[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.

HeuristicsUsed

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

[AM_20150616LGDFCA] Logarithms and Knowing that You are Correct

Question

Introduction

     This is a relatively straightforward question once the student has learned the rules of logarithms.  When I was first learning logarithms it took me quite some time to get used to the idea of “logs”.  Are they fallen trees?  So what are “logs”?  They are just the exponents or indices.  For example  2³ = 2 ´ 2 ´ 2 = 8  and we can write  log₂8 = 3.  Logarithm to base  2  of  8  is 3, because  3  is the index  i.e. to get  8  you need to multiply  2  by itself  3-fold.

     In general,  log_ba = x   Û   a = b^x.  Why?  Because that is exactly what logarithm means!  One way to remember this definition is to imagine: if you transport the log to the other side of the equation, the log drops off and you get the base  b  propping up the  x.  You can also do it the other way round.  If the base  b  of a power moves to the other side, it becomes a  “log”  with base  b.  [active mnemonics]

     What about the “common logarithm” lg?  It is the logarithm with base  b = 10.  In the days before pocket calculators were prevalent, students used books and slide-rules with logarithms of base  10  for multiplying and dividing large numbers.  Base 10 logarithms are still commonly used in today for the Richter Scale (in seismology, to measure earthquakes), for decibels (to compare the loudness of sounds or gain / loss in amplifiers), for pH (measurement of acidity / alkalinity in Chemistry) .... etc.  The aforementioned rule works exactly the same way, with  b = 10.

                      lg a = x   Û   a = 10^x                 (lg means log₁₀)

Note that in many calculators, their “log” button is for  lg  or logarithm of base 10.

Solution

Checking Your Answer

     The person who posted this question on Facebook got  33 333 333.3  as his answer, but did not realise that his answer is the same as the “model” answer, which is given to three significant figures in standard scientific notation.  Many students have the habit of checking their answers against the “model” answer usually given at the back of the book or worksheet, which may sometimes be wrong!  Anyway, in tests and examinations, you do not have the luxury of checking your answers like this.  In real life, if an engineer makes a calculation mistake, buildings may collapse and people die.  It is better to make it a habit to check your answers on your own and to know and be sure that you are correct.  One way to do this is to substitute the value of  x  back into the original equation to see if it works.  Nowadays, many models of calculators have a “store” function indicated by a button labelled with “STO” or an arrow “®” or something like that.  You can store the value into a variable (or memory location) like  X  and then key in something like  “log(3X) ”  and see whether you get  9  or something close.  Be aware that calculators can have rounding errors. 

Notations for “log”

     School students are taught to use “lg” to mean “log₁₀”  and  “ln”  to mean the natural logarithm “log_e”  where the special number  e = 2.7182818284 ...  discovered by the visually impaired but brilliant mathematician Euler.  Many calculators take “log” to mean “lg”  or  “log₁₀”.  For adult working professionals, “log” (without indication of the base) usually depends on what field they are in, or on the topic being discussed.  As mentioned before, base 10 is used for Richter scale, decibels and pH.  Computer scientists tend to use base  2  because of the binary system.  For rate of reaction (chemistry) or radioactive decay (chemistry / physics), the natural logarithm “ln” is often used.  In school, for the purposes of learning, we make the logarithm bases explicit.  Do not simply write “log”.  Write “lg”, “ln” or “log₂” or “log₇” or “log_b” (for whatever  b  is).  Note also that the letter “l” in all these notations is not the letter “i” or “I”, but it is the smaller case “L” (for logarithms).

Heuristics Used

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Levels Additional Mathematics

* International Baccalaureate (IB) Mathematics (revision)

* other syllabuses that involve logarithms and exponentials

[H2_20150512SSSTRR] Finding a Recurrence Relation for Terms in a Series

Question

Introduction

     This question pertains to the relationship between the partial sums of a series and its terms.  I am not sure if all the junior colleges teach this explicitly, but students are expected to know or be able to observe this relationship.  Let us follow our nose and focus on the first part first.

Reminders

     For the series  u₁ + u₂ + ¼ + u_n–1 + u_n + ¼  ,  the n^th partial sum

                    S_n = u₁ + u₂ + ¼ + u_n–1 + u_n

                 S_n–1 = u₁ + u₂ + ¼ + u_n–1

Taking the difference, we see that

                    u_n = S_n – S_n–1

Innocuous looking, this is actually a very powerful formula.  It is applicable to all sequences and series (not only for arithmetic and geometric series).  That means this formula can always be used!

     Another thing to note is that sequences  u_n  and partial sums  S_n  (which are themselves another sequence) behave like functions.  [In advanced mathematics, they are in fact defined as functions with domain as the positive integers.]  What this means is that  S_n-1  has the same formula as  S_n  except that  n  is replaced with  (n – 1). 

Solution

Checking

     Actually, the question setter forgot that the formula works for  n > 1. 

     OK, let us check whether the formula really works.  We know that  u₁ = 3.  Let us tabulate and compare the recursive formula with the explicit formula.  You can do this on a piece of rough paper.

n	recursive un = f(un–1)	explicit un = 3´2n–1
1	u1 = 3	3´21–1 =  3
2	u2 = 2´  3   = 6	3´22–1 =  6
3	u3 = 2´  6  = 12	3´23–1 = 12
4	u4 = 2´12 = 24	3´24–1 = 24
5	u5 = 2´24 = 48	3´25–1 = 48

Challenge

     What if the question wanted a recurrence relation for  S_n?

  

Heuristics Used

H04. Look for pattern(s)

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels, H2 Mathematics 

* International Baccalaureate Mathematics 

* other syllabuses that involve sequences and series

[H2_ACJC2000P1Q15b] Adders use Logs to Multiply

Question

Introduction

     Here is a question on arithmetic progressions, and not the first question of its type.  As you know, the schools in Singapore mimic questions from the GCE ‘A’ Levels, as well as from one another.  Before we go into the solution, let us go through some things you need to know.

Prerequisites

Solution

Remarks

     You might observe that the argument of the logarithm,  pq^n–1,  forms a geometric progression.  Indeed, any logarithm of a geometric progression will form an arithmetic progression.  However, this is not something that you should memorise.  Just stick to the basic principles and work it out.  Mathematics is not about memorisation.  It is about observing and understanding links between things.  If you want to memorise, ask: Why do adders like to live among logs?  Answer: That’s the way they multiply.  J

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

* GCE ‘A’ Levels, H2 Mathematics 

* International Baccalaureate Mathematics 

* other syllabuses that involve logarithms, arithmetic and geometric progressions

JCCDQBHWHCN029 Complex Numbers: Exponential Polar Form

[original source unknown]

Introduction

     Just as in my previous article, we are unable to trace the original source of this question.  This question contains an interesting link between logarithms and complex arguments.  For those interested, this link will be explored deeper at university level in the topic of Complex Analysis.

Prelude Stage 1:  Understanding the Problem

What is given?

     We are given  z  which is some quotient expression with complex numbers in rectangular form.  The denominator is the square of some complex number.  The numerator is actually  -1 + i  in a “reversed” form (be careful!).

What are you supposed to do / find?

     Find the modulus and prove that the argument is ` 5pi /12 `.

     Note that exact answers (using fractions, surds, logarithms, … etc but not decimals) are required, as with most maths questions at JC level.  You cannot use your calculator to show the steps of this problem, although you could numerically check your answers if you want.

What topic / sub-topic is this question on?

     Modulus and argument of complex numbers, cyclic roots (n^th roots of a complex number), exponential (polar) form.

Prelude Stage 2:  Planning the Method of Attack

How shall we do this?  What options are available?  Is there a short-cut?

     We could expand and simplify the bottom part, and then rationalise the denominator using the technique of complex conjugates.  After further simplification, we get something in rectangular form (something like ??? + ??? i).  Then find the modulus and the argument.  This method is do-able, but it is pretty tedious.

     We could also convert everything first to polar form, which is more suited for multiplication, division and powers.  There are two flavours of polar forms:
     (i)   trigonometric  r (cos q + i sin q) and

     (ii)  exponential  re^iq

I prefer to use the exponential polar form because it is more compact and yet it contains the same information  (r and q).  Then we can use the formulas for manipulating exponential polar forms.  This seems to be the ideal method, because we can “kill two birds with one stone”: we can handle the modulus as well as the argument calculations at one go.

Prelude Stage 3:  Execution

figure 2 - converting to polar form

     One should try to sketch or mentally visualize the complex numbers  i – 1  and  Ö3 + i  on Argand diagrams.  These complex numbers form special triangles with special angles (30º, 45º, 60º) with the horizontal axis, giving exact values for the moduli and arguments.

i – 1  forms a right-angled triangle with sides 1 and 1 and hypotenuse Ö2, making 45º with the horizontal in the left upper quadrant.  The angle turned from the real positive axis is thus 135º or 3p /4  radians

Ö3 + i  forms a right-angled triangle that makes  30º  or  p /6  radians with the positive real axis.

For checking, you can use your calculator to get the arguments, but you will get decimal answers.  After dividing by  p, you may convert the result to a fraction.  Thus you can verify the exact fraction of  p  for both arguments.

Now we proceed with the exponential polar form calculations.

figure 3 - solving the prelude

     Line #1 is the original expression which we convert to exponential form (line #2) as worked out earlier.  In line #3, we do modulus and argument (i.e. angle) calculations in a single step, with the modulus on the left of the ‘e’, and the argument on the right in brackets after the ‘i’.  The subtraction ‘–’ in the argument is because of the fraction/division in previous line.  Anything that appears below the fraction line will have their arguments negated.  The  p/6  is multiplied by 2 because of the squaring in the denominator.  All these are in accordance to these powerful exponential form rules:

     ·     ` r_1 e^(i theta_1) \cdot r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1+ theta_2))  `

     ·     ` r_1 e^(i theta_1) divide r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1- theta_2))  `

     ·     ` (r e^(i theta) )^n = (r^n) e^(i (n theta))  ` 

     After simplifying, we obtain line #4.  We should check that ` (5pi)/(12) ` is in the range

(-p, p] the principal range for complex arguments.  If not, we would need to adjust by adding or subtracting some multiple of  2p.  Finally, we just pick out the modulus (line #5) and the argument (line #6) to answer the question properly.

Prelude Stage 4:  Evaluation

Are we done for this part?  Is there another way to do this?

     Yes, we have answered the question as required, and we did it by combining the calculation with the proof.  We would also have done these separately, but it would have been longer to write out the solution.  This solution is short and sweet.

Is the answer correct?

     We can use the graphing calculator to verify that the modulus is correct.  Calculate and store the expression as ‘Z’.  Then we subtract Ö2/4  from the modulus (or absolute value) of ‘Z’ to see if we get zero.  This would show that those two expressions are equal.  The picture below shows how you can do it using Texas Instruments TI-84.  You can do something similar using the Casio fx-9860G.

figure 4 – numerically verifying the answer using a calculator

Yes!  Got it!!  As for the proof of the argument  5p/12, we just need to check that our steps were logical.

Part (i)  -- Piece of Cake

     This cyclic roots part of the question is solved by the standard method.  It is fairly straightforward, although the numbers are a bit ugly.  I shall just present my solution here below.

figure 5 – solution to part (i)

Part (ii) Stage 1:  Understanding the Problem

What are we supposed to find?

  

     We are to find the exact values of  a  and  b,  hence almost certainly, no decimal answers are accepted.  [This is because exact answers almost always entail an infinite number of decimal places, which our calculators are not able to supply.]  Although the  z³  term looks like we are need to solve for  z (as would be the usual case), let us remember that the  z  is actually given and known.  That is why we need to read the question carefully to avoid getting ensnared.

Part (ii) Stage 2:  Planning the Method of Attack

     The question looks a bit unusual, but do not be frightened.  We observed that the RHS involves an exponential expression, which suggests the use of the exponential polar form.  As for the LHS, since  z  is known,  we can actually calculate  z³.  In what form?  Rectangular or Polar?  In exponential polar form, as suggested by the RHS and also because this form is compact and especially convenient for the calculation of powers.  Let’s try this approach and see.

Part (ii) Stage 3:  Execution

figure 6 – solution to part (ii)

     For the LHS, we use the information taken from the earlier prelude to re-express  z  in polar form (line #2).  The key observation is that on the RHS, the + in the exponential index changes to a multiplication as we split  e^a+ib  into two parts.  Treating  e^a  as one entity (learn to observe and recognise ‘chunks’ as one), we see that the RHS is actually in exponential polar form  re^iq,  with  r = e^a  and  q = b.  In line #3, we apply the exponential law  (re^iq)ⁿ = r ⁿ e^{i (nq)}.  However, we note that  ` (5pi)/(4) `  is outside the required range.  It is more than  p.  So we need to adjust it by subtracting 2p.  We need to add or subtract as many times  2p  as necessary to keep the argument between  -p  and  p.  Here, we subtract just  2p.  [ Remark: although  ` (5pi)/(4) ` is  not the same as ` -(3pi)/(4) `,  ` e^(i(5pi)/(4)) ` is  exactly the same as ` e^(-i(3pi)/(4)) `, because they refer to the same complex number i.e. they are represented by the same point on the Argand diagram.  The difference is how much you turn clockwise or anti-clockwise from the positive real axis to get to that point.  ]

     After doing this adjustment (line #4), we are all set up to compare the moduli (highlighted in green) and the arguments (highlighted in pink) on both sides (lines #5 to #7).  We are using the fact that if  ` r_1 e^(i theta_1) = r_2 e^(i theta_2) `  then  ` r_1 = r_2 `  and    ` theta_1 = theta_2 `,  provided both  q ₁  and  q ₂  lie in the range  (-p, p].  Remember that in maths, it is useful to be able to recognise chunks.  In this context, the relevant chunks are those that give the moduli and the arguments, as highlighted.  To uncover  a,  we take logarithms (line #6).  Line #8 just writes the answer nicely.

Part (ii) Stage 4:  Evaluation

Are we done?  Is the answer correct?

     Yes, we have found the values of  a  and  b  and  b  is in the correct range,  as required.  We can verify the answers numerically using the calculator.  For example, like this:-

figure 7 – checking part (ii) answers

     Take full advantage of the graphing calculator’s ability to store complex numbers into variables like ‘Z’, ‘A’ and ‘B’.

Part (ii) Stage 5:  Reflection

What did we learn by solving this problem?

     We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems.  It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously.  It is convenient for manipulating products, quotients and powers.  We learned to apply the following rules for manipulating the exponential polar form of complex numbers.

     ·     ` r_1 e^(i theta_1) \cdot r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1+ theta_2))  `

     ·     ` r_1 e^(i theta_1) divide r_2 e^(i theta_2) = (r_1 r_2) e^(i (theta_1- theta_2))  `

     ·     ` (r e^(i theta) )^n = (r^n) e^(i (n theta))  `

     ·     if  ` r_1 e^(i theta_1) = r_2 e^(i theta_2) `  then  ` r_1 = r_2 `  and    ` theta_1 = theta_2 `,
            provided both  q ₁  and  q ₂  lie in the range  (-p, p]

     Using the last formula, we learned to compare moduli and arguments to solve for unknowns.

     [ Remark: Because of Euler’s Formula e^iq = cos q + i sin q, the idea behind De Moivre’s Theorem  (cos q + i sin q)ⁿ = cos (nq) + i sin (nq)  is actually subsumed by the third formula which implies that  (e^iq)ⁿ = e^{i( nq)}  when  r = 1.]

     We applied metacognition (self-monitoring and self-checking) to all the 5 stages of the maths problem-solving process.  We used heuristics like making comparisons, making observations, recognising relevant chunks, comparing.  We also practised being careful by checking all our answers and verifying that we answered the question in the form required, and verifying that the complex-argument answers are within the correct range.

If you encountered a similar problem in future, would you be able to solve it?

     [Please say “Yes!”, but make sure you can do it.]

Can you explain to a friend how to solve this sort of maths problem?

Can you set a similar question for yourself or your friend to solve?