Chapter 8
TAYLOR’S FORMULA. EXTREMES
We will extend in this chapter Taylor’s formula to scalar functions with many variables of the form f : A IRp IR. We remind that in the case p=1, if A is an interval, fCn+1(A) and a,xA, then it exists between a and x so that:
n
k
n k
a d f x a
n a x f d k x
f
0
1
) ( )! 1 (
1 ) ( ! 1 )
( (*)
We will show that even for p1, in certain conditions, a similar formula exists, so we can approximate in a neighbourhood V
V
a the function f with Taylor’s polynomial function Tn (of n degree):
n
k
k a
n d f x a
k x
T x f
0
) ( ! 1 )
( )
( .
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
Let f : A IRp IR be a real function of p real variables,
p1.
Definition 1. If f is differentiable n times in the point aA
then the polynomial function Tn: IRp IR
A x
( ),
! 1 ... ) ( ! 1
1 ) ( )
( d f x a
n a
x f d a f x
T n
a a
is called Taylor’s polynomial function of degree n associated to the function f in the point a; denoting da0 f f(a), then we can write
n
k
k a
n d f x a
k x
T
0
) ( ! 1 )
( .
The function:
Rn: A IR, Rn(x)=f(x)Tn(x), xA
is called remainderof order n, and the relationship
f(x)= Tn(x)+ Rn(x), xA
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
) ( ...
) (
1 1
a f dx x dx
x f
d
k
p p k
a
;
if a=(a1,…,ap), x=(x1,…,xp), then:
) ( )
( ... )
( ) (
) (
1 1
1 f a
x a x x
a x a
x f d
k
p p p k
a
.
In order to extend the formula (*) valid on the interval A for the case
p1 we need two new notions: the one of segment and the other of convex set.
Definition 2. Let x,y IRp. We call a segment which binds the points x and y the set:
x,y
xt(y-x)IRp t
0,1
.The set
p
pI I
x) -t(y
x R R
0,1
,y t
x
is called an open segment. If for any x,yAIRp, [x,y] A, we say that A is a convex set.
Theorem 1(Taylor’s formula). Let AIRp be a convex set,
) variables and the preceding Remark we obtain:
))
Therefore,
a)
As a consequence:
)
and Taylor’s formula is proved.
Remark 1. For n=0, from Taylor’s formula we can write
is called Mac-Laurin’s formula. equation:
S : z = f(x,y), (x,y)A.
The approximation of order 0:
U
M(x,y,f(x,y)) is replaced with the point M0(x,y,f(a,b)) which belongs to a part of the parallel plane with xOy of equation:
S0 : z = T0(a,b) = f(a,b), (x,y)A,
The approximation of order 1:
U , (x,y) f(x-a,y-b) d
f(a,b) (x,y)
T
f(x,y) 1 (a,b)
is more precise, because it presumes the substitution of the point M
with the point M1(x,y,T1(x,y)) from the part of the plane tangent in the point P(a,b,f(a,b)) on the surface S:
S1 : z = T1(x,y), (x,y)A. (see Fig.2.).
The approximation of second order:
U , (x,y) f(x-a,y-b) d
f(x-a,y-b) d
f(a,b) (x,y)
T
f(x,y) (a,b) (a,b)
2 2
2 1
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
M2(x,y,T2(x,y)) belonging to the surface of equation:
S2 : z = T2(x,y), (x,y)A,
surface tangent in P(a,b,f(a,b)) on the surface S and which has as plane tangent in P the plane S1(see Fig.3.).
M0
M z
y
x
0
z=T0(x,y)
(a,b,f(a,b))
U
(a,b,0)
Example 1. Expand the polynomial function
f(x,y,z) = x3+y3+xyzz2+xy+xz+1
in powers of x1, y+1 and z.
M
M1 S1
z
y
x
0
z=T1(x,y)
(a,b,f(a,b))
(a,b,0)
Fig.2.
S2 z
y
x
0
z=T2(x,y)
(a,b,f(a,b))
(a,b,0) A
Solution. We will use Taylor’s development of the function f
therefore:
vector from the linear space IR3[x,y,z] of the polynomials with three variables of degree smaller or equal with three formulated in the
Example 2. Using Taylor’s formula of order three compute the approximate value of the number (0.9)2.1.
' ( , ) ' ( , )
,with certitude that the approximation of the first order:
975
2. LOCAL EXTREMA
Let f : AIRpIR, p1.
Definition 3. The point aA is called local extreme point (relative) of function f if there exists a neighbourhood V
Va
such that „the increase” of function f:E(x)=f(x)f(a)
keeps a constant sigh on the set AV; if
E(x)0, xAV
we say that a is a point of local maximum, (or relative) for f and we write fmax=f(a); if
E(x)0, xAV
the point a carries the name of local minimum (or relative) of the function f and we write fmin=f(a).
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 a
I is an extreme point of the function f : I IR IR and f is differentiable at a, then f'(a)0 or, equivalently, daf=0). Fermat’s theorem admits a generalization for functions with several variables.
Theorem 2(of Fermat - necessary conditions for local extreme). Let f : AIRpIR, p1. If aÅ is an extreme point of the function f and f is differentiable in a, then
daf=0, or, equivalently, (a) , k , p
x f
k
1
0
.
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 r0 such that E(x)= f(x)f(a) keeps a constant sign for all
xS(a,r)A. Let s be an arbitrary versor from IRp and the auxiliary function
ts) f(a g(t) r
r
g:( , )IR, .
Then
) ( ) ( ) ( ) 0 ( )
(t g f a ts f a E a ts
keeps a constant sign for t(r,r), because atsS(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 functions), from Fermat’s theorem it results that '(0) ( )0
a
s f
g ;
therefore ( )0
a s f
, for any versor s from IRp; particularly, assigning
s=ek, k 1, p (vectors of the canonical basis in IRp) we obtain
, p , k (a) x
f
k
1 0
, or daf=0.
Definition 4. Let f : AIRpIR, p1 and aÅ.
If f is differentiable at a, and daf=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.
, p , k )
,...x (x x
f
p k
1 0
1
.
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
f : IR2IR, 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 ,
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:
Lemma 1. Let : IRpIR be a quadratic form.
(a) If is positively defined (i.e. (x)0, for any x IRp \ {0}) there exists m0 such that (x)m x 2, for x IRp .
(b) If is undefined then there exist s1, s2 IRp \ {0 such that (ts1)0 and (ts2)0, for any t IR*.
Proof. (a). Let S
xIRp | x 1
. ThenS 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 m be the minim of on S; of course m0 because is positively defined and:
(x)m, for any xS (1) Now, let x IRp \ {0}; then
S x
x
1
and 1 12 (x)
x x
x
;
consequently, from (1) results that (x)m x , for any x IRp
.
(b) Let Bc be the canonical basis from IRp. As is of quadratic form, there exists an orthogonal basis B in which has canonical form. Let T=(bij) be the transformation matrix from Bc to B. Then:2 2
2 2 2 1 1
1,..., ) ...
(x xp y y pyp
(2) where
, ,...,p
, i x b y
p
j j ij
i 12
1
, ,...,p
Then, according to (2) we obtain:
p
j i
j j i i
ij x x a x a
a x x
1 ,
2 ( )( )( )
2 )
(
,when xa.
Then, from (1) and (4) results that:
) ( )
( 2 1 ) ( )
( )
(x f a d f x a d2f x a x a x
f a a .
It remains to prove that lim ( )0
a x
x . Successively applying the
inequality of modules, respectively the inequality of Cauchy-Buniakowsky-Schwarz (see example 14, cap.5) we obtain:
p
j i
j j i i
ij x x a x a
a x x
1 ,
) )(
( ) ( 1
)
(
p
j i
ij p
j i
j j i i p
j i
ij x x a x a x
a
x , 1 , 1
2 2
1 ,
2
2 ( ) ( ) ( ) ( )
1
, and from (3) we obtain: lim ( )0
a x
x .
Theorem 3 (sufficient conditions for extremes). Let
f : AIRpIR be a function of class C2 and aA a stationary point for
f. If da f
2 is positively (negatively) defined, then
a is a local minimum point (respectively maximum), for the function f. If da f
2
is undefined then a is not an extreme point for the function f.
Proof. Because a is a stationary point it results that
0
f
da ; according to Lemma 2 it follows that for any xA
) ( )
( 2 1 ) ( )
(x f a d2f x a x a 2 x
f a (1)
and
) ( 0 ) (
lim x a
a
x .
1. We assume that da f
2
is positively defined; according to lemma 1 results that:
m0 such that da2f(xa)m xa 2 (2)
for any x IRp. From (1) and (2) results that: 2
) ( 2 ) ( )
(x f a m x x a
f
for any xA. But lim ( )0
a x
x and m0, so there exists V
Va
suchthat:
0 ) ( 2 x
m
, for any xV (4) From (3) and from (4) results that
f(x) f(a) 0 for any xA V,
which means a is a local minimum of the function f.
2. If da2f is negatively defined, and g=-f, then da2g is positively defined and according to the precedent point, there exists
V
V
a such that:g(x) - g(a) = f(x) + f(a) 0, for any x V A , which means a is a local maximum of the function f.
3. We assume that da2f is undefined. According to lemma 1(b) results that there are two directions s1 , s2 IRp-{0} such that:
da2f (s1t) > 0 and da2f (s2t) < 0 , for any tIR* (5) Taking x = a + ts1 and y = a + ts2, because aÅ, there is a neighbourhood V0 of the point 0IR such that x,yA, for tV0. From (1) we obtain:
f(x) - f(a) = t2(
2 1
da2f(s1) + (x)) (6)
f(y) - f(a) = t2(
2 1
da2f(s2) + (y)). (7) But
0
lim
t (x) = 0 = limt0 (y),
so there exists a neighbourhood U of the point 0IR, UV0, such that:
2 1d
a2f(s1) + (x) > 0 and da2f(s2) + (y) < 0 (8) for tU. Finally, from (8),(6) and (7) results that for tU:
f(x)f(a)>0 and f(y)f(a)<0,
which means the point is not a local extreme of the function f.
Hf(a)=
p j
p i j i
a x x
f
, 1
, 1 2
) (
.
We remind that if the eigenvalues of 1,2,…,p are greater (smaller) or equal to 0, then da2f is positively (negatively) defined, and if there exists i>0 and k < 0 it is undefined. Also, for determining the nature of the point a we can apply Sylvester’s criterion:
Denoting with 1= ''2( ) 1 a
fx ,2=
) ( ) (
) ( )
(
'' ''
'' ''
2 2 2
1
2 1 2
1
a f a f
a f a f
x x
x
x x x
,…, p=detHf(a), then:
(a) if zk0, k=1,p, then a is a local minimum;
(b) if 1<0 , 2>0 , 3<0,…., (1)p p >0, then z is a local maximum for z;
(c) if in the equalities from (a), respectively (b) there is a z such that zk=0 the method does not determine the nature of the point a;
(d) in the rest da2f is undefined, so a is not an extreme for the function z.
Example 4. Let us determine the local extremes of the function
f : IR2IR, 2 13
2 3 2 2 3 3 ) , (
2 3 2
3
x x x y y y
y x
f .
Because fC2(IR2) results that we can determine all the extreme points through the described method:
Step 1. Determine the stationary points from the system:
0 2 )
, ( `
0 2 3 )
, ( `
2 2 2
y y y x f
x x y x f
y x
,
so the points (1,1), (1,-2), (2,1) and (2,-2) are, according to Fermat’s theorem, possible extreme points.
Hf(x,y)=
1 2 0
0 3 2
y x
,
and
1(x,y)=2x3, 2(x,y)=(2x3)(2y1).
In (1,1), z1=1 and 2=1, hence this point is not a local extreme for f.
In the point (1, 2), 1 = 1 < 0, 2 = 5 > 0, so (1,-2) is a local maximum and fmax=f(1, 2)=79/6.
Because 1(2,1) = 1, 2(2,1) = 1 > 0 it results that (2,1) is a local minimum for f and fmin=23/2.
At last, in (2, 2) we have 1= 1<0, 2= 5<0, so the point (2, 2) is not an extreme point of the function f.
Remark. If in a stationary point a of the function f the second differential is the null function and f is of class Cn, n3 then we use a superior order development in Taylor’s formula; if the first non-zero partial derivatives are of odd degree, then the stationary point is not an extreme for f.
Example 5. Determine the local extremes of the function
f : IR3 IR, f(x,y,z)=x4+y4+z4- 4xyz.
Solution. Determine first the stationary points solving the
system:
0 ) (
4 ) , , ( `
0 ) (
4 ) , , ( `
0 ) (
4 ) , , ( `
3 3 3
xy z z y x f
zx y z
y x f
yz x z
y x f
z y x
So the points (0,0,0), (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) are possible extreme points. The Hessian of the function f in the current point is:
Hf(x,y,z)=4
2 2 2
3 3 3
z x y
x y z
y z x
.
(a). Because Hf(0,0,0) is the null matrix, d2(0,0,0)f=0. We will study the sign of the 3rd order differential in (0,0,0). Because
2 2 2
we have
d3f = 24(xdx3+ydy3+zdz3-dxdydz), so d3(0,0,0)f = 24dxdydz, which is a ternary form which changes sign in any neighbourhood of the origin; therefore (0,0,0) is not an extreme for f.
(b). Because
Hf(1,1,1)=4
3 1 1
1 3 1
1 1 3
,
we have 1=12, 2=842, 3=1643>0 and according to Sylvester’s criterion, (1,1,1) is a local minimum, and fmin=f(1,1,1)=-1.
(c). Since f(x,y,z)=f(y,z,x)=f(z,x,y), to study the nature of the remaining stationary points it is sufficient to analyse the nature of the point (1,-1,-1). Because
Hf(1,-1,-1)=4
3 1 1
1 3 1
1 1 3
,
we have 1 = 12, 2 = 842, 3 = 1643 > 0; it results that the three points are local minimums for f and fmin= f(1,1,1)= f(1,1,1)=
f(1,1,1)= 1.
Example 6. From all straight parallelepipeds for which the sum of the lengths of the sides is equal to a>0 determine that of which the volume is maximum.
Solution. Let x,y,z be the lengths of the sides of such a parallelepiped; then its volume is:
V(x,y,z) = xyz, V: A IR,
where A={(x,y,z) IR3| x + y + z = a, x,y,z > 0}. The maximum value of the function V coincides with that of the function
f(x,y) = V(x,y,a-x-y) = xy(a – x – y ),
where f is of class C2 on
B = {(x,y) IR2| x , y > 0 , x + y a}.
So the only stationary point is
maximum.
We consider the function:
g :B={(x,y)IR2x,y 0, x+ya} IR, g(x,y)=xy(a–x–y).
0 is constant; analogously on OD; on CD we have x+y=a, x[0,a]
and k(x)=g(x,a-x)=0. Therefore the image of the function g is
Im g=g( B )=
27 , 0
3
a
and its absolute maximum is attained in
the interior ; so
3 , 3
a a
is an absolute maximum for f, and
3 , 3 , 3
a a a
is a point of absolute maximum for V; consequently, the parallelepiped we are searching for is the cube of side length
3
a
.
Remark. If f : A IRp IR is a continuous function and A is a compact set, then its absolute extremes are determined as follows:
- determine local extremes on Å,
- determine local extremes on the frontier Fr A;
then the absolute maximum is the greatest local maximum, and the global minimum is the smallest local minimum (of the ones determined at (a) and (b)). For functions with two variables the problem of determining frontier extremes is reduced, as in the previous example, to determining the extremes of single variable functions. In the case of three variable functions the determination of frontier extremes comes back to the study of extremes of two variable functions, etc. We will handle this problem in the consecrated section of conditional extremes.
Example 7. Determine the absolute extremes of the function f(x,y) = xy on the triangular domain of vertices O(0,0), A(1,0), B(0,2).
Solution. The definition domain of the function f is
D={(x,y) IR2 | x,y 0, x + y / 2 1}.
D
0 y
x A
1. On the interior of the set D, hence on the set
D ={(x,y)|x,y > 0, 2x + y <2} the system:
0 `
0 `
x f
y f
y x
has a unique solution: (0,0)
D. Not having stationary points it results that on
D f has no local extremes.
2. On the frontier Fr D=OABO we consider the cases:
2.1. On the segment [OA] we have y = 0 and x[0,1]. The function g(x) = f(x,0) = 0 is constant on [0,1].
2.2. On the segment [OB] we have x = 0 and y[0,2]. The function h(y) = f(0,y) = 0 is constant on [0.2].
2.3. On the open segment AB we have y = 2 - 2x, x(0,1). Then the function k(x) = f(x,2-2x) = 2x(1-x) has a maximum in x =
2 1
, and kmax = k(
2 1
) = 2 1
.
Therefore fmin = f(x,0) = f(0,y) = 0, for x[0,1], respectively y[0,2] and fmax = f(
2 1
,1) = 2 1
, and
Im f = f(D) = [0, 2 1
].
Remark. Let us suppose that the conditions from the theorem of implicit functions are verified such that the equation
x y z
f Af , , 0, : IR3 IR, fC2(A)
defines the function zz
x,y in a neighbourhood of the pointz
Example 8. Determine the extremes of the function
x ySolution. Of course, the function
, ,
2 2 2
2 2 2 2 solution of the systemwhich admit the point (0,0) as stationary point, and z1
0,0 1 and
0,0 12
z .
But
f
x
2(x,y,z)4(x2+y2+z2)+8x2+2,f
xy
(x,y,z)8xy and 2y
f
(x,y,z) 4(x2+y2+ z2)+8y2, andf
x
2(0,0,1) 6,f
xy
(0,0,1) 0,
0,0, 1
42
y
f and
f
z2 (0,0,1) 6. Therefore the Hessian of thefunction z1 is:
3 2 0
0 1 0
, 0
1
z
H ,
1
1
Ä and
3 2
2
Ä , so the function z1 has a local maximum in (0,0) and z1 max z1(0,0)1.
Analogously
3 2 0
0 1 0 , 0
2
z
H ,
and Ä1 10, 0 3 2
2
Ä ; therefore the function z2 has a local minimum in the origin and z2 min z2(0,0)1.
3. CONDITIONAL EXTREMUM
Definition 5. Let f : A IRp IR and B A. We say that the point aB is a local extreme of the function f related to the set B
(conditional extreme of B, w.r.t. the set B) if there exists VVa such that the expression E(x)f(x)f(a) keeps a constant sign on the set
B
V ; if E
x 0 on VB, a is a local minimum of the function frelated to B; and if E
x 0 on V B, a is a local maximum related to B. The relative extremum w.r.t. a subset B A (meaning the extremum of the restriction function fB) are called conditionalextremum, or relativeextremum.
with pnm variables: x1,...,xn,y1,...,ym; writing x(x1,...,xn),
y(y1,...,ym), we shall try to find the extremum of the function
f : A IRn+m IR,
f(x,y) f(x1,...,xn,y1,...,ym) related to the set
x y Ag x y j m
B , j , 0, 1, ,
where gj:A IR, j
1,2,...,m
. The equations that have solutionsdefining the set B:
x, y
0gj , j1,m (*)
are usually called coupling equations (or constraint equations, or, simply, constraints), and the conditional local extremum of the function f related to the set B – extremum with constraints (or relative extremum).
Remark. If the constraints (*) can be explicitly given, meaning that there exist the functions y:C IRnIR, such that
x,yj x
B, j1,m and gj
x,y1
x ,...,ym
x
0, xC, j1,m,then the problem of finding the conditional extremum of the function
f is just that of hunting out the extremum of the function of n
variables h:C IR,
x xn
f
x xn y
x xn
ym
x xn
h 1,..., 1,..., , 1 1,..., ,..., 1,..., .This method of replacing the variables was already used in example 7, where the constraint x yza has led, by replacing the variable zax y in the function V, to finding the extremum of a function with two variables.
This remark suggests that we should impose to the system (*) the conditions of the theorem of implicit functions in such a way that the constraints (*) define the functions yj yj
x , j1,m. So,Fermat`s theorem (Theorem 2) shall provide necessary existence conditions of the conditional extremum, and theorem 3 shall provide sufficiently conditions and even techniques of setting the nature of these extremum. J.P. Lagrange was the one who made these findings, which led him to elaborating an ingenious practical process of finding the extremum with constraints.
Lagrange
’
s multipliers method
Theorem 4.(necessary conditions for conditional extremum). Let f,gj :AIR
n+m IR,
m
j1, functions of class C1 on A. If
a,b A, where a
a1,...,an
, b
b1,...,bm
is a point of local extreme of the function f with the constraints (*), and:
,...,
, 0 (**),...,
1
1 a b
y y D
g g D
m m
Then there exist the numbers 1,2,...,m IR named Lagrange’s multipliers such that the point (a,b) is a stationary point of the function
A
F: IR, F f
1g1
2g2 ...
mgm.Proof. The determinant of the matrix of the system
m
j
j k
j
k
y b a y g b
a y
f
1
,
, , k 1,m (1)
is
,...,
, 0,...,
1
1 a b
y y D
g g D
m
m ; it follows that the system has a unique
solution y1
1,y2
2,...,ym
m, so from (1) we obtain:
m
j k
j j k
b a y g b
a y
f
1
0 ,
, , j1,m; (2) (a,b) being an extreme point having the coupling equations (*) we have
a,b 0gj , gj C
A1
, j1,m
and
,...,
, 0,...,
1
1 a b
y y D
g g D
m
m ;
agreeing with the theorem of implicit functions there exists VVa and a unique function y y
x , yC1
V implicitly defined by thesystem (*), meaning that:
x,y x
0gj , xV , j1,m (3)
and y
a b. Differentiating with respect to xiin (3) it results:
m
k j
j
x x y x y x y g x
y x x g
0 ,
Taking xa, in (4), since y
a b, we shall obtain: and the theorem is fully demonstrated.Definition 6. Let f, gj : A IRn+m IR, , j1,m be functions of class C2 on A, where the functions gj verify (.
Then L:A IRm IRn+2m IR, defined through:
x,y,ë
f
x,y ëg
x,y ë g
x,y ... ë g
x,yfor any (x,y)A and any
1,...,
m
IRm is called Lagrange’s function, and the variables
,
,...,
m2
1 IR are called Lagrange’s multipliers.
According to the previous theorem the points of extreme of the function f with the constraints (*, if exist, are found through the solutions of the system of n+2m unknowns
x
1,
x
2,...,
x
n,m
y
y
y
1,
2,...,
, 1,2,...,m:m j L
m k y
L n i x
L
j k
i
, 1 , 0 ,
, 1 , 0 ,
, 1 ,
0
(***) Let us remark the fact that if
a,b,0
is a solution of this system (so stationary point for L), and the functions g1,g2,...,gmverify **, then d(a,b)gj=0 and
a,b dy , j ,m yg ...
dy a,b y g dx a,b x g ... dx a,b x g
m m
j
j n n
j j
1 0
1 1
1 1
(****)
is a linear compatible determined system, which, through Cramer’s rule, allows the find of the projections dy1,..,dym related to the projections dx1,..,dxn. This finding allows us to give sufficiently conditions for the determination of the conditional extremum of the function f and to elucidate their nature through the study of a quadratic form of n variables.
Theorem 5 (sufficient conditions for conditional extremum). Let f,gjC2
A, j 1,mand
a,b,0
a stationary point of Lagrange’s functions which verifies ** and the functions:
x y L
x y
f
x y g
x y x y A Fm
j j
j
, , , ,
, , ,
1 0
0
.
Let, also, : IRn IR the quadratic form found through replacing the projections dy1,...,dym according to the projections
n dx
dx1,..., (resulted from the system ****)) in d 2ab F
(maximum) of the function f with the constraints *; if f is undefined then (a,b) is not a point of conditional extreme for the function f.
Proof. Let B
x,y A|gj
x,y 0,j1,m
. From *** it results that
a bgj , =0, j1,m, so (a,b)B.
Then, for any
x,y B,E
x,y F
x,y F
a,b f
x,y f a,b ,so the local extremum of the function F are coincident with the local extremum of the function f with the constraints *.
On the other hand, how
0
, ,b
a is a stationary point of Lagrange’s function, from *** it results that:
a
b
i
,n
, k
,m
y
L
b
a
x
L
k i
1
1
,
,
,
,
0 0
0,
and
0
so
a
b
i
,n
x
F
b
a
x
g
b
a
x
f
i i
j m
j i
1
,
0
,
,
,
1
0 j
a
b
k
,m
y
F
b
a
y
g
b
a
y
f
k k
i m
j k
1
,
0
,
,
,
1
0 j
,meaning that (a,b) is a stationary point for the function F.
Finally, how f, gj are functions of class C2, it results that
A CF 2
and, in agreement with theorem 3, we can set the nature of the point (a,b) with the help d 2a,b F; but in quadratic form d 2a,b F of
n+m variables the projections dyj,j1,m are, as solutions of the
system ****, linear combinations of the projections dx1,...,dxn; so the signature of the form d 2a,b Fcoincides with that of the quadratic form of n variables. Therefore, from theorem 3 it results that if is positively (negatively) defined, then (a,b) is a point of minimum (maximum) local for f with the constraints *, and if is undefined, then (a,b) is not a conditional extreme of the function f.
A j m Cg
f, j , 1,
2
, then the method of Lagrange’s multipliers of determination the extremum points of the function f
with the constraints * means operating the following stages: 1. We form Lagrange’s function j
m
j j
g
f
L
1
and we determinate it’s stationary points solving the system ***. Then, for each stationary point
a,b,0
which verifies ** we solve the following steps:2. We form the auxiliary function
,
, ,0
,y x L y x
F x,yA
and we calculate d 2a,bF (of 2m variables).
3. We solve the linear system **** according to the projections dy1,...,dym then we replace them in d 2 F
b
a, obtaining the
quadratic form of n variables.
4. We analyse the signature of the quadratic form (reducing it to a canonical form through the theorem of Gauss, or Jacobi, or by Sylvester’s criterion) and we conclude the following: if it is positively (negatively) defined then (a,b) is a local minimum (maximum) of the function f with the constraints *, and if it is not defined, then (a,b) is not a point of extreme of the function f with the constraints *.
Example 9. Determine the image of the function:
f : A={(x,y) 2 | x2 + y2≤ 1} ,f
x,y x y13.Solution. The function f is continuous on the compact
A IR2; hence the image f(A) is compact in IR (theorem 15, chap.5) and f attains its bounds (theorem 16, chap.1). We shall solely determinate these bounds. How fx' 10, on the interior
( , ) ( 2 2 1
x y x y
A |
there are no stationary points, so neither extremum (Fermat`s theorem); it follows that the extremum are attained on the frontier of the set A,
which means that we have to determinate the extremum of the function f with the constraint 2 2 10
y
x . Let
IR be Lagrange’s multiplier; Lagrange’s function is:
x,y,
x y13 (x2 y2 1)L
,and the system (***) is:
0 2 1
x
Lx , Ly 12
y0, 1 02
2
x y
L ,
so we obtain
2 1
x ,
2 1
y and 1
4 1 2 2
. For
2 1
we have x
2 1 ,
2
1
y and for
2 1
we
obtain
2 1
x and
2 1
y .
We calculate the second degree differential of the function:
x,y L x,y,
F where
is constant (2 1
or
2 1
). We
obtain 2 2
2 2
.dy dx F
d By differentiation the constraint
0 1
2
2
y
x we obtain xdx+ydy=0 and how y= -x 0, dx=dy; so the quadratic form is defined through
2
24 4 dx h h
h
.
If
2 1
,
22 2 h
h
is positively defined, so (
2 1
,
2 1
) is an absolute minimum for f and fmin=f(
2 1
,
2 1
)=13- 2; if
2 1
, is negatively defined, (
2 1
,
2 1
) is a global maximum for f and fmax=13+ 2. As a consequence, the image of the function
f is f(A)= 13- 2,13+ 2.
Example 10. Find the extreme values of function
xyz z y x f IR IR
f : 3 , ( , , )
constrained by the equations: . 8 ,
5
y z xy yz zx
Solution. Let be Lagrange’s multipliers. Then Lagrange’s function is:
x,y,z, ,
xyz
x yz5
xy yzzx8
First, let us remark that the system is symmetric in x,y,z and that, by adding the first three relations we obtain 8+310=0. With these conclusions, the stationary points of the function L are:
-for 4, 2,
x,y,z
2,2,1
, 2,1,2
,1,2,2
constant; we obtain:
By differentiating the constraints we have:dx+dy+dz=0 and (y+z)dx+(z+x)dy+(x+y)dz=0 (1) (2,1,2), (1,2,2) are points of minimum of the function f with the given constraints and fmin= 4.
Example 11. Determine the distance between the surface of equation S : x2 + y2 + z2 = 1 to the right of the equations
D : y = x, z= x+2.
Solution. The distance from a point M (x, y, z) S to a point
P( u, v, w) D defines a function of six variables:
d (P, M) = g (x, y, z, u, v, w) =
xu
2 yv
2 zw
2 (1) Our problem consists in determination of the absolute minimum of the function g with the constraints:x2 + y2 + z2– 1 + 0; v – u = 0 and w – u –2 = 0 (2)
Because the associated Lagrange function has nine variables (all the six of the function plus three multipliers) and the derivatives are complicated, we shall try to simplify the problem. Let us first remark that the point of minimum of the function g coincides with that of the function:
f (x, y, z, u) = (x - u)2 + (y - u)2 + (z – u - 2)2 (3)
with the only constraint
x2 + y2 + z2 –1 = 0. (4)
In this case, Lagrange’s function is:
L (x, y, z, u, ) = (x - u)2 + (y - v)2 + (z – u - 2)2 +
+ (x2 + y2 + z2 - 1) (5) Let us determinate the stationary points for L from the system:
L’x = 2 [( + 1)x - u] = 0,
L’y = 2[( + 1)y - u] = 0,
L’z = 2[( + 1)z – u - 2] = 0,
L’u = 2(x + y + z - 3u –2) = 0,
L’ = x2 + y2 + z2– 1 = 0.
Solving this system we obtain:
3 (6)2 2 1 ,
1 3
4 ,
) 1 ( 3
2 ,
3
2
z
y x
u
By differentiating the constraint (4), from (6) it results:
dx dy
dz
2 1
(7) We are interested solely in the minimum of the auxiliary function
where is a constant. By differentiation, we shall obtain, successively:
dF=2[(x-u)(dx-du)+(y-u)(dy-du)+(z-u-2)(dz-du)+ (xdx+ydy+zdz)],
d2F = 2[(dx-du)2 + (dy-du)2 + (dz-du)2 + (dx2+dy2+dz2)] =
= 2[(+1) (dx2+dy2+dz2)+3du2 -2du (dx + dy + dz)] ) 7 (
) 7
(2
du dy dx du
dxdy dy
dx 5 2 3 3
5 4
1 2 2 2
. (9) According to Sylvester’s Criterion, for the quadratic form (9) to be positively defined it is necessary that 0
2 1 5
1
. Due to this from (6) we obtain:
3 2 ,
6 2 ,
6 1 ,
3 2 2
1
x y z u
(10)
From a geometrical point of view, the problem is uniquely determinate; it follows that the distance from the sphere S to the right is, according to formulas (10), (1) and (2):
2 2 3
3 5 3
4 , 3 2 , 3 2 , 6 2 , 6 1 , 6 1 )
,
(
d
D S
d .
4. SOLVED PROBLEMS
Exercise 12. Expand the function f defined by:
- y x - xy x
f(x,y) 4 2 2
in powers of x1 and y2.
Solution. By using the Taylor’s formula with a1 and b2
we obtain that:
) , y f( x-d
k! (x,y) T
f(x,y) ( ,- )
k
2 1 1
2 1 4
0
4
, since R4(x ,y) 0.
But f(1 ,-2)1-4120,
dy xdx xydy
dx y dx x
df 4 3 2 2 2
,
dx dy dy
dx dy
dx dx - f
,
Consequently:
Exercise 14. Expand into Mac Laurin`s series the function:
xy
Solution. We remark that
Since the both series are absolutely convergent we obtain:
)
Find its stationary points.
If
x,y is a stationary point for z then:We shall examine the two cases. We consider first a (-2,0) . We have
for all (x,y) U. Therefore (-2,0) is a stationary point for this implicit function.
The analogue demonstration for a=(
; Fermat`s theorem, the only candidate for local extreme. But
d2(0,0)f = (2dy – 6dx)dx + (2dx – 6dy)dy = -6dx2 + 4dxdy -4dy2
is a negative defined quadratic form; so we have a local maximum and fmax = f(0,0) = 10.
The partial derivative:
fy
so, according to the implicit functions theorem, it follows that: 3
But for both functions
3
Exercise 18. Find the extremum for the functions: (a) f : IR2 IR, f(x) = x3– 3x +y2;
(b) g : D IR, g(x,y) = f(x,y) and D = { (x,y) |x2 + y2 1 }. Solution. (a) The stationary points are solutions for:
hence the Hessian of the function f in these stationary points are:
results that the bounds of the function g are attained on the frontier of D.
Case 2: On the frontier of the set D:
FrD = { (x,y) IR2| x2 + y2 -1 = 0}.
This is a typical case of conditional extreme. We construct the auxiliary function of Lagrange:
L(x,y,ë) = g(x,y) + ë(x2 + y2 -1) = x3 -3x + y2 + ë(x2 + y2 -1). We find the stationary points of the function L by solving the system:
The only admissible solutions are
. 10 2 25 6 1 y , 6
10 1 x , 6
10 1 x ,
1 1 2 1
1
Using the relation xdx + ydy = 0, we find for F(x,y) = L(x,y,1), that
0 F
d(2x1,y1) and d F 0 2
) y , x
( 2 2 and so (x1, y1) is maximum point and
(x2, y2) is a minimum point.
5. EXERCISES
Exercise 1. Develop after the powers of x + 1, y + 1, z –1 the
polynomials:
(a) f(x, y) = x3 + y3– 3 x y + 13
(b) f(x, y, z) = x3 + y3 - z3 + 3x2y – 3xy + z2– z + 22.
Answers. (a) f (x, y) = 8 + 6(x + 1) + 6(y + 1) – 3 (x + 1)2– 3(y
+ +1)2– 3(x + +1)(y + 1) + (x +1)3 + (y + 1)3;
(b) f(x, y, z) = 13 + 12(x + 1) + 15(y + 1) – 2 (z - 1) – 6(x + 1)2–
-3 (y + 1)2 – 2 (z - 1)2 – 9(x + 1)(y + 1) +(x + 1)3 + (y + 1)3 – (z - 1)3 +
+6(x + 1)2(y + 1).
Exercise 2. Write the second degree Taylor polynomial for: (a) f(x, y) = ln(x2 + x + y2) in the point (0,1)
(b) f(x,y,z) = ex arctan (y + z) in the point (0,0,1)
Answers. (a) T2(x,y) = 2 1
x2– 2xy – y2 +3x – 3;
(b) T2(x,y,z)= 8
x2–y2–z2 +
2
1
xy+
2
1
xz – 2yz+ x
2 1 4
+
2
5
y+
2
5
z+
4
-2
3
.
Exercise 3.Using second-degree Taylor’s formula show that: (a)
1.013 1.99
3 2.9851(b) 1.033 0.98 1.0081
Exercise 4. Determine
E1=(1.1)1.2 E2=(0.99)1.01+(1.01)1.001+ +(1.001)0.99 and
E3 = (1.01)0.99 + (0.99)0.999 + (0.999)1.01 with three exact decimals.
Answers. E 1 1,121; E2 3,001; E 3 2,999.
Exercise 5. Determine the local extremum of the functions (a) f(x,y) = x3– 3xy2 + x2– 2xy + 3y2 + 2y +
27 370 (b) f(x,y) = 2x3– y3 +(y-x)2
(c) f(x,y) = ex (cos y – x) + cos y
(d) f(x,y,z) = x2 + y2 + z 2 – xy + x – 2z
(e) f(x,y,z) = x 3 + y 2 + z2 + 12xy + 2z
(f) f(x,y,z) = x +
y z x y x
2 2
4 2
(g) f(x,y,z) =
y x
z x z
y z y
x
, x,y,z >0
(h) f(x,y,z) = x3 + y3– z3 – x2y + 3z2 + 13
Answer. (a) f min= 13;
3 1 , 3
1
f
(b) fmin= ;
27 2 12 17 3
2 2 , 3
1
2
f (c) fmax=f(0,2k)=2, kZ;
(d) fmin = ;
3 4 1 , 3 1 , 3
2
f (e) fmin = f(24, -144, -1) = -6913; (f) does not have extremum; (g) fmin=f(a,a,a) = ,a 0;
2 3
(h) f min= f(0,0,0)=13, fmax = f(0,0,2) = 17.
Exercise 6. Determine the points of local extreme of the functions y = y(x) and x = x(y) defined implicitly by the equations:
(a) x3 + y3– 3x2y – 3 = 0
Answer: (a) y min = y1(0) =3 3; y max = y2 (-2) = -1; xmax =
x1
3 3 3 3; xmax = x2(-1)=1; (b) ymin = y(0) = 0.Exercise 7. Determine the points of local extreme of the implicit functions z = z(x,y) defined implicitly by the equations:
(a) x2 + y2 + z2– 4 z = 0
(b) x2 + y2 + z2 – xz – yz + 2x + 2y + 2z + 2=0
(c) x2 + y2 + z3 + z = 0.
Answer. (a) zmin=z1(0,0)=0, zmax=z2(0,0)=4; (b)zmin=z1
3 6,3 6
442 6,zmax=z2
63, 63
2 64;(c) z max = z (0,0) = 0.
Exercise 8. Study the local extremum of the functions of class C2implicitly defined by the equation x2–2x + y2 + z + ez = 0.
Answer: zmax = z(1,0)=0; the other implicit functions defined by the equation (z = z(x,y), x = x(y,z), y = y(z,x)) do not admit local extremum.
Exercise 9. Study the local extremum of the functions of class C2 implicitly defined by the equation:
(x2+y2+z2)2= a2-x2-z2, aIR*.
Answer. Out of the functions z = z(x,y), defined by the equation, two of them admit extremum:
zmin=z1(0,0)= 2
1 4 1
,2
1 2
a zmax=z2(0,0)= 2
1 4 1
2
1 2
a ;
analogously, for x = x(y,z),xmin= 2
1 4 1
0,0,2 1
1 2
x
a
xmax=x2(0,0)= 2
1 4 1
;2
1 2
a from the functions y = y(x,z) two of them admit extremum: ymin=y1(0,0)= a and ymax=y2(0,0)= a , for
