Polya's problem-solving approach

wee_ws SG

Hi,

The following has been extracted from:

http://wik.ed.uiuc.edu/index.php/P%C3%B3lya%2C_George

Polya's four steps to solving a problem (especially helpful for mathematics):

Step 1: Understand the problem

Step 2: Devise or make a plan

Step 3: Carry out the plan

Step 4: Look back at the completed solution

Do read the sub-points under each step and strive to apply the ideas in your mathematical problem-solving.

Thanks!

Cheers,
Wen Shih

wee_ws SG

Hi,

We look at a mathematical example of applying Polya's problem solving steps. It is suited for H2 mathematics.

Consider the question: Solve 3 (dy/dx)^2 + 2x (dy/dx) = 1.

Step 1: Understand the problem

We need to solve a differential equation.

Step 2: Devise or make a plan

We need to consider whether we could solve it by direct integration, by variable separable, by substitution or by other non-conventional means.

Upon close examination, we notice that the expression is a quadratic equation involving dy/dx. So we may apply the formula to find the roots of a quadratic equation.

Step 3: Carry out the plan

So dy/dx = (1/3) { -1 + sqrt (x^2 + 3) } or (1/3) { -1 - sqrt (x^2 + 3) }.

Integrating directly, we obtain

y = -(1/3) x + (1/3) integral sqrt (x^2 + 3) dx

or -(1/3) x - (1/3) integral sqrt (x^2 + 3) dx.

What remains is for us to deal with integral sqrt (x^2 + 3) dx, which could be found by making use of the substitution x = sqrt (3) (tan u). The remaining steps are routine and I urge H2 maths students to attempt.

Step 4: Look back at the completed solution

We ask the question of whether the above steps are sound, logical and free of errors. We may also make use of each solution obtained, differentiate it and substitute it into LHS of the given differential equation to see if we could reach RHS (i.e. 1).

Thank you for reading!

Cheers,
Wen Shih

wee_ws SG

Hi,

This is the second example:

The function f satisfies the identity f(x) + f(y) = f(x + y) -- (*)

for all x and y.

Show that the identity 2 f(x) = f(2x) holds and deduce that f ''(0) = 0.

By considering the Maclaurin series for f(x), find the most general function that satisfies (*).

Do take some time and patience to read on. It is an interesting mathematical exercise, which I have enjoyed tremendously whilst solving it.

Step 1: Understand the problem

We want to find the most general function f.

Before that, we know that f satisfies the property (*).

Also, we need to solve a smaller problem, i.e. to show that 2 f(x) = f(2x) holds and deduce that f ''(0) = 0.

Step 2: Devise or make a plan

Let's solve the smaller problem first.

We may need to carry out repeated differentiation since the concept of Maclaurin series has been asked for.

Step 3: Carry out the plan

LHS = 2 f(x) = f(x) + f(x)

and when we compare with f(x) + f(y) in (*), we observe that a suitable substitution for y is x.

So 2 f(x) = f(x) + f(x) = f(x + x) = f(2x) = RHS, and we are done. First victory!

f ''(0) = 0 is to be deduced, so we will need to apply what we have just shown and to make use of repeated differentiation as we have planned earlier in step 2.

Differentiating both sides with respect to x, we have

2 f '(x) = 2 f '(2x) -- (1)

Differentiating (1) again with respect to x, we have

2 f ''(x) = 2^2 f ''(2x) -- (2)

When x = 0, we have

2 f ''(0) = 2^2 f ''(0)

=> 2 f ''(0) = 0

=> f ''(0) = 0, and we are also done. Another small victory and one step closer to the solution!

Since we need to consider the Maclaurin series for f(x), we will have to investigate further the other coefficients in the expansion of f(x). What is f(0), f '(0) and in general f^(n) (0)?

First, what is f(0)? Since we know that 2 f(x) = f(2x), it is true that 2 f(0) = f(0) which leads to f(0) = 0.

Next, what is f '(0)? From (1), 2 f '(x) = 2 f '(2x), so it is true that 2 f '(0) = 2 f '(0) which does not say anything other than f '(0) = k where k is some real constant.

Finally, what is the nth derivative of f when x = 0? From observing (1) and (2), we can see a pattern, i.e. 2 f^(n) (x) = 2^n f^(n) (2x). LHS refers to 2 times the nth derivative of f(x) and RHS means (2 power n) times the nth derivative of f(2x).

Knowing this, it is valid to say that 2 f^(n) (0) = 2^n f^(n) (0)

=> (2^n - 2) f^(n) (0) = 0

=> f^(n) (0) = 0

Now, we are ready to find the Macluarin series for f(x). It is simply f(x) = kx, since we have investigated and concluded that f(0) = 0, f ''(0) and f^(n) (0) = 0 and f '(0) = k.

Phew, quite some work we have done!

Step 4: Look back at the completed solution

With our solution f(x) = kx, we check if property (*) can be satisfied, i.e.

f(x + y) = k(x + y) = kx + ky = f(x) + f(y) for all x and y.

Indeed f(x) = kx is the most general function that satisfies (*). Final sweet victory!

Thanks for reading!

Cheers,
Wen Shih

eagle SG

lol

I saw this printed on my AJC student's tutorial

SBS2601D SG

Isn't this how we do our maths problems sub-consciously?

wee_ws SG

Isn't this how we do our maths problems sub-consciously?

Hi,

The sub-conscious (unseen) process of problem-solving could be made more explicit, concrete and accessible to students, as my experience with students show that they give up easily at the slightest difficulty. The reason for giving up can be attributed to a lack of thinking in a systematic fashion, which could be developed over time. Thanks!

Cheers,
Wen Shih

SBS2601D SG

Hi,

Cheers,
Wen Shih

I see...look forward to seeing more such threads!

wee_ws SG

lol

I saw this printed on my AJC student's tutorial

Hi,

Is it applied in AJC? It will be good if the school follows through with the practice. Thanks!

Cheers,
Wen Shih

wee_ws SG

Hi,

It is interesting for me to find out today from my students that the school highlights only the main steps, without mentioning the details.

I will attempt to go further to describe the details and follow up with an example that is of A-level standard as the earlier two examples may be a little tough.

Step 1: Understand the problem

What am I given?
What do I need to solve?

Step 2: Devise or make a plan

What approaches are available for me to solve?
Have I solved similar problems before?
Are there smaller problems for me to solve?

Step 3: Carry out the plan

I will write out the steps of my solution.

Step 4: Look back at the completed solution

Is my final answer a reasonable one?
Are my steps clear, logical and error-proof?
Can I check the answer with calculator?
Can I substitute back?
Can I obtain the same answer via another approach?

Now let's look at an example involving differential equations:

The curve C satisfies the equation y '' = sin^3 x. It is given that C cuts the y-axis at -1 and passes through the point (4/pi, 2/9). Find y in terms of x.

Step 1: Understand the problem

What am I given?

1. y '' = sin^3 x

2. Points (0, -1) and (4/pi, 2/9) lie on C.

What do I need to solve?

1. Find the particular solution, because I am given particular points.

2. Express y in terms of x, because it is required by the question.

Step 2: Devise or make a plan

What approaches are available for me to solve?

  They are direct integration, variable separable, substitution which I have been exposed to.

Have I solved similar problems before?

  Perhaps, I may have seen it before in lectures, tutorials, tests, exams, etc.

  Are there smaller problems for me to solve?

1. Yes, I need to find dy/dx before I can find y in terms of x.

2. The expression sin^3 x may need to be simplified by applying a suitable trigonometric identity before I can integrate it.

3. Since I am given two points that lie on C, I will need to find constants in the particular solution.

Step 3: Carry out the plan

I will write out the steps of the solution as follow.

y '' = sin^3 x
       = (sin x)(sin^2 x), by Pythagorean identity
       = (sin x)(1 - cos^2 x)
       = sin x - (sin x)(cos^2 x)

I recognise that -(sin x)(cos^2 x) looks like f '(x) . {f(x)}^n, so I can integrate the expression directly.

Integrating both sides wrt x,
y ' = -cos x + 1/3 cos^3 x + c, so I have solved a smaller problem.

y ' = -cos x + 1/3 (cos x)(1 - sin^2 x) + c, following the same approach earlier.
= -cos x + 1/3 cos x - 1/3 (cos x)(sin^2 x) + c
= -2/3 cos x - 1/3 (cos x)(sin^2 x) + c

Integrating both sides wrt x again,
y = -2/3 sin x - 1/3 (1/3 sin^3 x) + cx + d
= -2/3 sin x - 1/9 sin^3 x + cx + d, so I have obtained a general solution and another smaller problem has been tackled.

Since (0, -1) lies on C, d = -1.

Now y = -2/3 sin x - 1/9 sin^3 x + cx - 1.

Since (4/pi, 2/9) lies on C, -2/3 - 1/9 + (4/pi) c - 1 = 2/9
=> c = pi/2

Finally, the particular solution is y = -2/3 sin x - 1/9 sin^3 x + (pi/2) x - 1.

Step 4: Look back at the completed solution

Is my final answer a reasonable one?

It is, because the expression is trigonometric given that the original expression is trigonometric.

Are my steps clear, logical and error-proof?

Yes, though I must be careful with minus signs.

Can I substitute back?

Yes, I can check using the two points that lie on C.

Can I obtain the same answer via another approach?

No, but I can differentiate y twice and see if I could obtain the expression

y '' = sin^3 x,

though it may be time-consuming in the exam. I may do it if it is a tutorial question or a question that I'm doing during my revision.

I hope this example convinces students that the Polya's method is effective, especially in its steps 1 & 2. By carrying out the first two steps rigorously, it will bring about the solution naturally.

Thanks for reading!

Cheers,
Wen Shih

wee_ws SG

Hi,

The following is a list of sources and types of error in students' problem solving:

Inaccuracy in reading

Inaccuracy in thinking

Weakness in problem analysis; inactiveness

Lack of perseverance

Source: www.cirtl.net

Thanks!

Cheers,
Wen Shih