Optimization:ASpecial
VarietyofEquilibriumAnalysis
Ch.9 9.1 OptimumValuesandExtremeValues
9.2 RelativeMaximumandMinimum:
FirstDerivativeTest
9.3 SecondandHigherDerivatives
9.4 SecondDerivativeTest
9.5 DigressiononMaclaurinandTaylor
Series
9.6NthDerivativeTestforRelative
ExtremumofaFunctionofOneVariable
1
9.1 OptimumValuesandExtremeValues
Goalvs.nongoalequilibrium
Intheoptimizationprocess,weneedto
identifytheobjectivefunctiontooptimize.
Intheobjectivefunctionthedependent
variablerepresentstheobjectof
maximizationorminimization
=PQC(Q)
2
9.2 RelativeMaximumandMinimum:
FirstDerivativeTest
9.21Relativeversusabsoluteextremum
9.22Firstderivativetest
3
9.2 Critical & stationary values
Thecriticalvalueofxisthevaluex0if
f(x0)=0
Astationaryvalueofyisf(x0)
Astationarypointisthepointwith
coordinatesx0andf(x0)
Astationarypointiscoordinateofthe
extremum
4
9.22Firstderivativetest
Thefirstorderconditionornecessarycondition
forextremaisthatf'(x*)=0andthevalueof
f(x*)is:
Arelativemaximumifthederivativef'(x)
changesitssignfrompositivetonegativefromthe
immediateleftofthepointx*toitsimmediate
right.(firstderivativetestforamax.)
y
A
x*
f '(x*) = 0
5
9.22Firstderivativetest
Thefirstorderconditionornecessarycondition
forextremaisthatf'(x*)=0andthevalueof
f(x*)is:
Arelativeminimumiff'(x*)changesitssign
fromnegativetopositivefromtheimmediateleft
ofx0toitsimmediateright.(firstderivativetestof
y
min.)
B
x*
f '(x*)=0
x
6
9.22Firstderivativetest
Thefirstorderconditionornecessaryconditionfor
extremaisthatf'(x*)=0andthevalueoff(x*)is:
Neitherarelativemaximanorarelativeminimaif
f'(x)hasthesamesignonboththeimmediateleftand
rightofpointx0.(firstderivativetestforpointof
inflection)
y
D
x*
f '(x*) = 0
x
7
9.2 Example 1 p. 225
y = f ( x) = x 3 12 x 2 + 36 x + 8
primative function
f ' ( x) = 3 x 2 24 x + 36 = 0
1st derivative
(
)
f ' ( x) = 3 x 2 8 x + 12 = 0
( x 2) ( x 6) = 0
left
f (1.5) = 38.375
f ' (1.5) = 6.75
f (5.5) = 9.375
f ' (5.5) = 5.25
*
x1 = 2;
()
y * = f xi*
f (2) = 40
f ' ( 2) = 0
f (6) = 8
f ' (6) = 0
*
x2 = 6 extrema
right
f 2.5 = 38.625
f ' (2.5) = 5.25
f (6.5) = 9.625
f ' (6.5) = 6.75
max
( +,1)
min
( ,+ )
8
primitive function and 1st & 2nd derivatives
1)
f ( x) = x 3 12 x 2 + 36 x + 8
2)
f ( x) = 3 x 2 24 x + 36
f ( x) = x 2 8 x + 12 = 0
( x = 2) ( x = 6)
3)
f ( x) = 6 x 24 = 0
x=4
9
9.2 Example 1 (neg) p. 225
y=
(
f ( x) = x 3 12 x 2 + 36 x + 8
)
f ' ( x) = 3 x 2 + 24 x 36 = 0
(
)
primative function
1st derivative
f ' ( x) = 3 x 2 8 x + 12 = 0
( x 2) ( x 6) = 0
left
*
x1 = 2;
y = f ( xi* )
*
x2 = 6 extrema
right
f (1.5) = 38.4
f ' (1.5) = 6.8
f (2) = 40
f ' ( 2) = 0
f ( 2.5) = 38.6
f ' (2.5) = 5.3
min
( , + )
f (5.5) = 9.4
f ' (5.5) = 5.3
f (6) = 8
f ' (6) = 0
f (6.5) = 9.6
f ' (6.5) = 6.8
max
( +,1)
10
primitive function and 1st & 2nd derivatives
(
1)
f ( x ) = x 3 12 x 2 + 36 x + 8
2)
f ( x ) = 3 x 2 24 x + 36
(
f ( x ) = ( x
2
)
8 x +12 = 0
( x 2 )( x 6 ) = 0
3)
)
)
f ( x ) = ( 6 x 24 ) = 0
x =4
11
9.2 Example 2 p. 226
AC = Q 2 5Q + 8
f / ( Q ) = 2Q 5 = 0
Q* = 5 / 2 = 2.5
left
f ( 2.4) = 1.76
f ' (2.4) = 0.2
primative function
1st derivative
extrema
()
AC = f Q*
f ( 2.5) = 1.75
f / ( 2.5) = 0
right
f ( 2.6) = 1.76 relative min
f ' (2.6) = 0.2
( ,+ )
12
plots: primitive convex function, as well as
1st & 2nd derivatives
2
1)
AC = Q 5Q + 8
2)
MC = 2Q 5
MC = 2Q 5 = 0
Q=5 2
3)
MC = 2
13
9.2 Smile test
1) R = 1200Q 2Q 2
primative function
dR
2)
= 1200 4Q = 0
dQ
1st derivative
3) 4Q = 1200; Q * = 300
extrema
d 2R
4)
= 4
2
dQ
2nd derivative
if f // ( x ) < 0, then relative maximum
14
Primitive function and 1st & 2nd
derivatives
1)
R = 1200Q 2Q 2
2)
MR = 1200 4Q
1200 4Q = 0
Q = 300
3)
MR = 4
15
9.3 SecondandHigherDerivatives
9.31Derivativeofaderivative
9.32Interpretationofthesecondderivative
9.33Anapplication
16
9.31Derivativeofaderivative
Giveny=f(x)
Thefirstderivativef'(x)ordy/dxisitselfa
functionofx,itshouldbedifferentiablewith
respecttox,providedthatitiscontinuousand
smooth.
Theresultofthisdifferentiationisknownasthe
secondderivativeofthefunctionfandisdenoted
asf''(x)ord2y/dx2.
Thesecondderivativecanbedifferentiatedwith
respecttoxagaintoproduceathirdderivative,
f'''(x)andsoontof(n)(x)ordny/dxn
17
9.3 Example
1)
2)
3)
4)
R = f ( Q ) = 1200 2Q 2
f (Q) = 1200 4Q
f ( Q ) = 4
f ( Q ) = 0
primative function
1st derivative
2nd derivative
3rd derivative
18
Primitive function and 1st & 2nd
derivatives
1)
R = 1200Q 2Q 2
2)
MR = 1200 4Q
1200 4Q = 0
Q = 300
3)
MR = 4
19
Example 1, p. 228
1)
y = f ( x) = 4 x 4 x 3 + 17 x 2 + 3 x 1
primative function
2)
f ' ( x ) = 16 x 3 3 x 2 + 34 x + 3
1st derivative
3)
f " ( x) = 48 x 2 6 x + 34
2nd derivative
4)
f (3) ( x ) = 96 x 6
3rd derivative
5)
6)
f ( 4) ( x ) = 96
f ( 5) ( x ) 0
4th derivative
5th derivative
20
9.32Interpretationofthesecondderivative
f'(x)measurestherateofchangeofa
function
e.g.,whethertheslopeisincreasingor
decreasing
f''(x)measurestherateofchangeintherate
ofchangeofafunction
e.g.,whethertheslopeisincreasingor
decreasingatanincreasingordecreasingrate
howthecurvetendstobenditself(p.230)
21
9.33Thesmiletest
()
x* < 0,
If f
*
then x is a relative maximum
()
If f x* > 0,
then x* is a relative minimum
22
9.3 Example
1)
TR = 1200Q 2Q 2
primative function
2)
MR = 1200 4Q = 0
1st derivative
3)
Q* = 300
MR = 4
extrema
4)
5)
MR < 0
2nd derivative
maximum
23
9.4 SecondDerivativeTest
9.41Necessaryandsufficientconditions
9.42Conditionsforprofitmaximization
9.43Coefficientsofacubictotalcost
function
9.44Upwardslopingmarginalrevenue
curve
24
9.41Necessaryandsufficientconditions
Thezeroslopeconditionisanecessary
conditionandsinceitisfoundwiththefirst
derivative,werefertoitasa1storder
condition.
Thesignofthesecondderivativeis
sufficienttoestablishthestationaryvaluein
questionasarelativeminimumiff"(x0)>0,
the2ndorderconditionorrelativemaximum
25
iff"(x0)<0.
9.42Profitfunction:Example3,p.238
1)
TR = 1200Q 2Q 2
revenue function
2)
TC = Q 3 61.25Q 2 + 1528.5Q + 2000
cost function
3)
= TR-TC = 1200Q 2Q 2 Q 3 61.25Q 2 + 1528.5Q + 2000 profit function
(
= Q 3 + 59.25Q 2 328.5Q 2000
)
simplified
5)
= 3Q 2 + 118.5Q 328.5 = 0
1st derivative
6)
*
Q2 = 36.5
quadratic eq. extrema
7)
*
Q1 = 3
= 6Q + 118.5
8)
(3) = 100.5
(36.5) = 100.5
solving 2nd der. w/ extrema
9)
*
Q1 > 0 min
*
Q2 < 0 max
applying the smile test
()
2nd derivative
()
26
9.4 Quadratic equation
Given a quadratic function in the form
ax + bx + c = 0
the roots of the quadratic function are
2
(
(
b b 4ac
x ,x =
2a
*
1
*
2
)
2
)
1
2
27
9.42Profitfunction:example3,p.238
revenue and cost functions
1)
TR = 1200Q 2Q 2
2)
TC = Q 3 61.25Q 2 + 1528.5Q + 2000
profit function
3)
= TR-TC = Q 3 + 59.25Q 2 328.5Q 2000
1st derivative of profit function
4)
= 3Q 2 + 118.5Q 328.5 = 0
5)
*
*
Q1 = 3 Q2 = 36.5
2nd derivative of profit function
6)
= 6Q + 118.5
7)
(3) = 100.5
(36.5) = 100.5
applying the smile test
8)
()
()
*
*
Q1 > 0 min Q2 < 0 max
28
9.4 Imperfect Competition, Example 4, p. 240
Total revenue function
1)
(
)
TR AR * Q = 8000 23Q + 1.1Q 2 0.018Q 3 Q
29
Example 4, p. 240
TR and 1st , 2nd & 3rd derivatives
Average and total revenue functions
1)
AR = f ( Q ) = 8000 23Q + 1.1Q 2 0.018Q 3
4)
MR = 8000 23Q + 1.1Q 2 0.018Q 3 (1)
2)
TR = f ( Q ) Q
Marginal revenue and the product rule
3)
MR = f (Q )Q + Qf (Q )
(
(
+Q 23 + 2.2Q .054Q 2
)
)
= 8000 46Q + 3.3Q 2 0.0.72Q 3
2nd derivative
5)
MR = f ( MR )
6)
7)
MR = 46 + 6.6Q 0.216Q 2 = 0
*
Q1 = 10.76
*
Q2 = 19.79
3rd derivitive
8)
MR = f ( MR )
9)
MR (Q) = 6.6 0.432Q
10)
MR (10.76) = 1.95
MR" (19.79) = 1.94
smile test
11)
*
MR (Q1 ) > 0 min.
*
MR" (Q2 ) < 0 max.
30
9.4 Strict concavity
Strictlyconcave:ifwepickanypairofpointsM
andNonitscurveandjointthembyastraight
line,thelinesegmentMNmustlieentirelybelow
thecurve,exceptatpointsMN.
Astrictlyconcavecurvecannevercontainalinear
segmentanywhere
Test:iff"(x)isnegativeforallx,thenstrictly
concave.
31
9.5 DigressiononMaclaurinandTaylor
Series
9.51Maclaurinseriesofapolynomial
function
9.5-2 Taylorseriesofapolynomialfunctions
9.53Expansionofanarbitraryfunction
9.54Lagrangeformoftheremainder
32
9.5-2 Taylorseriesofapolynomialfunctions
Taylor series for a polynominal function, the wt. sum of its derivatives
f ( x0 )
f / ( x0 )
f // ( x0 )
f (n) ( x0 )
0
1
2
( x x0 ) +
( x x0 ) +
( x x0 ) + +
( x x0 ) n
f ( x) =
0!
1!
2!
n!
Taylor series for an arbitrary function, any function by the wt. sum of its derivatives
f ( x0 )
f / ( x0 )
f // ( x0 )
f (n) ( x0 )
0
1
2
( x x0 ) =
( x x0 ) +
( x x0 ) + +
( x x0 ) n + R
f ( x)
0!
1!
2!
n!
f (n+1 ) ( p )
( x x0 ) n+1
Where R =
( n + 1)!
If n = 0, then
rise
run = rise > < 0
run
where p is a point on the curve between x and x 0
f ( x) f ( x0 ) = R = f / ( p ) ( x x0 ) =
whose tangent is parallel to the secant from x and x 0
33
8.1 Differentials
d
1)
y = f ( x ) = lim y x
derivative
x 0
dx
2)
y x f ( x ) =
as x 0 then 0
3)
y x = f ( x ) +
4)
y = f ( x ) x + x
5)
y = f ( x ) x
as finite x 0
6)
dy = f ( x ) dx
differential (see 8.4)
where f ( x ) is the factor of proportionality
A first - order Taylor series approximation of y1 is :
7)
f ( x1 ) = f ( x0 ) + f ( x0 ) ( x1 x0 )
8)
y = f ( x0 ) x
34
Difference Quotient, Derivative &
Differential
f(x)
y=f(x)
f(x0+ x)
B
x
y
f(x0)
D
A
x0
x
f(x)
f(x0)
C
x
x
x0+ x
35
Exercise 9.5-2(a): Geometric series
Find the first five terms of Maclaurin the series,
f // ( x0 )
f (3) ( x0 )
f ( n ) ( x0 )
2
3
( x x0 ) +
( x x0 ) + +
( x x0 ) n
f ( x ) = f ( x0 ) + f ( x0 )( x x0 ) +
2!
3!
n!
i.e., choose n equals 4 and let x 0 = 0 for
/
1
1 x
2
f / ( x) = ( 1) ( 1) (1 x )
= (1 x )
1
f ( x0 ) = (1 - 0 )
= (1 x )
2
f / ( x0 ) = (1 0)
f // ( x) = ( 2) ( 1) (1 x )
= 2(1 x )
3
f // ( x0 ) = 2(1 0 )
= 6(1 x )
4
f (3) ( x0 ) = 6(1 0)
f(x)
=
3
f (3) ( x) = (3)(2)( 1) (1 x )
4
f ( 4) ( x) = (4)(3)(2)( 1) (1 x )
5
= 24(1 x )
5
1
=1
2
=1
3
=2
4
f ( 4 ) ( x0 ) = 24(1 0 )
5
=6
= 24
2
( x 0) 2 + 6 ( x 0) 3 + 24 ( x 0) 4 +
2
6
24
1
a
1
2
3
4
n
n
f ( x ) = 1 + x + x + x + x + ... = x =
ax =
ax n = a
1
1 x
1 x
1 x
n =0
n =0
n =1
Geometric series : each term is obtained from the proceeding one by multipling
f ( x ) = 1 + 1( x 0) +
it by x, convergent if x < 1
Chiang & Wainwright p. 243 and Stewart, Calculus, pp 610 - 11
36
y = (1 a )
1
where a = 0 0.9
2
3
4
y = 1+ a + a + a + a +
37
5.7LeontiefInputOutputModels
Structure of an input-output model
Miller & Blair, p. 102
A1 A 1 = A0
AA-1 = I
A0 = I
Taylor series approximation (Chiang & Wainwright, p. 250, 9.5 - 2(a))
(1 x )
1
=
x n = 1 + x + x 2 + ... + x n 1 if x < 1 scalar algebra
n =0
( I A) 1 = A n = I + A + A2 + ... + A n 1
by analogy
n =0
2
1 0 0.15 0.25 0.15 0.25
1.2541 0.3300
=
+ 0.20 0.05 + 0.20 0.05 + ... 0.2640 1.1221
0 1
Ch. 5b Linear Models and Matrix
Algebra 5.5 - 5.8
38
The stock (V) - flow (v) problem :
Given interest rate (r) prove that
v
v
v
v
+
+ ... +
=
(1 + r ) (1 + r ) 2
(1 + r ) n r
v
v
v
V=
+
+ ... +
+ ...
(1 + r ) (1 + r ) 2
(1 + r ) n
V=
1
1
1
= v
+
+ ... +
+ ...
(1 + r ) (1 + r ) 2
(1 + r ) n
1
Let x =
and applying the geometric series
1+ r
1
V = v x + x2 + + xn
= v
1
1 x
1
1
= v
1
= v
1
1
1+ r 1
1
1+ r
1+ r
1+ r
1+ r r
1
= v
1
= v
= v
r
r
r
1
v
1
1
V = v
+
+ ... +
+ ... =
(1 + r ) (1 + r ) 2
r
(1 + r ) n
(
)
Walter Nicholson, Microeconomic Theory, 1972, p. 381,
39
Diewerts Quadratic Lemma
1)
Y j = f ( X ij , j )
2)
f ( x1 ) = f ( x0 ) + f ( x0 )( x1 x0 ) + ( f ( x0 ) / 2!)( x1 x0 ) 2 + ... + ( f n ( x0 ) / n!)(n t ) n + Rn ( x1 , x0 )
3)
Let
3)
ln Y1 ln Y0 =
f ( x ) = ln Y
and
f ( X
i
4)
i0
x = ln x
) ( ln X i1 ln X i 0 ) + 1 2 f ( X i 0 ) ( ln X i1 ln X i 0 )
f ( X i 0 ) = ln Y / ln X i = ( Y X i ) ( X i 0 Y0 ) = Pi 0 X i 0 Py 0Y0 = Pi 0 X i 0
(
2
P
i0 X 0
= si 0
i
)
5)
Py 0 MPPxi = Pi 0
6)
Y X i = Pi 0 Py 0
7)
f ( X i 0 ) = f ( f ( X i 0 ) ) = si ln X i = ( si1 si 0 ) ( ln X i1 ln X i 0 )
8)
ln Y1 ln Y0 =
=
s
i
i0
i
9)
s
ln ( Y1 Y0 ) =
Y1 Y0 =
( ln X i1 ln X i 0 ) + 1 2 ( si1 si 0 ) ( ln X i1 ln X i 0 ) ( ln X i1 ln X i 0 ) 2
i
( ln X i1 ln X i 0 ) + 1 2( si1 si 0 ) ( ln X i1 ln X i 0 ) = 1 2( si1 + si 0 ) ln( X i1
1 2 ( s
i
10)
i0
( X
i1
i
i0
(
)
+ si1 ) ln X i1 X i 0 + ln (1 + )
X i 0 ) 1 2( si 0 + si1 ) (1 + )
X i0 )
i
(Cooke and Sundquist, SJAE, July 1991)
i
40
Diewerts Quadratic Lemma
Let
i = set of inputs in the production of a commodity in the economy,
where e.g., i = KLEFMA inputs of capital, labor, enery,
fertilizer, materials, and planted acres
j = set of geographic regions such that 0 base region and 1 is the first region
Yj = the average yield per acre of commodity Y in region j
Pyj = the price of output Y in region j
X ij = the average quantity of input X i in region j
Pij = the price of input i in region j
s ij = the share of wage expenditures in sector i in region j
j = the productivity index in region j relative to the base region
41
Diewerts Quadratic Lemma
1)
W j = f (Wij , j )
2)
f ( x) = f (a ) + f ( a )( x a ) + ( f ( a ) / 2!)( x a ) 2 + ... + ( f n (a ) / n!)(n t ) n + Rn ( x, a )
3)
ln W1 ln W0 =
f (Wi 0 ) ( ln Wi1 ln Wi 0 ) +
i
(
)(
1 2 f (Wi 0 ) ( ln Wi1 ln Wi 0 )
2
)
4)
f (Wi 0 ) = ln W / ln Wi = W Wi Wi 0 W0 = N i 0Wi 0 N 0W0 = si 0
5)
Wj =
W N
ij
Nj
ij
i
6)
W Wi = N ij N j
7)
f (Wi 0 ) = f ( f (Wi 0 ) ) = si ln Wi
8)
f (a )( x a ) + ( f ( a ) / 2!)( x a) 2 =
=
9)
s
(
i
i0
i
i0
( ln Wi1 ln Wi 0 ) + 1 2( si
i
ln Wi ) ( ln Wi1 ln Wi 0 ) 2
( ln Wi1 ln Wi 0 ) + 1 2( si1 si 0 ) ( ln Wi1 ln Wi 0 ) = 1 2( si1 + si 0 ) ln(Wi1
i
) 1 2 ( si0 + si1 ) ln(Wi1
ln W1 W0 =
i
10)
s
W1 W0 =
(W
i1
)
Wi 0 + ln (1 + )
Wi 0 )
i
Wi 0 ) 1 2( si 0 + si1 ) (1 + )
i
42
Diewerts Quadratic Lemma
Let
i = set of standard industrial classifications for the sectors in the economy,
where e.g., i = 1 and 2 for agriculture and non - agriculture sectors
j = set of geographic regions of the U.S. where j = 0 to 9
such that 0 is the U.S. and 1 is the first region within the nation
W j = the average wage per job in region j
Wij = the average wage per job in sector i in region j
N ij = the employment in sector i in region j
Sij = the share of jobs in sector i in region j
s ij = the share of wage expenditures in sector i in region j
j = the industry mix coefficient in region j
43
9.5 Maclaurin Series of a Polynomial Function
f ( x) = a0 + a1 x + a2 x 2 + a3 x 3 + ... + an x n
primative function
f / ( x) = a1 + 2a2 x + 3a3 x 2 + ... + nan x n 1
1st derivative
f // ( x) = 2a2 + 6a3 x + ... + n(n 1)an x n 2
2 nd derivative
f /// ( x) = 6a3 + ... + n(n 1)(n 2)an x n 3
3rd derivative
f n ( x) = n(n 1)(n 2)(n 3) (3)(2)(1) an
n th derivative
Evaluating each function at x = 0, simplifying & solving for the coefficient
f (0) = a0
f (0) = 0! a0
a0 = f (0) 0!
f / (0) = a1
f / (0) = 1!a1
a1 = f / (0) 1!
f // (0) = 2a2
f // (0) = 2!a2
a2 = f // (0) 2!
f /// (0) = 6a3
f /// (0) = 3!a3
a3 = f /// (0) 3!
f n (0) = n(n 1)(n 2)(n 3) (3)(2)(1)an an = f n (0) n!
Substituti ng the value of the coefficients into the primative function
f ( 0) 0
f / ( 0) 1
f // ( 0 ) 2
f (n) ( 0 ) n
( x) +
( x) +
( x ) + +
( x)
f(x) =
0!
1!
2!
n!
44
9.5 Maclaurin Series of a Polynomial Function
Given
f(x) = 2 + 4 x + 3 x 2
f( 0 ) = 2
f /(x) = 4 + 6 x
f /( 0 ) = 4
f //(x) = 6
f //( 0 ) = 6
f ( 0) 0
f / ( 0) 1
f // ( 0 ) 2
f (n) ( 0 ) n
( x) +
( x) +
( x ) + +
( x)
f(x) =
0!
1!
2!
n!
2
( x ) 0 + 4 ( x )1 + 6 ( x ) 2 = 2 + 4 x + 3x 2
f(x) =
0!
1!
2!
45
9.5 Taylor Series of a Polynomial Function
Let x = x0 + where x 0 is a chosen point and is the deviation from x 0
f ( x) g ( ) = a0 + a1 ( x0 + ) + a2 ( x0 + )
2
f / ( x) g / ( ) = a1 + 2a2 ( x0 + ) + 3a3 ( x0 + )
primitive function
2
1st derivative
f // ( x) g // ( ) = 2a2 + 6a3 ( x0 + )
2 nd derivative
f /// ( x) g /// ( ) = 6a3
3rd derivative
At = 0, we know from the Maclaurin series expansion that
g ( 0)
g / ( 0) 1
g // ( 0 )
g (n) ( 0 )
0
2
( ) +
( ) +
( ) + +
( ) n
g ( ) =
0!
1!
2!
n!
Therefore x = x0
and
= x-x0 = 0.
Since g ( ) f(x), then g( 0 ) = f ( x) = f(x0 ) and
f ( x0 )
f / ( x0 )
f // ( x0 )
0
1
( x x0 ) +
( x x0 ) +
( x x0 ) 2 +
f ( x) =
0!
1!
2!
f (n) ( x0 )
( x x0 ) n
+
n!
46
9.5 Taylor Series of a Polynomial Function
Expand to quadratic function below around x 0 = 1, with n = 1
f ( x) = 5 + 2 x + x 2
f (1) = 8
f / ( x) = 2 + 2 x
f / (1) = 4
f // ( x) = 2
f // (1) = 2
f ( x0 )
f / ( x0 )
f // ( x0 )
0
1
( x x0 ) +
( x x0 ) +
( x x0 ) 2 +
f ( x) =
0!
1!
2!
f (n) ( x0 )
( x x0 ) n
+
n!
f ( x0 )
f / ( x0 )
0
( x x0 ) +
( x x0 )1
f ( x) =
0!
1!
8
4
0
( x 1) + ( x 1)1 = 8 + 4( x 1) + R = 4 + 4 x + R
f ( x) =
0!
1!
f // ( x0 )
( x x0 ) 2
R=
2!
47
9.5 Taylor Series of a Polynomial Function
Expand to quadratic function below around x 0 = 1, with n = 1
f ( x) = 5 + 2 x + x 2
f (1) = 8
f / ( x) = 2 + 2 x
f / (1) = 4
f // ( x) = 2
f // (1) = 2
f ( x0 )
f / ( x0 )
f // ( x0 )
0
1
( x x0 ) +
( x x0 ) +
( x x0 ) 2 +
f ( x) =
0!
1!
2!
f (n) ( x0 )
( x x0 ) n
+
n!
f ( x0 )
f / ( x0 )
f // ( x0 )
0
1
( x x0 ) +
( x x0 ) +
( x x0 ) 2
f ( x) =
0!
1!
2!
8
( x 1) 0 + 4 ( x 1)1 + 2 ( x x0 ) 2
f ( x) =
0!
1!
2!
= 8 + 4( x 1) + 1( x 1) 2
= 4 + 4x + x2 2x +1
= 5 + 2x + x2
48
9.54Lagrangeformoftheremainder
Y = ( 7-x)4
primitive function
Y = -4( 7-x)
1 deriative
Y = 0 at x = 7
the critical value
/
/
3
st
*
Y (7) = 12( 7-x) = 0
2 deriviative
Y (3) (7) = -24( 7-x) = 0
3rd derivative
Y ( 4 ) (7) = 24
4 th derivative
//
2
nd
Because first nonzero derivative Y ( n ) is even (4)
and Y (4) > 0 (24), critical value is a min.
49
9.54Lagrangeformoftheremainder
Y = ( 7-x)4
primitive function
Y ( 4) = 24
4 th derivative
decision rule :
n is even (4) and > 0
therefore a minimum
50
9.6 NthDerivativeTestforRelative
ExtremumofaFunctionofOneVariable
9.61Taylorexpansionandrelativeextremum
9.62Somespecificcases
9.63Nthderivativetest
51
9.6Taylorexpansionandrelativeextremum
A function f(x) attains a relative max (min) value at x0 if
f(x) -f(x0 ) is neg. (pos.) for values of x in the immediate
neighborhood of x0 (the critical value) both to its left and right
Taylor series approximation for a small change in x
f(x) - f(x0 ) =
f
/
( x0 ) ( x x )1
0
+
f
//
( x0 ) ( x x ) 2
0
1!
2!
At the max., min., or inflection, f / (x0 ) = 0
+R
and if f //(x0 ) = 0, and if f ( 3)(x0 ) = 0,
then f(x) - f(x0 ) = R
What is the sign of R for the first nonzero derivative?
52
9.62Somespecificcases
53
9.63Nthderivativetest
If the first derivative of a function f(x) at x0 is f
/
( x0 ) =
0 and
if the first nonzero derivative value at x 0 encountered in
successive derivation is that of the N th derivative, f (n)(x0 ) 0,
then the stationary value f ( x0 ) will be :
a relative max if N is even and f (n)(x0 ) < 0
a relative min if N is even and f (n)(x0 ) > 0
an inflection point if N is odd
54