Non-routine mathematical problem-solving

• wee_ws

  4 Mar 10, 06:37

  Hi,

    Q1 Can we find three distinct integers a_1, a_2, a_3 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 = 1?

    Q2 Can we find four distinct integers a_1, a_2, a_3, a_4 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 + 1 / a_4 = 1?

    Q3 Can we find five distinct integers a_1, a_2, a_3, a_4, a_5 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 + 1 / a_4 + 1 / a_5 = 1?

    Q4 Is there an observable pattern that may allow us to generalise the result involving n (n > 3) distinct integers, i.e.

           1 / a_1 + 1 / a_2 + 1 / a_3 + ... + 1 / a_n = 1?

    Try it in your spare time. Thanks in advance.

  Cheers,
  Wen Shih

• wee_ws

  14 Mar 10, 08:23

  Hi,

    Q1: 1/2 + 1/3 + 1/6 = 1 is one of many possibilities. However this solution leads to an interesting observation we shall see later.

    Q2: 1/2 + 1/4 + 1/6 + 1/12 = 1.

    Q3: 1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1.

    Q4: We may observe a repetitive 2-step pattern after going through the earlier parts, i.e.

    1. Take half of the previous sum.

    2. Add 1/2 in front.

    To illustrate, look at the sums in Q2 and Q3, i.e.

    1/2 + 1/2 { 1/2 + 1/3 + 1/6 } = 1,
    where 1/2 + 1/3 + 1/6 came from Q1;

    and 1/2 + 1/2 { 1/2 + 1/4 + 1/6 + 1/12 } = 1,
    where 1/2 + 1/4 + 1/6 + 1/12 came from Q2.

    Now we are ready to define a recurrence relation, i.e.

    u_1 = 1/2 + 1/3 + 1/6,

    u_{n + 1} = 1/2 + 1/2 u_n, where n >= 1.

    This recurrence relation will produce a sequence of 1's.

    Thanks for reading!

  Cheers,
  Wen Shih

• wee_ws

  14 Mar 10, 08:31

  Hi,

    Here's another interesting question. Consider the use of mathematical induction.

    Let a be a real number such that sin a + cos a is a rational number. Prove that

    (sin a)^n + (cos a)^n is a rational number

    for all natural number n.

    Thanks!

  Cheers,
  Wen Shih

• Forbiddensinner

  14 Mar 10, 09:00

  Originally posted by wee_ws:

  Hi,

    Q1 Can we find three distinct integers a_1, a_2, a_3 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 = 1?

    Q2 Can we find four distinct integers a_1, a_2, a_3, a_4 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 + 1 / a_4 = 1?

    Q3 Can we find five distinct integers a_1, a_2, a_3, a_4, a_5 such that the sum of their reciprocals

           1 / a_1 + 1 / a_2 + 1 / a_3 + 1 / a_4 + 1 / a_5 = 1?

    Q4 Is there an observable pattern that may allow us to generalise the result involving n (n > 3) distinct integers, i.e.

           1 / a_1 + 1 / a_2 + 1 / a_3 + ... + 1 / a_n = 1?

    Try it in your spare time. Thanks in advance.

  Cheers,
  Wen Shih

  Sir, you may want to change it to 'positive distinct integers'.

  With just the word integers, students can "cheat" and put 1/(-2) + 1/(2) + 1/(1) = 1 

  The same goes for any of the equation with an odd number of terms.

• wee_ws

  14 Mar 10, 11:03

  Hi,

    Thanks for your suggestion :)

    I did think about it, but I decided to leave the problem open to extensive investigation.

  Cheers,
  Wen Shih

• wee_ws

  21 Mar 10, 09:31

  Hi,

    Let's discuss the question on induction, which was posed on 14 Mar.

    First, we investigate some specific cases.

    When n = 2, (sin a)^2 + (cos a)^2 = 1 \in Q, the set of rational numbers.

    When n = 3,
    (sin a)^3 + (cos a)^3 = (sin a + cos a) [ (sin a)^2 + (cos a)^2 ]
                                           - (sin a)(cos a)(sin a + cos a),

    which is not clear to us whether it will be a rational number, since the issue lies with

    (sin a)(cos a).

    Now (sin a)(cos a) = 1/2 [ (sin a + cos a)^2 - {(sin a)^2 + (cos a)^2} ] \in Q.

    So we are certain that

    (sin a)^3 + (cos a)^3 \in Q.

    When n = 4,
    (sin a)^4 + (cos a)^4 = (sin a + cos a) [ (sin a)^3 + (cos a)^3 ]
                                           - (sin a)(cos a) [ (sin a)^2 + (cos a)^2 ],

    which is a rational number since the cases where n = 1, 2, 3 are true and (sin a)(cos a) is rational from the previous part.

    In general,

    (sin a)^{k+1} + (cos a)^{k+1} = (sin a + cos a) [ (sin a)^k + (cos a)^k ]
                                                        - (sin a)(cos a) [ (sin a)^{k-1} + (cos a)^{k-1} ]

    holds, based on the pattern we have seen thus far.

    We are now well-positioned to carry out induction, which will be elaborated in the next posting.

    Thanks for reading.

  Cheers,
  Wen Shih 

• wee_ws

  28 Mar 10, 08:26

  Hi,

    Let P(n) be the statement:
    (sin a)^n + (cos a)^n is a rational number, where n = 1, 2, 3, ... .

    P(1) and P(2) are true.

    Assume P(k - 1) and P(k) are true for some natural numbers k - 1 and k, i.e.,
    (sin a)^{k - 1} + (cos a)^{k - 1} \in Q and
    (sin a)^k + (cos a)^k \in Q.

    We show that P(k + 1) is true.

    From the last posting, we know that

    (sin a)^{k+1} + (cos a)^{k+1} = (sin a + cos a) [ (sin a)^k + (cos a)^k ]
                                                        - (sin a)(cos a) [ (sin a)^{k - 1} + (cos a)^{k - 1} ] and

    (sin a)(cos a) \in Q.

    So (sin a)^{k+1} + (cos a)^{k+1} \in Q and hence P(k + 1) holds if P(k - 1) and P(k) are true.

    Since
    1. P(1) and P(2) are true, and
    2. P(k - 1) and P(k) are true => P(k + 1) is true,
    P(n) is true by induction.

    Thanks for reading.

  Cheers,
  Wen Shih

• wee_ws

  28 Mar 10, 08:46

  Hi,

    This is a question about solving an unusual differential equation.

    The function h(x), where x \in R (the real set) and h(x) > 0, satisfies

    sqrt{ integral of h(x) wrt x } = integral of sqrt{ h(x) } wrt x.

    (a) By substituting h(x) = (dy/dx)^2, show that

          dy/dx = 2(y + c),

          where c is a constant.

    (b) Hence find a general expression for y in terms of x.

    (c) Given that h(0) = 1, find h(x).

    Give it a try!

  Cheers,
  Wen Shih

• wee_ws

  21 Apr 10, 08:04

  Hi,

    Using the substitution, we obtain

    sqrt{ integral of (dy/dx)^2 wrt x } = integral of sqrt{ (dy/dx)^2 } wrt x

                                                         = integral of (dy/dx) wrt x, since h(x) > 0.

    => integral of (dy/dx)^2 wrt x = (y + c)^2, where c is a constant.

    Differentiating both sides wrt x,

    (dy/dx)^2 = 2(y + c)(dy/dx)

    => (dy/dx) [ (dy/dx) - 2(y + c) ] = 0.

    Since h(x) > 0, (dy/dx) - 2(y + c) = 0.

    So dy/dx = 2(y + c).

    It is easy to find the general solution via the method of variable separable:

    y = A e^(2x) - c, where A is a constant.

    Now h(x) = 4(y + c)^2 = 4A e^(4x).

    Since h(0) = 1, 4A = 1, hence h(x) = e^(4x).

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  21 Apr 10, 08:32

  Hi,

    This is a problem on probability.

    Prove that, for any real numbers x and y, x^2 + y^2 >= 2xy.

    Carol has two bags of sweets. The first bag contains a red sweets and b blue sweets, whereas the second bag contains b red sweets and a blue sweets, where a and b are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot exceed the probability that they are of different colours.

    Thanks.

  Cheers,
  Wen Shih

• SBS261P

  10 May 10, 14:46

  Observe that (x-y)^2 = x^2 - 2xy + y^2 >= 0 for all real values of x and y.

  The result follows.

   

  P (same colour sweets) = a/(b+a) x b / (b+a) x 2 = 2ab / (b+a)

  P (different colour sweets) = a^2 / (b+a) + b^2 / (b+a) = (a^2 + b^2) / (b+a)

  Using the first inequality that we have proven, it follows that

  a^2 + b^2 >= 2ab

  (a^2 + b^2) / (b+a)  >= 2ab / (b+a)

  hence P (different colour sweets) >= P (same colour sweets)

  and it follows that P (same colour sweets) cannot exceed P (different colour sweets)

• wee_ws

  10 May 10, 17:18

  Hi,

    Bravo, you got it! Thanks for responding!

  Cheers,
  Wen Shih

• wee_ws

  18 May 10, 22:44

  Hi,

    This has been motivated by a tutorial question I saw from a JC.

    Without expanding, find the derivative of x(x + 1)(x + 2)(x + 3)(x + 4).

    One may begin with exploring the derivatives of

    x(x + 1), x(x + 1)(x + 2), and so on, to observe a systematic way of differentiating via the product rule.

    In the same spirit, can you find the derivative of ln(ln(ln(ln(x^2 + 1)))).

    Thanks!

  Cheers,
  Wen Shih

• wee_ws

  14 Jun 10, 10:10

  Hi,

    My colleague at NIE contributed these two interesting questions. He says that knowledge of O-level Additional Mathematics content will be sufficient to solve them. Do give these a try to give your brain a good workout!

  Q1 Find the exact value of cube_root [ 9 + sqrt(80) ] + cube_root [ 9 - sqrt(80) ].

  Q2 Find exactly the roots of the equation x^4 - 16x - 12 = 0.

    Thanks.

  Cheers,
  Wen Shih

• ~PEPPER~

  16 Jun 10, 10:49

  All these are IQ questions?

• wee_ws

  16 Jun 10, 12:56

  Hi,

    Concrete mathematical knowledge is required to solve the two questions. I believe IQ-based questions just need general logic to address them. Thanks.

  Cheers,
  Wen Shih

• wee_ws

  20 Jun 10, 18:33

  Hi,

    On another thread (http://www.sgforums.com/forums/2297/topics/401368), this question was raised: Find the integral of sqrt[1 + e^(2x)] dx by the use of a suitable substitution.

    One possible (traditional) substitution is to let e^x = tan u. The working, however, becomes complex later.

    A colleague of mine, Dr Lee TY (thanks a lot!), came up with this instead: let u = sqrt[1 + e^(2x)]. The steps are less tedious, as described below.

    u = sqrt[1 + e^(2x) ], so u^2 = 1 + e^(2x).

    Differentiating with respect to x, 2u (du/dx) = 2e^(2x)

    => du/dx = 2(u^2 - 1) / (2u) = (u^2 - 1) / u.

    Now integral sqrt[1 + e^(2x)] dx = integral u dx/du du

    = integral u^2 / (u^2 - 1) du

    = integral [ 1 + 1/(u^2 - 1) ] du

    = u + 1/2 ln |(u - 1)/(u + 1)| + c, from which we substitute back sqrt[1 + e^(2x) ].

    It is non-routine in that the substitution used is the entire expression.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  19 Jul 10, 23:12

  Hi,

    Q2 Find exactly the roots of the equation x^4 - 16x - 12 = 0.

    One may consider x^4 - 16x - 12 = (x^2 + ax + b)(x^2 + cx + d), where a, b, c and d are to be found.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  19 Jul 10, 23:15

  Hi,

    Here is an easy question about divisibility:

    For what values of k will 7(k - 1)(10k - 3) be divisible by 14?

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  30 Jul 10, 09:21

  Hi,

    We will require (k - 1)(10k - 3) to be an even number.

    Consider 10k - 3. This can be rewritten as 2(5k - 1) - 1, which is the form of an odd number. So 10k - 3 is always odd regardless of k.

    Now (k - 1) must always be even, so k is a member of the set of odd numbers.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  30 Jul 10, 09:33

  Hi,

    Vector-based proofs are sometimes elegant in showing geometrical properties.

    Consider the proof of the theorem: The medians of a triangle are concurrent.

    This site explains what a median is:

    http://www.mathopenref.com/trianglemedians.html

    The next site explains the steps of a proof using vectors (scroll down to the theorem):

    http://www.wmueller.com/home/papers/triangles.html

    Essentially, one

    1. uses the vector equation of a straight line,

    2. compares two expressions of the same point,

    to prove the property.

    For the keen learner, there is a paper by N J Lord that says more about vector-based geometrical proofs. Read it here:

    http://mathdl.maa.org/images/cms_upload/N00284._J._Lord.pdf

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  30 Jul 10, 09:57

  Hi,

    Here is a nice proof question involving vectors and the need for one to have a good understanding of geometry:

    Let v be any arbitrary vector.

    (i) Show that the components of v with reference to i, j and k are v.i, v.j and v.k respectively. Deduce that

                                               (v.i)^2 + (v.j)^2 + (v.k)^2 = |v|^2.

  Hint: Let v = (x, y, z). What is v.i, v.j, v.k?

    (ii) Now suppose that v is nonzero and that alpha, beta and gamma are the angles between v and the vectors i, j and k respectively. Show that

                                 (cos alpha)^2 + (cos beta)^2 + (cos gamma)^2 = 1.

  Hint: Consider the definition of scalar product.

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  31 Jul 10, 17:59

  Hi,

    This is an interesting letter that discusses about the use of a system of linear equations to balance chemical equations:

    http://pubs.acs.org/doi/pdf/10.1021/ed009p358

    Enjoy.

  Cheers,
  Wen Shih

• wee_ws

  12 Aug 10, 13:06

  Hi,

    This interesting problem came from a discussion forum on the Internet:

    Suppose that u = g(x) is differentiable at x = 1 and that y = f(u) is differentiable at u = g(1). If the graph of y = f(g(x)) has a horizontal tangent at x = 1, can we conclude anything about the tangent to the graph of g at x = 1 or the tangent to the graph of f at u = g(1)? Give reasons for your answer.

    Ponder over it :)

  Cheers,
  Wen Shih

• wee_ws

  17 Sep 10, 11:44

  Hi,

    Consider this problem involving Binomial distribution and probability:

    The probability that a particular make of light bulb is faulty is 0.2. The light bulbs are packed in boxes of 12. A buyer accepts a consignment of 50 boxes, if, when he chooses two boxes at random, he finds that they contain no more than one faulty light bulb altogether. Find the probability that he will accept the consignment.

    The usual approach is given below:

  1. Let X be the number of faulty bulbs in a box of 12.

      Then X ~ B(12, 0.2).
   
  2. For the two boxes picked, there are three possible cases:
   
       (i) both boxes 1 & 2 come with zero faulty bulbs,
   
       (ii) box 1 has zero faulty bulbs and box 2 has 1 faulty bulb.
   
       (iii) box 1 has 1 faulty bulb and box 2 has zero faulty bulb.
   
       Use multiplication and addition principles of probability.

    Another shorter approach we may consider is:

  1. Let X be the number of faulty bulbs in two boxes.

      Then X ~ B(24, 0.2).

  2.  The required probability is simply P(X <= 1).

    The shorter approach comes from the application of The Discrete Convolution Formula Theorem. Using this theorem, we have the following result which is not covered in H2 Statistics:

    If X_i ~ B(n, p) and X_1, X_2, ..., X_k are independent random variables, then

    X_1 + X_2 + ... + X_k ~ B(kn, p).

    In particular, we have used the theorem on the Poisson (where the sum of Poisson variables is Poisson) and Normal (where the sum of Normal variables is Normal) distributions.

    Thanks.

  Cheers,
  Wen Shih