![[Part 1]](http://s1.studyres.com/store/data/008795920_1-0c9d44c6bff0d29348d19d3efc363e24-300x300.png)
[Part 1]
... (1.1) very quickly. The first of these uses the Stirling numbers of the second kind, and the second uses the Eulerian numbers. Both results give (1.1) as series of binomial coefficients in n, rather than directly as polynomials expressed explicitly in powers of n. For many purposes of computation an ...
... (1.1) very quickly. The first of these uses the Stirling numbers of the second kind, and the second uses the Eulerian numbers. Both results give (1.1) as series of binomial coefficients in n, rather than directly as polynomials expressed explicitly in powers of n. For many purposes of computation an ...
5-1
... 5.1.4.4.2. Definition of Integer Subtraction: For all integers a, b, and c, a – b = c if and only if c + b = a 5.1.4.4.3. Theorem: Subtracting an Integer by adding the Opposite – For all integers a and b, a – b = a + (-b). That is, to subtract an integer, add its opposite. 5.1.4.5. Procedures for Su ...
... 5.1.4.4.2. Definition of Integer Subtraction: For all integers a, b, and c, a – b = c if and only if c + b = a 5.1.4.4.3. Theorem: Subtracting an Integer by adding the Opposite – For all integers a and b, a – b = a + (-b). That is, to subtract an integer, add its opposite. 5.1.4.5. Procedures for Su ...
EXAMPLE 5 Using Deductive Reasoning to Prove a Conjecture
... SOLUTION We’ll pick a few numbers at random whose last two digits are divisible by 3, then divide them by 3, and see if there’s a remainder. ...
... SOLUTION We’ll pick a few numbers at random whose last two digits are divisible by 3, then divide them by 3, and see if there’s a remainder. ...
Problem Set 2 Solutions
... This is false because the roots of this quadratic equation are x = ± −1 = ±i, which are imaginary. (c) (∃m ∈ N)(m2 < 1). There exists a natural number m such that m2 < 1. This is false because there are no natural numbers smaller than 1 and 12 6< 1. 2. For each of the following, use a counterexample ...
... This is false because the roots of this quadratic equation are x = ± −1 = ±i, which are imaginary. (c) (∃m ∈ N)(m2 < 1). There exists a natural number m such that m2 < 1. This is false because there are no natural numbers smaller than 1 and 12 6< 1. 2. For each of the following, use a counterexample ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.