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August 6

What is the addition equivalent of a factorial?

Hi all. I know that factorials are equal to a number times all the counting numbers less than it, ie 5! is 5*4*3*2*1. I was wondering what the equivalent of this is in addition, ie 6 would be 6+5+4+3+2+1. Is there a specific sign, or how would it be represented? The simpler, the better.

Much help appreciated ! Xhin Give Back Our Membership! 00:58, 6 August 2007 (UTC)

Triangular numbers, expressed T(n) or T_n for 1+2+...+n. PrimeHunter 01:05, 6 August 2007 (UTC)[reply]

Note that T_n = n·(n+1)/2. This has a simple expression and the notation T_n is much less common than n!. PrimeHunter 01:08, 6 August 2007 (UTC)[reply]

Very rarely I've seen the notation n? instead of n! --Oskar 05:33, 6 August 2007 (UTC)[reply]

Standard notation is

{n+1} \choose 2

see binomial coefficient. Bo Jacoby 08:14, 6 August 2007 (UTC).[reply]

No, it is not. We are not talking about a binomial coefficent, but about a triangular number that just happens to be computable as a binomial coefficient. No one would look at your proposed notation and think "Oh, a triangular number". It's bad enough that you try to inject your idiosyncratic notations into articles; please do not confuse unsuspecting questioners on the reference desk. --KSmrq^T 11:05, 6 August 2007 (UTC)[reply]

Quite the contrary. Everybody aquainted with binomial coefficients recognizes

n \choose 2

as a triangular number. :) Bo Jacoby 12:24, 6 August 2007 (UTC).[reply]

Recognizes that it's the formula for it, but not the notation for it. And usually they don't, seeing as binomial coefficients have other uses. There's a difference between a formula for something and a notation for something. Gscshoyru 12:29, 6 August 2007 (UTC)[reply]

Is 2x a formula for the double of x, or is it a notation? Having a simple formula, no special notation is needed. Bo Jacoby 12:48, 6 August 2007 (UTC).[reply]

So, you'd rather write out the giant, clumsy formula for the nth Fibonacci number than a simple F_n or Fib(n) or the like? Gscshoyru 12:51, 6 August 2007 (UTC)[reply]

I wrote, a simple formula, not a giant, clumsy formula. It is of course a matter of taste, but generally I prefer standard notation rather than special or ad hoc notation, 2x rather than double(x),

n \choose 2

rather than T_n−1. Bo Jacoby 13:45, 6 August 2007 (UTC).[reply]

Prefer what you like but confine it to your desktop; do not inject your notations into Wikipedia pretending they are common practice in mathematics. Your claim above that "Standard notation is …" is false and unhelpful. --KSmrq^T 18:36, 6 August 2007 (UTC)[reply]

Okay, at this point I believe all parties have made their positions sufficiently clear. Now could you all please stop, or at least take it somewhere outside the refdesk? Please? —Ilmari Karonen (talk) 18:41, 6 August 2007 (UTC)[reply]

I've seen a notation

\triangle n

(or sometimes with

n

inside the triangle, which is apparently unsettable with the TeX-subset supported by wikipedia). Cf The Art of the Infinite by Robert Kaplan. Donald Hosek 18:51, 6 August 2007 (UTC)[reply]

Pfff.

\triangle \!\!\!\!n

You mean

? (Stolen shamelessly from Steinhaus-Moser notation, where it's a bit bigger than T_n.) Confusing Manifestation 01:37, 7 August 2007 (UTC)[reply]

Using multiple \!'s in that fashion is, to mind an awful hack. I had wanted to do an \llap, or even \hbox to0pt{$n$\hss} and then a bit of kerning to position the n, but this is apparently not allowed in the TeX-subset here. properly speaking, I would have done something along the lines of \newdimen\@twidth \@twidth=\wd\hbox{$\triangle$} \def\tnum#1{\mathord{\raise0.1em\hbox to\@twidth{\hss\small#1\hss}\kern-\@twidth\triangle}} (perhaps with a few small adjustments after seeing how it comes out on the page). Donald Hosek 16:59, 7 August 2007 (UTC)[reply]

Yes, but it's not unsettable, is it? It just looks bad. :)

2^100

I was recently challenged to find the last digit of 2¹⁰⁰. Modular arithmetic was my first thought, but seemed too easy, so I went for finding patterns. I discovered that

$2^{4n}\equiv 6{\text{ (mod }}10)$

$2^{4n+1}\equiv 2{\text{ (mod }}10)$

$2^{4n+2}\equiv 4{\text{ (mod }}10)$

$2^{4n+3}\equiv 8{\text{ (mod }}10)$

This shows that 2¹⁰⁰ ends in 6. I was quite pleased with myself, as the result was confirmed by mod 10 arithmetic on the original number/expression/whatever it's called. However I noticed that the first generalisation falls apart irretrievably when $n=0$ . Does this mean that my method was not a good one to choose or just that 0 is a bad value to use to check whether everything works alright? asyndeton 20:27, 6 August 2007 (UTC)[reply]

The sequence of positive powers of 2 is (2,4,8,16,32,...). Modulo 10 you get the sequence (2,4,8,6,2,...). It is periodic, and so your discovery is OK for positive values of n. But you cannot generally divide by 2 modulo 10, so you cannot extend the sequence to the left. Bo Jacoby 20:41, 6 August 2007 (UTC).[reply]

I'm sorry, I wasn't clear. When I said I used modulo 10, I was summarising - I did the following

$2^{100}=(2^{5})^{20}$

$2^{5}\equiv 2{\text{ (mod }}10)$

$2^{20}=(2^{5})^{4}$

$2^{5}\equiv 2{\text{ (mod }}10)$

$2^{4}=16$

$16\equiv 6{\text{ (mod }}10)$ asyndeton 20:56, 6 August 2007 (UTC)[reply]

Trivially, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32. Modulo 10 (balanced) the sequence is 2, 4, −2, −4, 2. Before we even begin to compute we know periodicity is unavoidable; now we know the period. Since 2×2⁴ is congruent to 2, induction shows that 2×(2⁴)ⁿ is congruent to 2, and that k×2×(2⁴)ⁿ is congruent to k×2, assuming n ≥ 0.

The pattern is really there; but induction only allows us to increase n. Although we are perfectly able to divide 2 by 2 to get 1, our induction cannot start with 2⁰. Why? Because no other power of 2 is conguent to 1. This is the well-known "rho" pattern of (eventual) periodicity, where the "tail" of the rho (ρ) is a prefix that leads into a cycle. (Another example is repeating decimal fractions.)

In may be helpful to contrast this with powers of 3 modulo 10. We have 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81. Modulo 10 (balanced) the sequence is 3, −1, −3, 1. Unlike 2, which shares a factor with 10, here we have GCD(3,10) = 1; thus some power of 3 must be conguent to 1. And if 3×3ⁿ is congruent to 1, then 3ⁿ is a multiplicative inverse for 3, modulo 10.

a	↓	aⁿ (mod 10)
n	→	1	2	3	4	5
1	●	1
2		2	4	−2	−4	2
3	●	3	−1	−3	1
4		4	−4	4
5		5	5
−4		−4	−4
−3	●	−3	−1	3	1
−2		−2	4	2	−4	−2
−1	●	−1	1

We see hints of interesting (known) mathematics; exploration is encouraged. --KSmrq^T 22:42, 6 August 2007 (UTC)[reply]

Another way of looking at it is to look at the powers of 2 modulo 2 and modulo 5. If we know the congruence classes modulo these two numbers, we know what the result is modulo 10 (by the Chinese remainder theorem). To start with the modulo 5 part, we find the remainders (1, 2, 4, 3) repeated endlessly, with the period 4 neatly conforming to Fermat's little theorem. For modulo 2 we have (1, 0, 0, 0, 0, ...), in which the initial 1 explains why 2⁰ is different from the other powers of the form 2⁴ⁿ. --Lambiam 01:35, 7 August 2007 (UTC)[reply]

How lawyers make decisions?

Original title: Have the mathematicians made any formal study regarding how lawyers make decisions?

This morning, I got a letter from a good friend asking the above question. I searched Math Sci Net with keywords lawyers and decicisions but did not get anything useful. So I am forwarding his question to this site. The following is what he wrote too: The criteria adopted by lawyers to make decisions include “balance of probability”, and “beyond reasonable doubt”. Are there any formal mathematic studies, which can be applied to analysis how to evaluate decisions, eg “Fussy Logic” or “Artificial Intelligence”? Thank you in advance for your help. twma 23:45, 6 August 2007 (UTC)[reply]

Fussy logic seems just what lawyers need, but presumably you mean Fuzzy logic, which may be closer to what they use. --Lambiam 01:38, 7 August 2007 (UTC)[reply]

Decisions 'beyond reasonable doubt' are made by judges rather than by lawyers. See also Burden of proof and decision theory. Bo Jacoby 05:20, 7 August 2007 (UTC).[reply]

More precisely, by the trier of fact, which may be a judge, but also a jury, depending on the case and the system of law. But, while the trier of fact decides on the facts of the case, based on the evidence presented, in criminal cases a lawyer for the defendant will decide on the possible strategies to be used, based on an assessment of, among other things, the likelihood that the trier of facts will find against their client in important issues for determining guilt and the probable punishment. This includes an assessment of the probability that the evidence presented by the prosecution will establish the alleged facts of such issues with the burden of proof that is in force, which may well be that they are proved beyond reasonable doubt. --Lambiam 05:38, 7 August 2007 (UTC)[reply]

You are giving me an impression that a court is NOT a place to serve justice but a place for the rich and the strong to prosecute the poor and the weak. The title is too long. Am I allowed to shorten it? twma 06:33, 7 August 2007 (UTC)[reply]

What makes you think such a thing? I entirely agree with Anatole France, as quoted in our article on Law: "In its majestic equality the law forbids rich and poor alike to sleep under bridges, beg in the streets and steal loaves of bread." If you shorten the title, I suggest to leave out the word "formal", which tends to be used by mathematicians for things that are much more formal than your friend may be interested in.

I do not want to argue whether the ideal world has been carried out ideally in the practical realistic world. For example, the system did allow you NOT to have signature in the above paragraph. The title has been shorten. Promise that no long titles will be used again in the future as long as I am aware of the problem. twma 09:19, 7 August 2007 (UTC)[reply]

The article "Statistics and the Law" (Intuitive views of evidence may be altered by mathematical analysis), by Mary W. Gray, Mathematics Magazine, vol. 56, no. 2 (March 1983), pp. 67-81, may be useful. -- Four Dog Night 16:21, 9 August 2007 (UTC)[reply]

Thank you. Will look up soon. twma 09:05, 10 August 2007 (UTC)[reply]