Research on Application of Infinite Series(1st Sem Project)

8/9/2019 Research on Application of Infinite Series(1st Sem Project)

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In this project, we have discussed the three applicationsof infinite series. For this purpose, firstly, Taylor’s serieshas been presented. Then as a particular case,Maclaurin’s series has been obtained for expanding afunction in infinite series. Secondly, we have derived

solution of differential equation at an ordinary point inthe form of infinite series. Then, we have discussed

Fourier series expansion of a periodic function. As aspecial case, Fourier series of even and odd periodicfunctions have been obtained.


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INTRODUCTION


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In the 5 th Century, Zeno of Elea offered arguments crucially

divisible—for example, that for any distance there is such aperson in history to show that the concept of infinity is pro

An example is the famous series from Zeno's dichotomy an

After Zeno, Gottfried Leibniz started working over infiniteIn mathematics, the Leibniz formula for π, states that:

Using summation notation:

The infinite series above is called the Leibniz series. Howeexhibits sub linear convergence. Calculating π to 10 correct

requires about five billion terms because 101 102 1k

−<

+

for

on the notion that space and time are infinitely

hing as half that distance, and so on—Zeno was the firstlematical.d its mathematical representation:

eries.

er, Leibniz's formula converges extremely slowly: itdecimal places using direct summation of the series

1010 1

2

−> .


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After Gottfried Leibniz , Sir Isaac Newton and JosepIn numerical analysis, Newton's method (also known as thesuccessively better approximations to the roots (or zeroes)

The Newton–Raphson method in one variable is implemeGiven a function ƒ defined over the real x, and its derivativfunction f. Provided the function satisfies all the assumptioapproximation x1 is

From India, mathematician Srinivasa Ramanujan had alsoRamanujan’s Summation. Ramanujan summation is a tech

Although the Ramanujan summation of a divergent series i which make it mathematically useful in the study of diverge

undefined. Ramanujan's Summation is:

where C is a constant specific to the series. .

Raphson together started working over infinite series.Newton–Raphson method) , is a method for findingf a real-valued function.

ted as follows: ƒ', we begin with a first guess x0 for a root of thes made in the derivation of the formula, a better

orked over infinite series, which is known to us asique for assigning a value to infinite divergent series.not a sum in the traditional sense, it has propertiest infinite series, for which conventional summation is


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TAYLOR’S INFINITE SERIES


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Let us defined a function f on [a, b] such that

i) The (n-1)-th derivative(n 1)

( ) f x−

is continuous on[a, a+h],ii) The n-th derivative ( ) ( )n f x exist on (a, a+h).

Then according to the Taylor’s theorem we can expand the function f about the point x a= as follows :

2 ( 1)( 1) ( )( ) ( ) '( ) ''( ) ... ( ) ( )

1! 2! ( 1)! !

n nn nh h h h f a h f a f a f a f a f a h

n nθ

−−+ = + + + + + +

− where 0 1θ < <

or, 2 ( 1) ( 1)( ) ( ) '( ) ''( ) ... ( )1! 2! ( 1)!

nn

n

h h h f a h f a f a f a f a R

n

−

−+ = + + + + +−

Where( )

( ) ( )!

nn

n

h R x f a hn

θ = +

is the remainder or the error term of the expansion.

Now if the remainder term (x) 0n R → when n → ∞ ,then we obtain

2

( ) ( ) '( ) ''( ) ... ( ) ...1! 2! !

nnh h h f a h f a f a f a f a

n+ = + + + + + (1)

which is called the Taylor’s Infinite Series .

Now putting i.e.,a h x h x a+ = = − in (1), we obtain

2( )( ) ( ) ( )( ) ( ) '( ) ''( ) ... ( ) ...

1! 2! !

nn x a x a x a f x f a f a f a f a

n

− − −= + + + + +

which is the expansion of ( ) f x with respect to a .


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Expansion of sine series

Let ( ) sin f x x=

So (2n) ( ) ( 1) sinn f x x= − and (2n 1) ( ) ( 1) cosn f x x+ = − for any integer n.

Therefore(2n)

( ) ( 1) / 24n

f

π = −

and(2n 1)

( ) ( 1) / 24n

f

π + = −.

Hence the Taylor’s expansion of ( ) sin f x x= about4

x π

= is given by

2 3

34 4 4

4 4 4 4

( ) ( ) ( )sin ( ) '( ) ''( ) ( ) ...1! 2! !

x x x x f f f f n

π π π

π π π π − − −= + + + +

Or,1 1 1 1

sin ( ) ( ) ( ) ...4 4 42 2 2 2

2 3

(1!) (2 !) (3!) x x x x

π π π = + − − − +− −

1 1 1 1( ) ( ) ( ) ...

4 4 42

2 31

(1!) (2!) (3!)[ ] x x x

π π π = + − − − +− −


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MACLAURIN’S INFINITE SERIES


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If we put 0a = and h x= in the Taylor’s theorem, then for 0 1θ < <

( 1)2( 1) ( )

( ) (0) '(0) ''(0) ... (0) ( )1! 2! (n 1)! n!

n n x x x xn n f x f f f f f xθ

−−

= + + + + +−

which holds wheni) The (n-1)-th derivative (n 1) ( ) f x− is continuous on [0,x] .

ii) The n-th derivative ( ) ( )n f x exist on (0, x).

Here ( )(x) ( )n!

nn

n

x R f xθ = is the remainder of the expansion.

Now if ( ) f x possesses derivatives of all order in [0, x] and Maclaurin’s remainder term (x) 0n R → as

n → ∞

then2

( )( ) (0) '(0) ''(0) ... (0) ...1! 2! n!

nn x x x f x f f f f = + + + + +

is called the Maclaurin’s Infinite Series .


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Expansion of sine series

For ( ) sin f x x= , we have

(2n) (0) 0 f = and

(2n 1) ( ) ( 1) n f x+ = − for any integer n.

Hence the Maclaurin’s expansion of ( ) sin f x x= is given by

2 33sin (0) '(0) ''(0) (0) ...

1! 2! 2! x x x

x f f f f = + + + +

or,

2 3 4 5

sin . ...1 .0 ( 1) .0 11! 2 ! 3! 4 ! 5 !

0 x x x x x x

= + + +− − + +

3 5 7

...1! 3! 5! 7! x x x x

= − + − +


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SERIES SOLUTION OF ORDINARY

DIFFERENTIAL EQUATION


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Let us consider a second order linear differential equation of the form2

0 1 22( ) ( ) ( ) y 0d y dy

p x p x p xdx dx

+ + = (1)

Where 0 1 2( ), ( ) and ( ) p x p x p x are given functions of x over a domain S.

Eq. (1) can be written as2

2 ( ) ( ) y 0d y dy p x q xdx dx

+ + = (2)

Where 10

( )( )

( ) p x

p x p x

= and 20

( )q( )

( ) p x

x p x

= .

Ordinary point:Definition: A point x S ∈ at which 0 ( ) 0 p x ≠ that is ( ) p x and q( ) x are both analytic is called an ordinary point of Eq.

(2), other points of S will be called singular points.


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Solution at an ordinary point:

Let 0 x be an ordinary point of2

2 ( ) ( ) y 0d y dy

p x q xdx dx

+ + = (3)

Over a domain S.Let 0a and 1a be two arbitrary constants; then there exists a unique function ( ) y x which satisfies the differential equation

(3) which is regular in a certain neighbourhood of 0 x and which satisfies the initial condition 0 0( ) y x a= .

Without any loss of generality, we may take 0 0 x = then ( ) p x and q( ) x are regular in a neighbourhood x R< of the origin.

In this neighbourhood ( ) p x and q( ) x are expressed in the form

0 0

( ) , ( )m mm mm m

p x p x q x q x∞ ∞

= =

= =∑ ∑ (4)

Now, we have to obtain a series solution of (3) in the form

20 1 2( ) y x a a x a x= + + + ⋯ i.e.,

0

( )m mm

y x a x∞

=

=

∑ (5)

where 0 1 2, , ,a a a ⋯ are certain constants.

Let us assume that the term by term differentiation is allowed.

1

0

mm

m

dyma x

dx

∞−

=

∴ = ∑ and2

22

0

( 1) mmm

d ym m a x

dx

∞−

=

= −∑ (6)


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Now substituting (4),(5), and (6) in (3),we have

2 1

0 0 0 0 0

( 1) ( )( ) ( )( ) 0m m m mm m m m m mm m m m m

m m a x p x ma x q x a x∞ ∞ ∞ ∞ ∞

− −

= = = = =

− + + =∑ ∑ ∑ ∑ ∑

Or,2 1

0 0 0 0 0

( 1) ( )( ) ( )( )m m m mm m m m m mm m m m m

m m a x p x ma x q x a x∞ ∞ ∞ ∞ ∞

− −

= = = = =

− − = +∑ ∑ ∑ ∑ ∑

Now, equating the coefficients of like powers of x we have

2 0 0 0 0

3 2 0 1 1 1 0 0 1

1

2.1.

3.2. 2

.................................................................

.................................................................

( 1) ( 1) ( 2)m m

a p a q a

a a p q p a q a q

m m a m a m−

− = +

− = + + +

− − = − + −2 1 0 2...........m ma p a q− −+ +

(7)

From equations (7) we can successively determine 2 3 4, , ,a a a ⋯ in terms of 0 1anda a uniquely if 2 3 4, , ,a a a ⋯ arelinear combinations of 0 1anda a only.


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ExampleLet a differential equation

2

2 4 0d y

y

dx

− = (1)

And power series solution be2 3

0 1 2 3( ) ............................ y x c c x c x c x= + + + + (2)2

42 3 42 2 6 12 ............................

d yc c x c x

dx∴ = + + + (3)

Substituting (2), (3) in (1) we obtain2 2 3

2 3 4 0 1 2 3

22 0 3 1 4 2

(2 6 12 .......) 4( ......) 0

(2 4 ) (6 4 ) (12 4 ) ..................... 0

c c x c x c c x c x c x

c c c c x c c x

+ + + − + + + + =

∴ − + − + − + = Equating the coefficients of various power of x to zero obtain

2 0 3 1 4 2 0 5 3 12 1 2 1 22 , , , ,......3 3 3 5 15c c c c c c c c c c= = = = = =

Therefore the solution is

2 3 4 50 1 0 1 0 1

2 4 3 50 1

2 2 2( ) 2 .........................

3 3 15

2 2 2[1 2 ........] [ ......]3 3 15

y x c c x c x c x c x c x

c x x c x x x

= + + + + + +

= + + + + + + +

In terms of standard functions, we can write y(x) as2 2 2 2 2 20 1( ) [ ] [ ]

2 4 x x x x x xc c y x e e e e Ae Be− − −= + + − = +

Where, 0 1[2 ] / 4 A c c= + and 0 1[2 ] / 4 B c c= −


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FOURIER SERIES


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Let f be a periodic function of period 2l defined on [ , ]l l− , that is ( 2 ) ( ) f x l f x+ = . Hence it can be expanded in an

infinite orthogonal series in terms of the trigonometric functions. Let the series be written as

01 2 1 2

2 2( ) [ cos( ) cos( ) ...] [ sin( ) sin( ) ... ]

2a x x x x

f x a a b bl l l l

π π π π = + + + + + +

= 01

[ cos( ) sin( )]2 n nn

a n x n xa b

l l

π π ∞

=

+ +∑ (2)

Integrating both sides of (2) term by term on the interval [ , ]l l−

, we obtain0

1

( ) [ cos( ) sin( ) ]2

l l l l

n nl l l ln

a n x n x f x dx dx a dx b dx

l l

π π ∞

− − − −=

= + +∑∫ ∫ ∫ ∫

Or, 0( )l

l f x dx la

−=∫ as cos( ) sin( ) 0

l l

l l

n x n xdx dx

l l

π π

− −= =∫ ∫

Or, 0 1 ( )l la f x dxl −= ∫

Now, multiplying both side of (2) by cos( )mx

l

π and then integrating term by term on the interval [ , ]l l− , we obtain

0

1

( ) cos( ) cos( ) [ cos( ) cos( ) sin( ) cos( ) ]

2

l l l l

n nl l l ln

amx mx n x mx n x mx f x dx dx a dx b dx

l l l l l l

π π π π π π ∞

− − − −

=

= + +∑∫ ∫ ∫ ∫

Or,

( ) cos( )l

ml

mx f x dx la

l

π

−=∫ [as cos( ) sin( ) 0

l l

l l

n x n xdx dx

l l

π π

− −= =∫ ∫ 2 2cos ( ) sin ( )

l l

l l

n x n xdx dx l

l l

π π

− −= =∫ ∫ ]

Therefore, 1 ( ) cos( )l

m l

mxa f x dx

l l

π

−= ∫


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Similarly multiplying both sides of (2) by sin( )mx

l

π and then integrating term by term on the interval [ , ]l l− , we obtain

0

1

( ) sin( ) sin( ) [ cos( ) sin( ) sin( ) sin( ) ]2

l l l l

n nl l l ln

am x m x n x m x n x m x f x dx dx a dx b dx

l l l l l l

π π π π π π ∞

− − − −=

= + +∑∫ ∫ ∫ ∫

Or, ( ) sin( )l

ml

m x f x dx lb

l

π

−=∫

Or,1

( )sin( )l

m l

m xb f x dx

l l

π

−= ∫

The orthogonal series for ( ) f x , given in (2) is called the Fourier series and the coefficients 0 , , ( 1, 2,3, )n na a b n = … are called the Fourier

coefficients on [ , ]l l− .

These coefficients can be expressed as

0

1( )

l

la f x dx

l −= ∫ (3)

1( ) cos( )

l

m l

n xa f x dx

l l

π

−

=

∫ (4)

1( ) sin( )

l

m l

nxb f x dx

l l

π

−= ∫ (5)

These are called the Euler’s formulas.

If the period of the function is 2 π , that is [- π , π ], that is 0

1

( ) [ cos( ) sin( )]2

n n

n

a f x a nx b nx

∞

=

= + +∑ , then the Euler’s formulas are simplified as

0

1( )a f x dx

π

π π −= ∫ (6)

1( ) cosma f x nxdx

π

π π −= ∫ (7)

1 ( )sinmb f x nxdxπ

π π −= ∫ (8)


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Note-1:

If ( ) f x is a even function on [ , ]l l− that is ( ) ( ), f x f x− = l x l− ≤ ≤ .

Then 01

( ) cos( )2 nn

a n x f x a

l

π ∞

=

= + ∑

where 0 02

( )

l

a f x dxl= ∫ and 02

( ) cos( )

l

n

n xa f x dxl l

π

= ∫ Example:

Let 2( ) f x x= , where 2 2 x− ≤ ≤ .

Now, as ( ) f x is an even function, therefore

22

0 00

2 8( )

3

la f x dx x dx

l= = =∫ ∫

And2

22 2 2 20 0

2 16 16( 1)( ) cos( ) cos( ) cos2

nl

n n x n xa f x dx x dx nl l n n

π π π

π π

−= = = =∫ ∫

Therefore, the Fourier series is given by

2 2 2 21 1

8 / 3 16( 1) 4 16 ( 1)( ) cos( ) cos( )

2 2 3 2

n n

n n

n x n x f x

n n

π π

π π

∞ ∞

= =

− −= + = +∑ ∑


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Note-2:

If ( ) f x is a odd function on [ , ]l l− that is ( ) ( ), f x f x− = − l x l− ≤ ≤ .

Then,1

( ) sin( )nn

n x f x b

l

π ∞

=

= ∑

where 02

( )sin( )

l

n

n xb f x dxl l

π =

∫ Example

Let ( ) f x x= , where xπ π − ≤ ≤ .

Now, as ( ) f x is an odd function, therefore

0 0

2 2 2( )sin( ) sin( ) cos

l

n

n x n xb f x dx x dx n

l l n

π π π π

π π = = = −∫ ∫

Therefore, the Fourier series is given by

1 1

2( ) sin( ) cos sin( )n

n n

n x f x b n nx

l n

π π

∞ ∞

= =

= = −∑ ∑


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CONCLUSION


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Infinite series plays an important role in mathematics,physics and engineering. There are so many physicalproblems which cannot be solved analytically. Thesolutions of those problems can be derived in the form

of infinite series.Infinite series has several applications. In this project,

only four applications of infinite series have beendiscussed.


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1) Advanced Engineering Mathematics—R .K. Jain &

S. R. K. Iyengar

2) Higher Engineering Mathematics--Dr. B. S. Grewal

3) Differential Equations--J. G. Chakravorty & P. R. Ghosh

4) Engineering Mathematics (Vol-1) --B. K. Pal & K. Das

5) Advanced Engineering Mathematics--E. Kreyzig

Documents

Research on Application of Infinite Series(1st Sem Project)