[AP_Calculus_20150516LIM] Large abyss with no Limit?

Question

Introduction

     This looks like a scary limit to calculate.  Indeed, it is quite tricky, as  x  goes towards negative infinity and there is a square root.  However, we can often work out the correct answer informally first.

Informal Solution (intuitive approach)

Formal Solution

Final Remark

     Having the informal solution (which can be worked out on rough paper or even mentally) gives us an idea of the answer.  We can counter-check our formal solution with the informal solution for confirmation.

Suitable Levels

* Advanced Placement (AP) Calculus AB & BC

* University / college level calculus

* other syllabuses that involve limits

[H2_20150512GSS] A Square Spiral in Geometric Progression

Question

Introduction

     This is a type of question that has become quite common in H2 Mathematics at Junior College level.  I was not told where this question came from, but I seem to get a bout of déjà vu.  It is quite a common practice for the Singapore junior colleges to mimic questions from the GCE ‘A’ Levels and/or one another.

Solution

Summary

     The keys to solving this question are:-

          (1)  if a coordinate frame is needed but not given it the question, we invent our own.

          (2)  see whether we need to care about the direction or not, and then apply the
                 geometric series formula.

          (3)  When applying the geometric series formula, always identify correctly the first term,
                 the common ratio, and the number of terms to be summed.

Heuristics Used

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 ‘A’ Level H2 Mathematics

* other syllabuses that involve geometric series

[JCH2BXQNSR_20150429] Binomial Expansion for a Quotient

Question

Solution

Comment

     Note in the above working that  3(x + kx²)² = 3(x² + 2kx³ + k²x⁴),  but since we do not need the  x³  and  x⁴  terms, we omit them and just write “3(x² + ... )”.

Heuristics Used

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* revision for GCE ‘A’ Level H2 Mathematics

* AP Calculus

* other syllabuses infinite binomial series

[Pri20150330CPF] How to #Compare #Fractions

Comparing Positive Fractions

     How do you compare positive fractions, which are taught in primary (elementary) school?  For example, which is bigger:  ⁵/₆  or  ³/₄ ?

The “orthodox” method is to put them both to a common denominator.  The Lowest Common Multiple (LCM) of  6  and  4  is  12.  Multiplying the left fraction by  ²/₂  and the right fraction with  ³/₃  gives, respectively,  ¹⁰/₁₂  and  ⁹/₁₂.

Since  ¹⁰/₁₂  >  ⁹/₁₂,  we conclude that  ⁵/₆  >  ³/₄.

Another Method

     Here is a “short-cut” that I learned from a schoolmate in primary school.  Basically you “cross-multiply”: multiply the left numerator with the right denominator, and multiply the right numerator with the left denominator, and then compare the products so formed.  That will give you the correct inequality or equality sign (viz. ‘<’, ‘=’ or ‘>’).

As we can see, since  20 > 18,  we conclude that  ⁵/₆  >  ³/₄. 

     Does this method work?  Yes, definitely.  You can try it out with a few pairs of fractions and you can see for yourself that it is so.  Is this method legit?  Why does it work?    I give a formal proof of the method below.

Further Discussion

     Use the above method with care.  Some school teachers may not accept the method not because it is not correct, but it sounds “dubious” to them because they have not heard of it or they are not able to prove it for themselves.  Pupils can use this short-cut to give them a quick look-ahead to certain questions, and as a back-up to check their answer after using the Lowest Common Denominator method.  In questions that ask pupils to arrange a few fractions in ascending and descending order, this “crossing method” may give some speed advantage if done carefully.

     In primary school, pupils focus on positive fractions.  In secondary school, negative numbers and fractions are introduced.  Does the above trick work for negative fractions?  Was my theorem and proof above carefully phrased enough to cover the negative fractions? 

     Note that this method works for comparing two fractions at a time only.  Sometimes this cross-multiplying gives rather big numbers.  In that case, it is better to multiply each fractions by the LCM of their denominators.  Essentially this is the same as the orthodox method, except that we do not write the denominators.  Can the above proof be extended to cover this new short-cut?  What do you think?