Note on the transcendental equation with three
unknowns
q2f(z)−4 = rx−P′(t) + qP(t)(y+ 2) ±rx−P′(t)−qP(t)(y+ 2)
Abstract: Let P:= P(t) be a non square polynomial and f:= f(z) be a bijective application over Z.
Using the method of continuous fractions, we consider, in this paper, the number of integer solutions of
transcendental equation
q2f(z)−4 = rx−P′(t) + qP(t)(y+ 2) ±rx−P′(t)−qP(t)(y+ 2)
under the condition that
x2−P(t)y2−2P′(t)x+ 4P(t)y+ (P′(t))2−4P(t)−1=0.
We extend a previous result given by A. S. Sriram and P. Veeramallan.
Key-Words: Transcendental equation, Integer solutions, Diophantine Equation, Pell equation,
Polynomial, Bijection, Continued fraction, Recurrence relation.
1 Introduction
An algebraic equation is an equation of the form
f(x) = 0, where f(x) is entirely a polynomial
in x, such as x5−x3+x2−1 = 0. However,
if f(x) contains trigonometrical, arithmetic, or
exponential terms, it is referred to as a tran-
scendental equation, such as xex−2 = 0 and
xlog10x−1.2 = 0.
Transcendental equations are widely used in
science and engineering because they enable the
modeling and simulation of physical phenomena.
The importance of Transcendental equations is
demonstrated by the fact that they can be found
practically everywhere in mathematical analysis,
which has aided in the development of theoretical
sciences and the development of new technologies.
Transcendental equations enable the analysis of
mechanical vibration in the field of physics [1, 2],
the analysis of alternating current in electrical cir-
cuits [1, 2], electro magnetics theory [3, 4], quan-
tum mechanics [5, 6], digital signal processing [7,
8], and the modeling of wave heat conduction [9,
10].
In the case of logarithms, another application
for transcendental functions is the creation of phase
and magnitude plots in Bode analysis [2], [9], [10].
Transcendental functions enabled the develop-
ment of mathematical tools for analysis such as
Fourier [11], [12], and Laplace transform [1], [13].
Furthermore, hyperbolic functions are important
in mathematical analysis for science and engineer-
ing; for example, in civil engineering, they have
applications in the study and design of catenary
forms in chains and cables for suspended bridges
[14], in electrical engineering for the design of free
hanging electric power cables [15], [16], [17], and
in naval and civil engineering for the modeling of
sea wave behavior [18].
The determination of transcendental function
roots is a problem that appears in a wide range
of engineering applications (For more details, one
can see [3, 4, 5]). There are numerous numerical
approaches available for approximating the solu-
tion to any desired level of accuracy. In terms of
practicality, such root discovery algorithms are
often simple to implement and provide an ade-
quate method for obtaining root values. How-
ever, having an exact mathematical solution to
the problem under investigation is sometimes ad-
vantageous. For example, analytical derivatives
for uncertainty analyses and sensitivity studies
can be developed using an explicit expression for
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DOI: 10.37394/23206.2022.21.42
Amara Chandoul
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AMARA CHANDOUL
Higher Institute of Informatics and Multimedia of Sfax, Sfax University, Sfax, TUNISIA
5HFHLYHG-XO\5HYLVHG$SULO$FFHSWHG0D\3XEOLVKHG-XQH
the root. Analytical derivatives, in many circum-
stances, provide far more insight into the problem
than numerical derivatives.
Explicit expressions can also be used to en-
sure that approximation root seeking techniques
are convergent. Haji-Sheikh and Beck presented,
in 2000, a closed formulas for many analytical
heat transfer problems and detailed their applica-
tions [8]. Using Cauchy’s integral theorem from
complex analysis, Luck Rogelio, and James W.
Stevens presented, in [9], a straightforward way
for formulating accurate explicit solutions for the
roots of analytic Transcendental equations. Their
method was presented along with various exam-
ples.
In [6], Rogelio Luck, Gregory J. Zdaniuk, and
Heejin Cho described a method for finding a poly-
nomial equation with the same roots as a tran-
scendental equation and solving it for all of its
roots within a bounded region. Using Cauchy’s
integral theorem for complex variables, the pro-
posed method transforms the problem of finding
the roots of a transcendental equation into an
equivalent problem of finding the roots of a poly-
nomial equation with exactly the same roots.
The coefficients of the polynomial form an in-
teresting vector that lies in the null space of a
Hankel matrix made up of the Fourier series co-
efficients of the inverse of the original transcen-
dental equation.
The explicit solution can then be easily ob-
tained by employing the complex fast Fourier trans-
form.
In this paper, we aim to discover integer solu-
tions for the given transcendental equation using
the continued fraction method. For a non-square
polynomial P(t),we consider the Diophantine equa-
tion H(x, y, P (t)) = x2−P(t)y2−2P′(t)x+4P(t)y+
(P′(t))2−4P(t) = 1 and then the Pell equation
x2−P(t)y2 = 1. We also go over some of the most
important characteristics of the simple continued
fraction.
In the detailed, I referred to [1, 2] in the ex-
tensive, but I considered [7] as the major source
of inspiration that we will give a great generaliza-
tion of its results.
In the following, definitions and results needed
in our paper.
2 Method
Consider the transcendental equations
E+:= p2f(z)−4 = qx−P′(t) + pP(t)(y+ 2)
+qx−P′(t)−pP(t)(y+ 2)
and
E−:= p2f(z)−4 = qx−P′(t) + pP(t)(y+ 2)
−qx−P′(t)−pP(t)(y+ 2)
verifying H(x, y, P (t)) = x2−P(t)y2−2P′(t)x+
4P(t)y+(P′(t))2−4P(t) = 1, such that P:= P(t)
be a non square polynomial and f:= f(z) be a
bijective application over Z.
We get by squaring both sides of (E+and E−)
respectively
f(z)−2 = x−P′(t)
±p(x−P′(t))2−P(t)(y+ 2)2(1)
Then, we have
f(z)−2 = x−P′(t)±qH(x, y, P (t)) (2)
and thus x, y are given from the Diophantine equa-
tion
H(x, y, P (t)) = 1 (3)
and
z=f−1(x−P′(t)+2±1)
That is
z=f−1(x−P′(t) + 3) or z =f−1(x−P′(t) + 1).
Note that the resolution of (3), in its current
form, was described in [chandoul], by transform-
ing it into a Pell equation which can be easily
solved. To get it, it had set
T:u=x−P′(t)
v=y+ 2 (4)
we get,
T(3) := f
(3) : u2−P(t)v2= 1
This Pell equation is always known to be solved.
Its solutions are related to the pP(t)’s continued
fraction expansion.
We will be looking at continued fraction expan-
sions of pP(t), where P(t) is a non-square. In
fact, the following theorem summarizes a very in-
teresting form of continued fractions.
Theorem 1 Let P(t)be a non square polyno-
mial. Then
qP(t) = a0;a1, a2,···, al,2a0,
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Proof. Let A1the polynomial part of pP(t).
Then deg(pP(t) + A1)>1 and deg(−pP(t) +
A1)<1. Thus pP(t)+A1, is a reduced quadratic
irrational with degree of polynomial part is less
or equal to 2deg(A1). We can confirm that
qP(t) + A1= [2A1, A2,···, An]
for some n. which is equivalent to
qP(t) + A1= [2A1, A2,···, An,2A1]
consequently
qP(t)=[A1, A2,···, An,2A1]
3 Main result
The following theorem express our main finding.
we consider the result without giving their proof
since it can be proved by induction as similat to
that of Theorems in [1, 2, 7] were proved.
Theorem 2 Let pP(t) = a0;a1, a2,···, al,2a0
be the continuied fraction expansion of period length
l, of pP(t), where P(t)is a non square polyno-
mial and let pn
qn
its nth convergent. Then the fol-
lowing assertions holds
(1) The fundamental solutions of (E+) is (x1, y1, z1),
such as
x1=pl−1+P′(t),
y1=ql−1+ 2, if l is even
z1=f−1(pl−1+ 3)
and
x1=p2l−1+P′(t),
y1=q2l−1+ 2, if l is odd
z1=f−1(p2l−1+ 3)
(2) The fundamental solutions of (E−) is (x1, y1, z1),
such as
x1=pl−1+P′(t),
y1=ql−1+ 2, if l is even
z1=f−1(pl−1+ 1)
and
x1=p2l−1+P′(t),
y1=q2l−1+ 2, if l is odd
z1=f−1(p2l−1+ 1)
(3) Define the sequence
{(xn, yn, zn)}n≥1={(un+P′(t), vn+2, f−1(un+2±1))},
where {(un, vn)}defined by
un=u1un−1+ (a0u1+α)vn−1
∀n≥2,
vn=v1un−1+ (a0v1+β)vn−1
with α =xl−2and β =xl−2, if l is even.
and
un=u1un−1+ (a0u1+α)vn−1
∀n≥2,
vn=v1un−1+ (a0v1+β)vn−1
with η =x2l−2and δ =x2l−2, , if l is odd.
Then (xn, yn, zn)is a solution of (E+,), re-
spectively of (E−). So (E+,) respectively (E−)
have infinitely many integer solutions (xn, yn, zn)∈
Z3.
(4) The solutions (xn, yn, zn)of (E+,) satisfy the
recurrence relations
xk=u1xk−1+ (a0u1+α)yn−1−u1(2a0+P′(t))
−2α+P′(t), if l iseven
yk=v1xk−1+ (a0v1+β)yn−1−v1(2a0+P′(t))
−2β+ 2
zk=f−1(u1xk−1+ (a0u1+α)yn−1−u1(2a0
+P′(t)) −2α) + 3),
∀k≥2, if lis even,
and
xk=u1xk−1+ (a0u1+α)yn−1−u1(2a0+P′(t))
−2α+P′(t), if l iseven
yk=v1xk−1+ (a0v1+β)yn−1−v1(2a0+P′(t))
−2β+ 2
zk=f−1(u1xk−1+ (a0u1+α)yn−1−u1(2a0
+P′(t)) −2α) + 3),
∀k≥2, if lis odd.
(5) The solutions (xn, yn, zn)of (E−), satisfy the
recurrence relations
xk=u1xk−1+ (a0u1+α)yn−1−u1(2a0+P′(t))
−2α+P′(t), if l iseven
yk=v1xk−1+ (a0v1+β)yn−1−v1(2a0+P′(t))
−2β+ 2
zk=f−1(u1xk−1+ (a0u1+α)yn−1−u1(2a0
+P′(t)) −2α) + 1),
∀k≥2, if lis even,
and
xk=u1xk−1+ (a0u1+α)yn−1−u1(2a0+P′(t))
−2α+P′(t), if l iseven
yk=v1xk−1+ (a0v1+β)yn−1−v1(2a0+P′(t))
−2β+ 2
zk=f−1(u1xk−1+ (a0u1+α)yn−1−u1(2a0
+P′(t)) −2α) + 1),
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∀k≥2, if lis odd.
4 Examples
Remark 1 to deduce all the results given in [7],
just take f:= f(z) = z,P(t) := Cand set
v=y+ 2.Extract first (x, y, v)and then deduce
(x, y, z).
Example 1: Let f(z) = z+ 3 and P(t) = t4+
4t3+6t2+4t+2,It is easy to verify that f−1(z) =
z−3 and
qP(t) = ht2+ 2t+ 1; 2t2+ 4t+ 2i.
It can be seen that P(t) become D(n) = n2+ 1
with n=t2+ 2t+ 1. Then the transcendental
equations become
E+:= √2z+ 2 = qx−2n+√n2+ 1(y+ 2)
+qx−2n−√n2+ 1(y+ 2)
and
E−:= √2z+ 2 = qx−2n+√n2+ 1(y+ 2)
−qx−2n−√n2+ 1(y+ 2)
.
We have qD(n) = n; 2n.
So, the fundamental solution of E+is (x1, y1, z1) =
(2n2+ 2n+ 1,2n+ 2,2n2+ 1) and the other so-
lutions are given, for k≥2,by
xk= (2n2+ 1)xk−1+ (2n3+ 2n)yk−1
−8n3−4n
yk= 2nxk−1+ (2n2+ 1)yk−1−8n2
+2n−2
zk= (2n2+ 1)xk−1+ (2n3+ 2n)yk−1−8n3
−6n
Similarly, the fundamental solution of E−is
(x1, y1, z1) = (2n2+ 2n+ 1,2n+ 2,2n2−1)
and the other solutions are given, for k≥2,by
xk= (2n2+ 1)xk−1+ (2n3+ 2n)yk−1
−8n3−4n
yk= 2nxk−1+ (2n2+ 1)yk−1−8n2
+2n−2
zk= (2n2+ 1)xk−1+ (2n3+ 2n)yk−1−8n3
−6n−2
Further, for t= 1, P(t) = 17 and pP(t) = 4; 8.
So, (x1, y1, z1) = (65,10,62) is then the funda-
mental solution of E+.and the other solutions
are given, for k≥2,by
xk= 33xk−1+ 136yk−1−1296
yk= 8xk−1+ 33yk−1−320
zk= 33xk−1+ 136yk−1−1296
Similarly, (x1, y1, z1) = (65,10,62) is then the
fundamental solution of E+.and the other so-
lutions are given, for k≥2,by
xk= 33xk−1+ 136yk−1−1296
yk= 8xk−1+ 33yk−1−320
zk= 33xk−1+ 136yk−1−1294
Example 2: Let f(z) = z+ 3 and P(t) = t6+
t+ 1,a nonsquare polynomial in F3. It is easy to
verify that
pP(t) = [t3,2t2+t+ 2, t + 1, t + 1,2t, 2t
+2,2t, t + 1, t + 1,2t2+t+ 2,2t3]
and then we deduce the solutions.
5 Conclusion
Using the continued fraction method, we found
all feasible non-negative integer solutions to the
given transcendental equation in this study. It’s
also noteworthy to note that all of the results in
[7], can be derived easily from our theorem, it is
in fact a case of our general result.
This extension enable us to solve a wide range
of equations. On the integer solutions of Dio-
phantine and Pell equations, we also deduce var-
ious recurrence relations.
Another advantage of our findings is that the
procedure may be run on a computer, allowing us
to retrieve all of the answers after inserting the
coecients and verifying the method’s criteria.
Furthermore, it is important to note that the
proposed method can be extended to solve many
different types of transcendental equations, with
the goal of expanding the set of resolvable tran-
scendental equations and making them available
for use in various fields of science and engineering.
Finally, it will be interesting to include these new
functions in commercial or open source mathe-
matical software such as Maple, Mathematica,
Matlab, and GNU Octave.
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