Edinburgh Mathematical Notes最新文献_第9页

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300002743

H. Richmond

1. Imagine that a number of straight lines, coplanar and concurrent, are united so as to form as it were a rigid frame. Imagine that thiB frame is moved (continuously) in the plane in such a way that two selected lines always touch two cycloids traced in the plane. Then it will be found that Every line of the frame will move so as to envelope a cycloid. Isoptic (and orthoptic) are names used by Charles Taylor for loci on which two tangents of curves intersect at a constant angle. The cycloids form a family of curves in which each two members have the same isoptic locus; they may therefore be described as forming an isoptic family. Isoptic loci are of no great importance or interest. Our aim here is to investigate this and other instances in which curves of a uniform type are enveloped by the various lines of a rigid frame.

1. 想象一下，一些共面且平行的直线连在一起，形成一个刚性框架。假设这个坐标系在平面上(连续地)移动，两条选定的线总是接触平面上的两条摆线。然后将发现，框架的每条线都将移动，以包围摆线。等角(和正交)是查尔斯·泰勒用来表示曲线的两条切线以恒定角度相交的轨迹的名称。摆线形成了一个曲线族，其中每两个成员都有相同的等光轨迹;因此，它们可以被描述为形成一个等视光族。等视基因座并不十分重要或有趣。我们在这里的目的是研究这种情况和其他情况，在这种情况下，均匀类型的曲线被刚性框架的各种线条所包围。

引用次数: 0

George David Birkhoff 乔治·大卫·伯克霍夫

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300002755

I. M. H. Etherington

引用次数: 0

Inequalities for solutions of linear differential equations a contribution to the theory of servomechanisms 线性微分方程解的不等式对伺服机构理论的贡献

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300002986

Hans Bückner

Consider the n th order differential equation where the coefficients c v are real constants and f is a real function continuous in the interval a ≦ x ≦ b . The following theorem will be proved in §4: If the characteristic equation of (I) has no purely imaginary roots, then a particular integral η ( x ) can always be found which satisfies the inequality where C is a certain function of the c v only and M is the maximum of |f|. In particular we may take C = 1 if all roots of the characteristic equation are real.

考虑n阶微分方程，其中系数cv是实常数，f是在区间a≦x≦b内连续的实函数。我们将在§4中证明下列定理:如果(I)的特征方程没有纯虚根，则总能找到一个特殊的积分η (x)满足不等式，其中C仅是C v的某个函数，M是f的最大值。特别地，如果特征方程的所有根都是实数，我们可以取C = 1。

引用次数: 2

On the Newton-Raphson method of Approximation 关于牛顿-拉夫森近似法

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300000082

H. Richmond

simple proof of the theorem that a skew symmetric determinant of even order is a perfect square. This can be done by examination of the way in which cycles of two indices may be agglomerated in cycles of more indices, but it would hardly seem to be so simple as Sylvester believed. One can, however, easily enumerate the terms in the square root, the Pfaffian. For the squared terms in the skew determinant correspond exclusively to permutations containing cycles of two indices only, since (if) connotes — ay ciji, or a?.. Thus we have to find in how many ways 2m indices may be put into m such cycles. For first cycle take 1 and any a from the remaining 2m 1 indices; for second cycle take the next surviving index in natural order and any b from the remaining 2m 3; and so proceed. The number of terms in the Pfaffian is thus (2m 1)(2m 3)(2m 5) ... 5.3.1, a factorial composed of odd numbers. This is a well-known result.

偶阶的斜对称行列式是完全平方的简单证明。这可以通过检验两个指数的周期可能聚集在更多指数的周期中的方式来实现，但这似乎很难像西尔维斯特所相信的那样简单。然而，我们可以很容易地列举出平方根中的项。因为偏行列式中的平方项只对应于包含两个指标循环的排列，因为(if)意味着- ay ciji，或a?…因此，我们必须找出有多少种方法可以将2m个指标放入m个这样的循环中。对于第一个循环，从剩下的2m个指标中取1和任意a;对于第二个循环，按自然顺序取下一个幸存的指标，并从剩余的2m 3中取任意b;继续。Pfaffian中的项数是(2m 1)(2m 3)(2m 5)…5.3.1，由奇数组成的阶乘。这是一个众所周知的结果。

引用次数: 0

On pedal tetrahedra 脚踏四面体

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300002561

R. Robinson

P 1 , = -o (rq — sp), p which gives the result required. Since my Solution 2 was, quite unintentionally, rather unfair to the method of partial derivatives, I feel that I ought to draw attention to this shorter solution. The fact that the above solution is merely shorter than the one which I gave does not however detract from the practical advantages of the differential method. Any experienced teacher knows that the step which presents real difficulty to the beginner is the obtaining of equation (1) above. Although in the case of the example which I happened to choose for illustration (and it may not have been the best for the purpose) the above solution by partial derivatives happens to be quite as short as the solution by differentials, the fact remains that, while the technique of differentiation, when once understood, is almost " fool-proof," the pitfalls for the beginner in the solution given above are well known to every teacher of the subject. While the solution of a problem by partial derivatives may be quite a difficult piece of manipulation, exactly the same technique is required for the solution of a problem by differentials, however simple or complicated the problem in question may happen to be.

P 1 = - 0 (rq - sp) P，它给出了所需的结果。由于我的解2，无意中，对偏导数法不公平，我觉得我应该把注意力集中在这个短解上。上面的解仅仅比我给出的解短，但这并不减损微分法的实际优点。任何有经验的老师都知道，对于初学者来说，真正困难的一步是得到上面的公式(1)。虽然在我碰巧选择的例子中(它可能不是最好的)，上面的偏导数解恰好和微分解一样短，但事实是，虽然微分的技术一旦被理解，几乎是“万无一失”的，但上面给出的解决方案对于初学者来说的陷阱是每个老师都知道的。虽然用偏导数来解决问题可能是一个相当困难的操作，但用微分来解决问题也需要完全相同的技术，无论问题是简单还是复杂。

引用次数: 0

Dirichlet's Integrals 狄利克雷函数的积分

Edinburgh Mathematical Notes

Pub Date : 1900-01-01 DOI: 10.1017/S0950184300000124

L. Mordell

which is exactly what it would be using the theorem. But the theorem is true for n = 1 and n = 2, so it is true for n = 3, and hence similarly for all positive integral values of n. Applications, (i) Expansion for 2 cos n 6 in terms of 2 cos 9. Let x = cos 6 + i sin 9, y = cos 9 — i sin 9; then x + y = 2 cos 0 and xy = 1. Also x = cos n 0 + i sin w # ; y — cos n 9 — i sin n 0 so that £" + y = 2 cos n 9. Substituting these in the theorem, we obtain

这正是它使用theorem的方法。但是theorem对n = 1和n = 2是真实的，所以它对n = 3是真实的，对所有积极的应用价值也是相似的，(i) Expansion for 2 cos n n 6 in the terms of 2 cos 9。让x = cos 6 + i sin 9, y = cos 9——i sin 9;然后x + y = 2 cos 0和xy = 1。x =因为n + 0 i sin w;y——因为n 9—i sin 0所以这£“9 + y = 2,因为n。把这些应用到理论中，我们得到了答案

引用次数: 1