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May 13

The Number Between 0 and 1 That Doesn't Exist

\int _{0}^{1 \over 2\pi }{\left|\sin {1 \over x}\right|}dx

Since this number neither oscillates ( because of the presence of the absolute value function ) nor grows unrestrained towards infinity ( being obviously smaller than ${\tfrac {1}{2\pi }}$ ), it would logically follow that it converges towards a finite value on the ( 0, ${\tfrac {1}{2\pi }}$ ) interval, because it is an ever-growing and upper bound sum of positive numbers. Not having a limit in 0 does not change or influence this fact. But what is its exact or approximate value ? — 79.113.209.157 (talk) 02:29, 13 May 2013 (UTC)[reply]

It seems to be around 0.1005. 96.46.198.58 (talk) 03:41, 13 May 2013 (UTC)[reply]

P.S. Given that the average value of |sin x| is 2/π, it's not surprising that the answer is close to 1/π². But I imagine it would take a good deal of work to bound the difference by hand. 96.46.198.58 (talk) 03:53, 13 May 2013 (UTC)[reply]

This is difficult and unstable to evaluate numerically. You can convert the equation into the equivalent summation:

\int _{0}^{1 \over 2\pi }{\left|\sin {1 \over x}\right|}dx=\sum _{k=1}^{\infty }{2Ci(2k+1)-Ci(2k)-Ci(2k+2)}

where Ci(x) is the cosine integral. That probably doesn't help at all but it gets rid of the absolute value... — Preceding unsigned comment added by 24.255.30.187 (talk) 07:51, 13 May 2013 (UTC)[reply]

The integral is also equal to ${1 \over 2\pi }\int _{0}^{1}\left[-{1 \over (1-t)^{2}}-{1 \over (2-t)^{2}}-{1 \over t^{2}}-{1 \over (1+t)^{2}}+\pi ^{2}\csc ^{2}(\pi t)\right]\sin(\pi t)\,dt\approx 0.100466.$ Not sure how this can be simplified further. 96.46.198.58 (talk) 10:10, 13 May 2013 (UTC)[reply]

More digits: 0.1004663596222... — 24.255.30.187 (talk) 22:22, 13 May 2013 (UTC)[reply]

I was sort of thinking more of an approach along the lines of

\int _{0}^{1 \over \pi }{\left|\sin({\tfrac {1}{x}})\right|}dx=\sum _{k=1}^{\infty }\int _{1 \over (k+1)\pi }^{1 \over k\pi }{\left|\sin({\tfrac {1}{x}})\right|}dx

in the hopes that we might possibly get something like

{\tfrac {\operatorname {Area\ of} \ \sin({1 \over x})\ \operatorname {between} \ {1 \over k\pi }\ \operatorname {and} \ {1 \over (k+1)\pi }}{\operatorname {Area\ of\ the\ rectangle\ in\ which\ it\ is\ inscribed^{*}} }}=

either some constant, or some simpler function in k. [* which area is equal to

{\tfrac {1}{k(k+1)\pi }}

] — 79.113.243.187 (talk) 22:35, 13 May 2013 (UTC)[reply]

Indeed ! The limit of the above-described ratio approaches

{\tfrac {2}{\pi }}

as k approaches infinity, meaning

\int _{0}^{1 \over \pi }{\left|\sin {1 \over x}\right|}dx\approx {\frac {2}{\pi ^{2}}}

! — 79.113.243.187 (talk) 23:05, 13 May 2013 (UTC)[reply]

Okay, the integral can be rewritten as

\int _{0}^{\pi /2}[\csc ^{2}x-1/x^{2}]\sin x\,dx-\int _{\pi /2}^{\pi }(\sin x)/x^{2}\,dx+\int _{\pi }^{2\pi }(\sin x)/x^{2}\,dx.

According to Wolfram Alpha, this is

\mathrm {Ci} (2\pi )-2\mathrm {Ci} (\pi )-1+\gamma +\log 2\approx 0.1004663596222811304000035025372332217305344978423192,

where γ is the Euler constant and Ci is the cosine integral. 96.46.198.58 (talk) 01:15, 14 May 2013 (UTC)[reply]

Resolved

Knight moves on an infinite board.

Does anyone have any suggestions or solutions to the problem of how many squares on an infinite chessboard that a Knight can reach in exactly k moves? In no more than k moves?Naraht (talk) 16:11, 13 May 2013 (UTC)[reply]

In 0 moves it can reach just 1 square: the starting one.

In 1 move it can reach 8 squares, surrounding the starting position.

In the second move it can step back of forward or aside, so for n ≥ 2 it can reach (2n + 1)² squares which are at most n columns or rows apart from the starting position.

CiaPan (talk) 16:57, 13 May 2013 (UTC)[reply]

It is a bit more complicated than that. As far as I can see you need to enumerate them by hand for the first 5 moves but after that you get a convex figure filled with either all black or all white squares reached so after that it is simpler. Dmcq (talk) 19:50, 13 May 2013 (UTC)[reply]

This is a nice example of multivariate generating functions. Consider the polynomial

(xy^{2}+xy^{-2}+x^{2}y+x^{2}y^{-1}+x^{-1}y^{2}+x^{-1}y^{-2}+x^{-2}y+x^{-2}y^{-1})^{k}

. Then the coefficient of

x^{a}y^{b}

is the number of ways of reaching square

(a,b)

in 'k' moves. So the number of terms in the expansion is your answer. HTH, Robinh (talk) 21:24, 13 May 2013 (UTC)[reply]

And for "not more than 'k' moves", add a '1' to the term in brackets. Robinh (talk) 03:09, 14 May 2013 (UTC)[reply]

Oops, it was 'Knight' and I've read 'King'

--CiaPan (talk) 07:02, 14 May 2013 (UTC)[reply]

chord question

How many triads (3 note chords) can I get from the chromatic scale (12 tones).

Is it 12! / 3! 9! =220 ?

Hovever CEG, GEC are not the same so I am not sure if my formula is correct.

Ap-uk (talk) 20:46, 13 May 2013 (UTC)[reply]

Unfortunately the word "chord" has more than one meaning, so it isn't clear what you are asking. It might help if you would explain why you do not consider CEG and GEC the same. Looie496 (talk) 20:59, 13 May 2013 (UTC)[reply]

Actually I think inversions of these musical chords would be the same in this context, its just that someone posted an article about there being 110 three note chords in the chromatic scale , I think its 220 . — Preceding unsigned comment added by Ap-uk (talk • contribs) 21:24, 13 May 2013 (UTC)[reply]

As Looie mentioned, the number of chords depends on what you consider to be a distinct chord. For example, if you consider the A major chord starting on A440 to be different than the A major chord starting on A220, then you'll have many more chords than if you consider them the same. If you consider any arbitrary assortment of note names to be the same chord regardless of ordering, then the 12-choose-3 combination formula is appropriate. But typically chord aren't thought of a just a collection of notes, rather they're typically viewed as a fundamental note followed by two intervals (with chords differing only in which octave they start in being viewed as the same). How many chords depends on how loosely you allow the intervals to go. Unrestricted, and we're back to effectively the exhaustive list of pitch combiations. If you limit it to progressive sub-octave intervals (e.g. a major chord is a minor third stacked on top of a major third), you have a choice of 12 fundamental notes, 11 sub-octave intervals for the next higher note, and then 11 sub-octave intervals for the note above that, giving 1452 chords. This classifies C₄/G₄/C₅ as a "chord", which might not be what you want. Another definition would permit you the choice of 12 fundamental notes, but limit you to sub-octave intervals off of the fundamental (E.g. a major chord is a major third and a perfect fifth.) In this case you'd have 12 choices of fundamental notes, with 11-choose-2 choices of the interval combinations. This would result in 660 chords, including rather bizzare ones like C/B♭/B. If you had further restrictions on which intervals or notes could be in the chord for it to actually count as a chord, then then number gets further reduced. -- 71.35.111.68 (talk) 06:41, 14 May 2013 (UTC)[reply]