Mathematics

Publication Date: January 23 , 2018          Number of Pages: 19
In this paper , we present solutions to three problems; listed below
  as PROBLEMS 1 , 2 , and 3. The first two problems; PROBLEMS1 and 2;
    Are among five "Klamkin  Quickies" , published in the February 2004
    issue of the journal CRUX MATHEMATICORUM; issueNo1 ,
      Volume 30 , page17.
      The collection of 5 KLAMKIN QUICKIES was proposed  by
        (the late and legendary)  Murray S.Klamkin; who was , among other
          things , a coach/trainer/teacher  for the Canadian Teams that
            participated in the International Mathematical Olympiads  in the
              1980s , 1990s , and 2000s.
                                          PROBLEM1
                    This is KLAMKIN QUICKIE #3 , in a set of five.
                      Determine all integral solutions of  x^3 + y^3 = z^6 + 3.
  In our solution to PROBLEM1 in Section2; pages4-5; we argue
      modulo 9 to show that the given equation has no integer solutions.
                                                PROBLEM1
        This is KLAMKIN QUICKIE #4 , in a set of five.
          Determine the  maximum value of tanx + tany ,  where
                (1 + sinx )( 1 + siny ) = (cosx)(cosy).
          We present a solution to PROBLEM2 in Section3; pages 6-15.
          We show that the pairs that satisfy the condition (or equation) in
            PROBLEM2; and for which pairs both tanx and tany are both defined; are precisely the pairs that satisfy three conditions:
                1) x + y =2n(pi) , n an integer.
                2) x not of the form , k(pi) + (pi)/2; k an integer.
        And , 3) Likewise , y not of the form,  m(pi) + (pi)/2.
        From these three conditions , it follows that for all such pairs (x , y);
              the sum  tanx + tany is equal to zero; thus , 0 is the answer to
                                PROBLEM2.
      The third problem in this paper , PROBLEM3; is a problem featured in
          the J.I.R. McKnight Problems Contest 1994; and it is published in the  March 2000 issue of CRUX MATHEMATICORUM; issue No2 , Volume26 , on page10.
                                                PROBLEM3
            J.I.R. McKnight Problems Contest 1994
        Problem#4: Solve for x:  0<(or=) x <(or=) (pi)/2 .
8(sinx)(cosx) - (square root(6))(sinx) - (square root(6))(cosx) + 1 = 0.
In our solution in Section4 , pages16-19; we show that the given equation  has exactly two solutions:
        x = (pi)/12 ,  and 5(pi)/12. (radians)
    In degrees ; the two angles that satisfy the above equation are 15 and  75.              ;

Blood flow-through segments of large arteries of man, between adjacent bifurcations, can be modeled as pulsatile flow in tapered converging tubes, of small angle of convergence, up to 2 deg. Assuming linearity, rigid tube and homogeneous Newtonian fluid, the physiological flow field is governed by the Navier-Stokes equation with dominant nonlinear and unsteady terms. Analytical solution of this problem is presented based on an integral method technique. The solution shows that even for small tapering the flow pattern is markedly different from the flow obtained for a uniform tube. The periodic shear stresses at the wall and pressure gradients increase both in their mean value and amplitude with increased distance downstream. These results are highly significant in the process of atherogenesis.

Macroscopic elastic moduli governing the incremental deformations of lung parenchyma are calculated on the basis of a model for an individual lung element in the shape of a regular dodecahedron. Elastic stiffness within the element is provided by pin-jointed tension members along the edges of the dodecahedron, surface tension is incorporated into its pentagonal faces, and the influence of transpulmonary pressure is simulated by an externally applied hydrostatic tension. The analysis is based on a variational statement of nonlinear structural mechanics, and the results show how the moduli depend on the effective inflation pressure, the constitutive behavior of the idealized truss members, and the surface-area dependent surface tension. The theory is discussed in the light of available experimental information. A more general analysis is needed to account for the effects of structural as well as surface-tension hysteresis.

ABSTRACT Macroscopic elastic moduli governing the incremental deformation of pressurized lungs are calculated on the basis of a micromechanical study of an individual lung element (alveolus). The spongy lung tissue (parenchyma) is presumed to be in an initial state of isotropic tension, and the lung alveolus is idealized as a pin-jointed truss in the shape of a regular dodecahedron. The analysis is based on a variational statement of nonlinear structural mechanics, and the results show how the moduli depend on the initial tension and the assumed constitutive behavior of the idealized truss members.

Intrathoracic pressure variations are currently proposed as the main flow-generation mechanisms in standard and modified cardiopulmonary resuscitation (CPR) techniques. A method of changing pressure within the thorax and abdomen without any degree of heart compression was developed and tested in dogs. Intrathoracic and abdominal pressure waves were induced by cyclic inflation and deflation of the lungs and of perithoracic and periabdominal balloons. Various modes of CPR, depending on the rate of cycling, the use of a periabdominal balloon inflation, and a delay between the abdominal and thoracic pressure waves, were studied during ventricular fibrillation. During artificial systole (high intrathoracic pressure phase), the pressure which developed in the right ventricle (96.7 +/- 20.5 mmHg) was higher than the pressure in the aorta (89.3 +/- 20.5 mmHg, p less than 0.001). In artificial diastole (low intrathoracic pressure phase), the right ventricular pressure (11.7 +/- 2.6 mmHg) was lower than the aortic pressure (17.5 +/- 3.3 mmHg, p less than 0.001). The average flow in the carotid artery was 21.7 +/- 7.8 ml . min-1, which was 18 +/- 6% of the baseline carotid flow before CPR. Three different factors were found to improve the efficiency of CPR: periabdominal balloon inflation simultaneous with the intrathoracic pressure waves; increased frequency of the pressure waves from 60 to 100 cycles per minute; and inflation of the periabdominal balloon 50 to 100 ms before the thoracic balloon. Blood-gas and acid-base balance analysis during CPR revealed well-oxygenated arterial blood with a marked respiratory alkalosis and a slowly developing metabolic acidosis.(ABSTRACT TRUNCATED AT 250 WORDS)

The current-voltage relations obtained by integrating the Nernst-Planck equations for a variety of energy profiles are obtained. A simple and approximate method for comparing these relations is described. The method is based on using a linearized transform of current-voltage relations for an Eyring single barrier model. A parameter, gamma, related to the location of the single barrier in the Eyring model, and to the shape of the barrier in other models, is readily obtained from the slopes of the linearized relations. It is then a simple matter to determine whether a given current-voltage relation allows discrimination between any particular energy profiles. The results show that the equivalent Eyring model does not always place the peak energy barrier in the same position as other models and that quite large errors in the assignment of position may be made if such a model is used. The results are also used to test the ability of some experimental current-voltage diagrams to discriminate between various energy profiles.

On the aim and properties of-algebra, in this paper we introduce the two concepts algebra and algebra metric space as well as introduce concept convergent and Cauchy sequence in space and to study the existence of fixed point theorems with contraction condition and algebra expansion on this space.

Publication Date:January15 , 2018                          Number of Pages: 19
In this paper , we study and solve the following problem , by this author.
(a) Find all pairs (r , n) , where r is a positive rational number , n an integer;
        And  such that    r + 1/r = n .
  (b) Determine all triples  (r , n , k)  such that ,
                                    r + 1/r = n/(2^k) ;  where r is a positive rational
        number , k a positive integer; and n an odd positive integer.
  There are five results in this paper , stated on pages 3-7 of the
    introduction of this paper.
    According to Result4 , part(a) of PROBLEM1 has a unique solution:
              (r , n) = (1 , 2) , is the only such  pair.
    And , according to Result5 , there are two families of triples (r , n , k)
                that make up the solution set of part(b) of PROBLEM1:
      Family F1: The triples of the form , (r , n , k)= (1/(2^k) , 4^k + 1 , k) ;
And  Family F2: The triples of the form: (r , n , k)=(2^k , 4^k + 1 , k).
      Where in both cases , k can be any positive integer.

Publication Date: January15 , 2018                Number of  Pages: 33
In this paper , we study and solve a Mathematical Olympiad  number
theory problem. The following problem , stated below as PROBLEM1; was
featured in the 37th Austrian Mathematical Olympiad held in 2006.
                                            PROBLEM1
      37th AUSTRIAN MATHEMATICAL OLYMPIAD
          National Competition(Final Round , Part2)    Day2(June1 , 2006)
      Problem#4: Determine all rational numbers x such that
                1 + (105)(2^x)  is the square of a rational number.
The above problem is published in the September 2009 issue of yhe
          journal CRUX MATHEMATICORUM; issue No5 , Vol.35 , page292.
There are seven  results in this paper. All seven results , Results1-7 ,
      are listed in the introduction of this paper; see pages 4-9.
      It follows from Result2(iii) , Result4 , Result6 , and Result7; that
        PROBLEM1  has exactly  five solutions:
              x = - 8 , - 6 , -4 , 3 , and 4.
  If we let  f(x) = 1 + (105)(2^x) ,  then:
        f(- 8) = (19/16)^2  , f(- 6) = (13/8)^2 ,  f(- 4) = (11/4)^2 ,
            f(3) = (29)^2  ,  and f(4) = (41)^2.