Non-routine mathematical problem-solving

last SG · 2010-09-17T11:44:13+00:00

Hi, Interesting results can result from a very simple one. Read: http://johnmayhk.wordpress.com/2011/03/17/just-a-question-on-square-root/ Enjoy and be th...

wee_ws SG

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 SG

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 SG

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 SG

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 SG

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 SG

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 SG

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 SG

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.

Give it a try!

Cheers,
Wen Shih

wee_ws SG

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 SG

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 SG

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 SG

Hi,

Bravo, you got it! Thanks for responding!

Cheers,
Wen Shih

wee_ws SG

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 SG

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~ SG

All these are IQ questions?

wee_ws SG

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 SG

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 SG

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 SG

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 SG

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