euler’s number

Euler’s Number

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A Problem in Compound InterestLet us say we have 100 Rs. invested @ 12% p.a. interest for 12 months.

Clearly at the end of the 12 months, we get 112 Rs.

Now instead if we invest it @ 12% p.a. interest, but compound the interest at the end of first 6 months and then at the end of the 12 months:

At the end of the first 6 months, the interest is 6 Rs and the principal is 106 Rs.

At the end of the next 6 months the interest is 106*(6/100) = 6.36 Rs.

The total at the end of 12 months = 112.36 Rs.

Now instead if we invest it @ 12% p.a. interest, but compound the interest at the end of 3 months, then at the end of 6 months, then at the end of 9 months and then at the end of 12 months.

At the end of the 12 months the interest can be computed as = 112.550881 Rs.

We can continue this exercise by compounding at the end of every month, every week, every day, every hour and so on

The question is what would happen if we kept on compounding at shorter and shorter intervals. How much would be the principal?

Compound Interest for N Intervals

We can setup a calculation table as follows:Rate of Interest p.a. = 12%;Amount invested: 100 Rs.

Divide the year into N equal intervals; the interest at the end of each interval being (12/N) %

At the end of the 1st Interval Interest= Amount =

At the end of the 2nd Interval Interest= Amount =

At the end of the 3rd Interval Interest= Amount =

At the end of the nth Interval Amount =

In general, if Amount=P, R=(% Annual Interest)/100 Amount =

What if N is very large?The question is, what happens if we make N very large in the formula

What is

We will show that:

This e is called Euler’s Numbere being defined as:

RN

NPe

N

RP

1lim

n

n ne

11lim

?

One would expect to make much more money if it was compounded more frequently – what if we compounded it daily, for example?And here is the surprise: its not that much more money!

Its just about 2.718 times, if R, the annual rate of interest was 100% - that is, you doubled the money annually at simple interest.

Expression for eExpanding using the Binomial Theorem

Letting n→∞ each of the terms 1/n, 2/n, … (r-1)/n vanish to 0. We then get:

Note that:

From our definition of f(x): which is our definition of e! e=

...!

11...

11

...!3

21

11

!2

11

!11 32 rx

rnr

nxnnxnx

...!

11...

11

...!3

21

11

!2

11

!11)( 32limlim r

n

n

n

xr

nr

nx

nnx

nx

n

xxf

Evaluating eNow that we have defined an expression for e, does this expression yield any result?

What if the value of e keeps on increasing as we add together more and more terms?

...1

...1

...4

1

3

1

2

11lim

nrS

n

...!

1...

!

1...!4

1

!3

1

!2

1

!1

11lim

nre

n

As we add more and more terms of series for e together, the terms themselves become “very small” since they go as 1/r!

So we are adding together “incredibly small” positive numbers and intuitively one we may believe that the total sum of these positive numbers of the series would still be a finite

number (even if we were to “somehow” add all the infinite terms)!

For example, for r=10, the term (1/10!)=2.8x10-7.

Note that:

For example, the sum of the terms of this series actually keeps on increasing as we add more and more terms, even though the terms become smaller and small as r increases.

But, we cannot trust such an argument!

It so happens that the series sum for e, as we have defined, does converge to a finite value.

2<e<3We compare the respective terms of the two series:

The first 3 terms are the same.

The 4th term of S>4th term of e

The 5th term of S>5th term of e and so on for all subsequent terms

Recognising S as a geometric series from the second term onwards, we get S=3 and hence e<3Thus we get e<SThus : 2<e<3 and yet it continuously increases since each of its term is positive

This is the criteria for e to converge to a definite value between 2 and 3The value of e up to 12 decimal places is: 2.718281828459

Calculating eJust to illustrate, let us calculate e using the formula:

...!

1...

!

1...!4

1

!3

1

!2

1

!1

11lim

nre

n

The tables give the value of the series for e when evaluated for n terms:

n = No of Terms Value

1 1.000000000000 2 2.000000000000 3 2.500000000000 4 2.666666666667 5 2.708333333333 6 2.716666666667 7 2.718055555556 8 2.718253968254 9 2.718278769841 10 2.718281525573 11 2.718281801146 12 2.718281826198 13 2.718281828286 14 2.718281828447 15 2.718281828458 16 2.718281828459 17 2.718281828459 18 2.718281828459 19 2.718281828459 20 2.718281828459 - 5 10 15 20 25

-

0.500000000000

1.000000000000

1.500000000000

2.000000000000

2.500000000000

3.000000000000

Value of the Series / # of Terms

Notice that the additions due to the 17th term onwards do not make any difference to the 9th digit after the

decimal. The graph depicts the same pictorially.

e is Irrational!Assume e is rational. Then: Since 2<e<3, p, q can be assumed >1

Now, the left hand side of the equation is a positive integer. On the right hand side, the terms in the first parentheses are all positive integers. Hence it must be that:

is an integer.

Hence: is an integer.

Now consider:Comparing S and S’ term by term we see that S<S’

However, Hence, S’ is a geometric series with ratio

Hence: q>1 ⟹ S’<1; and S<S’ ⟹ S<1 But S should be a positive integer!

Hence by contradiction we show that e cannot be

rational.

(Multiply by q!)

(Re-arranging)

Each term of the series addition for e is a rational number; the addition of two rational numbers is another rational number, but the addition of infinite rational numbers leads to an irrational number. Very counter intuitive!!

Extending the Definition of eWe defined that: Nn

ne

n

n

,

11lim This definition can be extended as: Rx

xe

x

x

,

11lim

For any given x, there will be some n such that n<x<(n+1) and hence)1(1

11

11

11

11

11

1

11

11

1

1

n

n

x

x

n

nnxnnxn

n

nn

x

xn

n

n

1

11

11

1

11

1

)1(

11

1

ee

nn

n

nn

n

n

ne

n

n nn

111

1lim

)1(

11

1lim

11

1

)1(

11

1lim

)1(

1

11lim

een

nnnn

n

nnn

.1

11lim.

11lim

11.

11lim

Note that:

We thus get thatx

x

11

is between 2 numbers which both converge to e as x, n→∞Thus by the famous “squeeze theorem” of limits: Rx

x

xxe

,

11lim

Note that we can also rewrite as: Rxxe xx

,1lim1

0

Calculating We have defined𝑒= lim

𝑥→∞ (1+ 1𝑥 )𝑥

𝑒𝑟=( lim𝑥→∞ (1+ 1𝑥 )𝑥)𝑟

=( lim𝑥→∞ (1+ 1𝑥 )𝑥𝑟)Hence:

Applying the Binomial Theorem for real numbers as exponents:

lim𝑥→∞ (1+ 1𝑥 )

𝑥𝑟

= lim𝑥→∞ (1+𝑥𝑟1! ( 1𝑥 )

1

+(𝑥𝑟 ) (𝑥𝑟 −1 )

2 ! ( 1𝑥 )2

+𝑥𝑟 (𝑥𝑟−1 ) (𝑥𝑟 −2 )

3 ! ( 1𝑥 )3

+…)¿ lim𝑥→∞

(1+ 𝑟1 !

+(𝑟 )(𝑟 − 1

𝑥)

2 !+𝑟 (𝑟 − 1

𝑥)(𝑟 − 2

𝑥)

3 !+𝑟 (𝑟 − 1

𝑥)(𝑟 − 2

𝑥)(𝑟 − 3

𝑥)

4 !+…)

𝑒𝑟=(1+ 𝑟1 !

+𝑟 2

2 !+𝑟3

3!+𝑟 4

4 !+…)=∑

𝑘=0

∞ 𝑟 𝑘

𝑘 !Leading to:

Note that for series evaluates to 1 as it should since

And for we get back the familiar: 𝑒=𝑒1=(1+ 11 !+ 12!+ 13 !+ 14 !+…)

Differentiating

• Given:

• Thus:

is the only number to possess this property!

Calculating

• For , we know that, – This is based on the very definition of the log

function• Thus: • Using the series expansion of

• This is a very useful way to calculate in general– Of course we need to know the

Calculating • We now have a means to define what happens when a real number has

a complex number as an exponent• Firstly consider: where • We DEFINE using the infinite series representation of for real exponents

Since

• It is however not enough to simply define in this manner – it has to be ensured that:– The infinite series CONVERGES to a value, and,– The basic property of exponents: remains valid

• Then we can extend all the methods of manipulation of real exponents to complex exponents

Calculating • Consider . where . Then:

• We can write this product using the distributive law as:

… • Collecting all the terms with : • Collecting all the terms with : • In fact, collecting all the terms with will yield: • Thus, by our DEFINITION of for being a multiple of • Hence: . remains valid for complex exponents

Verify this!!

Calculating

• If we notice carefully, the proof previous proof of . is equally applicable for any quantities as long as:– Operations of addition and multiplication are defined – These operations are Commutative and Associative– The operation of multiplication distributes over addition– Thus could very well be any real of complex number– So, even if we were to choose A, where is a square matrix, we

could actually compute and get a square matrix as the result!!• Clearly the set of complex numbers satisfies all the above

properties and hence we can say:• where is any complex number

– The set of complex numbers of course includes all real numbers

Does Exist?

• We need final step: to ensure that the infinite series for CONVERGES

• Earlier We mentioned that:

• The two series within the braces are respectively DEFINED as the Cosine and Sine functions and we will show that they converge for any value of so that:

• Since Sine and Cosine series converge, too converges• And since we can see that converges for all

– Of course , always converges for NOTE: Cosine and Sine are DEFINED as infinite series, without recourse to Trigonometry or

Analytic Geometry

Convergence of Sine and Cosine

• Ratio Test for Convergence: Let where is the term of the series. If the series converges absolutely

• Convergence of : Calculating the Ratio

• Hence, the Sine function converges for all • Similarly, the Cosine function, converges for

all

• Both Sine and Cosine are continuous functions since each of the terms of their series are continuous. Also:

For , each term in the brackets is for as also

For , all the terms in the bracket are positive and some hand calculation will show that

• Thus is a continuous function such that: • Therefore there must exist , DEFINED as the smallest

positive number such that and

Properties of Sine and Cosine - 1

NOTE: is defined purely in algebraic terms, without reference to Geometry/Trigonometry

• We have DEFINED: and C and

• The Cosine sequence consists of even powers of the variable and the Sine sequence consists of odd powers, hence: and

• Differentiating each term of Sine sequence successively yield terms of the Cosine sequence; terms of Cosine sequence similarly yield the negative of the Sine functions: and

Properties of Sine and Cosine-2

NOTE: The properties of Sine and Cosine are derived without recourse to Trigonometry or Analytic Geometry

• Also: • Equating the real and imaginary parts of

and :• and



•

• Applying the properties derived so far: (by DEFINITION of ).

• , and If , but by definition, is the smallest positive root of the Sine

function so this cannot be So it must be that


• We prove that: , by induction: Case: [Defn of ] Let + Thus, by Induction, for any integer

• The “any integer ” is true because: • and

• Also:

Combining with: and


• Now, • Thus, is periodic with period

• Thus, is periodic with period – We have utilised the earlier results that and


• The famous limit: can be shown as:

• Also: for small • Hence: for small

– For example, if , the … evaluates to around which is around 4% error against the more accurate value for

𝑒𝑖 𝜋+1=0

thus giving us the expression:

This is a beautiful equation, relating 5 fundamental

mathematical quantities

Representing Complex Numbers with eWe usually represent Complex numbers as: c=a+ib Since: eiθ=(cos θ) +i(sin θ)

We can express: ireirribac sincos Where: sin,cos rbra

We can always find a pair of values r,θ such that this is true.

22222222 )sin(cos)sin()cos( rrrrba

22 bar

a

b

a

b 1tantancos

sin Thus: Also:

Hence: ireibac Where:

a

bbar

rbra

122 tan,

sin,cos

This is particularly useful when we want to multiply 2 complex numbers:)(

2121212121.. iii errerercc

Logarithmic Function

)log()log()log( baab

We assume that we can find a function, named log(x), such that for a,b ∈ R, a,b>0:

)log()log(log bab

a

)log(log abab 0)1log(

We can prove that log(x) is the inverse of the function y=rx for some real number r>1Let a=rp then p=log(a) and b=rq then q=log(b)Now, a.b=rp.rq = rp+q ⟹ log(ab)=p+q=log(a)+log(b)Also: a/b=rp/rq = rp-q ⟹ log(a/b)=p-q=log(a)–log(b)

Also consider: r(b.log(a))=(rlog(a))b=ab Since ab=r(b.log(a)) by definition of the log function: log(ab)=b.log(a)

That is: if p=rq then we saylog(p)=q

Also, since 1=r0, log(1)=0A very usual choice of r is the number 10. Also, another convenient choice or r is the number e!

So, when we say log(x) or log10(x), it implies r=10;when we mean to say r=e, we write ln(x) or loge(x)

The number r is called the base of the function log(x)Finally, it is evident that: logr(r)=1for any choice of r>0

Derivative of the Logarithmic Function

)log()log()log( baab

h

x

xhhhhh x

h

xx

h

h

x

xhx

h

hx

hx

h

xhxx

dx

d1loglim

11log

1lim

1loglim

loglim

)log()log(lim)log(

00000

xe

xuxx

h

x

u

u

h

x

xh

1)log(

111limlog

11loglim

10

The function loge(x) [base r=e] as the following properties, for any a,b>0)log()log(log ba

b

a

)log(log abab 0)1log( 1)log( e

We find the derivate of loge(x):

For any given value of x, h/x→0Furthermore:

uhhxu 0;

Hence, the derivative of loge(x) is:x

xdx

d 1)log(

By the chain rule of differentiation for any u(x): dx

du

udx

duu

du

du

dx

d 1).log()log(

Since logr(r)=1 for any r

What is the big deal about e and log?01ie is considered a beautiful since it ties up the 4 mathematical

entities 0, 1, π, and e in a simple equation.

)...()()(3

3

2

2xxxx e

dx

de

dx

de

dx

de is true only for e and for no other number.

e crops up in the mathematical description of several physical phenomena:

Radioactive decay of an element

Discharge of a capacitor in an electric circuit

Cooling of a hot body by radiation

Liquid draining our from a tank under gravity

Calculation of compound interest Population growth models

A Question!

• We have defined as the small positive number such that and of course and as infinite series

• Geometry and Trigonometry defines using circles and triangles

• Which one is the “real” definition?– What is real about a definition?!!

• Why is it that should have same value as when defined using a circle, as when defined using an infinite series?

euler’s number

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