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FIRST-DEGREE EQUATIONS AND INEQUALITIES

In this chapter, we will develop certain techniques that help solve problems stated in words. These techniques involve rewriting problems in the form of symbols. For example, the stated problem

"Find a number which, when added to 3, yields 7"

may be written as:

3 + ? = 7, 3 + n = 7, 3 + x = 1

and so on, where the symbols ?, n, and x represent the number we want to find. We call such shorthand versions of stated problems equations, or symbolic sentences. Equations such as x + 3 = 7 are first-degree equations, since the variable has an exponent of 1. The terms to the left of an equals sign make up the left-hand member of the equation; those to the right make up the right-hand member. Thus, in the equation x + 3 = 7, the left-hand member is x + 3 and the right-hand member is 7.

SOLVING EQUATIONS

Equations may be true or false, just as word sentences may be true or false. The equation:

3 + x = 7

will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result.

Example 1 Determine if the value 3 is a solution of the equation

4x - 2 = 3x + 1

Solution We substitute the value 3 for x in the equation and see if the left-hand member equals the right-hand member.

4(3) - 2 = 3(3) + 1

12 - 2 = 9 + 1

10 = 10

Ans. 3 is a solution.

The first-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can be determined by inspection.

Example 2 Find the solution of each equation by inspection.

a. x + 5 = 12
b. 4 · x = -20

Solutions a. 7 is the solution since 7 + 5 = 12.
b. -5 is the solution since 4(-5) = -20.

SOLVING EQUATIONS USING ADDITION AND SUBTRACTION PROPERTIES

In Section 3.1 we solved some simple first-degree equations by inspection. However, the solutions of most equations are not immediately evident by inspection. Hence, we need some mathematical "tools" for solving equations.

EQUIVALENT EQUATIONS

Equivalent equations are equations that have identical solutions. Thus,

3x + 3 = x + 13, 3x = x + 10, 2x = 10, and x = 5

are equivalent equations, because 5 is the only solution of each of them. Notice in the equation 3x + 3 = x + 13, the solution 5 is not evident by inspection but in the equation x = 5, the solution 5 is evident by inspection. In solving any equation, we transform a given equation whose solution may not be obvious to an equivalent equation whose solution is easily noted.

The following property, sometimes called the addition-subtraction property, is one way that we can generate equivalent equations.

If the same quantity is added to or subtracted from both members of an equation, the resulting equation is equivalent to the original equation.

In symbols,

a - b, a + c = b + c, and a - c = b - c

are equivalent equations.

Example 1 Write an equation equivalent to

x + 3 = 7

by subtracting 3 from each member.

Solution Subtracting 3 from each member yields

x + 3 - 3 = 7 - 3

or

x = 4

Notice that x + 3 = 7 and x = 4 are equivalent equations since the solution is the same for both, namely 4. The next example shows how we can generate equivalent equations by first simplifying one or both members of an equation.

Example 2 Write an equation equivalent to

4x- 2-3x = 4 + 6

by combining like terms and then by adding 2 to each member.

Combining like terms yields

x - 2 = 10

Adding 2 to each member yields

x-2+2 =10+2

x = 12

To solve an equation, we use the addition-subtraction property to transform a given equation to an equivalent equation of the form x = a, from which we can find the solution by inspection.

Example 3 Solve 2x + 1 = x - 2.

We want to obtain an equivalent equation in which all terms containing x are in one member and all terms not containing x are in the other. If we first add -1 to (or subtract 1 from) each member, we get

2x + 1- 1 = x - 2- 1

2x = x - 3

If we now add -x to (or subtract x from) each member, we get

2x-x = x - 3 - x

x = -3

where the solution -3 is obvious.

The solution of the original equation is the number -3; however, the answer is often displayed in the form of the equation x = -3.

Since each equation obtained in the process is equivalent to the original equation, -3 is also a solution of 2x + 1 = x - 2. In the above example, we can check the solution by substituting - 3 for x in the original equation

2(-3) + 1 = (-3) - 2

-5 = -5

The symmetric property of equality is also helpful in the solution of equations. This property states

If a = b then b = a

This enables us to interchange the members of an equation whenever we please without having to be concerned with any changes of sign. Thus,

If 4 = x + 2 then x + 2 = 4

If x + 3 = 2x - 5 then 2x - 5 = x + 3

If d = rt then rt = d

There may be several different ways to apply the addition property above. Sometimes one method is better than another, and in some cases, the symmetric property of equality is also helpful.

Example 4 Solve 2x = 3x - 9. (1)

Solution If we first add -3x to each member, we get

2x - 3x = 3x - 9 - 3x

-x = -9

where the variable has a negative coefficient. Although we can see by inspection that the solution is 9, because -(9) = -9, we can avoid the negative coefficient by adding -2x and +9 to each member of Equation (1). In this case, we get

2x-2x + 9 = 3x- 9-2x+ 9

9 = x

from which the solution 9 is obvious. If we wish, we can write the last equation as x = 9 by the symmetric property of equality.

SOLVING EQUATIONS USING THE DIVISION PROPERTY

Consider the equation

3x = 12

The solution to this equation is 4. Also, note that if we divide each member of the equation by 3, we obtain the equations

whose solution is also 4. In general, we have the following property, which is sometimes called the division property.

If both members of an equation are divided by the same (nonzero) quantity, the resulting equation is equivalent to the original equation.

In symbols,

are equivalent equations.

Example 1 Write an equation equivalent to

-4x = 12

by dividing each member by -4.

Solution Dividing both members by -4 yields

In solving equations, we use the above property to produce equivalent equations in which the variable has a coefficient of 1.

Example 2 Solve 3y + 2y = 20.

We first combine like terms to get

5y = 20

Then, dividing each member by 5, we obtain

In the next example, we use the addition-subtraction property and the division property to solve an equation.

Example 3 Solve 4x + 7 = x - 2.

Solution First, we add -x and -7 to each member to get

4x + 7 - x - 7 = x - 2 - x - 1

Next, combining like terms yields

3x = -9

Last, we divide each member by 3 to obtain

SOLVING EQUATIONS USING THE MULTIPLICATION PROPERTY

Consider the equation

The solution to this equation is 12. Also, note that if we multiply each member of the equation by 4, we obtain the equations

whose solution is also 12. In general, we have the following property, which is sometimes called the multiplication property.

If both members of an equation are multiplied by the same nonzero quantity, the resulting equation Is equivalent to the original equation.

In symbols,

a = b and a·c = b·c (c ≠ 0)

are equivalent equations.

Example 1 Write an equivalent equation to

by multiplying each member by 6.

Solution Multiplying each member by 6 yields

In solving equations, we use the above property to produce equivalent equations that are free of fractions.

Example 2 Solve

Solution First, multiply each member by 5 to get

Now, divide each member by 3,

Example 3 Solve .

Solution First, simplify above the fraction bar to get

Next, multiply each member by 3 to obtain

Last, dividing each member by 5 yields

FURTHER SOLUTIONS OF EQUATIONS

Now we know all the techniques needed to solve most first-degree equations. There is no specific order in which the properties should be applied. Any one or more of the following steps listed on page 102 may be appropriate.

Steps to solve first-degree equations:

  1. Combine like terms in each member of an equation.
  2. Using the addition or subtraction property, write the equation with all terms containing the unknown in one member and all terms not containing the unknown in the other.
  3. Combine like terms in each member.
  4. Use the multiplication property to remove fractions.
  5. Use the division property to obtain a coefficient of 1 for the variable.

Example 1 Solve 5x - 7 = 2x - 4x + 14.

Solution First, we combine like terms, 2x - 4x, to yield

5x - 7 = -2x + 14

Next, we add +2x and +7 to each member and combine like terms to get

5x - 7 + 2x + 7 = -2x + 14 + 2x + 1

7x = 21

Finally, we divide each member by 7 to obtain

In the next example, we simplify above the fraction bar before applying the properties that we have been studying.

Example 2 Solve

Solution First, we combine like terms, 4x - 2x, to get

Then we add -3 to each member and simplify

Next, we multiply each member by 3 to obtain

Finally, we divide each member by 2 to get

SOLVING FORMULAS

Equations that involve variables for the measures of two or more physical quantities are called formulas. We can solve for any one of the variables in a formula if the values of the other variables are known. We substitute the known values in the formula and solve for the unknown variable by the methods we used in the preceding sections.

Example 1 In the formula d = rt, find t if d = 24 and r = 3.

Solution We can solve for t by substituting 24 for d and 3 for r. That is,

d = rt

(24) = (3)t

8 = t

It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others. We use the same methods demonstrated in the preceding sections.

Example 2 In the formula d = rt, solve for t in terms of r and d.

Solution We may solve for t in terms of r and d by dividing both members by r to yield

from which, by the symmetric law,

In the above example, we solved for t by applying the division property to generate an equivalent equation. Sometimes, it is necessary to apply more than one such property.

Example 3 In the equation ax + b = c, solve for x in terms of a, b and c.

Solution We can solve for x by first adding -b to each member to get

then dividing each member by a, we have

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Linear equations with one unknown

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     3*x-1-(1/2)=0 

Step by step solution :

Step  1  :

1 Simplify — 2

Equation at the end of step  1  :

1 (3x - 1) - — = 0 2

Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  2  as the denominator :

3x - 1 (3x - 1) • 2 3x - 1 = —————— = ———————————— 1 2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 2.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(3x-1) • 2 - (1) 6x - 3 ———————————————— = —————— 2 2

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   6x - 3  =   3 • (2x - 1) 

Equation at the end of step  3  :

3 • (2x - 1) ———————————— = 0 2

Step  4  :

When a fraction equals zero :

 4.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

3•(2x-1) ———————— • 2 = 0 • 2 2

Now, on the left hand side, the  2  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   3  •  (2x-1)  = 0

Equations which are never true :

 4.2      Solve :    3   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      Solve  :    2x-1 = 0 

 Add  1  to both sides of the equation : 
                      2x = 1
Divide both sides of the equation by 2:
                     x = 1/2 = 0.500

One solution was found :

                   x = 1/2 = 0.500
Sours: https://www.tiger-algebra.com/drill/3x-1=1/2/

Collatz conjecture

Open problem on 3x+1 and x/2 functions

Directed graphshowing the orbitsof small numbers under the Collatz map, skipping even numbers. The Collatz conjecture states that all paths eventually lead to 1.

The Collatz conjecture is a conjecture in mathematics that concerns sequences defined as follows: start with any positive integern. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half of the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

The conjecture is named after Lothar Collatz, who introduced the idea in 1937, two years after receiving his doctorate.[1] It is also known as the 3n + 1 problem, 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.[2][4] The sequence of numbers involved is sometimes 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),[5][6] or as wondrous numbers.[7]

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered US$500 for its solution.[9]Jeffrey Lagarias stated in 2010 that the Collatz conjecture "is an extraordinarily difficult problem, completely out of reach of present day mathematics."[10]

Statement of the problem[edit]

Numbers from 1 to 9999 and their corresponding total stopping time
Histogram of total stopping times for the numbers 1 to 108. Total stopping time is on the xaxis, frequency on the yaxis.
Histogram of total stopping times for the numbers 1 to 109. Total stopping time is on the xaxis, frequency on the yaxis.
Iteration time for inputs of 2 to 107.
Total Stopping Time: numbers up to 250, 1000, 4000, 20000, 100000, 500000
Total stopping time of numbers up to 250, 1000, 4000, 20000, 100000, 500000

Consider the following operation on an arbitrary positive integer:

  • If the number is even, divide it by two.
  • If the number is odd, triple it and add one.

In modular arithmetic notation, define the functionf as follows:

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}}\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}

Now form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

In notation:

a_{i}={\begin{cases}n&{\text{for }}i=0\\f(a_{i-1})&{\text{for }}i>0\end{cases}}

(that is: ai is the value of f applied to n recursively i times; ai = fi(n)).

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

The smallest i such that ai < a0 is called the stopping time of n. Similarly, the smallest k such that ak = 1 is called the total stopping time of n.[3] If one of the indexes i or k doesn't exist, we say that the stopping time or the total stopping time, respectively, is infinite.

The Collatz conjecture asserts that the total stopping time of every n is finite. It is also equivalent to saying that every n ≥ 2 has a finite stopping time.

Since 3n + 1 is even whenever n is odd, one may instead use the "shortcut" form of the Collatz function:

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}},\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}
This definition yields smaller values for the stopping time and total stopping time without changing the overall dynamics of the process.

Empirical data[edit]

For instance, starting with n = 12 and applying the function f without "shortcut", one gets the sequence 12, 6, 3, 10, 5, 16, 8, 4, 2, 1.

The number n = 19 takes longer to reach 1: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

The sequence for n = 27, listed and graphed below, takes 111 steps (41 steps through odd numbers, in bold), climbing as high as 9232 before descending to 1.

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (sequence A008884 in the OEIS)
Collatz5.svg

Numbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with:

1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, ... (sequence A006877 in the OEIS).

The starting values whose maximum trajectory point is greater than that of any smaller starting value are as follows:

1, 2, 3, 7, 15, 27, 255, 447, 639, 703, 1819, 4255, 4591, 9663, 20895, 26623, 31911, 60975, 77671, 113383, 138367, 159487, 270271, 665215, 704511, ... (sequence A006884 in the OEIS)

Number of steps for n to reach 1 are

0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, ... (sequence A006577 in the OEIS)

The starting value having the largest total stopping time while being

less than 10 is 9, which has 19 steps,
less than 100 is 97, which has 118 steps,
less than 1000 is 871, which has 178 steps,
less than 104 is 6171, which has 261 steps,
less than 105 is 77031, which has 350 steps,
less than 106 is 837799, which has 524 steps,
less than 107 is 8400511, which has 685 steps,
less than 108 is 63728127, which has 949 steps,
less than 109 is 670617279, which has 986 steps,
less than 1010 is 9780657630, which has 1132 steps,[11]
less than 1011 is 75128138247, which has 1228 steps,
less than 1012 is 989345275647, which has 1348 steps,[12]

These numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit. As an example, 9780657631 has 1132 steps, as does 9780657630.

The starting values having the smallest total stopping time with respect to their number of digits (in base 2) are the powers of two since 2n is halved n times to reach 1, and is never increased.

Visualizations[edit]

  • Directed graph showing the orbits of the first 1000 numbers.

  • The x axis represents starting number, the y axis represents the highest number reached during the chain to 1. This plot shows a restricted y axis: some x values produce intermediates as high as 2.7×107 (for x = 9663)

  • The tree of all the numbers having fewer than 20 steps.

Supporting arguments[edit]

Although the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.

Experimental evidence[edit]

As of 2020[update], the conjecture has been checked by computer for all starting values up to 268 ≈ 2.95×1020. All initial values tested so far eventually end in the repeating cycle (4; 2; 1) of period 3.[13]

This computer evidence is not sufficient to prove that the conjecture is true for all starting values. As in the case of some disproved conjectures, like the Pólya conjecture, counterexamples might be found when considering very large numbers.

However, such verifications may have other implications. For example, one can derive additional constraints on the period and structural form of a non-trivial cycle.[14][15][16]

A probabilistic heuristic[edit]

If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average 3/4 of the previous one.[17] (More precisely, the geometric mean of the ratios of outcomes is 3/4.) This yields a heuristic argument that every Hailstone sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has two division steps for every multiplication step for almost all 2-adic starting values.)

Stopping times[edit]

As proven by Riho Terras, almost every positive integer has a finite stopping time.[18] In other words, almost every Collatz sequence reaches a point that is strictly below its initial value. The proof is based on the distribution of parity vectors and uses the central limit theorem.

In 2019, Terence Tao considerably improved this result by showing, using logarithmic density, that almost all Collatz orbits are descending below any given function of the starting point, provided that this function diverges to infinity, no matter how slowly. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades."[19][20]

Lower bounds[edit]

In a computer-aided proof, Krasikov and Lagarias showed that the number of integers in the interval [1,x] that eventually reach 1 is at least equal to x0.84 for all sufficiently large x.[21]

Cycles[edit]

In this part, consider the shortcut form of the Collatz function

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}},\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}

A cycle is a sequence (a0, a1, ..., aq) of distinct positive integers where f(a0) = a1, f(a1) = a2, ..., and f(aq) = a0.

The only known cycle is (1,2) of period 2, called the trivial cycle.

Cycle length[edit]

The length of a non-trivial cycle is known to be at least 17087915.[15] In fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form

{\displaystyle p=301994a+17087915b+85137581c}

where a, b and c are non-negative integers, b ≥ 1 and ac = 0. This result is based on the continued fraction expansion of ln 3/ln 2.

A similar reasoning that accounts for the recent verification of the conjecture up to 268 leads to the improved lower bound 114208327604 (or 186265759595 without the "shortcut"). This lower bound is consistent with the above result, since 114208327604 = 17087915 × 361 + 85137581 × 1269.

k-cycles[edit]

A k-cycle is a cycle that can be partitioned into 2k contiguous subsequences: k increasing sequences of odd numbers alternating with k decreasing sequences of even numbers.[16] For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a 1-cycle.

Steiner (1977) proved that there is no 1-cycle other than the trivial (1; 2).[22] Simons (2004) used Steiner's method to prove that there is no 2-cycle.[23] Simons & de Weger (2005) extended this proof up to 68-cycles: there is no k-cycle up to k = 68.[16] For each k beyond 68, this method gives an upper bound for the smallest term of a k-cycle: for example, if there is a 77-cycle, then at least one element of the cycle is less than 38137×250.[16] Along with the verification of the conjecture up to 268, this implies the nonexistence of a non-trivial k-cycle up to k = 77.[13] As exhaustive computer searches continue, larger k values may be ruled out. To state the argument more intuitively: we need not look for cycles that have at most 77 circuits, where each circuit consists of consecutive ups followed by consecutive downs.

Other formulations of the conjecture[edit]

In reverse[edit]

The first 21 levels of the Collatz graphgenerated in bottom-up fashion. The graph includes all numbers with an orbit length of 21 or less.

There is another approach to prove the conjecture, which considers the bottom-up method of growing the so-called Collatz graph. The Collatz graph is a graph defined by the inverse relation

{\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1,2,3,5\\[4px]\left\{2n,{\frac {n-1}{3}}\right\}&{\text{if }}n\equiv 4\end{cases}}{\pmod {6}}.}

So, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads backwards to all positive integers. For any integer n, n ≡ 1 (mod 2)if and only if3n + 1 ≡ 4 (mod 6). Equivalently, n − 1/3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of this article).

When the relation 3n + 1 of the function f is replaced by the common substitute "shortcut" relation 3n + 1/2, the Collatz graph is defined by the inverse relation,

{\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1\\[4px]\left\{2n,{\frac {2n-1}{3}}\right\}&{\text{if }}n\equiv 2\end{cases}}{\pmod {3}}.}

For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1/2 ≡ 2 (mod 3). Equivalently, 2n − 1/3 ≡ 1 (mod 2) if and only if n ≡ 2 (mod 3). Conjecturally, this inverse relation forms a tree except for a 1–2 loop (the inverse of the 1–2 loop of the function f(n) revised as indicated above).

Alternatively, replace the 3n + 1 with n′/H(n′) where n′ = 3n + 1 and H(n′) is the highest power of 2 that divides n′ (with no remainder). The resulting function f maps from odd numbers to odd numbers. Now suppose that for some odd number n, applying this operation k times yields the number 1 (that is, fk(n) = 1). Then in binary, the number n can be written as the concatenation of stringswkwk−1w1 where each wh is a finite and contiguous extract from the representation of 1/3h.[24] The representation of n therefore holds the repetends of 1/3h, where each repetend is optionally rotated and then replicated up to a finite number of bits. It is only in binary that this occurs.[25] Conjecturally, every binary string s that ends with a '1' can be reached by a representation of this form (where we may add or delete leading '0's to s).

As an abstract machine that computes in base two[edit]

Repeated applications of the Collatz function can be represented as an abstract machine that handles strings of bits. The machine will perform the following three steps on any odd number until only one 1 remains:

  1. Append 1 to the (right) end of the number in binary (giving 2n + 1);
  2. Add this to the original number by binary addition (giving 2n + 1 + n = 3n + 1);
  3. Remove all trailing 0s (that is, repeatedly divide by 2 until the result is odd).

Example[edit]

The starting number 7 is written in base two as 111. The resulting Collatz sequence is:

111 1111 101110111 10001100011 110111011 1011011 1

As a parity sequence[edit]

For this section, consider the Collatz function in the slightly modified form

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1\end{cases}}{\pmod {2}}.}

This can be done because when n is odd, 3n + 1 is always even.

If P(…) is the parity of a number, that is P(2n) = 0 and P(2n + 1) = 1, then we can define the Collatz parity sequence (or parity vector) for a number n as pi = P(ai), where a0 = n, and ai+1 = f(ai).

Which operation is performed, 3n + 1/2 or n/2, depends on the parity. The parity sequence is the same as the sequence of operations.

Using this form for f(n), it can be shown that the parity sequences for two numbers m and n will agree in the first k terms if and only if m and n are equivalent modulo 2k. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Hailstone cycles, then their corresponding parity cycles must be different.[3][18]

Applying the f function k times to the number n = 2ka + b will give the result 3ca + d, where d is the result of applying the f function k times to b, and c is how many increases were encountered during that sequence. For example, for 25a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 33a + 2; for 22a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so the result is 3a + 1. When b is 2k − 1 then there will be k rises and the result will be 2 × 3ka − 1. The factor of 3 multiplying a is independent of the value of a; it depends only on the behavior of b. This allows one to predict that certain forms of numbers will always lead to a smaller number after a certain number of iterations: for example, 4a + 1 becomes 3a + 1 after two applications of f and 16a + 3 becomes 9a + 2 after 4 applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a.

As a tag system[edit]

For the Collatz function in the form

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1.\end{cases}}{\pmod {2}}}

Hailstone sequences can be computed by the extremely simple 2-tag system with production rules

abc, ba, caaa.

In this system, the positive integer n is represented by a string of n copies of a, and iteration of the tag operation halts on any word of length less than 2. (Adapted from De Mol.)

The Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a as the initial word, eventually halts (see Tag system#Example: Computation of Collatz sequences for a worked example).

Extensions to larger domains[edit]

Iterating on all integers[edit]

An extension to the Collatz conjecture is to include all integers, not just positive integers. Leaving aside the cycle 0 → 0 which cannot be entered from outside, there are a total of four known cycles, which all nonzero integers seem to eventually fall into under iteration of f. These cycles are listed here, starting with the well-known cycle for positive n:

Odd values are listed in large bold. Each cycle is listed with its member of least absolute value (which is always odd) first.

CycleOdd-value cycle lengthFull cycle length
1 → 4 → 2 → 1...13
−1 → −2 → −1...12
−5 → −14 → −7 → −20 → −10 → −5...25
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17...718


The generalized Collatz conjecture is the assertion that every integer, under iteration by f, eventually falls into one of the four cycles above or the cycle 0 → 0. The 0 → 0 cycle is often regarded as "trivial" by the argument, as it is only included for the sake of completeness.

Iterating on rationals with odd denominators[edit]

The Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms. The number is taken to be 'odd' or 'even' according to whether its numerator is odd or even. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added. A closely related fact is that the Collatz map extends to the ring of 2-adic integers, which contains the ring of rationals with odd denominators as a subring.

When using the "shortcut" definition of the Collatz map, it is known that any periodic parity sequence is generated by exactly one rational.[26] Conversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture[3]).

If a parity cycle has length n and includes odd numbers exactly m times at indices k0 < ⋯ < km−1, then the unique rational which generates immediately and periodically this parity cycle is

{\frac {3^{m-1}2^{k_{0}}+\cdots +3^{0}2^{k_{m-1}}}{2^{n}-3^{m}}}.

 

 

 

 

(1)

For example, the parity cycle (1 0 1 1 0 0 1) has length 7 and four odd terms at indices 0, 2, 3, and 6. It is repeatedly generated by the fraction

{\displaystyle {\frac {3^{3}2^{0}+3^{2}2^{2}+3^{1}2^{3}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {151}{47}}}

as the latter leads to the rational cycle

{\displaystyle {\frac {151}{47}}\rightarrow {\frac {250}{47}}\rightarrow {\frac {125}{47}}\rightarrow {\frac {211}{47}}\rightarrow {\frac {340}{47}}\rightarrow {\frac {170}{47}}\rightarrow {\frac {85}{47}}\rightarrow {\frac {151}{47}}.}

Any cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions. For instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction

{\displaystyle {\frac {3^{3}2^{1}+3^{2}2^{2}+3^{1}2^{5}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {250}{47}}.}

For a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles. As an illustration of this, the parity cycle (1 1 0 0 1 1 0 0) and its sub-cycle (1 1 0 0) are associated to the same fraction 5/7 when reduced to lowest terms.

In this context, assuming the validity of the Collatz conjecture implies that (1 0) and (0 1) are the only parity cycles generated by positive whole numbers (1 and 2, respectively).

If the odd denominator d of a rational is not a multiple of 3, then all the iterates have the same denominator and the sequence of numerators can be obtained by applying the "3n + d " generalization[27] of the Collatz function

{\displaystyle T_{d}(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}},\\[4px]{\frac {3x+d}{2}}&{\text{if }}x\equiv 1{\pmod {2}}.\end{cases}}}

2-adic extension[edit]

The function

{\displaystyle T(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}}\\[4px]{\frac {3x+1}{2}}&{\text{if }}x\equiv 1{\pmod {2}}\end{cases}}}

is well-defined on the ring ℤ2 of 2-adic integers, where it is continuous and measure-preserving with respect to the 2-adic measure. Moreover, its dynamics is known to be ergodic.[3]

Define the parity vector function Q acting on ℤ2 as

{\displaystyle Q(x)=\sum _{k=0}^{\infty }\left(T^{k}(x)\mod 2\right)2^{k}}.

The function Q is a 2-adic isometry.[28] Consequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that almost all trajectories are acyclic in \mathbb {Z} _{2}.

An equivalent formulation of the Collatz conjecture is that

{\displaystyle Q\left(\mathbb {Z} ^{+}\right)\subset {\tfrac {1}{3}}\mathbb {Z} .}

Iterating on real or complex numbers[edit]

Cobweb plotof the orbit 10-5-8-4-2-1-2-1-2-1-etc. in the real extension of the Collatz map (optimized by replacing "3n + 1" with "(3n + 1)/2")

The Collatz map (with shortcut) can be viewed as the restriction to the integers of the smoothmap

{\displaystyle f(z)={\frac {1}{2}}z\cos ^{2}\left({\frac {\pi }{2}}z\right)+{\frac {3z+1}{2}}\sin ^{2}\left({\frac {\pi }{2}}z\right).}
Collatz map fractalin a neighbourhood of the real line

The iterations of this map on the real line lead to a dynamical system, further investigated by Chamberland.[29] He showed that the conjecture does not hold for positive real numbers since there are infinitely many fixed points, as well as orbits escaping monotonically to infinity. The function f has two attracting cycles of period 2, (1; 2) and (1.1925...; 2.1386...). Moreover, the set of unbounded orbits is conjectured to be of measure 0.

Letherman, Schleicher, and Wood extended the study to the complex plane, where most of the points have orbits that diverge to infinity (colored region on the illustration).[30] The boundary between the colored region and the black components, namely the Julia set of f, is a fractal pattern, sometimes called the "Collatz fractal".

Optimizations[edit]

Time–space tradeoff[edit]

The section As a parity sequence above gives a way to speed up simulation of the sequence. To jump ahead k steps on each iteration (using the f function from that section), break up the current number into two parts, b (the k least significant bits, interpreted as an integer), and a (the rest of the bits as an integer). The result of jumping ahead k steps can be found as:

fk(2ka + b) = 3c(b)a + d(b).

The c (or better 3c) and d arrays are precalculated for all possible k-bit numbers b, where d(b) is the result of applying the f function k times to b, and c(b) is the number of odd numbers encountered on the way.[31] For example, if k = 5, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using:

c(0...31) = { 0, 3, 2, 2, 2, 2, 2, 4, 1, 4, 1, 3, 2, 2, 3, 4, 1, 2, 3, 3, 1, 1, 3, 3, 2, 3, 2, 4, 3, 3, 4, 5 }
d(0...31) = { 0, 2, 1, 1, 2, 2, 2, 20, 1, 26, 1, 10, 4, 4, 13, 40, 2, 5, 17, 17, 2, 2, 20, 20, 8, 22, 8, 71, 26, 26, 80, 242 }.

This requires 2kprecomputation and storage to speed up the resulting calculation by a factor of k, a space–time tradeoff.

Modular restrictions[edit]

For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration, used by Tomás Oliveira e Silva in his computational confirmations of the Collatz conjecture up to large values of n. If, for some given b and k, the inequality

fk(2ka + b) = 3c(b)a + d(b) < 2ka + b

holds for all a, then the first counterexample, if it exists, cannot be b modulo 2k.[14] For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f2(4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting value is as good as checking an entire congruence class. As k increases, the search only needs to check those residues b that are not eliminated by lower values of k. Only an exponentially small fraction of the residues survive.[32] For example, the only surviving residues mod 32 are 7, 15, 27, and 31.

Syracuse function[edit]

If k is an odd integer, then 3k + 1 is even, so 3k + 1 = 2ak′ with k′ odd and a ≥ 1. The Syracuse function is the function f from the set I of odd integers into itself, for which f(k) = k′ (sequence A075677 in the OEIS).

Some properties of the Syracuse function are:

  • For all kI, f(4k + 1) = f(k). (Because 3(4k + 1) + 1 = 12k + 4 = 4(3k + 1).)
  • In more generality: For all p ≥ 1 and odd h, fp − 1(2ph − 1) = 2 × 3p − 1h − 1. (Here fp − 1 is function iteration notation.)
  • For all odd h, f(2h − 1) ≤ 3h − 1/2

The Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer n ≥ 1 such that fn(k) = 1.

Undecidable generalizations[edit]

In 1972, John Horton Conway proved that a natural generalization of the Collatz problem is algorithmically undecidable.[33]

Specifically, he considered functions of the form

{\displaystyle {g(n)=a_{i}n+b_{i}}\;\;,{\text{where}}\;{n\equiv i{\pmod {P}}}}

and a0, b0, ..., aP − 1, bP − 1 are rational numbers which are so chosen that g(n) is always an integer.

The standard Collatz function is given by P = 2, a0 = 1/2, b0 = 0, a1 = 3, b1 = 1. Conway proved that the problem:

Given g and n, does the sequence of iterates gk(n) reach 1?

is undecidable, by representing the halting problem in this way.

Closer to the Collatz problem is the following universally quantified problem:

Given g does the sequence of iterates gk(n) reach 1, for all n > 0?

Modifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one). Kurtz and Simon[34] proved that the above problem is, in fact, undecidable and even higher in the arithmetical hierarchy, specifically Π0
2-complete. This hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.[35]

In popular culture[edit]

In the movie Incendies, a graduate student in pure mathematics explains the Collatz conjecture to a group of undergraduates. She puts her studies on hold for a time to address some unresolved questions about her family's past. Late in the movie, the Collatz conjecture turns out to have foreshadowed a disturbing and difficult discovery that she makes about her family.[36][37]

See also[edit]

Further reading[edit]

  • The Ultimate Challenge: The 3x + 1 problem[10] published in 2010 by the American Mathematical Society, edited by J.C. Lagarias, is a compendium of information on the Collatz conjecture, methods of approaching it, and generalizations. It includes two survey papers by the editor, and five by other authors, concerning the history of the problem, generalizations, statistical approaches, and results from the theory of computation. It also includes reprints of early papers on the subject (including a paper by Lothar Collatz).

References[edit]

  1. ^O'Connor, J.J.; Robertson, E.F. (2006). "Lothar Collatz". St Andrews University School of Mathematics and Statistics, Scotland.
  2. ^Maddux, Cleborne D.; Johnson, D. Lamont (1997). Logo: A Retrospective. New York: Haworth Press. p. 160. ISBN .
  3. ^ abcdefgLagarias, Jeffrey C. (1985). "The 3x + 1 problem and its generalizations". The American Mathematical Monthly. 92 (1): 3–23. doi:10.1080/00029890.1985.11971528. JSTOR 2322189.
  4. ^According to Lagarias (1985),[3] p. 4, the name "Syracuse problem" was proposed by Hasse in the 1950s, during a visit to Syracuse University.
  5. ^Pickover, Clifford A. (2001). Wonders of Numbers. Oxford: Oxford University Press. pp. 116–118. ISBN .
  6. ^"Hailstone Number". MathWorld. Wolfram Research.
  7. ^Hofstadter, Douglas R. (1979). Gödel, Escher, Bach. New York: Basic Books. pp. 400–2. ISBN .
  8. ^Guy, Richard K. (2004). ""E17: Permutation Sequences"". Unsolved problems in number theory (3rd ed.). Springer-Verlag. pp. 336–7. ISBN . Zbl 1058.11001.
  9. ^Guy, R. K. (1983). "Don't try to solve these problems". Amer. Math. Monthly. 90 (1): 35–41. doi:10.2307/2975688. JSTOR 2975688. By this Erdos means that there aren't powerful tools for manipulating such objects.
  10. ^ abLagarias, Jeffrey C., ed. (2010). The Ultimate Challenge: the 3x + 1 problem. American Mathematical Society. ISBN . Zbl 1253.11003.
  11. ^Leavens, Gary T.; Vermeulen, Mike (December 1992). "3x + 1 Search Programs". Computers & Mathematics with Applications. 24 (11): 79–99. doi:10.1016/0898-1221(92)90034-F.
  12. ^Roosendaal, Eric. "3x+1 Delay Records". Retrieved 14 March 2020. (Note: "Delay records" are total stopping time records.)
  13. ^ abBarina, David (2020). "Convergence verification of the Collatz problem". The Journal of Supercomputing. 77 (3): 2681–2688. doi:10.1007/s11227-020-03368-x. S2CID 220294340.
  14. ^ abGarner, Lynn E. (1981). "On the Collatz 3n + 1 Algorithm". Proceedings of the American Mathematical Society. 82 (1): 19–22. doi:10.1090/S0002-9939-1981-0603593-2. JSTOR 2044308.
  15. ^ abEliahou, Shalom (1993). "The 3x + 1 problem: new lower bounds on nontrivial cycle lengths". Discrete Mathematics. 118 (1): 45–56. doi:10.1016/0012-365X(93)90052-U.
  16. ^ abcdSimons, J.; de Weger, B. (2005). "Theoretical and computational bounds for m-cycles of the 3n + 1 problem"(PDF). Acta Arithmetica. 117 (1): 51–70. Bibcode:2005AcAri.117...51S. doi:10.4064/aa117-1-3.
  17. ^Lagarias (1985),[3] section "A heuristic argument".
  18. ^ abTerras, Riho (1976). "A stopping time problem on the positive integers"(PDF). Acta Arithmetica. 30 (3): 241–252. doi:10.4064/aa-30-3-241-252. MR 0568274.
  19. ^Tao, Terence (10 September 2019). "Almost all Collatz orbits attain almost bounded values". What's new. Retrieved 11 September 2019.
  20. ^Hartnett, Kevin. "Mathematician Proves Huge Result on 'Dangerous' Problem". Quanta Magazine. Retrieved 26 December 2019.
  21. ^Krasikov, Ilia; Lagarias, Jeffrey C. (2003). "Bounds for the 3x + 1 problem using difference inequalities". Acta Arithmetica. 109 (3): 237–258. arXiv:math/0205002. Bibcode:2003AcAri.109..237K. doi:10.4064/aa109-3-4. MR 1980260. S2CID 18467460.
  22. ^Steiner, R. P. (1977). "A theorem on the syracuse problem". Proceedings of the 7th Manitoba Conference on Numerical Mathematics. pp. 553–9. MR 0535032.
  23. ^Simons, John L. (2005). "On the nonexistence of 2-cycles for the 3x + 1 problem". Math. Comp. 74: 1565–72. Bibcode:2005MaCom..74.1565S. doi:10.1090/s0025-5718-04-01728-4. MR 2137019.
  24. ^Colussi, Livio (9 September 2011). "The convergence classes of Collatz function". Theoretical Computer Science. 412 (39): 5409–5419. doi:10.1016/j.tcs.2011.05.056.
  25. ^Hew, Patrick Chisan (7 March 2016). "Working in binary protects the repetends of 1/3h: Comment on Colussi's 'The convergence classes of Collatz function'". Theoretical Computer Science. 618: 135–141. doi:10.1016/j.tcs.2015.12.033.
  26. ^Lagarias, Jeffrey (1990). "The set of rational cycles for the 3x+1 problem". Acta Arithmetica. 56 (1): 33–53. doi:10.4064/aa-56-1-33-53. ISSN 0065-1036.
  27. ^Belaga, Edward G.; Mignotte, Maurice (1998). "Embedding the 3x+1 Conjecture in a 3x+d Context". Experimental Mathematics. 7 (2): 145–151. doi:10.1080/10586458.1998.10504364.
  28. ^Lagarias, Jeffrey C.; Bernstein, Daniel J. (1996). "The 3x + 1 Conjugacy Map". Canadian Journal of Mathematics. 48 (6): 1154–1169. doi:10.4153/CJM-1996-060-x. ISSN 0008-414X.
  29. ^Chamberland, Marc (1996). "A continuous extension of the 3x + 1 problem to the real line". Dynam. Contin. Discrete Impuls Systems. 2 (4): 495–509.
  30. ^Letherman, Simon; Schleicher, Dierk; Wood, Reg (1999). "The (3n + 1)-Problem and Holomorphic Dynamics". Experimental Mathematics. 8 (3): 241–252. doi:10.1080/10586458.1999.10504402.
  31. ^Scollo, Giuseppe (2007). "Looking for Class Records in the 3x + 1 Problem by means of the COMETA Grid Infrastructure"(PDF). Grid Open Days at the University of Palermo.
  32. ^Lagarias (1985),[3] Theorem D.
  33. ^Conway, John H. (1972). "Unpredictable Iterations". Proc. 1972 Number Theory Conf., Univ. Colorado, Boulder. pp. 49–52.
  34. ^Kurtz, Stuart A.; Simon, Janos (2007). "The Undecidability of the Generalized Collatz Problem". In Cai, J.-Y.; Cooper, S. B.; Zhu, H. (eds.). Proceedings of the 4th International Conference on Theory and Applications of Models of Computation, TAMC 2007, held in Shanghai, China in May 2007. pp. 542–553. doi:10.1007/978-3-540-72504-6_49. ISBN . As PDF
  35. ^Ben-Amram, Amir M. (2015). "Mortality of iterated piecewise affine functions over the integers: Decidability and complexity". Computability. 1 (1): 19–56. doi:10.3233/COM-150032.
  36. ^Emmer, Michele (2012). Imagine Math: Between Culture and Mathematics. Springer Publishing. pp. 260–264. ISBN .
  37. ^Mazmanian, Adam (19 May 2011). "MOVIE REVIEW: 'Incendies'". The Washington Times. Retrieved 7 December 2019.

External links[edit]

Sours: https://en.wikipedia.org/wiki/Collatz_conjecture

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