The reduction of quadratic forms to the normal form with applications for production functions

Mathematical and Quantative Methods

The Reduction of Quadratic Forms to the Normal Form

with Applications for Production Functions

Catalin Angelo Ioan1, Gina Ioan2

Abstract: The article treats the reduction of quadratic forms to the normal form by Gauss's method taking in discussing various determinants whose behavior will determine its nature. Applications of this method are illustrated for the most common production functions: Cobb-Douglas and CES.

Keywords: quadratic form; production function; Cobb-Douglas; CES

JEL Classification: E17; E27

1. Introduction

Let consider the quadratic form: H:RnR, H(x)=



 n 1 j , i j i ijxx a x=(x1,...,xn)R n . The quadratic form H is called positive definite if H(x)>0 x0, negative definite if H(x)<0 x0, positive semi-definite if H(x)0 xRn and x0R

n -{0} such that H(x0)=0, negative semi-definite if H(x)0 xR n and x0R n -{0} such that H(x0)=0, semi-definite if x1,x2R n such that H(x1)H(x2)<0.

It is known that if a quadratic form is positive (negative) (semi) definite in a basis then it retains that character in any other basis.

We say that the quadratic form H has the normal form if there is a basis B of Rn

where H(x)=



 m 1 i 2 i iy b xB=(y1,...,yn), mn.

It follows from the above that, being given the normal form of H (whatever the process by which this is achieved), H is positive definite if and only if bi0, i=

1

Associate Professor, PhD, Danubius University of Galati, Faculty of Economic Sciences, Romania, Address: 3 Galati Blvd, Galati, Romania, Tel.: +40372 361 102, fax: +40372 361 290, Corresponding author: [email protected].

2

Assistant Professor, PhD in progress, Danubius University of Galati, Faculty of Economic Sciences, Romania, Address: 3 Galati Blvd, Galati, Romania, Tel.: +40372 361 102, fax: +40372 361 290, e-mail: [email protected].

n ,

1 , negative definite if and only if bi0, i= n1, , positive semi-definite if and only if mn and bi0, i= m1, , negative semi-definite if and only if mn and bi0,

i=1,m, semi-definite if and only if ij=1,n such that bi0, bj0.

Relative to bringing a quadratic form to the normal expression are essentially three big methods.

Jacobi’s Method

Considering the matrix associated to the quadratic form [H]=

 

aij _i,j1,nMn(R), let

i= ii 1 i i 1 11 a a a a     

, i=1,n - the principal diagonal determinants.

If i0 i= n1, then there is a basis B=f1,...,fn obtained from the canonical basis through a triangular matrix, such that the normal expression of H is:

2 n n 1 n 2 2 2 1 2 1 1 y ... y y 1 ) x ( H           _.

The method is limited by the fact that the determinants obtained from the first i rows and columns must be non-zero. If i= n1, such that k0 k=1,i1 (considering 0=1 we can assume that the condition is always satisfied) and i=0, then it will investigate all determinants of the form: i-1,p= pp 1 -i p p1 p 1 -i 1 -i 1 i 1 1 -i 1p 1 -i 1 11 a a ... a a a ... a ... ... ... ... a a ... a 

with pi. If such a determinant is non-null, then after

the change of variables (which is basically a renumbering of variables):

          i p p i k k x y x y p i, k n, 1, k , x y

the condition that i0 is satisfied.

From the Jacobi‟s method is obtained that if i0 i= n1, then H is positive

definite, and if

(-1)ii0 i= n1, then H is negative definite.

The essential drawback of Jacobi‟s method is that all determinants i must be non-null (regardless of any renumbering). The method also does not specify the nature

of quadratic form when i= n1, such that k=0 k= n1, , ki (obviously after possible renumbering).

The Eigenvalues Method

Considering the associated matrix of H, let the characteristic polynomial P()=det(A-In). It is shown that: P()=(-1)

n (n-1 n-1 +2 n-2 -...+(-1)nn) where k is the sum of diagonal minors of order k of the matrix A.

Considering the characteristic equation: P()=0, its roots are called the eigenvalues of the matrix A. For an eigenvalue , the vector vRn such that: Av=v is called eigenvector corresponding to .

It is shown that the eigenvalues of a symmetric matrix are real. Considering the basis B consisting of eigenvectors corresponding to the eigenvalues 1,...,n we get: [H]B=             n 1 ... 0 ... ... ... 0 ... from where: H(x)₁y₁2₂y2₂..._ny2_n.

The eigenvalues method appears, at first sight, much better to determine the nature of quadratic form in the sense that H is positive definite if and only if i0, i=1,n , negative definite if and only if i0, i= n1, , positive semi-definite if and only if mn and i0, i= m1, (therefore there are also null eigenvalues, but those non-null are strictly positive), negative semi-definite if and only if mn and bi0, i=

m ,

1 (therefore there are also null eigenvalues, but those non-null are strictly negative), semi-definite if and only if ij= n1, such that i0, j0 (so there are at least two eigenvalues of sign contrary).

This method presents also a key deficiency, consisting in the difficulty of solving the characteristic equation (of n-th degree).

The Gauss Method

The Gauss method identifies the terms of the form a_iix_i2 and builds a perfect square that contains all occurrences of the variable xi. The process is continued on the remaining quadratic form. If there is no term of the form a_iix_i2, then it identifies a mixed term: aijxixj with aij0. If no such term appears, the process ends. If so, it is considered a change of variable of the form: xi=yi+yj, xj=yi-yj obtaining new square terms and the process continues as above.

2. Preliminary Results

Be a square symmetric matrix A=

          nn 1 n n 1 11 a ... a ... ... ... a ... a Mn(R).

Let note, as above, k=

kk 1 k k 1 11 a a a a     

, k=1,n - the prinicipal diagonal determinants and define the appropriate determination of k board with the row i and the column

j as: k,ij= ij ik 1 i kj kk 1 k j 1 k 1 11 a a ... a a a ... a ... ... ... ... a a ... a .

It is noted that, due to the symmetry of the matrix A, we have: k,ij=k,ji. We also consider that: 0,ij=aij. We will note below:

k,=      k 1 k kk 1 k 1 k 1 11 ... a ... a ... ... ... ... a ... a where =



₁,...,_k



t, =



₁,...,_k



tRk, R k,=             k 1 k 1 k k kk 1 k 1 1 k 1 11 ... ... a ... a ... ... ... ... ... a ... a where =



₁,...,_k



t, =



₁,...,_k



t, =





t k 1,...,  , =



₁,...,_k



tRk, ,,,R

pq the minor of apq from the matrix

          kk 1 k k 1 11 a ... a ... ... ... a ... a

pq,rs the determinant of the matrix

          kk 1 k k 1 11 a ... a ... ... ... a ... a

obtained by deleting the rows p and q and the columns r and s. For k=2 we define _pq,rs=1.

It is noted that due to symmetry, we have: k,=k, and k,.

The proofs of the following two lemmas are absolutely trivial, following the Laplace development of appropriate determinants.

Lemma 2.1 k,=

 

k k 1 s , r rs s r 1 s r 1   



   Lemma 2.2

 

 

ps k k 1 s , p p p s p p s s p k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p , k 1 1 _{ }                                    _{ }          Lemma 2.3

Be the vectors =



₁,...,_k



t, =



₁,...,_k



t, =



₁,...,_k



t, =



₁,...,_k



tRk and ,,,R. Then: k,k,-k,k,-kk,=

 







     _{     }          k s r q p 1 s , r , q , p k rs , pq qr ps qs pr q q p p s r s r s r q p 1 Proof

From lemmas 2.1 and 2.2 it follows: k,k,-k,k,-kk,=

 

 

_      _{  }       _{    }





      k k 1 s , r rs s r 1 s r k k 1 v , u uv v u 1 v u 1 1

 

 

 

 

                                                                                              k ps k 1 s , p p p s p p s s p k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k k k 1 s , r rs r s 1 s r k k 1 v , u uv v u 1 r u 1 1 1 1

 





  



 

 

 

                                                       











                k 1 s , p ps k p p s s p k 1 s , p ps k p p s s p k s r q p 1 s , r , q , p rs , pq k s r s r q q p p s r q p k 1 s , r rs r s s r s r s r 1 s r k k 1 v , u , s , r uv rs v s s v r u v u s r 1 1 1 1 1

 





 

 

_                                                      







           k 1 s , p ps s p p s s p s p p p s p p s s p k k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k 1 v , u , s , r uv rs v s s v r u v u s r 1 1 1

 

 

                     

_



         k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k 1 s , r , q , p qs pr s r s r q p s r q p 1 1

 

 

 

 

                                         









                 k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k 1 s , r , p ps pr s r s r p p s r k q p 1 s , r , q , p qs pr s r s r q p s r q p k q p 1 s , r , q , p qs pr s r s r q p s r q p 1 1 1 1

 

 

 

 

 

                                                   











                     k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k s r 1 s , r , p ps pr s r s r p p s r k s r 1 s , r , p ps pr s r s r p p s r k p q 1 s , r , q , p ps qr s r s r p q s r q p k q p 1 s , r , q , p qs pr s r s r q p s r q p 1 1 1 1 1

 





 

 

 

                                              









                k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k r s 1 s , r , p pr ps r s r s p p s r k s r 1 s , r , p ps pr s r s r p p s r k q p 1 s , r , q , p ps qr p q qs pr q p s r s r s r q p 1 1 1 1

 





 

                          

_



          k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k q p 1 s , r , q , p ps qr p q qs pr q p s r s r s r q p 1 1

 





 





                                                              k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k s r q p 1 s , r , q , p ps qr p q qs pr q p s r s r s r q p k s r q p 1 s , r , q , p ps qr p q qs pr q p s r s r s r q p 1 1 1  





 





                                                              k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k r s q p 1 s , r , q , p pr qs p q qr ps q p r s r s s r q p k s r q p 1 s , r , q , p ps qr p q qs pr q p s r s r s r q p 1 1 1

 





 

                                    





          k k s r q p 1 s , r , q , p rs , pq s r s r q q p p s r q p k s r q p,q,r,s 1 p pr qs p q qr ps q p ps qr p q qs pr q p s r s r s r q p 1 1

 





 

                                       k k s r q p,q,r,s1 p rs , pq s r s r q q p p s r q p k s r q p,q,r,s1 p qr ps qs pr q q p p s r s r s r q p 1 1

 







     _{     }          k s r q p 1 s , r , q , p k rs , pq qr ps qs pr q q p p s r s r s r q p 1 Corollary 2.1

Be the vectors =



₁,...,_k



t, =



₁,...,_k



t, =



₁,...,_k



tRk and ,,,R. Then: k,k,-k,k,-kk,=

 







     _{     }          k s r q p 1 r , q , s , p k rs , pq qr ps pr qs q p q p s r s r r q s p 1 Proof

Lemma 2.4 k rs , pq qr ps pr qs      =0, pq, rs, p,q,r,s= k1, , k2. Proof Let P(k): _qs_pr_ps_qr_pq,rs_k=0, pq, rs, p,q,r,s= k1, . For k=2, let 2= 22 21 12 11 a a a a - symmetrical. We have: ₂₂₁₁₁₂₂₁_12,12₂ =a11a22-a - 122 (a11a22-a )=0. ₁₂2 For k=3, let 3= 33 32 31 23 22 21 13 12 11 a a a a a a a a a

- symmetrical. We have 9 variants: p=1, q=2, r=1, s=2; p=1, q=3, r=1, s=2; p=2, q=3, r=1, s=2; p=1, q=2, r=1, s=3; p=1, q=3, r=1, s=3; p=2, q=3, r=1, s=3; p=1, q=2, r=2, s=3; p=1, q=3, r=2, s=3; p=2, q=3, r=2, s=3

We will proof the equality for on variant, for the others doing analogously. Let, for example: p=1, q=2, r=1, s=3. Then:

3 13 , 12 21 13 11 23      =

















   



       33 2 12 11 2 23 22 2 13 13 23 12 33 22 11 23 23 13 33 12 13 22 23 12 2 23 33 22 13 12 23 11 a a a a a a a a a 2 a a a a a a a a a a a a a a a a a a a 0 a a a a a a a a a a a 2 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a 33 23 2 12 3 23 11 23 22 2 13 2 23 13 12 33 23 22 11 23 22 2 13 33 22 13 12 2 23 13 12 33 23 2 12 2 23 13 12 33 22 13 12 3 23 11 33 23 22 11             

Therefore, P(3) is true. Assuming P(i) true i=3,k, let P(k+1): 1 k rs , pq qr ps pr qs       =0, pq, rs, p,q,r,s=1,k1.

Let k-1=pq,rs obtained from k+1 by removing the rows p and q and the columns r and s.

Considering in Lemma 2.3: =the column r, = the column s, =the row p, = the row q of k+1, =apr, =aps, =aqr, =aqs it follows first:

_{k1,}=(-1)r+s+p+qk+1 k-1,=(-1) r+p_ sq k-1,=(-1)s+qrp k-1,=(-1) r+q_ ps k-1,=(-1) p+s_ qr

From Lemma 2.3, it follows:

(-1)r+s+p+q



_qs_pr_ps_qr_pq,rs_k1



=k-1,k-1,-k-1,k-1,-k-1k1,=

 







      _{     }          k s r q p 1 s , r , q , p 1 k rs , pq qr ps qs pr q q p p s r s r s r q p

1 =0 from the induction

hypothesis (where ij are appropriates minors of k-1. Corollary 2.2

Be the vectors =



₁,...,_k



t, =



₁,...,_k



t, =



₁,...,_k



t, =



₁,...,_k



tRk and ,,,R, k2. Then: k,k,-k,k,-kk,=0.

Proof

It follows from lemmas 2.3 and 2.4.

Lemma 2.5 ij , 1 k k j 1 k , k i 1 k , k 1 k ij , k           i,jk+2 k2.

Proof

For k=1 we will prove directly. We have therefore: ij , 1 k k j 1 k , k i 1 k , k 1 k ij , k           = ij 2 i 1 i j 2 22 21 j 1 12 11 11 j 2 21 j 1 11 i 2 21 i 1 11 22 21 12 11 ij 1 i j 1 11 a a a a a a a a a a a a a a a a a a a a a a a a a a   =



















    



       ij 2 12 j 2 i 2 11 i 1 22 j 1 j 2 i 1 12 j 1 i 2 12 ij 22 11 11 12 j 1 j 2 11 12 i 1 i 2 11 2 12 22 11 j 1 i 1 ij 11 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a 0 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a ij 2 12 11 j 2 i 2 2 11 i 1 22 j 1 11 j 2 i 1 12 11 j 1 i 2 12 11 ij 22 2 11 j 1 i 1 2 12 i 1 j 2 12 11 j 1 i 2 12 11 j 2 i 2 2 11 j 1 i 1 2 12 j 1 i 1 22 11 ij 2 12 11 ij 22 2 11              

From corollary 2.1, for =



a_1k1,...,_kk1



t, =



a1j,...,akj



t, =





t ik 1 i ,...,a a Rk, =ak+1 k+1, =ak+1 j, =ai k+1, =aij we have: ij , 1 k k j 1 k , k i 1 k , k 1 k ij , k           =

 







         _{     }  k s r q p 1 r , q , s , p k rs , pq qr ps pr qs qj pj 1 k q 1 k p is ir 1 k s 1 k r r q s p a a a a a a a a 1 .

From Lemma 2.4, we have _qs_pr_ps_qr_pq,rs_k=0, pq, rs, p,q,r,s= k1, , k2.

3. A New Approach of the Gauss Method

Suppose, first, that (after a possible renumbering) a110. We have:

H(x)=





    n 2 j , i j i ij n 2 j j j 1 1 2 1 11x 2x a x a x x a =









                            n 2 j , i j i ij 2 n 2 j j 11 j 1 11 2 n 2 j j 11 j 1 n 2 j j 11 j 1 1 2 1 11 x a x x a a a x a a x a a x 2 x a =







            n 2 j , i j i ij n 2 j , i j i j 1 i 1 11 2 n 2 j j j 1 1 11 11 x x a x x a a a 1 x a x a a 1 =







   n 2 j , i j i j 1 i 1 ij 11 11 2 1 11 x x a a a a a 1 y a 1

where we performed the change of variables: y1=



  n 2 j j j 1 1 11x a x a the others

remaining the same.

From the above it follows that if 10 then: H(x)=



     n 2 j , i j i ij , 1 1 2 1 1 x x 1 y 1 . Let P(k): H(x)=



              n 1 k j , i j i ij , k k 2 k k 1 k 2 2 2 1 2 1 1 x x 1 y 1 ... y 1 y 1 , i0 i= k1, Since P(1) is true, suppose P(k) true. If i=k1,n such that: k,ii0 then, after a possible renumbering, we assume: k,k+1 k+1=k+10.

Considering



     n 1 k j , i j i ij , k k x x 1 ) x ( ' H , we get:             

_

_

        n 2 k j , i j i ij , k n 2 k i i i 1 k , k 1 k 2 1 k 1 k k x x x x 2 x 1 ) x ( ' H =              





          n 2 k j , i j i 1 k ij , k n 2 k i i 1 k i 1 k , k 1 k 2 1 k k 1 k _{x 2x x x x =}                                                       n 2 k j , i j i 1 k ij , k n 2 k j , i j i 1 k j 1 k , k 1 k i 1 k , k 2 n 2 k i i 1 k i 1 k , k n 2 k i i 1 k i 1 k , k 1 k 2 1 k k 1 k _{x 2x x x xx xx} =

                                





             n 2 k j , i j i 1 k j 1 k , k 1 k i 1 k , k 1 k ij , k 2 n 2 k i i 1 k i 1 k , k 1 k k 1 k _{x x xx =}





_                  _{  }      



 



     n 2 k j , i j i j 1 k , k i 1 k , k 1 k ij , k 2 n 2 k i i i 1 k , k 1 k 1 k 1 k k x x x x 1 =



                  n 2 k j , i j i k j 1 k , k i 1 k , k 1 k ij , k 1 k 2 1 k 1 k k x x 1 y 1

where we performed the change of variable: yk+1=



        n 2 k i i i 1 k , k 1 k 1 k x x .

From Lemma 5, _k,ij_k₁_k,k_1i_k,k_1j_k_k_1,ij i,jk+2 k2 therefore:



            n 2 k j , i j i ij , 1 k 1 k 2 1 k 1 k k x x 1 y 1 ) x ( ' H . Therefore: H(x)=



                 n 2 k j , i j i ij , 1 k 1 k 2 1 k 1 k k 2 2 2 1 2 1 1 x x 1 y 1 ... y 1 y 1

and P(k+1) is also true.

Suppose now that 1,...,k0 and all k,ii=0 i=k1,n.

We have therefore: H(x)=



              n 1 k j , i j i ij , k k 2 k k 1 k 2 2 2 1 2 1 1 x x 1 y 1 ... y 1 y 1 . We have now two cases:

1. If _k,ij=0 i,j=k1,n then the algorithm ends and the normal form of H is:

H(x)= 2k k 1 k 2 2 2 1 2 1 1 y 1 ... y 1 y 1         _

Considering the matrix of changing of the canonical basis to a new basis: _BB

c

M =S -1

S=                                        1 ... 0 0 ... 0 0 ... ... ... ... ... ... ... 0 ... 1 0 ... 0 0 ... ... 0 0 ... ... ... ... ... ... ... ... ... 0 ... ... kn , 1 k 1 k k , 1 k k n 2 , 1 1 k 2 , 1 k 2 , 1 2 n 1 , 0 1 k 1 , 0 k 1 , 0 12 , 0 1

we have: x=M_BcBy from where H(x)=x

 

H x

c B t =yM

 

H B MBBy t B B t c c c =y

 

HBy t

In the new basis:

 

H B=                                   0 ... 0 0 ... 0 0 ... ... ... ... ... ... ... 0 ... 0 0 ... 0 0 0 ... 0 1 ... 0 0 ... ... ... ... ... ... ... 0 ... 0 0 ... 1 0 0 ... 0 0 ... 0 1 k 1 k 2 1 1

2. If pq such that k,pq0,kn-2, let the transformation:









                    q p q q p p x x 2 z x x 2 z with    0, 0, 0. We have therefore: xp= 2 z z_p _q  and xq= 2 z zp q  . After replacing, the term k,pqxpxq becomes:

2 z z 2 z zp q p q pq , k        =







p q



2 q 2 p pq , k z z z z 4     

and thus proceed as above.

We assume (again after a possible renumbering) that: k,k+1 k+20 therefore: xk+1= 2 z z_k1 _k2  and xk+2= 2 z z_k1 _k2 

. For the form:



    n 1 k j , i j i ij , k x x ) x ( ' H it

        

_

_

_

              n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 1 k j 1 k , k 2 k 1 k 2 k 1 k , k x x 2 x x 2 x x xx 2 ) x ( ' H









                    







                  n 3 k j , i j i ij , k n 3 k j j 2 k 1 k j 2 k , k n 3 k j j 2 k 1 k j 1 k , k 2 k 1 k 2 2 k 2 1 k 2 k 1 k , k x x x 2 z z 2 x 2 z z 2 z z z z 4 2















                                                    n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 1 k j 2 k , k n 3 k j j 2 k j 1 k , k n 3 k j j 1 k j 1 k , k 2 k 1 k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 1 k 2 k 1 k , k x x x z x z x z x z z z 2 z 2 z 2

Proceeding as above, it follows:





                                











                          n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 2 k j 1 k , k 2 2 k 2 k 1 k , k n 3 k j j j 1 k , k n 3 k j j j 2 k , k 2 k 2 k 1 k , k 1 k 2 1 k 2 k 1 k , k x x x z x z z 2 x x z 2 z z 2 ) x ( ' H























                                                                                                   n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 2 k j 1 k , k 2 2 k 2 k 1 k , k 2 n 3 k j j j 1 k , k n 3 k j j j 2 k , k 2 k 2 k 1 k , k 2 k 1 k , k 2 n 3 k j j j 1 k , k n 3 k j j j 2 k , k 2 k 2 k 1 k , k 2 k 1 k , k 1 k 2 k 1 k , k x x x z x z z 2 x x z 2 2 1 x x z 2 1 z 2

With the variable transformation:

yk+1=





_                   

_

_

            n 3 k j j j 1 k , k n 3 k j j j 2 k , k 2 k 2 k 1 k , k 2 k 1 k , k 1 k z x x 2 1 z we get:

                                                                     n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 2 k j 1 k , k 2 2 k 2 k 1 k , k 2 n 3 k j j j 1 k , k n 3 k j j j 2 k , k 2 k 2 k 1 k , k 2 k 1 k , k 2 1 k 2 k 1 k , k x x x z x z z 2 x x z 2 2 1 y 2 ) x ( ' H













                                               

















                                             n 3 k j , i j i ij , k n 3 k j j 2 k j 2 k , k n 3 k j j 2 k j 1 k , k 2 2 k 2 k 1 k , k n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k n 3 k j 2 k j j 1 k , k n 3 k j 2 k j j 2 k , k n 3 k j , i j i j 1 k , k i 1 k , k 2 k 1 k , k n 3 k j , i j i j 2 k , k i 2 k , k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x x z x z z 2 x x 1 z x 2 1 z x 2 1 x x 2 x x 2 z 8 y 2











          



                                               n 3 k j , i j i j 2 k , k i 1 k , k j 1 k , k i 1 k , k 2 j 2 k , k i 2 k , k 2 2 k 1 k , k ij , k 2 k 1 k , k n 3 k j 2 k j j 2 k , k j 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x 2 2 2 1 z x 2 z 8 y 2





_

_



















                                                                                        n 3 k j , i j i j 2 k , k i 1 k , k j 1 k , k i 1 k , k 2 j 2 k , k i 2 k , k 2 2 k 1 k , k ij , k 2 k 1 k , k 2 n 3 k j j j 2 k , k j 1 k , k 2 k 1 k , k 2 n 3 k j j j 2 k , k j 1 k , k 2 k 1 k , k 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x 2 2 2 1 x 2 1 x 2 z 8 y 2

Also, with the variable transformation:





_{ } 







 



    _{}   n 3 k j j j 2 k , k j 1 k , k 2 k 1 k , k 2 k 2 k x 2 z y it follows:







      



          



                n 3 k j , i j i j 1 k , k i 2 k , k j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x 3 2 2 1 y 8 y 2











              _{    }           n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x 2 1 y 8 y 2









                               





                    n 3 k i 2 i i 2 k , k i 1 k , k n j i,j k 3 i j i i 2 k , k j 1 k , k j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x 2 x x 2 y 8 y 2











                                  n j i 3 k j , i j i i 2 k , k j 1 k , k j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k 2 2 k 2 k 1 k , k 2 2 1 k 2 k 1 k , k x x 2 y 8 y 2

The form H becomes therefore:











                                           n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k k 2 2 k k 2 k 1 k , k 2 2 1 k k 2 k 1 k , k 2 k k 1 k 2 2 2 1 2 1 1 x x 2 1 y 8 y 2 y 1 ... y 1 y 1 ) x ( H

In particular, for ==== 2sign

 

_k _k it follows:

 

 







                                      n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k k 2 2 k 2 k 1 k , k k 2 1 k 2 k 1 k , k k 2 k k 1 k 2 2 2 1 2 1 1 x x 2 1 y sign y sign y 1 ... y 1 y 1 ) x ( H As:

 

 

 

 

                        2 k k k 1 k k k 2 k 2 k k k 1 k k k 1 k x sign 2 1 x sign 2 1 z x sign 2 1 x sign 2 1 z it follows: yk+1=

 



 



 

_{ } 







 



            n 3 k j j j 1 k , k j 2 k , k 2 k 1 k , k k k 2 k k k 1 k k k x sign 2 1 x sign 2 1 x sign 2 1

 



 



 

_{ } 



 



             n 3 k j j j 1 k , k j 2 k , k 2 k 1 k , k k k 2 k k k 1 k k k 2 k x sign 2 1 x sign 2 1 x sign 2 1 y

Considering the matrix of changing of the canonical basis to a new basis: _BB c M =S -1 where: S=

 

 

 

 

 

 

 

 

                                                                                                                          nn 3 k n n 3 k 3 k 3 k 2 k 1 k , k k k n 1 k , k n 2 k , k 2 k 1 k , k k k 3 k 1 k , k 3 k 2 k , k k k k k 2 k 1 k , k k k n 1 k , k n 2 k , k 2 k 1 k , k k k 3 k 1 k , k 3 k 2 k , k k k k k kn , 1 k 3 k k , 1 k 2 k k , 1 k 1 k k , 1 k k 2n , 1 3 k 2 , 1 2 k 2 , 1 1 k 2 , 1 k 2 , 1 2 1n , 0 3 k 1 , 0 2 k 1 , 0 1 k 1 , 0 k 1 , 0 12 , 0 1 b ... b 0 0 0 ... 0 0 ... ... ... ... ... ... ... ... ... b ... b 0 0 0 ... 0 0 sign 2 ... sign 2 sign 2 1 sign 2 1 0 ... 0 0 sign 2 ... sign 2 sign 2 1 sign 2 1 0 ... 0 0 ... ... 0 0 ... ... ... ... ... ... ... ... ... ... ... 0 ... ...

we have: x=M_BcBy from where H(x)=x

 

HB_cx t =yM

 

H B MBBy t B B t c c c =y

 

HBy t . In the new basis:

 B

H =

 

 

                                                     nn 3 k n n 3 k 3 k 3 k 2 k 1 k , k k 2 k 1 k , k k k 1 k 2 1 1 c ... c 0 0 0 ... 0 0 ... ... ... ... ... ... ... ... ... c ... c 0 0 0 ... 0 0 0 ... 0 sign 0 0 ... 0 0 0 ... 0 0 sign 0 ... 0 0 0 ... 0 0 0 1 ... 0 0 ... ... ... ... ... ... ... ... ... 0 ... 0 0 0 0 ... 1 0 0 ... 0 0 0 0 ... 0 1 where cij= 2 k 1 k , k k j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2              , i,j=k3,n.

Theorem 3.1

Given the quadratic form H:RnR, H(x)=



 n 1 j , i j i ijxx a x=(x1,...,xn)R n we have: 1. If 1,...,n0 then the normal form of H is:

H(x)= 2n n 1 n 2 2 2 1 2 1 1 y 1 ... y 1 y 1         _ where: yk=



      n 1 k i i ki , 1 k k kx x , k=1,n, 0=1.

2. If 1,...,k0, k,ij=0 i,j=k1,n then the normal form of H is:

H(x)= 2_k k 1 k 2 2 2 1 2 1 1 y 1 ... y 1 y 1         _ where: yk=



      n 1 k i i ki , 1 k k kx x , k= n1, , 0=1.

3. If 1,...,k0, k,ii=0 i=k1,n and ij=k1,n such that k,ij0 then the normal form of H is:

 

y sign

 

y H' (x') sign y 1 ... y 1 y 1 ) x ( H n 2 j ij , k k 2 i ij , k k 2 k k 1 k 2 2 2 1 2 1 1                  , x'



x_k3,...,x_n



where yp=



      n 1 p i i pi , 1 p p px x , p= k1, , 0=1, yk+1=

 

_{ } 







 



       n 3 k j j j 1 k , k j 2 k , k 2 k 1 k , k k k 1 k x sign 2 1 z ,

 

_{ } 







 



         n 3 k j j j 1 k , k j 2 k , k 2 k 1 k , k k k 2 k 2 k x sign 2 1 z y

and H '_n is the normal form of







                n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k k x x 2 1 ) x ( ' H .

Corollary 3.1

Given the quadratic form H:RnR, H(x)=



 n 1 j , i j i ijxx a x=(x1,...,xn)R n it follows (after a possible renumbering):

1. The quadratic form is positive definite if and only if: 1,...,n0;

2. The quadratic form is negative definite if and only if: (-1)kk0 k=1,n; 3. The quadratic form is positive semi-definite if and only if: k=1,n such that: 1,...,k0 and k,ij=0 i,j=k1,n;

4. The quadratic form is negative semi-definite if and only if: k= n1, such that: (-1)ii0, i= k1, and k,ij=0 i,j=k1,n;

5. The quadratic form is semi-definite if and only if: a. 1,...,n0, but do not meet 1 or 2;

b. k= n1, such that 1,...,k0 not meeting 1 or 2 and i,j=k1,n: k,ij=0; c. k= n1, such that 1,...,k0, k,ii=0 i=k1,n and i,j=k1,n such that

k,ij0.

We ask now the question what happens to the coefficient Cij of xixj from H . 'n We have Cij=k,ijk,k1k22k,k1ik,k2j.

How Cij=Cji it follows: k,k1ik,k2j= k,k1jk,k2i.

Noting ij the determinant obtaind by board k with the columns k+2 and j and the rows k+1 and i, from corollary 2, it follows:

(1) _k,ij_k,k2k1_k,ik1_{k, jk2}_k_ij ik+2, jk+3 In particular, for i=j, we have:

(2) _k,ii_k,k_2k₁_k,ik₁_k,ik₂_k_ii ik+3

How _k,ii=0, it follows: (3) _k,ik1_k,ik2_k_ii From Lemma 5:

(4) _k,ij_k1_k,k1i_k,k1j _k_k1,ij i,jk+2 k2 How k+1=k,k+1 k+1=0, we get from (4):

(5) _k,k_1i_k,k_1j _k_k_1,ij

In particular, for i=j: (6) 2_k,k_1i _k_k_1,ii For i=k+2 în (6): (7) 2_k,k1k2_k_{k1,k2k2}_k_k2 therefore implicitly k+20. From (5), for j=k+2: (8) _k,k_1i_k,k_1k₂_k_k_1,k_2i

We have now, from (3) and (5):

(9) k 1,ij jj 2 k 2 j 1 k , k j 2 k , k i 1 k , k       _{   } From (6) and (9): jj ij , 1 k k jj , 1 k j 2 k , k i 1 k , k       _{   } therefore: (10) jj , 1 k jj ij , 1 k k j 2 k , k i 1 k , k     _        if _k1,jj0

If _k_1,jj=0 then, from (6) it follows: k,k1j=0, and from (5) it follows: k1,ij=0. Also, from (3) it follows: _jj=0.

Therefore, if _k1,jj0 then: Cij=_k,ij_k,k_1k₂2_k,k_1i_k,k_2j= jj , 1 k jj ij , 1 k k 2 k 1 k , k jj , 1 k ij , k 2              and, in particular: Cjj=2kjj

and if _k_1,jj=0 then: Cij=_k,ij_k,k_1k₂ and, in particular: Cjj=0. Therefore:







                n 3 k j , i j i j 2 k , k i 1 k , k 2 k 1 k , k ij , k 2 k 1 k , k k x x 2 1 ) x ( ' H =                       





                 n 0 3 k j , i j i 2 k 1 k , k ij , k n 0 3 k j , i j i jj , 1 k jj ij , 1 k k 2 k 1 k , k jj , 1 k ij , k 2 k 1 k , k k jj , 1 k jj , 1 k x x x x 2 1 =                   





            n 0 3 k j , i j i jj , 1 k jj ij , 1 k k n 3 k j , i j i 2 k 1 k , k ij , k 2 k 1 k , k k jj , 1 k x x x x 2 =                 





          n 0 3 k j , i j i jj , 1 k jj ij , 1 k 2 k 1 k , k n 3 k j , i j i ij , k k jj , 1 k x x 1 x x 1 2 4. Bordered Matrices

Let the bordered matrix: HB=

                nn 2 n 1 n n n 2 22 21 2 n 1 12 11 1 n 2 1 a ... a a b ... ... ... ... ... a ... a a b a ... a a b b ... b b 0 and B_k=                 kk 2 k 1 k k k 2 22 21 2 k 1 12 11 1 k 2 1 a ... a a b ... ... ... ... ... a ... a a b a ... a a b b ... b b 0 , k= n1, .

From Lemma 2.1, we have: B_k=



 

   _  k 1 s , r rs s r 1 s r b b

1 where rs is the appropriate minor of aij from the matrix of H(x)=



 n 1 j , i j i ijxx a .

Let consider the quadratic form: HB_k(b)=



 

  _   k 1 s , r s r rs s r b b 1 and Bb_k =

 

 

k 1 _{k1 kk} k 1 1 k 11 ... 1 ... ... ... 1 ...         . Since

 

 

                            kk 1 k 1 k k 1 1 k 11 kk 1 k k 1 11 ... 1 ... ... ... 1 ... a ... a ... ... ... a ... a =kIk it follows: kBbk = k k  from where: Bb_k =k_k1 if k0.

If 1,...,k0 then PMk(R), invertible, such that:

Pt           kk 1 k k 1 11 a ... a ... ... ... a ... a P=                          1 k k 2 1 1 1 ... 0 0 ... ... ... ... 0 ... 1 0 0 ... 0 1

We obtain after few computations:

 

 

 

_                                       k 1 k 2 1 1 k t 1 kk 1 k 1 k k 1 1 k 11 1 ... 0 0 ... ... ... ... 0 ... 0 0 ... 0 P ... 1 ... ... ... 1 ... P

Therefore, for 1,...,k0 the normal form of HB_k(b) is: )

b (

HB_k =_k



₁c₁2 ₁₂c₂2..._k1_kc_k2



where c=(c1,...,ck) are the coordinates of b in the new basis.

As a result of these relations, it follows that if: 1,...,k0 then:  B_k 0. If (-1) i_

i0

i= k1, then: sign(B_k)=sign(k) therefore (-1) k B

k

Analogously the things happen if 1,...,p0 and _p,ij=0 i,j=p1,k. In this case, the normal form of HB_k(b) is:

HB(b)=_p



₁c₁2 ₁₂c₂2..._p1_pc_p2



where c=(c1,...,ck) are the coordinates of b in the new basis.

As a result of these relations, it follows that if: 1,...,p0 and p,ij=0 i,j=p1,k then:  B_k 0. If (-1)ii0 i=1,p then: sign(B_k)=sign(k) therefore (-1)k B_k 0. If H is semi-definite, the problem is more complicated. So, in the case of expression B_k=



 

   _  k 1 s , r rs s r 1 s r b b

1 if B_k is semi-definite then b‟, b”Rk such that B_k(b')0, B_k(b")0. Difficult issue arises where for another determinant B_s, sk, which signs depending on the values of b is not strictly determined, the values b‟, b”Rk are not necessarily the same as in the case of B_k.

5. The Convexity of the Functions

We present in this section some of the remarkable results of concavity or quasi-concavity of functions.

Definition 5.1 A subset DRn is called convex if x,yD [0,1] x+(1-)yD.

From definition, it follows that D is convex if and only if for any two points x,yD, the segment [x,y]D.

Definition 5.2 A function f:DRnR is called convex if x,yD [0,1] follows

f(x+(1-)y)f(x)+(1-)f(y).

Definition 5.3 A function f:DRnR is called concave if x,yD [0,1] follows

f(x+(1-)y)f(x)+(1-)f(y).

From the definitions, it follows that a function is convex (concave) if and only if for any segment [x,y]D the values of the restriction function is under (above) or on the chord determined by the values of the function on the extremities of its.

Definition 5.4 A function f:DRnR is called strictly convex if x,yD (0,1) follows

f(x+(1-)y)f(x)+(1-)f(y).

Definition 5.5 A function f:DRnR is called strictly concave if x,yD (0,1) follows

f(x+(1-)y)f(x)+(1-)f(y).

From these definitions, it follows that a function is strictly convex (concave) if and only if for any segment [x,y]D the values of the restriction function is under (above) the chord determined by the values of the function on the extremities of its.

Definition 5.6 A function f:DRnR, D – convex, is called quasiconvex if x,yD [0,1] then: f(x+(1-)y)max(f(x),f(y)).

Definition 5.7 A function f:DRnR, D – convex, is called quasiconcave if x,yD [0,1] then: f(x+(1-)y)min(f(x),f(y)).

From the definitions, it follows that a function is quasiconvex (quasiconcave) if and only if for any segment [x,y]D the values of the restriction function is under (above) the maximum (minimum) level registered by the function at the ends.

Definition 5.8 A function f:DRnR, D – convex, is called strictlyquasiconvex

if xyD (0,1) then: f(x+(1-)y)max(f(x),f(y)).

Definition 5.9 A function f:DRnR, D – convex, is called strictlyquasiconcave

if xyD (0,1) then: f(x+(1-)y)min(f(x),f(y)).

From the definitions, it follows that a function is strictly quasiconvex (quasiconcave) if and only if for any segment [x,y]D the values of the restriction function is strictly under (above) the maximum (minimum) level registered by the function at the ends.

Theorem 5.1 If A function f:DRnR, D – convex, is quasiconvex (quasiconcave, convex, concave) then –f is quasiconcave (quasiconvex, concave, convex).

After this theorem, where not explicitly stated, we state the results only for concave functions, ie quasi-concave.

Theorem 5.2 A function f:DRnR, D – convex, is quasiconcave (quasiconvex) if and only if

f-1[a,) (f-1(-,a]) is convex aR.

Theorem 5.3 If a function f:DRnR, D – convex, is concave then it is quasiconcave.

Theorem 5.4 If a function f:DRnR, D – convex, is quasiconcave then f is quasiconcave 0.

Theorem 5.5 If the functions fk:DRnR, D – convex, k= m1, , are quasiconvex then pi0, i= m1, the function f=max(p1f1,...,pmfm) is also quasiconvex.

Theorem 5.6 If the functions fk:DR n_

R, D – convex, k=1,m, are quasiconcave then pi0, i= m1, the function f=min(p1f1,...,pmfm) is also quasiconcave.

Theorem 5.7 If the function f:DRnR, D – convex, is quasiconvex (quasiconcave), and g:RR is increasing, the function gf:DR is quasiconvex

(quasiconcave).

Theorem 5.8 If the function f:DRnR, D – convex, is of class C1(D) then it is concave (strictly concave) if and only if:







       n 1 i i i i y x ) y ( x f ) ( ) y ( f ) x ( f x,yD

Theorem 5.9 If the function f:DRnR, D – convex, is of class C1(D) then it is convex (strictly convex) if and only if:







       n 1 i i i i y x ) y ( x f ) ( ) y ( f ) x ( f x,yD

Theorem 5.10 If the function f:DRnR, D – convex, is of class C1(D) then it is quasiconcave (strictly quasiconcave) if and only if:

f(x)f(y)  (y)



x y



( )0 x f n 1 i i i i     



 x,yD

Theorem 5.11 If the function f:DRnR, D – convex, is of class C1(D) then it is quasiconvex (strictly quasiconvex) if and only if:

f(x)f(y)  (x)



x y



( )0 x f n 1 i i i i     



 x,yD

Definition 5.6 A function f:DRnR, D – convex, fC1(D) is called

pseudoconvex if it is quasiconcave and f(x)f(y)  (y)



x y



0 x f n 1 i i i i    



 x,yD.

Definition 5.7 A function f:DRnR, D – convex, fC1(D) is called pseudo-concave if it is quasiconvex and f(x)f(y)  (x)



x y



0

x f n 1 i i i i    



 x,yD. Suppose, in what follows, that f:DRnR, D – convex, is of class C2(D). Let x0D. From Taylor series expansion:





_



 









                n 1 j , i j 0 j i 0 i 0 0 j i 2 n 1 i i 0 i 0 i 0 (x x x )x x x x x x f 2 1 x x ) x ( x f ) x ( f ) x ( f , (0,1) or otherwise, for x=x0+h:





             n 1 j , i j i 0 j i 2 n 1 i i 0 i 0 0 (x h)hh x x f 2 1 h ) x ( x f ) x ( f ) h x ( f , (0,1)

We can write this:





             n 1 j , i j i 0 j i 2 n 1 i i 0 i 0 0 (x h)h h x x f 2 1 h ) x ( x f ) x ( f ) h x ( f , (0,1)

From Theorem 5.8 it follows that f is concave (strictly concave) if and only if



      n 1 j , i j i 0 j i 2 h h ) h x ( x x f 0 (0).

Like a conclusion, if d2f is negative semi-definite then f is concave.

Conversely, if f is concave, suppose that d2f is not negative-semi-definite. In this case, x‟D such that:



   n 1 j , i j i j i 2 h h ) ' x ( x x f _

0. Because the function f is of class C2(D) it follows that VV(x‟) such that:



   n 1 j , i j i j i 2 h h ) x ( x x f 0 xV. Let r0 such that the n-sphere of center x‟ and radius r: B(x‟,r)={xRn xx' r}V. Let now xB(x‟,r) and h=x-x‟. We have:





            n 1 j , i j i j i 2 n 1 i i i h h ) h ' x ( x x f 2 1 h ) ' x ( x f ) ' x ( f ) x ( f , (0,1)

Because x'hx'  h  h  h  xx' r it follows



      n 1 j , i j i j i 2 h h ) h ' x ( x x f 0 therefore:



     n 1 i i i h ) ' x ( x f ) ' x ( f ) x ( f 0 which

contradicts the fact that the function is concave.

The proof is analogous in the case of convexity. Therefore:

Theorem 5.12 If the function f:DRnR, D – convex, is of class C2(D) then it is concave (convex) if and only if d2f is negative (positive) semi-definite.

Suppose now that d2f is negative definite. In this case, we have:



      n 1 j , i j i 0 j i 2 h h ) h x ( x x f _ 0, (0,1) therefore:



      n 1 i i 0 i 0 0 (x )h x f ) x ( f ) h x ( f

0. Therefore, the function is strictly concave. Analogously is shown for strictly convex functions.

Theorem 5.13 If the function f:DRnR, D – convex, is of class C2(D) then if d2f is negative (positive) definite, the function is strictly concave (strictly convex). The reciprocal question is: if f is strictly concave then d2f is defined negatively? The answer is unfortunately negative, meaning that d2f is negative semi-definite. Now consider a function f:DR+

n_

R, D – convex, fC2(D), the bordered hessian matrix: ) f ( HB =                                                               2 n 2 2 n 2 1 n 2 n n 2 2 2 2 2 1 2 2 2 n 1 2 2 1 2 2 1 2 1 n 2 1 x f ... x x f x x f x f ... ... ... ... ... x x f ... x f x x f x f x x f ... x x f x f x f x f ... x f x f 0

B k  = 2 k 2 2 k 2 1 k 2 k k 2 2 2 2 2 1 2 2 2 k 1 2 2 1 2 2 1 2 1 k 2 1 x f ... x x f x x f x f ... ... ... ... ... x x f ... x f x x f x f x x f ... x x f x f x f x f ... x f x f 0                                     , k=1,n

Theorem 5.14 If the function f:DR+ n_

R, D – convex, fC2(D) is quasiconcave then:  B₁ 0,  B₂ 0,  B₃ 0,... (the determinants signs being alternate).

Theorem 5.15 In order that the function f:DR+ n_

R, D – convex, fC2(D) be quasiconcave is sufficient that: B₁<0, B₂>0, B₃<0,... (the determinants signs being alternate).

Theorem 5.16 If the function f:DR+ n_

R, D – convex, fC2(D) is quasiconvex then:  B₁ 0,  B₂ 0,  B₃ 0,...,  B_n 0.

Theorem 5.17 In order that the function f:DR+nR, D – convex, fC2(D) be quasiconvex is sufficient that: B₁<0, B₂<0, B₃<0,..., B_n<0.

Remark 5.2 From Section 3, we have seen that if f is concave (convex, strictly concave, strictly convex) then (-1)kk0 (k0, (-1)

k_

k0, k0). From Section 4, it follows that the function is quasiconcave (quasiconvex, strictly quasiconcave, strictly quasiconvex).

6. The Convexity Analysis of Production Functions

6.1. The Cobb-Douglas Function

The Cobb-Douglas function has the following expression:

f:DRn-{0}R+, (x1,...,xn)f(x1,...,xn)=Ax₁1...x_nnR+(x1,...,xn)D, AR+,

1,...,n0

Computing the partial derivatives of first and second order, we get:

i i n 1 i 1 i x x f x ... x ... Ax ' f 1 i n i        _ i= n1,

j i j i n 1 j 1 i 1 j i x x x x f x ... x ... x ... Ax " f 1 i j n j i            _ ij=1,n







₂



i i i n 2 i 1 i i x x x f 1 x ... x ... Ax 1 " f 1 i n i i            _ i=1,n The Hessian matrix is:

Hf=









                        2 n n n n 1 n 1 n 1 n 1 2 1 1 1 x f 1 ... x x f ... ... ... x x f ... x f 1 We have now: k=

 

      _{ }  

_

_

      k 1 i i k 1 i i 2 k k 2 k 1 k k 1 x ... x A 1 1 k _{, k= n1, .} k,ij=

 

       k 1 2 k 3 1 k j 3 1 k i 2 1 k 1 1 k k 1 i i j i k _{A x} 1 _...x i _...x j _{...x k} 1                  _   

_

, k=1,n, ij, i,jk+1

We note first that k,ij0, k= n1, , ij, i,jk+1.

Because i0, i= n1, it follows: sign(k)=

 



     _{ } 

_

 k 1 i i k 1 1 sign .

We get therefore that:



 

      _{ } 

_

 k 1 i i k 1 1

sign 0, k= n1, implies that f is strictly convex. We have however, for k=1: 1-10, and for k=2: 1-1-20 therefore: 11, 1+21 which conflicts with i0, i= n1, . Therefore, the Cobb-Douglas function cannot be strictly convex.        _

_

_  k 1 i i 1

k=1,n (after a possible renumbering), k=even such that:



   k 1 i i 1 0 or k,p= n ,

1 , k,p=odd such that



   k 1 i i 1 0 and



   p 1 i i

1 0 then f has a saddle point; k=1,n (after a possible renumbering) such that:



   p 1 i i 1 0 p=1,k, but



   s 1 i i

1 =0 s=k1,n (this thing, because the fact that i0 cannot occur only for



   n 1 i i 1 =0) then: oif

 

      _{ } 

_

 k 1 i i k 1 1

sign 0, k=1,n1 implies the fact that f is convex. In this case, for k=1: 1-10 therefore 11, the equality



  n 1 i i=1 cannot occur; oif       _

_

_  k 1 i i 1

sign 0, k=1,n1 implies the fact that f is concave. In particular, for the Cobb-Douglas function: f(x1,...,xn)=Ax₁1x₂2

 

, 1,20, we have:

1+21implies the fact that f is strictly concave;

1+21 implies the fact that f has saddle points, therefore it is not convex and not concave;

1+2=1 implies the fact that f is concave. 6.2. The CES Function

The CES function has the following expression: f:DRn_-{0}R+, (x1,...,xn)f(x1,...,xn)=   _       

_

1 n 1 i i ix R+(x1,...,xn)D, ,1...,n0, 0,1,



  n 1 i i=1





                      _n 1 k k k 1 i i 1 1 n 1 k k k 1 i i x x f x x x ' f i i=1,n









₂ n 1 k k k 1 j 1 i j i 2 1 n 1 k k k 1 j 1 i j i x x x f x x 1 x x x 1 " f j i                         





              _ ij=1,n









₂ n 1 k k k i i n 1 k k k 2 i i i i n 1 k k k 2 1 n 1 k k k 2 i i x x x f x x x 1 x x x x 1 " f _{i i}              _{ }           _{ }           









                i= n1,

The Hessian matrix is:

Hf=   _{   }                                                                _{ }                                                     _{ }                                                     _{ }                                                                         2 n 1 s s s n n n 1 s s s 2 n n 2 n 1 s s s 1 n 1 2 n 2 2 n 1 s s s 1 n 1 1 n 1 2 n 1 s s s 1 n 1 2 n 2 2 n 1 s s s 2 2 n 1 s s s 2 2 2 2 n 1 s s s 1 2 1 1 2 1 2 n 1 s s s 1 n 1 1 n 1 2 n 1 s s s 1 2 1 1 2 1 2 n 1 s s s 1 1 n 1 s s s 2 1 1 x f x x x 1 ... x f x x 1 x f x x 1 ... ... ... ... x f x x 1 ... x f x x x 1 x f x x 1 x f x x 1 ... x f x x 1 x f x x x 1 We have now: k=





1 k n 1 s s s n 1 k s s s k 1 s 2 s k 1 s s k k x x x 1 f                     









, k= n1, (where the last sum in the numerator is 0





2 k n 1 s s s 1 k 1 j 1 i 2 k 1 s i 1 k k 1 s s j i ij , k x f x x x 1                                  







0, k=1,n, ij, i,jk+1

If 1 then k0, k=1,n1 and n=0. In this case, the function is convex (non strictly).