Thursday, October 29, 2015

[S2_20151029QFCS] Chunking and Piggy-back

Question

Introduction
We have seen here, here and here that recognising chunks is useful in mathematics.  It is an example of a having what Carolyn Kieran calls a structural view of algebra, which is a type of pattern recognition.  In this article, I show how, using chunking and substitution, we can piggy-back or ride on solutions of a simpler equation to obtain solutions of a more complicated equation.
The solution for part (a) of is easy enough.  Just factorise (or AmE. “factor”)
(y + 2)(y – 7) = 0
Hence                 y + 2 = 0   or   y – 7 = 0
y = -2   or   y = 7
This is Standard Operating Procedure.  But part (b) seems like a monster of an equation.  Oh dear!  What shall we do?

Observation
Can you observe anything appearing more than once?
Now do you notice any chunks that are repeated?  Once you can see the connection, you can make a substitution  y = x3 – 1  and then make use of the answer in part (a).  We  copy and paste the chunk (shown in green) into the previous solution and proceed from there.

Solution to part (b)
x3 – 1 = -2     or     x3 – 1 = 7
x3 = -1     or           x3 = 8
x = -1     or            x = 2
Solved!

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Lower Secondary Mathematics (Sec 2 ~ Grade 8-9)
any syllabus that includes algebraic factorisation (factoring) and substitution

anyone who is interested in and ready for algebra

[U_20151029ITCX] Roger Cotes’ Integral of the Reciprocal of x^n – 1

Question

Introduction
The book VisualComplex Analysis by Tristan Needham recounts the story of RogerCotes who considered the above problem.  Without ostensibly using complex numbers, Cotes discovered a geometrical principle that helped to factorise the denominator  xn – 1,  and hence decompose the above integral.
In this article, I am going to “cheat” by using complex numbers to split up the denominator.  The fact that the denominator splits completely into a product of simple linear factors makes it easy to decompose the integrand into partial fractions.  Once this is done, I can single out the one or two fractions with purely real linear denominators.  Then I can pair up the conjugate fractions to get fractions with real quadratic denominators.  In other words, I apply a divide-and-conquer strategy, splitting up a big problem into smaller problems (Heuristics!).  Then I collect all the partial answers together to form my final answer.

Solution

Remarks
Note that I have only used real integration, not complex integration.  Complex numbers are used only to derive the various algebraic fractions.

H02. Use a diagram / model   (mentally: imagine roots of unity in a circle)
H04. Look for pattern(s)
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
University / college level calculus
high school students very strong and interested in calculus and complex numbers
* anybody who loves a challenging calculus problem and complex numbers

Wednesday, October 28, 2015

Question

Introduction
This question was most probably taken from an Integrated Programme (IP) school in Singapore.  For your information, students in Integrated Programme schools do not take the GCE ‘O’ Levels, and each IP school is free to design its own individual curriculum.  In practice, they incorporate the mainstream GCE ‘O’ Level topics, as well as additional topics, and/or teach topics in advance, and may call their syllabuses by different names.  They tend to set more challenging questions than the mainstream schools, which are already targetting their internal examination standards above the ‘O’ Levels.  In other words, they tend to cram in more, but never less.  Part (d) tests approximate change, which had been taken out of the mainstream syllabus at this time of writing.
Note that the IP schools tend to be schools that traditionally attract the academically best students from each cohort.  Even before the IP programme was introduced, these schools were already setting harder questions.  Anyway, this blog welcomes everybody from all around the world who is willing to learn, regardless of the type of school they are from, even home-schoolers and independent learners!  Let us see how we can employ re-usable tactics to tackle this question.

Get to the root of the matter, fast!
The question is on the applications of differential calculus on a quadratic curve (a parabola).  Observe that the equation of the curve is given in completed square form.  From the equation, can you spot the line of symmetry (centre line) and the  y-intercept immediately (like within 5 seconds)?  [You need to know all the basic facts at your finger tips and make observations.]
The line of symmetry always passes through the maximum or (in this case) minimum point.  We know that the minimum is when the squared term  (x – 3)2  is zero.  So
the line of symmetry is  x = 3.
To get the  y-intercept, put  x = 0.  This gives  y = (-3)2 = 9.  So
C = (0, 9)  and equation of  CD is  y = 9
Since  PQ = 2k,  distance from  P  to the centre line = k.  All the above are basic observations that should be carried out mentally within one minute and you should be able to mark the diagram with pencil notes (shown above in blue).  Once this is done, let us get on to the real business.

Solution

Final Remarks
For part (c), they have already told you it’s maximum, so you do not need to prove that it is maximum.  However, intuitively it is obvious there is a maximum: imagine if  k = 0  or  k = 3, then we get very thin rectangles with area zero.  As  k  increases from  0, the area gets bigger and after that shrinks towards zero again.
Actually, was this problem really so difficult?  What did we do to solve it?  Let’s review
· Know all basic facts and skills thoroughly at immediate recall (e.g. extremum of parabola lies on line of symmetry, how to spot that from completed square form, how to find y-intercept)
· Use heuristics: e.g. make observations.  Use simple facts you already know (e.g. subtract lengths, substitute values of  x  to find  y  which is “height” have x-axis etc,).  Practice positive psychology: instead of worrying, write down everything you can deduce.  Then try to find connections.
· Apply the formulas
· For part (d), if it is not in your syllabus, do not worry about it.  But if you are curious or feel the itch to learn more, it is also not too difficult.  See the boxed formulas in the solution above.  Just remember that the ratio of small changes  DA/Dk  is approximately equal to the derivative  dA/dk.  You can detach the  Dk,  bring it to the other side of the equation, and that allows you to approximate  DA.

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
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
· GCE ‘O’ Level Additional Mathematics
· other syllabuses that involve applications of differentiation

· anyone who is interested in calculus!

Tuesday, October 27, 2015

[S2_20151027ACS] Completing the Square

Question

Introduction
Many students in Singapore are taught to the method of “completing the square” in a rote fashion.  For example, to handle a quadratic expression like  x2 – 6x + 8,  take half of the coefficient of the  x term, put it with  x, square that resulting binomial, and then also subtract the square of that same number, like this:
x2 6x + 8 = (x 3)2 – (3)2 + 8 = (x – 3)2 –1
Note that the sign inside the squared binomial always follows the sign of the  x  term.  Just follow the procedure!  It works!
Does the student understand why this works?  And, by the way, does this always work?  What if the coefficient of  x2  is not  1?  Well, we need to pull out that coefficient.  Many teachers teach pulling out that coefficient from all three terms  e.g.
3x2 + 12x + 5 = 3[x2 + 4x + 5/3] = 3[(x + 2)2 – (2)2 + 5/3] = 3[(x + 2)27/3]  = 3(x + 2)2 – 7
A better way to do this is to just pull it out from the first two terms:
3x2 + 12x + 5 = 3[x2 + 4x] + 5  = 3[(x + 2)2 – (2)2] + 5  = 3(x + 2)2 – 7

Now it seems that as the questions get harder and harder, students need to memorise more and more procedures by rote.  And what if they encounter a question such as the one featured above?

What now?
Let us go back to basics.  The square-of-sum identity is
and the square-of-difference identity is
Instead of memorising procedures (not that these are wrong in themselves), why not use the above identities and think backwards (which is a type of heuristic)?  You can even make it a game of “filling in the blanks”, as shown below.

Solution

According to the identities, the two green patches must be the same and likewise the orange patches must be equal.  Working from the right end, we figure out that the orange patch must be  2  since  22 = 4  or  Ö4 = 2 .  Once we know this, we try to figure out the green space by matching the middle  y  terms:  2( ? y)(2)  =  32y.  So the unknown number (?)  must be  32 ¸ 4 = 8.  So the green spaces must be filled with  8y.  Since  (8y)2 = 64y2, we deduce that  k = 64.  Bingo!

Learning Points
· You definitely need to know formulas and procedures, but ...
· There is no holy grail of mathematics, but ...
· heuristics (e.g. thinking backwards, pattern matching) are powerful problem solving tactics.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
· Lower Secondary Mathematics
· other syllabuses that involve algebraic identities and completing the square
· anyone who loves a challenge!