2 ) r 1 + ( 2 k 2 k 2 r 1 ) b 0 2 k 2 r + 1 ( b 1 2 ) r + + ( 2 k 3 ) b 0 5 ( b 1 2 ) k 2 + ( 2 k 1 ) b 0 3 ( b 1 2 ) k 1

( 2 k 2 k 1 ) b 0 2 k 1 ( b 1 2 ) + ( 2 k 2 k 3 ) b 0 2 k 3 ( b 1 2 ) 2 ( 2 k 2 k 5 ) b 0 2 k 5 ( b 1 2 ) 3 + + ( 2 k 2 k 2 r + 1 ) b 0 2 k 2 r + 1 ( b 1 2 ) r + ( 2 k 2 k 2 r 1 ) b 0 2 k 2 r 1 ( b 1 2 ) r + 1 + + ( 2 k 3 ) b 0 3 ( b 1 2 ) k 1 + ( 2 k 1 ) b 0 ( b 1 2 ) k ]

= b 1 i [ b 0 2 k + 1 [ 2 + ( 2 k 2 k 1 ) ] b 0 2 k 1 ( b 1 2 ) [ 2 ( 2 k 2 k 2 ) + ( 2 k 2 k 3 ) + ( 2 k 2 k 1 ) ] + b 0 2 k 3 ( b 1 2 ) 2 [ 2 ( 2 k 2 k 4 ) + ( 2 k 2 k 5 ) + ( 2 k 2 k 3 ) ] + + b 0 2 k 2 r + 1 ( b 1 2 ) r [ 2 ( 2 k 2 k 2 r ) + ( 2 k 2 k 2 r 1 ) + ( 2 k 2 k 2 r + 1 ) ] + + b 0 3 ( b 1 2 ) k 1 [ 2 ( 2 k 2 ) + ( 2 k 1 ) + ( 2 k 3 ) ] + b 0 ( b 1 2 ) k [ 2 + ( 2 k 1 ) ] ]

= b 1 i [ b 0 2 k + 1 ( 2 k + 2 2 k + 1 ) b 0 2 k 1 ( b 1 2 ) ( 2 k + 2 2 k 1 ) + b 0 2 k 3 ( b 1 2 ) 2 ( 2 k + 2 2 k 3 ) + + b 0 2 k 2 r + 1 ( b 1 2 ) r ( 2 k + 2 2 k 2 r + 1 ) + + b 0 3 ( b 1 2 ) k 1 ( 2 k + 2 3 ) + b 0 ( b 1 2 ) k ( 2 k + 2 1 ) ] = b 1 i j = 0 k ( 2 k + 2 2 k 2 j + 1 ) b 0 2 k 2 j + 1 ( b 1 2 ) j

By the similar way, we can prove (4).

3. Results

We can use (3) and (4) to compute the powers of a quaternion number and the powers of an octonion number.

To compute the power of a , replace b 0 2 by a 0 2 and b 1 2 by a 1 2 + a 2 2 + + a 7 2 in (3) and (4),

Re [ a n ] = j = 0 n / 2 ( n n 2 j ) a 0 n 2 j [ ( a 1 2 + + a 7 2 ) ] j Im [ a n ] = ( a a 0 ) j = 0 n / 2 1 ( n n 2 j 1 ) a 0 n 2 j 1 [ ( a 1 2 + + a 7 2 ) ] j (7)

Re [ a n ] = j = 0 ( n 1 ) / 2 ( n n 2 j ) a 0 n 2 j [ ( a 1 2 + + a 7 2 ) ] j Im [ a n ] = ( a a 0 ) j = 0 ( n 1 ) / 2 ( n n 2 j 1 ) a 0 n 2 j 1 [ ( a 1 2 + + a 7 2 ) ] j (8)

(7) and (8) give ( a 0 + a 1 i 1 + a 2 i 2 + + a 7 i 7 ) n when n is an even number and n is an odd number respectively.

Of course the proof is similar to that one we used to prove the powers formulas for a complex number, but to make it clearer, we will write A as:

A = ω ( a ) + υ ( a ) (9)

where

ω ( a ) = [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 1 a 0 0 0 0 0 0 0 a 2 0 a 0 0 0 0 0 0 a 3 0 0 a 0 0 0 0 0 a 4 0 0 0 a 0 0 0 0 a 5 0 0 0 0 a 0 0 0 a 6 0 0 0 0 0 a 0 0 a 7 0 0 0 0 0 0 a 0 ] ,

υ ( a ) = [ 0 0 0 0 0 0 0 0 0 0 a 3 a 2 a 5 a 4 a 7 a 6 0 a 3 0 a 1 a 6 a 7 a 4 a 5 0 a 2 a 1 0 a 7 a 6 a 5 a 4 0 a 5 a 6 a 7 0 a 1 a 2 a 3 0 a 4 a 7 a 6 a 1 0 a 3 a 2 0 a 7 a 4 a 5 a 2 a 3 0 a 1 0 a 6 a 5 a 4 a 3 a 2 a 1 0 ]

Since ω ( a ) will play an important role in our proof, we call it the fundamental matrix. The following proposition presents the main properties of the fundamental matrix ω ( a ) .

3.1. Proposition

Let a T = [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 ] T , a ¯ be the conjugate of a , λ be a real number, and I 8 be the identity matrix. Then

(a) A 2 = ω ( a ) a T

(b) ω ( a n ) = ω n ( a )

(c) ω ( a ¯ ) = ω T ( a )

(d) ω ( λ a ) = λ ω ( a )

(e) ω ( 1 ) = I 8

Proof:

(a) a 2 = [ ω ( a ) + υ ( a ) ] a T = ω ( a ) a T , (since υ ( a ) a T = 0 ).

In general a n + 1 = a a n , a n + 1 = ω ( a ) ( a n ) T , (n is a real number).

(c) a a ¯ = ω ( a ) a ¯ T = a 0 2 + a 1 2 + + a 7 2 = a 2

The verification of remaining propositions is straightforward.

Form proposition (a), a 2 can be computed from:

ω ( a ) a T = [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 1 a 0 0 0 0 0 0 0 a 2 0 a 0 0 0 0 0 0 a 3 0 0 a 0 0 0 0 0 a 4 0 0 0 a 0 0 0 0 a 5 0 0 0 0 a 0 0 0 a 6 0 0 0 0 0 a 0 0 a 7 0 0 0 0 0 0 a 0 ] [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 ] = [ a 0 2 ( a 1 2 + + a 7 2 ) 2 a a 0 1 2 a 0 a 2 2 a 0 a 3 2 a 0 a 4 2 a 0 a 5 2 a 0 a 6 2 a 0 a 7 ]

a 2 = a 0 2 ( a 1 2 + + a 7 2 ) + 2 a 0 ( a 1 i 1 + + a 7 i 7 )

So, multiplying the first row of ω ( a ) by ( a ) T gives Re [ a 2 ] , and to obtain Im [ a 2 ] we just multiply any row (except the first row) of ω ( a ) by a T because the multiplication result will be similar for all other rows, therefore Im [ a 2 ] will be summation of these results.

In general, since a 2 k + 2 = a 2 a 2 k , we will obtain Re [ a 2 k + 2 ] by multiply the first row of ω ( a 2 ) by ( a 2 k ) T , and to obtain Im [ a 2 k + 2 ] we multiply any row (except the first row) of ω ( a 2 ) by ( a 2 k ) T .

In case Re [ a ] = 0 ( a is a pure octonion) and Im ( a ) = a 1 i 1 + a 2 i 2 + a 3 i 3 + a 4 i 4 + a 5 i 5 + a 6 i 6 + a 7 i 7 0 , if n is an even number then:

Re [ a n ] = [ ( a 1 2 + + a 7 2 ) ] n / 2 Im [ a n ] = 0 (10)

And if n is an odd number then:

Re [ a n ] = 0 Im [ a n ] = a [ ( a 1 2 + + a 7 2 ) ] ( n 1 ) / 2 (11)

3.2. Example

Let a = i 2 + 2 i 3 3 i 6 + i 7

A = [ 0 0 1 2 0 0 3 1 0 0 2 1 0 0 1 3 1 2 0 0 3 1 0 0 2 1 0 0 1 3 0 0 0 0 3 1 0 0 1 2 0 0 1 3 0 0 2 1 3 1 0 0 1 2 0 0 1 3 0 0 2 1 0 0 ]

From (10),

Re [ a 6 ] = [ ( 15 ) ] 3 = 3375

Im [ a 6 ] = 0

So, a 6 = 3375

A 6 = [ 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 0 0 0 0 0 0 0 0 3375 ]

From (11),

Re [ a 5 ] = 0

Im [ a 5 ] = ( i 2 + 2 i 3 3 i 6 + i 7 ) [ ( 15 ) ] 2 = 225 ( i 2 + 2 i 3 3 i 6 + i 7 )

A 5 = [ 0 0 225 450 0 0 675 225 0 0 450 225 0 0 225 675 225 450 0 0 675 225 0 0 450 225 0 0 225 675 0 0 0 0 675 225 0 0 225 450 0 0 225 675 0 0 450 225 675 225 0 0 225 450 0 0 225 675 0 0 450 225 0 0 ]

3.3. Example

To compare our formulas with the De Moivre’s formula and Euler’s formula that were used in [7] to find a 7 for a = 1 + i 2 + i 4 + i 6 , take

A = [ 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 ]

From (8),

Re [ a 7 ] = j = 0 3 ( 7 7 2 j ) [ ( 3 ) ] j = 1 ( 7 5 ) ( 3 ) + ( 7 3 ) ( 9 ) ( 7 1 ) ( 27 ) = 64

Im [ a 7 ] = ( i 2 + i 4 + i 6 ) j = 0 3 ( 7 6 2 j ) [ ( 3 ) ] j = ( i 2 + i 4 + i 6 ) [ ( 7 6 ) ( 7 4 ) ( 3 ) + ( 7 2 ) ( 9 ) ( 27 ) ] = 64 ( i 2 + i 4 + i 6 )

So, a 7 = 64 ( 1 + i 2 + i 4 + i 6 ) , and

A 7 = [ 64 0 64 0 64 0 64 0 0 64 0 64 0 64 0 64 64 0 64 0 64 0 64 0 0 64 0 64 0 64 0 64 64 0 64 0 64 0 64 0 0 64 0 64 0 64 0 64 64 0 64 0 64 0 64 0 0 64 0 64 0 64 0 64 ]

Also let us compute a 8 .

From (7),

Re [ a 8 ] = j = 0 4 ( 8 8 2 j ) [ ( 3 ) ] j = 1 ( 8 6 ) ( 3 ) + ( 8 4 ) ( 9 ) ( 8 2 ) ( 27 ) + ( 81 ) = 128

Im [ a 8 ] = ( i 2 + i 4 + i 6 ) j = 0 3 ( 8 7 2 j ) [ ( 3 ) ] j = ( i 2 + i 4 + i 6 ) [ 8 ( 8 5 ) ( 3 ) + ( 8 3 ) ( 9 ) ( 8 ) ( 27 ) ] = 128 ( i 2 + i 4 + i 6 )

So, a 8 = 128 ( 1 + i 2 + i 4 + i 6 ) , and

A 8 = [ 128 0 128 0 128 0 128 0 0 128 0 128 0 128 0 128 128 0 128 0 128 0 128 0 0 128 0 128 0 128 0 128 128 0 128 0 128 0 128 0 0 128 0 128 0 128 0 128 128 0 128 0 128 0 128 0 0 128 0 128 0 128 0 128 ]

4. Conclusion

The formulas presented in this work are more suitable for computing the powers of an octonion number (the powers of matrices representing an octonion number). These formulas which are derived from binomial expansion also can be used to compute the power of a quaternion number (the powers of matrices representing a quaternion number), and a complex number (the powers of matrices representing a complex number).

Cite this paper: Ahmed, W. (2021) Powers of Octonions. Applied Mathematics, 12, 75-84. doi: 10.4236/am.2021.122006.
References

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[4]   Graves, R.P. (1882) Life of Sir William Rowan Hamilton. Hodges Figgis, Dublin.

[5]   Cayley, A. (1845) On Jacobi’s Elliptic Functions, in Reply to the Rev. Brice Bornwin; and on Quaternions. Philosophical Magazine Series, 26, 208-211.
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[7]   Bektaş, Ö. and Yüce, S. (2019) De Moivre’s and Euler’s Formulas for the Matrices of Octonions. Proceedings of the National Academy of Sciences, India Section A: Physical Sciences, 89, 113-127.
https://doi.org/10.1007/s40010-017-0388-z

 
 
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