Do complex numbers really exist?
Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are?
This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people actually ask this question.
The answer is always 5
Start by thinking of a number.
Take away your original number.
Try this with different numbers and wow your audience with this simple math magic trick.
I know how much money you got!
Without giving you any information, ask the students to count the value of a random set of coins and write the amount on a piece of paper. Then ask them to follow the next steps-
Double the amount.
Add the first odd prime number to the new total.
Multiply the result by 1/4 of 20.
Subtract the lowest common multiple of 2 and 3.
For the final answer - Take off the last digit and you will be able to guess how much the coins are worth!
Guess their age and shoe size!
Ask your child to follow the given directions but not present any of the calculations to you -
Ask them to write down their age
Multiply it by 1/5 of 100.
Add on today's date (e.g. 2 if it's the 2nd of the month).
Multiply by 20% of 25.
Now add on your shoe size (if it's a half size round to a whole number).
Finally, subtract 5 times today's date.
Ask them to reveal the final answer -
The hundreds are the age and the remaining digits are the shoe size. If for instance, somebody shows you 1206, there are 12 hundred - the age, and the remaining digits 06 (or 6) denote their shoe size.
The correct answer is always 37!
To perform this trick, ask any of your friends to choose a three-digit number with the same digits.
For example, 222
Add the digits together. So, 2+ 2 + 2 = 6
Divide the original number with this sum. So, 222 / 6 = 37.
Learn how to add 2-digit numbers at the drop of a hat!
Get a firm understanding of the basic principles of tens & unit places, and you will be able to add two-digit numbers at a lightning-fast speed.
For example, take two, two-digit numbers,
Then, split the 2nd number into tens & units, making it 79 = 70 + 9.
After that ,move up up the tens addition, which is 57 + 70 = 137.
To get the final answer, you must add the left units place digit, which is 137 + 9 = 146.
Predict any number!
To start with this math magic trick, choose any number like 22
Subtract one from it, 22 &ndash 1 = 21
Multiply it with three. (21 x 3 = 63)
Add 12 to this number. (63 + 12 = 75)
Divide it with three. (75 / 3 = 25)
Now add 5 to the above number. (25 + 5 &ndash 30)
Finally, ask your peers to subtract the original number from the above sum.
The answer will always be 8. (30 &ndash 22 = 8)
Turn six digits into three
To do this trick, take any three-digit number and write it twice to make a six-digit number.
Then divide the number by 7 which is 371371/7 = 53,053
Then, Divide it by 11 which is 53,053 / 11 = 4823
Then, Divide it by 13 which is 4823/13 = 371
The answer is the three-digit number. Therefore, 371
Choose any number, your end product will only be 2!
To perform this trick ask your friend to think of a number, let&rsquos say 8
The 11 Rule
By using this trick you can learn to quickly multiply any number with 11 mentally.
Choose any number and separate the two digits in your mind.
Add the two digits together.
Place the number from Step 2 between the two digits.
If the number from Step 2 is greater than 9, but keep the one's digit in the space and carry the ten's digit.
Examples: 72 x 11 = 792.
57 x 11 = 5 _ 7, but 5 + 7 = 12, so put 2 in the space and add the 1 to the 5 to get 627.
Multiplication at your fingertips!
Did you know that you can effectively multiply any number of digits using your hands?
A simple way to do the "9" multiplication table is to place both hands in front of you with fingers and thumbs extended.
To multiply 9 by a number, fold down that number finger, counting from the left.
For example - to multiply 9 by 5, fold down the fifth finger from the left. Count fingers on either side of the "fold" to get the answer. In this case, the answer is 45.
Consider the following operation on an arbitrary positive integer:
In modular arithmetic notation, define the function f as follows:
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.
(that is: ai is the value of f applied to n recursively i times ai = f i (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 .  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
For instance, starting with n = 12 , 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)
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 10 4 is 6171, which has 261 steps, less than 10 5 is 77 031 , which has 350 steps, less than 10 6 is 837 799 , which has 524 steps, less than 10 7 is 8 400 511 , which has 685 steps, less than 10 8 is 63 728 127 , which has 949 steps, less than 10 9 is 670 617 279 , which has 986 steps, less than 10 10 is 9 780 657 630 , which has 1132 steps,  less than 10 11 is 75 128 138 247 , which has 1228 steps, less than 10 12 is 989 345 275 647 , which has 1348 steps, less than 10 13 is 7 887 663 552 367 , which has 1563 steps, less than 10 14 is 80 867 137 596 217 , which has 1662 steps, less than 10 15 is 942 488 749 153 153 , which has 1862 steps, less than 10 16 is 7 579 309 213 675 935 , which has 1958 steps, less than 10 17 is 93 571 393 692 802 302 , which has 2091 steps and less than 10 18 is 931 386 509 544 713 451 , which has 2283 steps. 
These numbers are the lowest ones with the indicated step count, but not necessarily the only ones below the given limit. As an example, 9 780 657 631 has 1132 steps, as does 9 780 657 630 .
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 2 n is halved n times to reach 1, and is never increased.
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 × 10 7 (for x = 9663 )
The tree of all the numbers having fewer than 20 steps (click to enlarge).
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 2 68 ≈ 2.95×10 20 . All initial values tested so far eventually end in the repeating cycle (421) of period 3. 
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.   
A probabilistic heuristic Edit
Stopping times Edit
As proved by Terras, almost every positive integer n has a finite stopping time.  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."  
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 x 0.84 for all sufficiently large x . 
In this part, consider the shortcut form of the Collatz function
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 17 087 915 .  In fact, Eliahou (1993) proved that the period p of any non-trivial cycle is of the form
p = 301994 a + 17087915 b + 85137581 c
A similar reasoning that accounts for the recent verification of the conjecture up to 2 68 leads to the improved lower bound 114 208 327 604 (or 186 265 759 595 without the "shortcut"). This lower bound is consistent with the above result, since 114 208 327 604 = 17 087 915 × 361 + 85 137 581 × 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.  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 (12) .  Simons (2004) used Steiner's method to prove that there is no 2-cycle.  Simons & de Weger (2005) extended this proof up to 68-cycles: there is no k -cycle up to k = 68 .  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 × 2 50 .  Along with the verification of the conjecture up to 2 68 , this implies the nonexistence of a non-trivial k -cycle up to k = 77 .  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.
In reverse Edit
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
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:
- Append 1 to the (right) end of the number in binary (giving 2n + 1 )
- Add this to the original number by binary addition (giving 2n + 1 + n = 3n + 1 )
- Remove all trailing "0"s (i.e. repeatedly divide by two until the result is odd).
The starting number 7 is written in base two as 111. The resulting Collatz sequence is:
As a parity sequence Edit
For this section, consider the Collatz function in the slightly modified form
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) .
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 2 k . 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.  
Applying the f function k times to the number n = 2 k a + b will give the result 3 c a + 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 (e.g. for 2 5 a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 3 3 a + 2 for 2 2 a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so the result is 3a + 1 ). When b is 2 k − 1 then there will be k rises and the result will be 2 × 3 k a − 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, e.g. 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
Hailstone sequences can be computed by the extremely simple 2-tag system with production rules
a → bc , b → a , c → aaa .
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).
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 4 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.
|Cycle||Odd-value cycle length||Full cycle length|
|1 → 4 → 2 → 1 .||1||3|
|−1 → −2 → −1 .||1||2|
|−5 → −14 → −7 → −20 → −10 → −5 .||2||5|
|−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 .||7||18|
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.  Conversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture  ).
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
Common Number Sets
There are sets of numbers that are used so often they have special names and symbols:
The whole numbers from 1 upwards. (Or from 0 upwards in some fields of mathematics). Read More ->
The whole numbers, <1,2,3. >negative whole numbers <. -3,-2,-1>and zero <0>. So the set is
(Z is from the German "Zahlen" meaning numbers, because I is used for the set of imaginary numbers). Read More ->
The numbers you can make by dividing one integer by another (but not dividing by zero). In other words fractions. Read More ->
Q is for "quotient" (because R is used for the set of real numbers).
Examples: 3/2 (=1.5), 8/4 (=2), 136/100 (=1.36), -1/1000 (=-0.001)
(Q is from the Italian "Quoziente" meaning Quotient, the result of dividing one number by another.)
Any real number that is not a Rational Number. Read More ->
Any number that is a solution to a polynomial equation with rational coefficients.
Includes all Rational Numbers, and some Irrational Numbers. Read More ->
Any number that is not an Algebraic Number
Examples of transcendental numbers include &pi and e. Read More ->
All Rational and Irrational numbers. They can also be positive, negative or zero.
Includes the Algebraic Numbers and Transcendental Numbers.
A simple way to think about the Real Numbers is: any point anywhere on the number line (not just the whole numbers).
They are called "Real" numbers because they are not Imaginary Numbers. Read More ->
Numbers that when squared give a negative result.
If you square a real number you always get a positive, or zero, result. For example 2×2=4, and (-2)×(-2)=4 also, so "imaginary" numbers can seem impossible, but they are still useful!
Examples: &radic(-9) (=3i), 6i, -5.2i
The "unit" imaginary numbers is &radic(-1) (the square root of minus one), and its symbol is i, or sometimes j.
A combination of a real and an imaginary number in the form a + bi, where a and b are real, and i is imaginary.
The values a and b can be zero, so the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers.
Examples: 1 + i, 2 - 6i, -5.2i, 4
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Notable Properties of Specific Numbers
These are some numbers with notable properties. (Most of the less notable properties are listed here.) Other people have compiled similar lists, but this is my list — it includes the numbers that I think are important (-:
A few rules I used in this list:
Everything can be understood by a typical undergraduate college student.
If multiple numbers have a shared property, that property is described under one "representative" number with that property. I try to choose the smallest representative that is not also cited for another property.
When a given number has more than one type of property, the properties are listed in this order:
1. Purely mathematical properties unrelated to the use of base 10 (example: 137 is prime.)
2. Base-10-specific mathematical properties (example: 137 is prime remove the "1": 37 is also prime remove the "3": 7 is also prime)
3. Things related to the physical world but outside human culture (example: 137 is close to the reciprocal of the fine-structure constant, once thought to be exact but later found to be closer to 137.036. )
4. All other properties (example: 137 has often been given a somewhat mystical significance due to its proximity to the fine-structure constant, most famously by Eddington)
Due to blatant personal bias, I only give one entry each to complex, imaginary, negative numbers and zero, devoting all the rest (27 pages) to positive real numbers. I also have a bit of an integer bias but that hasn't had such a severe effect. A little more about complex numbers, quaternions and so on, is here.
This page is meant to counteract the forces of Munafo's Laws of Mathematics. If you see room for improvement, let me know!
One of the square roots of i .
When I was about 12 years old, my step-brother gave me a question to pass the time: If i is the square root of -1, what is the square root of i ? . I had already seen a drawing of the complex plane, so I used it to look for useful patterns and noticed pretty quickly that the powers of i go in a circle. I estimated the square root of i to be about 0.7 + 0.7 i .
I can't remember why I didn't get the exact answer: either I didn't know trigonometry or the Pythagorean theorem, or how to solve multivariable equations, or perhaps was just tired of doing maths (I had clearly hit on Euler's formula and there's a good chance that contemplating the powers of 1+ i would have led me all the way through base- i logarithms and De Moivre's formula to the complex exponential function).
But you don't need that to find the square root of i . All you need to do is treat i as some kind of unknown value with the special property that any i 2 can be changed into a -1. You also need the idea of solving equations with coefficients and variables, and the square root of i is something of the form "a+b i ". Then you can find the square root of i by solving the equation:
Expand the (a+b i ) 2 in the normal way to get a 2 + 2ab i + b 2 i 2 , and then change the i 2 to -1:
Then just put the real parts together:
Since the real coordinate of the left side has to be equal to the real coordinate of the right, and likewise for the imaginary coordinates, we have two simultaneous equations in two variables:
From the first equation a 2 -b 2 = 0, we get a=b substituting this into the other equation we get 2a 2 = 1, and a=۫/√ 2 and this is also the value of b. Thus, the original desired square root of i is a+b i = (1+ i )/√ 2 (or the negative of this).
(This is the only complex number with its own entry in this collection, mainly because it's the only one I've had much interest in see the "blatant personal bias" note above :-).
The unit of imaginary numbers, and one of the square roots of -1.
(This is the only imaginary number with its own entry in this collection, mainly because it stands out way above the rest in notability. In addition, non-real numbers don't seem to interest me much. )
-1 is the "first" negative number, unless you define "first" to be "lowest".
In "two's complement" representation used in computers to store integers (within a fixed range), numbers are stored in base 2 (binary) with separate base-2 digits in different "bits" of a register. Negative numbers have a 1 in the highest position of the register. The value of -1 is represented by 1's in all positions, which is the same as what you'd get if you wrote a program to compute
and let it go long enough to overflow.
As it turns out, that series sum can be treated as an example of the general series sum
As discussed in the entry for 1/2, the sum is equal to 1/(1- x ), but that is valid only when | x | x =2 and use the formula anyway, we get 1/(1- x ) = 1/(1-2) = -1, which is the same as the two's complement interpretation.
(I do not have many entries for negative numbers, as they do not interest so much. Perhaps I still relate to numbers in terms of counting things like "the 27 sheep on that hill" or "the 40320 permutations of the Loughborough tower bells".)
The (in)famous sum of the positive integers:
1 + 2 + 3 + 4 + 5 + 6 + 7 + . = 1 /12
In the 19 th century, new techniques (Cesaro, Abel) were developed to tame some of the infinite series sums that do not converge normally. Examples are shown in the entries for 1/4 and 1/2. But these techniques alone are not enough to handle the infinite series sum:
C = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + .
This sum diverges monotonically (increases towards infinity, without ever taking a step in the negative direction) and Cesaro/Abel will not work.
Euler had to deal with it when performing analytic continuation on what is now called the Riemann zeta function:
Zeta( s ) = 1 - s + 2 - s + 3 - s + 4 - s + .
Euler had s = -1, which gives Zeta( s ) = 1 + 2 + 3 + 4 + . Euler's approach was to express it as a linear combination of itself with an existing Cesaro- or Abel-summable series, namely the 1-2+3-4+. =1/4 series, but by Euler's considerably easier diffentiation method:
C = 1 + 2 + 3 + 4 + 5 + 6 + 7 + .
= 1 + (4-2) + 3 + (8-4) + 5 + (12-6) + 7 + .
= 1 - 2 + 3 - 4 + 5 - 6 + 7 + . + 4 (1 + 2 + 3 + . )
Cesaro/Abel and Euler's method both give a sum 1/(1+1) 2 = 1 /4 for the first part so we have
The value of the Riemann Zeta function with argument of -1 is -1/12. As described by John Baez 100 :
The numbers 12 and 24 play a central role in mathematics thanks to a series of "coincidences" that is just beginning to be understood. One of the first hints of this fact was Euler's bizarre "proof" that
which he obtained before Abel declared that "divergent series are the invention of the devil". Euler's formula can now be understood rigorously in terms of the Riemann zeta function, and in physics it explains why bosonic strings work best in 26=24+2 dimensions.
Baez, at the end of his "24" lecture, indicates that the significance of 24 is connected to the fact that there are two ways to construct a lattice on the plane with rotational symmetry: one with 4-fold rotational symmetry and another with 6-fold rotational symmetry — and 4×6=24. A connection between zeta(-1)=-1/12 and symmetry of the plane makes more sense in light of how the Zeta function is computed for general complex arguments. Also, the least common multiple of 4 and 6 is 12.
See also the zeta values 1.202056. and 1.644934. .
Srinivasa Ramanujan also explained 1 + 2 + 3 + 4 + . = -1 /12, but in a more general way than Euler. He used a new analytic continuation of the Riemann zeta function.
In Ramanujan's 1913 letter to G.H. Hardy, the as-yet-undiscovered Indian mathematicical genius listed many of his discoveries and derivations. In section XI he stated:
I have got theorems on divergent series, theorems to calculate the convergent values corresponding to the divergent series, viz.
1 3 + 2 3 + 3 3 + 4 3 + . = 1 /120,
Theorems to calculate such values for any given series (say: 1 - 1 2 + 2 2 - 3 2 + 4 2 - 5 2 + . ), and the meaning of such values.
In modern notation we append (ℜ) to the end of such a series sum, to signify Ramanujan summation:
The Ramanujan sum defines a function f ( x ) whose values for integer x are the terms in the series being summed. Then
1 + 2 + 3 + 4 + . n (ℜ)
= Sigma k =1 n f ( k )
= Integral x =0 n f ( x ) + Sigma k =1 ∞ Bk / k ! ( f ( k -1) ( n )- f ( k -1) (0)) + R
where " f ( k -1) " is the ( k -1) th derivative of f (). Hardy and Ramanjuan considered just the parts of this that do not depend on n :
For a converging series, f ( x ) would approach a limit as x approaches infinity, and this would give a value that is equal to the sum of the infinite series. In our case the f ( x ) diverges, and the series sum is infinite, but this Hardy-Ramanujan sum is not. f (0) is 0, and the 1 st derivative is constant f '( x ) = 1, and all higher derivatives are zero, so it reduces to just
B 2 is the second Bernoulli number which is 1/6, so we get -1/12.
The word "zero" is the only number name in English that can be traced back to Arabic (صِفر ʂifr "nothing", "cipher" which became zefiro in Italian, later contracted by removing the fi ). The word came with the symbol, at around the same time the western Arabic numerals came to Europe. 44 , 105
The practice of using a symbol to hold the place of another digit when there is no value in that place (such as the 0 in 107 indicating there are no 10's) goes back to 5 th -century India, where it was called shunya or Śūnyatā 107 .
(This is the only zero number with its own entry in this collection, mainly because a field can have only one additive identity.)
This is the Planck time in seconds it is related to quantum mechanics. According to the Wikipedia article Planck time, "Within the framework of the laws of physics as we understand them today, for times less than one Planck time apart, we can neither measure nor detect any change" . One could think of it as "the shortest measurable period of time", and for any purpose within the real world (if one believes in Quantum mechanics), any two events that are separated by less than this amount of time can be considered simultaneous.
It takes light (traveling at the speed of light) this long to travel one Planck length unit, which itself is much smaller than a proton, electron or any particle whose size is known.
This is the Planck length in meters it is related to quantum mechanics. The best interpretation for most people is that the Planck length is the smallest measurable length, or the smallest length that has any relevance to events that we can observe. This uses the CODATA 2014 value 50 . See also 5.390×10 -44 and 299792458.
The "reduced" Planck constant in joule-seconds, from CODATA 2014 values 50 .
This is the Planck constant in joule-seconds, from CODATA 2014 values 50 . This gives the proportion between the energy of a photon and its wavelength.
As of 1 st May 2019, the Planck constant (in joule-seconds) is defined to be exactly this value, in order to define the kilogram in terms of observable properties of nature. The definition reads:
The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10 -34 when expressed in the unit J s, which is equal to kg m 2 s -1 , where the metre and the second are defined in terms of c and δνCs.
where c is the speed of light by the existing (since 1967) definition (see 299792458) and δνCs is the unperturbed ground-state hyperfine transition frequency of Caesium-133 (see 9192631770).
The mass of an electron in kilograms, from CODATA 2014 values 50 . See also 206.786.
The mass of a proton in kilograms, from CODATA 2014 values 50 .
The mass of a neutron in kilograms, from CODATA 2014 values 50 .
The approximate time (in seconds) it takes light to traverse the width of a proton.
Value of the Boltzmann constant by the old (pre-2019) definition, as given in CODATA 2014 50 . This value was based on experimental observations and also upon the definition of the Kelvin, which was determined by measuring the temperature of the triple point of water and defining the Kelvin so that the triple point temperature comes out to 273.16 K. For the current (2019 and later) definition see 1.380649×10 -23 .
The Boltzmann constant (in Joules per degree Kelvin) by the 2019 redefinition, which reads:
The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be 1.380649×10 -23 when expressed in the unit J K -1 , which is equal to kg m 2 s -2 K -1 , where the kilogram, metre and second are defined in terms of h , c and δνCs.
where h is the Planck constant by the new definition (see 6.62607015×10 -34 ), c is the speed of light by the existing (since 1967) definition (see 299792458) and δνCs is the unperturbed ground-state hyperfine transition frequency of Caesium-133 (see 9192631770).
The quantum of electric charge in coulombs (one third of the electron charge), based on from CODATA 2014 values 50 . Protons, electrons and quarks all have charges that are a (positive or negative) integer multiple of this value.
The elementary charge or "unit charge", the charge of an electron in coulombs, from CODATA 2014 values 50 . This is no longer considered the smallest quantum of charge, now that matter is known to be composed largely of quarks which have charges in multiples of a quantum that is exactly 1/3 this value.
As of the 1 st May 2019, the elementary charge is not measured in terms of coulombs instead it is defined to be exactly 1.602176634×10 -19 coulombs in other words, the coulomb is now defined in terms of the elementary charge.
The reciprocal value of the coulomb in units of the elementary charge, by the new (2019) definition.
In 2019 the International System of Units (SI) was updated to define its seven base units in a way that defines all seven of them in terms of observable properties of nature, which are given arbitrary numerical values in terms of the base units:
The ampere, symbol A, is the SI unit of electric current. It is defined by taking the fixed numerical value of the elementary charge e to be 1.602176634×10 -19 when expressed in the unit C, which is equal to A s, where the second is defined in terms of δνCs.
where δνCs is the unperturbed ground-state hyperfine transition frequency of Caesium-133 (see 9192631770).
Approximate "size" of a proton 71 , in meters (based on its "charge radius" of 0.875 femtometers). "Size" is a pretty vague concept for particles, and different definitions are needed for different problems. See 10 40 .
The vacuum permittivity constant in farads per meter, using the old (pre-2019) definitions of the vacuum permeability (see 4π/10 7 ) and the (still current) definition of the speed of light (see 299792458). In older times this was called the "permittivity of free space". Due to a combination of standard definitions, notably the exact definition of the speed of light, this constant is exactly equal to 10 7 /(4 π 299792458 2 ) = 625000/22468879468420441π farads per metre.
In 2019 and later, the vacuum permittivity is something needing to be computed based on measurement. The greatest uncertainty contributing to its value is the measurement of the fine structure constant.
The gravitational constant in cubic meters per kilogram second squared, from CODATA 2014 values 50 . This is one of the most important physical constants in physics, notably cosmology and efforts towards unifying relativity with quantum mechanics. It is also one of the most difficult constants to measure. See also 1.32712442099(10)×10 20 .
The Planck mass in kilograms, using CODATA 2014 values. The Planck mass is related to the speed of light, the Planck constant, and the gravitational constant by the formula Mp = √ hc /2π G .
The constant 4π/10 7 that appears in the old (pre-2019) definition of the "magnetic constant" or vacuum permeability. It is related to the old definition of ampere, which stated that if exactly one ampere of current flows in two straight parallel conductors of infinite length 1 meter apart, the force produced would be 2×10 -7 newton per meter of length. This derives from an older definition stating that a similar setup with the wires one centimetre apart would produce a force of 2 dynes per centimetre of length (one dyne is 10 -5 newtons).
The fine-structure constant, as given by CODATA 2014 (see 50 ). The "(17)" is the error range. See the 137.035. page for history and details.
There are a few "coincidences" regarding multiples of 1/127:
e /π = 0.865255. &asymp 110/127 = 0.866141.
√ 3 = 1.732050. &asymp 220/127 = 1.732283.
π = 3.141592. &asymp 399/127 = 3.141732.
√ 62 = 7.874007. &asymp 1000/127 = 7.874015.
e π = 23.140692. &asymp 2939/127 = 23.141732.
There are a few more for 1/7. The √ 62 coincidence is discussed in the √ 62 entry, and the π and e π ones go together (see e π ).
This is the eccentricity of the orbit of the Earth-Moon barycentre at epoch J2000 the value is currently decreasing at a rate of about 0.00000044 per year, mostly due to the influence of other planets. The Moon is massive enough and far enough to shift the Earth itself a few thousand km away from the barycentre. See also 0.054900.
The version of the Gaussian gravitational constant computed by Simon Newcomb in 1895.
The "Gaussian gravitational constant" k , as originally calculated by Gauss, related to the Gaussian year Δ t by the formula Δ t = 2π/ k . The value was later replaced by the Newcomb value 0.01720209814, but in 1938 (and again in 1976) the IAU adopted the original Gauss value.
Mean eccentricity of the Moon's orbit — the average variation in the distance of the Moon at perigee (closest point to the Earth) and apogee. Due to the influence of the Sun's gravity the actual eccentricity varies a large amount, going as low as about 0.047 and as high as about 0.070 also the ellipse precesses a full circle every 9 years (see 27.554549878). The eccentricity is greatest when the perigee and apogee coincide with new and full moon. At such times the Moon's distance varies by a total of 14%, and its apparent size (area in sky) varies by 30% when the size at apogee is compared to the size at perigee. This means that the brightness of the full moon varies by 30% over the course of the year. In 2004 the brightest full moon was the one on July 2 nd due to the orbit's precession the brightest full moon in 2006 was a couple months later, Oct 6 th .
This change in size is a little too small for people to notice from casual observation (except in solar eclipses, when the Moon sometimes covers the whole sun but at other times produces an annular eclipse). But the eccentricity is large enough to cause major differences in the Moon's speed moving through the sky from one day to the next. When the Moon is near perigee it can move as much as 16.5 degrees in a day when near apogee it moves only 12 degrees the mean is 13.2. The cumulative effect of this is that the moon can appear as much as 22 degrees to the east or west of where it would be if the orbit were circular, enough to cause the phases to happen as much as 1.6 days ahead of or behind the prediction made from an ideal circular orbit. It also affects the libration (the apparent "wobbling" of the Moon that enables us to see a little bit of the far side of the moon depending on when you look).
This is the lowest value of z for which the infinite power tower
converges to a finite value. (The highest such value is e (1/ e ) = 1.444667. see that entry for more).
This is the Kepler–Bouwkamp constant, related to a geometrical construction of concentric inscribed circles and polygons. Start with a unit circle (a circle with radius 1). Inscribe an equilateral triangle inside the circle, then inscribe a circle inside the triangle. The radius of the smaller circle will be cos(π/3) = 1/2. Now inscribe a square inside that circle, and a circle inside the square this even smaller circle has radius cos(π/3)×cos(π/4) = √ 1/8 . Continue inscribing with a pentagon, hexagon, and every successive regular polygon. The circles get smaller but they do not go all the way down to zero the limit is this number, about 10/87.
The fraction 1/7 is the simplest example of a fraction with a repeating decimal that has an interesting pattern. See the 7 article for some of its interesting properties.
Reader C. Lucian points out that many of the well-known constants can be approximated by multiples of 1/7:
gamma = 0.5772156. &asymp 4/7 = 0.571428.
e /π = 0.865255. &asymp 6/7 = 0.857142.
√ 2 = 1.414213. &asymp 10/7 = 1.428571.
√ 3 = 1.732050. &asymp 12/7 = 1.714285.
e = 2.7182818. &asymp 19/7 = 2.714285.
π = 3.1415926. &asymp 22/7 = 3.142857.
e π = 23.140692. &asymp 162/7 = 23.142857.
These are mostly all coincidences without any other explanation, except as noted in the entries for √ 2 and e π . See also 1/127.
This is the integral of sin(1/*x), from 0 to 1. Mathematica or Wolfram Alpha will give more digits: 0.5040670619­0692837198­9856117741­1482296249­8502821263­9170871433­1675557800­7436618361­6051791560­4457297012.
A reader suggested to me the idea that some people might define "zillion" as "a 1 followed by a zillion zeros". This is kind of like the definition of googolplex but contradicts itself, in that no matter what value you pick for X , 10 X is bigger than X .
However, this is actually only true if we limit X to be an integer (or a real number). If X is allowed to be a complex number, then the equation 10 X = X has infinitely many solutions.
Using Wolfram Alpha, put in "10^x=x" and you will get:
x &asymp -0.434294481903251827651 Wn (-2.30258509299404568402)
with a note describing Wk as the "product log function", which is related to the Lambert W function (see 2.50618. ). This function is also available in Wolfram Alpha (or in Mathematica) using the name " ProductLog [ k , x ]" where k is any integer and x is the argument. So if we put in "-0.434294481903251827651 * ProductLog[1, -2.30258509299404568402]", we get:
0.529480508259063653364. - 3.34271620208278281864. i
Finally, put in "10^(0.529480508259063653364 - 3.34271620208278281864 * i)" and get:
0.52948050825906365335. - 3.3427162020827828186. i
If we used -2 as the initial argument of ProductLog , we get 0.5294805+3.342716 i , and in general all the solutions occur as complex conjugate pairs. Other solutions include x =-0.119194. ۪.750583. i and x =0.787783. ۰.083768. i .
In light of the fact that the -illion numbers are all powers of 1000, another reader suggested that one should do the above starting with 10 (3 X +3) = X . This leads to similar results, with one of the first roots being:
-0.88063650680345718868. - 2.10395020077170002545. i
The first fraction in Conway's FRACTRAN program ( page 147) that finds all the prime numbers. The complete program is 17 /91, 78 /85, 19 /51, 23 /38, 29 /33, 77 /29, 95 /23, 77 /19, 1 /17, 11 /13, 13 /11, 15 /2, 1 /7, 55 /1. To "run" the program: starting with X =2, find the first fraction N / D in the sequence for which XN / D is an integer. Use this value NX / D as the new value of X , then repeat. Every time X is set to a power of 2, you've found a prime number, and they will occur in sequence: 2 2 , 2 3 , 2 5 , 2 7 , 2 11 and so on. It's not very efficient though — it takes 19 steps to find the first prime, 69 for the second, then 281, 710, 2375 . (Sloane's A7547).
This is e -π/2 , which is also equal to i i . (Because e ix = cos( x ) + i sin( x ), e i π/2 = i , and therefore i i = ( e i π/2 ) i = e i 2 π/2 = e -π/2 .)
The Cesaro sum of the alternating-diverging infinite series sum:
which can be used to derive the Euler/Ramanujan "infamous" sum 1 + 2 + 3 + 4 + . = -1/12.
The first-order Cesaro method is illustrated in the entry for 1/2. Here we'll apply the method twice. We start with the terms of the infinite series:
This has the partial sums:
These diverge and are unbounded both above and below. The sum of the first n terms of that series is:
The average of the first n terms of A o ( n ) is A'( n )/ n :
( C ,1)-sum = A'( n )/ n : 1, 0, 2/3, 0, 3/5, 0, 4/7, .
This is not converging but offers hope in that (like 1-1+1-. ) it manages to at least remain bounded from above as well as below. The even terms are all 0 while the odd terms approach 1/2.
Let's take successive averages of this sequence: the Cesaro sum of the Cesaro sum. The sum of the first n terms of the above "( C ,1)-sum" is
1, 1, 5/3, 5/3, 34/15, 34/15, 298/105, .
and successive averages are just these over n :
1, 1/2, 5/9, 5/12, 34/75, 34/90, 298/735, .
which converge on 1/4, though it may be a bit tough to see here. This isn't actually how Cesaro defined the 2 rd order method. Instead, he put the sum of the first n terms of A'( n ) in the numerator:
and the binomial coefficients n C 2 (the triangular numbers), called "E''( n )", in the denominator:
E''( n ) : 1, 3, 6, 10, 15, 21, 28, 35, .
The second-order averages by Cesaro's method are:
( C ,2)-sum = A''( n )/ C ( n ,2) : 1, 1/3, 3/6, 3/10, 6/15, 6/21, 10/28, .
and these also converge on 1 /4. Adding this way makes it easier to see because e.g. for even n we can let h be n /2 and we get:
A''( n )/ C ( n ,2) = h C 2 / 2 h C 2
= ( h ( h -1)/2) / (2 h (2 h -1)/2)
= ( h 2 - h )/(4 h 2 -2 h )
= (1/2) (2 h 2 - h - h 2 )/(2 h 2 - h )
= (1/2) ((2 h 2 - h )/(2 h 2 - h ) - h 2 /(2 h 2 - h ))
= 1/2 - 1/2 ( h 2 /(2 h 2 - h ))
The part " h 2 /(2 h 2 - h )" clearly converges on 1/2, so the whole thing converges to 1/2 - 1/4.
This sum of 1 /4 appears as "1/(1+1) 2 " in Ramanujan's notebook. That can be derived by noting that 1-1+1-1+1-1+. has the 1 st -order Cesaro sum 1/2, and then doing this:
(1 - 1 + 1 - 1 + 1 - . ) 2
= (1 - 1 + 1 - 1 + 1 - . )×(1 - 1 + 1 - 1 + 1 - . )
= 1 + (-1×1 + 1×-1) + (1×1 + -1×-1 + 1×1) + (-1×1 + 1×-1 + -1×1 + 1×-1) + .
= 1 - 2 + 3 - 4 + .
So the sum of 1 - 2 + 3 - 4 + 5 - 6 + 7 - . must be the square of the sum of 1 - 1 + 1 - 1 + 1 - . which is the square of 1/2, which is 1/4.
There is another, perhaps easier, way to get the same answer. Start with this infinite series sum and assume it has a value, here called C :
Subtract the second from the first:
C - Cx = 1
C (1- x ) = 1
C = 1/(1- x )
If x is something like 1/2, it's easy to see that the sum 1 + 1/2 + 1/4 + 1/8 + . is 2, and 1/(1- x ) = 1/(1-1/2) is also 2, so the derivation is valid. But if x were, say, -1, then we'd get 1 - 1 + 1 - 1 + 1 - . = 1/2, whichis discussed in the entry for 1/2. Euler didn't worry about strict convergence and just went ahead with:
1 + x + x 2 + x 3 + x 4 + . = 1/(1- x )
Let's differentiate both sides!
1 + 2 x + 3 x 2 + 4 x 3 + . = 1/(1- x ) 2
If x =-1 we have the desired sum:
and again the answer is 1/4.
This is an infinite product of (1-2 - N ) for all N . This is also the product of (1- x N ) with x =1/2. Euler showed that in the general case, this infinite product can be reduced to the much easier-to-calculate infinite sum 1 - x - x 2 + x 5 + x 7 - x 12 - x 15 + x 22 + x 26 - x 35 - x 40 + . where the exponents are the pentagonal numbers N (3 N -1)/2 (for both positive and negative N ), Sloane's A1318. 30
This is Gottfried Helms' Lucas-Lehmer constant " LucLeh " see 1.38991066352414. for more.
1/3 is the simplest non-dyadic rational, and the simplest with a non-terminating decimal in base 10.
1/3 is the "Ramanujan sum" of the non-converging infinite series sum of -2 n :
Even though we're not allowed to, we could try to apply the series sum formula:
which converges the normal way only when -1 x x = -2 and the sum would be 1/(1-(-2)) = 1/3.
This is "Artin's Constant", the product (1-1/2)(1-1/6)(1-1/20). (1-1/( p ( p -1))) for all prime p . It relates to the conjecture regarding the "density" of primes p for which 1/ p has a as a primitive root, where a meets the conditions of OEIS sequence A85397. This includes 10, meaning that about 30% of primes have a reciprocal with a decimal expansion that repeats every p -1 digits the first two are 7 and 17.
has a value that is very close to, but not exactly, π/8. From Bernard Mares, Jr. via Bailey et al.  more on MathWorld at Infinite Cosine Product Integral.
If you take a string of 1's and 0's and follow it by its complement (the same string with 1's switched to 0's and vice versa) you get a string twice as long. If you repeat the process forever (starting with 0 as the initial string) you get the sequence
and if you make this a binary fraction 0.0110100110010110. 2 the equivalent in base 10 is 0.41245403364. and is called the Thue-Morse constant or the parity constant . Its value is given by a ratio of infinite products:
4 K = 2 - PRODUCT[2 2 n -1] / PRODUCT[2 2 n ]
= 2 - (1 × 3 × 15 × 255 × 65535 × . )/(2 × 4 × 16 × 256 × 65536 × . )
The Cesaro sum of the simplest Cesaro-summable infinite series sum:
The Cesaro sum technique is a generalisation of the definition of an infinite series sum as the limit of its partial sums. To illustrate the principle, let's consider an infinite sum that actually does converge in the normal way:
this has the partial sums:
which can easily be seen (and proven, by mathematical induction) to converge to 2. Cesaro considered the series of averages (arithmetic means) of the first N partial sums:
1, (1 + 3/2)/2, (1 + 3/2 + 7/4)/3, (1 + 3/2 + 7/4 + 15/8)/4, ..
1, 5/4, 17/12, 49/32, 129/80, 321/128, 769/448, .
which also converges on 2, though more slowly. This technique of averaging the first n partial sums can yield an answer for infinite series whose partial sums taken individually so not converge. Start with:
This doesn't converge, but let's take the average of the first n of these. The sums of the first n of these (for n =1, 2, 3, . ) are:
So the average of the first n partial sums are:
which converges on 1/2. See 1/4 for an example of 2 rd order Cesaro summation, and -1/12 to see Ramanujan's extension.
Ramanujan's notebook, when discussing the -1/12 series, uses "1/(1+1) 2 ", which suggests that he viewed the sum 1 - 1 + 1 - 1 + 1 - 1 + 1 - . to be "1/(1+1)". This can be derived from the generalisation of the series sum:
which converges the normal way only when | x | x = -1 we'd get "1 - 1 + 1 - 1 + 1 - 1 + 1 - . = 1/(1- x ) = 1/(1+1)". So the value 1/2 can be "justified" in two ways.
The odds of losing a game of chance. Flip a coin: if you get heads , your score increases by π, if you get tails , your score diminishes by 1. Repeat as many times as you wish — but if your score ever goes negative, you lose. Assuming the player keeps playing indefinitely (motivated by the temptation of getting an ever-higher score), what are the odds of losing?
The answer is given by a series sum: 1/2 + 1/2 5 + 4/2 9 + 22/2 13 + 140/2 17 + 969/2 21 + 7084/2 25 + 53820/2 29 + 420732/2 34 + . (numerators in Sloane's A181784) which adds up to 0.5436433121.
A more sophisticated analysis using rational numbers like 355/113 converges on the answer more quickly, giving 0.54364331210052407755147385529445. (see ).
This is the Omega constant, which satisfies each of these simple equations (all equivalent):
Thus it is sort of like the golden ratio. In the above equations, if e is replaced with any number bigger than 1 (and "ln" by the corresponding logarithm) and you get another "Omega" constant. For example:
if 2 x =1/ x , then x =0.6411857445.
if π x =1/ x , then x =0.5393434988.
if 4 x =1/ x , then x =1/2
if 10 x =1/ x , then x =0.3990129782.
if 27 x =1/ x , then x =1/3
if 10000000000 x =1/ x , then x =1/10
(the Euler-Mascheroni constant )
This is the Euler-Mascheroni constant, commonly designated by the Greek letter gamma . It is defined in the following way. Consider the sum:
S n = 1 + 1/2 + 1/3 + 1/4 + 1/5 + . + 1/ n
The sequence starts 1, 1.5, 1.833333. 2.083333. etc. As n approaches infinity, the sum approaches ln( n ) + gamma . Numberphile has a video about this constant: The mystery of 0.577.
Here are some not-particularly-significant approximations to gamma :
1/(√ π - 1/25) = 0.5772159526.
gamma = 0.5772156649.
1/(1+ 1/√ 10 ) 2 = 0.5772153925.
One of the infinite sums in Ramanujan's 1913 letter to G.H. Hardy, section XI:
see -1/12 for a simpler example.
this sum diverges, but a partial sum can be contemplated:
0 + 1 + 2 + . + n
= SUM i in [0.. n ] f ( i )
(where f ( x ) = x )
= - f (0)/2 + i INTEGRAL0..∞ ( f ( it )- f (- it )) / ( e 2π t - 1) dt
in this specific example we get
Value of the infinite series sum
It is (1-√ 2 ) times the Riemann zeta function of 1/2. More digits: 0.604898643421630370247265914. (Sloane's sequence A113024). Oddly, though the series sum converges to a reasonably small finite value, if you square the series sum:
and sum the terms in the needed order:
the magnitudes of the parenthesised parts keep growing, so the series sum diverges. However, there clearly is a sum, and technuqies such as Cesaro summation (see the entry for 1/4) can be used to evaluate it and get the proper answer, It is for sums like this that Cesaro summation is really needed. (The case for 1/4 is a bit harder to argue.)
The golden ratio (reciprocal form): see 1.618033. .
The Buffon's needle problem involves estimating the probability that a randomly-placed line segment of some given length will cross one of a a set of parallel lines spaced some fixed distance apart. If the length of the line segment is the same as the spacing between lines, the probability is 2/π.
This is the lowest point in the function y = x x . See also 1.444667. .
The natural logarithm of 2, written "ln(2)". See 69.3147. and 72.
ln(2) is the value of this infinite series sum:
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + .
= 1/2 + 1/12 + 1/30 + 1/56 + .
This is called a "conditionally convergent series" because the series converges if added up in the way shown above, but if you rearrange the terms:
1 + 1/3 + 1/5 + 1/7 + . - (1/2 + 1/4 + 1/6 + 1/8 + . )
then you have two series that do not converge and an undefined "infinity minus infinity".
You can create a long string of 1's and 0's by using "substitution rules" and iterating from a small starting string like 0 or 1. If you use the rule:
and start with 0, you get 1, 10, 101, 10110, 10110101, 1011010110110, . where each string is the previous one followed by the one before that (Sloane's A36299 or A61107). The limit of this is an infinite string of 1's and 0's which you can make this into a binary fraction: 0.1011010110110. 2, you get this constant (0.709803. in base 10) which is called the Rabbit Constant . It has some special relationships to the Fibonacci sequence:
- In the iteration described above, the number of digits in each string is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, .
- Expressed as a continued fraction, the constant is 0 + 1/(2 0 + 1/(2 1 + 1/(2 1 + 1/(2 2 + 1/(2 3 + 1/(2 5 + 1/(2 8 + . ))))))) where the exponents of 2 are the Fibonacci numbers.
- If you take all the multiples of the Golden Ratio 0.618033 and round them down to integers, you get 1, 3, 4, 6, 8, 9, 11, 12, . These numbers tell you where the 1's in the binary fraction are.
If you leave off the first two binary digits (10) you get 110101101101011010110110101. the bit pattern generated by a Turing machine at the end of the Turing machine Google Doodle. As a fraction (0.1101011. ) it is 0.8392137714451.
Value of x such that x =cos( x ), using radians as the unit of angle. You can find the value with a scientific calculator just by putting in any reasonably close number and hitting the cosine key over and over again. Here are a few more digits: 0.7390851332151606416553120876738734040134117589007574649656. 26
This is 3 - √ 5 , and is related to a sequence of Grafting numbers found by Matt Parker. With more precision, it is: 0.76393­20225­00210­30359­08263­31268­72376­45593­81640­38847.
Take an odd number of digits after the decimal point, add 1, and you get a Grafting number. For example, 76393+1 = 76394. The sequence of numbers derived this way starts: 8, 764, 76394, 7639321, 763932023, 76393202251, 7639320225003, .
A fiendishly engaging approximation to the answer to the "infinite resistor network" problem in xkcd 356, which introduced the world to the sport of "nerd sniping". See ries and 0.773239. .
The answer to a fiendishly engaging "infinite resistor network" problem in xkcd 356, which introduced the world to the sport of "nerd sniping" 90 . See also 0.636619. and 0.772453. .
This number, on a early calculator with 7-segment display, says "hello" when seen upside-down:
This is INTEGRAL0..1 x x d x , which is curiously equal to - SIGMAi..inf (- n ) - n , which was proven by Bernoulli. With more digits, it is 0.78343051071213440705926438652697546940768199014. It shares (with 1.291285. the nickname "sophomore's dream".
This is 0.1101011011010110101101101011011010110101101101011010110110. in binary, and is the slightly different version of the Rabbit constant generated by a Turing machine Google Doodle from June 2012. More digits: 0.8392137714451652585671495977783023880500088230714420678280105786051.
Decimal value of the "regular paperfolding sequence" 1 1 0 1 100 1 1100100 1 110110001100100 1 1101100111001000110110001100100 . converted to a binary fraction. This sequence of 1's and 0's gives the left and right turns as one walks along a dragon curve. It is the sum of 8 2 k /(2 2 k +2 -1) for all k &ge0, a series sum that gives twice as many digits with each additional term.
The minimum value of the Gamma function with positive real arguments. The Gamma function is the continuous analogue of the factorial function. This is Gamma(1.461632144968. ). (For more digits of both, see OEIS sequences A30171 and A30169.)
This is 1/2 of the square root of π. It is Gamma (3/2), and is sometimes also called (1/2)!, the factorial of 1/2.
This is Gamma (5/4), or "the factorial of 1/4". While some Gamma function values, like 0.886226. and 1.329340. , have simple formulas involving just π to a rational power, this one is a lot more complicated. It is π to the power of 3/4, divided by (√ 2 + 4 √ 2 ), times the sum of an infinite series for an elliptic function.
This is (4+4√ 2 )/(5+4√ 2 ), and is the best density achievable by packing equal-sized regular octagons in the plane. Notably, it is a bit smaller than 0.906899. , the density achievable with circles.
This is π/12, the density achievable by packing equal-sized circles in a plane. See also 0.906163. .
Catalan's constant, which can be defined by:
G = 1 - 1/3 2 + 1/5 2 - 1/7 2 + 1/9 2 - .
If you have a 2 n × 2 n checkerboard and a supply of 2 n 2 dominoes that are just large enough to cover two squares of the checkerboard, how many ways are there to cover the whole board with the dominoes? For large n , the answer is closely approximated by
This is the cube root of ( 5 √ 27 - 5 √ 2 ). Bill Gosper discovered the following identity, which is remarkable because the left side only has powers of 2 and 3, but the right side has a power of 5 in the denominator 108 :
( 5 √ 27 - 5 √ 2 ) (1/3) = ( 5 √ 8 5 √ 9 + 5 √ 4 - 5 √ 2 5 √ 27 + 5 √ 3 ) / 3 √ 25
(3 (3/5) -2 (1/5) ) (1/3) = (- 2 (1/5) 3 (3/5) + 2 (3/5) 3 (2/5) + 3 (1/5) + 2 (2/5) ) / 5 (2/3)
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Math 309 serves as the culmination of the Math 307-8-9 program in linear analysis. It combines analytic tools from Math 307 (e.g., general solutions and initial value problems for differential equations complex numbers and exponentials) with concepts and methods from Math 308 (e.g., eigenvalues and eigenvectors linear independence, bases, and orthogonal bases). These tools are applied to the solution and qualitative study of linear systems of ordinary differential equations and of boundary value problems for the classical partial differential equations (heat, wave, Laplace). For the latter, separation of variables is used to generate Fourier series solutions.
This syllabus indicates the topics to be covered instructors may vary the order and emphasis. In the (shorter) winter and spring quarters, choose the smaller number of lectures where a range is given to allow time for a midterm and a second midterm or several short quizzes.
1. Linear systems of ODE's (12-13 lectures)
- §§7.1-7.3: Linear algebra (3 lectures)
Review of eigenvalues and eigenvectors, extension to complex matrices, spaces of vector-valued functions, matrix differential equations
- §§7.4-7.8: Solving homogeneous linear systems (6 lectures)
- §7.9: Inhomogeneous equations (2 lectures)
Two or three methods from this section
- §9.1: Summary of phase plane portraits for linear systems (1 lecture)
- Optional: §§9.2-9.3: Stability of critical points for autonomous nonlinear systems (0-1 lecture)
2. Fourier series and boundary value problems (13-15 lectures)
Understanding LTE with MATLAB: From Mathematical Modeling to Simulation and Prototyping
The LTE (Long Term Evolution) and LTE-Advanced are among the latest mobile communications standards, designed to realize the dream of a truly global, fast, all-IP-based, secure broadband mobile access technology.
This book examines the Physical Layer (PHY) of the LTE standards by incorporating three conceptual elements: an overview of the theory behind key enabling technologies a concise discussion regarding standard specifications and the MATLAB® algorithms needed to simulate the standard.
The use of MATLAB®, a widely used technical computing language, is one of the distinguishing features of this book. Through a series of MATLAB® programs, the author explores each of the enabling technologies, pedagogically synthesizes an LTE PHY system model, and evaluates system performance at each stage. Following this step-by-step process, readers will achieve deeper understanding of LTE concepts and specifications through simulations.
• Accessible, intuitive, and progressive one of the few books to focus primarily on the modeling, simulation, and implementation of the LTE PHY standard
• Includes case studies and testbenches in MATLAB®, which build knowledge gradually and incrementally until a functional specification for the LTE PHY is attained
• Accompanying Web site includes all MATLAB® programs, together with PowerPoint slides and other illustrative examples
Dr Houman Zarrinkoub has served as a development manager and now as a senior product manager with MathWorks, based in Massachusetts, USA. Within his 12 years at MathWorks, he has been responsible for multiple signal processing and communications software tools. Prior to MathWorks, he was a research scientist in the Wireless Group at Nortel Networks, where he contributed to multiple standardization projects for 3G mobile technologies. He has been awarded multiple patents on topics related to computer simulations. He holds a BSc degree in Electrical Engineering from McGill University and MSc and PhD degrees in Telecommunications from the Institut Nationale de la Recherche Scientifique, in Canada.
10.8: Complex numbers - Mathematics
Posted February 27, 2018
This month’s CORE Excellence in Education Blog focuses on the importance of mathematical fluency and number sense as a critical foundation for entering mathematics in middle and high school.
(By Mary Buck, Senior Educational Services Consultant for CORE, with Dean Ballard, Director of Mathematics at CORE)
As a mathematics teacher, coach, and consultant I have had the opportunity to work with K-12 students and teachers across the United States, Guam and Japan. The one similarity across all countries is that learning new concepts in math is built on a solid foundation of understanding previous concepts and skills. In my work, the most heart-breaking thing that I observe is students at any grade struggling with new concepts mostly because they cannot do the simple calculations embedded in the work. In many Algebra 1 classes I have painfully watched students count out multiplication facts on their fingers or write out multiples using addition.
Let’s face it, teaching mathematics is challenging at all levels. Each year is filled with new content that is built on prior learning thus, the time spent teaching basic fluencies is time lost focusing on new concepts. I recently observed my grandniece (whom my sister will tell you is brilliant) count on her fingers 4 + 1. Since she had just completed 1 st grade, I was mortified! I recognize the importance of different counting strategies for building understanding and working toward fluency. However, by the end of first grade my niece should have had 4 + 1 on the tip of her tongue not on the tip of her fingers.
This example illustrates why I am on a mission to improve number sense and fluency in all schools. To this end, the IES Practice Guide, Assisting Students Struggling with Mathematics (2009), recommends providing 10 minutes of daily practice to strengthen needed fluency with facts and procedures (pp. 79-83). Hattie, Fisher and Frey (2017) support this recommendation by documenting that spaced practice, repeated practice of previously learned knowledge over “a long period of time,” has a high effect size of 0.71 (p. 129). In addition, the National Mathematics Advisory Panel (2008) found that fluency with whole numbers and fractions are part of a critical foundation for learning Algebra (pp. 17-18). When I was in high school, it was noted that Algebra 1 was the gate keeper to upper division mathematics. If students weren’t successful in Algebra 1, they would not take higher level math classes. I contend that lack of fluency and number sense are the gate keepers to middle school and high school mathematics.
So, what is number sense and what is fluency? The Common Core Standards (CCSSM) describe procedural fluency as “skill in carrying out procedures flexibly, accurately, efficiently and appropriately” (2010). Number sense is making sense of numbers – understanding numbers and how they work together. For example, in the primary grades students understand how numbers can be broken apart and put together when they explore and build fluency with concepts such as how to make 10 and how to break up 12 into 10 and 2. Multiplication builds on this foundation and with multiplication students are introduced to the distributive property: 8 x 12 is the same as 8(10 + 2) = (8 x 10) + (8 x 2). Students apply understanding of these concepts to do mental math, make sense of the standard algorithm for multidigit multiplication, and apply the distributive property in work with algebraic expressions. We want students to build fluency and number sense so that they can continue to understand and solve more complex problems at each new grade level. If students cannot do the “arithmetic” or calculations easily it is difficult for them to solve problems requiring simple calculations. This difficulty is caused by the amount of working memory we have available to us when solving problems. If our working memory is busy trying to figure out the computation, it is more difficult for us to make sense of the higher level mathematical concepts being grappled with at that moment. Similarly, when reading, if students struggle to decode the words, then they will have difficulty understanding the larger meaning of the text. Hence, it is important that students have automaticity with number facts:
Our ability to think would be limited indeed if there were not ways to overcome the space constraint of working memory. One of the more important mechanisms is the development of automaticity. When cognitive processes . . . become automatic, they demand very little space in working memory, they occur rapidly, and they often occur without conscious effort.
Understanding number relationships and different techniques for putting numbers together and taking them apart are essential parts of building number sense. Automaticity with number facts is equally essential for continued growth in mathematics. As teachers, I believe it is our responsibility to make sure that we are doing everything within our power to ensure students leave our mathematics classroom at the end of the year proficient in the fluency requirements as outlined in the CCSSM and included in all state standards. Anytime a child leaves a grade level without having the required fluency, the mathematical gap begins and continues to grow throughout the years as students fall further behind in their automaticity of facts and understanding of numbers. As we continue to teach mathematics, let’s make sure we are doing our part to teach the fluencies needed at each grade level.
Gersten, R., Beckmann, S., Clarke, B., Foegen, A., Marsh, L., Star, J. R., & Witzel, B. (2009). Assisting Students Struggling with Mathematics: Response to Intervention (RtI) for Elementary and Middle Schools (NCEE 2009-4060 ed.). Washington, DC: National Center for Education Evaluation and Regional Assistance, Institute of Education Sciences, U.S. Department of Education. Retrieved from https://ies.ed.gov/ncee/wwc/PracticeGuide/2
Hattie, J., Fisher, D., & Frey, N. (2017). Visible Learning for Mathematics. Thousand Oaks, CA: Corwin.
National Governors Association Center for Best Practices & Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics. Washington, DC: Authors.
National Mathematics Advisory Panel. (2008). Foundations for Success: The Final Report of the National Mathematics Advisory Panel. Washington, DC: U.S. Department of Education. Retrieved from https://www2.ed.gov/about/bdscomm/list/mathpanel/report/final-report.pdf
Willingham, D. T. (2004, Spring). Practice Makes Perfect—but Only If You Practice Beyond the Point of Perfection [Electronic version]. American Educator. doi: https://www.aft.org/periodical/american-educator/spring-2004/ask-cognitive-scientist
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