Determining the Existence of any Intersection Between a
Logarithmic Function and a Polynomial
Yizhou (Ethan) Xu
Mathematics
Supervisor: Ms. Kobbi
Sir Winston Churchill Secondary School
May 2016
Word Count: 3975
Acknowledgements: Special thanks to Ms. Kobbi and Dr. Mark
Krusemeyer who has provided helpful insights throughout this research
Abstract
This research paper aims to find mathematical formulae that can be used to predict the
existence of intersections between a logarithmic function and a polynomial. To narrow the focus
of this paper, several assumptions have been made: intersections between the two functions at
infinity are not considered as intersections, all coefficients of the two functions are real numbers
other than Β± β, and the base of the logarithmic function is greater than 1. Case analysis was used
as the main approach. The degrees of the two functions were varied in each case, and their
behaviour was analyzed. For each case, I established an inequality between the coefficients of the
two functions for determining if there are intersections between the functions.
Word Count: 119
Table of Contents
Introduction------------------------------------------------------------------------------------1
1. deg (f) = 0 or deg (g) = 0------------------------------------------------------------------3
2. deg (f) = 1, deg (g) = 1---------------------------------------------------------------------3
3. deg (f) = 2, deg (g) = 1---------------------------------------------------------------------6
4. deg (f) = 1, deg (g) = 2--------------------------------------------------------------------10
5. deg (f) = 2, deg (g) = 2-------------------------------------------------------------------12
6. Further observation-----------------------------------------------------------------------21
7. Conclusion---------------------------------------------------------------------------------22
Bibliography----------------------------------------------------------------------------------24
1
Introduction
My research question asks: are there mathematical formulae to predict whether the
equation, logπ π(π₯) = π(π₯), where π(π₯) and π(π₯) are polynomials, has any solutions.
The reason why I chose to investigate this question is that, from my past algebraic problem-
solving experiences, I often found it difficult to determine whether a solution exists for a given
equation. In order to solve this problem, I strive to find a quicker and more convenient way for
solving these questions, which can vastly increase the efficiency.
In my research paper, I have limited the scope of my investigation to a specific type of
equation: one between a logarithmic function and a polynomial function. There may be ways to
determine if intersections exist for other types of equations.
My research integrates three fields of mathematics: calculus, algebra, and geometry. Calculus
will be mainly employed to calculate the slopes of logπ π(π₯) and π(π₯) at a given point in order
to determine when the two functions do not intersect. Finding the slopes of logπ π(π₯) and π(π₯)
will be very helpful for determining the x-value at which these functions lie tangent to each other.
I will use algebra to manipulate the two functions and their derivatives to yield a restriction on
the coefficients of π(π₯) and π(π₯) such that there exists no intersections between them.
Geometry will be mainly used to predict the behaviour of logπ π(π₯) and π(π₯) in order to
determine how the restrictions on the coefficients of these functions should be placed such that
they do not intersect.
2
I will first limit the degrees of the functions π(π₯) and π(π₯) to 1 in order to gain a better
understanding of my investigation. Afterwards, I will vary the degrees of these functions to
generalize my results so that they can be applied to more scenarios.
The solution to my research question will provide a relatively simple way to solve these
equations. Its significance lies in the fact that it is especially useful for those who have not yet
learned calculus to solve for these similar questions, since my formulae do not require any
understanding of calculus or any other area of mathematics beyond high school algebra.
Considering that the formulae I find can decrease the number of operations one has to make and
thus shorten the calculation time for a given equation in the form of logπ π(π₯) = π(π₯), where
π(π₯) and π(π₯) are polynomials, my solution can potentially be very useful for efficiently
solving these similar equations.
To proceed with my investigation, I will first make several assumptions that can simplify my
research and narrow my focus:
1. Despite the theory that two curves can intersect at infinity, I would not consider
intersection of logπ π(π₯) and π(π₯) at π₯ = β or π₯ = ββ to be an intersection.
2. All coefficients of π(π₯) and π(π₯) must be real numbers other than Β± β.
3. logπ π(π₯) has base π > 1.
3
1. deg (f) = 0 or deg (g) = 0
I will discuss the two cases for which deg (f) = 0 or deg (g) = 0. When deg (f) = 0,
the two sides of the equation turn into a polynomial and a constant, which may be solved if
deg (g) < 5. When deg (g) = 0, the equation, under transformations, can still be written in
the form of a polynomial being equal to a constant term, which means that these two scenarios
are similar. These cases do not require a formula to determine whether logπ π(π₯) and
π(π₯) intersect since they can both be simplified to a polynomial equation.
2. deg (f) = 1, deg (g) = 1
Let π(π₯) = ππ₯ + c and π(π₯) = ππ₯ + π, where π, c, π, π β R and π, π β 0.
Figure 1. ππππ π(π₯) = log (10x + 3)
4
The shape of this curve, as well as the shape of any other functions in the form of
logπ(ππ₯ + π) for π > 1 and π > 0, is concave down. Each of these functions has one vertical
asymptote at the point where x = βπ
π and approaches β when x approaches β. The slope of this
type of functions approximates β as x approaches βπ
π and stays at 0 as x approaches β. For
π < 0, these functions remain concave down, but approaches β when x approaches βπ
π .
For π > 0 or π < 0, π(π₯) has to be strictly greater than logπ π(π₯) for all real values of
x in the domain of logπ π(π₯) in order to not intersect with logπ π(π₯). Function π(π₯) must
intersect with logπ π(π₯) if π(π₯) is set to be less than logπ π(π₯) for all x in the domain of
logπ π(π₯) since logπ π(π₯) approaches ββ as π₯ approaches βπ
π.
When π > 0, the leading coefficient of π(π₯), d, must be greater than 0 such that no
intersection exists between logπ π(π₯) and π(π₯). This is due to the fact that logπ π(π₯)
approaches β when x approaches β for π > 0, and if π < 0, then π(π₯) approaches
ββ when x approaches β. An intersection must exist in this case. Similarly, when π < 0,
logπ π(π₯) and π(π₯) have no intersection only if π < 0.
I will set restriction on the constant term of π(π₯) to determine whether logπ π(π₯) and
π(π₯) intersect. This way gives me the advantage of vertically translating π(π₯) through
manipulating the value of e without changing the shape of π(π₯), and has the nice property of
providing an inequality on the value of e.
Now, I suppose that the values for a, b, c, d, e are already given and it is my task to
determine whether an intersection exists between logπ π(π₯) and π(π₯).
5
In order to solve this question, I envision the function logπ π(π₯) and π(π₯) = ππ₯ + π to be
tangent to each other, Since logπ π(π₯) is concave down and π(π₯) is linear, there is only one
intersection between these two functions, which is at the point of tangency. The reason that
logπ π(π₯) and π(π₯) do not intersect at any points other than the point of tangency is that, for x-
value greater or less than the x-value at the intersection point, the slope of π(π₯) will be strictly
greater or less than the slope of logπ π(π₯) respectively, assuming that b > 01. As a result, the
value of logπ π(π₯) will be less than π(π₯) for all values of x other than the point of tangency in
the domain of logπ π(π₯) (βRooftop Theoremβ).
By increasing the value of e and thus translating π(π₯) upwards, the intersection point
between these two functions disappears. Therefore, if e is greater than the value for which π(π₯)
lies tangent to logπ π(π₯), there is no intersection between these two functions. I will use this
property to set a restriction on e such that the two functions do not intersect.
From the formula for calculating the derivative of a logarithmic function,
π
ππ₯logπ π(π₯) =
π
ln π (ππ₯ + π)
Suppose there exists a value π₯1 such that:
π
ππ₯logπ π(π₯1) =
π
ππ₯π(π₯)
Then, we can differentiate both functions to determine π₯1:
1 The case in which π < 0 is similar to π > 0, and it can be shown that logπ π(π₯) does not
intersect π(π₯) other than the point of tangency.
6
π
ln π β (ππ₯1 + π)= π
π
ln π β π= ππ₯1 + π
π₯1 =1
ln π β πβ
π
π
If logπ π(π₯1) < π(π₯1), there would be no intersection between the two functions:
logπ[ π(π₯1) + π] < π(π₯1) + π
logπ [π (1
ln π β πβ
π
π) + π] < π (
1
ln π β πβ
π
π) + π
π > logπ(π
ln π β π) β
1
ln π+
ππ
π
In this case, for π(π₯) and π(π₯) coefficients a, b, c, d, e that satisfy this inequality, there is
no real solution for logπ π(π₯) = π(π₯).
3. deg (f) = 2, deg (g) = 1
Let π(π₯) = ππ₯2 + ππ₯ + π and π(π₯) = ππ₯ + π, where b, c, d, e, f β R and b, e β 0.
There are two different behaviour of logπ π(π₯) for π > 0 and π < 0 . Here are two
examples of logπ π(π₯) for each of the two cases:
7
Figure 2. ππππ π(π₯) = πππ( π₯2 + 10π₯ + 1)
Figure 3. ππππ π(π₯) = πππ(β π₯2 + 10π₯ + 1)
For π > 0, logπ π(π₯) approaches β when x approaches β. For π < 0, logπ π(π₯) has
two vertical asymptotes at π₯ = βπΒ±βπ2β4ππ
2π, because π(π₯) is positive between its two roots.
It is helpful to note that π₯ β β if π2 β 4ππ < 0, and consequently there would be no intersection
8
between logπ π(π₯) and π(π₯). In these two scenarios, the shapes of logπ π(π₯) greatly differ
from each other in their end behaviour.
When π > 0, there must be an intersection between logπ π(π₯) and π(π₯), since π(π₯)
approaches β on one infinity side and ββ on the other, whereas logπ π(π₯) approaches β on
both sides. Thus, there may exist no intersection between these two functions if and only if
π < 0.
Lemma 1. The shape of ππππ π(π₯) when b < 0 and π2 β 4ππ > 0 is concave down.
Due to the fact that the concavity of a function does not change as it undergoes horizontal or
vertical shift, translation, expansion, or compression, if logπ(π + ππ₯2) is concave down for a
given real number π > 0, then logπ π(π₯) has to be concave down. The function
logπ(π + ππ₯2) can be further transformed into logπ π + logπ(1 +π
ππ₯2), and through horizontal
expansion or, this function transforms into logπ π + logπ(1 β π₯2), for logπ π as a real constant.
Given that a function is concave down if and only if its second derivative is less than 0, it is
important to calculate the second derivative of logπ(1 β π₯2). This is the end of the proof.
The tangent lines of concave down functions do not intersect with the functions at any points
other than their point of tangency. An upward vertical translation of π(π₯), as a result, removes
the intersection between these two functions if π(π₯) is tangent to logπ π(π₯).
According to the formula for calculating the derivative of logπ π(π₯),
9
π
ππ₯logπ ππ₯2 + ππ₯ + π =
2ππ₯ + π
ln π β (ππ₯2 + ππ₯ + π)
Suppose there exists a number π₯1 in the domain of π(π₯) such that:
π
ππ₯logπ π(π₯1) =
π
ππ₯π(π₯1)
2ππ₯1 + π
ln π β (ππ₯12 + ππ₯1 + π)
= π
(2π
ln π) π₯1 +
π
ln π= πππ₯1
2 + πππ₯1 + ππ
πππ₯12 + (ππ β
2π
ln π) π₯1 + (ππ β
π
ln π) = 0
π₯12 + (
π
πβ
2
ln π β π) π₯1 + (
π
πβ
π
ππ ln π) = 0
π₯1 =(
2π β ln π
βππ
) Β± β(ππ
β2
ln π β π)
2
β 4 (ππ
βπ
ππ ln π)
2
π₯1 =2π β ππ ln π Β± βπ2π2ln2π + 4π2 β 4πππ2ln2π
2ππ ln π
There are two possible values of π₯1, but one of the two is not in the domain of logπ π(π₯)
since the two functions would only intersect once in the given conditions. Let π₯1 that is within the
domain of logπ π(π₯) be Ξ±.
In order for logπ π(π₯) not to intersect with π(π₯), logπ π(Ξ±) must be strictly less than π(Ξ±)
due to the fact that logπ π(π₯) approaches ββ at its vertical asymptotes. It follows that:
logπ( ππΌ2 + ππΌ + π) < ππΌ + π
π > logπ(ππΌ2 + ππΌ + π) β ππΌ
Therefore, if π < 0 and π2 β 4ππ > 0 and the above inequality is satisfied, there exists no
solution for logπ π(π₯) = π(π₯) for deg (f) = 2 and deg (g) = 1.
10
4. deg (f) = 1, deg (g) = 2
Let π(π₯) = ππ₯ + π and π(π₯) = ππ₯2 + ππ₯ + π, where b, c, d, e, f β R and b, d β 0.
Lemma 2. Considering the end behaviour of ππππ π(π₯) and π(π₯), the leading coefficient
of π(π₯), π, must be greater than 0 if there exists no intersection between the two functions.
If logπ π(π₯) does not intersect π(π₯), then logπ π(π₯) must be either greater or smaller than
π(π₯) for all π₯ in the domain of logπ π(π₯). However, since, if π < 0, π(π₯) would reach ββ
as x approaches β, logπ π(π₯) cannot be entirely below π(π₯) for π < 0. It cannot be entirely
above π(π₯) either due to the fact that it approaches ββ for some real value of π₯. This is the
end of the proof.
It is now clear to see that logπ π(π₯) does not intersect π(π₯) if and only if π > 0. Ideally,
an upward vertical translation by a certain value on π(π₯) can reduce the original two
intersections between the two functions to one point of tangency, such that any other upward
vertical translation of π(π₯) would remove all intersections between them. My goal is to find
such a value.
Lemma 3. The intersection between the concave up function π(π₯) and the concave down
function ππππ π(π₯) is only one point if and only if the two functions are tangent to each other.
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Let π₯1 be the point of tangency between the two functions. Then, π
ππ₯logπ π(π₯1) =
π
ππ₯π(π₯1). For β π₯: π₯ > π₯1,
π
ππ₯π(π₯) >
π
ππ₯π(π₯1), while
π
ππ₯logπ π(π₯) <
π
ππ₯logπ π(π₯1) due to
the properties of concave up and concave down functions. Thus, π
ππ₯π(π₯) >
π
ππ₯logπ π(π₯) for
β π₯: π₯ > π₯1. Similarly, we can deduce that π
ππ₯π(π₯) <
π
ππ₯logπ π(π₯) for β π₯ < π₯1. Since
logπ π(π₯1) = π(π₯1), π(π₯) > logπ π(π₯) for βπ₯: π₯ β π₯1. This is the end of the proof.
In order to locate that potential point of tangency, it is important to first locate the point at
which π
ππ₯logπ π(π₯) =
π
ππ₯π(π₯). Suppose the two functions have the same slope at π₯1:
π
ππ₯logπ π(π₯1) =
π
ππ₯π(π₯1)
π
ln π β (ππ₯1 + π)= 2ππ₯1 + π
π
ln π= (2ππ₯1 + π)(ππ₯1 + π)
2πππ₯12 + (2ππ + ππ)π₯1 + ππ β
π
ln π= 0
π₯12 + (
π
π+
π
2π) π₯1 + (
ππ
2ππβ
1
2π ln π) = 0
π₯1 =β
ππ
βπ
2πΒ± β(
ππ
+π
2π)2 + 4(
12π ln π
βππ
2ππ)
2
π₯1 =β2ππ ln π β ππ ln π Β± β4π2π2ln2π + π2π2ln2π + 8π2π ln π β 4ππππln2π
4ππ ln π
There are two possible values of π₯1 , but one is not in the domain of logπ π(π₯) since
logπ π(π₯) and π(π₯) intersect only once if π < 0 and the two functions are tangent to each other.
Let the value π₯1 that is within the domain of logπ π(π₯) to be Ξ±.
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logπ π(π₯) must be strictly less than π(π₯), which can be satisfied by letting π(π₯) be greater
than logπ π(π₯) at π₯1. An inequality between the two functions follows:
logπ(ππΌ + π) < ππΌ2 + ππΌ + π
π > logπ(ππΌ + π) β ππΌ2 β ππΌ
5. deg (f) = 2, deg (g) = 2
Let π(π₯) = ππ₯2 + ππ₯ + π and π(π₯) = ππ₯2 + ππ₯ + π, where b, c, d, e, f, g β R and
b, e β 0.
5.1 π > π
When π > 0, there are two possible shapes of logπ π(π₯). One shape consists of two concave
down functions separated by a region in which π(π₯) β€ 0, as shown in Figure 2. It approaches
ββ when x approaches βπΒ±βπ2β4ππ
2π. This shape is distinguished from the other shape by the
fact that it satisfies the inequality π2 β₯ 4ππ. Let us denote the shape that satisfies the inequality
π2 β₯ 4ππ to be the first shape, and the other shape to be the second shape.
5.1.1 The first shape
Lemma 4. logπ π(π₯) consists of two concave down curves if π > 0 and π2 β₯ 4ππ.
13
It is useful to note that the concavity of logπ π(π₯) is preserved after horizontal and vertical
transformations. As a result, we can rewrite logπ π(π₯) as log(π₯2 + ππ₯ + π), with π2 β₯ 4π,
and then further transform the function to log[(π₯ +1
2π)
2
+πβ1
4π2]. The resulting function, after
horizontal shift, is log(π₯2 β π) for some positive real number k, which is equivalent to
log (1
ππ₯2 β 1) + log π. After vertical translation and horizontal expansion or compression, this
function ultimately becomes log(π₯2 β 1). π2
ππ₯2 log(π₯2 β 1) = π(β2π₯2β2)
(π₯2+1)2 for some positive real
number r. From this equation, we can see that π2
ππ₯2 log(π₯2 β 1) < 0. Therefore, log(π₯2 β 1)
consists of two concave down curves, which implies that the first shape of logπ π(π₯) also has
two concave down curves. This is the end of the proof.
The first shape that consists of two separate concave down curves cannot be discussed using
the aforementioned vertical translation of π(π₯). Since this shape has two separate curves, π(π₯)
may be tangent to one curve of this shape but intersect at two points with the other curve, such
that when vertical translation is applied to π(π₯), there still exist intersections between logπ π(π₯)
and π(π₯). A slight modification is needed for this method.
First, I will determine the lines of reflection for the two curves. The function π(π₯) reaches
minimum value at the vertex, where π₯ =βπ
2π, and it is symmetrical on either side of the line of
reflection. This means that logπ π(π₯) will also be symmetrical at π₯ =βπ
2π.
14
Lemma 5. a symmetrical pair of non-intersecting concave down curves and a symmetrical
concave up curve tangent to one of the concave down curves, both with reflection symmetry
parallel to the y-axis, intersect only at the tangent point if and only if the reflection line of the
concave up curve is on the same side as the tangent point according to the y-axis.
Without the loss of generality, we assume that the reflection line of the symmetrical concave
up curve is on the left side of the reflection line of the two concave down curves. By applying a
horizontal shift of the right side concave down curve to the left by a certain number of units, we
can create another concave down curve such that it is symmetrical to the left side concave down
curve at the reflection line of the concave up curve. If the concave up curve is tangent to the left
side concave down curve, then it must also be tangent to the new right side concave down curve
after the horizontal shift. Due to the fact that the concave up curve would intersect the right side
concave down curve from the left, it would intersect with the transformed concave down curve
for two times before it intersects with the original right side concave down curve. However, since
the concave up curve intersects with the transformed concave down curve only once, there would
be no intersection between the concave up curve and the right side concave down curve. This is
the end of the proof.
This lemma indicates that if we choose the tangent point to be on the same side as the vertex
of π(π₯) with respect to the reflection line of logπ π(π₯), then there will only be one point of
intersection between the two functions. This fact proves to be extremely useful by providing
information about the lower bound on the value of the constant term g in π(π₯) such that
15
logπ π(π₯) and π(π₯) do not intersect.
5.1.2 The second shape
The second shape is a continuous function that resembles the first shape, but is different in
that it is not entirely concave down. Here is an example of the second shape:
Figure 4. ππππ π(π₯) = πππ(π₯2 + 5π₯ + 12)
In the second shape, there exist two inflection points at which the slope of the function
changes from decreasing to increasing and from increasing to decreasing. The situation in which
the second shape would occur is when the equation π(π₯) = 0 has no real solution.
Essentially, there needs to be a minimum value on the constant value of π(π₯) such that
logπ π(π₯) does not intersect with π(π₯).
The difficulty of this case lies in the fact that the function logπ π(π₯) will no longer be just a
concave down curve and cannot be dealt with using the same method. To simplify this problem, I
16
will apply case analysis by examining the two distinct shapes of logπ π(π₯) for π > 0
individually.
The second shape is a continuous, non-concave down function with two inflection points,
and we can prove this fact through algebraic manipulation.
Lemma 6. ππππ π(π₯) = ππ₯2 + ππ₯ + π has two inflection points if b > 0 and π2 < 4ππ,
with b, c, d being real numbers.
logπ π(π₯) does not change its general shape after horizontal or vertical shift, translation,
expansion, or compression. The function can be reduced to logπ(1 + π₯2). π2
ππ₯2 logπ(1 + π₯2) =
2β2π₯2
ln πβ(1+π₯2)2, which equals to 0 only if 2 β 2π₯2 = 0. Since the numerator, (2 β 2π₯2), is a
quadratic polynomial, there are at most two solutions to π2
ππ₯2 logπ(1 + π₯2) = 0. Thus, logπ π(π₯)
has at most 2 inflections points.
π
ππ₯logπ π(π₯) = 0 at the reflection line2, and gradually approaches 0 as x approaches β.
Therefore, there must exist at least one point on the right side of the reflection line such that the
π
ππ₯logπ π(π₯) changes from increasing to decreasing. Due to the fact that the second shape of
logπ π(π₯) is symmetrical, there exist at least two inflection points for logπ π(π₯). Therefore, the
second shape of logπ π(π₯) has two inflection points, one on each side of the reflection line. This
is the end of the proof.
2 The slope of a symmetrical function that is continuous on (ββ, β) and differentiable on (ββ, β) equals 0 at the reflection line,
17
π
ππ₯logπ π(π₯) approaches 0 as x approaches Β± β, and has a rotational symmetry around the
point on logπ π(π₯) at x =βπ
2π, where the slope of logπ π(π₯) = 0. An example of this derivative
is shown on the figure below:
Figure 5. π¦ =20π₯+5
10π₯2+5π₯+1
Note that the derivative of π(π₯) is a line with a positive slope.
As mentioned previously, the tangent point that will be taken must be on the same side as the
vertex of π(π₯) according to the reflection point at π₯ =βπ
2π . This means that, if the tangent point
taken is at the left side of π₯ =βπ
2π, then the derivative of π(π₯) must intersect with the x-axis left
to π₯ =βπ
2π as well.
Since the derivative of logπ π(π₯) has rotational symmetry around the point on logπ π(π₯) at
π₯ =βπ
2π, and the set of all lines with positive slope is unchanged after an 180Β° rotation around that
18
point, without the loss of generality, we can assume that the vertex of π(π₯) is on the left side of
π₯ =βπ
2π.
From the fact that the derivative of π(π₯) is a line with a positive slope, we can deduce that
the derivative of π(π₯) will only intersect once with the derivative of logπ π(π₯) on the left side
of π₯ =βπ
2π. This finding is very useful, since if there are two intersections between the derivative
of π(π₯) and logπ π(π₯) at the left hand side of π₯ =βπ
2π, then it is more difficult to determine
which tangent point to use as the boundary for the constant term, g, in π(π₯).
It is important to note that, if logπ π(π₯) and π(π₯) are tangent to each other and the vertex
of π(π₯) and the tangent point are on the same side with respect to π₯ =βπ
2π, then there will be no
intersection between logπ π(π₯) and π(π₯) on the other side of π₯ =βπ
2π.
What we have determined here is that the case of the second shape is congruent with the case
of the first one, and that it can be solved using the same method.
Due to the occurrence of the second shape of logπ π(π₯) for π > 0, the function π(π₯) is
allowed to open downwards such that there exists no intersection between logπ π(π₯) and π(π₯).
In other words, e, the leading coefficient of π(π₯), can be negative. The reason is that the second
shape of logπ π(π₯) does not approach negative infinity for any real values of x, and thus allows
π(π₯) to approach ββ when x approaches Β± β.
Using the same approach as before, we first try to determine the values of x for which the
derivative of logπ π(π₯) and the derivative of π(π₯) are the same. This question simplifies to
solving a polynomial equation with degree 3, which has a maximum of three real solutions. In
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other words, the derivative of logπ π(π₯) and the derivative of π(π₯) intersects at most three
times.
There are three possibilities for the intersections between the derivatives of the two
functions: one intersection, two intersections including a tangent point, and three intersections.
If there exists only one solution to the polynomial equation, which means that there exists
one intersection between the derivatives, we should choose that point as the tangent point
between logπ π(π₯) and π(π₯) and determine a restriction on the value of g, the constant term of
π(π₯) such that there exists no intersection between the two functions. The reason why this
method works is that to the left of the intersection of the two derivatives, the derivative of π(π₯)
is greater than that of logπ π(π₯), and to the right of the intersection, that of π(π₯) is less than
that of logπ π(π₯), which guarantees that π(π₯) will be less than logπ π(π₯) for all values of x
other than the value of x at the tangent point.
It is possible for the two derivatives to have two intersections, but one of the intersections
must be a tangent point between the derivatives. If logπ π(π₯) = π(π₯) at this tangent point
between the derivatives, the two functions cross each other at this point even though their slopes
are equal. This point is also called the inflection point of the two tangent curves. This point can
be disregarded in our investigation, since it does not provide us with information about the
restriction on the constant term g for logπ π(π₯) and π(π₯) to not intersect. The way to see if an
intersection between the derivatives is a tangent point is by calculating the second derivatives of
logπ π(π₯) and π(π₯). If the second derivatives are equal to each other at the intersection, then
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this intersection is indeed a tangent point. After knowing which one is the inflection point, we
would just use the other intersection and employ the same method previously mentioned.
If there are three intersections between the two derivatives, then we call the x-coordinates of
the intersections as π₯1, π₯2, and π₯3, with π₯1 > π₯2 > π₯3. Since there is no tangent point
between the derivatives in these three intersections, π
ππ₯logπ π(π₯) >
π
ππ₯π(π₯) for π₯ between π₯2
and π₯3, and π
ππ₯logπ π(π₯) <
π
ππ₯π(π₯) for π₯ between π₯1 and π₯2.
To find the tangent point between logπ π(π₯) and π(π₯) such that it is the only point of
intersection, we first assume each of the three points, π₯1, π₯2, and π₯3 to be that tangent point.
Through calculations same to the ones mentioned above, we would obtain three maximum values
on the value of g such that logπ π(π₯) and π(π₯) do not intersect. At this point, we choose the
least of the three values as the restriction on g.
5.2 π < π
The shape of logπ π(π₯) when π < 0 and π2 β 4ππ > 0 is concave down3. π(π₯) can be
either a concave up or concave down quadratic function.
If π(π₯) is concave up, which implies that π > 0, then there would be only one intersection
between logπ π(π₯) and π(π₯) if the two functions are tangent to each other4. It follows that:
π
ππ₯logπ π(π₯) =
π
ππ₯π(π₯)
2ππ₯ + π
ln π β (ππ₯2 + ππ₯ + π)= 2ππ₯ + π
3 Refer to the proof of the lemma 1 on page 8 4 Refer to the proof of the lemma 5 on page 10
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(2π
ln π) π₯ + (
π
ln π) = 2πππ₯3 + (2ππ + ππ)π₯2 + (2ππ + ππ)π₯ + ππ
2πππ₯3 + (2ππ + ππ)π₯2 + (2ππ + ππ β2π
ln π) π₯ + (ππ β
π
ln π) = 0
By applying the cubic formula, we can find at least one and at most three solutions. Only one
value would be within the domain of logπ π(π₯) due to the fact that there is at most one point for
which a concave up and a concave down function has the same slope. Let that value be Ξ±.
logπ(ππΌ2 + ππΌ + π) < ππΌ2 + ππΌ + π
π > logπ(ππΌ2 + ππΌ + π) β ππΌ2 β ππΌ
There would be no intersection between the two functions if this inequality is satisfied with
the given conditions.
6. Further observation
Lemma 7. Two functions, π(π₯) and π(π₯), that are differentiable on the interval of
(ββ, β) and continuous on the interval of (ββ, β) do not intersect only if
β π₯1 β R: π
ππ₯π(π₯1) =
π
ππ₯π(π₯1)
Given that π(π₯) and π(π₯) are two non-intersecting continuous and differentiable functions
with x β R, without the loss of generality, we can assume that π(π₯) is greater than π(π₯) for all
real values of x. By applying downward vertical translation on π(π₯), we can guarantee that π(π₯)
and π(π₯) would intersect. Suppose that after translating π(π₯) n units down, π(π₯) and π(π₯)
intersect for the first time. We can show that the intersections must be tangent points, by
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contradiction. If one intersection is not a tangent point, then either on the left side or the right side
of that intersection there must be an interval of x-values such that π(π₯) < π(π₯). This means that
if we can translate π(π₯) up by a certain unit such that π(π₯) still intersects with π(π₯),
contradicting the assumption that π(π₯) first intersects with π(π₯) after its translation by n units
downward. This is the end of the proof.
This is an observation that I have made from my research, and would have great contribution
to future investigation. The implication is that if I determine that it is possible for two functions
to not intersect, I only have to locate the x-values for which the two functions have the same
slope, and determine for which x-value would place the greatest restriction on the constant term
for one of the function, significantly simplifying my future research.
7. Conclusion
The purpose of my research is to find a more efficient way of determining the existence of
intersections between a logarithmic function and a polynomial. I have discussed four cases and
provided solutions to each of the cases with respect to my research question.
The method I have employed, which is analyzing the behaviour of functions and finding
tangent points between the two functions, unfortunately becomes ineffective when the degrees of
π(π₯) and π(π₯) begin to increase. The reason comes from the fact stated in the Abel-Ruffini
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Theorem: there exists no general algebraic solution to a polynomial equation with degree greater
than or equal to 5 (Ε»olΔ dek, 2000). It is important to note that this theorem does not claim that there
is no solution to polynomial equations of degree greater than or equal to 5.
The original way for locating a point at which the slopes of logπ π(π₯) and π(π₯) are equal
to each other would not work with this restriction. To let π
ππ₯π(π₯) be equal to
π
ππ₯π(π₯) would then
turn into a polynomial equation with degree of deg (f) + deg (g) β 1, and it must be strictly
less than 5. Equivalently, deg (f) + deg (g) must be smaller than 6 in order for our original
method to work. I left out 6 cases that may be solved using this method. I would not discuss them,
since they can all be solved using the similar way of analyzing functions and finding tangent points
between them.
There are other ways of solving a polynomial equation. One of which is the Newtonβs
method, which gives good approximation to the zeroes of a function. Another way is by graphing
the polynomial and see if it intersects with the x-axis. However, both of them can be time-
consuming.
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Bibliography
Rooftop Theorem for Concave down Functions. University of Southern California, Santa
Barbara. Retrieved from
http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/Rooftop.pdf
Ε»olΔ dek, Henryk. (2000). The Topological Proof of Abel-Ruffini Theorem. Journal of the
Juliusz Schauder Center, 16, 253-265.