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

Type 1 Portfolio: Matrix Binomials

Extracts from this document...

Introduction

Type 1 Portfolio: Matrix Binomials

Hudson Liao

12/7/2008


I was given the expression X = image00.pngimage00.png and Y = image73.pngimage73.png, where I calculate X2,X3,X4;Y2,Y3,Y4. Below I calculated X2 and made my way up to X4, where I also did the same with Y2 to Y4.

X2= image105.pngimage105.png

X3 or X2 * X1 = image117.pngimage117.pngimage129.pngimage129.png

X4 or X3 * X1 = image01.pngimage01.pngimage15.pngimage15.png

Y2 = image27.pngimage27.png

Y3 or Y2*Y1 = image43.pngimage43.png

Y4 or Y3+ * Y1 = image61.pngimage61.png

Xn

X1 = 1X = 20X

X2 = 2X = 21X

X3 = 4X = 22X

X4 = 8X = 23X

Now I am going to find and expression for: [Xn, Yn, (X+Y)n], by inputting different ‘n’ values. By doing this I can find a correlation between each variable.

Expression: Xn = 2(n-1) X

This general statement was found by finding a relationship through values from X1 to X4. In the Xn table, a pattern begins to form from 1X, 2X, 4X and 8X. If we simplify these numbers by using a constant value such as 1X = 20X we can find a general statement for this expression.

Yn

Y1= 1Y = 20Y

Y2= 2Y = 21Y

Y3= 4Y = 22Y

Y4= 8Y = 23Y

Expression: Yn = 2(n-1) Y

The same method to determine the general statement for the expression Xn = 2(n-1) X was also used for Yn = 2(n-1) Y.

(X+Y)n

2I

4I

8I

16I

...read more.

Middle

8Y

b=2

8Y

32Y

128Y

b=3

18Y

108Y

648Y

b=4

32Y

256Y

2048Y

By considering integer powers of A and B, find expression for An , Bn and (A+B)n

For the statement A= aX, I am going to determine a general formula by inputting different numbers for the constant ‘a’ and as well for the terms ‘n’.

An=(aX)n = image116.pngimage116.png

If I input a=1and n=2 into An=(aX)n, the resulting value would be 2X:

image118.png

image119.png

image120.png

However if I continue to input a=1  and change the terms ‘n’ to 3 and 4a pattern begins to form:

image121.pngimage121.pngimage122.pngimage122.png

image123.pngimage123.pngimage124.pngimage124.png

image125.pngimage125.pngimage126.pngimage126.png

By changing the terms n=2 up to n=4 ‘X’ increases each time from, 2X, 4X to 8X. The number of X’s that is being increased is resulted from this expression, “2n-1”. Therefore, we can convert the formula An=(aX)n to:

image127.pngimage127.png

image128.png

The same expression can be also used for the statement B=bY because both of the statements, A=aX and B=bY have the same pattern. The only difference between the two statements is that ‘X’ and ‘Y’ have different matrices. Therefore we just change, An  to Bn by:

image128.png

image130.png

If the same values that were inputted for An=aX to Bn=b

...read more.

Conclusion

‘n’ cannot be a negative number.

In this second example I am going let ‘n’ equal to a positive integer and the constants a and b equal zero:

image56.png

image57.png

image58.png

image59.png

The limitation for the expression  image60.pngimage60.pngis that ‘n’ cannot contain a negative exponent nor a decimal value or a fraction because if we multiply an exponent raised to a negative number it would make the value flip. However, both of the constants ‘a’ and ‘b’ can equal to any set of real numbers. Therefore the limitations and scope are:

image62.png

image63.png

Use an algebraic method to explain how you arrived at your general statement.

The general statement that came fromimage64.pngimage64.png  is  image65.pngimage65.png This general statement should equal to image66.pngimage66.png.

To prove that this general statement equals to image66.pngimage66.png, I am going to expand the equation image53.pngimage53.png by using only variables:

image25.png

image67.png

image68.png

image69.png

image70.png

image71.pngimage71.png.

=image72.pngimage72.png

image74.png

image75.png

image76.png

image77.png

Therefore the equation image60.pngimage60.pngequals with image78.pngimage78.png

However, the equation image60.pngimage60.pngwould not work unless it is proven by the binomial theorem.

image79.pngimage79.pngxn-kyk

image80.png

image81.pngimage81.png

=image82.pngimage82.png+2image83.pngimage83.png+image84.pngimage84.png

image85.png

image86.png

This calculations tells us that AB must equal to zero for this equation to work image87.pngimage87.png. As said before the only way the equation image53.pngimage53.png works is because image88.pngimage88.png equals to a zero matrix: image89.pngimage89.png.

In the end, the expression image53.pngimage53.png can be substituted into a different equation where (aX)n+(bY)n can be replaced as image90.pngimage90.png.

image31.png

image91.png

...read more.

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IA- Type 1 The Segments of a Polygon

    the value which is squared to form the denominator is one less than n and after it has been subtracted, the value is squared, hence, forming the equation above. Now moving onto the numerator, the following table helps analyze the situation better.

  2. Math Portfolio: trigonometry investigation (circle trig)

    The frequency is approximately 0.3183 when expressed to radian. When we observe some specific features of the above graphs, we can see how the value of b affects graphs of y=sinbx and y=cosbx.

  1. Math IA Type 1 In this task I will investigate the patterns in the ...

    Now I will see if a similar conjecture can be made for cubic polynomials. First I will look at a few examples to see if a pattern can be found but first it is important to redefine or rather, refine the equation of D because previously, there were 4

  2. Math IA - Matrix Binomials

    general equation listed above: In the sequence {1, 2, 4, 8}, a=1 r=2 However, we must note that this formula only gives us the progression for the scalar values which will be multiplied to the matrix, (X+Y) in order to yield the product (X+Y)n.

  1. Math Portfolio Type II Gold Medal heights

    Using technology a line of best fit of exponential nature was generated. The default equation of exponential function is however the parameters of the logarithmic function were already ladled so they have to be renamed in order to avoid confusion.

  2. Matrix Binomials IA

    M = M2 = = M3 = = M4 = = Using the expression: (A + B) n = 2 n-1 whereby a=2, b= -2, and taking n=3, we calculate (A+B) raised to the third power - (A+B)3 = 23-1 = 22 = 4 = So since we know from

  1. Matrix Binomials. In this Math Internal Assessment we will be dealing with matrices.

    b= n=8 B n =bn *Yn = b= n=2 B n =bn *Yn = b=n=5 B n =bn *Yn = The examples shown above show how the constants aand b can take on any value and still have a solution, only if n is a positive integer.

  2. I.B. Maths portfolio type 1 Matrices

    The result gathered was that in entries 'a' and 'd' which are number greater than one, follows this pattern: Mn = To show whether my assumption is right or wrong let us try a couple more value calculating it manually and then comparing it using the graphics calculator.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work