On generalisation

wee_ws SG

Dear students,

Mathematics is a discipline of patterns and due to this nature, the process of generalisation gets its way frequently into the subject.

In 2008 H2 maths exam paper 1, a question (Q9) was asked about generalising the behaviour of the graph of y = (ax + b)/(cx + e). The general features include:

1. If ad - bc is not equal to zero, then the graph has no turning points.

2. If ad - bc is equal to zero, then the graph is a horizontal line parallel to the x-axis.

Now, let's consider the graph of ax^2 + by^2 = c where a, b and c are non-zero constants. What general features (related to the syllabus content on graphing) can we observe?

1. If a, b > 0 and c = 1, then we obtain the graph of an ellipse.

For example, x^2 + 4y^2 = 9 is an ellipse with centre (0, 0), horizontal length of 6 units and vertical length of 3 units.

Note: Substituting convenient values for variables is a very practical strategy that I'd like to encourage students to carry out when investigating mathematical properties.

2. If a = b = 1 and c > 0, then we obtain the graph of a circle (whose centre is at the origin and radius of sqrt(c) units) which is a special case of an ellipse.

3. If a > 0, b < 0 and c = 1, then we obtain the graph of a hyperbola. The graph of a hyperbola can also be obtained by setting a < 0, b > 0 and c = 1.

For example, x^2 - 4y^2 = 1 is a hyperbola with centre (0, 0), oblique asymptotes y = x/2 and y = -x/2 and x-intercepts x = 1 and x = -1.

Next, we consider the graph of y = (ax^2 + bx + c)/(dx + e) where b, c, d, e are non-zero constants.

1. If a = 0, then we have the graph of a rectangular hyperbola.

2. If a is non-zero, then we have a graph with

- an oblique asymptote,

- a vertical asymptote,

- two turning points or no turning points.

There have been questions set in the past that required students to arrive at the condition involving the constants for which the graph has stationary points or none. Do look out for such questions and attempt them.

Cultivate the habit of generalising results to help you master mathematics smartly!

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

In another question (Q11) of 2008 exam paper 1, we saw yet another form of generalisation involving a system of linear equations.

Consider the equations:

2x - 5y + 3z = 3,

3x + 2y - 5z = -5,

5x + (lambda)y + 17z = mu.

Students were led to investigate different values of lambda and mu:

1. When lambda = -20.9 and mu = 16.6, the planes intersect at a unique point.

2. When lambda = -22 and mu = 17, the planes meet in a line.

3. When lambda = -22 and mu is not 17, the planes have no point in common.

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

Let me illustrate another application of generalisation involving loci with this question:

In an Argand diagram, the point A is represented by the fixed complex number a, where 0 < arg(a) < pi/2. Sketch on a single Argand diagram, the loci representing the complex numbers w and z for which |w - a| = |ia| and |z| = |z + 2a|.

If we analyse the question carefully, we see that

1. A lies in the first quadrant because we are given that 0 < arg(a) < pi/2.

2. |w - a| = |ia| = |i||a| = |a| represents a circle whose centre is at point A and having a radius of |a|, the distance OA. Take note of the steps we took to make the locus equation look like the standard form of a circle.

3. |z| = |z + 2a| = |z - (-2a)| represents the perpendicular bisector of OA', where A' is a point in the third quadrant and OA' = 2OA. Note also that A', O and A are collinear. Observe the steps we took to make the locus equation look like the standard form of the perpendicular bisector. It is a common error of students to ignore the sign and they end up drawing a perpendicular bisector passing through the wrong midpoint.

If it is not easy to observe these, then a practical way is to set A to be the point (1, 1) and things will be much clearer.

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

A fourth application of generalisation is the transformed graph of y = a f(bx + c) + d based on y = f(x).

This has been asked in 2007 exam paper 1 Q5: State a sequence of transformations which map the graph of y = 1/x to the graph of y = (2x + 7)/(x + 2).

Firstly, we express y = (2x + 7)/(x + 2) in terms of y = 1/x, since y = 1/x is the starting point. So y = 2 + 3/(x + 2).

Now we want to obtain the transformations involved to obtain the graph of y = 2 + 3 f(x) from the graph of y = f(x):

1. Scaling of factor 3, parallel to the y-axis.

2. Translation of 2 units, parallel to the y-axis.

Can the ordering be 2 followed by 1?

I have written more about transformations that one may be interested to study more deeply about:

http://www.freewebs.com/weews/graphtransformations.htm

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

Solution curves in differential equations is another type of generalisation.

In 2008 exam paper 1 Q4, we saw that y = 3/2 ln (x^2 + 1) + c is a general solution to the differential equation. With different values of c (say c = 0, c > 0 and c < 0), we can sketch members of the family of solution curves. In this case, we observe that all members will look alike in terms of shape. However, this may not so in general.

Consider the DE (x^2)(dy/dx) -2xy + 6 = 0 whose general solution is y = cx^2 + 2/x. If c = 0, we have the familiar rectangular hyperbola y = 2/x; otherwise we obtain very different graphs for c > 0 and c < 0. To see that, set c = 1, 2, 3 and c = -1, -2, -3 on your GC and compare the graphs.

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

Part 6 is about indefinite integrals at:

http://www.freewebs.com/weews/ongeneralisations.htm

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

Part 7 is about inequalities and is at the same website address as above.

Thanks!

P.S. The key objective of these articles is to enable students to elevate their understanding of mathematics beyond the mundane problem-solving so as to bring about a deeper appreciation of the subject in them.

Cheers,
Wen Shih

wee_ws SG

Hi,

I was going through the topic of geometric progression related to compound interest problems yesterday with my students and I saw yet another type of generalisation, which may be helpful for students to learn about. We shall discuss three cases.

Case 1 (one deposit): Suppose we deposit $10 on the first day of the month and the bank pays compound interest of 2% per month on the last day of each month. What is the amount at the end of n months?

It is always helpful to tabulate with "Month", "Start" and "End" columns to aid in arriving at a general formula. For the convenience of typing, I will use a linear representation instead.

Month: 1
Start: 10
End: 1.02(10)

Month: 2
Start: 1.02(10)
End: (1.02^2)(10)

Month: 3
Start: (1.02^2)(10)
End: (1.02^3)(10)

Month: n
End: (1.02^n)(10) <-- our desired answer!

Case 2 (multiple deposits): Suppose we deposit $10 on the first day of the month and the bank pays compound interest of 2% per month on the last day of each month. We put a further $10 into the account on the first day of each subsequent month. What is the amount at the end of n months?

Month: 1
Start: 10
End: 1.02(10)

Month: 2
Start: 1.02(10) + 10 <-- do not simplify
End: (1.02^2)(10) + 1.02(10)

Month: 3
Start: (1.02^2)(10) + 1.02(10) + 10
End: (1.02^3)(10) + (1.02^2)(10) + 1.02(10)

Month: n
End: (1.02^n)(10) + {1.02^(n - 1)}(10) + ... + 1.02(10) <-- our required answer, which is a GP sum.

Case 3 (multiple withdrawals): Suppose we deposit $100,000 at the beginning of the year and the bank pays compound interest of 10% per annum at the end of each year. After the interest is credited, we withdraw $12,000 immediately. We repeat the same action of withdrawal for subsequent years. What is the amount at the end of n withdrawals?

The column headings are "Year", "Start" and "End".

Year: 1
Start: 100,000
End: 1.1(100,000) - 12,000

Year: 2
Start: 1.1(100,000) - 12,000
End: (1.1^2)(100,000) - 1.1(12,000) - 12,000

Year: 3
Start: (1.1^2)(100,000) - 1.1(12,000) - 12,000
End: (1.1^3)(100,000) - (1.1^2)(12,000) - 1.1(12,000) - 12,000

Year: n
End: (1.1^n)(100,000) - {(1.1^(n - 1)}(12,000) - {1.1^(n - 2)}(12,000) - ... - 12,000, our required amount. Notice that

- {(1.1^(n - 1)}(12,000) - {1.1^(n - 2)}(12,000) - ... - 12,000 is a GP sum if we see it as

-12,000 {1.1^(n - 1) + 1.1^(n - 2) + ... + 1}.

Thank you.

Cheers,
Wen Shih

wee_ws SG

Hi,

Part 9 on the graphs of rational functions is up on my website. Thanks!

Cheers,
Wen Shih