1
12
4

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THE IMPORTANCE OF JAVASCRIPT
IN WEB DEVELOPMENT
A thesis written at
WAGNER COLLEGE
in partial fulfillment
of the requirements for the
degree of
BACHELOR OF SCIENCE IN ‘Computer Science’
by:
Nicolette Cunsolo
May 2020
�
In this thesis we discuss the history of Web Development and JavaScript. As well as the
importance that JavaScript has in Web Development. I believe that without JavaScript
webpages wouldn’t be the way that they are today. JavaScript adds a touch to them that other
languages wouldn’t be able to do. Over the last few decades we have seen so many advances
in the World Wide Web. One of the most popular creations has been the programming
language JavaScript. It created a way for users to be interactive with the webpage. It made it
easier for websites to ask for information to be input. For example, asking for emails, names,
phone number and address became a lot easier for the developers to add to the websites.
Throughout my paper you will see the major impact JavaScript has had on the web
development world.
�
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The Importance of Javascript in Web Development
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Markel Landa Mendi
CS 400: Thesis
Comparing Approximation Methods for Ordinary
Differential Equations
Numerical Analysis is a branch of mathematics that deals with
approximation of different types of function and equations. The goal is to get as
close to the real solution as possible without actually getting there. Such methods
are used in situations when, for example, getting the precise answer to a problem is
not required or if we desire to get an idea of the projection the curve of a given
function will take on a graph. With this in mind, some of the more challenging
functions to approximate are those of differential equations. A differential equation
is a mathematical equation that relates the variable or several variables of an
unknown equation to its derivatives of various orders. In mathematical
applications, the functions generally represent quantities of physical objects, the
derivatives their rates of change and the differential equation defines the
relationship between both. Such equations play a prominent role in engineering,
economics, biology… where they are mainly used to model the behavior of
complex systems. The three methods that I will be comparing in this paper are
Euler, Higher Order Taylor Method and RungeKutta Method.
�
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Comparing Approximation Methods for Ordinary Differential Equations
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Mathematics

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ON SOME EQUATIONS INVOLVING
EULER TOTIENT FUNCTION
A thesis written at
WAGNER COLLEGE
in partial fulfillment
of the requirements for the
degree of
BACHELOR OF SCIENCE IN MATHEMATICS
by:
MOHAMED LOTFI
May 2022
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
MOHAMED LOTFI
Abstract. We prove that if n is a solution to the equation s(n) = ϕ(n), where s(n)
is the sum of the proper divisors of n and ϕ(n) is Euler Totient function, then n is
either 2 or an odd square. For a solution n that is an odd square, we show that the
Y p
> φ, p is prime
number of its distinct prime divisors is greater than 2 and
p−1
pn
and φ is the golden ratio. Finally, we prove that the only solutions to the equation
d(n) = ϕ(n), where d(n) is the number of divisors of n, are 1, 3, 8, 10, 18, 24, 30.
1. Introduction
Different functions in Number Theory have been extensively studied during the last
two centuries. Among the most famous of these functions are Euler Totient function
ϕ(n), the number of divisors function d(n) and the sum of proper divisors function s(n).
In this paper, we will investigate the positive integer solutions to the two equations
ϕ(n) = d(n)
(1)
ϕ(n) = s(n)
(2)
and
1.1. Definitions. Given n = pα1 1 pα2 2 ... pαk k , where pi are distinct primes and αi are
positive integers for all 1 ≤ i ≤ k, k is a nonnegative integer.
We provide some definitions that are necessary for the rest of the paper:
Function I: Euler Totient Function ϕ(n).
ϕ : N → N equals the number of positive integers smaller than or equal to n that
are relatively prime to n
ϕ(n) = n (1 −
1
1
1
) (1 − ) ... (1 − )
p1
p2
pk
which is equivalent to
1
�2
MOHAMED LOTFI
ϕ(n) = pα1 1 −1 pα2 2 −1 ... pαk k −1 (p1 − 1)(p2 − 1) ... (pk − 1)
We will use the second definition for the rest of the paper.
Examples: ϕ(4) = 2 , ϕ(5) = 4 and ϕ(6) = 2
Function II: Number of divisors Function d(n).
d : N → N equals the number of positive integers greater than or equal to 1 that
divides n (including n itself), and
d(n) = (α1 + 1)(α2 + 1)...(αk + 1)
Examples: d(4) = 3 , d(5) = 2 and d(6) = 4
Function III: Sum of divisors Function σ(n).
σ : N → N equals the sum of positive divisors of n (including n).
σ(n) = (1 + p1 + p21 + ... + pα1 1 )(1 + p2 + p22 + ... + pα2 2 ) ... (1 + pk + p2k + ... + pαk k )
Examples: σ(4) = 7 , σ(5) = 6 and σ(6) = 12
Function IV: Sum of proper divisors Function s(n).
s : N → N equals the sum of proper divisors of n (i.e. the divisors of n excluding n
itself).
s(n) = σ(n) − n
Therefore, s(n) =
(1 + p1 + p21 + ... + pα1 1 )(1 + p2 + p22 + ... + pα2 2 )...(1 + pk + p2k + ... + pαk k ) − pα1 1 pα2 2 ... pαk k
Examples: s(4) = 3 , s(5) = 1 and s(6) = 6
All the previous functions are multiplicative, except for s(n). For more information
and additional results we refer the reader to [1] and [2].
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
3
2. The Main Results
We begin by considering equation (1).
As s(1) = 0 < 1 = ϕ(1), we consider only the possible solutions > 1.
Proposition 2.1. Of the positive integer solutions that belongs to one of the following
classes:
(i) even integers
(ii) odd integers whose prime factorization contains a prime with odd power
(iii) integers of the form p2α such that p is an odd prime
(iv) integers of the form p2α q 2β such that p and q are distinct odd primes
n = 2 is the only solution to equation (1).
We will prove that Class (i) yields only one solution, namely n = 2, and the rest of
the classes yields no solutions.
Class (i): Let n = 2m for some positive integer m. Note that both 1 and m divides
n. If m > 1, then
s(n) ≥ m + 1
On the other hand, half of the numbers between 1 and n inclusively are even. Hence,
at least half of them are not coprime with n. So
ϕ(n) ≤ m
Therefore, for n > 2,
s(n) > ϕ(n)
But note that
s(2) = 1 = ϕ(2)
Therefore, if n is even, then 2 is the only solution to equation (1), as desired.
Class (ii):
�4
MOHAMED LOTFI
Let n be an odd positive integer > 1 whose prime factorization is
pα1 1 pα2 2 ... pαk k ,
where pi are odd primes for all i, 1 ≤ i ≤ k
We will prove that if there exists j such that αj is odd, then
s(n) ̸= ϕ(n)
Recall that
ϕ(n) = pα1 1 −1 pα2 2 −1 ... pαk k −1 (p1 − 1)(p2 − 1) ... (pk − 1)
As n is odd > 1, then all of its prime divisors are odd, including p1 . Therefore
(p1 − 1) is even, which implies that ϕ(n) is even. Recall that
σ(n) = (1 + p1 + p21 + ... + pα1 1 )(1 + p2 + p22 + ... + pα2 2 ) ... (1 + pk + p2k + ... + pαk k )
Note that if α1 is odd, then
(1 + p1 + ... + pα1 1 )
is even, as it is a sum of an even number of odd numbers.
This implies that σ(n) is also even. But
s(n) = σ(n) − n
Hence, s(n) is odd. Therefore, s(n) and ϕ(n) have different parities, which implies that
s(n) ̸= ϕ(n)
as claimed.
Class (iii): Let n = p2α , where p is an odd prime and α is a positive integer. Using
the formulas of ϕ(n) and s(n),
ϕ(n) = p2α − p2α−1 and s(n) = σ(n) − n = 1 + p + ... + p2α−1
As p ≥ 3,
p2α − p2α−1 ≥ 3p2α−1 − p2α−1 = 2p2α−1
Now, if 2p2α−1 > 1 + p + ... + p2α−1 , then ϕ(n) > s(n) and we are done. Thus it
suffices to show that
2p2α−1 > 1 + p + ... + p2α−1
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
5
The above inequality can be rewritten as
p2α−1 > 1 + p + ... + p2α−2 =
p2α−1 − 1
p−1
As p − 1 > 1 (because p is odd prime),
p2α−1 − 1
< p2α−1 − 1 < p2α−1 ,
p−1
as desired.
To prove our claim about Class (iv), we have to develop some machinery.
k
1 2α2
p2 ... p2α
Lemma I. Let n = p2α
1
k , for distinct primes pi , i = 1 , 2 , ... , k. If
√
p1 p2 ... pk
1+ 5
<
,
(p1 − 1)(p2 − 1)...(pk − 1)
2
then s(n) < ϕ(n).
Proof. Consider the inequality
1 +1 2α2 +1
k +1
p2α
p2
... p2α
1
k
k −1
k
1 − 1 2α2 − 1
1 2α2
< p2α
p2
... p2α
(p1 − 1)(p2 − 1) ... (pk − 1)
− p2α
... p2α
1
1 p2
k
k
(p1 − 1)(p2 − 1) ... (pk − 1)
which we will call inequality (*).
Dividing both sides of the inequality by the RHS yields
p21 p22 ... p2k
p1 p2 ... pk
<1
2
2
2 −
(p1 − 1)(p2 − 1) ... (pk − 1)
(p1 − 1) (p2 − 1) ... (pk − 1)
Let x =
p1 p2 ... pk
. The inequality becomes
(p1 − 1)(p2 − 1) ... (pk − 1)
x2 − x < 1
or
x2 − x −1 < 0
Solving this inequality yields
x∈
or, equivalently,
√
√ !
1− 5 1+ 5
,
2
2
�6
MOHAMED LOTFI
p1 p2 ... pk
∈
(p1 − 1)(p2 − 1) ... (pk − 1)
As the
√
√ !
1− 5 1+ 5
,
2
2
p1 p2 ... pk
> 0,
(p1 − 1)(p2 − 1) ... (pk − 1)
√
p1 p2 ... pk
1+ 5
then if
<
, the inequality (*) holds.
(p1 − 1)(p2 − 1) ... (pk − 1)
2
But s(n) =
2αk
2α1 2α2
2α2
2
2
k
1
p2 ... p2α
(1 + p1 + p21 + ... + p2α
1 )(1 + p2 +p2 + ... + p2 ) ... (1 + pk +pk + ... + pk ) − p1
k
=
1 +1
2 +1
k +1
(p2α
− 1)(p2α
− 1) ... (p2α
− 1)
1
2
k
1 2α2
k
− p2α
... p2α
1 p2
k
(p1 − 1)(p2 − 1) ... (pk − 1)
<
k +1
1 +1 2α2 +1
p2α
p2
... p2α
1
k
1 2α2
k
− p2α
... p2α
1 p2
k
(p1 − 1)(p2 − 1) ... (pk − 1)
1 − 1 2α2 − 1
k −1
< p2α
p2
... p2α
(p1 − 1)(p2 − 1) ... (pk − 1)
1
k
= ϕ(n), as claimed.
□
Corollary 2.2. Given any positive number k > 1, there exists a prime P such that if
n is an odd prime with k divisors and all the prime divisors of n are greater than or
equal to P then s(n) < ϕ(n)
Proof. Denote by pn the nth prime number. Since the sequence (pn ) increases to +∞,
pn
the sequence an =
is strictly decreasing and convergent to 1.
pn − 1
√ !1/k
1+ 5
> 1, one can find pi1 < pi2 < ... pik such that
As
2
√ !1/k
p ik
pi2
pi1
1+ 5
< ... <
<
<
pik − 1
pi 2 − 1
pi1 − 1
2
Thus,
�
pi 1
pi1 − 1
��
pi 2
pi2 − 1
�
�
...
pik
pik − 1
�
√
1+ 5
<
2
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
2α
2α
7
2αi
Let P = pi1 and n = pi1 i1 pi2 i2 ... pik k . Then by Lemma I,
s(n) < ϕ(n)
as desired.
□
We will need the following elementary inequality
n
m
>
n−1
m−1
for integers m > n > 1 . Denote it by inequality (**).
Now we can finish the proof of Class (iv).
Let n = p2α q 2β such that p and q are distinct odd primes with p < q. So
ϕ(n) = p2α−1 q 2β−1 (p − 1)(q − 1)
and
s(n) = σ(n) − n = (1 + p + ... + p2α )(1 + q + ... + q 2β ) =
(p2α+1 − 1)(q 2β+1 − 1)
− p2α q 2β
(p − 1)(q − 1)
There are two cases:
Case I: p ≥ 5
Then,
pq
(5)(7)
35
<
=
(p − 1)(q − 1)
(4)(6)
24
by inequality (**).
√
35
1+ 5
As
<
, then by Lemma I
24
2
s(n) < ϕ(n)
Hence,
s(n) ̸= ϕ(n)
Case II: p = 3
if q ≥ 17, then
pq
(3)(17)
51
<
=
(p − 1)(q − 1)
(2)(16)
32
�8
MOHAMED LOTFI
again by inequality (**)
√
51
1+ 5
But
<
. Thus, by Lemma I, s(n) < ϕ(n)
32
2
Upon checking the pairs (p, q) = (3, 5), (3, 7), (3, 11), (3, 13), it turns out that none
of them provides a solution to
s(n) = ϕ(n)
By combining the two cases, it follows that no solution belongs to the Class (iv), as
claimed.
After considering all of the four classes, we can conclude that if the equation
s(n) = ϕ(n)
has a solution greater than 2, it must be an odd square such that the number of its
Y p
distinct prime divisors is greater than 2 and
> φ, where p is prime and φ is
p−1
pn
√
1+ 5
.
the famous golden ratio =
2
Remark 2.3. Lemma I provides a quick way to check if s(n) < ϕ(n) for square odd n,
and, surprisingly, this method depends only on the prime factors, not on their powers
(provided of course they are even).
We continue this section with equation (2).
Proposition 2.4. The only solutions to the equation
d(n) = ϕ(n)
are 1, 3, 8, 10, 18, 24, 30
The solution n = 1 is trivial, so assume that n > 1. We will prove some lemmas that
will help us in proving the proposition. For the rest of the paper, let α be a positive
integer, unless stated otherwise.
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
9
Lemma II: If p is a prime ≥ 5, then ϕ(pα ) > d(pα ) (i.e pα−1 (p − 1) > α + 1).
Proof. Note that pα−1 (p−1) ≥ 5α−1 ·4. We will prove by induction that 5α−1 ·4 > α+1
Base Case (α = 1): 4 > 1 + 1
Induction hypotheses: assume that 5k−1 · 4 > k + 1
So
5(5k−1 · 4) > 5(k + 1) > (k + 1) + 1
Hence,
5k+1 · 4 > (k + 1) + 1
completing the induction.
We conclude that
pα−1 (p − 1) ≥ 5α−1 · 4 > α + 1 for all α and prime p ≥ 5, as desired.
□
Lemma III:
(i) ϕ(3α ) = d(3α ) if α = 1 (i.e. 2 · 30 = 2 )
(ii) ϕ(3α ) > d(3α ) if α > 1 (i.e. 2 · 3α−1 > α + 1 )
Proof. (i) is a direct substitution. We will prove (ii) by induction on α (starting from
α = 2).
Obviously, (α = 2): 6 > 3
Assume that 2 · 3k−1 > k + 1, for some positive integer k ≥ 2. So
3(2 · 3k−1 ) > 3(k + 1) > k + 2
Thus,
2 · 3k > k + 2
Induction is complete, and the lemma is proved.
Lemma IV:
(i) ϕ(2α ) < d(2α ) if α = 1 or 2
(ii) ϕ(2α ) = d(2α ) if α = 3
□
�10
MOHAMED LOTFI
(iii) ϕ(2α ) > d(2α ) if α > 3 (i.e. 2α−1 > α + 1 )
Proof. (i) and (ii) are direct substitution. As we did with Lemmas II and III, we will
prove (iii) by induction on α (starting from α = 4)
Base case (α = 2): 8 > 5, which is true.
Assume that 2k−1 > k + 1, for some positive integer k ≥ 4. So
2k = 2(2k−1 ) > 2(k + 1) > k + 2
concluding the induction.
□
Lemma V: For every p > 3, then ϕ(pα ) ≥ 2d(pα ), with equality holds if and only
if p = 5 and α = 1.
Proof. We will prove by induction on α (starting from α = 2) that 5α−1 · 4 > 2(α + 1)
Obviously, 51 · 4 > 6
Assume that 5k−1 · 4 > 2(k + 1), for some k ≥ 2. Then
5(5k−1 · 4) > 10(k + 1) > 2(k + 2),
finishing the induction.
As a result,
ϕ(pα ) = pα−1 (p − 1) ≥ 5α−1 · 4 > 2(α + 1) = 2d(pα ) for k ≥ 2.
Thus, the equality in Lemma V holds if and only if p = 5 and α = 1.
□
Lemma VI: For any α ≥ 2, ϕ(3α ) ≥ 2d(3α ), with equality holds if and only if
α=2
Proof. Again, we will prove it by induction on α (starting from α = 3) that
3α−1 · 2 > 2(α + 1)
Base Case: 32 · 2 > 8
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
11
Assume that 3k−1 · 2 > 2(k + 1), for some k ≥ 3
Then,
3(3k−1 · 2) > 6(k + 1) > 2(k + 2)
So
3k · 2 > 2(k + 2)
finishing the induction.
Therefore, for α ≥ 3.
ϕ(3α ) = 3α−1 · 2 > 2(α + 1) = 2d(3α )
Thus, the equality in the Lemma holds if and only if α ≥ 2, as claimed.
□
We will use the lemmas to prove the Proposition. Let n = pα1 1 pα2 2 ... pαk k , where pi
are distinct primes for 1 ≤ i ≤ k, with p1 is the smallest prime, and αi is a positive
integer for 1 ≤ i ≤ k. There are three cases to consider:
First: If p1 > 3, no solution exists.
Proof: ϕ(n) =
ϕ(pα1 1 ) ϕ(pα2 2 ) ... ϕ(pαk k )
(as ϕ(n) is a multiplicative function)
> d(p1α1 ) d(pα2 2 ) ... d(pαk k )
= d(n)
(Lemma II)
(d(n) is multiplicative)
Hence, no solution exists, as claimed.
Second: If p1 = 3, then n = 3 is the only solution.
Proof: There are two subcases:
Subcase I: the number of distinct primes that divides n is greater than 1 (i.e k ̸= 1)
By Lemma III: ϕ(3α1 ) ≥ d(3α1 )
By Lemma II: ϕ(pα2 2 ) ... ϕ(pαk k ) > d(pα2 2 ) ... d(pαk k )
Combining the previous statements, we conclude that
ϕ(n) = ϕ(3α1 ) ϕ(pα2 2 ) ... ϕ(pαk k ) > d(3α1 ) d(pα2 2 ) ... d(pαk k ) = d(n)
Therefore, no solution exists.
�12
MOHAMED LOTFI
Subcase II: the number of distinct primes that divides n is 1. Thus, n = 3α1 .
By Lemma III, α1 must be 1. n = 3 works, as claimed.
Third: If p1 = 2, then there are four subcases to consider, based on the value of α1 .
Subcase I: α1 > 3 Then,
ϕ(n) = ϕ(2α1 ) ϕ(pα2 2 ) ... ϕ(pαk k ) > d(2α1 ) d(pα2 2 ) ... d(pαk k ) = d(n)
by Lemmas II, III and IV. No solution exists.
Subcase II: α1 = 3.
If there is a prime number greater than 3 that divides n, then, again,
ϕ(n) = ϕ(2α1 ) ϕ(pα2 2 ) ... ϕ(pαk k ) > d(2α1 ) d(pα2 2 ) ... d(pαk k ) = d(n)
by Lemmas II, III and IV. No solution exists.
Therefore, we can limit our investigation to the cases n = 23 and 23 · 3α2 .
Note that ϕ(23 ) = d(23 ), so 8 is a solution. By Lemma III,
ϕ(3α2 ) ≥ d(3α2 )
It follows that
ϕ(23 · 3α2 ) = ϕ(23 )ϕ(3α2 ) ≥ d(23 )d(3α2 ) = d(23 · 3α2 )
by Lemmas III and IV. Equality holds if and only if α2 = 1. Indeed, ϕ(24) = d(24).
So 24 is a solution.
Subcase III: α1 = 2
Let n = 22 · m, where m is not divisible by 2. So
ϕ(n) = ϕ(22 · m) = ϕ(22 )ϕ(m) = 2ϕ(m)
Similarly,
d(n) = d(22 · m) = d(22 )d(m) = 3d(m)
So our goal is to find the solutions of
2ϕ(m) = 3d(m)
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
13
If m is divisible by a prime > 3, let m = pα2 2 pα3 3 ...pαk k with, pj > 3 for some j,
2≤j≤k.
We claim that 2ϕ(m) > 3d(m). Note that
α
ϕ(m) = ϕ(pα2 2 ) ... ϕ(pj j ) ... ϕ(pαk k )
≥ d(pα2 2 ) ... 2d(pα3 3 ) ... d(pαk k ) (Lemmas III and V )
3
> d(pα2 2 ) d(pα3 3 ) ... d(pαk k )
2
3
= d(m)
2
Thus, ϕ(m) >
3
d(m), which is our claim.Therem must be equal to 3α , for some posi2
tive integer α.
Note that if α ≥ 2, then by Lemma VI,
3
d(m)
2
Hence, α = 1. However, n = 22 · 3 does not satisfy the equation ϕ(n) = d(n). Thus,
ϕ(m) ≥ 2d(m) >
this subcase yields no solutions.
Subcase IV: α1 = 1.
Let n = 2m, where m is odd. So
ϕ(n) = ϕ(2m) = ϕ(2)ϕ(m) = ϕ(m)
Similarly,
d(n) = d(2m) = d(2)d(m) = 2d(m)
Our goal is to solve ϕ(m) = 2d(m).
Again, if m is divisible a prime > 3, let m = pα2 2 pα3 3 ...pαk k , with pj > 3. Then by
Lemmas III and V,
α
ϕ(m) = ϕ(pα2 2 ) ... ϕ(pj j ) ... ϕ(pαk k )
α
≥ d(pα2 2 ) ... 2d(pj j ) ... d(pαk k )
= 2d(m)
The equality holds only if the greatest prime that divides m is 5. Otherwise, by
Lemma V, the inequality will be strict. Also, the power of 5 must be 1 (Lemma V
again).
�14
MOHAMED LOTFI
This leaves us with two possibilities: m = 3α2 · 5, for some positive integer α2 ,
or m = 5.
Let m = 3α2 · 5. Note that for α2 ≥ 2,
ϕ(m) = ϕ(3α2 · 5) = ϕ(3α2 ) ϕ(5)
≥ 2 · 2d(3α2 ) d(5) (Lemma VI and ϕ(5) = 2d(5))
= 4d(3α2 ) d(5) = 4d(3α2 · 5) = 4d(m) > d(m).
Thus, we are left with only two possible solutions m = 31 · 5 or 5.
By checking these values, we know that they actually work, giving the solutions:
n = 2 · 3 · 5 and n = 2 · 5
Finally, we consider m = 3α2 , for some positive integer α2 . If α2 > 2, by Lemma VI,
ϕ(3α2 ) > 2d(3α2 )
So m = 3 or 32 . By checking, only m = 32 works. So n = 2 · 32 is a solution.
Having considered all the possible cases, we can conclude that the only solutions to
the equation ϕ(n) = d(n) are n = 1, 3, 8, 10, 18, 24, 30. (Q.E.D)
Acknowledgement
First and foremost, I would like to praise Allah the Almighty for His blessing given
to me during my study and in completing this thesis. Second, I would like to express
all my thanks and gratitude to my advisor Professor Adrian Ionescu for his constant
support and advice during my years of study at the Wagner College. I am deeply
indebted to Professors Florin Pop and Zohreh Shahvar for their guidance during much
of my undergraduate work. I would also like to thank the faculty and staff at Wagner
College, Department of Mathematics and Computer Science, for the help they have
given me. I would like to thank all of my family for their support and understanding
for the time spent away from them.
�ON SOME EQUATIONS INVOLVING EULER TOTIENT FUNCTION
15
References
[1] Apostol, T. M. Introduction to analytic number theory, Springer, 2011.
[2] Sandor, J., Mitrinovic, D., Crstici, B. Handbook of Number Theory, Vol. I, Springer,
2005.
Wagner College, Staten Island, NY 10301, U.S.A.
Email address: mohamed.lotfi@wagner.edu
�
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Mohamed Lotfi
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Mathematics

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PDF Text
Text
Seunghyun (Stewart) Noh
Professor Adrian Ionescu
CS400Senior RFT
Data Mining and Its Security
Pros
1.
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Provides info. based on previous purchases to project those who will respond to the
novel advertising operations like direct mail, online marketing, among others. (Sellers
can vend lucrative merchandise to targeted consumers). Retailers collect vast volumes
of data through years and predict the product sales trends, habits of customer purchase
and their preferences.
Provides relevant statistics about loans and reporting of credit. This will help banks to
identify fake credit card dealings and safeguard credit card owners.
Provides analytics to improve health care and reduce costs, guarantee patients get
proper care at the correct time and appropriate place, processes are developed with
supporting evidence.
Provides business/market owner to scope customers’ purchasing behaviour.
Provides discerning information from the available facts in the learning setting and
can help to predict students’ future learning behaviour.
Provides discerning the arrays in the multifaceted manufacturing process. Can be used
to generate the relationship between product design, product assortment, and
consumer necessities.
Provides campaigns to understand their product and service consumers. Also, the
relationship between the retailers and customers in built, and customers retained
through customer relationship management.
Provides ease on the process of fraud recognition from being cumbersome and
timeconsuming.
Provides averting intrusion including user verification, circumventing programming
blunders, and information fortification.
Provides the law enforcers to identify criminal suspects, and customize data
excavating skills to probe crimes, observe communication among the suspects.
Provides help to segment customers in the preference of their needs.
Provides analyzing financial transactions as to pick out patterns on money laundering.
Help business owners to use information to modify their products as per the
consumer’s desires.
Provides researchers to accelerate the process of analyzing the data, cleaning data,
preprocessing, and integration of databases.
Provides extracting useful knowledge from massive bioinformatics databases and
other associated lifescience areas.
Provides all sorts of information regarding the response of a consumer and in defining
consumer groups.
Provides evidence about selling promoted products online and ultimately reduce the
charges of the product and its services.
Cons
1. Requires advance skilled specialists since they are the only person who is able to
understand the different arrays and associations whose pattern connotation and the
user must make valid.
2. Often infringes on the confidentiality of the user and it puts at risk the user’s secretive
information which is very critical nowadays.
�3. Lack sufficient security systems to protect enormous amount of information about
people online that most companies collected through their emails and google search.
Which will open a window for hackers.
4. Some data based on previous purchase on big consumers like companies could cause
enterprises to be used against others.
5. During the data collection, a lot of irrelevant information may be obtained and this
will cause overwhelming and timeconsuming for the amount of data.
6. The technology and tools required in data mining and result interpretation consume
many resources and this will be a great cost at the implementation stage.
7. Perfection of the data mining process is yet to be developed and this will cause
possible inaccuracy of data and bad outcome from the decisionmaking process.
�
Dublin Core
The Dublin Core metadata element set is common to all Omeka records, including items, files, and collections. For more information see, http://dublincore.org/documents/dces/.
Title
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Senior Presentations Archive
Description
An account of the resource
This archive contains materials from Wagner’s annual ‘Senior Presentations.’ This event honors outstanding students from each discipline who completed their Senior Learning Community project with excellence. The work is representative of Wagner’s highest standards, and is exemplary of the diversity of subject matter, publicfacing scholarship, and civicminded professionalism our students have attained through their four years here. These students were specially invited to present their work in a formal setting, traditionally the day of Baccalaureate. Students are encouraged to present their work in a format appropriate for their discipline, and so, the presentations vary in their format. Some might be in the form of a short video, or paper abstracts, while others might be posters or music clips. We expect this archive to serve as a resource for generations to come. Congratulations to our Seniors!
Date
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2017 
Rights Holder
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Wagner College, Staten Island, NY
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Date Digital
2020
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Abstract
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Noh, Seunghyun (Stewart)
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5/1/2020
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Data Mining and Its Security
Contributor
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Professor Adrian Ionescu
Computer Science
Type
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text
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2 pages
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eng
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Wagner College, Staten Island, NY
Computer Science