NEW IDENTITIES FOR THE ARCTAN FUNCTION

[Pages:10]Journal of Mathematical Analysis ISSN: 2217-3412, URL: Volume 3 Issue 3(2012), Pages 1-10.

NEW IDENTITIES FOR THE ARCTAN FUNCTION

ANTHONY SOFO, JORGE CIMADEVILLA VILLACORTA

Abstract. We consider particular algebraic identities and by the use of the Beta function we derive an infinite set of binomial series for the arctangent function. Setting specific parameter values we obtain a number of new identities for .

1. Introduction

In 1671 James Geogory, a Scottish mathematician discovered the arctangent series, see [19]

(-1)

2 +1

arctan =

;

2 +1

=0

-1 < 1,

(1.1)

and when = 1 we obtain the Leibniz series of 1674, hence the slow converging series for . Using his transformation formula [1], in1755 Euler, [8], discovered

4 2 +1

arctan

=

=0 (1 + 2) +1 (

( + 1)

2

+ 1 ),

.

(1.2)

Euler's works on infinite series are extensive and the paper by Varadarajan, [20] dealing with this subject is worthwhile consulting for the interested reader. There are also some other works dealing with the expansion of the arctan function. Recently Chen, [7], obtained the nice identity

( arctan )

(-1)

2

( - 1)

=!

2+

=0

(1.3)

where

(0) = 1 and

( - 1)

() =

, for = 1, 2, 3...,

2+

=0

2000 Mathematics Subject Classification. Primary 05A10, 11B65, Secondary 05A19, 11Y16. Key words and phrases. ArcTangent functions; Binomial series; Pi series; hypergeometric functions. c 2010 Ilirias Publications, Prishtin?e, Kosov?e. Submitted May 31, 2012. Published October 2, 2012.

1

2

ANTHONY SOFO, JORGE CIMADEVILLA VILLACORTA

and when = 1, (1.3) reduces to (1.1). Also, in a similar vein, Adegoke and Layeni, [2] obtained the derivative of the arctan function, viz:

(-1) -1 ( - 1)! (

( 1 ))

(arctan ) =

sin arcsin

,

(1 + 2) 2

1+ 2

for = 1, 2, 3...,

(1.4)

however Lampret, [12] noted that (1.4) is incorrect due to some errors in the anal-

ysis. Choosing specific parameter values in (1.1), (1.2) and (1.3) we can obtain

various series representations for . In this paper we shall present two algebraic

identities from which, through integration and the application of the classical Beta

function, we obtain new series expansions of the arctan function. By appropriate

selection of the parameter we obtain new representations of . We remind the

reader of the following notation which will be useful throughout this paper. The

generalized hypergeometric representation

[, ], is defined as, see [13]

where for

[ 1, 2, ..., 1, 2, ...,

]

(

=

1)

( 2) ... (

)

=0 ( 1) ( 2) ... ( ) !

-0 , ( ) is Pochhammer's symbol defined by

(

)

{ =

( + ) ( )

=

( + 1)....( +

- 1), for

,

1, for = 0

(1.5) (1.6)

we also note that

()

( +

) - , for

>

=

.

()

(

1 +)

-

,

for

(1.7)

Here

-

0

denotes

the

set

of

non

positive

integers

and

the

Gamma

and

Beta

functions

are defined respectively as

( ) =

0

-1 -

and

1

( , )= ( , )=

-1 (1 - ) -1

0

( )( ) =

( + )

forThe numbers and are zero or positive integers (interpreting an empty product

as 1) and we assume, for simplicity, that the variable , the numerator parameters

1, 2, ..., and the denominator parameters 1, 2, ..., take on complex values

provided that no zeros appear in the denominator of

[, ] , that is / -0 ; =

1, 2, 3, ..., . Hence if a numerator parameter is zero or a negative integer then the

hypergeometric series

[, ] terminates, since, see [18]

{

0, >

(- ) =

(-1) (-

)! ! ,

0

;

.

The following Lemmas and Theorems are the main results presented in this paper.

ARCTAN FUNCTION

3

2. the main results The following Lemma will be useful in the proof of the main theorem.

Lemma 2.1. Let {0} , := 1, 2, 3..., and {-1}, then

[ -1

] (- )

(- ) +1

(1 + )

(- ) +

= 1+

1+

1+

=0

(1 - ) .

(2.1)

Proof. Consider the left hand side of (2.1), then

[ -1

] (- )

[ 1 - (- ) (- ) ]

(1 + )

(- ) +

= (1 + )

+

1+

1+

1+

=0

(1 + ) (- ) = 1 - (- ) +

1+

and the result follows.

Theorem 2.2. Let {0} , := 1, 2, 3..., and , then

arctan

(-1)

2 ( +1)

!

(

-1

=

(1 + 2)

(2

=0

=0

(-1) 2

+2

+2

+ 1) (

+

1 2

+

(-1) ( +1) 2( + + ) !

+ =0 (1 + 2) +1 (2

+2 +2

+ 1) (

+

1 2

+

. )

) )

(2.2)

Proof. From (2.1) and using Lemma 2.1, we replace the variables : 2, : 2 and rewrite as

1 1 + ( )2

=

-1 =0

(-1)

(

)2

+

(-1) ( 1+

2

)2

1 + (-1) +1( )2 +2 2

1+ 2

(1 - 2)

now we expand the denominator as a geometric series, so that

1

(-1) (

+2)

2 ( +1)

2

1 + ( )2 = =0

(1 + 2)

(1 -

2)

-1 =0

(-1)

(

)2

.

+

(-1) ( 1+

2

)2

4

ANTHONY SOFO, JORGE CIMADEVILLA VILLACORTA

Rearranging and integrating both sides with respect to [0, 1] gives us

1

1 0 1+(

)2

=

(-1)

-1 =0

(-1)

2

2(

+ ) (1 - 2)

2 ( +1)

0

(1 + 2)

=0

+

(-1) 1+

2 2

1

2 ( +1) (1 - 2)

,

0

we notice the integrals can be represented as Beta functions, hence

arctan

(-1)

-1 (-1)

2

2 ( +1)

=0

2

( + 1,

+

+

1)

2

=

(1 + 2)

=0

+

(-1) 1+

2 2

( + 1,

+

+

1)

2

-1 (-1)

2

! (

+

+

1 2

)

(-1)

2 ( +1)

=0

2 (

+

+

+

3 2

)

=

=0

(1 + 2)

+ (-1)

(1+

2 ! ( 2) ( +

+

+

1 2

)

+

+

3 2

)

-1 (-1) 2

(

+

1 2

)

(-1)

2 ( +1) !

( ) =0 (2 +1)

+

3 2

( +1)

=

.

(1 + 2)

=0

(-1)

+

(1+ 2) (2

2( +1) (

+

1 2

)

) +

3 2

(

+1)

Using the relational properties (1.6) and (1.7) of the Pochhammer function we can display

arctan

-1

(-1) 2

=

(-1)

2 ( +1)

(1 + 2)

=0

!

=0 (2

+

(1+ 2) (2

+2

+2

+1)

(

+

1 2

+

)

(-1) 2

+2 +2

+1) (

+

1 2

+

)

and the result (2.2) follows. We can also rewrite (2.2) in more familiar form

arctan

-1

=

2 (-1)

2?

=0

(-1)

2 ( +1) 4 ( 2

+2 ) +

=0 ( + 1) (1 + 2) ( 2 (

+ + + 1) ) ( + + +1

+ + +1 ) +

ARCTAN FUNCTION

5

+

=0 (

2 (-1) ( +1) 2( + + ) 4 ( 2

+2 ) +

+ 1) (1 +

2)

(

+1

(

+ 1) ( + 1) ) ( 2 ( + 1) ( + 1) ) .

( + 1)

( + 1) ( + 1)

Remark. For special values of the series (2.2) does not converge to faster than the classical Ramanujan series or the BBP- type formulas, [21]. However (2.2) is another new interesting representation of the arctan function.

We now highlight some examples. Example 1. For = 1, we have the general representation

(-1)

=

4

2

=0

-1

(-1)

!

=0 (2

+2

+2

+1)

(

+

1 2

+

)

+

2 (2

(-1)

+2 +2

+1) (

+

1 2

+

)

Example 2. For = 0 the first term on the right hand side of (2.2) vanishes, therefore

arctan

=

=0

2

!

(1)

20

(1 +

2) +1

(3)

2

,

1

[ 1, 1

]

2

= 1+ 2 2 1

3 1+ 2

2

using the properties of the Pochhammer and Gamma functions we recover Euler's identity (1.2). Where [, ] is the hypergeometric function as defined in (1.5) and follows by the consideration of

+1 where

4 2 +1

= (1 +

2) +1 (

( + 1)

2

+ 1 ).

Example 3. For = 1

arctan

(-1)

4

[ !

1

2

]

=

(1 + 2)

=0

(4

+ 1) (

+

1)

2

- (1 +

2) (4

+ 3) (

+

3)

2

(-1) 4 !

[

1

2 (2 + 1)

]

=

=0 (1 + 2)

(

+

1)

2

(4

- + 1) (1 +

2) (4

+ 3) (4

+ 1)

(-1)

4 4 (2 2 ( + 1) + 4 + 3)

=

=0 (1 +

2) +1 (4

+ 1) (4

( + 3)

4 2

)

6

ANTHONY SOFO, JORGE CIMADEVILLA VILLACORTA

=

(2 2 + 3)

[

1 2

,

1,

1

3 (1 + 2) 3 2

5 4

,

7 4

4]

- 4 (1 +

2)

4 4 ( 2 + 2)

- 105 (1 +

2)2 3

2

[

3 2

,

2,

2

9 4

,

11 4

]

4

- 4 (1 +

2)

=

(2 2 + 3) 3 (1 + 2) 4

3

1 2

,

1,

1,

4 2(

2 +7 2 +2)

5 4

,

7 4

,

2 2(

2 +3 2 +2)

4

- 4 (1 +

2)

and when = 1,

(-1) 4 (6 + 5)

= 4

=0 2 +1 (4

+ 1) (4

( + 3)

4 2

).

Example 4. For = 2

arctan

6!

=

(1 + 2)

1

(6

+1)(2

+

1 2

)

2

-

(6

+3)(2

+

3 2

)

4

=0

-

(1+

2 )(6

+5)(2

+

5 2

)

2

6

4

+1

(

4 2

)

(, )

=

=0 (1 +

2) +1 (

+ 3) (

+ 2) (

( + 1)

6 3

+6 )( 3 +3 )

+3

2

where ( , ) = (1 + 2) (6 + 5) (6 + 3)- 2 (1 + 2) (4 + 1) (6 + 5)+ 4 (4 + 1) (4 + 3) .

For = 1

= 4

=0 2

4

(

+1

4 2

) (40 2 + 60 + 23)

(

+ 3) (

+ 2) (

( + 1)

6 3

+6 )( 3 +3 )

+3

2

Example 5. In the case of = 3 and = 1 , from (2.2) we have, omitting the

3

algebraic details,

3

(-1)

(6 3

) (1640 3 + 3148 2 + 1934 + 381)

= 2

=0

33

(8 4

)( 4 3

)

.

(8 + 7) (8 + 5) (8 + 3) (8 + 1)

There are many interesting representations of , see for example [3], [4], [5], [6], [9], [10], [11], [14], [15], [16], [17], and the references therein.

ARCTAN FUNCTION

7

Remark. From any of these five examples or from (2.2), it is possible to obtain the derivatives of the arctan function. From example 2, for = 1, 2, 3, ....

!

(arctan ) =

(3)

=0 2

(

)

2 +1

(1 + 2) +1

!

=

(3)

=0 2

(

(

(-1)

+

) )

2 +2 +1

=0

!

(

=

(3)

(-1)

+

)

( !

2

+2

+1 )

2 +2 +1-

=0 2

=0

=

!

! 2 +1- ( 2 + 1 )

[

(3)

32

=0

2

+ 1,

+ 1,

+

3 2

] - 2 , for 1.

+ 1 - 2,

+

3 2

-

2

The next Lemma introduces a new generalized algebraic identity, similar to Lemma 2.1, and following the ideas used in Theorem 2.2 allows us to develop some new arctan identities.

Lemma 2.3. Let {0} , := 1, 2, 3..., and {-1}, also let { } =0 be a set of positive real numbers, then

(1 +

[

)

=0

-

]

+ (-1) ( )

=1

=0

(2.3)

=

=0

+ +1 (-1) -

=0

Proof. The proof follows the same idea as used in Lemma 2.1. Expand the left

hand side of (2.3) collect terms, hence we are done.

Theorem 2.4. Let {0} , := 1, 2, 3..., and , also let { } =0 be a set of positive real numbers, then

arctan

=

(-1)

1(

1

+

=1

(-1) (

)2

-

=0

2

=0

=0

=0

2) ?

( 2 2 ( +1) [ =0 (-1)

[ =0

2]

2] )

-

(2.4)

Proof. Using Lemma 2.3, we replace the variables : 2, : 2 and rewrite as

1 1 + ( )2 =

1 + =1(-1) (

)2

-

=0

2

2

=0

1 + 2 2( +1) =0(-1)

2

=0

2 -

8

ANTHONY SOFO, JORGE CIMADEVILLA VILLACORTA

if we now expand the denominator as a geometric series and integrate both sides

with respect to [0, 1] we obtain the identity (2.4).

By appropriate choices of the set of real positive numbers { } =0 it is possible to explicitly integrate (2.4) therefore producing a set of new identities for the arctan

function. Two examples are highlighted.

EXAMPLE 6. For = 2, put 0 = 1, 1 = 2, 2 = 1, therefore

arctan

1

=

=0

1- (

)2(1+2 2)

(1+ 2)2

+

( (1+

)4 2 )2

1+

2 6(1- )2 2

(1+ 2)2

(-1)

6

= =0 (1 +

2)2

1

=0

2

(1 -

2)2

[ (

1-

)2 (1 + 2 2)

( )4 ]

(1 + 2)2

+ (1 +

2)2

,

the integrals, as in Theorem 2.2, are special cases of the Beta function and therefore

may be evaluated to produce

arctan

(-1)

!

6

(

1

) - +

1 2

2

+1

(1+

2(1+2 2)

( ) 2)2

+

3 2

2

+1

=

=0 (1 + 2)2

4

+ ( ) (1+ 2)2

+

5 2

2

+1

(-1) 6 42 ( 2 ) ( , )

=

=0 (1 +

2)2 +2 3 (6

+ 1) (6

( + 5)

6 3

)( 3

)

where ( , ) = 8 4 ( + 1) + 5 2 (6 + 5) + 3 (6 + 1) . In hypergeometric form

arctan

=

4

[

5 2

,

1,

1,

1 2

5 (1 + 2)2 4 3

11 6

,

9 6

,

7 6

46

]

- 27 (1 +

2)2

4+5 2+3

+ 3 (1 +

2)2

3

[

2

1,

1,

1 2

7 6

,

5 6

46

]

- 27 (1 +

2)2

where

46 27(1+ 2)2

1. When

= 1

3

1

=

4 3 =0 3

(-1) ( 2

( 6 )( 3 3

) (130 + 109)

)

.

(6 + 1) (6 + 5)

EXAMPLE 7. For

1

arctan =

=0

= 3, put 0 = 1, 1 = 2 = 3, 3 = 1, therefore

1- (

)2(1+3 2+3

(1+ 2)3

4) + (

)4(1+3 2)

(1+ 2)2

-

( (1+

)6 2 )3

1+

2 8(1- )2 3

(1+ 2)3

.

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