JournalPi MU E PS~LON-' METHOD OF SECANTS FOR THE SOLUTION OF EQUATIONS
J. C. Peck, University of Southwestern Louisiana
One of the standard methods for finding the solution of equations f (x) = 0 is the "cobweb" or "spiral staircase" method of iteration. This method works fairly well if conditions are "just right;" however, if these conditions are not properly satisfied, the method is uneconomical to apply or may even fail to converge to a root.
Using the fact that a straight line approximates a curve over a small interval, a variation of the method of iteration can be derived which eliminates many of the conditions which must be satisfied for the standard method. The new method assumes that fie curve is nearly linear over a small interval and can be approximated by a secant line over this interval.
Before discussing the method of secants it is best that we give a short summary of the standard method of iteration. Let us assume that we have an equation f(x) = 0 with real roots. This equation can be expressed as: gb) = hb) where f (x) = g(x) - h (x). We next form the two equations: y = g(x) and y = 1-ix). (11 The value of x at the intersection of the graphs of these two equations is the value of x that makes f (x) = 0 in the original equation. (Fig. 1) Now the problem becomes a problem of finding the solution to the simultaneous equations (1).
Let us now assume that we have an approximation a to the root of the equation f (x) = 0. We now wish to improve this approximation. Our first step in refining this approximation is to solve one of the equations (1) for x. Solving y = h (x) for x yields Fig. 1 x = H(y). We now have the two equations: and
We then substitute our approximation a into (2) and determine y~ such that y, = g(a). (Fig. 2) This value of y is then substituted into (3) and a new value of x is determined. This new value of x is again substituted into (2) and the process is continued until two successive values of x are within the desired tolerance.
This method works very well for many cases; however, unless special care is taken in selection of the equations (I), it also fails for many cases. It is quite possible to choose these two equations in such a 146
manner that convergence will occur at a slow rate (Fig. 3) or will not - The slope of EF is:
Fig. 2 Fig. 3 Fig. 4 Fig. 5 h(j) - h(a) (fa = j-a
Consider a portion of Fig. 6. (Fig. 7) Other expressions for slopes M, and Ma are:
Fig. 7 z = (fay a root a root L. -Mix = (fay. occur at all. (Fig. 4) Care must also be exercised when deciding which one of the equations (1) is to be solved for x. If the wrong equation is chosen, Adding Max to both sides yields: divergence will occur. (Fig. 5) (fax - Mix = Hay + (fax The following method is a variation of the method of iteration which forces convergence to occur much more rapidly and also finds roots of many equations that the normal iterative method fails to determine.
The method begins to differ from the normal iterative method explained NOW is the fraction of the distance from C to F where the perpen- above at the point where one of the equations (1) is solved for x. With X+Y this new method either equation (1) can be solved for x and convergence will occur. Solving y = h(x) for x yields: dicular from 0 is drawn. This ratio will tell us how far along the line we should "slide" to take our next approximation for the root. Since the distance from C to F is the same as the distance from a to j, and our next approximation a' will be:
We shall refer to the graph of (4) as curve 1 and the graph of (5) as curve at = a+ A(j-a). - (fa -h We then repeat the process by using a' to find g(al) which is used to Substituting our initial approximation a into our original equations find H(g(al)) which we can again call j. These results can then be y = g(x) and y = h (x) , we get substituted back into (6) to yield a new a'. This process can be expressed two points: C(a, g(a)) on curve x=H(y) more compactly as: 1 and E(a,h (a)) on curve 2. y=h (XI (Fig. 6) Substituting g(a) into (5) yields a point F(j,g(a)) on curve 2, where j = H (g(a) ) . The ratio need not be calculated for every iteration; however, One more substitution of j into Via -Mi (4) yields a point D(j,g(j)) on curve 1. y=g (XI convergence will occur in fewer steps if the ratio is recalculated for each iteration. When the difference (H(g(ai ) ) - a; ) becomes sufficiently close One can easily see that the to zero, we know' that we have found the root. slope of CD is: We have observed only the case where the slopes are of opposite sign q(a) in the foregoing discussion. When the slopes are both positive, we have tt = g(l) - j-a Fig. 6 A GENERALIZATION OF THE STRUCTURE OF THE SENTENTIAL CALCULUS a graph somewhat like Fig. 8. In this case the ratio is a Ha - Ml Dennis Spellman, Temple University fraction with value greater than one so that when the difference (j - a) is multiplied by the ratio and added to a, our new approxi- x=H(Y) mation has a value greater than y=g(x) I. Introduction. j. A similar argument holds for the case where both slopes All formulae of the sentential calculus may be expressed in terms of are negative. conjunction, disjunction, and negation. If conjunction and disjunction are interpreted as operations defined on the truth-values "T" and "F" occurring in their truth-table definitions (or equivalently of substitute Several comparison tests were Fig. 8 run between the standard method of symbols, e.g. "1" and "0"), and if negation is considered a function of iteration and the method of secants these elements, then the resulting system is a Boolean algebra. In this variation of the standard method. paper we shall consider a system having a remnant of Boolean structure, as Using the FORTRAN language and an well as, containing a sub-system, the structure of which is identical with IBM 1620 computer for the calcula- that of the propositional calculus. tions, roots accurate to 8 decimal digits were sought for the equations x? - x - 30 = 0 and x? - ex = 0 . The methods compared as follows: We define the following operations on u : x? -x-30=0 Standard Method Secant Method g(x) = 6/x; h(x) = (x-l)/5 Root We make the following definition: Initial approximation Number of iterations If P â U, then p = 1 - p.
X' -eX=o 11. Interpretations g (u, @ , ) . g(x) = 2: h(x) = ex Root diverged Let p,q â U. Let p be the probability of P and q be the probability Initial approximation 3.00000000 of Q, and let P and Q be independent events. p @ q, p a q, and p can Number of iterations stopped at 97 be interpreted as the probability of P v Q, P A Q, and -P respectively where "V" represents disjunction, "A" represents conjunction, "-"represenzs negation. At the endpoints we have the operation tables below. In every case tested the method of secants converged more rapidly than did the standard method; however, special cases can be devised where this probably will not hold true. In general though, it appears that the method of secants is a definite improvement over the standard "cobweb" method of iteration.
We note also that 1 = 0 and 0 = 1. Consider the following tables: Theorem 1. F P. We see that the sub-system defined by the first set of "operation tables" Proof. 1 - (1 - P) = P- is indeed structurally identical to the sentential calculus.
p x = px --p,x â [0,11--can be interpreted graphically as a line segment having range [O,p] and passing through the points (0,O) and (1,p). AV The theorem follows from the properties of multiplication of real We see that the line segment can also numbers since p @ q = pq by definition. be characterized as the one originating from the vertex (0,O) of the unit square Theorem 3. = F @ q. [ (0,0), (1,O) , (1,l), (0,l) I and terminating Proof. = 1 - pq (1,~) at a height p on the opposite vertical F@q' = (1-p) + (1-q) - (1-p) (1-q) = 2 - p - q - (1-p-q+pq) side. (QP@~) 1-pq Mx . = F@q. (0,O) (q,O) (LO) Alternate Proof. p @ x = (1 - p) - x + p --p, x â [O, 11 -- can be interpreted graphically Case I. 0 Let p,q c U, i.e., 0 L p 1 and 0 L q ;1. ..O~pq/q~l Consider the function y = f (x) = E. Given any y â U, f: y - y; hence, .'. U is closed under @ f is onto U. Let Xi = Fa. 121-P>O . 1 - XI = 1 - x= .'.PC u* pe u . xi = xa OLq(1-P) -1-P .'. f is one-one. ... 0 LP - Proof. The theorem follows immediately from Theorem 2 since x - x is an isomorphism from (U, @) onto (U, @ ). Theorem 5. A. B. c. Proof. A. Let p=F,- a=S. From Theorem 3, 7 @ s = r-. IV. Properties. ^@q =I?= PCB = Fm '$" is to be read: "not identically equal to." P@P=P+P-pa =2p-Pa 34 P since 0 and @ are both commutative and associative, POP = pa ^ P generalized commutative and associative laws hold for both. pap=p+ (1-p) -p(l-p)=l-P+F?^l Under the isomorphism x - 57, the image of the generalized POP = ~(1- P) 7'0 product: pi (Â¥ --â (3 ft , is the generalized sum of the 1/2 @ (1/3 @ 1/4) = 1/4 images: pi @ - - @P.- Part A establishes that x - 5? is also an isomorphism from (U, @ onto (U, O 1. . like in part B, pi @ --â @p. =a(3 --- OR. We know (hat in a Boolean algebra a < n' =' a @b = 0. 1/3 ' 3/4 = 1- V p, 1 @ p = 1. 1 has no additive inverse. 1/2 @ x = 1 => x = 2f/V. .'. 1/3 3/4 7' 0 . 1/2 has no multiplicative inverse. In general, inverses do not exist. a .'.p'q- =^> pOq = 0 * We are familiar with the following three properties of the < relation: if p,q, r are real numbers then V. From Another Viewpoint. Another truth-functional connective is defined by the table below. P In Q The following theorems show that < is an inclusion-like relation in (U, @, 0). F T T F,F It is, perhaps, an unwieldy ingredient in the business of evaluating the truth-values of compound statements, but it is not at all artificial for it represents the exclusive sense of the word "or". Since all formulae of the propositional calculus can be expressed in the terms of conjunction and negation alone, they can certainly be expressed in terms of "exclusive disjunction", conjunction, and negation. Again we may consider " 111" and I should like to express my thanks to Dr. Leon Steinberg for his help 'A" as operatives defined on "T" and "F" and consider "-"as a function and encouragement. of these truth-values. Consider the tables below. REFERENCES 1. R. Le Blanc, Statistical and Inductive Probabilities, Prentice- all, I~c., Englewood cliffs, N. J., 1962. W. V. Quine, Mathematical Harper and Row, New York and EV~~S~O~I 2. a, 1962. We see that the truth-values under the operations " 111" and "A" are isomorphic to the integers modulo 2 and also to themselves under the operations "G" 3. I. M. Copi, Symbolic Loqic, The Macmillan company, New ~ork,1954. and "v" where "-"is the biconditional. Since I/(2l... is a Boolean rina with unit, the others are also. 4. N. H. McCoy, Introduction to Modern Alqebra, Allyn and Bacon, BOS~O~, 1960. We may make the following definitions of operations on U: p@q=p+q - 2pq P@q = P0q p@q=1-p-q+2pq P0q= P@a * I shall state without proof some of the properties of (U, 0,@) and (U, @, 0). U is closed with respect to all four operations. Theorem. = q" 8 p and P=F@q. Now we see that x - X is an isomorphism from (U, @, @) onto (U, @, 01, and vice-versa. Hence, we need explore the properties of only (U, 0, @I. Theorem. V p, q, r, Theorem. A. In general inverses do not exist. B. 6' does not distribute over 0. c. POP ft 0. In the same way that interpretations of p @q and p Dq are given. interpretations can be given to p @q and p @q. At the endpoints of (U, 0,@) we have the operation tables below. This sub-system is isomorphic to 1/(2) and the other two rings. A MYTHICAL HISTORY OF THE JARGON OF MATHEMATICS Of course, a plan of such divine simplicity worked marvelously. Billie McLaughlin, Portland State College One needs only to look at the work done in a small section of plane analytic geometry to realize the genius of these men. A retired butcher had found his way into mathematics. Having been a butcher all his life, he kept in Thousands of years ago mathematics was simply an ordinary occupation. practice with his cleaver by attacking wooden cones with it. One particular When a person said he was a mathematician, no one shuddered, or stuck his day, one of his words to use was "conic." "Cone," "conic:" there was a fingers in his ears, or gave him a look of thorough distaste. Rather, this resemblance, so he created the conic sections (curves created by slicing a announcement was met with giggles, snickers, and even chortles. Why should cone in various ways). He was quite inventive, and he used his day's quota the people not have laughed? All that mathematicians did was to sit around and requested three additional lists. (This won for him a solid gold cleaver drawing circles and triangles, muttering "&squared plus squared and the steer of his choice.) equals squared.'' It hardly seemed an appropriate pastime for the grossly retarded. Naturally, mathematicians felt they were fully competent, Out of the conic sections came the hyperbola, parabola, circle, ellipse, and the constant disparagement began to undermine their morale. Soon they a straight line, a pair of straight lines, and a point. All of these fig- were breaking out in spots and showing neurotic tendencies like tics, para- ures had common properties or properties common to at least some. The hyper- noia, persecution complexes, and disassociative reactions. Several were bola has all the qualities~axesof symmetry, one being conjugate, the other even more severely struck and became schizophrenic. Of course, their work transverse; two branches; two asymptotes: a latus rectum for either branch; was hampered by their lack of mental stability; and mathematics went into a and eccentricity. It also has two ridiculous definitions, neither of which catastrophic decline, causing it to fall into even greater disrepute than makes much sense. How often does a person feel compelled to talk about the before. locus of a point whose distance from two fixed points has a fixed difference? But the jargon was working. The definitions and properties sounded impres- This was almost the end of the field of mathematics. sive, perhaps made sense to a few, and baffled everyone else. However, one of the saner mathematicians, in a lucid moment, realized The men in calculus had a problem different from that in plane analytic that mathematics was doomed unless this group of pathetic psychotias re- geometry. They had been assigned to this branch because they had no ability ceived immediate help. So he took the group to the nearest accredited at all in drawing. All sections of geometry had refused to take them, and psychotherapist to see if therapy could not help them. these unartistic men were trying to create some form of mathematics which required no drawing. Weeks went by, and still they had nothing to call Fortunately, the therapist was able to help these wretches. He immedi- calculus but the name on their door. When they were asked to double as the ately put them in group therapy, and it was here that mathematics was actually janitorial staff, they decided they must attack the problem differently. saved. After a few sessions, they had gained insight enough into their prob- lem to consider a solution. Their problem (ignoring the extraneous factors Since none of the men could conceive of doing any mathematics without of incorrect toilet training, Oedipus complexes, and such) was that they a picture, they decided to borrow sketches from other fields. Perhaps if were suffering from deep insecurity and thwarted snobbery. Their profession these blighted artists could add a new line or two and give a reason for certainly gave them no prestige; they had either to change jobs or do some- doing so, they could claim that they had finally invented calculus. Anyone thing to mathematics. can draw a line with a ruler, so they tried it. They even found something to describe what they had done: they had taken the derivative of the equa- Thanks to the psychotherapist, they were able to devise a way to save tion of the figure and shown its geometric interpretation. their beloved profession. He had talked to them at one of the sessions (about their improvement, historians think), and no one understood a syl- Now that they had started, they were creating volumes of work. The lable. Hot damn! If only they could manage to garble their speech in the original derivative was called the first derivative because they had found same way, they could cure their problems1 second and third derivatives. Naturally, these were not simply first or second derivatives but first or second derivatives taken with respect to a At the very next session they began to work on what was to become a given variable. The derivative was obtained by differentiation, not by most successful jargon. Since they could not use jargon for the existing derivation. The mathematicians even founded a concept which they named concepts, they decided to start branching out, making up anything which 'the partial derivative," taken with respect not to a partial variable but seemed reas,onable; for now they would have jargon to hide any inconsisten- simply a given variable. cies. The few who had not yet recovered from their psychoses were put to work creating the new jargon since they spent most of their time babbling nonsense anyway. The men were so excited about actually creating calculus that they ELEMENTARY MATRICES EXPRESSED IN TERMS OF THE KRONECKER DELTA turned in their mops permanently and had a celebration. Feeling sublimely Mark E. Christie, Bowdoin College daring from all the wine, these dedicated mathematicans boldly created a concept even more ludicrous than that of differentiation~antidifferentiation. When they were sober, they realized that they had committed a ghastly error by simply putting a negating prefix in front of the name of the former con- The Kronecker Delta is defined as: 6,,=1if i=j, 6ij =0if if). cept and changed the latter's name to integration. Since they had been able It is possible to express an elementary matrix entirely in terms of the to differentiate all sorts of functions--algebraic, logarithmic, trigono- Kronecker Delta, using it to represent the elements of the matrix under con- metric~theymight integrate these functions. When they integrated they had sideration. The process of finding the precise K-Delta expression for a an integral, but they had all types of integrals. There was the definite given elementary matrix is essentially one of trial and error. It is known integral, the indefinite integral, also the improper, the iterated single, that the statement must contain the general term 6 and other terms to the double, and the triple integral. They even claimed that they could find ij' areas, volumes, first moments, moments of inertia, and work by integration. express any non-zero off diagonal elements and any diagonal elements equal to zero. Work in all branches of mathematics was as successful as it was in plane By way of example, consider the elementary matrix P defined as an analytic geometry and calculus. The jargon was created, and the resourceful PI mathematicians were able to find concepts for it. Much of the vernacular n x n matrix such that every element on the principle diagonal is 1 except is quite impressive today because it is either utterly misleading or means for the pth and qth rows. In these rows the diagonal element equals zero, absolutely nothing unless the listener has a Master's degree in mathematics. while the element in the pth row and column equals 1, as does that in A logarithm sounds like a pulsating piece of felled tree, the deleted neigh- the row and pth column. All other off diagonal elements equal zero. borhood a problem for the N.A.A.C.P. Who was the witch of Agnesi? Are you a surd? What is Lipschitz's condition? After a person casts out nines, does Consider the following expression written in terms of K-Deltas; he have a Baire function? Is society safe from the standard deviate? Do shrinking and stretching transformations hurt? Is the sheet of a surface usable on a bed? Is the class called Life Drawing a person's first Baire class? Do amicable numbers love each other? will the annihilator get you? To demonstrate that the above statement is true, it is necessary to show the following three properties: (1) that one can express any P by the It is quite apparent that jargon has absolutely solved the problem of PQ the ancient mathematicians. Since they unleashed their powerful vernacular expression; (2) that P P = I = [6 ] ; (3) that premultipli~ationof a on the public, the guffaws and chortles have been replaced by awful sighs PI Pq i j matrix A by P (in the K-Delta form) does change the pth and qth rows of A. and reverent silence. Who now would dare to laugh when a mathematician PI opens his mouth? Who understands but the smallest part of what a mathema- tician utters? Who listens? (1) consider P = [6ij + (6iq - 6^)(6 - 6qj)1: PI if i / p or q, then the expression equals [6ij], ~t is clear that our K-Delta form can express any P PI. ADDITION PROGRAM FOR TURING MACHINE J. Woeppel and M. Wunderlich State University of New York at Buffalo Expanding the second term: Many people consider that the Turing machine is the abstract proto- type of a digital computer. Therefore, anyone who plans to master computer science might find it useful to first become acquainted with this theoretical device conceived by A. M. Turing in 1937. [l] Clearly the expansion of the third term will yield the same result. Now factoring 6 .) and expanding the last term: (6iq - ^pj - q: For our purposes the Turing machine consists of an infinite tape divided into squares called "cells" and a device called the "sensing device" which is able to sense the symbol in the current cell, that is the cell where the sensing device is presently resting, and replace this symbol with another symbol or leave it as is. The symbols are E, Z, 0, ~ut6 = 6 = 0 aince q / p. and 1, though a person could introduce any finite number of symbols. The Pq tape can be moved to the right one cell at a time or to the left one cell Therefore, Ppq P pq l6ij1 - at a time, or left at the current cell. We assume that all data on the tape is given in binary notation with (3) using the notation of (2), we havei the lowest order bit to the extreme right of the word which may consist n n of an arbitrary number of cells (i.e., the usual order in which numbers P A = = + (biq 6 1 bqk) 1 Wik - ip (spk - are written). The tape is originally assumed to have all E's in the cells pq k=I k=l left of a given cell and a Z in the cell immediately to the right of this if i q or p, ten 2dikxj = (a 1, given cell. The symbols in these cells are never changed. The remaining + ij cells have either a 0 or a 1 in them and these symbols may be replaced by k=l either a 0 or a 1. The given cell also has either a 0 or a 1 in it which may be replaced by either a 0 or a 1. One could imagine the tape to appear as follows: 8 Sensing device ~huswe know that the K-Delta representation under consideration does indeed correctly express P EEEEEEEE Z pq- 1 move tape left move tape right 4 Note: Blank cells have a 0 or a 1 in them. The "given cell" is positioned at the "sensing device." We move the tape not the sensing device. Later on we will find it convenient to name the cells. It must be remembered that the machine does not recognize names for the cells; it is only for our convenience. The "given cell" above is called the "AC"; the cell containing "Z" (the next cell to the right) is called the "origin" or "zero cell"; the remaining cells are numbered 1, 2, 3, .. . to the right from the "origin." The following diagram illustrates this: We will now list the macro instructions and give their defining programs in a few cases: AC0 12 34 EEEEEEEE Z Move the tape to the left one location -and then execute instruction y. 1. IFL 0,~ Note: Blank cells have a 0 or a 1 in them. 2. IFL 1-Y 3. IFL Z,Y The "current symbol" is defined to be the symbol in the cell at the 4. IFL E,Y sensing device. Each cell is referred to as a "location." The numbered cells are referred to by their number, e.g. loc. 27; this is the 27th Since this is the first program which has been presented, we will cell right of the origin. When we say that we move the tape to loc. N, explain it briefly to the reader. We know when this program is executed we mean we position the tape so that loc. N is at the sensing device. that the tape is moved to the left one cell and instruction y is executed. since each symbol appears in one of the four instructions. A specific set of instructions which the machine is asked to execute is called a "program", and tFe n instructions of a given program are Move the tape to the right one location and then consecutively numbered 1, 2, 3, through n. execute instructions y. Following Arden's [2] method, we will have only three basic instruc- Transfer to instruction y. (Execute instruction y tions which the Turing machine can execute: next. ) 1. IFL ,2 (a) If the current symbol is x, move the tape one cell to the left 2. IFR ,y and execute instruction y. Otherwise, execute the next instruction. This instruction is denoted symbolically IFL x,y where x is a symbol (d) HALT The machine halts. (The state of the machine remains and y is an instruction number. the same indefinitely without affecting the tape; used to stop the machine, e.g. at the end of the (b) If the current symbol is x, move the tape one cell to the right program. ) and execute instruction y. Otherwise, execute the next instruction. This instruction is denoted symbolically IFR x,y where x is a symbol and 1. TRA ,1 y is an instruction number. 2. .... (c) If the current symbol is x, replace it with z, then execute the Replace the current symbol with x and then execute next instruction. Otherwise, just execute the next instruction. This the next instruction. (Note, the current symbol instruction is denoted symbolically IFS x,z where x and z are symbols. can not be z or E and x can not be 7 or E.1 1. IFS 0,x We have deviated slightly from A. M. Turing's [l] and subsequent 2. IFS l,X authors' definition of a Turing machine in order to make its analogy 3. *... with the modern day computer more apparent: of course, the machine described above is not a stored-program computer. For example, Turing Move the tape to the right N locations and then does not number his cells, thus he has no "locations." execute instruction y. A. M. Turing presents an argument in [I] that a machine such as 1. IFR ,2 described above is capable of performing any calculation that a computer 2. IFR ,3 is able to perform. So, let us see how we would add two four-digit binary 3. .... numbers together. To do this with only the three basic instructions requires one hundred or more instructions and is very difficult to follow. i. IFR , (i+l) To avoid this, we introduce "macro instructions." Each macro instruction (i+l). .... must be defined by a program of previously defined macro instructions or basic instructions. N. IFR , (N+l) (N+l). TRA ,Y Move the tape to the left M locations and then execute instruction y. 1. MTA ,2 5. IFS ,0 2. TRA 0,6 If the current symbol is x, execute instruction y. 3. MTP N,4 6.7. MTPIFS 0,y,1 Otherwise, execute the next instruction. 4. TRA 0,7 8. MTA ,5 Move the tape to loc. N and then execute instruction We shall explain this program briefly. First we move the tape to the Y. AC and test to see if there is a 0 there by instructions 1 and 2. If there is a 0 there, the left-hand column of the table is pertinent and we wish 1. TRA E,3 to leave the 0 in the AC and leave whatever symbol is in loc. N there. So 2. MTR 1,l we merely move the tape to the origin and execute instruction y (by 3. TRA z,5 4. MTL 1,3 executing Instruction 6). If there is not a 0 there, we go on to instructions 3 and 4 which move the tape to loc. N and test to see if 5. MTL N,y there is a 0 in loc. N. If there is, we wish to put a 0 into the AC and Move the tape to the AC and then execute instruction a 1 into loc. N. Instructions 7, 8, and 5, operating in that order, accomplish this. If there is not a 0 in the AC, we wish to put a 1 into Y. the AC and a 0 into loc. N. Instruction 5 alone accomplishes this since Put the symbol which is presently in loc. N into the there is already a 1 in the AC. AC, leave the tape positioned at the origin (loc. O), and then execute instruction y. Note, the symbol (b) ADN N,M Add the symbol in loc. N to the symbol in loc. M. in loc. N is not changed. If the sum is less than or equal to 1, the sum is placed in loc. N and the AC remains unchanged. If Put the symbol which is presently in the AC into the sum is greater than 1, a 1 is put into the AC loc. N, leave the tape positioned at the origin, and and a 0 into loc. N. In both cases the tape is left then execute instruction y. Note, the symbol in at the origin and the next instruction executed. the AC is not changed. 1. MTP M.2 6. CLA M, 8 1. MTA ,2 5. MTP 0,y 2. TRA 0,8 7. IFS ,1 2. TRA 0,6 6. MTP N,7 3. MTP N,4 8. MTP 0,9 3. MTP N,4 7. IFS ,0 4. TRA 0,7 9. .... 4. IFS ,1 8. MTP 0,y 5. IFS ,0 We will now introduce three instructions which give us the body of our The dots in instruction 9 indicate that the next instruction is to be addition program. The first instruction adds the contents of the AC and executed. the contents of a location. The second instruction adds the contents of two locations. The third is a combination of the first two and is the Again we will briefly explain the program. The first two instructions only instruction of the three which actually appears in the addition move the tape to loc. M and test for a 0. If there is a 0 there, the sum program. of the symbols in loc. N and loc. M is the symbol in loc. N, so we merely move the tape to the origin and execute the next instruction (by executing Let x be the symbol in loc. N and let z be the symbol Instruction 8). If there is a 1 in loc. M, we go on to instructions 3 in the AC. Add x and z, write it as a double bit and 4 which move the tape to loc. N and test for 0. If there is a 0 binary sum. Put the left-hand bit of the double there, the sum would be 1, so we transfer to instruction 7 which puts a bit sum obtained into the AC and put the right-hand 1 into loc. N and then we move the tape to the origin and execute the bit of the double bit sum obtained into loc. N. next instruction (by executing Instruction 8). If there is no 0 in loc. Leave the tape positioned at the origin and execute N, we wish to put a 1 into the AC and a 0 into loc. N. To do this we go 5 instruction y. to instruction which puts a 0 into loc. N, then instruction 6 brings the WI 1 in lor. M into the AC and we are finished. It must be noted that except in the case where there is a 1 in both loc. N and loc. M, the AC was left unaltered. (c) ANM N,M Add the symbol in the AC, the symbol in loc. N, and To subtract a number B from a number A, we would replace B by its comple- the symbol in loc. M together. This sum is less ment and add it to A. For example, if we assume that we are working with than 4. Express it in a binary form (two bits), a maximum of three-digit binary numbers, the complement of the binary put the left-hand bit into the AC and the right-hand number 0101 (5) is 1011 which is the first number, 0101, subtracted from bit into loc. N. Leave the tape at the origin and binary 10000 (16). The assumption that we are only working with three- execute the next instruction. digit binary numbers is required since we must have a way of telling whether a number is a complement quantity or not. Note, the four-digit 1. ADA N,2 representation of binary five is also binary eleven. For this purpose 2. ADN N,M we reserve the extreme left-hand bit of a four-bit word; if it is a 1, 3. .... the next three bits are a three-digit complement quantity; if it is a 0, the next three bits are a regular three-digit quantity. We could also To do this we use the first two instructions. Let all sums be expressed as double bits. The first instruction adds the symbol in the put negative numbers on the tape in their complement form. AC and the symbol in loc. N and puts the right-hand bit of the sum into loc. N and the left-hand bit of the sum into the AC. The second instruction We have not mentioned multiplication or division or decimal quantities. Multiplication and division could be taken care of by still more compli- adds the right-hand bit of the previous sum now in loc. N and the symbol cated programs using addition and subtraction as developed above and a in loc. M and puts the right-hand bit of this sum (the sum of the symbol shift instruction. Decimal quantities could be handled by the manner in now in loc. N and the symbol in loc. M) into loc. N. If the left-hand bit of this sum is 0, it does nothing to the AC; thus leaving the left- which the data was placed on the tape. That is, one could presume that hand bit (highest order bit) of the previous sum of the symbols in the the decimal point lies after the first ten places of all numbers. AC and loc. N undisturbed (there is no carry for this sum). If, on the Thus, it is reasonable to say that the Turing machine can perform any other hand, the left-hand bit of this sum (the sum of the symbol now in calculation that a computer can perform. But is a computer capable of loc. N and the symbol in loc. M) is 1, it puts a 1 into the AC. In this case, the left-hand bit of the previous sum, that is the sum of the symbols calculating everything that a Turing machine is able to calculate? in the AC and loc. N, is lost but it would have been 0 anyway because the Recently many authors including M. 0. Rabin and D. Scott [3] have come total sum, that is of the AC, loc. N, and loc. M, has to be less than 4. to the conclusion that the Turing machine with its infinite tape is too They base The following program adds a number A and a number B where A is in general to serve as an actual model for digital computers. locations N-3, N-2, N-1, N; and B is in locations M-3, M-2, M-1, M; and their argument on the fact that the Turing machine has an unlimited stores the results in N-4, N-3, N-2, N-1, N (where the highest order bit amount of memory space on its tape while an actual computer would have a is in N-3 or M-3 for A and B respectively). Note, N and M must be 4 limited amount of memory space. and IM -Nl > 4. 1. MTA ,2 5. ANM N-2, M-2 REFERENCES 2. IFS ,0 6. ANM N-3,M-3 3. ANM N,M 7. ST0 N-4,8 1. A. M. Turing, "On Computable Numbers, with an Application to the 4. ANM N-1.M-1 8. HALT Entscheidungsproblem, " Proc. London Math. Soc., vol. 42, pp. 230-265, 1937. Instructions 1 and 2 put a 0 into the AC since there is no initial carry. Instructions 3, 4, 5, and 6 add the ones' bit, twos' bit, fours' 2. B. W. Arden, "An Introduction to Digital Computing," Addison-Wesley, bit, and eights' bit of the binary numbers A and B and the previous carry p. 2-8, 1963. together respectively putting the lowest order bit of each sum into locations N, N-1, N-2, and N-3 respectively, and putting the highest order 3. M. 0. Rabin and D. Scott, "Finite Automata and Their ~ecisionProblems, bit of each sum, that is the carry, into the AC. Finally, instruction 7 in Sequential Machines: Selected Papers, ed. by Edward F. Moore, puts the final carry into location N-4. Thus the sum of A and B appears Addison-Wesley, pp. 63-91, 1964. in locations N-4, N-3, N-2, N-1, N. Note, by merely adding extra ANM'S we could increase the maximum magnitude of A and B. We have shown that a simple machine such as described above can perform This paper was presented at the National Pi Mu Epsilon Meeting at addition of positive integers. Subtraction could also be handled easily Cornell University, September 1, 1965, by James Woeppel. by writing an instruction which would replace a number by its complement. A MULTIPLICATION-FREE CHARACTERIZATION OF RECIPROCAL ADDITION Definition i. James Williams, Carleton College Let N be the set of positive integers; v n c N, lSa=a, (n+l)a=na+a all = a, a/(n+ 1) = a/n o a . Motivation. Axioms. Reciprocal addition, which we shall represent by the symbol "0, " is I. a+b=b+a; aob=boa defined, in terms of the real number system, by 11. (a +b) + c = a + (b+ c); (a oh) 0 c = a o (b o c) 111. a+O=a; aou=a IV. O1=0; u'=u; O#~#U+ (a+a1=0 and aoam=m) V. aoO=O and 1s commonly used with an ideal infinity element u with the properties VI. a + (a o b) = 2a o (2a + 4b) a o u = a and a + u = u. In such applications as parallel-series problems In electronics, for instance, the operations of addition and reciprocal Theorem L. V a, b â r, addition frequently occur together (the constants 0 and u corresponding to the two extremes of short and open circuit values) exclusive of the use of 1. (2a/2) = a multiplication; it is therefore of interest to develop an axiomatic charac- 2. 2a=O =+ a=O terization of + and o independently of the not~onof multiplication. 3. a=al + (a=O or a=u) 4. a+u=u We begin with a listing of some properties of R*, the real number system 5. aob=O + (a=O or b=O) with u and the operation 0 adjoined. a+b=u + (a=u or b=u) 6. r#(0) + O#u I. We use the mapping 7. aob=u + a'=b; a+b=O + a8=b 8. (a + b)' =a' +b'; (a 0 b)' =a' 0 b' to show that R* is a dual system: 9. 2(a/2) = a 10. 2 (a o b) = 2a o 2b; (a + b)/2 = a12 + b/2 (1) f (-a) = I/(-a) = -l/a = -f (a) -Proof -1.1. (2) f(a + b) = l/(a + b) = (l/a) (l/b)/(l/a + l/b) = l/a o l/b = f (a) o f (b) f(a 0 b) = l/(l/(l/a + l/b) = l/a + l/b = f (a) + f (b) Proof 1.2. (3) f (ah) = l/ab = (l/a) (l/b) = f (a)f (b) If 2a = 0, then The argument for special elements proceeds by definitial patching thus: (1.1; hyp, def; 111) -Proof -1.3. If a = a' and a#~,then 2a=a+a1=0 11. + and o are correlated independently of multiplication by the relation Therefore, ab a+ (aob) =a+- a+b (a, b, a+b # 0, U) -Proof -1.4. (1) If a = 0, then a+u=O+u=u. (2) (?ma! If r # u and r + s = 0, then r =r +0=r1+(r+s) We now proceed to investigate a system r with special elements o and = (r' + r) + s u, with a monary operation ' , and with binary operations + and o , =o+s=s. satisfying axioms I - VI for all elements a, b, c in r. 8 I* g,',3 'I 170 (3) (Lemma) If 2a =w, then -Proof -1.8. a=2ao2a=uow=w. (1.1, hyp, 111) If a = 0, then (4) (Lerana) If a # w, then (a o b)' = (0 ob)' = 0' = 0 = 0 ob' = 0' o b' (2a) ' = (2a) ' + 2 - 0 = a' o b'. (hy~,V, IV, V, IV, ~YP) = (2a)' + 2(a + a') (111, def; IV, hyp) If a # 0 and b # 0, then = ((2a)' + 2a) + 23' = 2a'. (11: IV, hyp; 111) (a o b) o (a' o b') = (a o a') o (bob') (5) If O#a#u, then =uou= w. (I, 11: IV, hy?: 111) a + (a o a') = 2a o (2a + 4a1), (VI) Therefore, (a o b) ' = a' o b' by Theorem 1.7. a + w = 2a o ((2a + 2a') + 2a0) (IV, hyp: def, 11) If a = u, then = 2a o ((2a + (2a) ') + (2a) ') (4) j (a+b)'= (w+b)' =ul =w=w+bl =LO' +bt = 2a o (2a) '. (IV, 3, hyp; 111) = a' + b'. (hyp, 1.4, IV, 1.4, IV, kypl (6) Therefore a # u + a + u = w. (185) If a # w and b # w, then (a b) (a' b') = (a a') (b b') If a = w, then a + u = 2u; suppose 2w # u, then + + + + + + =o+o=o. (I, 11: IV, hyp: 111) (7) u = 2w + w = 3w. (hyp, 6 with a = 2w; def) Therefore (a + b)' = a' + b' by Theorem 1.7. 2u=u+w=w+ (uou) (def, 111) = 2w 0 (2w + 44 (VI1 -Proof -1.9. a 2w 0 (3u + 3w) (def, 11) = 2w 0 (w + u) (7) suppose 0 # a # u : =2w02w=w (def, 1.1) a + (a o a1/2) = 2a o (2a + 4(aa/2)) (VI) a + (a o a') o a' = 2a o (23 + 4(a1/2)) (def, 11) Contradiction, therefore v a c I?, a + u = w. a + (w o a') = 2a o (2a + 4(a8/2)) (IV, ~YP) a + a' = 2a o (2a + 4!a1/2)) (111) -Proof - 1.5. 0 = 2a o (2a + 4(a8/2)) (IV, ~YP) If a o b = 0 and b # 0, then Therefore either 2a = 0 or 2a + 4 (a '12) = 0 by Theorem 1.5- a=aou=ao(bob') = (aob) ob' ~ut2a = 0 + a = 0 by Theorem 1.2, and a # 0, therefore = 0 o b' = 0. (111: IV, hyp: 11; hyp; V) 0 = 2a + 4(a8/2) = 2a + (4(a/2)) ' (1-8) Therefore, a o b = 0 + (a = 0 or b = 0). 2a = ((4(a/2)) ') ' = 4(a/2) = 2(2(a/2)) (1.7, def) Similarly, if a + b = w and b # u, then (2a)/2 = (2(2(a/2)))/2, a = 2(=/2). (I.1) a=a+o=a+ (b+bl) = (a+b) +bl Finally: = w+ b' = u. (111; IV, hyp; 11; hyp; 1.4) 2(0/2) = (0 o 0) + (0 o 0) = O+ 0 = 0, (def, 111, 111) 2(42) = (w o w) + (uo u) = w+ w = u. (def, 111, 1.6) -Proof -1.6. Proof 1-10. If r # [OJ, Z X < r: X # 0; suppose 0 = w, then -- o#x=x+o=x+~=~. PYP,111, hyp, 1.3) 2(a o b) = 2((2a)/2 0 (2b)/2) = 2((2a o 2b)/2) Contradiction, therefore r # [o] => 0 # u. = 2a o 2b. Proof 1.7. (a + b)/2 = (2(a/2) + 2(b/2)/2 -- = (2(a/2 + b/2))/2 If uob=u and r=[O), then a8=0=b. = a12 + b/2 (8) If a ob=w and J?# [o], then a#O since a=O + O=Oob=aob=u. Contradiction by Theorem 1.6. -Meta-Theorem. a* = a' ow=a'o (aob) = (a'oa) ob 4 I' is a dual system; that is, for each theorem T in r, the statement T* =wob=b (111; hyp; 11; IV, 8; 111) formed by replacing each occurrence of +, 0, 0 and u with 0, +, u and 0 (9) Similarly, if a + b = 0, and I? # [O], then a # u since (respectively) is also a theorem in I'. a=w =+ u+b=a+b=O. (1-3, ~YP,~YP) Contradiction by Theorem X.6- a' = a' + 0 = a' + (a + b) = (a' + a) + b =O+b=b (111: hyp; 11; IV, 9; 111) Lemma. -Proof. 1f rcsk-l, then (ka=O =3 a=O) + (ah=@ * a-d* Axioms I - IV are self-dual. Theorem 1.2 is the dual of Axiom V. The for suppose dual of Axiom VI is proved as follows: For any given x and y in let r, 1) ka # 0, 2) a/k = u, and 3) a # u, the11 a = x/2, b = y/4, then u=a o a/(k - 1) (dcf, liyp 2) 2a = 2(x/2) = x (1.9) a = (a/(k - 1))'= al/(k - 1) (1.7, 1.8) 4b = 2(2b) = 2(2(y/4)) t (k - 1)a = (k - 1)(a1/(k - 1)) =a = 2(2((y/2)/2)) = 2 (y/2) = Y. (r c s,'-,) (def, 11; hyp; def, 11; 1.9; 1.9) ka = (k - 1)a + a = a' +a= 0. (def; Iv, hyp 3) Contradiction by hywthesis 1. Making the appropriate substitutions in Axiom VI then gives Therefore (a # 0 + ka # 0) + (a/k = u + a = a). V x, y c r; (10) x/2 + (x/2 0 y/4) = x 0 (x + y) -----Basic Lemma for Theorem 11. k-1 The meta-theorem is completed by noting that if T is any theorem with V a c r, V k c N; If r c nn-l- (Sn n Tn) and ka = 0 + a = 0, a proof P, then T*, the dual of T, will have a proof P*, each step of which then r c Sk 1) Tk. is the dual of the corresponding step in P. We shall prove r c Sk and show r c Tk by a duality argument. We begin with proofs for special cases: At this pint it would be nice to know how close r comes to representing the notion of a field. In one direction, beginning with any field, we may If a = 0, then k(O/k) = (k-O)/k = 0, construct all of the axioms of just as we did for the real numbers, with r if a = u, then k(&) = (ku)/k = u. the one exception that the derivation of VI requires the principle 1.2: 2a = 0 + a = 0, thus eliminating fields such as 1/(2). In the other The cases for k = 2 and k = 4 follow directly from the fact direction, however, the answer seems less clear: Theorem I1 contains some that r c fl [s2, S4, SB, ---I, which in turn follows from Theorem partial results. 1.9 and Lemma 11. Definition g. We next prove the case for k = 3 as follows: if 3a # 0 and We classify r by writing for each k â N, a # u, then a/6 # u since if a16 = u, then r c Sk iff V a c r, k(a/k) = a a/3 = 2( (a/3)/2) = 2(a/6) = 2u = u. I? c Tk iff V a c r, (ka)/k = a. Contradiction by Lemma 14. 2a o a = a o (a + a) = a12 + (a12 o a/4) - a12 + a/6 (def; 10; def, 11) (2a o a) + a1/6 = a12 + a/6 + a'/6 = a12 (11; 1.6; IV, a/6 # 0; 111) 2(2a 0 a) + 2(a1/6) - 2(a/2) - a (11, def; 1.9) 2(2((2a 0 a)/2)) + 2( (a1/3)/2) = (1.9; def, 11) 4((2a)/2 0 312) + a1/3 = (def, 11; I; 1.9) 4(a/3) + 3'13 = (1.1, def) 3(a/3) + (a13 + (a/3) ') = (def, 11; 1.8) 3 (a131 = a (IV, 14, hyp; 111) Finally, by induction for any k > 4, choose 0 # a # u, then a/2(k - 2) # u by 13, since r c S2(k-2) by Theorem 1.9, hypothesis, and Lemma 11- (13) -. (k - 4)a' # 0 by 13 since r c Tk-4. r c Sp + (alp = a + a = a), since if alp = u, then (l7 â Sp, 1.4) a = p(a/p) = p u = u. (k - 4)a. .# by Theorem 1.5 since a # u by- hypthesis,- and Similarly, r c Tp + (pa = 0 =3 a = 0). therefore (k - 4)a/k # u by 14. --Proof 11.1. 17s Having dispensed with the prelminary deta~ls,we make the following Assume pa = 0 for all a c I+, for some prime integer p. computation: (23) (Lemma) We note that P is a prime order abelian group w~th a + [a o ((k - 4)a8)/2(k - 2)] = 2a o [2a + 4[((k - 4)a1)/2(k - 2111 respect to addition as a result of Axioms I-IV and the above [VI with b = ,((k - 4)a1)/2(k - 2)1 assumption. = 4[a/2 o [a12 + ((k - 4)a1)/2(k - 2111 (1.9, 1-10) = 4[a/2 o [(k - 2)(a/2(k - 2)) + ((k - 4)a9)/2(k - 2))1] (24) Considering [n] c I/(p) as an equivalence class of n, we (r c sk-2) define, V a c P, V [n] c 1/(p), = 4[a/2 o [(k - 2) (a/2(k - 2)) + (k - 4) ((a1/2)/(k - 2))Il [nla = na. (I. 10) = 4[a/2 o [(k - 2) (a/2(k - 2)) + (k - 4) ((a1/2)/(k - 2))11 [nla is well defined since if [n] = [m], then (n - m) = 0 (mod p) (12) and therefore (n - m)a = 0, na = ma. = 4[a/2 o [(k - 2)(a/2(k - 2)) + (k - 4)(a/2(k - 2))'Il (1.10, 1.8) Theorem 11.1 now follcws from the fact that any additlve p-order Abelian = 4[a/2 o 2(a/2(k - 2111 (IV, 20) group forms a vector space with respect to definition 24 of multiplication. = 4[a/2 o a/(k - 211 (def, 11; 1.9) Theref ore, -Proof - 11.2. a + [a o ((k - 4)a8)/2(k - 211 = 4(a/k) (def, 11) Assume P has p elements for some prime integer p. a + [((k - 4)a)/(k - 4) 0 ((k - 4)a1)/2(k - 2)1 = (r 6 Tke4) P a + ((k - 4)a8)/k= Then, with respect to addition, is a cyclic group, so that by (def, 11; IV, 21; 111) Theorem 11.1, it forms a one dimensional vector space with respect to a + [((k - 4)a)/kl ' = 4(a/k) (1.8) definition 24 of multipl~cation. To extend the scalar-field properties of I?, we choose an Adding ((k - 4)a)/k to both sides gives arbitrary normalizing constant a c I? - [o], and define a = 4(a/k) + ((k - 4)a)/k (def, 11; IV, 22; 111) x-y = [k(x)l [k(y)la where = k(a/k) (def, 11) [k(x)la = x, [k(y)la = y. k-1 Therefore, if l? c (Sn Tn) and ka = 0 a = 0, then nnZl n + We also define l/x by l/x = y +3 x-y = a. l? c Sk by 16, 17, 18, and 19. (Lemma) V n < p, V a c l?, (na)/n = n(a/n) = a since, by To arrive at l? c Tk, we note that the above proof takes the form "if P induction, we have first l? c (& 0 Ta ) as mentioned in Lemma k-1 and Q, then R" [where P is l? c 1, Tn), Q is ka = 0 + a = 0, and 15; then if l? c llnZl (Sn 1) Tn) for some k < p, then R is l? c Sk)]. Since P and P are identical, and since P and Q a # 0 + ka # 0 by Lemma 23 and therefore l? c Sk ri Tk by + Q* by 14, P and Q =+ P* and Q*; finally, P* and Q* =+ R* Lemma 15. The induction stops with k = p - 1. by the meta-theorem, so that P and Q + R* and l? â Tk. Definition s. I? = l? - [m]. (25, def i) (25, 24, 25) -Theorem -11. (n c bl, 27) With respect to suitable definitions of multiplication, Therefore l/(na) = a/n by definition 26. Finally, we choose x, y c P - (01 such that x' # y and compute 1. If for some prime integer p and for all a c l?, pa = 0, then P is a vector space over I/(p), the ring of integers modulo p. 2. If P has a prime number of elements p, then l? is derivable from I/(p) . 3. If V a c P and V n c N, na = 0 + a = 0, then P is a vector space over the field of .rational numbers. 176 So that 0 is alyebra~callydetermined by the field properties of and the conventions for special elements (e.g. a + u = u). RESEARCH PROBLEMS -Proof - 11.3. Proposed by GEORa BRAmR, University of Minnesota. Assume V a c l?, V k â N, ka = 0 + a = 0. The real-valued functions on (-a,-) is a semi-group, if we define the (29) We define V k â N, product of two such functions by (-k)a = ka', a/(-k) = at/k, 0.a = 0. (30) (Lemma) By analogy with Lemma 27 of the previous proof, we conclude a c (Sk (1 Tk), so that, in view of definition 29, V k c I, The function X acts as identity and each constant function c has the properties: (31) V a c p, V p, q c I, q # 0, we define (p/qJa = (pa)/q - The above multiplication by rational numbers is well defined since Q, #O, if R/% =&I%: if &, h, %, € andif Q,% If we denote the set of real-valued functions by S, and the set of constants then R = hql/%. by I, we have Moreover, lf we define f 2 g to mean that f(t) 2 g(t) for all t, then S is right partially ordered, i.e., Assuming the normal field properties for rational numbers, the f2h f xgkf xh. field-[vector space] correlation laws are proved as follows: (5) + Problem: Obta~nproperties of abstract right partially ordered semi-groups. (def i) What algebraic properties, in addition to (1) - (5), would constitute a characterization of the semi-group of real functions? What proprties would characterize the sub semi-groups? Proposed by SIM KAH, University of Minnesota. Examine, heuristically, the geometry of geodesics on a cylinder in order to obtain a set of axioms of incidence and order. Using these axioms, deductively develop a geometry and attempt to find models, other than the cylinder, to which this geometry applies. PROBLEM DEPARTMENT Determine the area of the crescent. Edited by M. S. Klamkin, Ford Scientific Laboratory This department welcomes problems believed to be new and, as a rule, demanding no greater ability in problem solving than that of the average member of the Fraternity, but occasionally we shall publish problems that should challenge the ability of the advanced undergraduate and/or candidate for the Master's Degree. Solutions of these problems should be submitted on separate signed sheets within four months after publication. Irtl. Proposed by DONALD W. CROWE, University of Wisconsin and M. S. KLAMKIN, An asterisk (*) placed beside a problem number indicates that the problem Ford Scientific Laboratory. was submitted without a solution. Determine a convex curve circumscribing a given triangle such that Address all communications concerning problems to M. S. Klamkin, Ford 1. The areas of the four regions (3-segments and a triangle) Scientific Laboratory, P. 0. Box 2053, Dearborn, Michigan 48121. formed are equal and 2. The curve has a minimum perimeter. PROBLEMS FOR SOLUTION SOLUTIONS 177. Proposed by C. S. VENKATARAMAN, Sree Kerala Varma College, Trichur, South India. lb5. Proposed by D. J. NEWMAN, Yeshiva University. 3 If s is the semi-perimeter and R, r, rl , ra, and r3 are the circum-, Express cos 0 as a rational function of sin 0 and cos3 9. in-, and ex-radii respectively of a triangle, prove that Solution by R. C. Gebhart, Parsippany, N. J. R 2s3.a r - rl ra r3 A Since 1. 4 COS" 0 = 3 9 30, 178. Proposed by K. S. MURRAY, New York City. cos + cos 2. 256 cos3 9 sin6 8 = 6 cos 0 - 8 cos 39 + 3 cos 79 - cos 98, A 3. 64 cos 9 sin6 0 = 5 cos 9 9 cos 39 5 cos 59 cos 79, Show that the centroid Of A ABC coincides with - + - 4. 256 c0s9 9 = 126 cos 9 + 84 cos 39 + 36 cos 59 + 9 cos 79 + COS 90 the centroid of A A'B'C82he= A', BB", and 6 5- 64 cos 0 cos 0 = 35 cos 9 21 cos 30 7 cos 59 + cos 79, C' are the midpoints of BC, CA, and AB, + + cos 9 is a linear combination of the left-hand sides of equations 1-5, respectively. 3 3 6 8 B i.e., cos A a A cos 6 + B cos 9 sin 9 + C cos 9 sin 9 + D cos" Generalize to higher dimensions. A' + E cos 0 cos6 0. It then follows that 179. Proposed by DONALD SCHROEDER, Seattle, washington. A=2, B=C=D=l, and E=-2. Whence, 3 3 It is well known that 2 cos 9 + (cos3 0) (sin 9)' + (cos3 @I3 cos 8 = 3 3s + 4' = 5', 1 - (sin 9)' + 2 (cos3 9)' lo 11' 12' = 13' 14'. + + + Also solved by K. S. Murray, P. Myers, and F. Zetto. Generalize the above by finding integers a satisfying m 2m .I. Proposed by M. S. KLAMKIN, Ford Scientific Laboratory. I.. (a+k)' = (a+kIa. k=O k=m+l 4 Given a centrosyinmetric strictly convex figure and an intersecting translation of it; show that there is only one common chord and that 180. Proposed by R. C. GEBHART, Parsippany, N. J. this chord is mutually bisected by the segment joining the centers- In the figure, AB = BC and LABC = 90. The arcs are both circular Solution by the proposer. with the inner one being tangent to AB at A and at C. 181 169. Proposed by JOE KONHAUSER, University of Minnesota. From an arbitrary point P (not a vertex) of an ellipse, lines are If the two ovals intersect in more than two points, drawn through the foci intersecting the ellipse in points Q and R. then each oval would contain at least three Prove that the line joining P to the point of intersection of the parallel and equal length chords. The length tangents to the ellipse at Q and R is the normal to the ellipse and direction being the length and direction at P. of the translation. This is impossible since the middle chord must be larger than Editorial Note: The proposer notes that he does not know the source the outer equal ones by strict convexity. of the problem and he has not been able to locate it in any of the books he has examined. If A and B denote any pair of centrosymmetric points of the Solution by s. Schuster, University of Minnesota. first oval and A' and B' the B' corresponding points of the s = QR is the polar of S. Then translated oval then AA'B'B t)(PT,US),- and M(QP,QT;QU,QS) is a parallelogram. It then is a harmonic set of lines. The follows that the entire figure A projectivity (defined by the conic) (both ovals) is cen'trosymmetric between the lines on Q and the about point M, the midpoint of lines on P makes the correspondences: the segment joining the two QP-PB', QT-m, QU-PR, QS-PQ. centers. Since D and E are the .'. a (PB', ,;PR, PQ). ~utsince only two double points, they must LA'PQÂ¥SLB'PRwe have PT i PB'. correspond to each other, i.e., Other interesting properties of the be centrosymmetric with respect figure are: to M. Note that this generalizes the well known result for two BB' i PQ, AA' L PR , intersecting congruent circles. A'B i BS and B'A i AS . Also solved by M. Wagner. Also solved by Paul Myers and the proposer. 168~roposedby JERRY TOWER, North High School (student), Columbus, Ohio. Editorial Note: ~t would be of interest to supply a nonprojective proof. Determine x asymptotically if 170. Proposed by C. S. VENKATARAMAM, Sree Kerala varma College, India. log x = n log log x. Prove that a triangle ABC is isosceles or right-angled if Solution by Sydney Spital, California State Polytechnic College. (1) a3 cos A + b3 cos B = abc. The given equation may be rewritten as The solutions by Paul J. Campbell, University of Dayton, and Marilyn Mantel, University of Nebraska were essentially identical and are x1" = log x given as follows: whose limiting solution is easily seen to be x = e. To determine x asymptotically, we let x = e(1 + 6) and assume (to be justified Replacing cos A and cos B by the law of cosines, i.e., ba c? aa aa c? ba subsequently) that 6 = 0 Substituting back into (I), yields + - = + - (1): cos A = 2bc , cos B 2ac ' (1) reduces to (after some algebraic manipulations) Thus, 1 a a a 6 = el1' - 1 + O(2). (a + b - c? ) (a - b la = 0, which implies the desired result. 1 Since el1' = 1 + O(;), our previous assumption is justified and the Sydney Spital, California State Polytechnic College and the proposer desired asymptotic behavior is given by use the projection formula x = el+l,n + ,,(I_a) = e (l+l/n) + 0 (-5 . a cos B + b cos A = C Also solved by Paul J. Campbell, K. S. Murray and F. Zetto. to obtain -BOOK -REVIEWS a a (a - b ) (to cos B - a cos A) = 0 Edited by which implies the desired result. Roy B. Deal, Oklahoma State University Also solved by H. Kaye, P. Myers, M. Wagner and F. Zetto. General Theory g Functions Inteqration. By Angus E. Taylor. New York, 171. Proposed by MURRAY S. KLAMKIN, Ford Scientific Laboratory. Blaisdell Publishing Company, 1965. xvi + 437 pp., $12.50. For 0 < 0 < v/2, it is well known that the inequality A rather thorough self-contained treatment of certain aspects of analysis, basically concerned with the development of Lebesgue measure, then the y > cos* 0 Lebesque integral, followed by treatment of the Daniel1 integral and then measure, establishing the inter-relatedness of the two, and providing the holds for m = 1. What is the smallest constant m for which it necessary topological background in the beginning with many interesting holds? related concepts such as signed measures, LP spaces, vector valued functions of bounded variation, and Stieltjes integrals developed throughout. Solution by the proposer. -Prelude -to Analysis. By Paul C. Rosenbloom and Seymour Schuster. Englewood It follows from the power series expansions that m ^$. We now show Cliffs, New Jersey, Prentice-Hall, 1966. xix + 473 pp. that it suffices to have m = - , i.e., A unique textbook in which a vast number of important mathematical concepts a are skillfully woven into pre-calculus material, using a variety of peda- sin 0 tan 0 > O3 . gogical techniques to establish these concepts at an elementary level. It Since the 1.h.s. and the r.h.s. of (1) are equal at 0 = 0, it does not contain trigonometry, however, as the authors feel that this should suffices to show that the derivative of the 1.h.s. is greater than be developed by the use of calculus. the derivative of the r.h.s. of (I), i.e., -A Concept of Limits. By Donald W. Hight. Englewmd Cliffs, New Jersey, (2) 2 sina 0 + tana 0 > 30' . Prentice-Hall, 1966. xii + 138 pp., $3.95. Similarly to establish (2), it suffices to show A brief but rather complete treatment of the elementary theorems of limits a of real sequences and real functions with some history and many examples and 4 sin 0 cos 0 + 2 tan 0 sec 0 > 60 , exercises designed to lead the reader with a high school algebra and trigo- nometric background to a thorough understanding of the fundamental concepts. 4 a a 4 cos 20 + 2 sec o + 4 tan o sec 6 > 6 . Linear Statistical Inference and its Applications. By C. Radhakrishna Rao. New York, John Wiley, 1965. xviii + 522 pp., $14.95. (4) follows since it can be rewritten as Perhaps the most comprehensive modern treatment of statistical theory and methods available, treating analysis of variance, estimation, large sample theory, statistical inference, multivariate analysis with many facets of Sydney Spital, California State Polytechnic College, establishes the the modern fundamental concepts of each, and a rather thorough treatment above result by showing that of mathematical prerequisites such as vector spaces, including Hilbert lo 0 csc 0) Spaces, probability and measure theory, and necessary excerpts from analysis. -1" = ;o; sec 0 It presupposes a mathematical maturity at about the advanced calculus level and some knowledge of elementary statistical theory and methods. is monotone decreasing on (0,77/2] and then the desired minimum value of m is given by Vector Analysis. By Barry Spain. Princeton, New Jersey, Van Nostrand, 1965. limit m(0) = ix + 114 pp., $5.95. 3 . 0 -0 A brief but thorough treatment of the mathematics of classical vector analysis Also solved by Paul J. Campbell, H. Kaye, Paul Myers, and Ricky Pollack. from a classical point of view with a brief introduction to classical tensor analysis, omitting all applications. Topoloqical Structures. By Wolfgang J. Thron. New York, Holt, Rinehart and Winston, Inc., 1966. x + 240 pp. A modern self-contained treatment of general topology, with historical and Henry E. Fettis and James C. Caslin. Ten Place Tables of the Jacob~an motivation remarks, which proceeds from a very elementary introduction to Elliptic Functions. Aerospace Research, United States Air Force, Wriqht- some rather difficult advanced theorems in twenty-three brief chapters. Patterson Air Force Base, Ohio, 1965. iv + 562 pp. Although the book is brief, there is an excellent bibliography and enough difficult but important theorems and exercises to warrant a year's study. Fisher and Ziebur. Calculus G Analytic Geometry. Englewood Cliffs, New In addition to the classical topics of general topology, there are modern Jersey, Prentice-Hall, 1965. xiii + 768 pp. treatments of quotient spaces, nets, filters, uniform spaces, proximity spaces, topological groups, and paracompactness and metrization. John N. Fujii. Numbers and Arithmetic. New York, Blaisdell Publishing Company, 1965. xi + 559 pp., $8.50. Introduction to Basic Fortran Proqraming Numerical Methods. By William Einar Hille. Analytic Function Theory, Volume 11. New York, Ginn and Prager. Company, 1962. xi1 + 496 pp. Based on lecture notes prepared for a one-semester course for students of J. Nagata. Modern Dimension Theory. New York, John Wiley, 1965. 259 pp., applied mathematics with limited mathematical background, covering Fortran $11.00. programming with applications to computing with polynomials, interpolation, quadrature, solution of equations and integration of ordinary differential L. B. Rall, editor. Error in Diqital Computation, Voiume 11. New York, equations. John Wiley, 1965. x + 288 pp., $6.75. Albert Einstein and the Cosmic World Order. By Cornelius Lanczos. New York, Seaton E. Smith, Jr. Explorations ,& Elementary Mathematics. Englewood Wiley-Interscience, 1965. vi + 139 pp. Cliffs, New Jersey, Prentice-Hall, 1966. xv + 280 pp., $5.95. An outgrowth of a public lecture series at the University of Michigan Hans F. Weinberger. A First Course in Partial Differential Equations a in the spring of 1962 called "The Place of Albert Einstein in the History Of Com~lexVariables Transform Methods. New York, Blaisdell Publishing Physics," written by one of the outstanding mathematics expositors and a Company, 1965. ix + 446 pp. former associate of Einstein. Herbert S. Wilf. calculus and Linear Alqebra. New York, Harcourt, Brace and World, Inc., 1966. xiv + 408 pp., $8.50. Modular Arithmetic. By Burton W. Jones. An excellent book for secondary school teachers, outstanding high school students, or for college students who might wish to use it as an elementary NOTE: All correspondence concerning reviews and all books for review secondary reference on modular systems. should be sent to PROFESSOR ROY B. DEAL, DEPARTMENT OF MATHEMATICS, OKLAHOMA STATE UNIVERSITY, STILLWATER, OKLAHOMA, 74075. CALIFORNIA ALPHA, University of California INITIATES Edward Brill Eugene Gross S. J. Morizumi Charles Joseph Burke Benjamin Halpern John Piepenbrink Tao-Geun Cho Thomas N. Hibbard Alan Radnitz Anna Maria DeRoe Hsiao Chung Hsieh George W. Raiuey, Jr. Richard W. 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