8 - derivación basada recursividad de duplex método square

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  • 8/12/2019 8 - Derivacin Basada Recursividad de Duplex Mtodo Square

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    Recursion based Derivation of Duplex Square MethodVitthal Jadhav

    Pune, Maharashtra, India.

    Email: [email protected]

    Abstract

    Duplex square method, introduced by Vedic mathematics, is well known faster method for

    squaring of any number. This paper derives duplex method using recursive approach.

    1. Method Derivation

    i

    i

    S ..................n n n

    S ..................n n

    S ( .................. ) Sn n n n n

    nLet a a a a a a

    in

    Then a a a a ai

    a a a a a a

    = + + + + + =

    = + + + + =

    = + + + + + = +

    ==

    S Sn n n

    a = +

    . (1)

    Squaring on both side, we get(S ) ( S ) ( ) * *S (S )

    n n n n n n na a a= + = + +

    i( S ) ( ) * * (S )

    n n n n

    na a a

    i= + +

    =

    (2)

    i( S = )

    n

    na

    i

    =

    Similarly,

    i

    i

    i

    ( S ) ( ) * * (S )n n n n

    ( S ) ( ) * * (S )n n n n

    ( S ) ( ) * *

    na a a

    i

    na a ai

    a a ai

    = + +

    = + +

    = +

    =

    =

    =

    i

    i

    (S )+

  • 8/12/2019 8 - Derivacin Basada Recursividad de Duplex Mtodo Square

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    Thus, general recurrance relation given as

    i( S ) ( ) * * (S ) 0 i n

    n i n i n i n i

    n ia a a

    i= + +

    =

    .(3)

    Also ( S ) ( )a=

    (4)

    Now using equation (3) and (4), we will expand equation (2) as below

    i

    i i

    ( S ) ( ) * * (S )n n n n

    ( S ) ( ) * * ( ) * * (S )n n n n n n

    na a a

    in n

    a a a a a ai i

    = + +

    = + + + +

    = = =

    i i i

    ( S ) ( ) * * ( ) * * ( ) * * (S )

    n n n n n n n n

    n n na a a a a a a a a

    i i i

    = + + + + + +

    = = =

    i

    i i

    i i

    ( S ) ( ) * * ( ) * * ( )n n n n n n

    * * ..... ( ) * * (a )n

    n na a a a a a a

    i in

    a a a a ai i

    = + + + +

    + + + + +

    = =

    = =

    i

    Rearranging, we get

    i i

    i i

    i i i i

    ( S ) ( ) ( ) ( ) ..... ( ) (a ) * * * *n n n n n n

    * * ..... * *n

    ( S ) ) ( * * * .....n n n n

    n na a a a a a a a

    i in

    a a a ai i

    n n n n(a a a a a a a

    i i i i

    = + + + + + + +

    + + +

    = + + + +

    = = = =

    = = = =

    i

    * )a ai

    n( S ) (a ) a a for 2 j ni jn i

    i i j

    + =

    = + =