1. TAYLOR ’ S FORMULA - 8.Taylor n dim

Chapter 8 TAYLOR’S FORMULA. EXTREMES We will extend in this chapter Taylor’s formula to scalar

  p

  functions with many variables of the form  IR. We

  f : A  IR n+1

  remind that in the case

  p=1, if A is an interval, f  C (A) and a,x 

  A,

  then it exists  between a and x so that: _n

  1 _{k n} 1  ₁     (*)

  f ( x ) d f ( x a ) d f ( x a ) _{a }  _{k  k ! ( n}  1 )!

  We will show that even for

  p1, in certain conditions, a similar

  formula exists, so we can approximate in a neighbourhood

  V the V a

  function (of

  

f with Taylor’s polynomial function T n degree):

_n n

  1 _k    .

  f ( x ) T ( x ) d f ( x a ) _{n a}  _{k } k !

  One of the most important problems in technique, economy, and so on is the one of optimisation of diverse processes; the optimisation of a process relies in finding the extremes (minimum or maximum) of a function  function that mathematically models the process  in certain conditions imposed to the variables  conditions that give the domain of definition of that function. This type of problems is handled by the theory of optimisation. We will approach only a few techniques of finding the extreme points (locale, sometimes even global) and their nature by using the formula (*).

1. TAYLOR’S FORMULA

  p

  Let  IR be a real function of

  f : A IR p real variables, p1.

  Definition 1. If f is differentiable n times in the point a  A

p

  then the polynomial function : IR  IR

  T n

  1 1 n        x A

  T ( x ) f ( a ) d f ( x a ) ... d f ( x a ), _{n a a}

  1 ! n ! is called Taylor’s polynomial function of degree n associated to the

  function  , then we can write f in the point a; denoting d f f ( a ) _n

_a

  1 _k   .

  T ( x ) d f ( x a ) _{n a}  _k  k !

  The function:

    R : A  IR, R (x)=f(x) T (x), x A n n n

  is called remainder of order

  n, and the relationship  f(x)= T n (x)+ R n (x), x A

  we call it Taylor’s formula of order n, or the Taylor’s development of

  the function f around the point a.

  Remark. Using the operator of differentiation d we have

  ( k )

    _k    

  ;

  d f  dx  ...  dx f ( a ) _{a 1 p}

     x  x _{1 p}

    if

  a=(a 1 ,…,a p ), x=(x 1 ,…,x p ), then: ( k )

    _k    

d f ( x  a )  ( x  a )  ...  ( x  a ) f ( a ) .

_{a 1 1 p p}

     x  x _{1 p}

    In order to extend the formula (*) valid on the interval

  A for the case

p1 we need two new notions: the one of segment and the other of

convex set.

p

  Definition 2. Let x,y IR . We call a segment which binds

  the points x and y the set: _p  x  t(y x)  I R  .

• The set _{p p} x t(y x)

    x , y  t   ,

  1 

• p

  I R

  I R

    x , y    t     ,

  1 

  

  is called an open segment. If for any  , 

  x,y AIR [x,y]

  A, we say

  that A is a convex set.

  p

  be a convex set,

  Theorem 1(Taylor’s formula). Let AIR n+1

  on   Å . For any 

  

f : A  IR be a function of class C A and a A x A

  exists (

  a,x) such that

  ) ( )! 1 (

  ... ) ( _{1 1 1} )) ( ( ) ( ... ) (

    

     

  ... )) ( ( ) ( ... )) ( ( ) ( ) ( ) ( ₁ ² _{2 1} ² _{1 1 1 1} ^''  

     .

     

   

     

       

  ) 1 ( _{1 1 1} a x t a f

x

a x x a x

_p

_{p p}

  F a x t _{p p p}

       a x t a x x f a x a x t a x f

  a x t a x f a x a x t a x f

    

   

       

     

     

       

 

  According to the rules of derivation of the functions of several variables and the preceding Remark we obtain:

  F F F F (0) and .

           

  F a x a x t _{p p p}

  , we need explicit values ) ( ),..., ( ), ( ' ^{) ) 1 ( ( ''}

   

     

   

    

      

  F a x t n _p

_{p p}

_n

  ) ( _{1 1 1 ) (} a x t a f x

a x

x

  … )) ( ( ) ( ... ) ( ) (

     .

     

    

      

      

  ) 2 ( _{1 1 1} a x t a f

x

a x x a x

_p

_{p p}

  )) ( ( ) ( ... ) (

  a x t a x f a x a x t a x x f a x a x _{p p p p p p}

     )) ( ( ) ( ... )) ( ( ) ( ) ( ... ₂ ² ₁ ² _{1 1}

    

       

    

     

    n n

  F F F x f

  1 ) ( !

  1 ,…,x p

  F is derivable n+1 times,

  it results that

  f is of class C n+1

          . Because

  F a x t a a x t a f a x t a f t

  IR defined by: )) ( ),..., ( ( )) ( ( ) ( _{1 1 1 p p p}

  F : [0,1] 

  function

  ) and the

  ), x=(x

  F (n+1)

  

1

,…,a p

  Proof. Let us consider a=(a

   .

             ^

  1 ) ( ) ( ^{1 2} a x f d n a x f d

n

a x f d a x f d a f x f ^{n n} _{a a a}

  1

  1 ) ( !

  2

  1 ... ) ( !

  with the derivative

  continuous. From Mac-Laurin’s formula it follows that for any

  F n F

n

  )! 1 (

         ^{n n}

   

  ) 1 ( ( ^{) ) 1 ( (}

  1 ) ( )

  1

  !

  1 ... ) ( '

  !

  1 ) (

  Because ) (

  t  [0,1] it exists  (0,1) so that:

         .

   

  n t

F

n t F t F t F t F ^{n n n n} 

  1 ) ( ) ( ^{) 1 ( 1 ) ( '' 2} F t

  !

  2 ) ( '

  ... ) ( !

  ) ( !

  ) ( )! 1 (

    

  ( n  1 )

    _{( n}    _{1 )} .

  F ( t )  ( x  a )  ...  ( x  a ) f ( a  t ( x  a )) ₁

_{1 p p}

     x  x ₁

    Therefore, _{'' 2 (n) n}

        

  F( ) f(a), F'( ) d f(x _{a a a}

  a), F ( ) d f(x a),...,F ( ) d f(x

  a) _{( n}   _{1 ) n 1}

  and .

  F (  )  d f ( x  a ), where   a   - (x

  a)  (a, x)

  

  As a consequence: _{n n}

  1 _{( k ) ( n} 1  _{1 ) k n}

  1 1  ₁ F (

1 ) f ( x ) F ( ) F (  ) d f ( x a ) d f ( x a )

      

_{a }   _{k k}

_{   }

k ! ( n 1 )! k ! ( n 1 )! and Taylor’s formula is proved.

  Remark 1. For n=0, from Taylor’s formula we can write _p

   f ,

  f(x)  f(a)  d f(x 

_{î k k}

  a)  (î )  (x  a ), î  (a,x)  _k

  x

_{1 k}

  equality witch represents a generalization (for

  p1) of Lagrange’s formula.

  Remark 2. The remainder of order n from Taylor’s formula:

  1  _{n 1} R ( x )  d f ( x  a ),   (a, x) _{n } ( n  1 )! allows us to evaluate the error from the approximation of order _n

  n:

  1 _k     ,

  f(x) T (x) d f(x _{n a}

  a), x A  _{k  k!}

  in case that the partial derivatives of order

  n+1 of the function f on A

  (which interfere in the differential expression of order

  n+1) are

  bounded by the same constant

  

M0:

_{p n  1 n  1 n}  ₁ M   M  p       . f ( x ) T ( x ) R ( x ) x a x a _{n n i i}

  

    ( n  1 )!  ( n  _{k 1} 1 )!

   

  Remark 3. If a=0=(0,...,0)A, Taylor’s formula: _n ( k )

     

  1  

      

  f ( x ) f ( ) x ... x f ( )

_{1 p}



      _k  k ! x x _{1 1 p}

   

   ( n 1 )

     

  1  

    

  x ... x f (  x ) _{1 p}

      

  ( n 1 )! x x _{1 p}  

  is called Mac-Laurin’s formula.

  Remark 4. In the case p=2 (using Newton’s binomial) the

  Taylor development of the function 

  f around the point (a,b) A (or in powers of   x a and y

  b) becomes:

  1   f  f 

  

f ( x , y )  f ( a , b )  ( a , b )( x  a )  ( a , b )( y  b ) 

    1 !  x  y _{2 2}   ₂ 

   1  f  f  f _{2 2}  ( a , b )( x  a )  ₂ 2 ( a , b )( x  a )( y  b )  ( a , b )( y  b )  ₂

    2 !  x  x  y  y

   _{n n}  1  f  _{k n k k}

  ,  ...  C ( a , b )( x  a ) ( y  b )  R ( x , y ) _{n n n  k k}

   n !   x  y _k

  where: _n  n  ₁ ¹ 1  f   _{k n 1 k k}

  R ( x , y )  C (  ,  )( x  a ) ( y  b ) , _{n n} 

_{1  }

_n

_{1 k k} 

  ( n  1 )!   x  y _k

             . a -

  ( x a ), b (y

  b), with (0,1) In this case the graphical representation of

  f is a surface of

  equation:

   S : z = f(x,y), (x,y) A.

  

2

² Let us suppose that  and f C (A), d f  0  d f U is _{( a , b ) ( a , b )}

  neighbourhood of the point

  (a,b) 

A.

  The approximation of order 0:

  f(x,y)  T (x,y)  f(a,b), (x,y)  U

  means, from a geometrical point of view, that if 

  (x,y) U the point

M(x,y,f(x,y)) is replaced with the point M (x,y,f(a,b)) which belongs to

  a part of the parallel plane with

  xOy of equation: 

  S : z = T (a,b) = f(a,b), (x,y)

  A,

  approximation which, in general, is unsatisfying even if the neighbourhood

  U is “small”. (see Fig.1.) z z=T (x,y) M (a,b,f(a,b)) M y

  U (a,b,0) x Fig.1.

  The approximation of order

  1:    

  f(x,y) T (x,y) f(a,b) d f(x-a,y-b) , (x,y) U _{1 (a,b)}

  is more precise, because it presumes the substitution of the point M with the point

  M (x,y,T (x,y)) from the part of the plane tangent in the

  1

  1

  point

  P(a,b,f(a,b)) on the surface S: 

  S 1 : z = T 1 (x,y), (x,y) A.

  (see Fig.2.).

  The approximation of second order:

  1 ₂

  f(x,y)  T (x,y)  f(a,b)  d f(x-a,y-b)  d f(x-a,y-b) , (x,y)  U _{2 (a,b) (a,b)}

  2 is more precise than the preceding; in this case the point

  

M(x,y,f(x,y)) on the graphic S, (x,y) U is replaced with the point

M

  2 (x,y,T 2 (x,y)) belonging to the surface of equation: S : z = T (x,y), (x,y) 

  A,

  2

  2

  surface tangent in

  P(a,b,f(a,b)) on the surface S and which has as plane tangent in (see Fig.3.). P the plane S

  1

  Example 1. Expand the polynomial function f(x,y,z) = x

  3

  3

  2

• +y
• +xyz  z
• +xy+x  z+1

  in powers of x  1, y+1 and z.

  M M ₁ S ₁ z y x z=T ^{1 (x,y)} (a,b,f(a,b))

  (a,b,0)

Fig.2.

  S ² z y x z=T ^{2 (x,y)} (a,b,f(a,b))

  (a,b,0) A Fig.3

  Solution. We will use Taylor’s development of the function f

  3

  around the point . Because

  a=(1, 1,1) IR f is a polynomial of third

  4

  degree

  d f=0, so R

3 (x,y,z)=0 and Taylor’s formula is:

f ( x , y , z )  f ( 1 ,  1 , 1 )  d f ( x  _{( 1 ,  1 , 1 )} 1 , y  1 , z  1 ) 

  1 ₂

  1

₃

 d f ( x  _{( 1 ,  1 , 1 ) (} 1 , y  1 , z  1 )  d f ( x  _{1 ,  1 , 1 )} 1 , y  1 , z  1 )

  2 _{2 2}

6 But

  ,

  df  ₂ 3 x dx  3 y dy  yzdx  zxdy  xydz  _{2 2} 2 zdz  ydx  xdy  dx  dz d f 

  6 xdx  6 ydy  ( zdy  ydz ) dx  ( zdx  xdz ) dy  ₂ ,  ( ydx  xdy ) dz  2 dz  2 dxdy and _{3 3 3} .

  d f 

  6 ( dx  dy  dxdydz ) Now we compute

     ,

  f(1,1,1)=2, d f _{2 2 ( 1 ,  1 , 1 ) 2} 2 dx ₂ 5 dy 4 dz

     ;

  d f  _{( 1 ,  1 , 1 )} 6 dx  6 dy  2 dz  4 dxdy 2 dydz 2 dzdx therefore: _{2 2 2} f ( x , y , z )  

  2  2 ( x  1 )  5 ( y  1 )  4 ( z  1 )  3 ( x  1 )  3 ( y  1 )  ( z  1 )   2 ( x  1 )( y  1 )  ( y  1 )( z  1 )  ( z  1 )( x  1 )  _{3 3} .  ( x  1 )  ( y  1 )  ( x  1 )( y  1 )( z  1 )

  Remark. The polynomial f from the preceding example is a

  vector from the linear space IR [

  3 x,y,z] of the polynomials with three

  variables of degree smaller or equal with three formulated in the

  2

  2

  2

  2

  2

  2

  2

  2

  2

  canonical base B ={1,

  x, y, z, xy, yz, x , y , z , x y, x z, y x, y z, z x, z y, c

  3

  3

  3

  [

  x , y , z , xyz} (dim IR x,y,z]=20). The development of f in powers of

  3   x 1, y+1, z 1 means in fact formulating this vector in the base

  2

  2 B={       

1, x 1, y+1, z 1, (x 1)(y+1), (y+1)(z 1), (x 1)(z 1), (x 1) , (y+1) ,

  2

  2

  2

  2

  2

  2 (z  1) , (x  1) (y+1), (x  1) (z  1), (y+1) (x  1), (y+1) (z  1), (z  1) (x  1),

  2

  3

  3

  3      (z 1) (y+1), (x 1) , (y+1) , (z 1) , (x 1) (y+1) (z 1)}.

  Example 2. Using Taylor’s formula of order three compute

  2.1 the approximate value of the number (0.9) . y

  Solution. We need the value of the function in the f(x,y)=x

  point

  (x,y)=(0.9; 2.1); we will choose as point (a,b) around which we will develop the function

  f in which we can compute precisely the

  function z and its partial derivatives and also, as close as possible from ( _{2 . 1} x,y). Let then be (a,b)=(1,2). Then:

  1 ² ( . 9 )  f ( x , y )  T ( x , y )  f ( _{3 (} 1 , 2 )  d f ( x  a , y  b )  d f ( x  a , y  b )  _{1 , 2 ) ( 1 , 2 )}

  2

  1 _{3 }

  1

  1

  

1

_{2 }

  1

  1

  1 _{3 }

  1

  1  d f ( x  a , y  b )  1  d f ,  d f ,  d f , _{( 1 , 2 ) ( 1 , 2 ) (}             _{1 , 2 ) ( 1 , 2 )}

  6

  10

  10

  

2

  10

  10

  6

  10

  10      

  As _y  _{1 y} ,

  d f  yx dx  x ln xdy|  _{( 2 y  1 , 2 ) ( 1 y  1 , 2 ) 2} 2 dx

     

  d f x dxdy y ( y

  1 ) x dx _{( 1 , 2 )}  _y    _{1 y}

_{1 y y}

₁  x ln x dx  yx dx  x ln xdy  ln xdy  x dxdy| 

     _{2 3 ( 1 , 2 ) 2}

  =   , and ,

  dxdy 

  2 dx  dxdy  2 dx ( dx dy ) d f  _{( 1 , 2 )} 7 dx dy _{2 . 1}

  2

  7 it follows that:     . ( . 9 ) 1 . 807 10 1000

  Let us observe that the approximation of order 0 is

  2.1

  2.1

  (0.9) T (x,y)=1, and the one of order two is (0.9) T

  1 (x,y)=0,8. In

  none of these cases we do not have an estimation of the error. If we want to compute an expression with a pre-established precision  we must find an .

  nIN so that R   _{n 4} Example 3. Let us compute with two exact N  4  .

  1 .

  9 _{2 1 4 1} decimals. We consider the function which we

  f ( x , y )  x y

  compute around the point (4; 1). Because

  N=f(x,y) where x=4.1 and y= 0.9, according to Taylor’s formula there exists (4; 4.1) and

  (0.9; 1) so that: _n

  1 _{k  }

  1

  

1

N  f ( 4 ; 1 )  d f ,   R ( x , y ) , _{( 4 ; 1 ) n}    _{k  1  } k !

  10

  

10

  1   _{n 1}

  1 1  where   . We must find an nIN

  R ( x , y ) d f  ,  _{n (  ,  )}

   ( n 1 )! 

  10 10 

   ₂

  minimal so that , where (

  R ( x , y )  _n 10 x,y)=(4.1; 0.9). Because:

  

  1 1 

  1

  1

        

R ( x , y ) d f  ,  f ' ( , ) f ' ( , ) _{(  ,  ) x y}

  

  10 10 

  10

  10

  1  f ' (  ,  )  f ' (  ,  ) ,  x y 

  10 _{1 1} 

  

  1 _{2 4}

  1

  1

  1

        and     , f ' ( , ) f ' ( , ) _{x y 4 3}

  2

  2

  4

  2

  

  3 then  and the approximation of order zero:

  R ( x , y )

  40 N

  T (x,y)=f(4.1)=2 does not look very satisfying. Let us try an estimation as precise as possible for the reminder of first order.

  Because

  1 ^{2 }

  1 1  1 ''

  1 2 '' 1 ''    ^{2       2   } R ( x , y ) d f  ,  f ( , ) f ( , ) f ( , ) _{1 (  ,  ) xy x 2 2 2 y}

  2

  10

  10

  2

  10

  10

  10  

  1 _{'' '' ''}   

  f ^{2 (  ,  )}

  2 f (  ,  ) f _xy ^{2 (  ,  )}

    _{x y}

  200 and _{3 1 ''} 

  1 _{2 4}

  1   ,

  f _x ^{2 (  ,  )  }

  4

  32 ₁  ₃  _''

  1 _{2 4}

  1

  1 ₄

  10

  5

      , f ( , )      _xy

  8

  8

  2

  9

  24 _{'' 1}  ₇ ⁷

  3 _{2 4}

  3 1  ₄ 10 

  3

  10

  5

              , f _y ^{2 ( , )  }

  2

  16

  16

  2

  9

  32

  9

  24   1 

  1

  10 5 

  1 It follows that     , so we can affirm

  R ( x , y )   ₁

  200

  32

  24 24 100   with certitude that the approximation of the first order:

  

  1 1 

  1

  1 1            

  N T ( x , y ) f ( _{1 (} 4 ; 1 ) d f  ,  _{4 ; 1 )} 2   2 . 025 1 . 975

  

  10 10  10 

  4 2  is an evaluation with two exact decimals of the number

  N.

2. LOCAL EXTREMA

  p

  Let f : AIR IR, p1.

  Definition 3. The point a  A is called local extreme point

(relative) of function  such

f if there exists a neighbourhood V

  V a

  that „the increase” of function

  f: 

  E(x)=f(x) f(a)

  keeps a constant sigh on the set 

  A V; if E(x)  0, x  A 

  V

  we say that

   a is a point of local maximum, (or relative) for f and we

  write f =f(a); if

  max   

  E(x) 0, x A

  V

  the point

  a carries the name of local minimum (or relative) of the function f and we write f =f(a). min

  If

  V=A, and the point a is a local extreme (minimum or

  maximum) we say that

  a is a global extreme (or absolute) of the

  function f.

  We saw that, in the case of functions with a single real variable, Fermat’s theorem offers necessary conditions of extreme _ (if  I is an extreme point of the function

  a f : I IR  IR and f is

  differentiable at  or, equivalently,

  a, then f ' ( a ) d f=0). Fermat’s a theorem admits a generalization for functions with several variables.

  Theorem 2(of Fermat - necessary conditions for local p extreme). Let f : AIR IR, p1. If a  Å is an extreme point

  of the function f and f is differentiable in a, then 

  

f

  . d f=0, or, equivalently, (a) , k

  1 , p

  a

  

  x

_k

Proof. Because the point a belongs to the interior of the set

  

A, and it is a local extreme of the function f, it follows that there

  exists 

  r0 such that E(x)= f(x) f(a) keeps a constant sign for all p

   

  and the auxiliary

  x S(a,r)

A. Let s be an arbitrary versor from IR

  function   IR ,   .

  g : ( r , r ) g(t) f(a ts)

  Then      

  g ( t ) g ( ) f ( a ts ) f ( a ) E ( a ts ) keeps a constant sign for    

  t ( r,r), because a ts S(a,r); this means

  that the function

  g has a local extreme in t=0. Because g is

  differentiable in

  t=0 (being a composite function of differentiable

  

  f

  functions), from Fermat’s theorem it results that   ;

  g ' ( ) ( a )

  

  s

  

  f p

  therefore  , for any versor ; particularly, assigning ( a ) s from IR

  

  s p

   (vectors of the canonical basis in IR ) we obtain

  s=e k , k

  1 , p 

  f

    , or

  (a) , k 1 , p d a f=0.

  

  x _k

p

Definition 4. Let  Å . f : AIR IR, p1 and a

  If

  f is differentiable at a, and d a f=0, the point a is called stationary point of the function f.

  If a is stationary point for f, or f is not differentiable in a, we say that

  a is a critical point of the function f (for function f).

  Remark 1. From Fermat’s theorem results that the interior

  points which are local extremes of the differentiable function

  f are

  found, if existing, within the solutions of the system of p equations with

  p unknowns.

  

  f   .

  (x ,...x ) , k _{1 p} 1 , p

  

  x _k Remark 2. As in the case p=1, Fermat’s theorem gives

  necessary, but not sufficient, conditions of existence of the local extreme points.

  For example, the function

  2 f : IR IR, f(x,y)=xy

  has null partial derivatives in (0,0), but the origin is not a local extreme point of the function

  f (f being a quadratic undefined form).

  So (0,0) is a stationary point (therefore critical) for

  f which is not a

  local extreme. But, for the function

  g : [0,)[0,)  IR, g(x,y)=xy ,

  the origin (0,0) is a critical point (

  g is not differentiable in (0,0)) which is not stationary, but is a global extreme point.

  Remark 3. We have seen that from single variable functions the second order’s differential sign in a stationary point gives us information over the nature of this point. In the case of several variable functions the second order differential (which is a quadratic form) sign in a stationary point will determine the nature of the respective point. We will use the following lemmas:

  p Lemma 1. Let : IR  IR be a quadratic form. p

  (a) If is positively defined (i.e. \ {0})

    (x)0, for any x IR ₂ p there exists , for . m0 such that  ( x  ) m x x IR p

  (b) If is undefined then there exist  IR \ {0 such

   s , s

  1

  2

• that  (ts )  0 and  (ts )  0, for any t IR .

  1

  2

_p

Proof. (a). Let

  IR . Then S  x  | x 

  1 S is a closed and

  

 

  bounded set, therefore a compact set, and is a continuous

  

  function, therefore  is bounded and reaches its edge on

  S (theorem

  16, cap.5). Let on

  m be the minim of  S; of course m0 because 

  is positively defined and:

     (x) m, for any x S (1) p

  Now, let \ {0}; then

  x IR

   

  1

  1

  1 and   ;

  x  S x  ( x )

₂

   

  x x x

   

  p consequently, from (1) results that  , for any .

  ( x  ) m x x IR

  p

  (b) Let be the canonical basis from IR . As  is of

  B c

  quadratic form, there exists an orthogonal basis  has

  B in which

  canonical form. Let to

  T=(b ij ) be the transformation matrix from B c B.

  Then: _{2 2 2} (2)

   ( x ,..., x )   y   y  ...   y _{1 p 1 1 2 2 p p}

  where _p (3)

  y  b x , i   _{i ij j} 1 , 2 ,...,p   _j  ₁

  and  ,  ,...,  are the eigenvalues of the attached matrix of the

  1 2 p

  quadratic form  in canonical base. We can assume (with an eventual permutation of the indexes) that  0 and  0 (the form 

  1

  2

  being undefined). The systems:

  (4)

  y  _{1 i} 1 , y  , i   2 , 3 ,...,p 

  (5)

  y  _{2 i} 1 , y  , i   1 , 3 , 4 ,...,p  have, according to (3), the matrix

  respectively, a unique solution. Let , respectively

  s  ( a ,..., a ) _{1 1 p} be the solutions of the systems (4), respectively (5). s  ( b ,..., b ) _{2 1 p}

  Then, according to (2) we obtain: and (6)

   ( s )     ( s )    _{1 1 2}

₂

• But is a quadratic form, so for any

   t IR _{2 2 2} and .

   ( ts )  t  ( s )   t   ( ts )  t   _{1 1 1 2 2}

2 Lemma 2. Let  

  f C (A) and a

  A. Then there exists a

  function  

   : AIR such that for any x

  A: ₂

  1 ₂         ,

  

f ( x ) f ( a ) d f ( x a ) d f ( x a ) x a ( x )

_{a a}

  2 where     . lim ( x ) ( a ) _{x a} 

  Proof. Let x  A \ {a}. According to Taylor’s formula there

  exists (

  a,x) such that:

  1 ₂      (1)

  f ( x ) f ( a ) d f ( x a ) d f ( x a ) _a 

  2 Let _{2 2}  

  f f ù

     (2) _ij

  (x) (î ) (a), i ,j

  1 , p     _'' x x x x _{i j i j}

  Because  and  

  f C ( A ) a when x _{x x i j}

  a, from (2) results: ù

    (3) lim (x) , i,j _ij 1 , p _{x a}  Then: _{2 2 '' '' p}

  d f ( x  a )  d f ( x  a )  f (  )  f ( a ) ( x  a )( x  a )  a  x x x x  i i j j _{i j i j}

  



_{i , j}



₁

  so, from (2) we obtain: _{2 2 p}

   (4) d f ( x  a )  d f ( x  a )  ( x )( x  a )( x  a )

   a ij i i j j



_{i , j}



₁ Let 

  (a)0 and

  p

  2

    ,when 

  ( x )  ( x )( x  a )( x  a ) x _{2 ij i i j j} a.