An Application Of Classical Polynomial Theory In Optimal Ladder Reclining Problem.
AN APPLICATION OF CLASSICAL POLYNOMIAL THEORY
IN OPTIMAL LADDER RECLINING PROBLEM
(An Example of Problem Solving in Teaching Mathematics)*
models on how applying this approaches in the classrooms still limited to most
mathematics educators. This paper is intended to fill this gap by demonstrating the
development “simple” mathematics theorems from a simple riddle-like problem, as a
result of a problem solving activity.
Asep K. Supriatna1 and Stanley P. Dewanto
Dept. Mathematics, Faculty of Mathematics & Natural Sciences, Universitas Padjadjaran
Km 21 Jatinangor Sumedang, Fax: 022-7794696
Abstract
The paper has twofold purposes. First, the paper demonstrates a stage-by-stage problem
solving which is an important process that should be experienced by all mathematics
students. It can be regarded as a model in teaching mathematics in the context of
meaningful learning and constructivism. Second, in this paper we consider an application
of classical polynomial theory in determining the existence of a ladder reclining problem.
A ladder with 5 meters length is supposed to be placed so that the upper edge of the ladder
reclines to a vertical wall while its lower edge touch the horizontal floor. It is also required
that a part of the ladder touches a corner of certain geometric object (in this case a cube
that has edges 1-meter length). We address the question on how many ways the ladder can
be placed to satisfy such condition. Surprisingly, it was found that there are only two ways
to place the ladder. Furthermore, in this paper we also give a necessary and sufficient
condition in order a ladder with length l can be placed in the intended way so that a part of
the ladder touches a part of a cube with edge k.
1.
Introduction
Problem solving capability is regarded as one among the most important
competences that should be mastered by all mathematics graduates. It is related to the
capability of applying mathematical concept in everyday and real life problems. Other
factors such as the ability of problem identification, choosing relevant factors, formulating
mathematics description and its analysis, and interpreting back the results of the
mathematical analysis into the real world are included in the problem solving capability.
This kind of competence can be developed through an appropriate experience in the
classroom activities on a regularly basis.
Many mathematics educators believe that the development of this competence can
be achieved by students who well experienced both in hands and minds on problems.
Knowing the real problems well, from where the mathematics problems come, will give a
meaningful learning for the students who did the mathematical exercises. This will help
them in understanding the mathematics and we believe that this will lead to increase their
curiousness in extending the problems at hand. We also believe this scientific curiousness
eventually will contribute to the development of a good transferable skill, namely
inovation.
Meaningful learning and constructivism -- through problem solving -- are among
the pedagogical jargons that increasingly popular to mathematics educators. However, the
*
Presented in the-XI I National Conference of Mathematics, Bali 23-27 July 2004
1
Corresponding e-mail address: asupriat@unpad.ac.id and
2.
Understanding Problem Solving
The term problem solving occurs in many different professions and disciplines and
has different meanings. Problem solving in mathematics has a more specific meaning, but
it still open to different interpretations. So, problem solving is an all-encompassing term
that means different things to different people at the same time, and different things to the
same person at different times. In this paper, problem solving is defined as the process of
applying previously acquired knowledge to new and unfamiliar non-routine situation [4].
Polya writes in How to Solve It that in studying modern heuristic, one “endeavors
to understand the process of solving problems, especially the mental operations typically
useful in this process” [5, p.129]. Under appropriate circumstances, one can learn to use
heuristic, with the result being a demonstrable improvement in their problem solving
performance.
Heuristic means a general strategy or suggestion, independent of any particular
topic or subject matter, that helps problem solver approach and understand a problem and
efficiently marshall their resources to solve it [6]. Following Polya, some important
heuristic in problem solving are [6]:
1. Analysing and Understanding a Problem:
a. Draw a diagram if at all possible.
b. Examine special cases.
c. Try to simplify it by using symmetry or “without loss of generality”.
2. Designing and Planning a Solution:
a. Plan solutions hierarchically.
b. Be able to explain at any point of solution what you are doing and why,
what you will do with the result of this operation.
3. Exploring Solutions to Difficult Problems:
a. Consider a variety of equivalent problems.
b. Consider slight modifications of the original problem.
c. Consider broad modifications of the original problem.
4. Verifying a Solution:
a. Use this specific test: does it use all the data? Conform to reasonable
estimates?
b. Use the general test: can it be obtained differently? Reduced to known
results? Can it be generate something?
The next section will show how these heuristic appear in a relatively simple
problem – a ladder reclining problem.
2
3.
The Ladder Reclining Problem
3.1.
Problem statement:
Suppose we have a ladder with 5 meters length and a cube with has edges 1-meter
length. It is supposed to be placed so that the upper edge of the ladder reclines to a vertical
wall while its lower edge touches the horizontal floor. It is also required that a part of the
ladder touches the corner of the cube (see Figure 1). To be specific, we want to know the
answers to the following questions:
1. How many possible ways to lean the ladder satisfying such condition?
2. What is the maximum height of the wall (1+a in the figure) that can be touched by the
ladder to satisfy such condition?
Here it is assumed that neither a n = 0 nor a 0 = 0 . If two neighbouring coefficients in the
sequence a n , a n −1 ,! , a1 , a0 have different signs then we call there is a sign change. The
following theorem is due to Descartes.
Theorem 1 (Descartes): The number of positive zeros of a polynomial f(x) is equal to the
number of its sign changes, or differs from it by a positive even number. The
number of negative zeros is obtained similarly from the number of sign changes
in the sequence of coefficients of the polynomial f(-x).
The sign of changes in the polynomial f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 is two and
in the polynomial f (− x) = x 4 − 2 x 3 − 23x 2 − 2 x + 1 is also two. It implies that the possible
number of the positive zeros (and also the negative zeros) is two and zero. Hence so far we
can only conclude that the number of possible ways to recline the ladder is two or none.
However, by trial and error, physically there is always one way to place the ladder. This
means that we know for sure that there will be two ways and only two ways to place the
ladder. In fact this can be shown analytically using the Sturm’s theorem by counting the
number of sign changes in the Sturm chain. Next we will define the Sturm chain.
Sturm Chain
We begin with the following fundamental theorem of algebra.
Theorem 2: Every polynomial (5) can be written as a product of its irreducible factors in
the form
f ( x) = c( x − α 1 ) r1 ( x − α 2 ) r2 ! ( x − α k ) rk ( x 2 + b1 x +c 1 ) s1 ! ( x 2 + bt x + c t ) st
with c, α i , bi , ci are real and r j , s j are positive integer constants.
Figure 1
3.2.
Next, we consider a polynomial that has roots exactly the same to the roots of f(x),
but with the simplest multiplicity,
6
ϕ ( x) = c( x − α 1 )( x − α 2 ) ! ( x − α k )( x 2 + b1 x + c1 ) ! ( x 2 + bt x + c t )
If ϕ ' ( x) denotes the first derivative of ϕ (x) , then it is the sum of terms in which in each
term one of the irreducible factors in (6) is differentiated according to the product rule of
differentiation. This means that ϕ (x) and ϕ ' ( x) does not have a common divisor except a
constant c. So that if we divide ϕ (x) by ϕ ' ( x) then we will have a polynomial q1 ( x) with
a remainder − ϕ 2 ( x) , which is a polynomial of a lower degree than ϕ ' ( x) . Divide ϕ ' ( x)
by ϕ 2 ( x) to obtain a new remainder and do the procedure repeatedly to obtain Sturm chain
ϕ ( x), ϕ ' ( x), ϕ 2 ( x) !ϕ r −1 ( x), ϕ r ( x) , with
ϕ ( x) = q1 ( x)ϕ ' ( x) − ϕ 2 ( x)
ϕ ' ( x) = q 2 ( x)ϕ 2 ( x) − ϕ 3 ( x)
The Existence of Solutions
To answer the first question we will use some classical polynomial theorem. To set
up the relationship between a and b we use Phytagorean theorem as follows:
1
(1 + a ) 2 + (1 + b) 2 = 25 .
On the other hand, we also have the following proportionality,
2
a 1
= .
1 b
Substituting (2) into (1) will obtain (1 + a ) 2 + (1 + 1 / a ) 2 = 25 or
3
a 4 + 2a 3 − 23a 2 + 2a + 1 = 0 .
Now, the problem of determining the number of possible ways to lean the ladder satisfying
such condition is equivalent to the problem of finding the number of positive real roots
(zeros) of the polynomial
4
f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 .
The Fundamental Theorem of Algebra guarantees that this polynomial has at least one root
and at most four roots in the complex field. In general, the number of positive real roots
can be determined by the Descartes’ rule of signs.
Consider the signs of the coefficients of the polynomial
5
f ( x) = a n x n + a n −1 x n −1 + ! + a1 x + a 0
The result of the previous procedure is the following theorem [see Gellert et al. (1977, p.
123) for the proof].
3
4
ϕ 2 ( x) = q3 ( x)ϕ 3 ( x) − ϕ 4 ( x)
!=!
ϕ r − 2 ( x) = q r −1 ( x)ϕ r −1 ( x) − ϕ r ( x)
ϕ r −1 ( x) = q r ( x)ϕ r ( x)
Theorem 3 (Sturm): Suppose that W (x) denotes the number of sign changes in the Sturm
chain. If ϕ (x) is a polynomial with only simple zeros, and if a < b , ϕ (a) ≠ 0
and ϕ (b) ≠ 0 , then W (a ) − W (b) is equal to the number of the zeros of the
polynomial ϕ (x) in the closed interval [a,b].
zeros for the quartic equation is found by L. Ferrari in the middle of the 16 century
(http://mathworld.wolfram/QuarticEquation.html).
To determine the end points of the closed interval [a,b] we use the following
theorem [see Gellert et al. (1977, p. 123) for the proof].
Theorem 4: If α i is the zeros of the polynomial (5) then
max(α 1 , α 2 ,!, α k ) < 1 + a n −1 + a n − 2 + ! + a1 + a 0 .
Using the Sturm chain we can determine the number of the positive zeros of the
polynomial (4), which is the solution of our original reclining problem. The resulting
closed interval is [-M,M] with M = 1 + 1 + 2 + − 23 + 2 + 1 = 30 . Hence, all the zeros are
in the closed interval [-30,30]. However, in our problem, the meaningful zeros must be
contained in the closed interval [0,5] or even smaller, e.g. [0,4] (see Figure 1). The
resulting Sturm chain with the signs of the values of the function ϕ (x) is given in the
following table.
Table 1: Sturm Chain
Sturm Chain
ϕ ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1
The sign of ϕ (x)
x=0
x=3
ϕ (0) = +
ϕ (3) = -
x=4
ϕ (3) = +
x=5
ϕ (5) = +
ϕ ' ( x) = 4 x 3 + 6 x 2 − 46 x + 2
ϕ ' (0) = +
ϕ ' (3) = +
ϕ ' (3) = +
ϕ ' (5) = +
ϕ 2 ( x) = 49 / 4 x 2 − 29 / 4 x − 3 / 4
ϕ 2 ( 0) = -
ϕ 2 (3) = +
ϕ 2 (3) = +
ϕ 2 (5) = +
ϕ 3 ( x) = 97968 / 2401x − 6032 / 2401 ϕ 3 (0) = -
ϕ 3 (3) = +
ϕ 3 (3) = +
ϕ 3 (5) = +
ϕ 4 ( x) = 1020425 / 887364
ϕ 4 (0) = +
ϕ 4 (3) = +
ϕ 4 (3) = +
ϕ 4 (5) = +
The number of sign changes
2
1
0
0
We see from the Table that W(0)=2 and W(5)=0. Hence there are 2-0=2 real zeros in the
closed interval [0,5]. Since W(3)=1 and W(4)=0, then we can also conclude that one of the
real zeros is in [0,3]. The other zero is in [3,4] and there is no zero in [4,5]. Finally, we
conclude that there are two different ways to recline the ladder in our problem. The
maximum height that can be achieved by the ladder is somewhere between 1+3 meters and
1+4 meters.
For comparison, to find the exact value of the zeros, we use Maple V and found
that the first zero is 3.838501161 and the second zero is 0.260518353. This confirms that
the maximum height of the vertical part of the ladder is in the interval [4,5] (4.838501161
meters to be exact). Maple V also found all the negative zeros, i.e. –0.168622783 and –
5.930396731. As predicted by the Sturm chain, all the four zeros are in the interval [30,30]. Figure 2 show all positive zeros found by Maple V. Analytically, the formula of the
5
Figure 2: The graph of the equation f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 and its positive zeros
3.3.
Extension of the Ladder Reclining Problem
We can extend the problem by doing the following simple experiment. By writing a
simple computer program we can construct the following table showing the relationship
among the length of the ladder (l), the length of the cube (q) and the number of ways in
which the ladder can be placed on (n).
Table 2: The relationship among l, q and n.
Length of the ladder (l)
Length of the cube Number of ways
(q)
(n)
6
1
2
5
1
2
4
1
2
3 (minimum l )
1
2
2
1
0
6 (minimum l)
2
2
5
2
0
9 (minimum l)
3
2
12 (minimum l)
4
2
15 (minimum l)
5
2
3*k (minimum l: conjecture) k
2 (conjecture)
We see from the table that there is a minimum bound for the length of the ladder in order
the problem has a non-trivial solution. We may make a conjecture that if the length of the
cube supporting the ladder, in order it can be placed in such a way as in our reclining
problem, then the minimum length of the ladder is 3k. This observation leads to the
following theorem.
Theorem 5 (Sufficient condition for positive solutions): If l ≥ 3q then there are exactly
two positive solutions to the ladder reclining problem.
6
Proof:
General equation for the height of the vertical part of the ladder with the length of
the ladder l and the length of the cube q is given by
q2
(q + x ) 2 + ( q + ) 2 = l 2 .
7
x
Or alternatively,
8
F ( x) = x 4 + 2qx 3 + (2q 2 − l 2 ) x 2 + 2q 3 x + q 4
According to Ferrari [2,3], the solution to this quartic equation is
z
x1, 2 ,3, 4 = − 1 2 q ± 1 2 y ± 1 2
y
y = q2 + l 2
9
z = −2q y + l y − 2q − 2ql
2
Consider
x
as
the
biggest
2
3
zeros.
If
2
l ≥ 3q and
q > 0 then
The condition l ≥ 3q is not a necessary condition in order the reclining problem has
two positive solutions. This can be seen by observing that the pair q = 1 and l = 2,85
produces a positive solution while l < 3q . Figure 4 shows the point-plot and contour-plot
of the positive zeros as a function q and l. There are some positive zeros if l > 2q roughly.
In fact, 2q 2 is the lowest bound of l in order the problem to have a positive solution.
This fact is summarised in the following theorem.
Theorem 5 (Necessary and sufficient condition for positive solutions): There are
exactly two positive solutions to the ladder reclining problem if and only if
l ≥ 2q 2 .
Proof:
• ⇒ Suppose that there is a positive solution, say x with x is real. This means
z = −2q 2 y + l 2 y − 2q 3 − 2ql 2 ≥ 0 .
we
have y = q 2 + l 2 ≥ 10q 2 and
z = l (l − 2q 2 )(l + 2q 2 ) ≥ 0 .
z = (−2q 2 + l 2 ) y − 2q (q 2 + l 2 ) ≥ (−2q 2 + 9q 2 ) 10q 2 − 2q (q 2 + 9q 2 ) = 7 10q 3 − 20q 3 > 0
Hence the third term of x is positive. The sum of the first two terms in x is also positive, i.e.
− 1 2 q + 1 2 y ≥ − 1 2 q + 1 2 10q = 1 2 ( 10 − 1)q > 0 .
This shows that x is positive.
Since all of the coefficients of F(x) are positive except 2q 2 − l 2 ≤ 2q 2 − 9q 2 < 0 , then the
number of sign changes in the coefficient of F(x) is two. This implies that the number of
positive zeros of F(x) either two or zero. Considering that we have found one of the zeros
then they must two zeros altogether. •
We conclude that l ≥ 2q 2 .
2
•⇐
Suppose l ≥ 2q 2 then
y = q 2 + l 2 ≥ q 2 + (2q 2 ) = 3q
and z ≥ 0 as can be seen as follows:
z = −2q 2 y + l 2 y − 2q 3 − 2ql 2
z ≥ −2 q 2 y + ( 2 q 2 ) 2 y − 2 q 3 − 2 q ( 2 q 2 ) 2
z ≥ (−2q 2 + 8q 2 ) y − (2q 3 + 16q 3 )
z ≥ (−2q 2 + 8q 2 )3q − (2q 3 + 16q 3 ) = 18q 3 − 18q 3 = 0 .
To show that x is positive, it is enough to show that the sum of the first two terms of x is
positive as follows.
− 1 2 q + 1 2 y ≥ − 1 2 q + 1 2 q 2 + 8q 2 = 1 2 ( 9 − 1)q = q > 0 .
As in the poof of the previous theorem there exist another positive solution.
These prove the theorem. •
Figure 4: The point-plot and contour-plot of the positive zeros as a function q and l.
7
8
4.
Concluding Remarks
1.
In this paper we have applied some principles of classical theory in polynomial to
solve a real world problem of a ladder reclining problem. Two simple theorems are
constructed. There are some possible venues to proceed for future exercises. For example,
by observing Figure 4, there are many pairs of (q.l) produce the same solutions. It is worth,
theoretically, to explore these pairs and to find the possible mathematical structure
describing the relationship of the pairs.
2.
Problem solving strategies are both complex and subtle, as we saw in section 3.
However, most of solutions may be obtained “based on principles that can be learned and
practiced” [7, p. ix]. Referring to important heuristics in problem solving described by
Polya (section 2), in terms of problem clarity, the example in this paper is self-understood,
since it has been already simple by only presenting a special case (the length of the ladder
is 5 meters) and it has provided a diagram representing the ladder as a line (section 3.1).
Hence, one may proceed to next step, designing and planning a solution. Using the known
relevant factors and conditions in section 3.1, one may find a polynomial describing the
relationship between the factors. In this stage presumed previously acquired knowledge is
applied to solve the polynomial problem (section 3.2). A solution is found and then it is
verified by comparing to the numerical result by the help of mathematical software. The
use of mathematical software also enables one to explore solution to a more difficult
problem (section 3.3). [In fact, the use of technology (e.g. mathematical software) is
encouraged by many institution [9] “to foster teaching and learning, increase student’
understanding of mathematical concepts, and prepare students for the use technology in
their (future) careers or their graduate study [8, p. 10]].
References:
[1].
Gellert, W et al. (1977). The VNR Concise Encyclopedia of Mathematics. Van
Nostrand Reinhold. New York.
[2]
(http://mathworld.wolfram/CubicEquation.html).
[3]
(http://mathworld.wolfram/QuarticEquation.html).
[4]
National Council of Supervisors of Mathematics (1977). Position Paper on Basic
Mathematical Skills. National Institute of Education. Washington, D.C.
[5]
Polya, G. (1957). How to Solve It. Princeton University Press. Princeton, N.J.
[6]
Schoenfeld, A.H. (1980). Heuristics in the Classroom. In Problem Solving in School
Mathematics. The National Council of Teachers Mathematics, Inc. Virginia.
[7]
Erickson, M.J. and J. Flowers (1999). Principles of Mathematics Problem Solving.
Prentice Hall. London.
[8]
The Mathematical Association of America (2003). Guidelines for Programs and
Departments in Undergraduate Mathematical Sciences (Revised Edition).
[9]
Pusat Kurikulum (2002). Kurikulum Berbasis Kompetensi. Balitbang Depdiknas.
9
IN OPTIMAL LADDER RECLINING PROBLEM
(An Example of Problem Solving in Teaching Mathematics)*
models on how applying this approaches in the classrooms still limited to most
mathematics educators. This paper is intended to fill this gap by demonstrating the
development “simple” mathematics theorems from a simple riddle-like problem, as a
result of a problem solving activity.
Asep K. Supriatna1 and Stanley P. Dewanto
Dept. Mathematics, Faculty of Mathematics & Natural Sciences, Universitas Padjadjaran
Km 21 Jatinangor Sumedang, Fax: 022-7794696
Abstract
The paper has twofold purposes. First, the paper demonstrates a stage-by-stage problem
solving which is an important process that should be experienced by all mathematics
students. It can be regarded as a model in teaching mathematics in the context of
meaningful learning and constructivism. Second, in this paper we consider an application
of classical polynomial theory in determining the existence of a ladder reclining problem.
A ladder with 5 meters length is supposed to be placed so that the upper edge of the ladder
reclines to a vertical wall while its lower edge touch the horizontal floor. It is also required
that a part of the ladder touches a corner of certain geometric object (in this case a cube
that has edges 1-meter length). We address the question on how many ways the ladder can
be placed to satisfy such condition. Surprisingly, it was found that there are only two ways
to place the ladder. Furthermore, in this paper we also give a necessary and sufficient
condition in order a ladder with length l can be placed in the intended way so that a part of
the ladder touches a part of a cube with edge k.
1.
Introduction
Problem solving capability is regarded as one among the most important
competences that should be mastered by all mathematics graduates. It is related to the
capability of applying mathematical concept in everyday and real life problems. Other
factors such as the ability of problem identification, choosing relevant factors, formulating
mathematics description and its analysis, and interpreting back the results of the
mathematical analysis into the real world are included in the problem solving capability.
This kind of competence can be developed through an appropriate experience in the
classroom activities on a regularly basis.
Many mathematics educators believe that the development of this competence can
be achieved by students who well experienced both in hands and minds on problems.
Knowing the real problems well, from where the mathematics problems come, will give a
meaningful learning for the students who did the mathematical exercises. This will help
them in understanding the mathematics and we believe that this will lead to increase their
curiousness in extending the problems at hand. We also believe this scientific curiousness
eventually will contribute to the development of a good transferable skill, namely
inovation.
Meaningful learning and constructivism -- through problem solving -- are among
the pedagogical jargons that increasingly popular to mathematics educators. However, the
*
Presented in the-XI I National Conference of Mathematics, Bali 23-27 July 2004
1
Corresponding e-mail address: asupriat@unpad.ac.id and
2.
Understanding Problem Solving
The term problem solving occurs in many different professions and disciplines and
has different meanings. Problem solving in mathematics has a more specific meaning, but
it still open to different interpretations. So, problem solving is an all-encompassing term
that means different things to different people at the same time, and different things to the
same person at different times. In this paper, problem solving is defined as the process of
applying previously acquired knowledge to new and unfamiliar non-routine situation [4].
Polya writes in How to Solve It that in studying modern heuristic, one “endeavors
to understand the process of solving problems, especially the mental operations typically
useful in this process” [5, p.129]. Under appropriate circumstances, one can learn to use
heuristic, with the result being a demonstrable improvement in their problem solving
performance.
Heuristic means a general strategy or suggestion, independent of any particular
topic or subject matter, that helps problem solver approach and understand a problem and
efficiently marshall their resources to solve it [6]. Following Polya, some important
heuristic in problem solving are [6]:
1. Analysing and Understanding a Problem:
a. Draw a diagram if at all possible.
b. Examine special cases.
c. Try to simplify it by using symmetry or “without loss of generality”.
2. Designing and Planning a Solution:
a. Plan solutions hierarchically.
b. Be able to explain at any point of solution what you are doing and why,
what you will do with the result of this operation.
3. Exploring Solutions to Difficult Problems:
a. Consider a variety of equivalent problems.
b. Consider slight modifications of the original problem.
c. Consider broad modifications of the original problem.
4. Verifying a Solution:
a. Use this specific test: does it use all the data? Conform to reasonable
estimates?
b. Use the general test: can it be obtained differently? Reduced to known
results? Can it be generate something?
The next section will show how these heuristic appear in a relatively simple
problem – a ladder reclining problem.
2
3.
The Ladder Reclining Problem
3.1.
Problem statement:
Suppose we have a ladder with 5 meters length and a cube with has edges 1-meter
length. It is supposed to be placed so that the upper edge of the ladder reclines to a vertical
wall while its lower edge touches the horizontal floor. It is also required that a part of the
ladder touches the corner of the cube (see Figure 1). To be specific, we want to know the
answers to the following questions:
1. How many possible ways to lean the ladder satisfying such condition?
2. What is the maximum height of the wall (1+a in the figure) that can be touched by the
ladder to satisfy such condition?
Here it is assumed that neither a n = 0 nor a 0 = 0 . If two neighbouring coefficients in the
sequence a n , a n −1 ,! , a1 , a0 have different signs then we call there is a sign change. The
following theorem is due to Descartes.
Theorem 1 (Descartes): The number of positive zeros of a polynomial f(x) is equal to the
number of its sign changes, or differs from it by a positive even number. The
number of negative zeros is obtained similarly from the number of sign changes
in the sequence of coefficients of the polynomial f(-x).
The sign of changes in the polynomial f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 is two and
in the polynomial f (− x) = x 4 − 2 x 3 − 23x 2 − 2 x + 1 is also two. It implies that the possible
number of the positive zeros (and also the negative zeros) is two and zero. Hence so far we
can only conclude that the number of possible ways to recline the ladder is two or none.
However, by trial and error, physically there is always one way to place the ladder. This
means that we know for sure that there will be two ways and only two ways to place the
ladder. In fact this can be shown analytically using the Sturm’s theorem by counting the
number of sign changes in the Sturm chain. Next we will define the Sturm chain.
Sturm Chain
We begin with the following fundamental theorem of algebra.
Theorem 2: Every polynomial (5) can be written as a product of its irreducible factors in
the form
f ( x) = c( x − α 1 ) r1 ( x − α 2 ) r2 ! ( x − α k ) rk ( x 2 + b1 x +c 1 ) s1 ! ( x 2 + bt x + c t ) st
with c, α i , bi , ci are real and r j , s j are positive integer constants.
Figure 1
3.2.
Next, we consider a polynomial that has roots exactly the same to the roots of f(x),
but with the simplest multiplicity,
6
ϕ ( x) = c( x − α 1 )( x − α 2 ) ! ( x − α k )( x 2 + b1 x + c1 ) ! ( x 2 + bt x + c t )
If ϕ ' ( x) denotes the first derivative of ϕ (x) , then it is the sum of terms in which in each
term one of the irreducible factors in (6) is differentiated according to the product rule of
differentiation. This means that ϕ (x) and ϕ ' ( x) does not have a common divisor except a
constant c. So that if we divide ϕ (x) by ϕ ' ( x) then we will have a polynomial q1 ( x) with
a remainder − ϕ 2 ( x) , which is a polynomial of a lower degree than ϕ ' ( x) . Divide ϕ ' ( x)
by ϕ 2 ( x) to obtain a new remainder and do the procedure repeatedly to obtain Sturm chain
ϕ ( x), ϕ ' ( x), ϕ 2 ( x) !ϕ r −1 ( x), ϕ r ( x) , with
ϕ ( x) = q1 ( x)ϕ ' ( x) − ϕ 2 ( x)
ϕ ' ( x) = q 2 ( x)ϕ 2 ( x) − ϕ 3 ( x)
The Existence of Solutions
To answer the first question we will use some classical polynomial theorem. To set
up the relationship between a and b we use Phytagorean theorem as follows:
1
(1 + a ) 2 + (1 + b) 2 = 25 .
On the other hand, we also have the following proportionality,
2
a 1
= .
1 b
Substituting (2) into (1) will obtain (1 + a ) 2 + (1 + 1 / a ) 2 = 25 or
3
a 4 + 2a 3 − 23a 2 + 2a + 1 = 0 .
Now, the problem of determining the number of possible ways to lean the ladder satisfying
such condition is equivalent to the problem of finding the number of positive real roots
(zeros) of the polynomial
4
f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 .
The Fundamental Theorem of Algebra guarantees that this polynomial has at least one root
and at most four roots in the complex field. In general, the number of positive real roots
can be determined by the Descartes’ rule of signs.
Consider the signs of the coefficients of the polynomial
5
f ( x) = a n x n + a n −1 x n −1 + ! + a1 x + a 0
The result of the previous procedure is the following theorem [see Gellert et al. (1977, p.
123) for the proof].
3
4
ϕ 2 ( x) = q3 ( x)ϕ 3 ( x) − ϕ 4 ( x)
!=!
ϕ r − 2 ( x) = q r −1 ( x)ϕ r −1 ( x) − ϕ r ( x)
ϕ r −1 ( x) = q r ( x)ϕ r ( x)
Theorem 3 (Sturm): Suppose that W (x) denotes the number of sign changes in the Sturm
chain. If ϕ (x) is a polynomial with only simple zeros, and if a < b , ϕ (a) ≠ 0
and ϕ (b) ≠ 0 , then W (a ) − W (b) is equal to the number of the zeros of the
polynomial ϕ (x) in the closed interval [a,b].
zeros for the quartic equation is found by L. Ferrari in the middle of the 16 century
(http://mathworld.wolfram/QuarticEquation.html).
To determine the end points of the closed interval [a,b] we use the following
theorem [see Gellert et al. (1977, p. 123) for the proof].
Theorem 4: If α i is the zeros of the polynomial (5) then
max(α 1 , α 2 ,!, α k ) < 1 + a n −1 + a n − 2 + ! + a1 + a 0 .
Using the Sturm chain we can determine the number of the positive zeros of the
polynomial (4), which is the solution of our original reclining problem. The resulting
closed interval is [-M,M] with M = 1 + 1 + 2 + − 23 + 2 + 1 = 30 . Hence, all the zeros are
in the closed interval [-30,30]. However, in our problem, the meaningful zeros must be
contained in the closed interval [0,5] or even smaller, e.g. [0,4] (see Figure 1). The
resulting Sturm chain with the signs of the values of the function ϕ (x) is given in the
following table.
Table 1: Sturm Chain
Sturm Chain
ϕ ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1
The sign of ϕ (x)
x=0
x=3
ϕ (0) = +
ϕ (3) = -
x=4
ϕ (3) = +
x=5
ϕ (5) = +
ϕ ' ( x) = 4 x 3 + 6 x 2 − 46 x + 2
ϕ ' (0) = +
ϕ ' (3) = +
ϕ ' (3) = +
ϕ ' (5) = +
ϕ 2 ( x) = 49 / 4 x 2 − 29 / 4 x − 3 / 4
ϕ 2 ( 0) = -
ϕ 2 (3) = +
ϕ 2 (3) = +
ϕ 2 (5) = +
ϕ 3 ( x) = 97968 / 2401x − 6032 / 2401 ϕ 3 (0) = -
ϕ 3 (3) = +
ϕ 3 (3) = +
ϕ 3 (5) = +
ϕ 4 ( x) = 1020425 / 887364
ϕ 4 (0) = +
ϕ 4 (3) = +
ϕ 4 (3) = +
ϕ 4 (5) = +
The number of sign changes
2
1
0
0
We see from the Table that W(0)=2 and W(5)=0. Hence there are 2-0=2 real zeros in the
closed interval [0,5]. Since W(3)=1 and W(4)=0, then we can also conclude that one of the
real zeros is in [0,3]. The other zero is in [3,4] and there is no zero in [4,5]. Finally, we
conclude that there are two different ways to recline the ladder in our problem. The
maximum height that can be achieved by the ladder is somewhere between 1+3 meters and
1+4 meters.
For comparison, to find the exact value of the zeros, we use Maple V and found
that the first zero is 3.838501161 and the second zero is 0.260518353. This confirms that
the maximum height of the vertical part of the ladder is in the interval [4,5] (4.838501161
meters to be exact). Maple V also found all the negative zeros, i.e. –0.168622783 and –
5.930396731. As predicted by the Sturm chain, all the four zeros are in the interval [30,30]. Figure 2 show all positive zeros found by Maple V. Analytically, the formula of the
5
Figure 2: The graph of the equation f ( x) = x 4 + 2 x 3 − 23x 2 + 2 x + 1 and its positive zeros
3.3.
Extension of the Ladder Reclining Problem
We can extend the problem by doing the following simple experiment. By writing a
simple computer program we can construct the following table showing the relationship
among the length of the ladder (l), the length of the cube (q) and the number of ways in
which the ladder can be placed on (n).
Table 2: The relationship among l, q and n.
Length of the ladder (l)
Length of the cube Number of ways
(q)
(n)
6
1
2
5
1
2
4
1
2
3 (minimum l )
1
2
2
1
0
6 (minimum l)
2
2
5
2
0
9 (minimum l)
3
2
12 (minimum l)
4
2
15 (minimum l)
5
2
3*k (minimum l: conjecture) k
2 (conjecture)
We see from the table that there is a minimum bound for the length of the ladder in order
the problem has a non-trivial solution. We may make a conjecture that if the length of the
cube supporting the ladder, in order it can be placed in such a way as in our reclining
problem, then the minimum length of the ladder is 3k. This observation leads to the
following theorem.
Theorem 5 (Sufficient condition for positive solutions): If l ≥ 3q then there are exactly
two positive solutions to the ladder reclining problem.
6
Proof:
General equation for the height of the vertical part of the ladder with the length of
the ladder l and the length of the cube q is given by
q2
(q + x ) 2 + ( q + ) 2 = l 2 .
7
x
Or alternatively,
8
F ( x) = x 4 + 2qx 3 + (2q 2 − l 2 ) x 2 + 2q 3 x + q 4
According to Ferrari [2,3], the solution to this quartic equation is
z
x1, 2 ,3, 4 = − 1 2 q ± 1 2 y ± 1 2
y
y = q2 + l 2
9
z = −2q y + l y − 2q − 2ql
2
Consider
x
as
the
biggest
2
3
zeros.
If
2
l ≥ 3q and
q > 0 then
The condition l ≥ 3q is not a necessary condition in order the reclining problem has
two positive solutions. This can be seen by observing that the pair q = 1 and l = 2,85
produces a positive solution while l < 3q . Figure 4 shows the point-plot and contour-plot
of the positive zeros as a function q and l. There are some positive zeros if l > 2q roughly.
In fact, 2q 2 is the lowest bound of l in order the problem to have a positive solution.
This fact is summarised in the following theorem.
Theorem 5 (Necessary and sufficient condition for positive solutions): There are
exactly two positive solutions to the ladder reclining problem if and only if
l ≥ 2q 2 .
Proof:
• ⇒ Suppose that there is a positive solution, say x with x is real. This means
z = −2q 2 y + l 2 y − 2q 3 − 2ql 2 ≥ 0 .
we
have y = q 2 + l 2 ≥ 10q 2 and
z = l (l − 2q 2 )(l + 2q 2 ) ≥ 0 .
z = (−2q 2 + l 2 ) y − 2q (q 2 + l 2 ) ≥ (−2q 2 + 9q 2 ) 10q 2 − 2q (q 2 + 9q 2 ) = 7 10q 3 − 20q 3 > 0
Hence the third term of x is positive. The sum of the first two terms in x is also positive, i.e.
− 1 2 q + 1 2 y ≥ − 1 2 q + 1 2 10q = 1 2 ( 10 − 1)q > 0 .
This shows that x is positive.
Since all of the coefficients of F(x) are positive except 2q 2 − l 2 ≤ 2q 2 − 9q 2 < 0 , then the
number of sign changes in the coefficient of F(x) is two. This implies that the number of
positive zeros of F(x) either two or zero. Considering that we have found one of the zeros
then they must two zeros altogether. •
We conclude that l ≥ 2q 2 .
2
•⇐
Suppose l ≥ 2q 2 then
y = q 2 + l 2 ≥ q 2 + (2q 2 ) = 3q
and z ≥ 0 as can be seen as follows:
z = −2q 2 y + l 2 y − 2q 3 − 2ql 2
z ≥ −2 q 2 y + ( 2 q 2 ) 2 y − 2 q 3 − 2 q ( 2 q 2 ) 2
z ≥ (−2q 2 + 8q 2 ) y − (2q 3 + 16q 3 )
z ≥ (−2q 2 + 8q 2 )3q − (2q 3 + 16q 3 ) = 18q 3 − 18q 3 = 0 .
To show that x is positive, it is enough to show that the sum of the first two terms of x is
positive as follows.
− 1 2 q + 1 2 y ≥ − 1 2 q + 1 2 q 2 + 8q 2 = 1 2 ( 9 − 1)q = q > 0 .
As in the poof of the previous theorem there exist another positive solution.
These prove the theorem. •
Figure 4: The point-plot and contour-plot of the positive zeros as a function q and l.
7
8
4.
Concluding Remarks
1.
In this paper we have applied some principles of classical theory in polynomial to
solve a real world problem of a ladder reclining problem. Two simple theorems are
constructed. There are some possible venues to proceed for future exercises. For example,
by observing Figure 4, there are many pairs of (q.l) produce the same solutions. It is worth,
theoretically, to explore these pairs and to find the possible mathematical structure
describing the relationship of the pairs.
2.
Problem solving strategies are both complex and subtle, as we saw in section 3.
However, most of solutions may be obtained “based on principles that can be learned and
practiced” [7, p. ix]. Referring to important heuristics in problem solving described by
Polya (section 2), in terms of problem clarity, the example in this paper is self-understood,
since it has been already simple by only presenting a special case (the length of the ladder
is 5 meters) and it has provided a diagram representing the ladder as a line (section 3.1).
Hence, one may proceed to next step, designing and planning a solution. Using the known
relevant factors and conditions in section 3.1, one may find a polynomial describing the
relationship between the factors. In this stage presumed previously acquired knowledge is
applied to solve the polynomial problem (section 3.2). A solution is found and then it is
verified by comparing to the numerical result by the help of mathematical software. The
use of mathematical software also enables one to explore solution to a more difficult
problem (section 3.3). [In fact, the use of technology (e.g. mathematical software) is
encouraged by many institution [9] “to foster teaching and learning, increase student’
understanding of mathematical concepts, and prepare students for the use technology in
their (future) careers or their graduate study [8, p. 10]].
References:
[1].
Gellert, W et al. (1977). The VNR Concise Encyclopedia of Mathematics. Van
Nostrand Reinhold. New York.
[2]
(http://mathworld.wolfram/CubicEquation.html).
[3]
(http://mathworld.wolfram/QuarticEquation.html).
[4]
National Council of Supervisors of Mathematics (1977). Position Paper on Basic
Mathematical Skills. National Institute of Education. Washington, D.C.
[5]
Polya, G. (1957). How to Solve It. Princeton University Press. Princeton, N.J.
[6]
Schoenfeld, A.H. (1980). Heuristics in the Classroom. In Problem Solving in School
Mathematics. The National Council of Teachers Mathematics, Inc. Virginia.
[7]
Erickson, M.J. and J. Flowers (1999). Principles of Mathematics Problem Solving.
Prentice Hall. London.
[8]
The Mathematical Association of America (2003). Guidelines for Programs and
Departments in Undergraduate Mathematical Sciences (Revised Edition).
[9]
Pusat Kurikulum (2002). Kurikulum Berbasis Kompetensi. Balitbang Depdiknas.
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