[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.
     

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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

[H2_Expository] Derivative of the Arcsecant function

Introduction

     This article explains how to differentiate, i.e. find the derivative of, the arcsecant function, which is seldom discussed in class.  If you look at the various syllabus outlines, sometimes they do not explicitly mention or imply the arcsecant, neither do they explicitly exclude this.  The important thing is: we should be able to figure it out from our basic knowledge, which is the whole point of mathematics.  The people who talk about and focus on mathematics syllabus content as if it is the only or most important thing are missing the point of mathematics education, and there are a plenty of these idiots around.  Do not follow them.

     The arcsecant, written as arcsec or sec^-1, is also known as the inverse secant function.  So arcsecant means that, given the value of a secant function, you want “the” angle whose secant is that given value.  The problem is there are many possible values.  Look at the graph below.

Defining the Inverse of the secant properly

     On the graph, a horizontal line can pass through (infinitely) many points.  Like any periodic trigonometric function, the secant function is not a one-to-one (a.k.a. “injective” or  “one-one”) function.  As such, it is not invertible.  However, we can restrict the function so that its domain is  [0, p] \ { ^p/₂ }.  This is highlighted in yellow on the graph.

     Remember that sec x = ¹/_{cos x}.  Basically we follow the principal values  [0, p]  of the arccosine function   except  ^p/₂  where the cosine is zero and its reciprocal the secant is undefined.       With this restriction on the domain, we get a one-one secant function, with range (-¥, -1] È [1, +¥).  We can now define the inverse function, and its graph is obtained by reflecting the above graph along the mirror line  y = x.  We get this:-

     Observe that in the yellow regions in both graphs, the gradients at the points are non-negative.  We have chosen the domain of the secant function, which is the same as the range of the arcsecant function, such that the derivatives will be non-negative.

Deriving the Derivative (refer to the“Onion” Method for differentiation)

Remarks

     Do not confuse  arccos y   with  (cos y)^-1.  They are not mean the same thing.

     In case you are wondering, the prefix “arc-” means the angle, which (if you use radians) is literally the same as the arc-length when the radius equals to  1.  In symbols, s = rq  with r = 1 means  s = q.  Although the term can be used with degrees or other units, when doing advanced mathematics like calculus, we would usually be using radians anyway. 

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* International Baccalaureate (IB) HL Mathematics

* Advanced Placement (AP) Calculus AB & BC

* University / College calculus

* other syllabuses that involve differentiation

* any learner interested in calculus

[AM_20151127DTCR] The “Onion” Method for Differentiation

Question

Introduction

     The differentiation of the secant function is not taught directly as part of the Additional Mathematics syllabus.  It can be derived from known facts.  I first show the standard application of the (extended) Chain Rule for novices, and then show a more effective way of applying the chain rule, which I called the “Onion Method”.  This looks like peeling onions or unpacking Matryoshka dolls (“Russian dolls”). 

Reminders

Solution 1  (for beginners)

Solution 2  (a more expedient way)

Final Remarks

     When we peel onions, we peel from the outer layer inwards.  Likewise, when we have a composite function, we differentiate from the outer layer first, and then work to the inner layers.  Every time we differentiate a layer, we write down the changed layer and then copy and paste everything within that layer.  With regular practice, this should become second nature.
     For another example of the “onion”, take a look at the derivative of the arcsecant function.

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

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* GCE ‘A’ Level H1 Mathematics

* GCE ‘A’ Level H2 Mathematics (revision)

* International Baccalaureate SL & HL Mathematics

* AP Calculus AB & BC

* other syllabuses that calculus

* anyone who loves to learn!