[JCH2CNEFTG_20150410] Exponential Half-Power Trick

Question

Introduction

     This is a complex-number question that appeared for ‘A’ Level Mathematics in November 1998.  Remember that mathematics is never out-dated.  Many Singapore schools keep this sort of questions in their question banks (for tutorial exercises, tests and examinations), in the hope that it becomes part of students’ repertoire.

     Students are expected to know the famous Euler’s Formula, one of the most beautiful formulas discovered by this visually-challenged but prolific mathematician.  It links the exponential function with trigonometry via the idea of angle rotation.  Adding e^if  with its reciprocal  e^-if (which is also its conjugate) gives a cosine expression, while subtracting gives a sine expression.

     The above question can be solved by rationalising the denominator and using heavy trigonometry and half-angle formulas.  There is nothing wrong with this approach.  I am going to illustrate a kewl approach, using what I call the exponential “half-power trick”.  Basically, whenever you see an expression like  1 ± e^{2kq i},  force out the factor  e^{kq i}.  This gives you either a sin or cos expression.  For example, e^{6q i} – 1 = e^{3q i} (e^{3q i} – e^{-3q i}) = i×2e^{3q i} sin q.

Solution

Observe that we have killed two birds with one stone.  At the last step, we just compare the real and imaginary parts to read off the answers.

Suitable Levels

* GCE ‘A’ Level H2 Mathematics (“Complex Numbers”)

* precocious students who love complex numbers

[H2PCRP20150406] Colouring The Pentagon (Combinatorics)

Introduction

     This is a question suitable for the mainstream Junior College students taking H2 mathematics, but is some primary school olympiad question from somewhere.  Whatever!  Mathematics is for everybody, young and old.  Anybody can solve this problem if  s/he makes observations and uses the right approach and thinking skills.

An Incisive Insight

     Although this pentagon is not a regular pentagon, the colouring scheme depends just on the order of colours on the edges.  We can start from one edge and see what colours are possible.  And then we can rotate the colouring scheme around.  [ We are breaking down and simplifying the problem. ] 

     Fiddling around with various possibilities, you might realise that:-

·  you cannot have three of the same colour going round the pentagon

·  you cannot have three sets of pairs of edges with the same colour.

·  you cannot two single colours and one double colour

There must be one single colour and two pairs of doubled colours.  All colouring schemes will have a  “12123”  colouring pattern going around in a loop.  The diagram below shows an example where colour 1 = yellow (Y), colour 2 = red (R)  and colour 3 = blue (B).

Do we need to consider a “21213”  pattern?  If  “12123”= “YRYRB”, we can later reassign colours, swapping R and Y to give 1=R and 2=Y and then “12123”=“RYRYB”.  So we have got that covered.  Let us worry about the reassignment later. 

Solution

     Observe that the position of the “3” (the single colour) can be rotated round the edges of the pentagon in  5  ways.

     Observe also that there are  3! = 3 ´ 2 ´ 1 = 6  ways to shuffle the colours i.e. assign colours 1, 2 and 3  to  Y, R and B.

The above two processes (rotation and shuffling) are independent of each other.  Rotation of the single colour can be done with or without the shuffling of colours.  Hence we can use the Multiplication Principle and calculate

          the total number of ways = 5 ´ 6 = 30.

Tada!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB Mathematics HL / SL

* Primary School Maths Olympiad

* other syllabuses that include combinatorics

[JCH2CNAGDH_20150402] Vector Rotation via Complex Numbers

Question

The points  D  and  G  in an Argand diagram represent the complex numbers  3i  and  6 + 7i  respectively.  DG  is in fact a diagonal of the square  DFGH.  Find the complex numbers represented by  F  and  H.

Introduction

     This is a question that tests students’ concept of the geometry complex numbers.  The key to solving this question is to understand that multiplying a complex number  re^iq  gives a stretching effect, as well as a rotating effect.  The stretch is by  r  times  and the rotation is by  q  radians anti-clockwise.  If  r = 1,  then there is just rotation and effectively no stretching.

Important Principles

Solution

[AJC Promo 2012 Q5] Range of Composite Functions with Inverse

Diagram 1.  Whole Question

     This is a question taken from Anderson Junior College, one of Singapore’s above-average junior colleges (in terms of the calibre of student intake).  By now this college is reputed to set the most difficult examination questions in Singapore.  It seems that they are trying to give the top junior colleges a run for their money, so to speak.  The question is difficult because it really tests students’ understanding of the concepts.  If you do not understand what is happening, you would be totally lost – even your graphing calculator (GC), the student’s favorite psychological crutch, would not be of much help.

Diagram 2.  Part (i) of Question

     Part (i) of the question tests students’ understanding of 1-to-1 functions (a.k.a. one-one functions or injective functions).  A function is invertible if and only if the function is one-one.  [ For the current A level H2 syllabus, it is assumed that the codomain is always the same as the range, so there is no need to worry about survjectivity. ]  Using the GC to graph the function  f   and to obtain the local maximum point, one sees that the required domain is  -2 < x < 0.  

Ans:  k = -2.   [ I am using black for explanations and blue for written answers. ]

     The domain is highlighted in yellow in the diagram below.  If the yellow region were to extend to the left beyond this point, it would be possible for a horizontal line to cross two points in the yellow region.  That would make the function  f  not 1-to-1 and hence not invertible.

Diagram 3.  Graph of f

Diagram 4.  Part (ii) of Question

For part (ii), we recall that a composite function exists if and only if the range of the first function (read from right to left) is a subset of the domain of the second function.
We require

Range of g  Í  Domain of f-1

The domain of  f^-1  (the second function) is actually the range of  f.  From the above diagram, the relevant part of  f(x)  goes from -¥  to  f(-2) = -2 + ln 4 = 2 ln 2 – 2 » -0.614.  The range of  g is everything from -1 downwards (see diagram below: imagine taking every possible point of  R, the domain of g on the x-axis and shooting them over to the y-axis).  We write

            Range of g     = (-¥, -1]
            Domain of f^-1 = (-¥, 2 ln 2 – 2]

Since Range of g Í Domain of f^-1,  therefore  f^-1g  exists.

Diagram 5.  Range of g

     Finding the range of composite functions is something that many students have difficulty with.  There are two methods: the direct method and the two-step method.  The direct method is usually difficult or infeasible.  In this case, finding range from the graph of  y = f^-1g(x) is practically impossible, because there is no simple formula for  f^-1.   

The two-step method: Step 1.  Find the range of the first function. Step 2.  Transfer this range to the x-axis of the graph of the second              function and map every point therein over to the y-axis.

     Step 1 has been done already.  We have found that  Range of g = (-¥, -1] . 

Diagram 6.  The Two-step Method

     For step 2, although the formula for  f^-1  is impossible to find (it’s a pretty nasty question, isn’t it?), we know that this graph is a reflection of the graph of  y = f(x)  (shown in blue on the diagram on the right) in the line  y = x, and we can sketch this (shown in red on the diagram on the right).  Now transfer the range of g from the y-axis of your first diagram over to the x-axis of this diagram on the right (shown in green).  Now imagine taking every possible point of this set and mapping it over to the y-axis of the second graph.  The problem is: how to find  f^-1(-1)  when we don’t even know the formula for f^-1?  (really evil problem, isn’t it?).  One way to deal with this is to make an educated guess for  f(what?) = -1. 

     Notice that f(-1) = -1.  Therefore  f^-1(-1) = -1.

     What if your intuition really sucks and you cannot make a guess?  The GC can come to your rescue.  Set up the graph of  y = f(x)  with the restricted domain and then intersect that with the graph of  y = -1.  The intesection is at  x = -1,  which means f(-1) = -1,  or  f^-1(-1) = -1,  as above.

Diagram 7.  Using Intersection on the GC

     Going back to diagram 6: From the range of g on the x-axis of the graph on the right, the points will land on every point from 0 down to f^-1(-1) = -1, including -1 but excluding 0 (because x-axis is an asymptote for  y = f^-1(x)).  We answer thus:-

     Range of  f^-1g = [-1, 0)

Diagram 8.  Part (iii) of Question

Part (iii) tests students understanding of increasing and decreasing functions.

f  is an increasing function means      whenever   a > b,   f(a) > f(b)   f  is an decreasing function means      whenever   a > b,   f(a) < f(b)

These are in fact the definitions of increasing and decreasing functions.  One can recognise an increasing function from its graph by the up slope (positive gradients) as you move from left to right.  For a decreasing function, the slope will be down as you move from left to right (negative gradients).  [An interesting note:  if  f  is a decreasing function,  then  f^-1  will also be a decreasing function.]

   In our case,  the graph of  f  is down-sloping, so it is a decreasing function. 

          Since  f  is a decreasing function,  whenever   a > b,   f(a) < f(b).

Once again we seem to have the pernicious problem of not knowing formula for  f^-1.  How to solve the inequality then?  Well, we can apply  f  to both sides of the inequality and the inequality reverses (because f is a decreasing function).  Note that  f  and  f^-1  “cancel”  as functions  i.e.  ff^-1(w) = w  for whatever the  w  is as long as it is well-defined.  Hence we proceed as follows

                                          f^-1g(x) > -1

                                        ff^-1g(x) < f(-1)

                                              g(x) < -1

                                          -1 - x² < -1

                                               - x² < 0

This latter inequality is the last trick on the question-setter sleeve desgined to unsettle the student.  How do you solve this inequality?  Do you need to equate or intersect with anything?  Anyway, what is the meaning of solve?  

To solve an inequality means to find all the possible values of  x  such that when you substitute each value into the inequality, the inequality becomes a true statement.

Note here that if we substituted  x = 0,  we would get  0 < 0, which is not true.  However, if we substituted any other real number,  x²  would always be a positive number, the LHS would always be a negative number, which is less than zero.  Conclusion:  x  can be any real number except 0.

Ans:  x Î R \ {0}

Note that the written solution (the parts typed in blue) is actually very short, although the explanation is rather long, because a lot of deep thinking is involved.

Reflection

Let us think back on the lessons learnt while solving this particularly difficult problem.

*  the reason why this problem seems difficult is because it tests students’ understanding of
    concepts (which most are weak in).  From experience with many cohorts of students, the
    JC teachers know what concepts students are weak in and they like to set questions that
    exploit the chinks in students armours.

*  remember the horizontal line test and domain restriction to get a 1-to-1 function, so that the
    function is invertible.

*  remember the condition for the existence of composite functions

*  inverse functions swap the domain and range with the original functions

*  the two-step method is recommended for finding range of composite functions

*  increasing functions preserve inequalities, while decreasing functions reverse
    inequalities.  You can recognise a decreasing function from the downward

    slope of its graph as you go from left to right.

Although you do not have the formula for  f^-1,

*  the value of  f^-1(-1)  can be found by intersecting the graph of  y = f(x)  with the horizontal line
     y = -1.

*  f  and  f^-1  “cancel” each other.

Finally,

*  what is the meaning of “solve an inequality”?

*  How to solve inequalities like  -x² < 0?  What about  -x² > 0?   x² > 0?

H2Maths VJC/2007/P2/Q9 The Power of Rephrasing

Introduction

     Today we discuss a question taken from one of the top junior colleges in Singapore’s 2007 preliminary examinations (final school internal examinations before the actual A-levels).  The seemingly difficult problems dissolve quickly using the right heuristics (rules-of-thumb / guidelines / problem-solving tactics).  Once the correct approach is determined, the calculation is simply a matter pressing buttons on the graphing calculator (e.g. TI-84 Plus or Casio fx-9860G).  This question can be tackled using heuristics.

     Try to guess what heuristic(s) will be useful in solving the above problem.  In a typical examination question, they will not tell you what topic or concepts are being tested in that question.  Try to determine what concepts are involved.

     Let’s tackle the question part by part.  As usual, we use the 5 step problem-solving process with metacognition (self-prompting, self-monitoring).

Part (i)

 

Part (i) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     That seems to be something to do with probability …

Do you understand the question?  Can you put it in your own words?

When Ai Wan (爱玩 “loves to play”?) rolls the die for the eight time, he gets exactly three ‘6’ to win the prize.  What are the chances of this event happening?

Part (i) Stage 2:  Planning the approach

Is there a heuristic you can use?

     How about “9. Restate the problem in another way” …

So how do you rephrase the question?

     “win on 8^th roll” means “two ‘sixes’ on the first 7 rolls” and “a ‘six’ on the 8^th roll”.

How will you proceed?

Break down the problem into parts (“11. Solve part of the problem”).

     (1) find probability of “two ‘sixes’ on the first 7 rolls”

     (2) find probability of “a ‘six’ on the 8^th roll” (this is easy.  Obviously  ¹/₆)

Then multiply them together, since these are independent.

How do you write out probability of “two ‘sixes’ on the first 7 rolls” in symbols?

P(X = 2)

Think backwards:  What is X?  How do you define it?

X  is a random variable.  It is the number of ‘sixes’ in the first 7 rolls.

What distribution does it follow?  How do you know?

It follows a Binomial Distribution: Because there are only two outcomes: either you get a ‘six’, or you don’t.  There is a fixed number of trials and these are independent, giving a constant probability. 

So what’s the number of trials (n)?

There are 7 rolls, so seven.  Each roll is a trial.

What is the probability of “success” (p)?

One out of six.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

This is just probability for a single value of  X, so it’s a p.d.f.  It can be found from the graphing calculator.  [ Or use the formula ⁿC_x p^x q^n–x = ⁷C₂ (¹/₆)²(⁵/₆)⁵ ]

Part (i) Stage 3:  Executing the plan

Part (i) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  It’s quite small, which tallies with my intuition, as getting three sixes in 8 rolls is highly unlikely.  If I get an answer like 0.4 something, I might smell a rat.  If I get a probability that is less than 0 or more than 1, then I know for sure it’s definitely wrong.

Part (ii)

Remark

     This problem is related to the Geometric Distribution, which is not in the syllabus.  Nevertheless it is solvable using the knowledge contained in the syllabus, so it is within the student’s reach.

Part (ii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     probably probability again … maybe Binomial or something related

Do you understand the question?  Can you put it in your own words?

Ai Ying (爱赢 “loves to win”?) already has got 2 ‘6’ and 2 ‘1’.  That means she needs just one more ‘6’.  She needs to throw exactly four more times (no more and no less).  What are the chances of this happening?

Part (ii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way” … but the problem still seems complicated because of the initial conditions (two “sixes” and two “ones” ) …

Is there a way to simplify the problem or another way to think about it?  Is there a feature about the situation … ?

     Ah!  Since all the trials are independent, what happens in the past does not affect the future.  This is the memoryless or forgetfulness property of the (Bernoulli) trials.  That means I don’t need to worry about the two “sixes” and two “ones”.  I can forget the past!  It is as though I can just start from a clean slate!

So what does it mean to say “Ai Ying needs to throw exactly four more times”?

     It means after throwing three more times, she does not get any six, and then she gets a six on the next throw.

How do you write this out mathematically?

P(S’, S’, S’, S)

What do you mean by S?  What is S’ ?

S  is the event of getting a ‘6’ in a roll of the die.  S’  means not getting a ‘6’.

Now, how do you proceed?

The chain of events can be broken down.  (“Split the problem into smaller parts”)

P(S) = ¹/₆.  P(S’) = 1 – ¹/₆ = ⁵/₆.  Since the four events are independent, I can just multiply their probabilities  all together!  This is easy!

Part (ii) Stage 3:  Executing the plan

 

Part (ii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  Again the answer is quite small, which tallies with my intuition.

Part (iii)

Part (iii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     Probability, Binomial Distribution and … er … rephrasing?

Do you understand the question?  Can you put it in your own words?

Ai Du (爱赢 “loves to win”?) cannot win in eight rolls, so he needs to roll some more.  What are the chances of this happening?

Part (iii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way”

So how do you restate the question?

 “requires more than eight rolls of the die to win a prize” means “after 8 rolls, he does not get three ‘sixes’”.

Which means?

Which means “after 8 rolls, he gets only 2 or less ‘sixes’”.

How do you write the required probability mathematically?

P(D < 2)

Think backwards:  What do you mean by D?

D  is a random variable that counts the number of ‘sixes’ within 8 rolls.

What distribution does it follow?  Why?

Binomial Distribution, again.  Same reasons as in part (i).

So what’s the number of trials (n)?  What is the probability of “success” (p)?

n = 8,  p = ¹/₆.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

Because it is a “<” probability, it’s a c.d.f, which can be found from the graphing calculator, or checking tables.  [ Calculation by hand is possible, but tedious.  Also, there is a recurrence formula that can help a bit, but this is out of syllabus. ]

Part (iii) Stage 3:  Executing the plan

 

Part (iii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  The answer is high, which tallies with my intuition.  Since it is hard to win, my gut feel is that you would very likely need more rolls of the die to win.

Stage 5:  Reflection

What did you learn from solving this problem?

     I learned that the memoryless (forgetfulness) property of Bernoulli trials (i.e. the type of trials involved in the Binomial distribution) is useful to allow me to disregard past events and simplify the problem.

     I learned that whenever I encounter a maths problem that seems difficult, I should not despair.  I should try the following heuristics:-

     * rephrasing the problem (heuristic 9)

     * breaking the problem into smaller parts (heuristic 11)

     * writing in mathematical notation (heuristic 13)

     * thinking backwards (heuristic 5)

     * simplify the problem (heuristic 10)

Rephrasing is particularly useful, as it helps to tackle all three parts of this exam question.  Even though part (ii) was on the fringes of the syllabus (just barely in the syllabus), we could still solve it by using heuristics, metacognition and using what we know already.

Conclusion

     Heuristics and metacognition will guide you when you seem to be in uncharted territory.  Use what you know (basic probability) to tackle what you do not know.  Do not be afraid.  Have faith.  May the Heuristics be with you!

 

JCCDQBHWHCB037(ii) Combinatorics : Grouping and Insertion Method

Introduction

     This question is from a source that does not credit the original source.  In the original question was poorly worded.  It did not have the words “the letters of the word”.  Instead of “each vowel must be separated”, it said “a vowel must be separated”, which is might mean there is just one such instance.  This is ambiguous.  I have taken the liberty to rephrase some parts of the question to make its meaning clearer.  In this article, I shall discuss only part (ii) of this question.

Stage 1:  Understanding the question

What is the given in the problem?  Can you organise the information?

     Though not absolutely necessary, it is helpful to draw a diagram that separate the letters into vowels and consonants and write the stack up the same letters in columns.

     There are  5  vowels (of which  O is repeated)  and  6  consonants (of which  N  and  S  are repeated).

Can you explain the problem in your own words?

     The letters of the word ‘CONNOISSEUR’ are re-arranged, which means that all the 11 letters are used.  Each vowel must be separated from another with exactly one consonant, which means that the letters must contain the pattern  “v c v c v c v c v” (where v = vowel, c = consonant).  Important: Note that the question does not say that the first letter must be a consonant.

Stage 2:  Planning

Have you seen a similar problem before?

     Yes, but this looks a bit more challenging.  There are more possibilities as first letter need not be a consonant.

What heuristics can you try?

     ·  Solve part of the problem

     ·  Split the problem into smaller problems

What topic-specific tactics can you try?

     The “v c v c v c v c v” pattern can be treated as a group (Grouping Method).  Since there are six consonants, there are two more “c”s (consonants) in the full pattern.  This looks like a problem that can use the Insertion Method.

Stage 3:  Execution

The number of ways to insert the group = ³C₁ = 3

     [these are the patterns “ccvcvcvcvcv”, “cvcvcvcvcvc” and “vcvcvcvcvcc”  ]

For each pattern,

     the vowels can be arranged in  ^5! / _2!  ways (division because there are 2 ‘O’)

     the consonants can be arranged in  ^6! / _{2! 2!}  ways (division because of 2 ‘N’ and 2 ‘S’)

Hence the total number of ways is

Stage 4:  Evaluation

Is the answer correct?

     Yes, the answer is correct.

Stage 5:  Reflection

What have we learned by solving this problem?

     We have learned once again that heuristics and metacognition are useful in solving mathematical problems.  Specifically, we have used the following heuristics:-

          ·  Drawing a diagram

          ·  Solve part of the problem

          ·  Split the problem into smaller problems

     We have also used the following techniques that are useful for combinatorical problems:-

          ·  Grouping Method

          ·  Insertion Method

          ·  Division Method (for dealing with repeated letters)

     It is also important to understand the problem correctly and not make unfounded assumptions.  If the wording is not clear, you may want to rephrase it.