[STEP3_2018_Q2] Sequence of Functions & Mathematical Induction

     This is a challenging problem on a sequence of functions, appearing as a Sixth Term Examination Papers (STEP) question.  STEP is the entrance exam for University of Cambridge and the University of Warwick undergraduate mathematics.  Some colleges and university departments may also require STEP.

     The student does not need to know that the question involves a Rodrigues type of formula.  However great facility in symbolic manipulation including algebra and calculus is needed, as this is what would be expected of students in a rigorous course involving mathematics or a related discipline.

     The first part is done via Differentiation using the Product Rule and the Chain Rule.

     For the mathematical induction proof in part (ii), the following is a rather standard way to begin.  You should do this even if you do not feel confident about the proof.  Just write it down, and worry later.  Say something like “RTP” (required to prove) or “to be proven” so as not to give the impression that you are making unproven assertions or making circular arguments, like what modern journalists and political activists are prone to do.

     The starting case is usually easier to handle.  Just follow your nose and differentiate using the Product Rule and the Chain Rule with  n = 1.

      The next part, the induction step, is the most challenging part.  The trick is to be clear about what is required and be observant.  There are no derivatives in the final required expression, and yet you should know that the earlier part of the question serves as a hint that you must use derivatives.  Using the induction hypothesis [IH], we end up with an expression that has two derivatives, which is a pain to do by hand.  So we repeatedly make use of [1] to convert back to some expression involving the function sequence, but not involving derivatives.  After some cancellation and simplification we finally complete the step.

     The following is the standard type of conclusion for mathematical induction proofs.  Just remember to write it in and earn the marks allocated.

     The last part is again challenging.  The key to solving it is to observe that  ‘x’  does not appear explicitly in the desired final expression.  So we proceed to try to eliminate the term that contains  ‘x’.  Examining the LHS would suggest the types of terms that we need formulas for, and upon subtraction, will kill off the term that contains  ‘x’.

     Ta da!  Done finally!

     To recap:  the strategies used to solve this question is observation, anticipation (know what you want at the ‘end of the rainbow’) and elimination (get rid of the unwanted term).  Needless to say, you would also need to be thoroughly familar with the standard ‘A’ level Further Maths stuff involving differentiation using the Product Rule and the Chain Rule, sequences and mathematical induction.
     

You are invited to join my group Effective and Elegant Mathematics on Facebook.

This article is suitable for

* GCE ‘A’ Level Further Mathematics

* Students doing STEP and/or students applying to study undergraduate mathematics in Cambridge / Oxford / Warwick

* other syllabuses calculus and sequences

* any learner who is interested

[OlymLSec20160201PHHE] Pigeonhole Principle and Harry’s emails

Problem / Question

Handsome Harry has a secret email account that only four friends know.  Today he received 8 emails in that account. Which of the following is certainly true? (A)  Harry received two emails from each friend. (B)  Harry cannot have received eight emails from one of his friends. (C)  Harry received at least one email from each friend. (D)  Harry received at least two emails from one of his friends (E)  Harry received at least two emails from 2 different friends.

Introduction

      This question is from some Kangaroo Mathematics Competition, which tests students on logic and not necessarily things from Singapore Mathematics syllabus. 

Solution

      (D)  Harry received at least two emails from one of his friends

Explanation

      This is an example of the Pigeonhole Principle.  Perhaps the easiest way to understand this is to imagine an array of pigeonholes with four columns (one for each of Harry’s friends) and pigeons (representing individual emails sent from the friends).  In the diagram below, I draw dots instead of pigeons.

As you can see, no matter how the eight dots / pigeons are placed, at least one of the friends will have at least two dots.  It is not possible for all the friends to have less than two emails.

Formal Proof

     We can use a proof by contradiction argument.  Suppose it were not true that Harry received at least two emails from one of his friends.  That would mean each of his  4  friends sent at most one email.  But then the total number of emails would be  4  or less.  This contradicts the given fact that Harry received  8  emails.  So this state of affairs is not possible.  Therefore, the opposite is true.  We conclude that Harry received at least two emails from one of his friends.

Final Remarks

      The Pigeonhole Principle is very useful in many situations, including computer science.  In general, if you have more objects (“pigeons”) than there are containers or slots (“pigeonholes”), one of the containers must have at least two of those objects.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels

* Lower Secondary Mathematics Competition / Olympiad

* other syllabuses that involve logic, combinatorics or Pigeonhole Principle

* any precocious or independent mathematics problem solver who is interested

[Pri_20151223WNSV] Unravelling Four Whole Numbers

Problem

If  A,  B,  C  and  D  are whole numbers such that  A × B = 8,  B × C = 28, C × D = 63,  B × D = 36,  find the values of    A,  B,  C  and  D.

Introduction

     This question seems to be taken from a secondary school textbook from a chapter on linear equations.  However, I think a good  primary school pupil could attempt this.

Strategy

     The key to solving the above problem is to make observations.  When you multiply up the first two equations, you get an  A,  a  C  and two  Bs  in the product.  Hmmm ... This doesn’t look promising ...  Ah!  But when you multiply the second and the third equations together, you get an  B,  a  D  and two  Cs  in the product.  This can cancel (via division) with the fourth equation which has one B  and one  D  in the product.

Solution

Remark

     Always cancel as much as possible, to avoid large numbers and reduce chances of making careless mistakes.

     By the way, a whole number is a non-negative (zero or positive) integer that does not contain any fractional part.   As such, the set of whole numbers is {0, 1, 2, 3, 4, ...}.  Thus we do not need to consider the negative square roots.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics (Challenge)

* Lower Secondary School Mathematics (Challenge)

* other syllabuses that involve whole numbers

* anyone game for a challenge

[AM_20151130IAXS] A Motif for the Absolutely Absolute

Problem

Introduction

     This question would pose a challenge for many students, although theoretically it is within reach of a good Additional Mathematics student (~ grade 10).  Graphs of both  sin x  and  cos x  are waves that oscillate up and down.  There are many pairs of vertical bars, indicating the absolute values or modulus, and these seem confusing.

Strategy

     Let us graph the functions  y = |cos x|   and  y = |sin x|.   Note that  ||sin x| – |cos x|| = ||cos x| – |sin x||.   The absolute difference of  |cos x|   and  |sin x|  is the difference between them ignoring the negative sign (if any) of the result.  And this is just the difference between the higher value and the lower value. 

Can you see any repeating patterns?  [H04]  Can you visualise the required area?  How many times is that of the basic pattern (known as “motif” in art)?  [H09, H10, H11]

Solution

Remarks

     Our total area is made up of four congruent pieces.  When  0 < x < ^p/₄,  cos x  is higher than  sin x.  That allows us to strip away all the absolute signs and do the calculation.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* challenge for GCE ‘O’ Additional Mathematics  * IB Mathematics SL HL

* GCE ‘A’ Level H2 Mathematics  * IB Mathematics HL

* AP Calculus AB & BC

* University / College calculus

* other syllabuses that involve integration and area

* whoever is game for a challenge in integration

[AM_20151127DF2D] Differentiation with Chunking and Elimination

Question

Introduction

     Although this looks like a differential equation question, the student is not required to solve the differential equation.  The requirement is just to derive the equation.  This would be a challenging question for secondary 4 (~ grade 10) students taking Additional Mathematics or their counterparts in Integrated Programme schools.

Strategy

     One way to do this is to differentiate the given equation once and again and just verify the equation by substitution.  The problem is that when we repeatedly apply the Product rule

the terms tend to sprawl.  A way to keep things neat is to try to recognise chunks and also use elimination.

Solution

Remarks

     After differentiating once, we notice that  10xe^2x   is twice of  5xe^2x,  and this allows the simplification in [1].  The second differentiation yields  10e^2x   which, we notice, is twice of  5e^2x.  We can get rid of that term.   Multiplying equation [1] by 2 gives  10e^2x  in equation [3],  which matches nicely with the same term in  [2].  So we can eliminate that term via elimination.  After that, we just need to rearrange things to get the final equation.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics, “Integrated Programme Mathematics”

* GCE ‘A’ Levels H2 Mathematics (revision)

* AP Calculus AB / BC (revision)

* University / College calculus (revision)

* other syllabuses that involve differentiation

* any learner interested in calculus

[S1_20151124AESR] Slanted Rectangle does not need Pythagoras

Question

Introduction

     This is another “Bonus Question” at a secondary level from somewhere that the question poser did not mention, but I guess it is most likely an Integrated Programme school in Singapore.  It is a beautifully crafted question.  The presence of a slant line seems to necessitate the usage of Pythagoras’ Theorem.  However, we have seen that Pythagoras’ Theorem can actually be avoided even in Primary (Elementary) School problems.  So a 10 year old kid with a rudimentary knowledge of algebra could do this.  Can you spot a short cut?

Making Observations

     Stare at the diagram for a while.  What do you observe?

Solution

             area of  DDB_nC_n =  ½  of the area of  AB_nC_nD.

              area of  DDB_nC_n =  ½  of the area of  DB_nPQ.

        \  area of DB_nPQ  =  area of AB_nC_nD = n cm².

HeuristicsUsed

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics

* challenge for Primary school Olympiad

* other syllabuses that involve areas and a tiny bit of algebra

* anyone game itching for a challenge

[AM_20151119] Anticipation and Bridging as Proof Tactics

Question

Introduction

     In a previous article, I have showed an “evil” tactic that can be used against “evil” questions.  Here is another “evil” trigonometric proof question.

     Many traditionalist school teachers insist on starting either from the LHS or the RHS, and working all the way to the other side.  Personally I do not mind any form of presentation as long as it is logical.  But not many students are able to do that.  People tend to fall into the trap of beginning a proof with the statement that they are supposed to prove in the first place.  This is called circular reasoning (or “begging the question” or petitio principii).  It is definitely a no-no.  So there is some advantage to sticking to the traditionalist mould.  The disadvantage is, of course, that it stifles creativity and this gives a misleading image of mathematics to the learner.

More “Evil” Tactics

     Now, if we do not want to “break the rules”, perhaps we can “bend the rules” a little.  On a piece of rough paper, or in your mind, secretly work from both sides and try to bridge them in the middle.  Let us compare

Think:  How are they similar?  How are they different?

As you can see, the LHS already has a preponderance of  cos 75°.  One of these  cos 75°  must somehow disappear.  The  RHS  has  4 sin 75°  which the LHS does not have.  So if we start from the LHS, we can use our magic wand [SV4] to create  4 sin 75°  out of thin air, remembering to divide by the same thing, so that the original value does not get changed.

Solution

Reflection

     Have you learned anything from this problem?  Let us review the heuristics used to help us solve this problem successfully.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Special Variants of Heuristics (Good for Trigonometric Proofs)

SV1   Work from both sides and try to bridge them in the middle [~ H04]

SV2   comparing (similarities/differences) what you have now with what you want [~ H05]

SV3   anticipate what will happen in the end and what you must do now [~ H05]

SV4   Magic Wand or create something out of nothing (无中生有) [~ H09]

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* IB SL & HL Mathematics (revision)

* other syllabuses that involve trigonometry

* just about anyone who is interested, really

[AM_20151119ISSD] Differences of Squares Hiding under Square Roots

Question

Introduction

     Here is another question involving surds.  As we know, surds are literally absurd, because they are irrational.  How to we do this one?

Strategy

Solution

Remark

     Reflect: What did you learn from solving this question?

     For another example of using the difference of squares formula, please look at this article.

Heuristics Used

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 ‘O’ Level Additional Mathematics (indices / surds)

* challenge for GCE ‘O’ Level “Elementary” Mathematics (indices)

* revision for IB Mathematics HL / SL

* other syllabuses that involve indices and/or surds
* any precocious or independent learner who is interested

[AM_20151117ISCP] Estimating a Crazy “Prosperous” surd to the Nearest Integer

Question

Introduction

     To the Chinese, the number  8  (八)  is considered to be auspicious, because it sounds like “prosper” (发) in the various Chinese languages/dialects.  But the LHS expression featured above seems too prosperous for comfort.  There is an explosion of  8s  coupled with eighth roots.  How to even handle that?

Strategy

     As usual, often one good tactic is to look for patterns or chunks.  [H04]  Can you see the sub-expressions containing the eighth roots (highlighted in green and pink)?

Do you notice any similarities between the two chunks?  Do you notice any difference(s)?

If we call the green chunk  a  and the pink chunk  b,  we can remove the roots by taking the eighth powers.  Then we get whole numbers, which are less complicated.

Solution

Remark

     The crux of the problem is the factorisation of  a⁸ – b⁸.  It is based on repeated application of the difference of squares    X² – Y² = (X + Y) (X – Y)  formula which schools expect students to know.
     For another example of using the difference of squares formula, please look at thisarticle.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* “IP Mathematics” so called
* challenge for students taking GCE ‘O’ Level Additional Mathematics
* other syllabuses that involve surds
* precocious kids who like to test themselves

[Pri20151117MSAS] MCQ tactic for the Area of a Hollow Square

Question

Solution 1

     Width of the smaller square  WX = (156 ¸ 4) cm   = 39 cm

     Width of the larger square     AB = (39 + 2´8) cm = 55 cm

     Area = (55² – 39²) cm² = 1504 cm² 

     Ans: (1)

Solution 2

     This is a Multiple Choice Question (MCQ).  Observe that the shaded area is an even number, because it is 8 cm width all around.  Since 1504 is the only even number among the options, (1) is the correct choice.

Remark

     No tedious calculation is needed!

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
* Primary School Mathematics
* other syllabuses that involve areas and perimeters
* anyone who loves his/her brain tickled

[S2_20150519XFCH] Chunking for Factorisation

Question

Introduction

     This is a question on factorisation (Am: factoring).  The key to doing this is chunking.  Can you see any chunk (maybe hidden) that is being repeated?  Take this out as the common factor and simplify.  Remember the square-of-differenceidentity  (a – b)² = a² – 2ab + b² .

Solution

Summary

     As in this question,  chunking is a very powerful tactic.  When you attempt to solve mathematics problems, open your eyes wide!

Heuristics Used

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (usually secondary 2 ~ grade 8 approx.)

* other syllabuses that involve algebraic factorisation and expansion

[S2_ABNPQX_20150502] Jack-in-the-Box of Fractions

Question

Introduction

     This number pattern question requires a lot of observation and algebra skills.  One also needs to understand how subscript notation like  F_n  works.

Solution

[S2_20150501AXC] 2011: Chunking and Substitution with Algebra

Question

Introduction

     This lower secondary algebra question seems complicated, doesn’t it?  Can you spot any chunk that is repeated, or almost the same?  This is one of the keys to solving the problem.  Another key that you need is the relevant algebraic identities and tricks.

Reminders

     First, let us review some of these useful formulas.                                        

This is the square-of-difference identity, illustrated here.

This swapping trick is illustrated here. 

     Looking back at the question, do you notice anything that is repeated?  Can you see any chunks that are the same or almost the same?  (n – 2011)  is almost the same as  (2012 – n)  isn’t it?  Whenever you see a repeated chunk, it is a good idea to substitute that chunk with another variable that you invent.  To name this new variable, you can use any letter that is not used before, so as not to conflict with existing letter(s).

Solution

Summary

     To solve the given problem, we have used the following:-

     (1)  square-of-difference identity

     (2)  swapping technique

     (3)  observation of repeated chunks

     (4)  using substitution with the chunks

Heuristics Used

H04. Look for pattern(s)  e.g. chunking, observation

H09. Restate the problem in another way  e.g. swapping, identities

H10. Simplify the problem e.g. substitution for chunks

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

· Lower Secondary Mathematics

· other syllabuses that involve whole numbers and ratios