What is Riemann`s Hypothesis? March 25, 2012 Draft
... Now 243,112,609 − 1 is quite a hefty number! Suppose someone came up to you saying “surely p = 243,112,609 − 1 is the largest prime number!” (which it is not) how might you convince that person that he or she is wrong? [11] Here is a neat—and, we hope, convincing—strategy to show there are prime num ...
... Now 243,112,609 − 1 is quite a hefty number! Suppose someone came up to you saying “surely p = 243,112,609 − 1 is the largest prime number!” (which it is not) how might you convince that person that he or she is wrong? [11] Here is a neat—and, we hope, convincing—strategy to show there are prime num ...
SUM OF TWO SQUARES Contents 1. Introduction 1 2. Preliminaries
... n = a2 + b2 for some nonnegative a, b ∈ Z. We deliberately include 0 as a possible value for a or b so that squares themselves will fall into this category, e.g., since 4 = 22 + 02 . In this paper, we are interested not only in characterizing the numbers that have a representation as the sum of two ...
... n = a2 + b2 for some nonnegative a, b ∈ Z. We deliberately include 0 as a possible value for a or b so that squares themselves will fall into this category, e.g., since 4 = 22 + 02 . In this paper, we are interested not only in characterizing the numbers that have a representation as the sum of two ...
UNSOLVED PROBLEMS SOME UNSOLVED PROBLEMS by In this
... In this paper I shall discuss some unsolved problems in number theory combinatorial analysis, set theory, elementary geometry, analysis and probability . The choice of problems is purely subjective, I discuss problems on which I worked myself or which interested me and it is certainly not claimed th ...
... In this paper I shall discuss some unsolved problems in number theory combinatorial analysis, set theory, elementary geometry, analysis and probability . The choice of problems is purely subjective, I discuss problems on which I worked myself or which interested me and it is certainly not claimed th ...
Modeling Prime Factorization with Bijections between Natural
... Conventional binary representation of aP natural number n describes it as the value of a polynomial P (x) = ki=0 ai xi with digits ai ∈ {0, 1} for x=2, i.e. n = P (2). If one rewrites P (x) using Horner’s scheme as Q(x) = a0 +x(a1 +x(a2 +...)), n can be represented as an iterated application of a fu ...
... Conventional binary representation of aP natural number n describes it as the value of a polynomial P (x) = ki=0 ai xi with digits ai ∈ {0, 1} for x=2, i.e. n = P (2). If one rewrites P (x) using Horner’s scheme as Q(x) = a0 +x(a1 +x(a2 +...)), n can be represented as an iterated application of a fu ...
calcuLec11 - United International College
... The pattern does suggest that we may be able to get useful information by finding a line that “best fits” the data in some meaningful way. One such procedure, called “leastsquares approximation”, require the approximating line to be positioned so that the sum of squares of vertical distances from th ...
... The pattern does suggest that we may be able to get useful information by finding a line that “best fits” the data in some meaningful way. One such procedure, called “leastsquares approximation”, require the approximating line to be positioned so that the sum of squares of vertical distances from th ...
3.4 Finite Limits at Points
... The previous example is a limit of a certain form, which is called the “0/0 form,” meaning that the function is a fraction in which the numerator and denominator both approach zero as x approaches the limit point. This is one of many indeterminate forms; knowing we have 0/0 form tells us nothing abo ...
... The previous example is a limit of a certain form, which is called the “0/0 form,” meaning that the function is a fraction in which the numerator and denominator both approach zero as x approaches the limit point. This is one of many indeterminate forms; knowing we have 0/0 form tells us nothing abo ...
Infinitesimal Calculus - gauge
... This statement requires imagining numbers that remain positive while they are shrinking to zero. In other words, it requires infinitesimals. But there are no infinitesimals on the real line, and it is difficult to imagine them on the real line. Consequently, at some point, the endless process of app ...
... This statement requires imagining numbers that remain positive while they are shrinking to zero. In other words, it requires infinitesimals. But there are no infinitesimals on the real line, and it is difficult to imagine them on the real line. Consequently, at some point, the endless process of app ...
Hammack 5: Logarithm Review
... Notice loga ( x) is negative if 0 < x < 1, Also loga ( x) tends to −∞ as x gets closer to 0. Take note that the domain of loga is all positive numbers (0, ∞) because this is the range of a x . Likewise the range of loga is the domain of a x , which is R. Also, because loga (1) = a (1) = 0, the x-in ...
... Notice loga ( x) is negative if 0 < x < 1, Also loga ( x) tends to −∞ as x gets closer to 0. Take note that the domain of loga is all positive numbers (0, ∞) because this is the range of a x . Likewise the range of loga is the domain of a x , which is R. Also, because loga (1) = a (1) = 0, the x-in ...
PARTITION STATISTICS EQUIDISTRIBUTED WITH THE NUMBER OF HOOK DIFFERENCE ONE CELLS
... is no smaller than the number of dominoes we needed to remove from the partition to obtain the 2-core. We conjecture that this condition is in fact unnecessary. Conjecture 1.6. The statistics h1,1 and a2 are equidistributed on the set of partitions of n with 2-core {k, k 1, · · · , 1} for all non-ne ...
... is no smaller than the number of dominoes we needed to remove from the partition to obtain the 2-core. We conjecture that this condition is in fact unnecessary. Conjecture 1.6. The statistics h1,1 and a2 are equidistributed on the set of partitions of n with 2-core {k, k 1, · · · , 1} for all non-ne ...
