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February 23

Is there a version of OEIS that also shows categories of number that don't show up when you type nothing but seq:xxxx?

If I type a low 4-digit prime it doesn't actually show all sequences that contain it since if I sort by sequence number the first hit's not A000040 or below. Is there something that's nearly exhaustive for natural numbers of this size? (it probably can't be completely exhaustive since some combinations of "natural numbers no bigger than this" and "sequences" would probably need prohibitive or infinite computing time to check) Sagittarian Milky Way (talk) 22:54, 23 February 2018 (UTC)[reply]

To clarify perhaps, if you type 4567 into the search you want to find all OEIS sequences which contain 4567. But A000040 (the sequence of primes) is not showing in the search results since the list stored in OEIS only goes to 271. Afaik you don't have an extended search option in OEIS which allows you to search beyond what's actually stored on the database, and OEIS is pretty unique so I'm pretty sure you won't find a similar database that does have that option. (I'm guessing such an option wouldn't be feasible even if you did make reasonable restrictions such as excluding hard to compute sequences.) Not sure why you want this, but you might try to find a different approach to the problem or try typing 4567 into Wolfram Alpha. --RDBury (talk) 00:36, 24 February 2018 (UTC)[reply]

Some of the sequences have an additional table of the first 1,000 or 10,000 terms, but I don't know if it searches them (probably not), and probably most sequences don't have the table. Bubba73 ^{You talkin' to me?} 02:04, 24 February 2018 (UTC)[reply]

February 25

Do these converge to nice numbers?

Do 1⁄(2*3) + 1⁄(3*5) + 1⁄(5*7) + ...

and 1⁄(2*3) + 1⁄(5*7) + 1⁄(11*13) + ...

(where the numbers in the denominators are primes) converge to known numbers? That is, I have their approximate numerical values - do they converge to something concise? Bubba73 ^{You talkin' to me?} 00:52, 25 February 2018 (UTC)[reply]

Not that I know of, but the 1st one is OEIS sequence A210473, and the 2nd one is related to OEIS sequence A089581. Iffy★Chat -- 10:45, 25 February 2018 (UTC)[reply]

Thanks. Bubba73 ^{You talkin' to me?} 17:41, 25 February 2018 (UTC)[reply]

I don't know whether you tried it, but if you have good numerical values, you can check the Inverse Symbolic Calculator to obtain a guess what the number's exact value could be. —Kusma (t·c) 20:26, 25 February 2018 (UTC)[reply]

The first sum (let's denote it

S

) can be calculated in the following way:

S=\sum \limits _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+3)}}-{\frac {1}{6}}

.

On the other hand:

S'=\sum \limits _{n=-\infty }^{\infty }{\frac {1}{(2n+1)(2n+3)}}=2*\sum \limits _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+3)}}-1

from where it is clear that

S={\frac {S'+1}{2}}-{\frac {1}{6}}

.

The value of

S'

can be calculated using the following integral

S'={\frac {1}{2i}}\oint \limits _{C}{\frac {\cot(\pi z)}{(2z+1)(2z+3)}}dz

,

where the closed contour

C

is chosen to go from

-\infty

to

\infty

below the real axis and back above it. The only poles of the function under the integral are those of cotangent. Therefore

S'=0

. The final result is

S=1/3

.

The second sum can be calculated in a similar way.

Ruslik_Zero 20:39, 26 February 2018 (UTC)[reply]

The denominators in the series are actually the primes. Bubba73 ^{You talkin' to me?} 20:47, 26 February 2018 (UTC)[reply]

... so the first sum is probably closer to 0.3 than to 0.33, but I can't prove it. Dbfirs 08:49, 27 February 2018 (UTC)[reply]

Also, I'm pretty sure you can avoid complex analysis and get S = 1/3 using telescoping sums. In case anyone cares

1⁄(1*3) + 1⁄(5*7) + 1⁄(9*11) + ... = π⁄8. — Preceding unsigned comment added by RDBury (talk • contribs) 10:51, 27 February 2018 (UTC)[reply]

@RDBury: Can you give a reference for that result? I’d like to add it to our article List of formulae involving π. Thanks. Loraof (talk) 17:07, 27 February 2018 (UTC)[reply]

See Leibniz formula for π#Convergence. --RDBury (talk) 10:53, 28 February 2018 (UTC)[reply]

Yes, numerically I got about 0.30109317 for the first one and about 0.21042575 for the second one. Bubba73 ^{You talkin' to me?} 15:40, 27 February 2018 (UTC)[reply]

Sorry, I missed that they are primes. But you should explain better how these sequences are constructed. Do they involve only prime pairs? Ruslik_Zero 18:13, 27 February 2018 (UTC)[reply]

No, consecutive primes. The first one: 1st and 2nd primes, then 2nd and 3rd, then 3rd and 4th, etc. The second one: 1st and 2nd primes, then 3rd and 4th, then 5th and 6th, etc. Bubba73 ^{You talkin' to me?} 18:24, 27 February 2018 (UTC)[reply]

February 27

Material conditional

I must have asked this question before: Is $p\to q\equiv \neg (p\land \neg q)$ "intuitively" defined, or can be proven by logic? (not truth-tables)

I found these 2 proofs here and here – are they valid? I do not quite understand them. יהודה שמחה ולדמן (talk) 00:18, 27 February 2018 (UTC)[reply]

Depending on which system of logic you're talking about, p→q might be something you define via a truth table, or as an abbreviation for an expression such as ~p∨q or ~(p∧~q), or → might be taken as a fundamental operation whose behavior is determined by axioms and rules of inference. The problem with something like Proof Wiki is that you can't necessarily pull a single proof out of context; a proof of a given statement depends on other statements which in turn depend on others. The proofs you linked to seem to be using a particular flavor of a subgenre of logic systems called natural deduction, and I'm guessing that the proofs are valid withing the context of one the sources listed as references. Specifically, the linked proofs seem to be taking the → symbol as fundamental with behavior determined by the Rule of Implication and Modus Ponens, but even within natural deduction it might be a defined symbol where the Rule of Implication and Modus Ponens are proved as theorems. So the short answer to the question is that it depends on which system of logic you're talking about. The good news is that different systems of propositional logic are equivalent in the sense that they prove the same statements, at least for classical systems. (In intuitionist logic e.g., the linked proofs are not valid.) But there can be a lot of variation in how you get there; you could build a system of logic built just from the ↓ (nor) operation, or a system using just truth tables with no normal proofs at all. --RDBury (talk) 10:36, 27 February 2018 (UTC)[reply]

The thing is, I understand pretty well why the

\land ,\lor

functions give their answers; it is called "by definition".

But it seems to me that

\to

is not "Well-defined". And now you tell me that Modus Ponens can be proven in some systems... does this ever end?

So I wonder: Is it

p\to q:=\neg (p\land \neg q)

, or is it

\equiv

? יהודה שמחה ולדמן (talk) 11:25, 27 February 2018 (UTC)[reply]

As an example of a similar situation, you could define an even number as one that can be written as x+x, or you could define an even number as one that is divisible by 2. They are two different definitions that turn out to be equivalent. Euclid used the first definition but most modern authors use the second. But it's like you're asking which is 'the' definition of even; the answer is it depends which book you're reading, but in the end it doesn't really matter because the two definitions give the same results. It's the same with logic; five books on logic will have five different sets of axioms, rules of inference, and definitions of the operations. The book here defines → in terms of ~ and ∨, another book might define it a different way, and another might define ∨ in terms of ~ and →. Again, it doesn't matter in the end because the same set of statements can be proved. --RDBury (talk) 12:43, 27 February 2018 (UTC)[reply]

OK, so what are the axioms and rules of inference for mathematical logic? יהודה שמחה ולדמן (talk) 13:41, 27 February 2018 (UTC)[reply]

As RDBury said above, that depends on which system of logic you have in mind. The axioms of propositional logic are different from the axioms of intuitionist logic, for example. Gandalf61 (talk) 16:06, 27 February 2018 (UTC)[reply]

I finally found a proof on youtube, but I keep getting a posting-error. יהודה שמחה ולדמן (talk) 00:29, 2 March 2018 (UTC)[reply]

Meter to French foot and vice versa

Our Units of measurement in France before the French Revolution says that in 1799 one meter was set to 443.296 French lines, or 3 feet 11.296 lines. That corresponds to ⁠9,000/27,706⁠ m. A note follows: This can be shown by noting that 27706 x 16 = 443296 and that 9 x 16 = 144, the number of lignes in a pied. I feel that it must be entirely correct, but I simply can't get it. How does it work? Could you elaborate, step by step, how we get from 1 m = 443.296 lines to 1 ft (144 lines) = ⁠9,000/27,706⁠ m? --Lüboslóv Yęzýkin (talk) 17:49, 27 February 2018 (UTC)[reply]

It's simple arithmetic. If 1 foot is 144 lines and 1 meter is 443.296 lines, then 1 foot is 144/443.296 meter. Multiply both numerator and denominator by 1,000 to get rid of the decimals: 144,000/443,296. Now divide both by the common factor of 16 and you have 9,000/27,706. That is what the footnote is clumsily trying to say. I have no idea why the remaining common factor of 2 was not eliminated to write the fraction in lowest terms as 4,500/13,853. --69.159.62.113 (talk) 00:44, 28 February 2018 (UTC)[reply]

One more point. One of the motivations for adopting metric in France was that each region had its own definitions of different units. So speaking of the "French foot" is wrong; it's only the "Paris foot" or the "French royal foot" or some such. In the article, look back a few paragraphs from where it mentions 9,000/27,706. --69.159.62.113 (talk) 00:47, 28 February 2018 (UTC)[reply]

March 1

Math homework question.

From a tenth grade math book.

A man and his wife walk up a moving escalator. The man walks twice as fast as his wife. When he reaches the top he has taken 28 steps. When she reaches the top she has taken 21 steps. How many steps are visible in the escalator at any one time?

Hint: It was in the "Fractional equations" section. — Preceding unsigned comment added by 2600:1700:2ED1:4020:AD71:72DF:DA0A:A11E (talk) 15:34, 1 March 2018 (UTC)[reply]

I'll give you a starting point. After the man gets to the top, he's covered 28 steps, but x1 steps have been skipped over because the escalator is moving. after the wife gets to the top, she's covered 21 steps, but x2 steps have been skipped over because the escalator is moving. As the escalator doesn't change, 28+x1 = 21+x2. Figure out the equation linking x1 and x2 together from the speeds of the 2 people, and the rest is simple algebra. Iffy★Chat -- 16:39, 1 March 2018 (UTC)[reply]

Here is a worked answer using lots of algebra.

A      answer aka number of steps visible on the elevator
V      velocity of elevator  (independent of speed of man or speed of wife)
L      Length of elevator
TM     time for man
TW     time for wife
SM     Speed for man
SW     Speed for wife

SM = 2 * SW

28 == A - V * TM
TM == L  / SM

21 == A - V * TW
TW == L / SW    

Putting it together

28 == A - V * L  / SM      next substitute SM with 2 * SW 
28 == A - V * L  /  ( 2 * SW )     call this AAA

21 == A - V *  L / SW      call this BBB

Next we calc AAA - BBB

7 == - V * L  /  ( 2 * SW ) -  ( - V *  L / SW )     Next we pull out the common factors  (V * L  / SW )
7 ==  (V * L  / SW ) * (-1/2 + 1 )
7 ==  (V * L  / SW ) * (1/2)                Next we multiply both sides by TWO
14 ==  (V * L  / SW ) 

recall BBB
21 == A - V *  L / SW        Next we substitute (V * L  / SW )  with 14
21 == A - 14                 Next we add 14 to BOTHSIDES
35 == A

110.22.20.252 (talk) 06:50, 2 March 2018 (UTC)[reply]

March 2