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An integral equation of the first kind, with kernel involving a hypergeometric function, is discussed. Conditions sufficient for uniqueness of solutions are given, then conditions necessary for existence of solutions. Conditions sufficient for existence of solutions, only a little stricter than the necessary conditions, are given; and with them two distinct forms of explicit solution. These two forms are associated at first with different ranges of the parameters, but their validity in the complementary ranges is also discussed. Before giving the existence theory a digression is made on a subsidiary integral equation.
Corresponding theorems for another integral equation resembling the main one are deduced from some of the previous theorems. Two more equations of similar form, less closely related, will be considered in another paper. Special cases of some of these four integral equations have been considered recently by Erdélyi, Higgins, Wimp and others.
In the consideration of Question 12612 appearing in the Educational Times for January of this year, proposed by the Rev. Dr. Haughton, F.R.S., of Trinity College, Dublin, the following Diophantine Equation suggests itself:
Let denote the class of finite-dimensional Lie algebras L (over a fixed, but arbitrary, field F) all of whose maximal subalgebras have codimension 1 in L. In (2) Barnes proved that the solvable algebras in are precisely the supersolvable ones. The purpose of this paper is to extend this result and to give a characterisation of all of the algebras in . Throughout we shall place no restrictions on the underlying field of the Lie algebra.
The paper is entitled Solutio facilis problematum quorumdam geometricorum difficillimorum, and is printed in Novi Commentarii Academiae Scientiarum Imperialis Petropolitanae, Tom. xi., pp. 103–123. The volume is for the year 1765; the title-page is dated 1767.
Let A(z) be a transcendental entire function and f1, f2 be linearly independent solutions of
We prove that if A(z) has Nevanlinna deficiency δ(0, A) = 1, then the exponent of convergence of E: = flf2 is infinite. The theorems that we prove here are similar to those in Bank, Laine and Langley [3].
The proposition that if AA′‖BB′ then BB′‖AA′ appears at first sight so simple that it might be regarded as almost intuitive. This is because we already think of parallelism as a symmetrical relationship between two straight lines, in accordance with Euclid's definition of parallels as “straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.” If we take along with this definition Euclid's fifth postulate, or Playfair's equivalent, it defines a unique line through a given point parallel to a given line; but, without the postulate, it cannot be assumed to define more than a class of lines, and a stricter definition is required.
Suppose is holomorphic in Δ = {z:|z|<l} and (an)∈lp where 1≦p≦2. We prove that for k=1,2,…, and almost every θ. This result is sharp in the following sense: Let p∈[1,2] and ε(r) be a positive function defined on [0,1] such that limr→1-ε(r)=0. Then there exists a function holomorphic in Δ with (an)∈lp such that
Gray and Mathews, in their treatise on Bessel Functions, define the function Kn(z) to be
We shall denote this function by Vn(z). This definition only holds when z is real, and R(n)≧0. The asymptotic expansion of the function is also given; but the proof, which is said to be troublesome and not very satisfactory, is omitted. Basset (Proc. Camb. Phil. Soc., Vol. 6) gives a similar definition of the function.
We prove that if Ω is a simple convergence set for continued fractions K(an/bn), then the closure of Ω is also such a convergence set. Actually, we prove more: every continued fraction K(an/bn) has a “neighbourhood” where rn>0 and sn>0, with the following property: Every continued fraction from {n} converges if and only if K(an/bn) converges.
Given sets of balls of different colours, in how many ways may they be arranged in line so that no two balls of the same colour shall come together.
If we have two colours only, and the same number ‘m of each colour, there are evidently two arrangements possible; if we have m, m – 1 respectively, only one arrangement is possible; if we have m, m – 2 ; in, m – 3, &c., no arrangement is possible. We may write these results