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Just Tilt Your Head: A Graphical Technique for Classifying Fixed Points 倾斜你的头:一种分类固定点的图形技术
Q4 Mathematics Pub Date : 2023-05-19 DOI: 10.1080/0025570X.2023.2204794
H. Diamond
Summary Iterations of the form are used throughout mathematics, and convergence (or not) of an iteration to a fixed point is always a central question. We present an informal graphical technique for resolving whether a fixed point in one dimension is attracting or repelling. Specifically, given a real-valued function f, we are determining whether or at a point for which . The former condition on classifies the fixed point as attracting, and the iteration as locally convergent; the latter corresponds to a repulsive fixed point, in which the iterations get further away from the fixed point no matter how close you start them off. The technique requires only that we be able to compute points on the graph of and not necessarily on an explicit functional form or its derivative.
形式的总结迭代在整个数学过程中都被使用,迭代收敛(或不收敛)到一个固定点总是一个中心问题。我们提出了一种非正式的图形技术来解决一维中的不动点是吸引还是排斥。具体地说,给定一个实值函数f,我们正在确定是否或在一个点上。前一个条件将不动点分类为吸引,迭代分类为局部收敛;后者对应于一个排斥不动点,在这个不动点中,无论你开始迭代有多近,迭代都会离不动点更远。该技术只要求我们能够计算的图上的点,而不一定是显式函数形式或其导数上的点。
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引用次数: 0
A Simple Proof that π > 3.05 π>3.05的一个简单证明
Q4 Mathematics Pub Date : 2023-05-17 DOI: 10.1080/0025570X.2023.2199695
Jake H. Lewis
Summary In classrooms where most students are simply told that , accept the fact, and move on, methods for finding lower or upper bound on are usually not taught. Here, I consider a University of Tokyo entrance exam problem: Prove that , I provide students with a simple, yet nontraditional, proof method. In particular, this method does not require a calculator (as in many exams), cumbersome circle geometry, direct use of calculus-based methods, or partial sums of any infinite series.
摘要在课堂上,大多数学生只是被告知,接受事实,然后继续前进,通常不会教授找到下限或上限的方法。在这里,我考虑东京大学入学考试的一个问题:证明这一点,我为学生提供了一种简单但非传统的证明方法。特别是,这种方法不需要计算器(就像在许多考试中一样)、繁琐的圆几何、直接使用基于微积分的方法或任何无穷级数的偏和。
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引用次数: 0
Drawing Cubes, Finding Integral Cubes, and Solving Cubics 绘制立方体,寻找积分立方体和解决立方体
Q4 Mathematics Pub Date : 2023-05-17 DOI: 10.1080/0025570X.2023.2205820
S. Northshield
Summary We present some variations on the theme of “cubes”. We start with right and wrong ways to draw cubes and orthogonal axes. From there we classify cubes in 3-space with integral coordinates. Further, we find that four complex numbers are the vertices of a drawing of a regular tetrahedron if their average of squares equals the square of their average. Tangentially, we find short proofs of the Siebeck-Marden theorem and a method for solving a cubic equation. Finally, we consider sets of complex numbers whose average of squares equals the square of their average.
摘要我们介绍了一些关于“立方体”主题的变体。我们从绘制立方体和正交轴的正确和错误方法开始。从那里我们用积分坐标对三空间中的立方体进行分类。此外,我们发现四个复数是正四面体图的顶点,如果它们的平方平均值等于它们的平均值的平方。切向地,我们发现了Siebeck-Marden定理的简短证明和求解三次方程的方法。最后,我们考虑复数集,其平方的平均值等于其平均值的平方。
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引用次数: 0
Another Simple Proof of Pascal’s Theorem 帕斯卡定理的另一个简单证明
Q4 Mathematics Pub Date : 2023-05-17 DOI: 10.1080/0025570X.2023.2199678
Q. H. Tran
Summary We give another simple proof of Pascal’s theorem, based on the use of directed angles and directly similar quadrilaterals.
基于有向角和直接相似四边形的使用,我们给出了Pascal定理的另一个简单证明。
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引用次数: 0
Clough’s Theorem 克劳夫定理
Q4 Mathematics Pub Date : 2023-05-17 DOI: 10.1080/0025570X.2023.2206298
R. Viglione
Summary We offer a visual proof of Clough’s theorem on the semiperimeter of an equilateral triangle.
摘要我们提供了关于等边三角形半周长的Clough定理的直观证明。
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引用次数: 0
Note on a Binomial Coefficient Divisor 关于一个二项式系数除法器的注记
Q4 Mathematics Pub Date : 2023-05-15 DOI: 10.1080/0025570X.2023.2203055
Matthew Just
Summary Let n be a positive integer, and k and integer with . Let g be the greatest common divisor of k and n. We use the cycle construction to give a combinatorial proof that the fraction n/g divides the binomial coefficient .
设n为正整数,k与整数。设g是k和n的最大公约数。我们利用循环构造给出了分数n/g能整除二项式系数的组合证明。
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引用次数: 0
When Additive and Multiplicative Inverses are the Same 当加法逆和乘法逆相等时
Q4 Mathematics Pub Date : 2023-05-11 DOI: 10.1080/0025570X.2023.2199700
Karen S. Briggs, Caylee R. Spivey
Summary A professor and her students were working through a routine problem on isomorphisms that required finding the multiplicative inverse of 43 modulo 50. The professor, however, simply asked for “the inverse” of 43 modulo 50 and got a surprising response. One student quickly answered that the inverse of 43 modulo 50 was 7. Thinking that the student was responding with the additive inverse, the professor restated her question and asked for the multiplicative inverse. The student checked her work and said that 7 is the multiplicative inverse of 43 mod 50. The observation that the additive and multiplicative inverse of an element could be the same led to a research project into the questions of when, why, and how frequently this phenomenon occurs. Answering these questions led to tools such asthe Chinese remainder theorem, the fundamental theorem of cyclic groups, and quadratic residues.
摘要一位教授和她的学生正在解决一个关于同构的常规问题,该问题需要找到43模50的乘法逆。然而,这位教授只是简单地问了43模50的“逆”,得到了令人惊讶的回答。一个学生很快回答说43的倒数取50是7。考虑到学生的回答是加法逆,教授重述了她的问题,并要求使用乘法逆。这位学生检查了她的工作,说7是43 mod 50的乘法逆。一个元素的加法逆和乘法逆可能是相同的,这一观察结果导致了一个研究项目,研究这种现象何时、为什么以及发生频率。回答这些问题产生了诸如中国余数定理、循环群基本定理和二次余数等工具。
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引用次数: 0
Generalizations of Bertrand’s Postulate to Sums of Any Number of Primes Bertrand命题对任意素数和的推广
Q4 Mathematics Pub Date : 2023-05-05 DOI: 10.1080/0025570X.2023.2231336
J. Cohen
Summary In 1845, Bertrand conjectured what became known as Bertrand’s postulate or the Bertrand-Chebyshev theorem: twice and prime strictly exceeds the next prime. Surprisingly, a stronger statement seems not to be well-known: the sum of any two consecutive primes strictly exceeds the next prime, except for the only equality . Our main theorem is a much more general result, perhaps not previously noticed, that compares sums of any number of primes. We prove this result using only the prime number theorem. We also give some numerical results and unanswered questions.
摘要1845年,Bertrand猜想了后来被称为Bertrand公设或Bertrand-Chebyshev定理的东西:两次和素数严格超过下一个素数。令人惊讶的是,一个更强的说法似乎并不为人所知:任何两个连续素数的和都严格超过下一个素数,除了唯一的相等。我们的主要定理是一个更普遍的结果,可能以前没有注意到,它比较了任何数量的素数的和。我们只用素数定理来证明这个结果。我们还给出了一些数值结果和未回答的问题。
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引用次数: 0
Beyond the Basel Problem: Part II 巴塞尔问题之外:第二部分
Q4 Mathematics Pub Date : 2023-05-02 DOI: 10.1080/0025570X.2023.2199674
K. Williams
Summary In the 18th century, Euler evaluated the sum in terms of the Bernoulli number for any positive integer k. In this article we evaluate the sum where a and b are integers satisfying and .
在18世纪,欧拉用伯努利数计算了任意正整数k的和。在本文中,我们计算了a和b为满足和的整数的和。
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引用次数: 0
Extending Galileo’s Ratio to Hex Numbers and Beyond 将伽利略比率扩展到十六进制数及其以后
Q4 Mathematics Pub Date : 2023-04-28 DOI: 10.1080/0025570X.2023.2199704
Rajib Mukherjee, Manishita Chakraborty
Summary In this article we are extending Galileo’s ratio to Hex numbers and beyond.
摘要在这篇文章中,我们将伽利略的比率扩展到十六进制数及以上。
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引用次数: 0
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Mathematics Magazine
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