Hi,
Applying reverse thinking in learning
--------------------------------------------------
Reverse Thinking is a method you can apply to any idea, any statement, any situation, any problem, any solution, any explanation, or any method. It works particularly well on ideas that we take for granted. Reverse Thinking assumes that the opposite way of understanding something - or the reverse idea, will be as interesting, if not more interesting than the first idea. It will certainly be more stimulating and likely to lead to more creative ways of thinking about things. Reverse Thinking is based on the insight that the conventional advice given to people who want to change their behaviour creates traps for them.
-- Extracted from "http://reversethinking.typepad.com/weblog/2007/09/what-is-reverse.html".
The Japanese used reverse thinking strategies, with great success, in the attack of Pearl Harbour. General Eisenhower's bold landing at Normandy beach, which ended WWII shortly, also demonstrated the apt application of reverse thinking contrary to conventional warfare.
Below are some reverse thinking strategies one can consider to maximise learning:
1. Conventional thinking: Everything in the content is important.
Reverse thinking: Some things in the content matter most of the time. I can identify my school's and the A-level exam trends of the recent few years.
2. Conventional thinking: Problem-solving is restricted to knowledge specific to the topic.
Reverse thinking: Problem-solving can involve application of knowledge of other topics.
3. Conventional thinking: I keep getting stuck with the questions.
Reverse thinking: I am able to recognise those situations when I tend to get stumped and I can find ways to resolve them effectively.
4. Conventional thinking: In a test or an exam, I have to complete a question in a fixed order.
Reverse thinking: I can attempt the parts that are within my ability if I am unable to do one portion. Exercising such flexibility allows me to maximise the opportunity to score marks whenever possible.
5. Conventional thinking: I keep making mistakes.
Reverse thinking: I am constantly reminded of the opportunity to learn from mistakes. I only need to be mindful of them, rather than become averse to or feel regretful over mistakes.
6. Conventional thinking: I am contented with getting correct answers.
Reverse thinking: I am also interested to find out how correctness can be checked and other methods there are to solve a question. I can go deeper to appreciate why the question was set in that manner, or even brainstorm about possible directions that the question may have taken.
7. Conventional thinking: I rely solely on given materials to learn.
Reverse thinking: I can customise the given materials to suit my unique learning needs. I can summarise important points in a way that makes sense to me. I can choose questions from other sources for further practice.
8. Conventional thinking: I learn too slowly.
Reverse thinking: I can find smarter ways to accelerate learning, humbly from capable classmates and by referring to helping websites. I am aware that I will need time to master concepts, so I need to be patient.
9. Conventional thinking: I can't do mathematics.
Reverse thinking: I can make up my mind to improve on this subject. I just need to keep trying, perhaps with the same level of resilience as Thomas Edison who finally succeeded, after almost two years of failed attempts, in finding the right lighting material for the light bulb. I know that every little effort counts, even though results do not show readily.
10. Conventional thinking: I face fear, worry and anxiety during a test or an exam.
Reverse thinking: Such a response is normal given the high stake of a test or an exam. I just try to focus my energy to conquer the paper - question by question, part by part - by writing whatever meaningful mathematical content I can think of.
11. Conventional thinking: I procrastinate whenever it has got to do with mathematics.
Reverse thinking: It is habitual to put off something I do not like. I will do mathematics first, without thinking too much about my dislike for it. Thinking while doing something is often harder than mere doing.
12. Conventional thinking: My teacher can't teach well.
Reverse thinking: I will not allow this to affect me in my effort to excel. I can consider other avenues of help to clarify the doubts I am having. I can also give feedback to my teacher in a tactful manner so that he/she will improve, benefitting my juniors in future.
Thanks in advance for reading.
Cheers,
Wen Shih
Hi,
Please read my new maths learning article that is related to statistics:
http://wenshih.files.wordpress.com/2010/02/tackling-an-area-of-difficulty-in-statistics-i.pdf
Thanks.
Cheers,
Wen Shih
Hi,
Students often write sqrt(x^2) = x, which is incomplete.
In fact, sqrt(x^2) = |x|, where
|x| = x if x >= 0
or |x| = -x if x < 0.
Think about the next five questions:
1. What is sqrt[ (x^2 - 2x + 1) ], if -1 <= x <= 1?
2. What is sqrt[ (x^2 + 2x + 1) ], if -1 <= x <= 1?
3. Solve 1/2 < x^2 < 3.
4. Solve 1/2 < e^(2x) < 3.
5. Make y the subject of the equation ln |y - 40| = kt + c.
Thanks.
Cheers,
Wen Shih
Hi,
Roots of quadratic equations is a topic within the O-level Additional Mathematics syllabus. We recall that if x = a and x = b are possible roots of a quadratic equation, we can form easily the quadratic equation which is given by
x^2 - (a + b)x + ab = 0.
The coefficient of x is the sum of roots and the constant is the product of roots. This fact is also applicable for complex roots.
We shall use this prior knowledge of roots of quadratic equations for the topic of complex numbers where one needs to find roots of a polynomial with real coefficients.
First, we state two important results involving complex numbers z and z* (the conjugate of z):
1. z + z* = 2 Re(z),
2. z . z* = |z|^2.
Now if z = z1 and z = z1* are possible complex roots, the quadratic equation involving z will be
z^2 - (z1 + z1*)z + (z1 . z1*), which can further be simplified to
z^2 - {2 Re(z1)} z + |z1|^2, using the earlier results.
Consider the polynomial equation 2z^3 + 5z^2 + 12z + 5 = 0, where -1 + 2i is a complex root. Another complex root is -1 - 2i. With the two complex roots, we can form the quadratic expression
z^2 - 2(-1)z + {sqrt(5)}^2 = z^2 + 2z + 5, since
Re(-1 + 2i) = -1 and |-1 + 2i| = sqrt(5).
The original polynomial expression can now be factorised quickly to
(z^2 + 2z + 5)(2z + 1),
from which we are able to find the remaining real root z = -1/2.
There may be situations where one is given a polynomial equation with unknown real coefficients to be found. Consider a slight modification to the previous example:
It is given that -1 + 2i is a root of 2z^3 + az^2 + bz + 5 = 0, where a and b are real constants. Find the values of a and b and write down the other roots of this equation.
Again, we find it practical to obtain the quadratic expression with the two complex roots, i.e., z^2 + 2z + 5. Then it is valid for us to say that
2z^3 + az^2 + bz + 5 = (z^2 + 2z + 5)(2z + 1),
from which the values of a and b can be found by comparing the coefficients of both sides.
One may generate other possibilities of two unknown coefficients appearing within the polynomial expression:
2z^3 + 5z^2 + az + b,
2z^3 + az^2 + 12z + b,
az^3 + 5z^2 + 12z + b,
etc.,
and still use the same approach of:
1. forming the quadratic expression,
2. factorising the polynomial expression,
3. comparing the coefficients of both sides,
to find the values of a and b.
Thank you for reading!
Cheers,
Wen Shih
Hi,
One type of question in mathematical induction deals with the need for the student to guess the formula by observing a sequence of terms. Guessing the formula is challenging if one is totally clueless about the direction to begin with. The way to tackle this issue effectively is to become aware of useful number patterns which I will now share with you.
Pattern 1: Even and odd numbers
Example 1:
u_1 = 2, u_2 = 4, u_3 = 6, u_4 = 8, ..., so u_n = 2n, where n >= 1.
Example 2:
u_1 = 1, u_2 = 3, u_3 = 5, u_4 = 7, ..., so u_n = 2n - 1, where n >= 1.
Pattern 2: Numbers following AP/GP
Example 3:
u_1 = 3, u_2 = 7, u_3 = 11, u_4 = 15, ..., so u_n = 3 + (n - 1)(4), where n >= 1, for a general term of an AP whose first term is 3 and common difference is 4.
Example 4:
u_1 = 1/2, u_2 = -1, u_3 = 2, u_4 = -4, ..., so u_n = (1/2)(-2)^(n - 1), where n >= 1, for a general term of a GP whose first term is 1/2 and common ratio is -2.
Pattern 3: Squares and cubes
Example 5:
u_1 = 1, u_2 = 9, u_3 = 25, u_4 = 49, ..., so u_n = (2n - 1)^2, where n >= 1, for a general term of a sequence of squares of odd numbers 1, 3, 5, 7, etc..
Example 6:
u_1 = 1, u_2 = 1/8, u_3 = 1/27, u_4 = 1/64, ..., so u_n = (1/n)^3, where n >= 1, for a general term of a sequence of reciprocals of positive integers 1, 2, 3, 4, etc..
Pattern 4: Powers and factorials
Example 7:
u_1 = 0 (or 2^0 - 1), u_2 = 1 (or 2^1 - 1), u_3 = 3 (or 2^2 - 1), u_4 = 7 (or 2^3 - 1), u_5 = 15 (or 2^4 - 1), ..., so u_n = 2^(n - 1) - 1, where n >= 1. Note that the power of 2 is one less than the subscript. Observe also, that a constant value of 1 is subtracted after calculating the power of 2.
Example 8:
u_1 = 1 (or 0!), u_2 = 1 (or 1!), u_3 = 2 (or 2!), u_4 = 6 (or 3!), u_5 = 24 (or 4!), u_6 = 120 (or 5!), ..., so u_n = (n - 1)!, where n >= 1. Notice the pattern: u_1 x 1 = u_2, u_2 x 2 = u_3, u_3 x 3 = u_4, u_4 x 4 = u_5, u_5 x 5 = u_6, and so on, which leads us to arrive at the possibility of factorials.
Pattern 5: Fractions
Example 9:
u_1 = 1/2, u_2 = 2/3, u_3 = 3/4, u_4 = 4/5, ..., so u_n = n/(n + 1), where n >= 1. We see that the numerators follow the subscripts and denominators are one more than numerators.
Pattern 6: Combinations of patterns 1 to 5
Example 10:
u_1 = 3/7, u_2 = 3/10, u_3 = 3/13, u_4 = 3/16, u_5 = 3/19, ..., so
u_n = 3 / [7 + (n - 1)(3)], where n >= 1. Numerators are fixed at the value 3 while denominators are terms of an AP whose first term is 7 and common difference is 3.
Thank you for reading, and may you have an easier time with conjectures from now on!
Cheers,
Wen Shih
Sorry guys,
I'm new to this sgforum page. Just realised I am not supposed to "advertise" in the homework section.
My sincere apologies.
Peter
Hi,
Guesses for 2010 H2 Mathematics exam:
1. Induction on a conjectured formula.
2. Area approximated by the sum of rectangles.
3. Graph of a derivative function.
4. Rates of change.
5. Differential equation involving substitution or formulation.
6. Vector algebra or vector geometry involving a given 3D-diagram.
7. Hypothesis testing involving an unknown (mu_0, x-bar, n, or alpha).
8. Permutations and combinations used to compute probabilities.
Feel free to add to this list, to benefit students who are preparing for the exams :) Thanks!
Cheers,
Wen Shih
Dear students,
It is a crucial skill for a student to know how to dissect the questions and gather the intention of the examiner as quickly as possible.
We shall look at an exam paper at:
staff.tpjcian.net/tay_boey_yiong/Maths/2010%20JC1%20SA%20(with%20answers).pdf
and analyse some of the questions in it.
Q1 This question is about binomial expansion and its concepts of expanding in increasing powers and finding an approximation by a suitable substitution.
Q2 This question focuses on the solving of an inequality by the algebraic approach. With the solution found, the student is next asked to consider a suitable substitution to solve a similar inequality.
Q3 This real-life problem requires the student to form a system of linear equations based on the information presented. With the unit prices found, one can then arrive at the utility bill for June.
Q4 This problem is about a recurrence relation in which the student is asked to determine its formula by observing some terms. The student is then assessed on summation concepts.
Q5 This question assesses the student on the method of differences. The difference of two similar expressions is first determined by the use of partial fractions. In the last part, the student is to determine the limit of convergence.
Q6 Tranformations of graphs is the theme of this question. Part (i) covers a sequence of two linear transformations. Part (ii) tests the student on the reciprocal graph y = 1/f(x). Part (iii) is about the graph of y^2 = f(x).
Q7 Concepts of functions are required to solve this problem. Part (a) looks at a quadratic function and assesses the student on graph sketching, the condition for its inverse to exist, and the definition of its inverse. Part (b) considers a logarithmic function and an exponential function whereby questions focus on the existence of a composite function, its definition and range.
As a form of practice, analyse the remaining questions 8 and 9.
Good luck in your coming exams!
Cheers,
Wen Shih
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Hi,
Some JC 1 students may be re-taking their promo exams in Dec or when school reopens. I have written some summaries in the hope that they will help you grasp key concepts faster. Please read article 22 at:
http://wenshih.wordpress.com/articles-about-learning-maths/
Have faith that you can make it, jiayou!
Cheers,
Wen Shih
Hi,
This is part 1 of Learning P&C I have written for some of my students:
To solve any P&C problem, we need to apply these important points:
1. Determine if it involves permutation or combination.
2. Identify any restriction and/or repetition.
3. Determine if addition and/or multiplication principle(s) should be used.
4. Determine if mutual exclusion principle should be used.
Example 1: Seven students are eligible for selection to a delegation of four
students to attend a seminar. Two of them will not attend together but each is prepared to attend in the absence of the other. How many ways can the delegation be chosen?
By point 1, the problem refers to combination, because of keywords like
"selection" and "chosen".
By point 2, restriction comes in the form of two students (say, A and B) not
attending the seminar together.
By point 3, addition principle should be used, because there are three
possible cases.
Suppose the two students are A and B.
Case 1: Student A and three others chosen from remaining five.
Number of ways = 5C3.
Case 2: Student B and three others chosen from remaining five.
Number of ways = 5C3.
Case 3: Simply choose four from remaining five.
Number of ways = 5C4.
By addition principle, total number of ways = 5C3 + 5C3 + 5C4 = 25.
Alternatively, we may consider the mutual exclusion principle mentioned in point 4:
Number of ways = number of ways without restriction - number of ways where A
and B are chosen,
that is,
number of ways = 7C4 - 5C2 = 25.
Thanks.
Cheers,
Wen Shih
Hi,
This is part 2 of Learning P&C.
Let's apply the previous points in another illustration.
Example 2:
Find the number of four-letter words that can be formed from the
letters of
the word TEACHERS. How many of these contain two E's?
By point 1, it
involves permutation since the order of letters matters.
By point 2,
restriction refers to the number of letters to be formed while
repetition
refers to the presence of two E's.
By point 3, we will use both
addition and multiplication principles. Addition
principle is used because
there are two cases.
Case 1: Four-letter words contain two EE's, which
is a restriction.
We apply the multiplication principle involving two
events E1 and E2. E1 is
the event of choosing two letters out of the those
that are not E's. E2 is the
event of ordering four letters of which two of
them are E's.
Number of ways for E1 to occur = 6C2.
Number of
ways for E2 to occur = 4! / 2!.
By multiplication principle, total
number of ways for case 1 = 6C2 x (4! / 2!) = 180.
Case 2: Four-letter words do not contain two EE's, which is also a form of
restriction.
Again, we use the multiplication principle involving
two events E1 and E2. E1
is the event of choosing two letters from seven
different types. E2 is the event of ordering four different
letters.
Number of ways for E1 to occur = 7C4.
Number of ways
for E2 to occur = 4!.
By multiplication principle, total number of ways
for case 2 = 7C4 x 4! = 840.
Finally by the addition principle, the two
cases give us a total of 180 + 840
= 1020 ways.
Note on my choice of
examples: Cambridge set a question in 2008 (similar to
example 1) involving
the selection of diplomats. It set a question in 2009
(similar to example 2)
involving word permutations.
Thanks.
Cheers,
Wen Shih
Hi,
This is part 3 of Learning P&C.
We now look at an example related to the arrangement of people in a line. This type of question was asked in the 2007 exam.
Example 3: A, B, and six persons stand in a line.
(a) Find the number of different possible orders.
(b) Find the number of different possible orders in which A and B are together.
(c) Find the number of different possible orders in which A and B are separated
by at least three persons.
By point 1, this problem is about permutations due to the keyword "different
possible orders".
By point 2, parts (b) and (c) involve restrictions which are respectively:
1. A and B are together;
2. A and B are separated by at least three persons.
We can solve part (a) directly:
number of ways = 8! = 40320.
We can use multiplication principle for part (b). Two events E1 and E2 are
involved. E1 denotes the event of arranging six persons and one group consisting of A and B. E2 denotes the event of arranging A and B within the group.
Number of ways for E1 = 7!.
Number of ways for E2 = 2!.
By multiplication principle, total number of ways = 7! x 2! = 10080.
For part (c), we will require both multiplication and addition principles.
To apply multiplication principle, we carry out three events:
E1 - Arrange six persons (in 6! ways);
E2 - Slot in A and B appropriately;
E3 - Arrange A and B (2! ways).
E2 will involve addition principle in which we could identify four cases.
Case 1: AXXXXXX
We notice that B may be slotted in via four ways, i.e.,
AXXXBXXX, AXXXXBXX, AXXXXXBX, and AXXXXXXB.
Case 2: XAXXXXX
We notice that B may be slotted in via three ways. i.e.,
XAXXXBXX, XAXXXXBX, and XAXXXXXB.
Case 3: XXAXXXX
See if you could identify two cases.
Case 4: XXXAXXX
See if you identify one case.
By multiplication principle,
number of ways = 6! x (4 + 3 + 2 + 1) x 2! = 14400.
Thanks.
Cheers,
Wen Shih
Hi,
This is part 4 of Learning P&C.
A common P&C problem involves circular arrangement. This type of question has
appeared in the 2007 examination.
Example 4: A team of two boys and two girls meet to discuss their project at a
round table with five fixed seats. Find the number of ways in which the members
of the team could be seated if
(a) they do not mind who they are sitting with,
(b) the two boys insist on sitting on adjacent seats.
Sitting problems are typically associated with ordering, so we have a
permutation problem.
In part (a), we arrange five items (two boys, two girls, and vacancy) without
any restriction. In part (b), we must consider two boys sitting next to each
other as a restriction.
We can solve part (a) directly:
number of ways = (5 - 1)! = 24.
In part (b), we may apply multiplication principle of two events E1 and E2. E1
refers to the event of arranging four items (two girls, 1 group consisting
of two boys, and vacancy). E2 refers to the event of arranging two boys within
the group.
Number of ways for E1 = (4 - 1)! = 6.
Number of ways for E2 = 2! = 2.
Total number of ways = 6 x 2 = 12.
Thanks.
Cheers,
Wen Shih
Hi,
This site by a HK teacher offers a nice set of practice questions on P&C:
http://johnmayhk.wordpress.com/2011/01/07/exercise-in-counting-balls-boxes/#more-6823
Enjoy!
Cheers,
Wen Shih
Hi,
This quote by Tony Alfonso is meaningful to be shared with you all:
When you miss a shot, never think of what you did wrong. Take the next shot thinking of what you must do right.
Often, when a student does badly in any assessment he/she becomes greatly affected and gets uptight when the next one comes. He/she is very likely to think of the past failure instead of taking the initiative to search actively for ways to right past errors or improve content understanding and problem-solving ability.
I hope all students will resolve to take the next assessment thinking of what they must do right in order to excel eventually to the grades they desire!
Thanks.
Cheers,
Wen Shih
Hi,
Statistics questions from H1 Maths exam have been more challenging (and interesting) than those from H2 Maths exam. I list some examples for the benefit of learning:
1. N2009/Q11(b)(ii) What do you understand by the term 'unbiased estimate'?
2. N2009/Q11(b)(iii) Find the set of values of alpha for which the null hypothesis will be rejected.
3. N2010/Q11(a) Draw a sketch of a possible scatter diagram of the data for which the product moment correlation coefficient is approximately -0.8.
4. N2010/Q10 Find the set of values within which the mean mass of this sample must lie for the owner's new claim to be accepted at the 5% level of significance.
5. N2008/10 Combining the two samples into a single sample, carry out a test, at the 10% significance level, of the same null hypothesis and alternative hypotheses.
Students taking H2 Maths may wish to consider attempting statistics questions from H1 Maths exam papers that are obtainable from their teachers.
Thanks and have a very happy Lunar New Year!
Cheers,
Wen Shih
Hi,
Maths is not linear, so learning it should be so.
Enjoy this presentation:
http://prezi.com/aww2hjfyil0u/math-is-not-linear/
Thanks.
Cheers,
Wen Shih
Hi,
This essay topic appeared in the 2010 GP exam:
Can mathematics be seen as anything more than a useful tool in everyday life?
Worth pondering ;)
Cheers,
Wen Shih
Hi,
This is a useful website about H2 Physics:
The wonderful creator is a current teacher at TJC. She hopes that the content she has put up will help students in their revision. I applaud her great effort and sincerity to benefit students' learning.
Thanks.
Cheers,
Wen Shih
Hi,
A common question about Complex Numbers is this:
Find the real and imaginary parts of e^(iθ) / { 1 + e^(iθ) }.
The usual method to obtain both parts at the same time is to consider:
1. multiplying the numerator by e^(-i θ/2),
2. multiplying the denominator by e^(-i θ/2).
Many students may not appreciate the reason for doing this.
The rationale for applying this technique is to enable the denominator to become real by the result
z + z* = 2 Re(z),
which, in our example, becomes
e^(i θ/2) + e^(-i θ/2) = 2 cos (θ/2).
Interestingly, alternative multipliers are possible:
A1. multiplying top and bottom by { 1 - e^(-iθ) },
A2. multiplying top and bottom by { 1 + e^(iθ) }.
In A1, we will obtain a denominator that is purely imaginary by the result
z - z* = 2i Im(z). Try and work the details on your own.
In A2, the denominator is purely real. Go try!
A1 and A2 may be practical in situations where the usual approach may not work, such as:
Find the real and imaginary parts of e^(iθ) / { 1 + ie^(iθ) }.
I would encourage you to go see the maths for yourself.
Thank you.
Cheers,
Wen Shih
Hi,
This article covers common errors in counting (permutations and combinations) problems:
www.nctm.org/eresources/view_media.asp?article_id=9112
Enjoy!
Cheers,
Wen Shih