Shilov Calculus of Rational Functions LML

8/13/2019 Shilov Calculus of Rational Functions LML

1/62


2/62


3/62


4/62


5/62

LITTLE M THEM TICS LI R RY

G ShilovC LCULUS

OF R TION LFUN CTIONSTranslated from the Russian

V Kisin

MIR PU LISHERSMOSCOW


6/62

First published 976Second printing 982

English translation Mir Publishers 982


7/62

Foreword Graphs Derivatives3 ntegr ls4 Solutions to Problems

ont nts

67 43748


8/62


9/62

r phsThough we assume that the reader is conversantwith graphs, we shall anyway remind the basic points.Let us draw two mutually perpendicular straight lines,one horizontal and one vertical, and denote by 0 theirintersection point. The horizontal line will be referred to asthe axis of abscissas and the vertical l ine-as the axis ordi-nates The point divides each line into two semi-axes,a positive and a negative one; the right-hand s emi-ax is ofabscissas and the upper semi-axis of ordinates are calledpositive, while the left-hand semi-axis of abscissas and thelower semi-axis of ordinates are called negative. We markthe positive semi-axes by arrows. Now the position of eachpoint M on the plane can be defined by a pair of numbers.To do this we drop perpendiculars from the point M ontoeach of the axes; the perpendiculars will cut on the axes thesegments OA and OB Fig. 1). The length of the segment OAtaken with the sign if A is located on the positive semiaxis and with the sign if it lies on the negative semiaxis, will be called the abscissa of the point M and will be

denoted by x Similarly, the length of the segment OB withthe same rule of signs) wi ll . be called the ordinate of thepoint M and denoted by y The numbers, x and y are calledthe coordinates of the point Each point on the plane isdetermined by coordinates. Points of the abscissa axis havethe ordinate equal to zero, while the points of th e ordinateaxis have zero abscissa. The origin of coordinates 0 ~ t l ipoint of intersection of axes) has both coordinates equal tozero. Conversely, if two arbitrary numbers x and y of anysigns aregiven, we can always plot, and this is very impor-

8 M

2*

o x

Fig. 1

A

7


10/62

tant exactly a unique point M with the abscissa x and theo r i ~ t e y to achieve this we have to layoff the segmentO = x on the abscissa axis and to erect a perpendicularAM = y signs being taken into account); the point M willbe the one sought for.Let the rule be given which indicates the operations thatshould be performed over the independent variable denotedby z) to obtain the value of the quantity of interest denoted by y .Each such rule defines in the language used by mathema-ticians, the quantity y as function of the independent variable x.I t can be said that a given function is just that specific ruleby which the values of yare obtained from the values of xFor instance, the formula

fy= 1 x2indicates that to obtain the values of y we have to squarethe independent variable x add it to unity and then divideunity by the obtained result. If x takes on some numericalvalue Xo then by virtue of our formula y takes on a certainvalue Yo. The values o and o define a point M 0 in the planeof the drawing. We can then replace Xo by another number Xl \and calculate by the formula the new value YI; the pairof numbers Xl YI defines a new point M I on the plane. Thegeometric locus of all points of the plane, whose ordinatesare related to abscissas by the given formula, is called thegraph of the corresponding function.Generally speaking, the set of graph points is infinite sothat we cannot hope to plot all of them without exception byusing the foregoing rule. But we shall not have to do thatIn most cases a certain moderate number of pointsIs sufficient for us to be able to realize the general shape of thegraph.The method of plotting a graph point-by-point consistsjust in plotting a certain number of graph pointsand in joining these points by as smooth a curve aspossible.As an example we shall consider the graph of the function

iY= z2 1


11/62

y

o oo oo t r t t ~ x

Fig. 2

0 x

Fig. 3

Let us compile the following table f 2 3 1 2 I 3 f 1/2 1/5 1/10 1/2 1/5 I 1/10

The first line l ists the values of 0 1 2 3 1 2 3 .As a rule integral values of are more useful for calcula-tions. The second line lists the corresponding values of calculated by formula (1). Plotting the correspondingpoints on the plane (Fig. 2) and connecting them by a smoothcurve we obtain the graph (Fig. 3).The rule of plotting a graph point-by-point is, as wehave seen, extremely simple and requires no science .Nevertheless, it may be for this very reason that blindadhering to this point-by-point rule may be fraught withserious errors,


12/62

y

o oo- - Q - - - - Y - - - + - - - - - t - - = - - - - + - - - . . . I o o f - - - - - I ' - - - - - - - : : ~ Fig. 4

- - - - o - - - - o - ~ = - - - + _ _ - _ _ - - - _ + _ - . . : . : : : : : ~ _ e ; _ _ - - 9 ' _ - - ' ~ x

Fig. 5Let us plot point-by-point the curve specified by the

equation 1Y= 3x2 1 2 2The table of x and y values corresponding to this equationis as follows

y

ot 1/4 11/12111/6761 1/4 I 1/121I1/676

The corresponding points on the plane are plotted inFig. 4 which is very similar to Fig. 2. Connecting the plottedpoints with a smooth curve we obtain the graph Fig. 5).It may seem that we could put the pen away and feel satisfied: the art of plotting graphs has been grasped But for thesake of a test let us calculate for some intermediate valueof x for example for x = 0.5. After performing the calcula-tions we obtain an unexpected result: y = 16. This is in10


13/62

y

~ ~ ~ X

Fig. 6 Fig. 7striking disagreement with our graph. And we cannot guarantee that calculation of y for other intermediate valuesof x-and their number is infinite-will not produce evengreater discrepancies. Unfortunately the method of tracingthe graph point-by-point proves rather unreliable.We shall discuss below another method of graph plottingwhich is more reliable in the sense of safeguarding us fromsurprises similar to one we have encountered above. Usingthis method we shall be able to plot the correct graph ofEq. 2). In this method-let us term it for instance bysuccessive operations w have to perform directly in thegraph all the operations which are written down in a givenformula, viz. addition subtraction multiplication division, etc.Let us consider a few simplest examples. We shall plota graph corresponding to the equation

y 3This equation expresses that all points of the curve of interest have equal abscissas and ordinates. The locus of thepoints for which ordinates are equal to abscissas is thebisectrix of the angle between positive semi-axes and of theangle between negative semi-axes Fig. 6).The graph corresponding to the equation y = witha coefficient is obtained from the foregoing graph bymultiplying each ordinate by the same number Let us setfor example 2; each ordin-ate of the foregoing graph

11


14/62

4 ~ X

.y

_ . . I . ~ X

Fig 8 Fig 9must be doubled so that as a result we obtain a straightline rising more steeply Fig 7). With each rightward stepalong the x-axis the line rises two steps up along the y axisBy the way this enables us to perform readily the plottingon squared paper. In the general case of the equationy kx with an arbitrary k a straight line is obtained Ifk > 0, then with each rightward step the line will r s ksteps up along the y axis If k < 0, the line will descend.Consider the formula

y kx bTo plot the corresponding graph we have to add to eachordinate of the already known line the same number b. Thiswill shift the straight line y == kx as a whole upward inthe plane by b units for b > 0; if b < 0 the original curvewill naturally be shifted downward). As a result we shallobtain a straight line parallel to the original one;it doesnot pass any more through the origin of coordinates butcuts on the ordinate axis the segment b Fig. 8).The number k is called the slope of a straight line y = kx + b we already mentioned that this number k showsby what number of steps the straight line moves upward pereach rightward step In other words, k is the tangent ofthe angle between the direction of the x-axis and the straightline y = kx b.The equation k ) Y X Xo Yo xcorresponds to the s traight line with the slope k i t passes

through the point x o Yo Fig. 9), since setting x Xugives y = Yo.12


15/62

y

Co l

08

J A//

... x 01 Fig. 10 Fig. 11

5)

Thus the graph of any first-degree polynomial in x isa straight line which is plotted according to the aforesaidrules. Let us pass over to the second-degree polynomials.

Consider the formulay x

It can be presented in the formy= where Yt = x

I n other words the required graph Till be obtained if eachordinate. of the already known line y x is squared. Letus find out what this should produce.Since 02 0, 12 ::=: 1, { 1 2 1, we obtain three reference points A B C Fig. 10). If x > 1, then x > x therefore to the right of the point B the curve wil l be above thebisectrix of the quadrant angle Fig. 11). If 0 < x < 1,then 0 < x < x; therefore the curve between the points Aand B will be under the bisectrix. Moreover we state thatas it approaches the point A the curve will enter an anglebounded above by the line y == kx however small k andbelow by the x-axis; indeed the inequality x < kx issatisfied for all x < k. This fact means that the sought-forcurve is tangent to the abscissa axis at the point 0 Fig. 12).Let us move nov; leftward along the x-axis from the poin t OWe know that the numbers and Then squared give3-742 13


16/62

y

A o

Fig. 12 Fig. 13

the same result a2 The ordinate of our curve will thereforebe the same both for x = a and for x = a In geometrical terms this means that the graph of the curve in the lefthand semi-plane can be obtained by reflection relative tothe ordinate axis of the curve already plotted in the right-hand semi-plane. We obtain the curve which is called the,parabola Fig. 13Now, fol lowing the same procedure, we sketch a morecomplicated curve

y = ax 6and a still more complicated one

y = ax2 b 7The first of these curves is obtained by multiplying allordinates of parabola 5 -we shall refer to i t as a referenceparabola by a number aIf a > 1 the curve will be similar to 5 but will risemore steeply Fig. 14 .If 0 a < 1 the curve will be less steep Fig. 15 , andwhen a < 0 its branches will turn downward Fig. 16 .Curve 7 will. be obtained from curve 6 by shifting itupward by a segment b if b > 0 Fig. 17 . If b 0, wehave to sh ift the curve downward F ig. 18 . All these curvesare also called parabolas.14


17/62


18/62


19/62

f ~ x

y

a

Fig. 20

2 3

Fig. 21

x

product is positive too. Between the points 2 and 3 oneof t he m ul ti pl ie rs is negative a nd t herefore the product isnegative. Two of the multipliers are negative between thepoints 1 and 2, so that the product is positive, etc. Weobtain the following distribution of signs Fig. 20 . TQ theright of the point 3 a ll m ult ip li er s w ill increase as increases, so that the product will increase greatly at that. To theleft of the point 0 a ll m ul ti pl ie rs increase their negativevalues and the product which is positive also rises steeply.

Now i t is not difficult to sketch the general shape of thegraph. Fig. 21 .So far we only used the operations of addition and multiplication. Now we supplement them with division. Let usplot the curve

1 9\Y= x To do this we s ha ll p lo t s ep ar at el y the graph of the numerator and that of the denominator.The graph for the numerator

l 1is the straight line parallel to the abscissa axis at the distan-ce 1 from it. The graph of the denominator

xis a reference parabola s hifted upw ar d by 1. The two graphsare shown in Fig. 22.Let us divide each ordinate of the numerator by the cor-responding i.e. taken for the same ordinate of the d C J O -

f7


20/62

- - - - - - - - - - - - - - - - - - - - ~ xo ig

o- - t - - - - + - - - - - - I - - - - - I - - - - 4 - - - - - - : j ~ - - - = ~ = = = - - - . x

ig

umer tor J ~ . . . . x

ig


21/62

y

~ t ~ Xo

Fig. 25

umer tor

Fig. 26minator. If = 0 then l 1, whence y 1. Forx 0 the numerator is less than the denominator and thequotient is less than unity. Since both the numerator andthe denominator are always positive the quotient is posi-tive too; hence the graph is enclosed within a band betweenthe abscissa axis and the line y = 1. When x increaseswithout limit the denominator also increases withoutlimit while the numerator remains constant; therefore thequotient approaches zero. All this results in the graph of thequotient shown in Fig. 23. The curve thus obtained is thesame as that plotted point by point Fig. 3 .In the method of graphical division special role is playedby those values of x with which the denominator becomeszero. If this is not accompanied by the numerator also turn-ing zero, the quotient becomes infinite. To realize themeaning of this expression let us plot the curve

y 10We already know the graphs of both the numerator and thedenominator Fig. 2 i . For x = 1 we have Yl = 1,whence y 1. For x > 1 the numerator is smaller than thedenominator and the quotien t is less than 1, similarly tothe foregoing example infinite rise of results in the quo-tient approaching zero and we obtain the part of the curvecorresponding to the values x > 1 Fig. 25 .Let us consider now the range of values between 0 and1. When approaches zero from the side of 1, the denomina-tor approaches zero while the numerator remains equal to 1.Therefore the quotient increases without limit exceeding

f9


22/62

Fig. 27

however large number given sufftciently small valuesof and we o bta in th e branch going off to infinity Fig. 26 .If 0 the denominator and consequently the wholequotient become negative. The general shape of the graph is presented in Fig. 27.

f lg. 2820

y

Fig. 29


23/62

Now we are ready to tackle theproblem of plotting the curve discussed at the beginning of thissection

y

Fig. 30

fY= 3x2 1 2 11Ve shall first. plot the graph ofthe denominator. The curve Yl 3x2 is the triple reference parabola Fig. 28 . Subtraction ofunity means shifting the graph byone unity downward Fig. 29 . Thecurve intersects the x-axis at twopoints which can easily be foundby setting 3x2 - 1 equal to zero:

X1 2 + Y1/3= + 0.577 Letus square the obtained. graph. Theordinates of the points Xl and X 2will remain equal to zero. All other ordinates will bepositive and the graph will be located above the abscissaaxis. At the point x 0 the ordinate will be equal to_1 2 0 = 1, and this is the maximum ordinate within theinterval from Xl to 2 Outside this interval the curve willsteeply rise upward on both sides of the plot Fig. 30 .The graph of the denominator is thus plotted. The dottedline in the same drawing shows the graph of the numerator Y2 = 1. Now we only have to divide the numerator bythe denominator. Since both the numerator and the denominator are always positive, the quotient will be posit ive andthe whole graph will be located above the abscissa axis.For x = 0 the numerator and the denominator are equal,and their ratio is equal to 1. Let us move rightward frompoint 0 along the abscissa axis. The numerator remainsequal to 1 while the denominator decreases; consequently,the quotient increases from its value 1. When we reach thepoint 2 = 0.577 ... , the denominator becomes equal tozero. This means that by this moment the value of thequotient will go to infinity Fig. 31 . To the right of thepoint 2 the denominator will rapidly change in a reverseway from the value 0 to 1 and then will grow unboundedly.The graph of the quotient, on the contrary, will return frominfinity to 1, will intersect the line y = 1 at the same point4-742 21


24/62

vy

enomin tor

~ o ~ ~ x

Fig. 31a

Fig. 32

as the graph of the denominator and will then approximatewithout limit to zero Fig. 32 .The pattern on the left-hand side of the ordinate axiswill be the same Fig. 33 .We have marked on this graph the points correspondingto integer values x 0, 1, 2, 3, 1 2 3 . These arethe very points which we selected while plotting the graphpoint-by-point in page 10. But the true shape of the graphdiffers considerably from that given in Fig. 5We see that in reality the curve does not decrease smoothlyfrom the value 1 for x 0 to the value 1/4 for x == 1 andto still smaller values, but instead goes up to infinity. I-Ierewe can also locate the point with the coordinates x 1/2,

y = 16 which could not be forced onto the former, erroneousgraph but fits perfectly the new, correct graph.We have discussed the simplest operations which can beperformed with graphs. More precisely, we started off withthe simplest equation y == x and then used the four arithmetic operations: addition subtraction multiplication anddivision.The functions y x obtained as a result of such operationsare presented in the form of a quotient of two polynomials

y x _ _ _x_} ~ a _ o _ a _ f x _ _ _ _ _ t ~ a n x nQ x bo btx o +bmxm

22


25/62


26/62


27/62

tYoI

y

Fig.

y

oFig. 35

I t is easy to imagine a specific situation in which sucha problem would arise. For instance let the graph of Fig. 34plot the cost of one ton of some product as function of dailyconsumption of electric energy. Low daily consumption ofenergy will mean very siow production of one ton of theproduct and the existence of permanent expenditures costof manpower etc. will result in high cost per ton of theproduct. High daily consumption of energy will shortenthe time of production of one ton of product but the costper ton will be increased due to the rise in the cost of theenergy consumed. This cost per ton will be minimum ata certain value of the daily energy consumption; we arenaturally very much interested to know what this minimumcost should be and what daily consumption of energy corresponds to. Below we shall discuss a similar problemp. 32 .To obtain an exact answer to the question on the positionof the point of local minimum new methods are requiredwhich lead us into the field of the calculus called the diffe-r nti l calculus.The idea of solving the suggested problem is as follows A tangent straight line can be drawn at each point of thegraph y x . The tangent straight line at the point A ofthe y x graph Fig. 36 is defined as the line a passingthrough the point A in such a way that the curve y x itself,as it approaches the point enters into any no matterhow small angle with the vertex in containing the straightline x and remains inside it. In Sec. 1 the abscissa axiswas said to be tangent to the reference parabola in just thissense. An arbitrary straight line passing through thepoint A is called the secant for the graph y x ; an anglewith the vertex in A and the bisectrix into which the curvein. the vicinity of the point A does not enter can always be

25


28/62

indicated for a secant not coinciding with the tangentFig. 37). We shall denote the sl.ope of the tangent at thepoint x, y by k = k x . This function k x is called thederivative of the function y x). Later we shall prove that ify x is a polynomial, then k x is also a polynomial, andif y x is a rational function, then k x is a rational func-tion too, and shall present precise rules for the calculation01 k x . Let us assume the function k x as already foundfor the specified function y x . At the sought-for point of thelocal minimum Xl , Yo) the tangent line X must be horizontal{reductio ad absurdum the proof by contradiction : wehave seen that the curve y = y x must enter into the anglehowever small containing the line if the line X is oblique,we can construct a small angle with x as its bisectrix,whose sides have slopes of the same sign Fig. 36) and,consequently, the curve y y x cannot have a local minimUl at xo Yo)] Therefore, at the point x = Xo of the localminimum, k xc = O. Thus we have the equation

k x =

yy

a

IIlYe,rI

x0 o xt ig. 3C Fig. 37

26

Generally speaking, this equation can have several solutions. Each of them defines the point xo, Yo) on the curvey = y x at which the tangent is horizontal; we must findall these solutions and among them single out that which isof interest to us. Thus, once the function k x is known, theproblem is reduced to solving the algebraic equation.We shall now pass to plotting the function k x . Firstof all let y = x 2 be our reference parabola. We want top


29/62


30/62

Fig. 40

thus it is apparent that with small deviations of x from X othe curve passes within the angle however small formed bythe straight lines

y = Yo+ 2xo + 8 x - xowhich is what will be the case if we assume that


31/62

we obtain by removing the brackets in Eq. 14) by theNewton formula* and by subtracting Eq. 13) from Eq. 14)L\y = a1+ 2a2xO + ... + n a n x ~ - l Llx+ t2 15

Further on, the slope of the secant is obtained by dividing by Ar. Since 8 2 : = 8 1 , its expression has the form

fly +2 + + n l= at a2x nanxo etThe complete equation of the secant is as follows:

y = at+ 2a2xO+.+ n a n x ~ - l + BI x - xo + oWhen we assume that = 0, 81 becomes zero and weobtain the equation of the tangent

Ytan = at+ 2a2xO+ . + n a n x ~ - l x-xo + oConsequently the expression for the slope of the tangentline isk=al 2a2xO . . l l a n X ~ - 1 16)

When Xo is fixed, k is a number; if Xo is varied, this numberwil l also vary and we shall obtain the function giving thevalues of corresponding slopes of tangents to the curvey = P x) at its various points. As we already said, thisfunction is the der ivat ive of P x); i t is denoted as P z .The obta ined formula can be written in the form

pi x = t+ 2a2x + n a n x n 1 16 )The rule of forming P x) out of P x) is quite simple:in the sum 13 each x k is replaced by kXk - l In par ticular, the der ivat ive of a constant i.e. of a func-tion which for all. values of x takes on the same value y = aois equal toO. In this case, however, this is also obviousgeometrically: the tangent to the plot of the function y = aois horizontal at each pointReturning to the general case, we shall .emphasize theequality stemming from Eq. 15):

p x + ~ x . P x) + P x) L\x + 8 2 17) The Newton formula: for any k and any u and vk k k -1u+v k=u k +- uk-IV+ uk-2v2+I 12k k -1 k -2 k k -1 k+ U k- 3v3l + u 2vk-2+- uvk 1-L uk123 I 12 1

29


32/62

Let us consider a similar problem about a tangent forthe general rational functionp (x)y = Q(x) 18

where P z and Q (x) are polynomials.Using Eq. 17 we obtain+ 8. P(xo+L\x) P x o ) + P x O ) ~ x ~ e 2 19o Y - Q x o ~ x } Q{xo)+Q (XO) J.X+82

Subtracting Eq. 18 from Eq. 19 , we obtainLl p (xo)+P (xo) ~ x 8 P (xo) =Y= Q (xo) Q (xo) L\x+ 82 Q (xo)_ [P (xo) Q(xo) - Q (xo)P (xo)] Ax+ 82 20- Q (xo) [Q (xo)+Q (xo) L\x+82

Hence the slope of the secant is~ y P (xo) Q xo}-Q (xo) P XO +Et 21I X = Q2(xo)

The complete equation of the secant takes the form_ P (xo) Q (xo}-Q 0) P XO)+81 _ +Y- Q2 (xo) 81 x Xo Yo

Let us assume that Q (xo) 0 and ~ x 0, we obtainthe equation of the tangent _ P (xo) Q (xo) - Q (xo) P (xo) +Ytan - Q2 (xo) X - Xo o

The slope of the tangent when x Xo is thus equal to, _ pI xc Q (xo) - Q (xo) P (xo)y Xo - Q2(xO}

This formula gives the rule for calculating the derivativeof the quotient r= pi Qo;.Q P (22)Let us consider several examples of applying all theobtained formulas.1. What two positive numbers, whose sum is equal to c,

yield the maximum product?This problem has an elementary algebraic solution. Notresorting to this solution, we shall use our general method.30


33/62


34/62


35/62


36/62

R

oFig. 42

C \I

Fig. 43Substituting Eq. 26) into Eq. 25), we find also the totalexpenditures for the most economical cruise:

R s 1 + Vq + yqc6. What is the tangent to the curve y = x3 for x == \Ve have y 3x2 which for x 0 obviously yieldsy = 0, so that the z-axis is the tangent Fig. 43).We see that in this case the tangent passes from one side

of the curve onto the opposite side: when x > 0 the curveis above the tangent while for x < 0 it goes under the tangentline. The points of the graph at which the tangent linepasses from one side of the curve to the other are called thepoints of infl ti n Fig. 44). Thus, the same value x 0determines the point of inflection for the family of curves

y Cx3 + xfor various values of C Fig. 45).

Indeed, we have y 0) 1, so that the equation of thetangent Iine drawn through the point 0, 0) takes the IormYtan=Z

therefore the d i ffcrenc.eY Ytan=Cx 3

alters sign when \ve pass from negative to positive values of xLet state the problem of finding the inflection pointof the graph of a given function y = f x .As can be seen from Fig. 45, if the curve passes from theposition below the tangent into that above the tangent34


37/62

as x increases -then its slope .fr {x) on both sides of the pointof tangency is greater than exactly at this point:y . x)> y xo)

x X oSimilarly if the curve passes from the posit ion above thetangent into the below-tangent position as x increases,then x) on both sides of the tangency point is less thanexactly at this point:

x) < xo)x =f;= X o

In the first case the inflection point is that of a local minimum of the function x), and in the second case-that ofa local maximum of this function. To find these points wemust first calculate the derivative of the function z),The function (y (x)) is called the second deriuatice of thefunction y x) and is denoted as y x). f next step is tofind the solution of the equation

y x) = Abscissas of all inflection points sought for are containedamong the solutions to this equation. Superfluous solu-

/_ - - - - - - - A : . . . - - - - - - - - - - j ~

_

Fig. 44 Fig. 4535


38/62


39/62

Fig. 47

16. Square pirces are cut c ~ \ ~ ~ asquare sheet f iron at its c ~ ~ : ~ ~ i : ~ r s andthe sheet Is then bent dl I1.g thedashed lines) so that the sheet ,is converted into a o x 0 pencd from p . h o ~ e(Fig. 47). At what size of the removedsquare pieces will the volume ofthe box be maximal?17. The formula for the derivativeof the function y = y f x is

y x2V f x

Having assumed this formula to be valid, sol ve the fol lowingproblem . l house is at a distance nfh km from a straight road(Fig. 48). A person has to walk to the road and then driveto the town which is at a distance measured along thestraight line of s km frcm the house, The speed of tho pede3t-rian is u and t I ~ 8 spood of the car is v Find the shortestroute.

Fig. .i818. How large should a sector cut out of a circle with theradius R be so that tho funnel rolled out of the remainder

would have maximum holding capacity?Numerous problems involving derivatives and theirapplication to various fields of rna thematics, physics, etc.can be found in popular books of problems.

3. IntegralsHow should we define the area of a plane figure bounded,

in the general case, by a non-rectilinear contour? Let usstart with the following premises:37


40/62

O : J I I ~

FIg. 49

1. The area of a rectangle with sides a and b is equal to bo 2. Areas of equal i.e. coinciding when superposed figures are equal.3. If the figure > is composed of figures


41/62


42/62


43/62

reciprocal to differentiat ion and called integration ~ ~function x) whose derivative is y x) is called the anti-derivative of y x , or the integral of y x). Let us note thatif we have determined one of the antiderivatives x)of the function y x) then any other function of the type x) C where C is constant is suitable as an antiderivative of the same function y x) since the derivative ofa constant is zero As we have mentioned, the sought-forfunction S x) drops to zero when :r a. Therefore, havingdetermined an antiderivative x), we can write for theunknown constant C the equation

S a) = F a) C = 0, whence C a)Finally, having found the antiderivative function F x , weshall write the sought-for equation in the forms x) = F x) F a)and, in particular,

S b) == F b) F a) 27which gives the solution of our problem or, more precisely,the solution is reduced to the problem of determining theantiderivative of a given function f x).

To make possible the application of the general result 27obtained above we must be able to determine the antiderivative. If the function y ==y x) is a polynomial

y ao alx anxnthen one of the antiderivatives can easily be written, viz .:

Therefore, no substantial difficulties are encountered incalculations of areas of figures bounded above by curvesof the type y P x), where P x) is a polynomial.Let, for example, y = y x) be a linear function Fig. 55varying on the segment a x b from the value p to thevalue q:

q p x -ab -a41


44/62


45/62

2. Let us calculate the areas 8 82 and 8 3 Fig. 57 ou -ded by the curve y = x x - 1 x - 2 x-3 and by the segments of the z axis from 0 to 1, from 1 to 2, and from 2 to 3.Here we havey = p x = x4 - 6x3 11x2 - 6xand its antiderivative is

F { x ~ 6x4 1ix3 _ 6x\ - 5 4 3 2Whence

f 3 11 19S t=F \1 -F O == S -2+T - 3= 3The minus sign indicates that the area 8 1 lies below thex-axis. Then

S 2 = F 2 ) - F 1 ) = ~ - 2 4 + 88 - 1 2 + ~ = i .5 3 30 30and, finally,Sa- F 3 -F 2 - 243- 243 99 27 i _ l- - 5 2 15 - 30

The last result coincides with 8 1; this could be predictedon the basis of symmetry3. Let us calculate the area S Fig. 58 under the curvey t/x2 between the lines x = 1 and, x = where N isa large number The antiderivative of y x = t/x2 is,y

c 8y

o f ~ . ~ . ~ XFig 56 Fig 57

43


46/62

y\ \ :::o 1

1Y=;2

Fig. 58

... xN

obviously F x ==- 1/x , and vie obtainS = F IV - F 1 1 - N 29

I t is interesting that nt the V J G V I ~ e V u [ J[l;ge, this area issmaller than 1. Formula 29 indicates th.at i t is logical toassign to the whole infinitely stretched figure, bounded belowby the z-axis, above by the curve y == x2 , on the left bythe segment of the line x == 1 and not bounded on the r ight -hand side, a finite area, namely the area equal to 1.We see that calculus of integrals based on tho calculusof derivatives enables us to solve in a unified manner a num-ber of problems on areas, which cannot he solved by meansof elementary mathematics. However, the problem of areacalculation is a comparatively particular problem, only onerealization of the general problem of finding the antideriva-tive function from its derivative. l many problems inmat.hem atics, mechanics, ph ysics, chemistry, biology arereduced to this general problem: the integra l calculus makesi t possible to solve by means of a general method a greatnumber of problems with most varied specific conditions butwi th common mathematical essence for instance, calculation of energy required for launching a satellite; finding thelaw that governs radioactive decay; quantitative analysisof the course of a chemical reaction or of proliferation ofbacteria . Not being also able to describe here all theseattractive applications we advise the reader to pay atten-tion to a monograph by G. Phillips Dif ferential equa-tions , containing a wide variety of problems concerningdiverse fields of science and technology and requiring appli-cation of the integral calculus.4. More careful approach will be necessary in the exampleto follow. The subject is the area S under the same curvey = x2 between the lines x a and x = b> a Fig. 59).44


47/62


48/62


49/62

y

Fig 61

y

Fig 6220. alculate the area between two curves Fig. 61 :

y = cxm X = dynRecommendation. Use Example 1 of Sec. 3.21. alculate the area bounded above by the linex y = 2 and below by the parabola y = x2 Fig. 62 .Recommendation. Express the area in question as the differencebetween two areas bounded on both sides by vertical lines.

22. The velocity gained by a falling body in the time telapsed after falling began is equal to gt g = 9.81 m s2 .What is the distance covered by the body during this time?Recommendation. Velocity is the derivative of the displacementwith respect to time.


50/62

olutions to roblems

A ( - ~ )I I I 2 4 I x2 1 21

Fig 63 To problem 1 Fig 64 To problem 2

Fig 66 To problem rIIIIII I I

- , . - - - - - - I - - - . . 3 I I r - - - - - 4 - - - - - - ~ ~ X

Fig 65 To problem 3

Fig 67 To problem 5


51/62

:v:I II(I Ir.I I I ,

I I

I/ III111III

\\\\\\\\\\\,

\ \\ \ \ \ ,

IIIIIIIIII /1 //A

/ II

y

ig 68 To problem 6 ig 69 To problem 7

y

ig 7 To problem 8 ig 71 To problem 9


52/62

o

- - -\.-

~ X///

////

////

Fig 72 To problem 10

ig 74 To problem 12

y

ig 73 To problem 11

ig 75 To problem 13

ig 76 To problem15 The height of the can must be equal to its diameter16 The sides of the removed squares make up 1/6 of the side ofthe whole square

50


53/62

7 The sine of the angle between the path of the pedestrian andthe normal to the road must be equal to ulv provided this ratio doesnot exceedY 1 - ~ f

Otherwise the shortest route is the path covered on foot alongthe straight line towards the town18. a n V i radian 930 19. S = 33 8

n l m20. S c 1 nmar m 1 n ~ 1 - m ~ 1 .21. S = 9 222 8 = gt 2/2.


54/62


55/62

OTH R OOKS FOR YOUR LI R RY

THE KINEMATIC METHOD INGEOMETRICAL PROBLEMSby YU.I. LYUBICH L SHORWhen solving a geometrical problem it is helpful to imagine what would happen to the ele-ments of the figure under consideration if someof its points started moving. The relationshipsbetween various geometrical objects may thenbecome clear graphically and the solution of theproblem may become obvious.The relationships between the magnitudes of segments angles and so on in geometrical figuresare usually more complicated than the relation-ships between their rates of change when the figure is deformed. Therefore in solving geometrical problems one may benefit from a theoryof velocities i.e from kinematics.This little book uses a number of examples toshow how kinematics can be applied to problems of elementary geometry and gives someproblems for independent solution. The neces-sary background information from kinematicsand vector algebra is given as a preliminary.The book is based on lectures given by the authors for school mathematics clubs at the KharkovState University named after A.M. Gorky. It isintended for high school students.


56/62

SOLVING EQU TIONS IN INTEGERSby A.O. GELFONThis book is devoted to one of the most interesting branchesof number theory the solution of equations in integers.The solution in integers of algebraic equations in more thanone unknown with integral coefficients is a most difficultproblem in the theory of numbers. The theoretical importanceof equations with integral coefficients is quite great as theyare closely connected with ny problems of number theory.Moreover these equations are sometimes encountered inphysics and so they are also important in practice. The elements of the theory of equations w t integral coefficientsas presented in this book are suitable for broadening themathematical outlook of high-school students and studentsof pedagogical institutes. Some of the main results in thetheory of the solution of equations in integers have beengiven and proofs of the theorems involved are supplied whenthey are sufficiently simple.


57/62

THE REMARKABLE CURVESby A I MARKUSHEVICHThis small booklet has gained considerable popularityamong mathematics enthusiasts It is based on a lec-ture delivered by the author to Moscow schoolboysin their early teens contains a description of a circleellipse hyperbola parabola Archimedian spiral andother curves The book has been revised and enlargedseveral times in keeping with the demands from readersbut the characteristic style of the book which is de-monstrative and lucid rather than deductive and dryhas been retainedThe booklet is intended for those who are interested inmathematics and possess a middle standard background


58/62


59/62

1_ _________


60/62

Little Mathematics Libraryc x

G E SHILOV

L ULUSOF R TION L FUN TIONS

Mir Publishers Moscow


61/62


62/62

The pamphlet "Calculus of Rational Functons" discusses

graphs of functions and the differential and integral calculi

as applied to the simplest class of functions, viz. the rationalfunctions of one variable. It is intended for pupils of senior

forms and first year students of colleges and universities.

Documents

Shilov Calculus of Rational Functions LML