Wikipedia:Reference desk/Archives/Mathematics/2016 March 29

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March 29

Complex number writen "s" (instead of "z") for the Riemann Zeta function ?

Good day to the thinkers on the history of mathematics. The most classical writing for the complex numbers is "z"; That is true in French, in English and in German. But when it comes to the Riemann zeta function, the most commonly used letter is "s". See [1]. Where does this strange shifting come from ? I thank you for for your thoughts. NB: I just asked the same question on the French Reference desk, will you beat the Froggies? Easy job!--Jojodesbatignoles (talk) 07:17, 29 March 2016 (UTC)

That's the notation used by Riemann himself in his 1859 publication On the Number of Primes Less Than a Given Magnitude:

In this paper Riemann defines his ζ function as the analytic continuation of Euler's.

$\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}$

The usage of s as a variable may well go back to Euler but I don't know. For what it's worth, as far as I could see,

H. M. Edwards (1974). Riemann's Zeta Function. Academic Press. ISBN 0-486-41740-9.

does not give an explanation for Riemann's choice of notation. —Tobias Bergemann (talk) 09:37, 29 March 2016 (UTC)

PS: Damn, too slow. —Tobias Bergemann (talk) 09:40, 29 March 2016 (UTC)

Indeed Tobias, I got a faster answer from the Frogies (my countrymen) but there are some explanations for this masterpiece: they got the question before the "post" in English and they got it at 9h am (I don't know where you write from). Thanks.--Jojodesbatignoles (talk) 09:53, 29 March 2016 (UTC)

mathmatical delimas abounding

If a sphere is 20/1000 of a mm in diameter and its container is 5cm , what is the maximum number of spheres possible? Now work this problem another way, what is the total radius of 500,000 of said spheres? Now, another math equation to compare this to is 20/1000 of a mm in diameters in a row, ³√500,000 on width, hight, length. Do you find may quandary(questioning)? Use the equation V=(4/3)πr³ in question number 2, which my omit the negative spaces but will illuminate what I have found in relation to question number 3. Find the volume of 500,000 spheres and then the radius of such a containment space (in which the volume has remained the same while shape may have become ambiguous, i.e. altering the shape of a water ballon does not change its containing volume without compression.). This all started due to potentially centuries old data on ovary cell count for oocytes(egg cells) which I am trying to refute. I was a trained as a medical lab tech, the opposing party is a physician. He gave question number three as his explanation of containing such a number (in acuality speaking on the proposed one million oocytes at birth, I broke it down to half of that given I could find the size of an ADULT ovarian fossa(space) as 5cm and an adult ovary at approximately 4cm x 2cm x 3cm. Using 5cm to give "the benefit of the doubt", a phrase we have where I am originally from in America.), I however give questions 1, 2, or "4" to refute this data. In doing such I find a delima in the difference between 20/1000 mm diameter in a cube like section of 100x100x100 and the actual radius using volume and the reversal of the volume equation leading to radius. Do the math and you will see. Feel free to drop a line on my talk page if you don't want to post the answers here. I needed math activiam help and this is where I was directed to. Math geeks unite! (38.129.255.217 (talk) 23:11, 29 March 2016 (UTC))

er? "If a sphere is 20/1000 of a mm in diameter and its container is 5cm". first question, is the container a sphere? 175.45.116.66 (talk) 23:13, 29 March 2016 (UTC)

1) You didn't list the shape of the container; is it also a sphere ?

2) For the maximum that could be packed in, you would want an arrangement like hexagonal close pack.

3) For the fewest, you would want an arrangement like primitive cubic.

4) A random arrangement will be irregular and somewhere between these two extremes.

See close-packing of equal spheres. Also, in the case of animal cells, they will squeeze into shapes that pack more tightly than spheres, if pressure is applied. StuRat (talk) 23:16, 29 March 2016 (UTC)

I was hoping some one else would run these numbers with me. The actual shape is ambiguous, most similar to a football but not as tall from lace side to bottom. In running the equations you may find the concern I have found in the numerical difference between the values found in question 2 and question 3, also you may find what I am considering a mathematical impossibility to be born with one million primordial follicles, though the doctor used the diameter of a primordial oocyte without surrounding follicle at 20μ, so I used that number in attempt to prove that that also was not possible. (Crlinformative (talk) 23:47, 29 March 2016 (UTC))

Your message reads like stream-of-consciousness writing. You are asking people to help you; it would be polite to make some effort to express yourself clearly. I strongly suggest you rewrite your comment using full sentences, complete ideas, numbering questions if you want to refer to them by number, etc. --JBL (talk) 23:33, 29 March 2016 (UTC)

I corrected the word is to in, took out a typed extra "a", and added a comma. At which Part do you get lost in it. I will be happy to clarify. (Crlinformative (talk) 23:49, 29 March 2016 (UTC))

Your message is a massive ramble. You say that you want to refute something, but there is no clear statement of any kind in it. Why don't you write out, carefully, the argument that you find problematic, and then indicate afterwards what parts of it you object to or what you believe it contradicts? --JBL (talk) 00:06, 30 March 2016 (UTC)

I will send you a message so others may still assist in the mathematics.(Crlinformative (talk) 00:14, 30 March 2016 (UTC))

I agree with JBL, that it's very difficult to tell what you are asking. I get a general idea of what it's about, but please show us all you math and where you have found an inconsistency with something published previously. Only then will we be able to "do the math". StuRat (talk) 01:13, 30 March 2016 (UTC)

Some clarification has been made on my talk page that might be the basis for common understanding. (I do not have more time tonight to deal with this, but I will take it up again in 12 or so hours if no one else has.) --JBL (talk) 02:31, 30 March 2016 (UTC)

Here is another way of looking at it: folliculogenesis Another way to assess the problem I encounter with the claims that a person is born with 1,000,000 primordial follicles in the ovaries is this: Look at image 3, this is declared a secondary follicle with an approximate diameter of 0.2mm. That is 1/1000 of the 2cm length in 4cm x 3cm x 2cm given as ovary size here on the page on ovary. This is less than ten times larger than size of a primordial follicle at 0.03mm. For the simplification of math lets say that the primordial follicle was even smaller being .02mm. That would make it 1/10,000 of the lesser length in the size of an adult ovary. That gives you the dimensions 2/10,000 x 1.5/10,000 x 1/10,000 of an ovary being the size of a primordial follicle. This yields the possibility of less than 30,000 total primordial follicles(being that they are actually .03 - .05mm in diameter) in an adult ovary if it contained solely primordial follicles. It does not however. Therefore, I conclude it to be mathematically impossible for an ovary to contain this primordial follicle cell count, and I postulate that they are actually formed inside the ovary just as with spermatogenesis. Do the math your self if you don't believe. Also try using V=(4/3)πr3 for volume and obtain a total volume of 1,000,000 primordial follicles (or 500,000 as per ovary) and then work backwards to gain a minimum required radius of such a volume r=3√((V÷(4/3))÷π). You will find the same error in cell count of primordial follicles related to required radial size if the ovary's dimensions were adjusted to a sphere shape 3x3x3. — Preceding unsigned comment added by Crlinformative (talk • contribs) 04:37, 30 March 2016 (UTC)

Are you, perhaps, thinking that volume scales with three times the length, and not the cube of the length? If you have a cubic box and pack it with tiny cubes 1/1,000 the size of the box, you would be able to fit 1,000³ = 1,000,000,000 = 10⁹ = 1 billion tiny cubes in your box. With other shapes, such as spheres, you do have to take void spaces into account. See our article on sphere packing. For equal spheres the densest packing uses approximately 74% of the volume. Random packing of equal spheres generally have a density around 64%. For finite spaces, edge conditions will reduce this a bit, but not by much for the scales you are asking about. -- ToE 16:25, 30 March 2016 (UTC)

You originally asked If a sphere is 20/1000 of a mm in diameter and its container is 5cm , what is the maximum number of spheres possible? Assuming that your 5cm container is a sphere, then the answer is (50mm / 0.02mm)³ · packing_density = 2,500³ · packing_density = 15,625,000,000 · packing_density = 10,000,000,000 = 10 billion for a packing density of 64%. So mathematically, there would be no problem packing in a measly half-million. -- ToE 16:38, 30 March 2016 (UTC)

TOE, the method you used already accounted for packing density, by assuming primitive cubic packing. If you had calculated the volume of the large and small spheres, and used that to figure out how many times the smaller volume fit into the large volume, then you would need to apply the packing density. StuRat (talk) 17:23, 30 March 2016 (UTC)

I'm not following you, Stu. (D_Large/D_Small)³ = V_Large/V_Small. -- ToE 18:39, 30 March 2016 (UTC)

If the shape of the units was cubes, that would be true, as each cube offset one side length from the next would touch an adjacent cube on all points of all faces. But with spheres, when you set them up as each offset by one diameter from the last, in all directions, you have them touching at only one tangent point each, with lots of empty space between them at all other points. This is what the primitive cubic packing density describes. So, you've already done it, and don't need to account for it again.

Put another way, the wasted space of a sphere inscribed in a cube applies to both the smaller units (in a primitive cubic arrangement) and the container, and at the same ratio, thus cancelling each other out. StuRat (talk) 00:57, 31 March 2016 (UTC)

The OP is asking about cells, and appears to believe that two numbers differ by orders of magnitude; the details of the geometry are almost certainly not relevant to the conclusion. --JBL (talk) 01:05, 31 March 2016 (UTC)

Yes, the packing density is a minor factor, but still should be considered for the best possible answer. StuRat (talk) 01:10, 31 March 2016 (UTC)

ToE's analysis was correct. Stu, you said "If you had calculated the volume of the large and small spheres, and used that to figure out how many times the smaller volume fit into the large volume, then you would need to apply the packing density." But that's exactly what he did. The volume of the large sphere is (4*pi/3)*R_large³ and the volume of each small sphere is (4*pi/3)*R_small³. The factor of (4*pi/3) cancels, so the ratio of the volumes is (R_large/R_small)³, which is of course the same if you replace radius by diameter. You still need to apply the packing density factor. The fact that there's a cube in the expression doesn't imply that cubic packing is involved, if that was your confusion. It may be a really poor packing with a packing density of 0.1, not a cubic packing at all. Where does that enter into the expression if you don't apply it at the end as ToE does? CodeTalker (talk) 02:53, 31 March 2016 (UTC)

OK, maybe I was looking at it wrong. My calcs:

Vcontainer = (4*pi/3)R3 = (4*pi/3)(25mm)3 = 65,450 mm3 
Vcell = (4*pi/3)r3 = (4*pi/3)(0.01mm)3 = 0.00000418879 mm3 
Vcontainer / Vcell =  15,625,036,538
15,625,036,538 × packing density = 15,625,036,538 × 0.64 = 10 billion

Looks like both methods agree. StuRat (talk) 03:21, 31 March 2016 (UTC)

Now lets try it using the 4cm × 2cm × 3cm dimensions for the ovary, and assume it to be a rectangular prism:

Vcontainer = 40mm × 20mm × 30mm = 24,000 mm3 
Vcell = (4*pi/3)r3 = (4*pi/3)(0.01mm)3 = 0.00000418879 mm3 
Vcontainer / Vcell =  5,729,578,231
5,729,578,231 × packing density = 5,729,578,231 × 0.64 = 3 and 2/3 billion

Figuring the corners in reality are rounded inward, we could maybe reduce that down to 3 billion or so. So, yes, a half million seems quite realistic, and in fact, rather sparsely populated. StuRat (talk) 03:31, 31 March 2016 (UTC)

According to the OP, discussion on StuRat's page resolves the mathematical issue in question to the OP's satisfaction. --JBL (talk) 13:41, 31 March 2016 (UTC)