CHAPTER 4 SUPPLEMENTARY EXERCISES 1. Simplify the following expressions
5.4 FINITE DIFFERENCES AND ANTIDIFFERENCES
FINITE DIFFERENCES AND ANTIDIFFERENCES 153 29. The current,I, (in amperes) in an electrical circuit isI=100∕RwhereRis the resis- tance (in ohms). Find the rate of change ofIwith respect toRwhen the resistance is 5 ohms.
30. Suppose that the pricep(in dollars) and the monthly salesx(in thousands of units) of a commodity satisfy 3p3+2x2=3800. Determine the rate at which sales are changing at a time whenx=20,p=10 and the price is falling at a rate of $1 per month.
31. Consider the Cobb– Douglas Production Function, 20x1∕4y3∕4 =1080. Use implicit differentiation to calculate dy
dx whenx=16 andy=81.
Similarly, one defines asecond difference, analogous to a second derivative, as Δ2f(x) = Δ(Δf(x)) = Δ(f(x+1) −f(x))
= [f(x+2) −f(x+1)] − [f(x+1) −f(x)]
=f(x+2) −2f(x+1) +f(x)
⧫Note that the use of Δas a “difference” differs from its frequent use as “a small change” in many calculus texts. Some texts attempt to overcome possible confusion by indicating a first difference by a capital E as in
E(f(x)) =f(x+1) −f(x) However, that choice is not without possible confusion.
Example 5.4.2 Second Finite Difference
If f(x) =x2 and Δf(x) =2x+1 find Δ2f(x) Solution:
Δ2f(x) = Δ(Δf(x)) = [2(x+1) +1] − (2x+1)
=2
Alternately, using the second difference formula directly, Δ2f(x) =f(x+2) −2f(x+1) +f(x)
= (x+2)2−2(x+1)2+x2
=x2+4x+4−2x2−4x−2+x2
=2
Example 5.4.3 First and Second Finite Differences (cont.)
If f(x) =ax, findΔf(x)andΔ2f(x) Solution:
Δf(x) =ax+1−ax=ax(a−1) Δ2f(x) = Δ(Δf(x)) = Δ(ax(a−1))
FINITE DIFFERENCES AND ANTIDIFFERENCES 155
=ax+1(a−1) −ax(a−1)
=ax(a−1) [a−1]
=ax(a−1)2
Generally, ther+1stdifference off(x) is defined as
Δr+1f(x) = Δ(Δrf(x)) = Δr[f(x+1) −f(x)] = Δrf(x+1) − Δrf(x)r, a positive integer where Δrf(x)is read as “delta to the rof f(x)” or the “rth difference off(x)”. Note that Δ1= Δ.
It is easily shown that
Δ(f(x) ±g(x)) = Δf(x) ± Δg(x) and
Δ(f(x)g(x)) =f(x+1)g(x+1) −f(x)g(x) Adding and subtractingf(x)g(x+1)to the right hand side above yield
f(x+1)g(x+1) −f(x)g(x+1) +f(x)g(x+1) −f(x)g(x)
=g(x+1)Δf(x) +f(x)Δg(x) Clearly,
Δ(c(f(x))) =cΔf(x)wherecis a constant.
For the difference of the quotient, Δ
(f(x) g(x)
)
= f(x+1) g(x+1)− f(x)
g(x) = g(x)f(x+1) −f(x)g(x+1) g(x+1)g(x)
= g(x)f(x+1) −g(x)f(x) +g(x)f(x) −f(x)g(x+1) g(x+1)g(x)
= g(x)Δf(x) −f(x)Δg(x) g(x+1)g(x)
Note the striking similarity of these results to those in the differential calculus.
Example 5.4.4 Difference Operations
Let f(x) =x2+3x+5 and g(x) =3x.
CalculateΔ[f(x) ±g(x)], Δ[f(x)g(x)], andΔ[f(x)∕g(x)].
Solution:
Firstly, calculateΔf(x)andΔg(x).
Δf(x) = [(x+1)2+3(x+1) +5] − [x2+3x+5] =2x+4 Δg(x) =3x+1−3x=3x(3−1) =2(3x)
Now, using the expressions in the text,
Δ[f(x) ±g(x)] = Δ[f(x)] ± Δ[g(x)] = (2x+4) ±2(3x) Δ[f(x)g(x)] =g(x+1)Δf(x) +f(x)Δg(x)
= [3x+1](2x+4) + (x2+3x+5)[2(3x)]
=3x[3(2x+4) +2(x2+3x+5)]
=3x[2x2+12x+22]
=3x(2)[x2+6x+11]
and
Δ[f(x)∕g(x)] =g(x)Δf(x) −f(x)Δg(x) g(x+1)g(x)
=3x(2x+4) − (x2+3x+5)(2)3x 3x+13x
=3x[(2x+4) −2(x2+3x+5)]
3x+13x
=−2x2−4x−6
3x+1 = −2(x2+2x+3) 3x+1
Maxima and Minima
Discrete functions have maxima and minima just as continuous ones. They are the respec- tive largest and smallest values in a region. Locating maxima and minima is usually more arduous in the discrete case, although the basic principle is the same as for differential calculus.
A minimum off(x) occurs atx∗if
f(x∗+1)>f(x∗)andf(x∗−1)>f(x∗).
That is, the function value is larger thanE(x∗)on each side ofx∗. This can also be expressed as
Δf(x∗−1)<0<Δf(x∗).
Iff(x∗)<f(x)for allx, then the minimum is local.
Iff(x∗)≤f(x)for allx, then the minimum is absolute or global.
FINITE DIFFERENCES AND ANTIDIFFERENCES 157 We state a sufficient, but not necessary, condition for a minimum:
f(x∗−1)<0<f(x∗)andΔ2f(x)≥0 for all x.
x* x x*
Minimum
f(x) f(x)
x Maximum
A similar analysis applies for a maximum off(x). There is a maximum atx∗if f(x∗)>f(x∗+1) and f(x∗)>f(x∗−1)
These can be expressed as
Δf(x∗)<0<f(x∗−1).
This is equivalent to a sufficiency condition for a maximum,Δ2f(x)≤0 for allx. Note the similarity of the second difference to the second derivative as a sufficiency condition.
Example 5.4.5 Finding Maxima and Minima
Examine f(x) =3x2−4x+1, x=0,1,2,… for maxima and minima.
Solution:
In this case,
Δf(x) = [3(x+1)2−4(x+1) +1] − [3x2−4x+1]
= [3x2+6x+3−4x−4+1] − [3x2−4x+1]
=6x−1 By trial and error,
f(0) =1, f(1) =0, f(2) =5, f(3) =16 so
Δf(0) = −1, Δf(1) =5, and Δf(2) =11
The change of sign inΔf(x)is a signal for passing through a maximum or a minimum.
In this case,Δf(0)<0<Δf(1)so x∗=1 locates the minimum of f(x), f(x∗) =f(1) =0.
Economic Applications−Inventory Policy
A significant concern, especially in manufacturing enterprises, is the maintenance of eco- nomically defensible inventories. Too much stock and costs rise for warehousing, deteri- oration, idle capital, and obsolescence. Stock too little and costs increase for production delays, lost sales, and idle workers. It is a vital managerial function to specify the trade-off of these “opposing” costs.
The figure illustrates Inventory vs. Time. An initial inventory,Q, is used to satisfy a fixed demand ofRitems per unit of time,t,(t=0, 1, 2, …, T)until depletion att=T when the cycle repeats.
0 Q
T Time
Inventory
t
There is a “holding cost” to maintain and store inventory. It usually varies directly with the amount and duration of inventory held. There is also a “set up cost” to initiate a cycle to produce inventory (if there were no setup cost in this example, there would be no need to hold inventory as items could be manufactured to meet demand).
The trade-off is clear! If holding costs are reduced by shorter production runs, the need for more frequent set ups increases and vice versa. Management seeks a balance that min- imizes the total cost per unit time.
As the inventory varies uniformly fromQat replenishment to zero on depletion at time T, the average inventory isQ/2. LetChbe the holding cost per unit of inventory per unit time. The holding cost in a cycle isChQT∕2. Also, letCsbe the setup cost per cycle.
The total cost per cycle isCs+ChQT∕2. The total cost per unit time,Z, is
Z= Cs T +ChQ
2 The total cost per unit time,Z, can also be written as
Z= CsR Q + ChQ
2 .
FINITE DIFFERENCES AND ANTIDIFFERENCES 159 The quest for a minimum begins with a first difference ofZ. That is,
ΔZ= CsR Q+1+Ch
2 (Q+1) −CsR Q − ChQ
2
=CsR
[Q− (Q+1) Q(Q+1)
] + Ch
2
= −CsR Q(Q+1)+Ch
2
SettingΔZ=0 and solving the resulting quadratic yield the roots
Q=
−1±
√
1+8RCs Ch 2 Clearly, the negative root is extraneous.
⧫Inventory policies are among the most important in the profitability of manufacturing enterprises. Large sums are spent in devising better inventory policies and it is the sub- ject of a large and sophisticated literature. There are actually professional societies for inventory-related personnel as a web search will attest. These examples only scratch the surface.
Example 5.4.6 Production Run
Suppose that the cost of initiating a production run is Cs=$100 and the unit holding cost is Ch=$1 item/day. The demand rate is R=2 items/day. Devise an optimal inventory policy.
Solution:
Use the positive root for Q, above, to locate the approximate value,
−1+
√
1+ (8)(2)(100) 1
2 =19.5
Next, calculate the value of Z at the nearest integer values.
At Q=20, ΔZ= −100 (20)(21)+ 1
2 =0.262 and at Q=19, ΔZ= −100
(19)(20)+ 1
2 = −0.263
The change in sign ofΔZ, as Q varies from 19 to 20, signals the passage through a minimum in this case.
Z(19) = 100(2)
19 +(1)(19)
2 =20.03 $∕day Z(20) = 100(2)
20 +(1)(19)
2 =20.00 $∕day
There is little choice between the two policies, Q=19 and Q=20 in this instance.
Antidifferences
Antidifferences, as the name suggests, is the inverse of differencing. It is analogous to the integral,the subject of the next chapter when the study of the “other” calculus resumes.
This finite counterpart usefully introduces the next chapter.
The antidifference off(x),F(x), is defined byf(x) = ΔF(x). It is also expressed as F(x) = Δ−1f(x) – the symbolΔ−1 denoting an antidifferencing operation. In words, the difference of the antidifference yields the original function.
⧫Caution! The−1 in Δ−1 is not an exponent! It is a commonly used mathematical notation signifying an inverse.
Example 5.4.7 Finding Antidifferences
Find the antidifferences of axand the permutationxPr. Hint:xPr= x!
(x−r)!
Solution:
We have already noted thatΔax= (a−1)axis the first finite difference of axso Δ−1ax= ax
(a−1).
For the permutation f(x) =xPr=x(x−1) · · · (x− (r−1))
ΔxPr= [(x+1)(x) · · · (x+1− (r−1))] − [x(x−1) · · · (x− (r−1))]
= [(x+1) − (x−r+1)]((x)(x−1) · · · (x+1− (r−1))
=r(xPr−1)
ThereforeΔxPr+1= (r+1)xPrso
Δ−1xPr= xPr+1 r+1
FINITE DIFFERENCES AND ANTIDIFFERENCES 161 Antidifferences are particularly useful when summing series and summations. Consider the sum
∑b x=a
f(x) =f(a) +f(a+1) +f(a+2) + … +f(b−1) +f(b). Substituting, the antidifference,F(x) yields
∑b x=a
f(x) =
∑b x=a
[F(x+1) −F(x)]
=F(a+1) +F(a+2) + … +F(b) +F(b+1) −F(a)
−F(a+1) − … −F(b−1) −F(b)
=F(b+1) −F(a)
Again, you will note the similarity to the integral calculus in the next chapter.
Example 5.4.8 Sum of the FirstnPositive Integers
Find the sum of the firstnpositive integers.
Solution:
We seek
∑n x=1
x. To evaluate the summation, F(x), by antidifferences, seek an antidifference F(x) such thatΔF(x) =x. As F(x) =x is a polynomial, try F(x) =ax2+bx+c. The coef- ficients a,b,c are determined so thatΔF(x) =x. That is
ΔF(x) = [a(x+1)2+b(x+1) +c] − [ax2+bx+c] =2ax+a+b ForΔF(x)to equal x requires 2ax=x and a=1∕2. With c=0,a+b=0 and a= −b=1∕2.
Therefore,
∑n x=1
x=
∑n x=1
Δ
(x2−x 2
)
= (n+1)2− (n+1)
2 − 1−1
2 = n(n+1) 2 You may recall this well-known algebraic result.
Summation by Parts
In many applications, challenging summations arise. Evaluating them can sometimes be simplified using asummation by parts.
Consider evaluation of a sum
∑b x=a
f(x)h(x)
The “trick” is to know an antidifference of eitherf(x) orh(x). Let us suppose that h(x) = Δg(x)whereg(x) is the antidifference ofh(x). Now, we seek to evaluate
∑b x=a
f(x)Δg(x) Earlier we had
∑b x=a
Δ[f(x)g(x)] =
∑b x=a
f(x)Δg(x) +
∑b x=a
g(x+1)Δf(x) Also, it can be shown (Exercise 5.4.12) that
∑b x=a
Δ[f(x)g(x)] =f(b+1)g(b+1) −f(a)g(a)
Substituting on the left side, rearranging, and solving for the term of interest
∑b x=a
f(x)Δg(x) =f(b+1)g(b+1) −f(a)g(a) −
∑b x=a
g(x+1)Δf(x)
Now, the first two terms require no summation and are easily obtained by simple substitu- tion. The last term does require a summation, and the central purpose in using a summation by parts is that this remaining summation is easier than the original one.
You will encounter the same idea in the integral calculus (see Chapter 7) where it is known asintegration by parts.
Example 5.4.9 Summation by Parts
Evaluate
∑3 x=1
x2Δx3. Solution:
Using the expression for a summation by parts:
∑3 x=1
x2Δx3= (3+1)2(3+1)3− (1)2(1)3−
∑3 x=1
{(x+1)3[(x+1)2−x2]}
=1024−1− [8(3) +27(5) +64(7)] =416 Enumerating, as a check,
∑3 x=1
x2Δx3=
∑3 x=1
{x2[(x+1)3−x3]} =
∑3 x=1
[3x4+3x3+x2]
= (3+3+1) + (48+24+4) + (243+81+9) =416
FINITE DIFFERENCES AND ANTIDIFFERENCES 163
Example 5.4.10 Another Summation by Parts
Evaluate
∑k x=1
xax. Solution:
The summation by parts formula does not apply directly. In an earlier example, we found thatΔax= (a−1)axsoΔ(ax∕(a−1)) =ax. Substituting for ax.
∑k x=1
xax=
∑k x=1
x (Δax
a−1 )
= ( 1
a−1 )∑k
x=1
xΔax
Set f(x) =x and g(x) =axand the summation by parts expression applies!
( 1 a−1
)∑k
x=1
xΔax= ( 1
a−1 ) [
f(k+1)g(k+1) −f(1)g(1) −
∑k x=1
g(x+1)Δf(x) ]
= ( 1
a−1 ) [
(k+1)ak+1−a−
∑k x=1
g(x+1)Δf(x) ]
= ( 1
a−1 ) [
(k+1)ak+1−a−
∑k x=1
ax+1 ]
= ( 1
a−1 ) [
(k+1)ak+1−a−a (a(
1−ak) 1−a
)]
Note thatΔf(x) = Δx= (x+1) −x=1 and that the last summation was evaluated as a geometric sum.
EXERCISES 5.4
1. Iff(x) =x3findΔf(x)andΔ2f(x).
2. Verify thatΔ ( ax
a−1 )
=ax.
3. Repeat Example 5.4.4 using an algebraic operation before differencing.
4. FindΔ(1∕(f(x))).
5. For the permutationnPr=n(n−1) … (n−r+1), a product ofrfactors, show that ΔnPr = (r)[nPr−1]for fixedr.
6. Show that
a) Δ{x} =1 b) Δ
(n r )
= ( n
r−1 )
c) Δ(an) =an(a−1) 7. Show that the condition for a maximum atx∗is
Δf(x∗)≤0≤Δf(x∗−1).
8. Repeat Example 5.4.9 by direct differencing.
9. Summation by parts
Verify the expression in the text whena=1 andb=3 starting with the differencing of
∑b x=a
f(x)g(x)for arbitraryf(x) andg(x).
10. Find the antidifferences of
a) 8x b) nP3 c) x4 d)
( n r−1
)
11. Use antidifferences and summations to evaluate a)
∑n x=1
x2 b)
∑n x=1
ax c)
∑n x=1
x3
12. Prove that
∑b x=a
Δ[f(x)g(x)] =f(b+1)g(b+1) −f(a)g(a). Hint: carry out and expand the first difference.
CHAPTER 5 SUPPLEMENTARY EXERCISES