[U_Complex20151117] Lagrange Identity for Complex Numbers

Question

Introduction

     I was revising my complex analysis just for fun (I had graduated almost 3 decades ago) and I came across this problem which is question 5 from page 9 of the book Complex Analysis by Ahlfors.  This book is pretty hard core for a first course in complex analysis, but this is not surprising since Lars Ahlfors was no less than a Field’s medallist.

     Intuitively, I knew that this identity looks like the vector identity

                                        |a · b|² = |a|² |b|² – |a ´ b|²

and if one strips away the |a|² |b|², it boils down to the Pythagorean Trigonometric Identity

                                         cos² q = 1 – sin² q

I thought if I could just define the appropriate dot and cross products (actually this can be done), I could solve it easily.  However, this itself requires proof.  I was begging the question.  In fact, it is precisely because of this Lagrange Identity (and the related Cauchy-Schwarz Inequality) that allows  cos q  and   sin q  to be meaningfully defined.

     OK, it looks like I have to do it the hard way.  The tricky part in the manipulation of those sums in sigma notation is to ensure, at each step, that I did not introduce any spurious terms, nor miss out any terms.  To simplify the notation, in what follows I shall assume that  i  and  j  are indices in the range of whole numbers  [n] = {1, ... , n}.

Solution

Remarks

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* University / College level Complex Analysis

* other syllabuses that involve Complex Analysis

* any independent learner game for a challenge

[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?

JCCDQBHWH_FN021(b) Range of Composite Functions

[original source unknown]

Introduction

     This question is from the usual book which did not credit the source.  It comes from some unknown junior college in an unknown year.  It is a challenging question because most students are poor at finding the range of composite functions.  Furthermore, this question has a little twist: you are given the range of the composite function, but you are required to solve for something.  So you need to, in a way, work backwards and/or use inequalities (another weak point for many students).

     Students are reminded of the right-to-left convention for functions in JC as well as GCE ‘A’ level examinations.  This means in ‘fg’ the  g  is done first before the  f.  There are some university professors who use a left-to-right convention, but here we do not.  So take note.

     Again metacogntion and heuristics are very important and I will illustrate their use.

Stage 1:  Understanding the Problem

What is this (part of a) question about?

range of composite functions, solving for unknown

What is given in the question?

The range of  fg.

What is the question asking for?

The value of  k  that leads to the given range.

Stage 2:  Planning the strategy

What heuristics do you think can be used for this question?

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

Can you recall the definition of the range of a function?  The range of a composite function?

Stage 3:  Execution

Any observations that can make your job simpler?

     Yes.  Observe that  g(x)  is a quadratic with positive  x²  coefficient.  So this is a parabola that looks like a happy smile.  To locate the minimum point, we can complete the square (a technique learnt in secondary school).

     g(x)  =  x² + 2x – 1  =  x² + 2x + 1² – 1² – 1  =  (x + 1) ² – 2

when  x = -1,  g(x) = -2.  The minimum point is (-1,-2).  So the range of  g  is all the

numbers from  -2  upwards.  i.e.  R_g = [-2,` \oo `).

So what now?

     With the two-stage method, suppose now we have

                  x ` \in ` R_g

That means?

                  x > -2

That means?

                  x + k + 1 > -2 + k + 1

Why do you do that?

     I want to slowly manipulate the LHS to get  ln(x + k + 1)  which is  f(x).  Continuing,

                  ln(x + k + 1) > ln(k – 1)

                                f(x) > ln(k – 1)

i.e.                            R_fg = [ln(k – 1), ` \oo `).

Why is there no switching in the inequality sign?

     The slope of the graph of  the natural logarithm is always positive (albeit getting less steep for increasing  x).  So applying  ‘ln’  on both sides does not change the inequality.

What is the clue again?

     We are told that  R_fg = [ln 3, ` \oo `).  Aha!  *epiphany*  *light bulbs flashing*

ln(k – 1) must be equal to ln 3!!!  This can be solved easily!

figure 1 – working forward and backwards

Stage 4:  Evaluation

Is the answer correct?

     Substituting  k = 4,  we see that    ln(x + k + 1) =  ln(x + 5)  and with  x > -2,  this will be  > ln 3  as given in the clue.

And why  x > -2?

     This is because g(x) > -2, which we knew  from completing the square.  We treat the  ‘g(x)’  as the  ‘x’  when applying  f,  because this is what  fg(x)  really means.

Stage 5:  Reflection

What did we learn from solving this question?

     We used metacognition to do self-monitor and self-questioning during the 5 stage problem-solving process.

     We used the following heuristics.

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

     We learned to apply the definition of the range of fg.  There are two possible methods: the one-stage method and the two-stage method.  The latter is usually better.

     In the two-stage method, the range of  fg  is found by first finding the range of  g  and then applying the function  f  to it.  After the first step of finding the inequality for  g(x),  we can simply use  x  in the formula for  f.  How?  We set  x  to be in the range of  g(x) from the previous step,  then slowly manipulate the inequality until the expression for  f(x)  appears.  This will give us the range of  fg.

     From the formula for the range of  fg,  we learned how to make use of the given clue to work backwards to find the unknown k.

     Difficult questions can be tackled by thinking systematically and logically, and using heuristics and metacognition.  Mathematics is hard, but it is fun after you have learned it.  If you have really learned it, you become more powerful because you can use the same technique to solve all kinds of problems in future.

Any of your own reflections?  Please post in the comments below.