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March 21

Questions in Unit-Conversions

1. How much Liters there are in 12.6cm^3? It's not homework! I just want to understand how would you even address such a question? (you can change the numbers if you want...)

2. Is there any formula to work with in this "type" of questions? (what about this formula?)

3. Does the aforementioned question is differs anyway from a question like "How much are 15.6ml from 90dl"? (seems like a typical division question doesn't it?) if indeed there is difference, what is it?

Thanks for all the help. 109.64.137.68 (talk) 20:52, 21 March 2014 (UTC)[reply]

One litre = 1000 cm³ = 1000 millilitres (ml) = 10 decilitres (dl). —Tamfang (talk) 21:13, 21 March 2014 (UTC)[reply]

The question "How much are 15.6ml from 90dl" sounds more like a badly-phrased subtraction question to me. Dbfirs 07:26, 22 March 2014 (UTC)[reply]

Concur... if it is meant as a subtraction, i.e. how much is left if 15.6 mL is removed from 90 dL, then noting that 1 dL = 100 mL, the answer is 9000 − 15.6 = 8984.4 mL = 89.844 dL, though 90 dL (2 sig. fig.) would also be a reasonable answer.

If it is meant to be a proportion, i.e. what proportion of 90 dL is 15.6 mL, then the answer is 15.6 / 9000 = 13 / 7500 as a fraction or 15.6 / 90 = 0.17 % (2 sig. fig.)

With the first question, remember that volume consists of three distance dimensions and so litres must equate to some region in cubic metres, and the litre is defined as 1 L = 1000 cm³ so 1 mL = 1 cm³. EdChem (talk) 07:53, 22 March 2014 (UTC)[reply]

What about the second and third questions. can someone please kindly take a look? 109.64.137.68 (talk) 17:52, 22 March 2014 (UTC)[reply]

How do the answers already given to the third question leave you unsatisfied?

The Youtube lesson is accurate. It could stand to say a bit more explicitly that the method works because the quantity in parenthesis is equivalent to 1. —Tamfang (talk) 00:25, 23 March 2014 (UTC)[reply]

Sorry, I confused. They did answer. How did this guy had 2.2 at the denominator ? How did he got it? 95.35.60.188 (talk) 07:59, 23 March 2014 (UTC)[reply]

1 kilogram = how many pounds? —Tamfang (talk) 08:10, 23 March 2014 (UTC)[reply]

Sorry I'm not familiar with ppunds and I dont have a way to check it. 95.35.60.26 (talk) 08:45, 23 March 2014 (UTC)[reply]

See Pound (mass), but they are not much used outside America and Britain. Dbfirs 09:06, 23 March 2014 (UTC)[reply]

March 22

OEIS text

I have great difficulty interpreting the comment text given with OEIS entries. Take for example http://oeis.org/A086089 which is " Decimal expansion of 3*sqrt(3)/(2*Pi). " and says:

"

OFFSET 0,1

COMMENTS Limiting ratio of areas in the disk-covering problem.

From Daniel Forgues, May 26 2010: (Start)

Consider: A060544, Centered 9-gonal (or nonagonal) numbers, starting with

a(1)=1, P_c(9, n), n >= 1. Every third triangular number, starting with a(1)=1, P(3, 3n-2), n >= 1. Then:

1/(sum_{n=0..infinity} 1/binomial(3n+2,2)) = 1/(sum_{n=1..infinity} 1/binomial(3n-1,2)) = 1/(sum_{n=1..infinity} 1/P_c(9,n)) = 1/(sum_{n=1..infinity} 1/P(3,3n-2)) = 1/(sum_{n=1..infinity} 1/A060544(n)) = this constant. (End)

Also, decimal expansion of product_{n>=1} (1 - 1/(3n)^2). [Bruno Berselli, Apr 02 2013]

LINKS Table of n, a(n) for n=0..101.

EXAMPLE 0.8269933431326880742669897474694541620960797205499609791990...

MATHEMATICA RealDigits[3 Sqrt[3]/(2 Pi), 10, 110]1 (* or, from the third comment: *) RealDigits[N[Product[1 - 1/(3 n)^2, {n, 1, Infinity}], 110]]1 (* Bruno Berselli, Apr 02 2013 *)

"

I understand that this series is that value (though why one would want that value I'm not sure...) and that the value is approximated in the EXAMPLE.

But why has it anything to do with disk covering? What are nonagonal numbers and why are they relevant? And what does the Consider part from 'start' to 'end' actually say - I can't read the formulae in this ascii presentation.

-- SGBailey (talk) 16:46, 22 March 2014 (UTC)[reply]

Disk covering problem links to http://mathworld.wolfram.com/DiskCoveringProblem.html which has the number at the bottom. We also have an article about centered nonagonal numbers: 1, 10, 28, 55, 91, ... Let s = sum of the reciprocals of the centered nonagonal numbers = 1/1 + 1/10 + 1/28 + 1/55 + 1/91 + ... Then 1/s = 3*sqrt(3)/(2*Pi). Mathematicians often want to know the sum of the reciprocals of a sequence. PrimeHunter (talk) 19:40, 22 March 2014 (UTC)[reply]

One thing is OEIS tries to use ascii for mathematical notation, which takes some getting used to. Another is that OEIS is a reference work for people who already know the subject, so it doesn't explain jargon or specialized notation. Also, like many reference works, it has information that is useful to some but gibberish to others, for example Mathematica scripts are cryptic unless you happen to be a Mathematica user. In other words, treat it like a buffet: Take what you understand and leave the rest. --RDBury (talk) 17:40, 23 March 2014 (UTC)[reply]

March 23

Odd function, possible to define cleanly?

In thinking about Monotonic Functions, I was wondering if it was possible to get a function that wasn't monotonicly increasing or decreasing in any subinterval. In other works, for this function f, that for all a and b in the domain in f, that there exists a c, such that f(c) is either greater than both f(a) and f(b) *or* f(c) is less than both f(a) and f(b). Also, can this function be continuous?Naraht (talk) 22:52, 23 March 2014 (UTC)[reply]

The constant function f(x)=0 is not monotonically increasing or decreasing, and in fact is not strictly increasing or decreasing at all. However, I don't think a function you describe (with those inequalities) can be continuous because every continuous function must obey the intermediate value theorem, which would preclude your property from holding on at least some finite interval. Another way to look at it (if f is differentiable) is that the derivative must be zero at every point where f is not monotonically increasing or decreasing, and the constant function is the only function with that property for all x. As said before, the constant function is not strictly increasing or decreasing anywhere.--Jasper Deng (talk) 00:48, 24 March 2014 (UTC)[reply]

OK, that makes sense. Does this give us any information on how to construct a function which is discontinuous?Naraht (talk) 01:50, 24 March 2014 (UTC)[reply]

The "in other words" bit is incorrect: being non-monotonic is not the same as that for every interval, some value of the function in the interval falls outside the range of its endpoints, only that every interval contains both increases and decreases. Ignoring that, real functions clearly exist for which there exists no interval on which it is monotonic. Example: The indicator function that selects rational numbers. The second question is tricker: does a continuous version of such a function exist? Yes, but I expect it to inherently be a fractal function. An example would be to take the function y=x over the domain

[0,1]

, divide the domain into thirds, and replace the line with three lines in a zig-zag of lines going from (0,0) to (1/3,3/2) to (2/3,1/3) to (1,1). Recursively replace every segment with a scaled version. This will be non-monotonic on every subinterval while being continuous, but will have subintervals for which the endpoints are both the minimum and maximum for the interval. —Quondum 02:13, 24 March 2014 (UTC)[reply]

Such continuous functions do exist, and are widely used. The classic example is the Wiener process. The study of such processes is of great importance in finance, physics and engineering. I should note that Quondum is correct in his comments on your "in other words...". It's not obvious to me that the type of function you require will always be fractal if it is to be continuous: I suspect that it is possible to construct a function that matches your criteria that is almost everywhere not self-similar, though I don't have a definition of such a function available - it would require some careful fiddling around to construct. Finally, it should be obvious that a function meeting your criteria can't be everywhere differentiable (follows instantly from your definition). RomanSpa (talk) 16:31, 24 March 2014 (UTC)[reply]

My use of the term fractal does need to be interpreted widely; statistical self-similarity must be included. Nevertheless, is probably better to ignore my mention of fractal due to my use thereof being ill-defined. —Quondum 18:46, 25 March 2014 (UTC)[reply]

The Weierstrass function is an example of continuous a function that is not monotonic on any interval. I guess you can prove using a Baire category argument that "most" continuous functions are nowhere monotonic. —Kusma (t·c) 20:59, 24 March 2014 (UTC)[reply]

The construction of the Blancmange curve is interesting, in that it's almost designed to fit the OP's request. The pictures give a pretty good sense of how it works, even if one doesn't want to wade through all the series descriptions. Also, Pathological_(mathematics)#Prevalence may be of interest, along the lines of Kuma's last sentence. SemanticMantis (talk) 21:18, 24 March 2014 (UTC) EDIT: Sorry, I was speaking from memory. I am now not sure if the Blancmange curve is nowhere monotone... SemanticMantis (talk) 21:26, 24 March 2014 (UTC)[reply]

The Weierstrass function is almost *exactly* what I was looking for, except, it only has that characteristic in finite ranges (say 0-2), but "stretching" that out shouldn't be that difficult.Naraht (talk) 14:36, 25 March 2014 (UTC)[reply]

Huh? It seems to fit your requirement that it be continuous and nonmonotonic on every subinterval of the entire domain. That it is a periodic function does not change this. —Quondum 18:46, 25 March 2014 (UTC)[reply]

Hmm. you are right on that one. I think however if it was *stretched* you would end up with a function where for every pair of values a & b, there would exist c1 and c2 such that f(c1) > f(a) and f(c1) > f(b) and f(c2) < f(a) and f(c2) < f(b) which is more strongly wierd in that regard.Naraht (talk) 10:27, 26 March 2014 (UTC)[reply]

March 24

March 25

Abstract Algebra Integral Domains

Let A be a finite integral domain. Prove the following: If A has characteristic 3, and 5·a=0, then a=0.

I have For the sake of contradiction, assume a≠0. 5·a=0 a+a+a+a+a=0 char(A)=5 which is a contradiction to the given statement char(A)=3 Therefore, a=0.

Any suggestions? I am not sure how to get a contradiction to a≠0. — Preceding unsigned comment added by Abstractminter (talk • contribs) 00:02, 25 March 2014 (UTC)[reply]

This looks like homework to me. A suitable hint would be: what is the definition of an integral domain? RomanSpa (talk) 02:46, 25 March 2014 (UTC)[reply]

Is the characteristic 3 or is it 0? I don't see how a finite ring can have characteristic zero. —Kusma (t·c) 08:26, 25 March 2014 (UTC)[reply]

Tribbonacci numbers

Wikipedia gives only one definition to the Tribbonacci numbers, but OEIS gives 3 different definitions. For what these definitions are, go to OEIS and type Tribonacci and the first 3 definitions are the 3 definitions it gives. Georgia guy (talk) 17:15, 25 March 2014 (UTC)[reply]

As far as I can see, the 3 OEIS sequences simply differ in the values of the first three terms (0,0,1), (1,1,1) or (0,1,0). The recurrence relationship is the same for all 3 sequences. Gandalf61 (talk) 17:27, 25 March 2014 (UTC)[reply]

But Wikipedia thinks thtat (0,0,1) is by definition the Tribbonacci sequence. Any thoughts?? Georgia guy (talk) 17:30, 25 March 2014 (UTC)[reply]

Exactly the same convention applies to Tribonacci sequences as to Fibonacci sequences. "The" Tribonacci sequence begins 0, 0, 1 but you can define "a" sequence beginning with any numbers you choose. Dbfirs 17:41, 25 March 2014 (UTC)[reply]

Unlike the Fibonacci sequence, where starting with 1, 0 or 0, 1 or 1, 1 only results in different offsets, in the Tribonacci sequence you get different numbers for different starting values, even when the starting values seem to be reasonable extensions of the Fibonacci definition. So the definition of "the" tribonacci numbers is somewhat vague. The WP article being referenced is poorly sourced and what sources are given are not always reliable, so there may be a certain amount of OR there that shouldn't be taken too seriously. OEIS is somewhat inconsistent as well sometimes referring to A000073 and sometimes to A001590 as "the" tribonacci numbers. I'm guessing there isn't really a consensus in the literature on what a tribonacci number actually is. --RDBury (talk) 21:39, 25 March 2014 (UTC)[reply]

I agree with everything that RDBury writes above except that he doesn't mention the possibility of starting a Fibonacci sequence with numbers other than "1, 0 or 0, 1 or 1, 1". Starting with other different values gives a different sequence that is often called a Fibonacci sequence, though possibly it ought to be called a Fibonacci-type sequence See 1, 4, 5, 9, 14, 23, 37, 60, 97 for example. Dbfirs 08:09, 26 March 2014 (UTC)[reply]

Using 1-s2/s1 as a measure of one-dimensionality

Let $s_{i}$ be the singular values of a nonzero $m\times n$ matrix $M$ with $m>n$ , sorted in descending order such that $s_{i}>s_{i+1}$ .

In particular $s_{1}$ is the largest singular value.

Then, the quantity $q=1-s_{2}/s_{1}$ is 1 if the columns of $M$ are linearly dependent, i.e. if the data in $M$ is purely one-dimensional. In any case $0\leq q\leq 1$ . High $q$ means that $M$ is nearly one-dimensional, low $q$ means a second dimension is also important (and maybe also a third, fourth ...).

Is this quantity known in the mathematical or statistical literature? Does it have a name?

82.69.98.189 (talk) 19:44, 25 March 2014 (UTC)[reply]

March 26

Distribution of points on a sphere

I've been using this algorithm for coming up with a uniform distribution of points on an n-sphere (Python notation): X = [random.gauss(0, 1) for i in range(n)], t = math.sqrt(sum([i**2 for i in X])), Y = [X[i] / t for i in range(n)], return Y. Does the resulting Y distribution have a name, or can a closed form be derived for it? 70.190.182.236 (talk) 20:39, 26 March 2014 (UTC)[reply]

This is the unique rotationally-invariant probability measure on the sphere. I'm not sure it has a name as such, but it can be obtained as a symmetry reduction of the Haar measure on the group of rotations SO(n). The fact that your probability distribution has this property follows from the fact that if

X_{1},\dots ,X_{n}

are iid (standard) Gaussian random variables, then the n-dimensional random variable

X=(X_{1},\dots ,X_{n})

is normally distributed with pdf

{\frac {1}{(2\pi \sigma ^{2})^{n/2}}}\exp(-x^{T}x/2\sigma ^{2})

, which is rotationally invariant. Sławomir Biały (talk) 21:02, 26 March 2014 (UTC)[reply]

Note that this topic was discussed here before (although not specifically what to name the method you describe): See Wikipedia:Reference_desk/Archives/Mathematics/2007_August_29#Distributing_points_on_a_sphere and Wikipedia:Reference_desk/Archives/Mathematics/2010_April_30#Even_distribution_on_a_sphere. StuRat (talk) 21:16, 26 March 2014 (UTC)[reply]

I think those are about optimally distributing multiple points on a sphere, whereas the OP is asking about the probability distribution of a single point chosen "uniformly at random" on a sphere. Sławomir Biały (talk) 21:20, 26 March 2014 (UTC)[reply]

March 27

Lottery probability

Hi

let's say in a lottery game one has to choose k numbers out of N. In such situation, it's obvious the probability to choose exactly the right k numbers is 1/C(N,k). But How do I find the chanaces to choose correctly just j (j<=k) of the k numbers (I am looking for the derivation of the answer)? Thanks, 212.179.61.122 (talk) 09:15, 27 March 2014 (UTC)[reply]

To be sure I understand: you choose k numbers, the lottery draws k numbers, and you win if you have at least j in common? In that case, let's start by finding the probability of having exactly j in common, which we'll do by counting the number of ways to have j in common. First, I select the common j from the k which were drawn: C(k, j). Then I fill in the remaining k - j from the undrawn N-k: C(N-k, k-j). So

{\frac {C(k,j)\cdot C(N-k,k-j)}{C(N,k)}}

is the probability of getting exactly j in common. If you want at least j in common, then sum from j to k:

\sum _{i=j}^{k}{\frac {C(k,i)\cdot C(N-k,k-i)}{C(N,k)}}

.--80.109.80.78 (talk) 09:59, 27 March 2014 (UTC)[reply]