

Hi,
A nice integration question from STEP 2011 paper I:
The number E is defined by E = integral {0, 1} (e^x) / (x + 1) dx.
Show that integral {0, 1} (xe^x) / (x + 1) dx = e  1  E,
and evaluate integral {0, 1} (x^2)(e^x) / (x + 1) dx in terms of e and E.
Thanks.
Cheers,
Wen Shih 
Hi,
Another STEP question regarding tangents to a curve:
This distinct points P and Q, with coordinates (ap^2, 2ap) and (aq^2, 2aq) respectively, lie on the curve y^2 = 4ax. The tangents to the curve at P and Q meet at the point T. Show that T has coordinates (apq, a(p + q)). You may assume that p and q are nonzeros.
Give it a try, thanks.
Cheers,
Wen Shih 
Hi,
Let us revisit a question from N2011 paper 1.
In Q4(ii)(b), we gave the following reason to explain why the approximation is not very good:
Series does not have sufficient number of terms.
An alternative (yet convincing) visual approach is to sketch the graphs of y = 1  3x^2 + 4x^4, y = (cos x)^6 and x = pi/4. From the sketch, it is easy for us to conclude the following:
In the vicinity of x = pi/4, we observe that 1  3x^2 + 4x^4 does not approximate (cos x)^6 very well. The result is that the area under the quartic curve is somewhat larger than the area under the cosine curve, giving us a poor approximation.
With the sketch, we could even go further and provide a crude value of an upper limit for the approximation to be good enough, i.e., 0.5 radians and below or pi/6 and below.
Thanks.
Cheers,
Wen ShihEdited by wee_ws 11 Dec `11, 6:57PM 
Hi students,
May you find these explanations helpful in reviewing your understanding of concepts:
1. http://www.slideshare.net/weews/howtofindtherangeofacompositefunction
2. http://www.slideshare.net/weews/howtofindtheequationofatransformedcurve
3. http://www.slideshare.net/weews/howtosolveproblemsinvolvingrelationshipsbetweenplanes
4. http://www.slideshare.net/weews/howtointegratecertainrationalexpressions
5. http://www.slideshare.net/weews/howtoproveasumresultinvolvingap
6. http://www.slideshare.net/weews/onewaytovisualisetherelationshipbetweenplanes
Thanks.
Cheers,
Wen ShihEdited by wee_ws 28 Dec `11, 7:58PM



Hi all,
this is my method of deriving f'(x) f(x) types of integration expressions without memorizing them at all.http://www.slideshare.net/momoeagle/simplifiedintegration
Thanks.
Edited by eagle 02 Jan `12, 10:10AM



Hi,
Graphs of ellipse, rectangular hyperbola and hyperbola can be sketched quite quickly if one were to follow a sequence of easy steps.
To sketch an ellipse:
1. Draw the x and yaxes.
2. Identify the centre.
3. Identify top, down, left, right corners with reference to the centre.
4. Join the corners up.For example, to draw the graph of (x^2)/6 + (y^2)/3 = 1, we:
 Identify the centre at the origin.
 Identify the corners (0, sqrt(3)), (0, sqrt(3)), (sqrt(6), 0), (sqrt(6), 0).
 Join the four points with a smooth eggshaped curve.To sketch a rectangular hyperbola:
1. Draw the axes.
2. Draw the vertical and horizontal asymptotes.
3. Find intercepts, if any, to identify one quadrant where the curve lies in.
4. Draw the curve in the other quadrant.For example, to draw the graph of y = (x  2)/(x + 2), we:
 Draw the lines x = 2 and x = 1.
 Locate the intercepts (0, 1) and (2, 0), so the curve lies in the 4th quadrant.
 Draw the curve in the 2nd quadrant.To sketch a hyperbola:
1. Draw the axes.
2. Draw the oblique asymptotes intersecting at the centre.
3. Identify the vertices.
4. Sketch the two branches of the curve.For example, to draw the graph of (1/25)(x  3)^2  (1/49)(y + 1)^2 = 1, we:
 Draw the oblique asymptotes y + 1 = (7/5)(x  3) and y + 1 = (7/5)(x  3) intersecting at (3, 1).
 Identify the vertices (2, 1) and (8, 1) which are obtained by subtracting/adding 5 units to the xcoordinate of the centre.
 Sketch the left and right branches of the curve.Thanks.
Cheers,
Wen ShihEdited by wee_ws 27 Jan `12, 10:08PM 
Hi,
Here is a checklist of concepts and skills for IB HL Maths on Mathematical Induction (syllabus content 1.4):
1. Prove results involving sums.
2. Prove results involving divisibility of numbers.
3. Prove results involving nth derivatives.
4. Prove results involving complex numbers raised to some power.
5. Prove results involving matrices raised to some power.
6. Formulate and verify a conjecture based on an observation of a pattern.
7. Provide a counterexample to refute a conjecture.I will write more about the other topics, i.e., matrices, complex numbers, vectors, and calculus soon. Thanks!
Cheers,
Wen ShihEdited by wee_ws 24 Feb `12, 8:09AM 
Hi,
A good way to truly understand the intricacies of counting/probability is to come up with different ways to solve the same problem. Consider this question:
Four marbles are randomly chosen, without replacement, from a bag of 18 marbles of which 3 are red, 6 are green and 9 are blue. Find the probability that the marbles chosen consist of at least one of each colour.
See if you could come up with at least 3 methods, thanks!
Cheers,
Wen Shih 
Hi,
In response to the last post, I shall elaborate the three methods.
Method 1: Using P&C concepts, we calculate the required probability,
P(2R, G, B) + P(R, 2G, B) + P(R, G, 2B)
= [ (3C2)(6C1)(9C1) + (3C1)(6C2)(9C1) + (3C1)(6C1)(9C2) ] / (18C4)
= 27/68.
Method 2: Via probability of individual events for each case identified previously, we obtain
(3/18)(2/17)(6/16)(9/15)(4!/2!) + (3/18)(6/17)(5/16)(9/15)(4!/2!)
+ (3/18)(6/17)(9/16)(8/15)(4!/2!)= 27/68.
Method 3: Via complementation principle, we have
1  P(only R or G or both)  P(only G or B or both)  P(only R or B or both)
+ P(all G) + P(all B)= 1  (9C4)/(18C4)  (15C4)/(18C4)  (12C4)/(18C4)
+ (6C4)/(18C4) + (9C4)/(18C4)= 27/68.
Thanks.
Cheers,
Wen ShihEdited by wee_ws 15 Apr `12, 7:58AM 
Hi,
A useful set of notes on P&C, for reference, at:
http://download.nos.org/srsec311new/L.No.07.pdf
Thanks.
Cheers,
Wen Shih 
Hi,
Is Mathematics difficult? Many learners think so and they struggle with the subject in their academic careers. I humbly offer five ways to effectively overcome the difficulties of learning Mathematics and eventually mastering it:
1. Change your mindset. If you think often that Mathematics is hard, it really becomes so as research has shown (read: http://www.psychologytoday.com/blog/ambigamy/201204/nobelprizewinnerweighsindontworrybehappy).
2. Get acquainted with question trends. Mathematics naturally contains full of patterns and that includes the typical but key concepts and skills that teachers always think about when they assess their students.
3. Reinterpret the technicalities of Mathematics. One way is to use simple English to translate the inintelligible content.
4. Learn collaboratively with peers. This helps to share your difficulties with those who may be facing in the same situation as you. At the same time, you are likely to learn about ways to improve your mathematical understanding from those who are more able than you.
5. Look deeply into the problems and generate possibilities. Are there alternatives to solving this problem? Why are certain pieces of information given? What if some pieces of information are not provided? What linkages between topics may be observed in this problem? Are there possible extensions to the problem? What variations can there be with this type of problem?
Thanks very much!
Cheers,
Wen Shih 
Hi,
A student who struggles to solve mathematical problems is a norm and a necessity.
On page 94 of How to Solve It, George Polya writes:
...a teacher wishing seriously to help the student should, first of all, stir up his curiosity, give him some desire to solve the problem. The teacher should also allow some time to the student to make up his mind, to settle down to his task.
Teaching to solve problems is education of the will. If the student had no opportunity in school to familiarize himself with the varying emotions of the struggle for the solution his mathematical education failed in the most vital point.Thanks.
Cheers,
Wen Shih 
Hi,
A nottoointimidating question about the induction on the sum of a trigonometric expression, as a form of practice (in preparation for JC1/2 MYE):
Prove by induction that
cos (3/2) + cos (5/2) + ... + cos {(2n1)/2} = {sin (n)  sin (1)} / {2 sin (1/2)}
for positive integers n >=2.Thanks.
Cheers,
Wen Shih 
Dear educators and students,
This excellent article discusses specific issues of mathematical difficulties that are worth pondering over to overcome them:
http://www.pbs.org/parents/strugglingtolearn/understanding_math.html
I will add some points related to the use of graphic calculators, from personal teaching experiences:
1. Students cannot recall the sequence of keystrokes.
2. Students become confused with similar calculator commands.
3. Students are unaware of inherent limitations of the graphic calculator.
4. Students are not proficient with window settings that would give sufficient graphical information.
5. Students are unaware of when and probably how the graphic calculator will aid in or solve a mathematical problem.
6. Students are not proficient in updating their calculators.
7. Students find it difficult to translate information from the graphic calculator to concrete mathematical steps.
Do feel free to share your experiences, to enrich this discussion :) Thanks.
Cheers,
Wen Shih 
Hi,
Is it possible to plot the roots of z^n = re^(alpha i) on a circle without the use of protractor?
I tackled this issue when I was trying to plot the roots and I realised I did not have the tool at hand. Then an idea of using geometrical construction came to me. I could use a geometrical result, a ruler and a pair of compasses to plot the roots.
Suppose we wish to represent the roots of z^6 = {4/sqrt(3)}^6 on an Argand diagram. The roots lie on a circle whose centre is at the origin and whose radius is 4/sqrt(3) units. The angle between adjacent roots is pi/3 radians (equivalently, 60 degrees).
The steps I took were:
1. Construct the circle with radius 4/sqrt(3) units at the origin with the pair of compasses.
2. Measure the length (60/360) x 2pi x {4/sqrt(3)} with the ruler.
3. Mark the positions of the six roots along the circumference of the circle with compasses.
From this, I learnt that facing a constraint is not necessarily a bad thing, it's how one could respond positively to it and gain something from it :)
Thanks and have a great Sunday!
Cheers,
Wen ShihEdited by wee_ws 24 Jun `12, 7:55AM 
Hi,
I learnt two lessons trying to tackle this recent question about system of linear equations that I encountered:
A, B and C have k marbles altogether. When A gives B 30 marbles and B gives C 12 marbles, the number of marbles A, B and C each has respectively is in the ratio 1 : 2 : 3. Find the least value of k, assuming that each of them has some marbles initially.Lesson 1: Is there a shorter solving method than the traditional one?
Let a, b, c be the number of marbles that A, B, C has respectively.
Total marbles: a + b + c = k  (1).
Since a  30 : b + 18 = 1 : 2, we obtain 2a  b = 78  (2).
Since a  30 : c + 12 = 1 : 3, we obtain 3a  c = 102  (3).
Solving (2) and (3) with GC, we obtain
a = 34 + (c/3), b = 10 + (2c/3).
Note that a + b = 24 + c => a + b  c = 24  (4).
(1) + (4) gives 2(a + b) = k + 24 => a + b = 12 + (k/2).
From (4), we obtain 12 + (k/2) = c + 24 => c = 12 + (k/2).
When c = 12 + (k/2), a = 30 + (k/6), b = 18 + (k/3).
Note that
 k must be divisible by 6
 c > 0 => k > 24
 b > 0 => k > 54.
The next k divisible by 6 is 60.
Check back: when k = 60, a = 40, b = 2, c = 18, a  30 : b + 18 : c + 12 = 10 : 20 : 30 = 1 : 2 : 3.
Thus, least k = 60.A little tedious, I would say :P
Lesson 2: Can I apply knowledge I have learnt in the past?
I decided to try and apply P6 model method and this made the solution (much) more efficient than before.
We have 6 parts in the end, equal to k, so each part is k/6. We deduce k is divisible by 6.
Initially, c = 3(k/6)  12. Since c > 0, we obtain k > 24.
Initially, b = 2(k/6)  18. Since b > 0, we obtain k > 54.
Initially, a = (k/6) + 30, so k > 0.
Least k = 60, since it is the next value of k divisible by 6.Thanks for reading and have a happy Youth Day!
Cheers,
Wen Shih
