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December 2

Number of Factorizations

Hi, I know that $\mathbb {Z} _{n}$ (ring of integers modulo n) is not a Unique Factorization Domain (for non-prime n), so there're numbers with non-unique factorization. So, my question is: how many factorizations modulo n exist for a number m? עברית (talk) 07:56, 2 December 2016 (UTC)[reply]

Perhaps I'm being dim, but do you have an example showing that Z_n isn't a UFD? --RDBury (talk) 00:27, 3 December 2016 (UTC)[reply]

They mean for instance 1=1x1=2x3=4x4 (mod 5). I don't know what the mean by factorizations though - one could have any number of factors like for instance 2x2x2x2 or 4x3x3. Dmcq (talk) 00:44, 3 December 2016 (UTC)[reply]

Z_n is not a domain (except if n is prime), so it can't be a UFD. That doesn't mean it doesn't have the unique factorization property though. But I just thought of an example: 6=2⋅3=2⋅3⋅3=2⋅3⋅3⋅3=2⋅3⋅3⋅3⋅3... (mod 12) so the number of factorizations is infinite. --RDBury (talk) 01:09, 3 December 2016 (UTC)[reply]

@Dmcq: Z₅ is a field, so all values (except 0) are units, and there are no primes. None of your examples really count as distinct "factorizations" in this context; factors of units are ignored. --Trovatore (talk) 01:22, 3 December 2016 (UTC)[reply]

Thanks. I was trying to figure out what they meant. Dmcq (talk) 11:46, 3 December 2016 (UTC)[reply]

Website that helps you make math questions

Is there a website that helps you make a math question like for example I want to make a math question about set theory. In this question, I want to make three circles Venn diagram connecting together and each circle has a number like 23 people like comedy genre movies only, 24 people like action genre movies only, 19 like horror genre movies only, 5 people like comedy and action movies but not horror, 6 people like action and horror movies but not comedy, and 10 people like comedy and horror movies but not action and in the middle of the diagram, 10 people like all three movie genres. Then, I make questions like what is A ∩ C ∩ H, A ∩ C ∪ H, and etc? Donmust90 (talk) 15:24, 2 December 2016 (UTC)Donmust90Donmust90 (talk) 15:24, 2 December 2016 (UTC)[reply]

I would approach this by Googling for blank Venn diagrams with the appropriate number of circles, then edit them to add the elements you want. You might consider just putting single letters and maybe digits in the diagram, allowing 26 or 36 elements (you can add Greek letters, etc., if you need more), and then use an index to map each to the full description. Otherwise, the Venn diagrams can get very crowded quickly. MS Paint allows you to add text to an existing pic, as do any number of software packages. StuRat (talk) 15:39, 2 December 2016 (UTC)[reply]

I use Ordinal Fractions in those situations.

100 C: comedy
200 no comedy
010 A: action
020 no action
001 H: horror
002 no horror
122 23
212 24
221 19
112 5 
211 6
121 10
111 A ∩ C ∩ H: 10

See [1]. Bo Jacoby (talk) 15:37, 5 December 2016 (UTC).[reply]

December 6

Vector Problem

If dU/dt=WxU and dV/dt=WxV, prove that d(UxV) /dt=Wx(UxV).
This is not a homework problem. 47.29.88.170 (talk) 08:52, 6 December 2016 (UTC)[reply]

You need the identity

\mathbf {V} \times (\mathbf {W} \times \mathbf {U} )+\mathbf {U} \times (\mathbf {V} \times \mathbf {W} )+\mathbf {W} \times (\mathbf {U} \times \mathbf {V} )=0

I can't immediately spot this on either the Vector algebra relations or Vector calculus identities page, but you can easily prove it by expanding each of the terms according to

\mathbf {U} \times (\mathbf {V} \times \mathbf {W} )=(\mathbf {U} \cdot \mathbf {W} )\mathbf {V} -(\mathbf {U} \cdot \mathbf {V} )\mathbf {W}

and observing that they then cancel. --catslash (talk) 10:26, 6 December 2016 (UTC)[reply]

See Triple product#Vector triple product (Jacobi identity) --catslash (talk) 10:33, 6 December 2016 (UTC)[reply]

This is not a homework problem. Then what is it? Tigraan^{Click here to contact me} 12:29, 6 December 2016 (UTC)[reply]

Angle quintisection

What formula is there for the sine of quintuple angle as a function of sine of simple angle? Can it be inverted to find out the sine of a fifth of an angle?--82.137.11.135 (talk) 14:09, 6 December 2016 (UTC)[reply]

[EDITED 17:28, 6 December 2016 (UTC) per RDBury] Thanks to List_of_trigonometric_identities#Angle_sum_and_difference_identities, one can compute $\sin(5\theta )=\sin(\theta )^{5}+5\sin(\theta )\cos(\theta )^{4}-10\sin(\theta )^{3}\cos(\theta )^{2}$ (I asked WolframAlpha, actually, as any lazy person would do). That formula ~~cannot be reformulated as a "normal" function purely of the sine (you can convert cosine in sine only if you know the sign), and~~ certainly not be inverted; by that I mean there is no function such that $\sin(\theta )=f(\sin(\theta )))$ (because $\sin(0)=\sin(\pi )$ so the right hand side results must be equal for these values, but $\sin(0/5)\neq \sin(\pi /5)$ ).

This is no guarantee of the possibility of "angle quintisection" (in the sense that you could construct the fifth of a given angle with straight lines and circles). See Angle_trisection#Proof_of_impossibility (I think a similar argument would apply to quintisection as well, but hunting down the polynomial's roots could prove harder). Tigraan^{Click here to contact me} 15:02, 6 December 2016 (UTC)[reply]

Slight copy error, it should be

\sin(5\theta )=\sin(\theta )^{5}+5\sin(\theta )\cos ^{4}(\theta )-10\sin ^{3}(\theta )\cos ^{2}(\theta ).

You then get a polynomial in sin using cos²θ=1-sin²θ. The proof of the impossibility of angle trisection actually works for any odd number. So angle n-section is only possible (for general angles) when n is a power of 2. In fact, one only has to prove that the regular n²-gon is impossible to construct. --RDBury (talk) 15:55, 6 December 2016 (UTC)[reply]

In other words, you have

\sin(5\theta )=16\sin ^{5}(\theta )-20\sin ^{3}(\theta )+5\sin(\theta )

.

However the OP didn't ask about geometric construction, only about extracting from this a formula for

\sin(\theta )

given

\sin(5\theta )

(equivalent to finding

\sin({\tfrac {\theta }{5}})

given

\sin \theta

).

The only real problem with this is that it requires solving a quintic polynomial, which is impossible by usual means - this is the Abel–Ruffini theorem. -- Meni Rosenfeld (talk) 16:17, 6 December 2016 (UTC)[reply]

True, I only mentioned the geometry thingie because of the thread title. (Note that although a third-degree polynomial can be factorized algebraically, it involves cubic roots which are non-constructible generally speaking.)

As for the Abel–Ruffini theorem, it only means there is no general factorization method for quintic polynomials (with "usual means"). Maybe this particular quintic polynomial,

16X^{5}-20X^{3}+5X-a

, can be factorized. I doubt it, because it would mean finding a factorization that works for every a. The proof section of the ABT says it remains valid for a given set of algebraically independent coefficients, but the set

\{16,0,-20,0,5,a\}

is not that (the first four coefficients are rational). Tigraan^{Click here to contact me} 17:28, 6 December 2016 (UTC)[reply]

Thanks for your answers. Is the situation similar for cosine case?--82.137.9.227 (talk) 19:56, 6 December 2016 (UTC)[reply]

What is the situation when applying a numerical root-finding algorithm for the mentioned quintic equation for every a?--82.137.9.227 (talk) 20:02, 6 December 2016 (UTC)[reply]

I made a mistake in the polynomial above, I've corrected it now. It doesn't materially change the results. I've also taken the liberty of updating Tigraan's response accordingly.
It's true that it doesn't immediately follow from the theorem that this particular polynomial is unsolvable, but it's almost certain.
For cosine it's quite similar - $\cos(5\theta )=16\cos ^{5}(\theta )-20\cos ^{3}(\theta )+5\cos(\theta )$ .
There's no real problem with solving it numerically, but you can only do that once a is given (if you do it with a symbolic a, you quickly get unwieldy expressions), and it might be easier to calculate the sine directly with Taylor series or whatnot. -- Meni Rosenfeld (talk) 11:16, 7 December 2016 (UTC)[reply]

(I'm an amateur here, so feel free to correct me.) I could easily have messed up typing all these numbers into Mathematica, but I just checked the Cayley resolvent, and it appears to have a rational root at

z=125/16

for all

a

. This would suggest that this is a family of rare quintics which are in fact solvable. 129.234.195.173 (talk) 16:20, 7 December 2016 (UTC)[reply]

A little more digging suggests that it's actually an example of de Moivre's quintic. 19:36, 7 December 2016 (UTC) — Preceding unsigned comment added by 129.234.195.173 (talk)

Interesting.

I double-checked the resolvent and it checks out. I only wonder why Mathematica won't return the roots in radical form.

I'm not an expert on this myself but I'd say Bo (in a comment below) is on to something, the solvability of this is probably connected to the fact that in complex numbers you can find it by simply taking a fifth root. -- Meni Rosenfeld (talk) 01:09, 8 December 2016 (UTC)[reply]

See Chebyshev polynomials. — 86.125.199.30 (talk) 00:45, 7 December 2016 (UTC)[reply]

Let $a=\cos(\theta )+i\sin(\theta )$ . The problem is to solve the quintic equation $z^{5}=a$ . The solution can be expressed by roots ( $z={\sqrt[{5}]{a}}$ ), but cannot be constructed by compas and straightedge. Bo Jacoby (talk) 22:39, 7 December 2016 (UTC).[reply]