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If m are homogeneous equations with nonvanishing determinant, all Aj would be zero-a contradiction. Before proceeding to the study of various specific problems connected with the zeros of polynomials, we shall find it useful to consider certain general theorems to which we shall make frequent reference. Since, how- ever, the point of view is largely that of the geometric theory of functions of a complex variable, we have preferred to use the title of the Geometry of the Zeros of a Polynomial in a Complex Variable. IkS 1. If so, we might ask how many zeros are contained in each Rk or in a specified subset of the Rk or, conversely, what subset of the R contains a prescribed number of zeros of f z.
In fact, C. Lewis expressed his political views on numerous occasions, and we need not infer his attitude towards Nazism or Communism by resorting to extraneous studies, since we are able to quote him directly. Detestation for any ethic which worships success is one of my chief reasons for disagreeing with most communists.
As for the idea of Equality, Lewis spoke often about it. Equality is for me in the same position as clothes. It is a result of the Fall and the remedy for it.
Any attempt to retrace the steps by which we have arrived at egalitarianism and to re-introduce the old authorities on the political level is for me as foolish as it would be to take off our clothes.
The Nazi and the Nudist make the same mistake. But it is the naked body, still there beneath the clothes of each one of us, which really lives. Such a character would accept help from Fascists, convinced that he and his friends are the leaven which could influence positively the dough of their party. Furthermore, he is tempted to accept help from the defenders of the capitalist status quo, whose commercial and imperial agenda does not retain even the least appearance of theism.
Starting off from the premise that a radical change is needed, Spartacus demands a leftist revolution and is willing to accept help from those on the left who openly declare themselves enemies of God Lewis, God The readers will have realized by now that the three names represent three different ideologies: Fascism, Conservatism and Communism.
Following in J. Another way would be that of conversion: Hidden in the heart of this striving for Liberty there was also a deep hatred of personal freedom. That invaluable man Rousseau first revealed it. From that starting point, via Hegel another indispensable propagandist on our side , we easily contrived both the Nazi and the Communist state.
Even in England we were pretty successful. I heard the other day that in that country a man could not, without a permit, cut down his own tree with his own axe, make it into planks with his own saw, and use the planks to build a tool shed in his own garden. On the contrary, by virtue of his ethical and political acuity, he often publicly drew attention to the fallacies and delusions which can accompany the subtle semantic mutations constantly emerging in discourse on education, freedom, politics, values, and so on.
The edition contains no information about the date or place of publication, but the terminus 1 C. Hooper, New York, Macmillan, , p. The available information implies that the book was translated and distributed by a Romanian Baptist organization based in the USA. Leana, Margareta and Veta. The translation was made by Livius Percy and edited by John Tipei. According to more plausible information, the initiative for translating and publishing the book came from a group of Baptist Christians in the United States, members of congregation in Chicago pastored by Petru Popovici.
Such books, printed in runs of about 3, copies, did not usually include all the customary bibliographic information —except for the printing house and year of publication. This was to protect individuals from retaliation by the Communist government. Once printed, the books were distributed to various organizations specializing in the clandestine dissemination of Christian literature within the Communist bloc. The version prepared for print before proved impossible to recover, so the initial handwritten draft of the translation was carefully reviewed with the help of Marinela Blaj, a collaborator of the publisher at the time.
I specialize in the diabolic. Not surprisingly, when Screwtape was published by Humanitas in , the interest in the old Logos version revived significantly. The low interest of the Evangelical market for C. However, the financial difficulties they encountered in recent years have left the project in an uncertain state. Any analysis of the reception of C. Packer, D. Lewis, Anatomia unei dureri, trad. The book includes includes 18 letters Lewis wrote to the author. Given the limitations of this study, I have chosen to mention only a few representative titles, without proposing an exhaustive presentation of the manner in which Lewis is quoted by these authors.
In recent years the blogosphere has played an important role in the diffusion of the information on C. Lewis among Romanian Evangelicals and beyond.
A thorough survey of the most visited Evangelical blogs would reveal hundreds of quotations from C. Lewis, recurrent references to his works, and announcements of books by or about him published or about to be published. Two of them, moreover, were published as individual volumes: Having translated Surprised by Joy himself, the author of this study can attest from experience the difficulties and satisfactions that come from translating a work of such high intellectual calibre.
More recently, Humanitas has enriched its Lewis series with three more releases: Macovescu, The book on the Psalms had a run of 2, copies, while the other two were produced in 4, copies each.
Screwtape, the most popular of C. Votaries of C. Lewis will find between the covers of this book not only a different version of The Abolition 1 R.
The Silver Chair. Professor Rodica Albu has given Lewis a pre-eminent place in her research, privileging two of his works: Her reading, in Communist Romania, of the first Narnia volume effectively turned the book into a powerful parable about a repressive regime which sought to impose its absolute and unchallenged control on all spheres of society and even on the mind and consciousness of man itself.
Lewis, but not to the extent it had when Lewis was made a headliner in the list of Humanitas Printing House. Band, Brendow, , pp. II, no. New technologies and formats such as the audiobook and ebook will undoubtedly contribute to enlarging the circle of readers who will discover with delight the clarity and rigour of his argumentation, the penetration of his intellect, and the richness of his imagination.
I cannot conclude this study without drawing the attention of the Romanian readers to the biography written by Alister McGrath1 Andreas Idreos Professor of Science and Religion at the University of Oxford and currently translated into Romanian by Natan Mladin. No doubt, when it is published, it will be of great interest to Lewis fans in Romania. Lewis in the Romanian cultural space, I believe it is useful to offer a bibliography for the use of researchers interested in looking further into the subject.
Lewis, Despre minuni. Cele patru iubiri. Problema durerii, trad. Lewis, Despre minuni, trad. Lewis, Problema durerii, trad. Lewis, Surprins de bucurie. Povestea unei convertiri, trad. McGrath, C. A Life: According to the information provided by the translator himself, the volume is going to be published either during the autumn of , or the spring of LEWIS Un vis, trad. Lewis, Nepotul magicianului Cronicile din Narnia, vol.
In other volumes: Abolirea omului, trad. Cuza, Cronicile din Narnia: Lewis to the author. About C. Albu, Rodica coord. Lewis, trad. Works Cited Printed sources: Victimele terorii comuniste. Keller, Timothy. Argument pentru Dumnezeu. Editura Kerigma, Lewis, C. New York: Harcourt, Essays on Theology and Ethics. Grand Rapids: Eerdmans, Fern-seed and Elephants and Other Essays on Christianity. Reflections on the Psalms.
The Screwtape Letters. London, Geoffrey Bles, Since this holds for all choices of circles Ck passing through pairs of vertices of H and containing H, all the zeros off' z must lie in the region common to all possible regions K', that is, in the polygon H.
Thus, as stated in Walsh [2 a ], Ths. If, however, the zeros of f z are fixed and not all collinear, no zero off' z other than a multiple zero of f z may lie on the polygon K or circle C or may lie in a certain neighborhood of each zero off z. In view of the similarity of forms 5,8 and 6,2 , we may state for infrapolynomials the following analogue to Lucas' Theorem 6,1 due to [Fejer 3].
Let E be a closed bounded pointset and let p be an infrapolynomial on E. IzI 5 1 and De Bruijn  in the general case, may be proved as follows. The zeros of the kth derivative f k z , 1 f of genus zero also contains all the zeros of f'. Use Ths. If r is the smallest number such that all zeros off', the derivative of a polynomial f, lie in Izl 6.
Then c; 0 0 for k: The zeros of Fin eq. With Z as in ex. Then all the zeros of F' lie in the convex hull H of the zeros of F; no zero lies on aH unless it is also a zero of F [Nagy 18]. If E is a closed convex set of more than one point, every polynomial having all its zeros on E is an infrapolynomial on E [Motzkin-Walsh 4]. Reread sec. If in Th. IzIs1 7. The critical points of a real polynomial. In the Lucas Theorem 6,1 we treated the zeros z5 of f z as independent parameters.
Obviously, if we impose some mutual restraints upon the z; , such as the requirements that the z; be symmetrical in a line or point, we may expect the locus of the zeros of f z to be a smaller region than that given by the Lucas Theorem. Let us in particular assume that f z is a real polynomial and thus that its non-real zeros occur in conjugate imaginary pairs.
Let us construct the circles whose diameters are the line-segments joining the pairs of conjugate imaginary zeros of f z. These circles we shall call the Jensen circles of f z. See Fig. We shall now state a theorem which was announced without proof by Jensen  in It was proved by Walsh  in and later by Echols  and Nagy .
Every non-real zero of the derivative of a real polynomial f z lies in or on at least one of the Jensen circles off z. In particular, if z is a non-real point outside all the Jensen circles, f' z 0 0, a result which proves the Jensen Theorem.
Actually, from the above expressions we may derive the following more specific result: If a real polynomial f has at least one real zero, each non-real critical point off lies interior to at least one Jensen circle off.
If f has no real zeros, each non-real critical point off lies either on all the Jensen circles off or interior to at least one Jensen circle off and exterior to at least one Jensen circle off.
In fact we have also proved the following more general result. If in eq. The Jensen Theorem supplements Rolle's Theorem in describing the location of the zeros off' relative to those of f. A theorem which describes the number of zeros off' is the following one due to Walsh . Let I: Let R be the configuration consisting of I and of the closed interiors of all the Jensen circles which intersect I. We shall prove this theorem with the aid of Th. Let us denote by K the boundary of the smallest rectangle which has sides parallel to the co-ordinate axes and which encloses R.
An immediate consequence of Th. Any closed interval of the real axis contains at most one zero of f' z if it contains no zero off z and if it is exterior to all the Jensen circles for f z An analogous theorem, due to Fekete and von Neumann , holds for infrapolynomials [see sec.
It is the following: Let the pointset E be symmetric in the real axis and let J denote the circles having as diameters the pairs of conjugate imaginary points of E. If p is a real infrapolynomial on E, then any non-real zero of p must be in or on at least one circle J.
Any non-real zero of p on E clearly satisfies Th. Finally, we may state results analogous to Th. Let G be the Green's function with pole at infinity, for an infinite region R having as boundary a finite set B of Jordan curves, which are symmetric in the real axis. Then every non-real critical point of G lies in or on at least one of the circles whose diameters join the pairs of symmetric points of B. Let E,,, A, A denote the ellipse having as minor axis the line-segment joining the pair of conjugate imaginary points A and A and as major axis a linesegment in' times as long as the minor axis.
If Z is a non-real critical point of a real polynomial f, then the equilateral hyperbola with vertices at Z and Z either passes through all the zeros off or separates them. Consider -3[F z ]. Under the conditions of ex. Let all the zeros of the real polynomial p lie in the strip S: Show that rk [Walsh 23]. Let Hk be the equilateral hyperbola with vertices at a,. Let all the zeros of the distance polynomial F in ex.
Then any zero of F' not on E lies in at least one of the spheres having as diameters the line-segments joining pairs of zeros of F symmetric in E [Nagy 18].
Use Th. Some generalizations. We ask now whether or not the Lucas Theorem Th. We shall first prove If K is a convex region which encloses all the zeros a. Let us now set a D].
It is important therefore that we determine the nature of the region S K, 'F. If K is an ellipse, then S K, 0 is an oval-shaped region bounded by a fourth-order curve.
If K is the line-segment AB in Fig. The region S K, 0 is, however, always star-shaped with respect to K. This fact is obvious when Q is also a point of K. We need therefore only consider the case that Q: Let us choose any point Q': Since the angle subtended at Q' by K cannot be less than ip', we infer that it is greater than 0 and that therefore Q' lies in the region S K, 0.
In view of this discussion we may restate Th. If all the zeros and poles of each rational function f, z entering in eq. Marden . For, if P: Let us denote by d1 and d2 the distances of Q1 and Q2 from P respectively and by w the angle formed by the ray PQ1 with the positive real axis. In other words, every point s of S K, 0 is a zero of at least one function F z of type 8,1. Like the Lucas Theorem, it has various physical interpretations. An example of such a field is the one due both to the charges carried by long straight wires at right angles to the z-plane and to the electromagnetic field induced by the currents flowing through these wires.
Another example is the velocity field in the two-dimensional flow due to a vortex-source obtained by placing a source and vortex at the same point. As another application of Th. Assume the contrary. If all the points at which a given pth degree polynomial f z assumes n given values c1, c2 , , c,, are enclosed in a convex region K, and if the m; where the m, satisfy 8,2. Let the m; be constants satisfying ineq. The function f maps H, n to 1, upon itself.
If, in ex. In the previous section 9.
Polynomial solutions of Lame's differential equation. In this section we shall extend the Lucas Theorem to systems of partial fraction sums. In the later case, the zero zk has an interpretation similar to that assigned to the zeros of 6,2. The term 2a; a; - 2J-1 in the conjugate imaginary of 9,3 may be regarded as the force upon a unit mass at the variable point zk due to the mass 2a; situated at the fixed point a,.
The term ; - zk -1 may be regarded as the force upon the unit mass at zk due to the unit mass at the variable point z,. In other words, the system 9,3 defines the zk as the points of equilibrium of n movable unit particles in a field due top fixed particles a, of mass akl2. We leave to the reader the physical interpretation of the tk.
For the general case we shall now prove a theorem due to Marden . Then at z1 circle C would subtend an angle A1 z1 of magnitude less than IT - 2y. Through z1 let us draw the circle F concentric with C and let us draw the line T tangent to P at z1. By the assumption concerning z1 , all the points z1 lie in or on the circle P and hence the quantities 2 - zl -1 are represented by vectors drawn from z1 to points on the side of T containing circle C. In short, both types of terms 2, - 2 -1 and a; a; - z1 -1 entering in eq.
This means according to Th. Since this result contradicts eq. With the first part of Th. Since we now know that all the zeros z; of S z lie in circle C', we may infer from Th. Let us assume concerning V z that its zero tl , farthest from the center of C, were outside C' and let us draw through t1 a circle F and its tangent T. By then repeating essentially the same reasoning as in the first part, we can show that our assumption concerning t1 implies the non-vanishing of the left side of eq.
In the case of real, positive a; , the part of Th. For this same case, Walsh  has given the following generalization of the Jensen Theorem Th. Let the a; in eq. Then no non- real zero of any Stieltjes polynomial having m pairs of non-real zeros may lie , p. In the proof of this theorem, we shall use two lemmas. The first is one which may be easily verified by elementary calculus; namely, The circles whose diameters are the vertical chords of the ellipse a lie in the closed interior of the ellipse E,,, a, a and have this ellipse as LEMMA 9,2a.
The second lemma is 9,7 then the following. LEMMA 9,2b. Except for the term z1 eq. That is, if zl were outside of the Jensen circles for all the a, , 1 Now, to prove Th. If, then, z2 is also exterior to all the Jensen circles E1 a, , a , it lies interior to, say, E1 z3, 23 , and so forth. Eventually, we must come to a value of k, k Instead of assuming that the a; are positive real numbers, let us suppose that the a; corresponding to a pair a,, a, form a conjugate imaginary pair.
We may then prove the following two theorems. If the a, and the corresponding a, are real or appear in conjugate imaginary pairs and if larg x,I Stieltjes polynomial and those of the corresponding Van Vleck polynomial lie in the smallest convex region which encloses both all the real points a, and all the ellipses having the pairs of points a; and a; as foci and having eccentricities equal to cos arg a.
Under the hypotheses of Th. The part which concerns the Stieltjes polynomials was first proved in Vuille . The theorem in its entirety was established in Marden . Under the hypothesis of ex. The movable particles attract one another according to the inverse distance law [Bocher 4].
The zeros of the Hermite polynomials Hn z are all real and distinct. Use ex. The derivative under linear transformations. We obtained these results largely by use of Th. We now wish to see what further generalizations, if any, may be derived by use of the method of conformal mapping. For instance, we know by virtue of Lucas' Theorem 6,2 that any circle C containing all the zeros of a polynomial f z also contains all the zeros of the derivativef' z of f z. Since we may map the closed interior of C conformally upon the closed exterior of a circle C', can we then infer that, if all the zeros of f z lie exterior to C', so do all the zeros of f' z?
Let us consider how to generalize Th. Let us inquire as to the choice of the m, necessary and sufficient for a finite Zk to be such a zero. The logarithmic derivative of F Z calculated from eq. This implies that not all m, may be positive.
In other words, under the general transformation 10,2 the zeros of the logarithmic derivative of a polynomial are not carried into the zeros of the logarithmic derivative of F Z.
The polynomial fi z is of degree at most n - 1. In order to associate the polar derivative fl z with a more familiar invariant, let us introduce the homogeneous co-ordinates 27 by substituting z intof z and fl z. This result provides further evidence of the invariant character of the polar derivative. If the zeros of a polynomial f z are symmetric in a line L, then between two successive zeros of f z on L lie an odd number of zeros of its derivative f z and any interval of L which contains all the zeros of f z lying on L also contains all the zeros of f' z lying on L.
Apply 10,7 to Rolle's Theorem. Let furthermore multiplicities qr at. Then a given zero z,'. Covariant force fields. At the fixed points P, of a unit sphere S let us place masses m; which repel attract if m; Let us denote by c P resultant force at P.
At the points z, let us place masses m, which repel attract if m, torial plane of S. Let O z be the resultant force upon a unit mass at z due to masses m, at the points z,. According to sec. The augmented system may, however, be considered as comprised of the p pairs of masses, m; at P, and -m; at N. According to Lem. Since for every j the circle C, lies on the sphere S, the resultant force ' P due to all p pairs is tangent to the sphere S. The points of equilibrium in the spherical force field project into the points of equilibrium in the corresponding plane force field.
Circular regions. Since fl z is a generalization of the ordinary derivative, its zeros may be expected to satisfy some invariant form of the Lucas Theorem 6,2 that any circle C containing all the zeros of f z also contains all the zeros off ' z. In order to find the corresponding theorem for the polar derivative, we need to consider the class of regions which includes the interior of a circle as a special case and which remains invariant under the transformation 10,2.
As is well known, this is the class of so-called circular regions, consisting of the closed interiors or exteriors of circles and the closed half-planes. In our subsequent work involving circular regions we shall find the following lemma very useful. Then the point w1 lies inside or outside the circle C' according as the circle C does or does not separate the two points Z and z1. This completes the proof of Lem. Using the above equations, prove the following. If the circle C passes through the point Z, then C' is a straight line passing through Z.
If the circle C passes through the point z1 but not through the point Z, then C' is a circle passing through the point w1.
Zeros of the polar derivative. We are now ready to state the invariant form of the Lucas Theorem Th. If all the zeros z; of the nth degree polynomial f z lie in a circular region C and if Z is any zero of 13,1. Because of the importance of Laguerre's theorem to our subsequent investigations, we shall give two proofs of it and also suggest a third in ex. The first proof will use the results of sec.
Let us assume that Z and are'both exterior to the region C. Since all the zeros of f z lie in C, it follows that f Z 0 0 and, hence, also Z s. Through Z a circle P may be drawn which separates the region C from the point.
The force 1b; at P due to the pair consisting of m, at F. The vectors D; are consequently all drawn from P to points on the same side of the tangent line to P at P. According to Th. This means that P cannot be an equilibrium point in the spherical field and that consequently Z cannot be an equilibrium point in the corresponding plane field. This contradiction to our assumption concerning Z proves the first part of Laguerre's Theorem.
To prove the second part of the theorem, let us assume first that a circle K through Z and has at least one z, in its interior, no z, in its exterior and the remaining z; on its circumference.
This corresponding circle K' through P and Q on the sphere then has at least one P, in its "interior", no P; in its "exterior" and the remaining Pi on its circumference. The forces Di are then directed from P along the tangent line to K' at P or to one side of this line and hence cannot sum to zero. This contradicts the hypothesis that Z is a zero of fl z and so at least one z; must be exterior to K.
Since a contradiction would also follow if K were assumed to have at least one zk in-its exterior and no z, in its interior, we conclude that K must separate the z, unless it passes through all of them. While the proof which we have just completed was based upon the properties of equilibrium points, our second proof of Laguerre's Theorem 13,1 will be based upon the properties of the centroid of a system of masses. If each particle w; in a system of positive masses m; lies in a circle C', then their centroid w also lies in C' and any line L through w either passes through all the w, or separates the w;.
This lemma is intuitively obvious. In order to prove it analytically, let us write eq. If circle C' did not contain w, it would subtend at w an angle A, 0 circle C'; using Lem. That is, not both Z and may lie exterior to C. In the second part of Laguerre's Theorem we know by hypothesis that Z is different from all the z;. Any circle K through Z and would transform into a line L through w, the centroid of the w;. Hence, either K passes through all the z, or it separates some z; from the remaining z;.
Thus, we have completed the proof of Laguerre's Theorem. In our discussion of Laguerre's Theorem, we have implied that is a given point and that the zeros Z of f1 Z were to be found. Thus C may be interpreted as the point at which all the mass must be concentrat in order to produce at Z the same resultant force as the system of masses m; at the points z;. That is, C may be interpreted as the center of force.
Based upon this interpretation, a theorem equivalent to Laguerre's Theorem has been given by Walsh [lb, p. The poles Sk may be equal or unequal. Like the kth ordinary derivative f k Z of f z , the kth polar derivative fk z is a polynomial of degree n - k.
Just as the position of the zeros of f k z may be determined by repeated application of the Lucas Theorem 6,2 see ex. If all the zeros of an nth degree polynomialf z lie in a circular region C and if none of the points 1 , 2, - - - , k k S n - 1 lies in region C, then each of the polar derivatives fl z , f2 z , , fk z , in the eqs.
For, by Laguerre's Theorem 13,1 , all the zeros of f1 z lie in C; hence, all those off2 z lie in C; hence, all those of f2 z lie in C; etc. Let us express the polar derivative fk z directly in terms of f z and 1, if from eqs. Gk - Z Let us put fk z in still another form which will more clearly show the relation of its coefficients to those of f z. For this purpose, let us write f z and fk z in the form n f z n-k. Substituting into eq. If all the points z; lie on a circle C, the following is true: The centroid of the zeros of the derivative of a polynomial f z is the same as the centroid of the zeros off z.
Let f z be an nth degree polynomial, t an arbitrary point for which f t f' t 0 0, and L an arbitrary line through t. Then, if at least one and at most p zeros off z , 1 zeros must lie on L [Nagy 6]. Let H: Then at least one zero off z lies in or on each circle through the points t and v [Fejer 2].
Then either at least one zero of f z lies inside CD or all the zeros off z lie on C,. Let polar co-ordinates r, 0 be introduced with pole at zo and with polar axis along a ray from zo through a root C of eq. If P has a k-fold zero at the origin, 0 5 k 5 n, and n - k zeros in C, then Q z has a k fold zero at the origin and n - k - 1 zeros in C [Ballieu 1]. Apply Th. The zeros of the distance polynomial given in ex.
Let K be a circular domain in the z-plane and S an arbitrary pointset in the If the nth degree polynomial f is such that f z E S for all z e K, then w-plane. Take C as the complement of K and apply Th. If the nth degree polynomial satisfies the conditions 1 f z 1 Generalization to abstract spaces. We now proceed to extend Laguerre's theorem Th. For this purpose we need to define an abstract homogeneous polynomial and its polar derivative as well as an analogue to a circular region.
If P,, x, y 0 0, P is said to be of degree n. We may then use eq. This form may be defined as the nth polar of P x when E is not necessarily finite dimensional. Having defined an abstract homogeneous polynomial and its polar form, we next introduce a concept equivalent to "circular region". Similarly, we may define a "circular region" in E by an inequality H x, x? We now state a generalization of Laguerre's Theorem Th. The question now arises as to when the class 91 of homogeneous polynomials not vanishing on E1 is an empty one.
It will be empty if E1 has a two dimensional linear subspace L.
For then with the use of suitable coordinates in L, any P x becomes a homogeneous binary polynomial and, as such in an algebraically closed field, it has at least one zero in L.
If 0 for some satisfying 14,14 , then the left side of 14,15 will be positive and hence the two dimensional space spanned by xo and must be negative definite will belong to E1. For 91 not to be empty, for every satisfying 14, That is to say, a necessary and sufficient condition for 91 not to be empty is that be negative definite for all a satisfying 14,14 [HSrmander 1].
For other generalizations of Laguerre's Theorem to abstract spaces, we refer the reader to Zervos . Thus also and thus is n! Use induction based upon 14,1. The nth polar may be written as 14,19 P x1, x2, Show that the right side of 14,19 is a symmetric n-linear form satisfying 14,9. Apolar polynomials. So far we have been concerned with the relative position of the zeros of certain pairs of polynomials. In Chapters I and II, the pair consisted of a polynomial and its ordinary derivative.
In Chapter III, the pair consisted of a polynomial and its polar derivative. We shall now apply the results obtained to the study of the comparative location of the zeros of other pairs or sets of related polynomials. We begin with a pair of so-called apolar polynomials. Let us denote by z1, z2 , , z,, the zeros of f z and by t1, S2 ,.. Two nth degree polynomials f z and g z are apolar if and only if the elementary symmetric functions s n, p of the zeros off z and the elementary symmetric functions a n, p of the zeros of g z satisfy the relation: If f z and g z are apolar polynomials and tf one of them has all its zeros in a circular region C, then the other will have at least one zero in C.
Let us prove this theorem on the assumption that all the zeros z1 , z2 , , z. In view of eqs. In other words, at least one of the zeros 1 , S2 , , n of g z must lie in any circular region C containing all the zeros of f z.
Similarly, at least one of the zeros z1, z2 , , zn of f z lies in any circular region containing all the zeros of g z. From Grace's Theorem, we may deduce at once the following result due to Takagi [I ]. If f z and g z are apolar polynomials, any convex region A enclosing all the zeros of f z must have at least one point in common with any convex region B enclosing all the zeros of g z.
For, if A and B had no point in common, we could separate them by means of a circle C enclosing say A, but not containing any zero of g z. This would contradict Grace's Theorem. Let 1 be a symmetric n-linear form of total degree z2 , , Zn and let C be a circular region containing the n points zio , z2o n in z1 , By the well-known theorem of is linear and symmetric in the z1, z2 , , zn may algebra, any function linear and symmetric in the variables z1 , z2 , be expressed as a linear combination of the elementary symmetric functions s n, p of these variables.
That is, we may find constants Bk so that D Z1, z2, Consequently, z1 0 , z 20 ,.. Conversely, as may be shown by a reversal of the above steps, Th.
In other words, as shown in Curtiss , Th. A result similar to Th. Grace's Theorem 15,3 was derived by repeated application of Laguerre's Theorem 13,1. Similarly, the following generalization of Grace's Theorem to abstract spaces, due to Hormander , may be derived by repeated application of Th. Let a homogeneous nth degree polynomial P x , its nth polar and a hermitian symmetric form H x, y be defined in a form P x1 , x2 , , vector space E with values in an algebraically closed field K.
A further generalization, also due to Hormander , is the following: Let P x , a homogeneous polynomial defined in a vector space E over a field K, assume values in a vector space G over K.
Let H x, y and E1 be defined as in Th. By a supportable set M c G we mean a set M that is not intersected by any hyperplane through the origin and any point E G - M. By Th. If all the zeros of F z lie in a circular region K, then at least one zero of G z lies in the circular region K' obtained on inverting K in the unit circle. If f z and g z are apolar polynomials with only real zeros, any interval A containing the zeros of f z must have at least one point in common with any interval B containing the zeros of g z.
Use Cor. If f and g are given as in eqs. Express f1 and g1 in terms of fi. If the polynomials f and g have the property specified in ex. Let B be a Banach space over C with norm Il. In order that there exist a transformation 10,2 which carries f and g of eqs. We shall now apply Ths. We shall first consider a result due to Szego [I]. This follows from Th. One of these will be a. If all the zeros of f z. Let A: It may be easily verified that E A, i; is bounded by a Pascal limacon and that, when A is an arbitrary bounded convex region, E A, is star-shaped with respect to C.
Another consequence of Th. Given the polynomials f, g and h in eqs. Using the symmetry of the hypotheses in f and g, we may show Ih z I This result completes the proof of Cor. They are generalizations of the results stated in exs.
The region A in Fig. This lemma follows from Th. If fl1 is a zero of g z and if the hypotheses of Th. This verifies Lem. If f z and g z satisfy the hypotheses of Th. That is, Lem. Now, to prove Th. Nk - Z , and compute f1 z from eqs. F'n - z , apply Lem. Letf z , g z and h z be the polynomials defined in Th. If all the zeros off z lie in the sector 9': The region Yin Fig. The necessary lemmas paralleling Lem.
If f z has only real zeros and g z has only real zeros of like sign, then the h z of eq. If f z has only real zeros with a sign of a and g z have only real zeros with a sign E', then the h z of eq. The results thereby obtained are due to the following: If the hypotheses of Th. By induction, all the zeros of the f,' z of eq. If all the zeros of f z are real and positive and if all the zeros of g z are real and exterior to the interval 0 16,19 Letf z , g z and h z be defined by eqs.
Let D z, L denote the distance of point z to line L and let al , a2 , F'1 , i32 , ' ' ' , n ; Y1 , Y2 , - - - , y,1 denote respectively the zeros of the f z , g z and h z defined by eqs. Then, if -1 [De Bruijn-Springer 2]. Take the K of ex. Use the Jensen Formula 16,20 as modified in Rosenbloom . The region E A, t is then star-shaped with respect to.
In Cor. Linear combinations of polynomials. Our next application of the theorems of sec. We shall assume that the zeros offk z lie in a circular region Ck. Unless otherwise specified, the region Ck will be bounded by a circle Ck with center ck and radius rk.
This result follows almost at once from Th. On the strength of Th. This leads to eq. To find all possible positions of , we must allow each a, to occupy all possible positions in its circular region C,. In other words, all the zeros of F z lie in the locus 17 as defined in Th. In order to find Pk , we shall need three lemmas which essentially concern the location of the centroid of a system of particles possessing real or complex masses. The first lemma is due to Walsh [lc, pp. All three lemmas are proved in Marden .
If the points a1 , a2 ,. In the case of exclusively positive real m,, we may deduce Lem. In other words, the locus of the point oc of eq. If the point al describes the closed exterior of the circle Cl but the remaining a, describe the closed interiors of the circles C; , then the locus of the point a of eq. Furthermore, this point al lies on or outside circle C, , whereas the remaining points a; lie on their respective circles C,.
That is, every point a of the locus lies in C and every point of C is a point of the locus. Since C' is the entire plane, so is C. If two or more of the a; vary over the closed exteriors of'their circles C; , then the locus of point a is the entire plane. For example, let us suppose that al varies over the closed exterior of the circle C, with center c, and radius r, and oc2 varies over the closed exterior of C2 while the remaining a; vary over the closed interiors of the circles C;.
Returning now to discussion of the locus of the points zk of eq. If in the notation of Th. For this same function h z , we may obtain an altogether different set of results if we write eqs.
In short, zk must lie on or outside E. If each zero of f1 z lies in or on the circle C1 , if each zero off2 z lies in a closed half-plane S satisfying the relation 91 z? By hypothesis, a, lies in C1 , a2 lies in S and a - Thus at least one zero of f1 z lies in the circle IzI 4. The trinomial eq. Thus, it has at least one root in the circle IzI 5.
An equivalent statement of the result in ex. Hint for last result: Study the variation of the Z-points as Z decreases continuously from A to 0 [Nagy 11 ]. Let u1, u2 , , u,, be n distinct points inside a circular region C and let v1 , v2 , , vn be n distinct points outside C. Apply Ths. AIA2 Let S1 a, 0, 0: Then, if all the zeros off lie in S1 a, 0, 0 , all the zeros of F lie in S2 a, 0 - a, 0 [Han-Kuipers 1].
We begin with a theorem due to Walsh . To prove this theorem, we shall assume Z to be any zero of h z ; i. An interesting special case under Th. When used in conjunction with Cor. Let the polynomials f z , g z and h z be defined by the eqs. We thereby derive a result due to Takagi . We have stated Cor. IkS 1. By Cor. This establishes the following result due to Takagi .
Thus far, we have considered linear combinations of a single polynomial and its derivative. Let us now study the linear combinations of the products [ f '' z f zn-' z ] of the derivatives of two given polynomials f1 z and f2 z.
The first result which we shall prove is the following one due to Walsh : Let Z be any zero of h z. In the third case Z is a point in or on the circle rk, as may be determined by use of Lem.
Conversely, if Z is any point of IF, it is a possible zero of h z. If, however, Z lies in rk , we may according to Lem. As an application of Th. If all the zeros of an nth degree polynomial f z lie in the circle C: Here , circle C1 is the same as circle C but circle C2 is merely the point z - 0.
Let F z and G z be polynomials which have all their zeros in the strip I3 z I 0. A two-circle theorem for polynomials. The Lucas Theorem which we developed in sec. Furthermore, as we remarked in sec. Let us now consider the class T of all polynomials f z of degree n which have n, zeros in or on a circle C, , n2 zeros in or on a circle C2, etc. Let us now determine the precise locus of the critical points of the polynomials of class T.
In Fig. It has with C, and C2 a common center of similitude and its center is the centroid of the system of two particles, one of mass n2 at c, and the other of mass n, at c2. This is an equation which is linear and symmetric in the zeros of fl z and in the zeros of f2 z.
Obviously the first Z is a point in C1 and the second Z is a point in C2. Thus we have proved that every zero Z off' z lies in at least one of the circles C1 , C2 and C. We thereby complete the proof of Th.
Concerning the number of zeros off' z , we may as in Walsh  deduce the following result. If the closed interiors of the circles C1 , C2 and C of Th. For, if S1 is any point in C1 and 2 any point in C2, then we may allow all the n1 zeros of f z in C1 to approach 1 along regular paths entirely in C1 and similarly allow all the n2 zeros of f z in C2 to approach 52 along regular paths in C2.
Thus 1 and i2 become zeros of f' z of the respective multiplicities nl - 1 and n2 - 1, the remaining zero off' z then being a point of C. During this process, no zero off ' z can enter or leave C1, C2 or C. Hence, the number of zeros in C1 , C2 and C was also originally n1 - 1, n2 - 1 and 1.
Let the circle C1 with center c1 and radius r1 enclose all the points in which a pth degree polynomial P z assumes the value A and let the circle C2 with center c2 and radius r2 enclose all the points in which P z assumes the value B.
This theorem is stated and proved in P61ya-Szego [1, vol. Substituting from eq. Let ml': If every zero of an nl-degree polynomial fl z lies in or on the circle Cl: If an nth degree polynomial f z has a k-fold zero at a point P and its remaining n - k zeros in a circular region C, then f' z has its zeros at P, in C and in a circular region C' formed by shrinking C towards P as center of similitude in the ratio 1: If C and C' have no point in common, they contain respectively n - k - 1 zeros and one zero of f' z [Walsh 1 b, p.
Let F z be an nth degree polynomial whose zeros are symmetric in the origin 0. Let 0 be a k-fold zero of F z and let all the other zeros of F z lie in the closed interior of an equilateral hyperbola H with center at 0. Apply ex. Given the disks Ck: Two-circle theorems for rational functions. The question raised in sec. Here the answer, also due to Walsh [lb, p. The corresponding quotient f z would then be constant and its derivative identically zero. We may therefore replace the interiors of circles C1 and C2 by arbitrary circular regions C1 and C2.
We may also introduce the binary forms n L. If each zero of the form q lies in a circular region Cl , 77 lies in a circular region C2 and if the regions C1 if each zero of the form and C2 have no points in common, then no finite zero of the jacobian of the two forms lies exterior to both regions C1 and C2.
Let E be a vector space over an algebraically closed field K of characteristic zero. We may prove Th. Thus M 0 0 and consequently' x1 , x 0 0 as was to be proved. Laguerre's Theorem Th. Let positive particles of total mass n be placed at certain points of a circular regions C1 on the unit sphere S and negative particles of total mass -n at certain points of a circular region C2 on S.
If the regions C1 and C2 have no common points, then no point on S exterior. Thus, obtain another proof of Th. Letf z be a polynomial of degree m and g z a polynomial of degree n 0 M. Let all the zeros of g z lie in circular region R bounded by a circle C.
The multiple zeros off2 are not critical points off. The general case. In generalization of secs. The results which we shall obtain are due to Marden  and . In applying Th. In short, when no 0, Th. For n o 0, 21,4 is the equation of a bicircular quartic, a result which for the m, positive integers and the C, interiors of circles coincides with the result due to Walsh . With subscripts 1, 2, 3 replacing 0, 1, 2 respectively, Fig. In this case, we must take the precaution that not all the regions C, have a point in common.
Proceeding now to the proof of Th. In the second case Z lies in the locus R described by the roots of equation 21,5 when o , C9 are allowed to vary independently over the circular regions Co , C, , , C,, , respectively.
Let us first choose Z as any fixed point which lies exterior to all the regions Then by Lem. C, Z I Hence, by Lem. Then, as the points , vary over the regions Cj, point w assumes the value of zero at least once. To complete the proof of Th. We shall now establish the following converse of Th. Then ineq. We also note that Lem.
In this case, therefore, we may, without difficulty, retrace the steps which lead to Lem. Using Lem. If p 0, may be satisfied only under one of the following two circumstances. Let us choose the region Co Walsh. Let each F, p be an n, degree distance polynomial [cf. Some important special cases. We shall now consider under Th. On the other hand, for this special case eq. These results may be summarized in the form of two theorems both due to Walsh .
If the points C11 C2, S3 varying independently have given circular regions as their loci, then any point z forming a constant cross-ratio with C1, C2 and b3 also has a circular region as its locus. Regarding Th. Without loss of generality, we may take 0 at the origin and take the centers c, , c2 and c3 of the circles on the x-axis.
If eqs. Denoting the roots of 22,8 by y1 and Y2, we have the relations from eq. Using eqs. In other words, the bicircular quartic 22,7 degenerates into the two circles r1 and r2 as required in Th. For further details, the reader is referred to Walsh lc, p. Let the polynomial f z of degree n; have all its zeros in the closed interior of C,.
Let C; denote the circles of radius r, which have their centers at the zeros of the logarithmic derivative of the g z of eq. Let the r in ex.
Let a, b, c1 , c2 , c3 , be real numbers and let each zf be a point in or on the circle C, with center at cf and with radius r. Then the zeros of the derivative of the entire function of genus zero or one Let C1, C2 Assume in eq. Let Ci , C2 , , C, denote the circles of radius r with centers at the zeros of s z. If no circle C,: The curve 21,16 reduces to one or more circles in the following cases: Let Cl: IzI r1 , C3: Polynomials with two given zeros.
In Chapters II and V we developed several theorems on the location of all the critical points of a polynomial f z when the location of all the zeros of f z is known. In the present chapter we shall investigate the extent to which the prescription of only some of the zeros off z fixes the location of some of the critical points off z.
A first result of this nature is the one which we may derive immediately from Rolle's Theorem by using eq. This result states that, if the zeros of a polynomial are symmetric in a line L, then between any pair of zeros lying on L may be found at least one zero of the derivative. We now ask whether or not, given two zeros of a polynomial f z , we may locate at least one zero off ' z even when no additional hypothesis such as that of symmetry in a line is placed upon the remaining zeros.
Since eq. This means not only that at least one zero of f' z lies in the circle C of Th. Let us now allow the points z1 and z2 to vary arbitrarily within circle IzI 5 R and inquire regarding the envelope see Fig. Any point on the circumference of C may be represented by the complex number: This is again the best result, as may be seen by choosing f: In other words, f z is univalent in C'.
In the notation of sec. Then E' cannot lie wholly interior or exterior to H [Motzkin-Walsh 2]. Mean-Value Theorems. We may derive results similar to those of sec. In the form stated below, these theorems were first proved by Marden  and , but in certain special cases they had been previously treated in Fekete  and Nagy .
Both theorems employ the notation S K, 0 as in sec. Let P z be an nth degree polynomial and let zl , z2 , , z' be any m points of a convex region K. We may similarly describe the location of point a. Let P z be an nth degree polynomial and let C: The proofs of both Ths. For example, to prove the first, let us write eq. Hence, by Th. We shall now apply Th. Let P z be an nth degree polynomial; let C: Let us denote by Q z an nth degree polynomial which assumes the same values at the points a and i.
If now we replace n by n - 1 in Th. Since K may be taken as the line-segment joining the points and? Let q be a positive continuous function on the finite interval I: In the notation of ex. Polynomials with p known zeros. As a generalization of Ths. Lucas' Th. Furthermore, as in Th. Let us apply Th.