IIT Kanpur Student's Summer Project Report on Euler's Contributions to Mathematics
1. Indian Institute of Technology
Kanpur
SUMMER PROJECT REPORT 2015
Book : Euler The Master Of Us All
Jitendra Thoury
Student of IITK
Roll No. -14290
Date : May29,2015
3. Summer Project
1 Introduction
”Mathematics lacks the tactile solidity of architecture.It is intangible,existing not in
stone and morter but in human imagination.Yet like architecture it is real and it has
its masters.”
Once the great mathematician of his era Laplace said that ”Read Euler,read Euler, Euler is the
master of us all.”
Why such a great mathematician said such statement? Because There is nearly no branch of
mathematics among them in which Euler doesn not give his valuable contribution. The book ”Euler
The Master Of Us All” contains some selected branches of mathematics and then gives us basic
introduction to Euler’s work on these branches of mathematics.The Book also have small biography of
Euler.He born in land of great mathematicians of his time ’Basel’(Switzerland) ;the land of Bernoulli
familyJakob Bernoulli, Johan Bernoulli and Denial Bernoulli.His full name is Leonhard Euler.
There are few things that we must note about Euler during reading this book. Euler was far from
infallible and his proofs are not so rigorous and do not follow recent mathematical boundations but
nobody can argue that his proof was wrong beacause Euler gave more than one proofs of most of the
results that he deduced and these were more rigorous.Most of his proofs were of analytical type.Euler
use infinite serieses as his best analysis tool and specially the logarithms.One of the interesting fact
about Euler is that “whenever Euler saw products he took logarithms.” Author of the book Prof.
Dolcini writes that “Euler wrote mathematics faster than most people can absorb it.”
The way of writting the book is so nice for a begginers and experinced all type of readers.The
way of explaining a topic is this that it first gives us the knowledge of mathematics that Euler got
from his predecessors about that topic and then expresses Euler way of resoning and demonstrates
his coclusions and results.And as the summary of the topic it introduces us from further research
on that topic and leaves some unsolved problems for mathematicians community.
There is no great prior knowledge required for reading this book except some basic courses of
mathematics that covered in first year by a student of mathematics or any engineering. So this book
can be read and learn by any student of mathematics or who have very basic knowledge of it.And
the author goes very cleverly in sequencing the chapters that we do not have to worry about such
things.
The book divided into eight chapters each of which related to a different branch of mathe-
matics.Each of the chapter have three subsections ’Prologue’-research before the Euler came,’Enter
Euler’-research in the field by Euler and last is ’Epilogue’-research in the field after Euler and further
challanges for mathematicians in the field.So this nice arranged book in this pattern.
The very first chapter of the book is from number theory mostly focuses on prefect numbers.The
study of prefect numbers starts from ancient greeks whose motivation was to find number which is
equal to sum of its all proper divisors.They found that these numbers are rare.Euclid gave a theorem
or say method for generating these numbers but that was limited to even prefect numbers.Ancient
Greek mathematicians could not find any odd prefect numbers.Euler improved the method of Euclid
by adding necessary condition.Chapter ends with introducing the reader by further research on the
odd prefect numbers.
Chapter second of the book is logarithms which have the series expansion of logarithms and their use
as a analysis tool.This was Euler’s favorite tool of analysis so chapter have some proofs from various
branches of mathematics.The third chapter is about the surprising infinite serieses.In this chapter
some open problems of Euler’s time is solved and introduces by some challanges in theory.Chapter
consists a very important theorem about relation of coefficients of polynomial with its roots.Next
chapter of the book is the Analytic number theory.Euler said it as the most difficult and profund
subject.Chapter mainly focuses on prime numbers and on some infinite serieses having prime numbers
.This gives us the good introduction to techniques of proving results.Complex Variables is the fifth
chapter of the book.In this we will see Euler’s infinite series expansion of sin(x) and cos(x) using
De Moivre’s Theorem.In the end we come in the touch of solution of some strange equations in
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4. Summer Project
the form of complex numbers.In the chapter Euler and Algebra author tries to see us the effort
of mathematicians in proving the fundamental theorem of algebra.In the beginning ,we touch the
thought of Japanese mathematicians aim to find out the roots of a polynomial in the terms of its
coefficients.But that proved after 200 years that fifth degree polynomial’s roots can not be found in
terms of their coefficients in general.Ends with the proof of the fundamental theorem of algebra.In
between the chapter we will find few tricks and method of finding roots for 3rd
and 4th degree
polynomials in terms of their coefficients.Geometry of Euler was of mixture of synthetic and analytic
type.Synthetic geometry is said to be pure geometry which uses only the ancients Greeks way or this
does not superimpose the triangle on a plane paper.But the analytic type of geometry superimposes
the triangle on a paper plane or use coordinate system to represent a geometric object.Euler gave the
proof of Heron formula (for triangle area) in two different ways ,one by using synthetic geometry and
other one by analytic geometry.Further we see the advantages of analytic geometry in reaching on
the Euler’s line.The last chapter of the book is combinatorics.There author makes changes slightly
in his pattern and he teaches us combinatorics by taking a problem and then by solving it.He gives
us the proof of formula for calculating number of derrangments of n items. And here the content of
the book ends but author also gives us the further study sources in the appendix.
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5. Summer Project
2 Prefect Numbers
2.1 Proper divisors :-
All divisors of a natural number (except that number) are called as proper divisors of the that
number.Ex.- All divisors of number 6 is 1,2,3,6 but proper divisors are 1,2,3.
Number of proper divisors :- If a number n can be written in the form
N = Pα1
1 .Pα2
2 ...Pαk
k where P1, P2...Pk are prime numbers and α1, α2...αk are non negative integers.
This is obvious that all the proper divisors will be generated by multiplication of any combination
of P1, P2...Pk because any number that not formed by these does not divides N.So the total number
of divisors can be found by multiplication theorem of permutation theory.
So any of the divisor will take the form Pβ1
1 .Pβ2
2 ...Pβk
k s.t. βi ∈ {0, 1, 2...αi } for all i ∈ {0, 1...k} So
now the total number of divisors = (α1 + 1).(α2 + 1)...(αk + 1).
That’s why proper divisors=(α1 + 1).(α2 + 1)...(αk + 1) -1.
Sum of all divisors(σ(N))=
(P
α1+1
1 −1)
(P1−1) .
(P
α2+1
2 −1)
(P2−1) ...
(P
αk+1
k
−1)
(Pk−1)
This follows by simply the geometric series sum.
The standard notation is σ.
So the sum of proper divisors = σ(N) - N.
2.2 Definition -
A whole number whose sum of all proper divisors equal to that number is called a prefect num-
ber.These prefect are rare.Ex.6,28.
So mathematically a number is prefect if and only if σ(N) − N = N ⇒ σ(N) = 2N.
Theorem 1.1 :- Euclid gave a way of generating prefect numbers.
Statement : If 2p
− 1 is a prime number then 2p−1
.(2p
− 1) will be a prefect number.
Such numbers are named as Merssene primes.
Proof : The proof follows simply by checking that this number is prefect or not.So for checking so
say N = 2p−1
.(2p
− 1). Then sum of its divisors,
σ(N) = (2p
− 1).(1 + 21
+ 22
+ ...2p−1
) = (2p
− 1).2p
= 2 × N
So N is a prefect number.
Note : 1.There is so much wastage of time in getting Merssene primes.So we want to get some
conditions on such p.By simply checking we can say if p is not a prime number then (2p
− 1) will
also not be prime number.
2. Two numbers m and n are said to be amicable if sum of the proper divisors of m is n and vice
versa.
σ(m) − m = n , σ(n) − n = m
σ(m) = σ(n) = m + n
Euler’s study of prefect numbers started from here.
3.Some properties of σ-
If a and b are two relative primes(no common factor except 1) then
σ(a × b) = σ(a) × σ(b).
Euler proved that Euclid’s sufficiency condition is also necessary condition for even prefect numbers.
Theorem 1.2(Euclid-Euler Theorem) :-
Statement : A number N is even prefect number if and only if N can be written in form N =
2p−1
.(2p
− 1) where 2p
− 1 is prime number.
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6. 2.3 Challanges in Theory :- Summer Project
Proof of this theorem follows from above property of σ.
2.3 Challanges in Theory :-
Prefect numbers known to today’s date are only even numbers and they are in finite numbers.So
there are two profound problems in theory of prefect numbers -
(i)Whether there are any odd prefect number or not.
(ii)Are there finite number of prefect numbers or they are infinitely many?
Also mathematicians could not prove the infinitude or finitude of Merssene primes.Problem
(i) would immediately follow from infinitude of Merssene primes.These problems proposed many
centuries ago and till now not answered.Mathematicians admitted to being stumped on the topic
for a large time but now they are running for finding conditions for odd prefect numbers assuming
that they exist.Great mathematicians J. J. Sylvester proved that if any odd prefect number exists
then it must have 3 different prime factors.
2.4 Further Research :-
Theorem 1.3 :- An odd prefect number must have at least 3 different prime factors.
Proof : case(1)(one prime factor)
Say N = Pα
(P is a prime number)
Then σ(N) = 2 × N
P α+1
P −1 = 2 × Pα
⇒ 2 × Pα
− Pα+1
= 1
Here LHS is divisible by Pα
but RHS is not .So such odd prefect is not possible.
Case(2)(two prime factors)
Say N = Pα
.Qβ
(P and Q are prime numbers)
Then, σ(N) = 2 × N = 2 × Pα
.Qβ
.
(1 + P + P2
+ ...Pα
).(1 + Q + Q2
+ ...Qβ
) = 2 × Pα
.Qβ
.
2 = (1 + 1
P + 1
P 2 + ... 1
P α ).(1 + 1
Q + 1
Q2 + ... 1
Qβ )
But RHS ≤ (1 + 1
3 + 1
9 + ... 1
3α ).(1 + 1
5 + ... 1
5β )
RHS ≤ (1 + 1
3 + 1
9 + ...).(1 + 1
5 + 1
25 + ...) ≤ 3
2 .5
4 < 2
So RHS = LHS
So such odd prefect number is not possible.
Recent Research - 1.An odd prefect can not be divisible by 105 = 3 × 5 × 7.
2. It must contain at least 8 different prime factors.
3. Smallest odd prefect number must exceed 103
00.
4. The second largest prime factor of an odd prefect number must exceeds 1000.
5. The sum of reciprocals of all odd prefect number is finite.
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7. Summer Project
3 Logarithms
Logarithms is Euler’s most favorite chapter and one of the newest subject of study included in this
book.Euler used logarithms in as a most important analysis tool in calculas,analytic number theory
and algebra etc.
Euler’s Definition of a Function -
A function of a variable quantity is an analytic expression composed in any way whatsoever of
the variable quantity and number or constant quantities.
−→ This is not a modern concept.Euler seemed to equate function with a formula. Before
Euler people Henry Briggs generated a log table of base 10 logarithm and that calculated the value
of base 10 log. of any number by linear interpolation.He calculated the values of logarirhms of
√
10,
√
10, ...8192 th square root of 10 and then approximated any desired value by these.
3.1 Logarithm and exponential series :-
Euler obtained the series expansions of exponentials and logarithms by using infinitesimal small and
infinitely large number concept.At that time no one could believe that such a concept could do such
a great research.
Series expansion for ax
- Let ω(> 0) be an infinitesimal small number so that,
aω
= 1 + (such that is also infinitesimal small)
So connect and ω by , = k × ω
Take j = x
ω (j is infinitely large and x is finite)
ax
= (aω
)j
= (1 + k · ω)j
So by binomial expansion ax
= 1 + j.k.x
j + j.(j−1)
2
k.x
j
2
+ ...
Now using approximations j.(j−1)
j2 = 1, j(j−1)(j−2)
j3 = 1 and so on. So ax
= 1 + k · x + k·x2
2.1 + ...
We can choose a base ‘a such that k = 1 and call that base is ‘e .
So now ex
= 1 + x
1! + x2
2! + x3
3! + ...
Series expansion of ln(1 + x) :- By previous result , ejω
= (1 + ω)j
ln(1 + ω)j
= j · ω
we can choose (1 + ω)j
=(1+x) so ln(1 + x) = j · ((1 + x)
1
j − 1)
Then by similar reasoning used above we get,
ln(1 + x) = x − x2
2 + x3
3 + ...
Clearly this series is convergent for x > 1 but it is not so easy to visualize that this series is also
convergent for 0 < x ≤ 1 because terms like x2
2 , x3
3 are growing rapidly.
So for visualizing this -
ln(1 − x) = −(x + x2
2 + x3
3 + ...)
so ln(1+x)
(1−x) = 2 × (x + x3
3 + x5
5 + ...)
Cosider f : (−1, 1) −→ (0, ∞)
So we can find that this series is convergent for all x ∈ (−1, 1).
Divergence of Harmonic Series - This is the Euler’s proof of divergence of harmonic series.
Putting X=1 in the expansion of ln(1 − x) we get-
ln(0) = −(1 + 1/2 + 1/3 + 1/4 + ...) ⇒ 1 + 1
2 + 1
3 + ... = ln(1
0 ) = ∞.
Hence proved that harmonic series diverges.
Euler-Mascheroni constant - Euler proved by repeating use of expansion of ln(1 + 1/n) that the
following limit exists,
γ(Euler-Mascheroni constant) = limn→∞ (
n
k=1(1/k) − ln(n + 1)) = 0.577218.
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8. Summer Project
This constant appears in many places in mathematics like pi.
4 Euler and Infinite Series
Jakob Bernoulli contributed in the study of infinite series as he gave exact sum of many infinite series
and prove some other’s divergence and convergence.Euler used these infinite series in his analysis.
He proposed a series called p-series -
Sp =
∞
k=1
1
kp = 1 + 1
2p + 1
3p + ...
For p=1 this series become harmonic series which diverges and for other values the sum of the series
is converging but not known at that time.Specially for p=2 the problem was called basel’s problem.
4.1 Euler’s solution of Basel’s problem -
He first approxiamted this series to a rapidly converging sequence.He used his favorite tool of analysis,
Consider, I =
1/2
0
−ln(1−t)
t dt
By doing series expansion and afterthat integrating we get,
I = 1
2 + 1
22·22 + 1
23·32 + ... =
∞
k=1
1
2k−1.k2 .
Now put z=1-t in I, I =
1/2
1
ln(z)
1−z dz
Now by doing binomial expansion of (1 − t)−1
and afterthat solving the integration we get,
I =
∞
k=1 1/k2
=
∞
k=1
1
2k−1.k2 + [ln(2)2
].
Exact sum of the series - Euler’s Assumption - Euler represented the infinite polynomial in
terms of their roots in same way as we do for finite polynomial.
P(x) = sinx
x = 1 − x2
3! + x4
5! − ...
As it is clear that ±nπ(n is any natural number) is a general solution of the P(x).
So P(x) = 1 − x2
π2 . 1 − x2
4π2 ....
By comparing coefficient of x2
in both the expansions of P(x) we get that ,
∞
k=1 1/k2
= π2
6 .
Proof of Wallis formula is very simple by using this representation,
simply by putting x=π/2 in the P(x) we get that-
2/π = 1.3.3.5.5.7.7....
2.2.4.4.6.6.8.8....
The sum of the p-series was still unknown for p > 2.However Euler found the expressions for exact
sum of series for even p’s.
Theorem 3.1
Statement :- If the nth
polynomial ,
f(y) = yn
− A · yn−1
+ B · yn−2
− C · yn−3
+ ... ± N
is factored as
f(y) = (y − r1).(y − r2)...(y − rn)
then
n
k=1
rk = A ,
n
k=1
(rk)2
= A ·
n
k=1
rk − 2B,
n
k=1
= A ·
n
k=1
(rk)2
− B ·
n
k=1
rk + 3C...and so on.
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9. 4.1 Euler’s solution of Basel’s problem - Summer Project
Observation - Consider a polynomial ,
R(x) = 1 − A · x2
+ B · x4
− ... ± N · x2n
can be written in the form
(1 − r1 · x2
) · (1 − r2 · x2
)...(1 − rn · x2
)
substitute y = 1
x2 so
1 − A.
1
y
+ B.
1
y2
− C.
1
y3
... ±
1
y2n
Now by multiplying both sides by yn
we get the same equation as in above theorem.So above results
can also be applied to R(x).
So again considering P(x),
A =
1
3!
, B =
1
5!
, C =
1
7!
... so on.
So
n
k=1
rk =
n
k=1
(kπ)2
=
1
3!
,
n
k=1
(rk)2
=
n
k=1
(kπ)4
= (
1
3!
)2
−
2
5!
⇒
∞
k=1
1
k4
=
π4
90
and so on.
This proof of Basel’s problem was not much acceptable at that time because of Euler’s assumption
so Euler gave two more rigorous proof of the problem.
Challanges : Sum of the p-series for odd p’s is still unknown and mathematicians are stumped on
this problem.
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10. Summer Project
5 Euler and Analytic Number Theory
Analytic number theory is said to be most difficult and profound branch of study in mathematics.This
is one of the most ancient branch of mathematics.Out of 13 volumes of the book Elements of Euclid
three volumes are devoted to number theory.In these volumes he described basic definitions and
basic theorems of number theory.
Euclid’s theorem 4.1 : No finite collection of prime numbers include them all.
Remarks : 1.There are infinitely many prime numbers.
2. There is at least one prime number between P(n) and 2 · 3 · 5 · 7...P(n) + 1 (here P(k) denotes
kth
prime).
All prime numbers are odd (except 2) so there are two families of primes,
4 × k + 1series − 5, 13, 17, 29...
4 × k − 1series − 3, 7, 11, 19, ...
Theorem 4.2 : There are infinitely many primes of 4 · k − 1 series.
Theorem 4.3 : A prime from 4 · k + 1 series can be uniquely written in the form a2
+ b2
(sum of
two prefect squares) but a prime from 4 · k − 1 series can not.
5.1 Enter Euler
Euler found sum of the many infinite series using his armours of analysis and properties of numbers.
He used his previous results to reach on conclusions.Ex.-
S =
1
15
+
1
63
+
1
80
+
1
255
+ ... =
7
4
−
π2
6
Proof - Ovserve that terms in the series are those reciprocal whose denominators are one less than
all prefect squares which simultaneously of other powers.Ex.16 = 42
= 24
, 81 = 34
= 92
Take the series,
π2
6
= 1 +
1
4
+
1
9
+
1
16
+
1
25
+
1
36
+
1
49
+
1
64
+
1
81
+
1
100
+
1
121
.
π2
6
−1 =
1
4
+
1
16
+
1
64
+ ... +
1
9
+
1
81
+
1
729
+ ... +
1
25
+
1
625
+ ... +
1
36
+
1
1296
+ ... +
1
7
+
1
49
+ ...
So ,
π2
6
− 1 =
1
3
+
1
8
+
1
24
+
1
35
+
1
48
+
1
49
... =
∞
k=1
1
k2 − 1
− S =
3
4
− S
S =
3
4
+ 1 −
π2
6
=
7
4
−
π2
6
Theorem 4.4 :
∞
k=1
1
ks
=
∞
p=2
1
1 − 1
ps
for s > 1 and p is prime.
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11. 5.1 Enter Euler Summer Project
This theorem states a great relation between primes numbers and p-series.We can prove this theorem
by continuosly eliminating the prime factors from p-series.
Take x = 1 +
1
2s
+
1
3s
+
1
4s
+
1
5s
+ ... (1)
x
2s
=
1
2s
+
1
4s
+
1
6s
+ ... (2)
By dividing eqn (1) and eqn (2), x · 1 −
1
2s
= 1 +
1
3s
+
1
5s
+
1
7s
+ ... (3)
x
3s
· 1 −
1
2s
=
1
3s
+
1
9s
+
1
15s
+ ... (4)
By dividing eqn (3) and eqn (4), x · 1 −
1
3s
(1 −
1
2s
) = 1 +
1
5S
+
1
7s
+
1
11s
+
1
13s
+ ... (5)
By continuing this process we will get
x ·
P
1 −
1
ps
= 1
So
x =
∞
k=1
1
ks
=
∞
p=2
1
1 − 1
ps
What about the density and distribution of prime numbers in integers ??This can be understand by
following theorem.
Theorem 4.5 : Statement -
p
1
p diverges.
Euler’s proof : Let M =
∞
k=1
1
k =
p
1
1− 1
p
ln(M) = −[ln 1 −
1
p
+ ln 1 −
1
3
+ ln 1 −
1
5
+ ...]
ln(M) =
1
2
+
1
2
×
1
22
+
1
3
×
1
23
+ ...
+
1
3
+
1
2
×
1
32
+
1
3
×
1
33
+ ...
+ 1
5 + 1
2 × 1
52 + 1
3 × 1
53 + ...
ln(M) = 1
2 + 1
3 + 1
5 + ... + 1
2 × 1
22 + 1
32 + 1
52 + ...
ln(M) = p
1
p + p
1
p2
2 + p
1
p3
3 + ... A = p
1
p ; B = p
1
p2 ; C = p
1
p3 ; and so on.
Lemma : L = B
2 + C
3 + D
4 + ... converges.
Observation-
∞
k=2
1
kn ≤
∞
1
1
xn dx
This observation simply follows the fact that
l+2
k=l+1
1
kn ≤
l+1
l
1
xn dx
So now -
∞
k=2
1
kn ≤ 1
n−1
Note that
p
1
pn ≤
p
1
p2 ≤
k=2
1
k2 ≤ 1.
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By using above results,
L = p
1
p2
2 + p
1
p3
3 + ... ≤ 1
2 ·
∞
k=2
1
k2 + 1
3 ·
∞
k=2
1
k3 + ... ≤ 1
2 + 1
3 · 1
2 + 1
4 · 1
3 + ...
So now L ≤ 1 , using this lemma Euler reached on the conclusion that if series M diverges then
series A would also diverge.Hence the sum of reciprocals of primes diverge to infinity.
6 Euler and Complex Variables
Journey of complex numbers begin with the problem of finding roots for equation x2
+ 1 = 0.If we
try to solve it we get x = ±
√
−1 .But the problem was that there is no such real number exists which
square is negative number.So the problem was discarded by saying that this equation is unsolvable
like other equation ex
+ 1 = 0 , cos(x) = 2 .But after some decades mathematician seemed that it
is unavoidable when dealing with real solution of cubic equation.
Theorem 5.1 : A real solution to the depressed cubic x3
= mx + n is given by
x =
n
2
+
n2
4
−
m3
27
1
3
+
n
2
−
n2
4
−
m3
27
1
3
Proof : Proof follows from letting the solution x = p
1
3 + q
1
3 and putting this in equation.By getting
the values of p and q from equation we can find solution.
Remarks : 1. A general cubic equation z3
+ az2
+ bZ + c = 0 can be tranformed into depressed
polynomial by putting z = x − a
3 .So now we can find the solutions of any cubic equation.
But the problem occured when mathematicians applied this theorem to equation x3
= 6x + 4.
By applying above theorem, x = 2 + 2
√
−1
1
3
+ 2 − 2
√
−1
1
3
.
But it was known that equation have three real roots.So mathematicians went deeper in the
subject and write theory for complex numbers and these number are used in solving many problems
in a very simple way.They said that this imaginary number
√
−1 can be treated as real number
while applying binary operations.
Theorem 5.1(De Moivre’s Theorem) : (cosθ ± isinθ)n
= cos(nθ) ± isin(nθ) .
Proof : Proof of the above theorem can be done simply by using mathematical induction.
Euler found nth
roots of any real and complex number using this theorem.Euler also used this the-
orem in finding the very famous series expansions of sin(x) and cos(x).
Theorem 5.2 :
cos(x) = 1 − x2
2! + x4
4! − x6
6! + ... and sin(x) = x − x3
3! + x5
5! − x7
7! + ...
Proof :
(cosθ + isinθ)n
= cos(nθ) + isin(nθ) (6)
(cosθ − isinθ)n
= cos(nθ) − isin(nθ) (7)
eqn(6) + eqn(7) ⇒ cos(nθ) =
(cosθ + isinθ)n
+ (cosθ − isinθ)n
2
. (8)
Take θ is very small and n is very large such that x = nθ is finite.
lim
θ→0
cos(θ) = 1 lim
θ→0
sin(θ)
θ = 1
So using binomial expansion in equation(3) we get,
10 IIT KANPUR
13. Summer Project
cos(x) = 1 −
n.(n − 1)
2!
·
x2
n2
+
n(n − 1)(n − 2)(n − 3)
4!
·
x4
4!
− ...
so
cos(x) = 1 −
x2
2!
+
x4
4!
−
x6
6!
+ ...
similarly we can obtain expression for sin(x).
Theorem 5.3 : For any real x,
eix
= cos(x) + isin(x)
. This is called Euler’s identity .
Proof : One proof simply follows from series expansions of sin(x) and cos(x).
Alternative proof by using calculas :
x = sin−1
y =
y
0
dv
√
1 − v2
put v = iz ,
x = i
0
−iy
dz
√
1 + z2
= iln( a + z2 + z)
y = sin(x) = iz so
√
1 + z2 = cos(x) Then x = iln(cos(x) − isin(x))
⇒ cos(x) + isin(x) = eix
.
Euler also gave solution of complex arguments of trigonometeric and logarithm functions.Euler found
out the logarithm of negative values of its argument.He found that there are infinitely many roots
exists for logarithm of any number as its argument.Further research makes complex numbers more
important to study.Using complex number we can prove the Fundamental Theorem of Algebra.
7 Euler and Algebra
According to Euler Algebra is the science which teaches how to determine unknown quantities by
means of those that are known.
Algebra is a branch of mathematics which uses complex number so frequently in its theory.
7.1 Eighteenth century and Algebra
There were two open problems in Algebra in eighteenth century.
1.First was to find out solutions of any nth
degree polynomial.
2.Second was to give a Fundamental theorem of Algebra (to prove or disprove it).
But there was no major progress in both of the problems.First one is remaining unsolved yet
today’s date and second one is solved in nineteenth century.Euler gave a method for solving quartry
equations but he could not give general method for nth
degree.
7.2 Euler and Fundamental theorem of Algebra
This was conjecture at that time that every polynomial with real coefficients can be written in the
product of real linear and real quadratic factors.(Fundamental theorem of Algebra)
11 IIT KANPUR
14. 7.3 Further research Summer Project
7.2.1 Theorem 6.1
Any quartic polynomial
P(x) = x4
+ Ax3
+ Bx2
+ Cx + D
where A,B,C,D are real ,can be decomposed in two real factors of second degree.
Proof : Transform P(x) in depressed quartic by tansformation x = y − A
4 .
So now cosider the new polynomia,
R(x) = x4
+ bx2
+ cx + d
case 1. c = 0
Now R(x) = x4
+ bx2
+ d .This is a quadratic in x2
.
(i)If b2
− 4d ≥ 0
then R(x) will have two real roots in x2
.So R(x) = x2
+ b−
√
b2−4d
2 · x2
+ b+
√
b2−4d
2 (ii)If b2
−4d ≤
0
4d ≥ b2
⇒ d ≥ 0 ⇒ 2
√
d − b ≥ 0 .
Observe that R(x) = (x2
+
√
d)2
− (x · 2
√
d − b)2
R(x) = (x2
+
√
d − x · 2
√
d − b)(x2
+
√
d + x · 2
√
d − b)
case 2. c = 0
Assume that we can write R(x) = (x2
+ ux + α)(x2
− ux + β)
If we succed in finding real u, α, β then we are done.
By comparing coefficints in R(x) we get,
b = α + β − u2
, c = u(β − α), d = αβ
as c = 0 ⇒ u = 0, (β − α) = 0
By eliminating α and β we get,
4d = (u2
+
c
u
+ b)(u2
−
c
u
+ b) ⇒ f(u) = u6
+ 2bu4
+ (b2
− 4d)u2
− c2
= 0
as u → ∞f(u) → ∞ and f(0) = −c2
≤ 0
So by IVP f(u) must have one real solution of u and from u we will get one pair of real values of
α β.
Remarks - 1.If P(x) is an odd degree polynomial then it must have one real solution.
P(x) = (x − a)Q(x) where a is a real number.
Now Q(x) is an even degree polynomial(degree one less from P(x)).
So if fourth degree polynomial decomposed in two quadratic factors and fifth degree also decomposed.
2.If we could prove decomposition for real polynomial of degree 2,4,8,...2n
... then we could prove for
any.
7.3 Further research
In nineteenth century Niels Abel proved that general fifth degree equation can not be solvable in the
form of algebric operations on the coefficients of equation.But the Fundamental theorem of Algebra
proved in nineteenth century by joint work of many mathematicians. Liouville’s Theorem - An
entire ,bounded complex function is constant.
Entire function - A function differentiable on all complex space.
Bounded function - If there ∃M ∈ such that |f(z)| ≤ M ∀Z ∈ Domain
12 IIT KANPUR
15. Summer Project
Lemma - If P(z) is not a constant complex polynomial then the equation P(z)=0 have at least one
solution.
Fundamental theorem of Algebra - Any nth
degree polynomial
P(z) = cnzn
+ cn−1zn−1
+ ...c0
can be factored into n complex linear factors.
Proof - By above lemma ,∃α1 ∈ C(complex space) such that
P(z) = (z − α1)Q(z)
∃α2 ∈ C such that
P(z) = (z − α1)(z − α2)R(z).
By repeating this process n times we get,
P(z) = cn(z − α1)(z − α2)...(z − αn)
. Hence the Fundamental theorem of Algebra has proved.
8 Euler and Combinatorics
Combinatorics is branch of Discrete Mathematics which primary obective is to count finite collection
of items.Ancient Indian mathematicians had a developed combinatorics theory than others.They had
a multiplication rule for counting the ways of doing any work. Multiplication Rule : If there
is a task which have two steps and first steps have n ways of completing it and second step have m
number of ways of completing it then the total number of ways in which the task can be completed
will be m × n. In most of the problems we use its more general case for n steps task for which the
formula can be obtaoned by induction formula
(k) = Nk × (k − 1)
here (k)= number of ways of completing first k steps of task and Nk=number of ways of doing
nth
step.
Problem 1.Numbers of subsets of r different items chosen from n distinguishable items?
Solution - Say C(n,r) are the number of ways of doing so.
We know number of permutations of r items chosen from n,= n!
r! (by multiplication rule).Now think
about it as a two step process.We can say that we first make a set of r objects and then arrange
them ,so by multiplication rule -
n(n − 1)(n − 2)...(n − r + 1) = C(n, r) · r!
Thus
C(n, r) =
n!
r!(n − r)!
.
Problem 2. Number of ways of choosing r items from n different type of items if there are plentiful
item of each type(basically we allow n items to be reused)?
Solution - We will struck this problem by an example.
Suppose there are four boxes which are different.They contain Red(R),Yellow(Y),White(W) and
Brown(B) balls plentiful.
Now we want to take 3 balls out of them.
One way is Red ,Red and White.We represent this by RR||W|. Here vertical bars represent the
divisions of boxes.
13 IIT KANPUR
16. Summer Project
Actually we do not have any connection with this that which color ball puts out because we are
sepearating boxes by order that first is Red and then Yellow and then White and last Brown.
So RR||W| ≡ XX||X|.
So we can move these 6 pieces randomly such that three of them are different from three other,so
number of ways =C(6,3) because if we choose any three then we are done.So by same reasoning we
can extend it to general problem such that there are n+r-1 total pieces and r X’s.Thus, the number
of ways=
C(n + r − 1, r).
Euler’s curious question : What is the total number of derangements of n different things?
Solution : Euler said,consider (n) is the number of derangements of n items.Ex.- (1) = 0, (2) =
1, (3) = 2, (4) = 9, (5) = 44etc.
Euler observed a relation between these numbers. Theorem 7.1 For n ≥ 3
(n) = (n − 1)[ (n − 1) + (n − 2)].
Proof : Suppose a,b,c,d... is correct arrangement.We can choose first item by (n-1) ways for de-
rangement. For a moment assume first item is b for simplicity.
case 1.If sequence starts with b,a... then we are obliged to derrange (n-2) items c,d,e...
(n − 2)
ways.
case 2.If second letter of sequence is a thenour aim is ultimately to derrange (n-1) letters a,c,d,e...
So
(n − 1)
ways.
So now total number of ways
(n) = (n − 1)[ (n − 1) + (n − 2)].
Theorem 7.2
(n) = n (n − 1) + (−1)n
.
Proof : Again Euler observed this pattern by his bright eyes.
(r) − r (r − 1) = −[ (r − 1) − (r − 1) (r − 2)].
Use repeatedly with r=n-1,n-2,n-3,...3 to get :-
(n) − n (n − 1) = (−1)n
.
Theorem 7.3
(n) = [1 −
1
1!
+
1
2!
−
1
3!
+ ...
(−1)n
n!
]
(n) = n (n − 1) + (−1)n−2
n (n − 1) = n(n − 1) (n − 2) − n(−1)n−3
...
n(n − 1)(n − 2)(n − 3)...3 (2) = n! (1) + n(n − 1)(n − 2)...3(−1)2
By adding them all we get the result.
14 IIT KANPUR
17. 8.1 Partition of a whole number Summer Project
8.1 Partition of a whole number
Representation of a number as the sum of other whole numbers.There are few special cases.
(i)Partition into different summands.
(ii)Partition into odd summands.
Theorem 7.4The number of partitions of a whole number as the sum of different summands are
same as the number of partition as the number of odd summands.
Proof : Let
Q(x) = (1 + x)(1 + x2
)(1 + x3
)...
The coefficients of xn
in the Q(x) represents the total number of ways in which n can be written as
the sum of different summands.Consider
R(x) =
1
(1 − x)(1 − x3)(1 − x5)...
R(x) = (1 + x + x2
+ x3
+ ...)(1 + x3
+ x6
+ x9
+ ...)(1 + x5
+ x1
0 + ...)...
Coefficient of xn
in R(x) is the total number of ways in which n can be written as the sum of odd
summands.
P(x) = (1 − x)(1 − x2
)(1 − x3
)(1 − x4
)...
P(x)Q(x) = (1 − x2
)(1 − x4
)(1 − x6
)...
1
Q(x)
=
P(x)
P(x)Q(x)
= (1 − x)(1 − x3
)(1 − x5
)...
Q(x) =
1
(1 − x)(1 − x3)(1 − x5)...
= R(x).
Hence proved.
15 IIT KANPUR
