Or
This expression supports the pattern observed.
2)
As observed from the previous question, the numbers inside the matrix are getting further apart exponentially. It is noticed that if the matrix is simplified by an exponentially growing scale factor, the elements in the matrix will remain within 2 of each. Therefore for the scale factor to increase by 2 it must exponentially increase by 2 also.
Calculations when n=3:
P3=3== 4
S3=3==
Calculations when n=4:
P4= 4= = 8
S4= 4= =8
Calculations when n=5:
P5=5= = 16
S5=5= = 16
Calculations when n=7:
P7=7==64
S7=7= =64
Calculations when n=10:
P10=10==512
S10=10= =512
Calculations when n=20:
P20=20==524288
S20===524288
It can be observed that all the matrices have numbers in them which are within 2 of each other.
P3=3== 4.
From the above example, a scale factor of 4 is obtained by the multiplication by 2 (one less than the exponent that is used to multiply the matrix)
2(3)-1= 4, this proves that the scale factor does indeed exist in the formula. when it came to the matrix part of the equation there was slightly more difficulty, however the numbers that were inside the matrix were divided by the scale factor which gave numbers within 2 of each other. Therefore the numbers would always be the same number, this would be true but for some reason they turned out to be 2 apart. Thus one of the elements in the expression would have to add one, and also subtract one, which is very closely related to geometric sequences. It was noticed that the numbers in matrix “P” would increase each time by a certain number. A and D, increase as follows; (3, 5, 9, 17…) while B and C increase as follows; (1, 3, 7, 15…). Both seem to follow a very similar pattern as numbers of A and D increase by exponents of 2 each time: (3+21=5, 5+22=9, 9+23=17) and the number: (1+21=3, 3+22=7, 7+23=15). The numbers were seen to have grown apart each time by 2. Therefore a similar expression would have to be made and an equation was derived to suit the pattern;
2n-1
This derivation was tested and proved through these following examples.
P3=3==2n-1=23-14.
Using this logic to form the above expression, it was realized that the same number would have to be multiplied by the same exponent. (Apart from them differing by 2)
This gave the formula:
Pn =2n-1.
To test for error in formula I plugged in all matrices for ‘P’
P5=5= = 16, 2n-1=25-1= 16
P7=7==64 =2n-1=27-1=64
P10=10==512=2n-1=210-1=512
P20=20==524288=
2n-1=220-1=524288
The formula works with all values of Matrix ‘P’ therefore the equation can be assumed will work for most values of ‘n’ if not all.
Matrix “S” was a much clearer pattern. It was evident that the expression for the matrix would have to be based on the same principle. The in the numbers A and D differ from the numbers from B and C by 2. It can also be noted that the numbers in the matrix are always larger than the ones in the P matrix. I manipulated the expression for P by plugging in numbers like 6 and 5 to obtain an expression for S but these to numbers were far too great for all intents and purposes. Therefore I made it smaller and approved 3 as the number for the expression in S.
2n-1.
Tested that the formula was correctly worked:
S3=3==4=2n-1=.23-1=4
S4= 4= =82n-1=.24-1=8
According to the question the Matrices P= and S=.
This pattern can be written as:
2n-1,
For the S values the pattern can be written as:
2n-1 .
3)
K= 1 M =
K= 2 P =
K= 3 S =
K= 4 =
Call this matrix D:
D2=2 = = = 2
D3=3= =4
D4=4 = = 8
As seen in matrices P and S, the matrix “D” has a similar pattern with the inside numbers being within 2 of each other. It is also noticed that if the sum of the numbers inside the bracket are added, it is always quadruple that of the previous solution. This is shown below:
D4=4 = 8 = 257+255+255+257 = 1024
D3=3= 4 = 65 + 63 + 63 +65 = 256 x 4 = 1024
Thus we can derive a equation for Rn :
Rn = 2n-1
Using R6, R7 and R10 we can test this expression in R.
By use of a graphic calculator (GDC) answered were determined to be:
R6 = 6 = = 32
R7= 7= 64
R10= 7= = 512
Now following the expression above:
R6 = 26-1 = 25 = 32
R7= 27-1 = 26 = 64
R10= 210-1 = 29 = 512
As a result of this above calculations the expression in terms of R is proven. As the value of K increased from 1-M, 2-P, 3-S and 4-R another pattern was observed. The pattern was found when the numbers inside the brackets were added together and as K increased so the multiplication factor increased from double to Triple and eventually quadruple. Thus a new equation can be formulated to read:
2n-1
4)
If we are to fully investigate the values of k and n, we must use various values one being negative, one being a real number and the other an integer.
For intents of investigating these values we will use K= -6, 0 and
When K=-6
By use of graphic display calculator:
-E2 2 = = 2
Applying the expression:
-E2= 22-1 = 2= 2
When K =0
=
Name this matrix F
By use of graphic display calculator:
-D3= 3 = = 4
Applying the expression:
-D3= 23-1 = 2= 4
When K =
= name this matrix T
By use of graphic display calculator:
-T3 = 3 = = 4
Applying the expression:
-T3= 23-1 = 4
This indicates that the expression is proved for all numbers including negative, real and integer numbers. Thus there are effectively no limitations for value K.
Therefore the scope of value K is in range of
-∞ ≤ K ≤ ∞ for ∑ R, Z, O numbers
To discover limitations of n we must use a value for n to prove the expression once again.
Let n=-2, 0 and
= name this matrix P
When n =-2
By use of graphic display calculator:
-P-2 = -2 = Err. Domain
Applying the expression:
-P-2 =2-2-1=
The graphic display calculator was unable to verify the expression but when done by applying the expression, the normal patterns are observed (the difference of the numbers in the bracket are 2)
When N=0
By use of graphic display calculator:
-P0 = 0 = =
Applying the expression:
-P0 =2-0-1=
When N=
By use of graphic display calculator:
== 8
Applying the expression:
==
Thus:
n= (-∞; ∞) ∑ R, O, Z numbers
It was seen that the graphic display calculator gave a domain error when putting in a negative value for ‘n’ however the using the general expression the patterns of difference by 2 for numbers inside the brackets. Therefore we can verify the expression acceptable by substituting a negative, integer and real numbers. K and N therefore will continue into infinity as all of their values work for M of our general expression:
2n-1
5) In this investigation of matrix powers, one can conclude that the calculator was one of the major limitations in finding my solutions which can be said to be technology is limiting. This was shown in my attempt to solve the expression for –P2 , which gave me a domain error. This can be explained by the calculator’s inefficiency to calculate numbers larger than a certain range. This is a perfect example of usefulness of expressions created to solve such inefficiencies. When considered in terms of k and n, we can say with fairly accurate readings that the expression was proven throughout the investigation by mean of integers, negative and real numbers. In this case I can say that my results hold strong.
