[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

[P6_20160217RTTU] Books on Bookshelves

Problem

Introduction

     Here we have a numerically challenging problem that involves ratios, and it ultimately reduces to an algebraic problem with two unknowns.  Nevertheless, we are spoilt for choice as regards to methods of solution:-

     (1)   Bar Diagram Modelling

     (2)   explicit letter-symbolic Algebra

     (3)   “p” and “u”  (parts and units)

     (4)   Distinguished Ratio Units

     Despite the fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us remember that it is only one of the ways of solving problem by diagramming, which is just one of the eleven Primary School heuristics recommended by the Singapore Ministry of Education.
     The methods have a lot in common, and they differ mainly in the form of presentation.  However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces.  One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes.  We need to be flexible and open-minded.  I present a solution using my own Distinguished Ratio Units.

Solution

Ans:  735 books

Commentary

     First off, we need to equalise the numerators of  ²/₅  and  1¹/₄ = ⁵/₄  and put them ratio form.   This is because the  “2”  in the  ²/₅  represents the same quantity as the  “5”  in  ⁵/₄.

We do this adjustment by multiplying the former through by  5  and the latter through by  2.  Thus we deduce that the original number of books in A and in B are  25  and  8  “heart” units respectively. 
     Next, we add on the  2  and  3  “triangle” units.  By doing a comparison, we can figure out that  1  “triangle” unit must be  45  more than  17  “heart” units.  So  2  “triangle” units must be equal to  34  “heart” units plus  90.  Replacing the  2  “triangle” units (shown in yellow) with their equivalent, we now know that  59  “heart” units plus 90 gives  444.  This allows us to figure out that  1  “heart” is actually  6.  Thus, we can work out what  1  “triangle” unit, and then what  5 “triangle” units are worth.

Final Remarks

     Due to the difficulty of the numbers, the solution presented above is about as streamlined as I can make it to be.  

     There is another variation that can be used – equalising the “triangle” units (akin to the technique of elimination in standard algebra).  What we do is we multiply the group with total  444  by  3  and to multiply the group with total  489  by  2.  This would give  6  triangle units on each side.  Then we can compare the “heart” units and continue from there.  This way of proceeding is not for those who fear 4-digit numbers.

     If there are nicer or more elegant ways to tackle this question, I would definitely love to hear from you.

Heuristics Used

H01. Act it out

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any problem solver who loves a challenge

[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_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

[S2_20151107SLFC] Simultaneous Linear Equations with Fractional Coefficients

Question 

Introduction

     Somebody said, “Dear Algebra, please stop asking me about your  x.  She is not coming back.  Don’t ask me  y!”

     Here we have here linear equations that appear to be more tricky than the usual fanfare.  This is because the coefficients of the unknowns  x  and  y  are fractions.  In school, students learn to solve these using the method of substitution and the method of elimination.  They tend to prefer to do it by the former method, because psychologically it seems easier to accept.  However, the latter matter is generally more effective and yields a shorter solution.  This question is like a fly trap for those who prefer the method of substitution.  You make either  x  or  y  the subject from one of the equations, and then substitute that into the other equation.  This yields a complicated algebraic fractions within algebraic fractions.  The more complicated your solution is, the higher chance there is for making careless mistakes.  It is a good idea that students get out of their comfort zones and adopt a new skill.  So how do we do it?

The smart tactic

     First, we need to clear away the fractions by multiplying through with the LCM of the denominators appearing in each equation.  For the first equation, LCM(4, 8) = 8.  Since  4  is a factor of  8,  8  is like a giant that absorbs the number  4.  OK, so we multiply the first equation through by  8.  For the second equation, we multiply every term by LCM (3, 2) = 6

Solution

Remarks

     As you can see, equations [3] and [4] both contain – 3y  and this can be eliminated via subtraction.  So the value of  x  comes out easily.  Now once  x  is known, one can find the  y!

For another example of simultaneous equations, please refer to this article.

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

* Lower Secondary Mathematics (Secondary 2)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra, linear simultaneous equations

[OlymUSec_20150412BDP] Guessing Cheryl’s Birthday

Question

Introduction

     This “Primary 5 mathematics” (actually an upper secondary Olympiad) logic puzzle has gone viral.  It has been making its rounds in various forums in Singapore and overseas, stumping adults and children alike.  It is actually a parody of an old puzzle.  Can it even be solved?  It seems that there is no information given by each parties that we can exploit.  Actually there is!  In a subtle way ...

Solution

     In the beginning, everybody knows that Albert knows only the month and Bernard knows only the numerical day of the month.

     When Albert tells us “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” he is leaking out information (from his knowledge of the month) that the day of the month appears more than once and cannot be (June 18 or May 19).  Actually, the original phrasing is more like “If I don’t know when Cheryl’s birthday is, then Bernard does not know too.”.  The person who set this question merely changed the names of the people and the dates, without appreciating the subtle but crucial difference between a statement of fact and an implication (an “if ... then ... ” statement). 

     Ruling out June 18 and May 19, we also know that Albert knows that the birthday month is neither June nor May.  Otherwise, how would he have been so confident in saying that he knows Bernard would not know Cheryl’s exact birthday?  So we can eliminate those months.

     Bernard acknowledges the above state of affairs and the embedded hint.  With the choice narrowed down and with his knowledge of the numerical date, he now knows Cheryl’s birthday.  Since we know that Bernard knows Cheryl’s birthday, we know that it cannot be a numerical date that appears more than once (otherwise he would not have been able to know).  So we can cross out July 14 and August 14.

     Now Albert would telepathically thank Bernard for this helpful hint.  Because now he is able to deduce Cheryl’s birthday with his knowledge of the month.  That would mean that this cannot be a month with two candidate dates.  We blot out the August dates and see for ourselves the only remaining possibility.

Conclusion: Cheryl’s birthday is  July 16.

Remarks

     This puzzle was solved using the process of elimination and analysing our knowledge of what each party knows and can know.  Thus we successively narrow down the possibilities until the answer becomes obvious.  Here we learn that

     knowledge of other people’s knowledge can itself give us knowledge. 

This principle is actually employed in cryptology (the use of secret codes) which finds applications in fields like banking, the military (cf. interesting story of how the German Enigma code was broken in WorldWar II) and communications.  As an example, radio communication can tell the enemy of troop positions and warn of an impending attack, and that is why radio silence is imployed as a precaution.  Sensitive information in certain organisations is restricted on a “need to know” basis.

Suitable Levels
* Upper Secondary Olympiad
* other syllabuses that involve knowledge or epistemology
* application of mathematical principles in real life
* for all people interested in logic puzzles

[IB-HL H&H_Rev6C Q11] Factors of a Complex Polynomial

Question

Introduction

     This question is taken from the Haese & Harris textbook for IB HL Mathematics and is rather challenging.  Here I present two solutions.  In the first solution, I use substitution to make a variable “disappear”.  [ heuristics: H10. Simplify the problem, H11. Solve part of the problem]  And that allowed me to crack the rest of them problem.  In the second solution, I rephrased the problem in terms of tangents to curves. [ heuristic H09. Restate the problem in another way ] This gives another angle from which to tackle the problem.

Solution 1

Solution 2

Thinking Back

     Both solutions are related in the sense that they hinge on some form of the relation maked as [*].  That led to an equation in  a.  Once a  is found,  k  can be found, and the solutions proceed similarly.  The equation [*] is a manifestation of the fact that for a repeated root, both P(x) and P’(x) share a common factor.