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December 30

What is the minimum amount of information to rotate a point in 3D Cartesian space?

Suppose you are given a 3D cartesian point {x,y,z} = {1,2,3}. What is the minimum amount of (additional) information you need to rotate that point in 3D space to another point in 3D space? For many years I thought all I need is

1. X coordinate of rotation origin
2. Y coordinate of rotation origin
3. Z coordinate of rotation origin
4. X coordinate of the normal vector
5. Y coordinate of the normal vector
6. Z coordinate of the normal vector
7. Angle of rotation in either degrees or radians
8. Rotation using the normal vector using which rule? Right Hand Rule or Left Hand Rule?

But I was mistaken, because I got the wrong answers to my problem. It took me a few days until I realized what my problem is.

I am missing the Cartesian Axis Type.

For example:

Cartesian Axis Type 1: X-axis is towards the observer, Y-axis is to the right (from observer's point of view), Z-axis is upwards (from observer's point of view)

Cartesian Axis Type 2: X-axis is to the right (from observer's point of view), Y-axis is upwards (from observer's point of view), Z-axis is away from the observer

Depending on which Cartesian Axis Type, you will get completely different answers.

So what is the minimum amount of information to rotate a point in 3D Cartesian space? Am I missing anything else? Ohanian (talk) 03:37, 30 December 2017 (UTC)

In application, you have a point to rotate (3 values), a point to rotate around (3 values), and an angle. That is all. The direction of rotation is implied by the angle. If I say rotate 30 degrees and you want to go the wrong way, then I will tell you to rotate 330 degrees. With those 7 values, you translate the two points so that the point to rotate around is the origin. Then, you rotate (a simple rotation matrix). Then, you reverse the original translation. It is three matrix multiplications, and is very standard in any graphics programming, though you normally plan to do the same thing more than once. So, you multiply the translation, rotation, and un-translation matrices together and then use that on the original point. 71.85.51.150 (talk) 03:48, 30 December 2017 (UTC)

The article on the Euclidean group might be useful here. --RDBury (talk) 07:05, 30 December 2017 (UTC)

As Gandalf explains below, in 3D space you rotate about a line, not about a point. And you need 4 values, rather than 3, to specify the line.

In 4D, by the way, you rotate about a plane. And you can also have a rotation about two planes simultaneously; since those intersect at a point, you could call it a rotation about the point, though you'll need several parameters to specify the rotation even once the point is given. -- Meni Rosenfeld (talk) 20:26, 30 December 2017 (UTC)

The minimum amount of information required to uniquely specify a rotation about a general axis in 3 dimensional Euclidean space is 5 real numbers - we say that there are five "degrees of freedom". One way to see this is as follows :

• The direction of the axis can be specified with two real numbers. If we use three numbers to define a vector then we have redundant information because vectors with the same direction but different lengths define the same axis direction - so the mapping from 3 dimensional vectors to axis directions is many-to-one. If we restrict ourselves to, say, vectors with length 1 then we only need two independent values.
• The location of the axis can be specified with two real numbers. Two not three because we only need to specify a point on any plane that is perpendicular to the axis direction, and these planes have two dimensions. Again, If we use three numbers to locate a specific point on the axis then we have redundant information because all the points in any line parallel to the axis direction will define the same axis location.
• The size of the rotation is specified by one more real number. If you are only concerned with the end state and not the route taken to get there then there is redundancy here as well, because you only really need to know the rotation size module 360 if measuring in degrees (or module 2π if using radians). But for some applications the total size of the rotation (whether it involves one full turn, two full turns etc.) may be significant.

The left/right handedness of the axes and the direction of a positive rotation are usually taken as conventions that are defined in advance. They are not an intrinsic part of the Euclidean space.

It may be convenient to use more than five parameters to specify a general rotation - for example, you could use 6 numbers to define a rotation about the origin together with a translation. But then these parameters will involve some redundancy - the mapping from tuples of parameter values to rotations will be many-to-one. Gandalf61 (talk) 11:38, 30 December 2017 (UTC)

I think people are confused, Assuming I am talking to aliens from another galaxy, how can I know what the convention is if all that the aliens gave me is {x,y,z}={1,2,3} what are the minimum amount of information those aliens need to provide me to rotate that point to another location in 3D space? You bet I need to know the handedness at the bare minimum. And I also need to know the Cartesian Axis Type. Ohanian (talk) 00:32, 31 December 2017 (UTC)

Your question is pretty vague. It doesn't really make much sense to "rotate one point to another point". To specify a rotation in 3 dimensions, you need to specify a point in SO(3) (the special orthogonal group over $\mathbb {R} ^{3}$). This is itself a three-dimensional manifold, and so 3 numbers suffice. Then again, as sets, euclidean 3-space has the same cardinality as the line, so you can encode the entire rotation in a single real number. That probably isn't what you mean, but "pieces of information" is pretty vague. Most likely, it's the dimension of SO(3) that's really what you're after (and like I said, it's three). –Deacon Vorbis (carbon • videos) 01:31, 31 December 2017 (UTC)

After rereading this, I guess you want to specify the center of the rotation too. In that case, yes, it would be 6-dimensional instead. –Deacon Vorbis (carbon • videos) 02:55, 31 December 2017 (UTC)

Rotations in 3 dimensions don't have a unique centre - they have an axis, a line of points unchanged by the rotation. By choosing a point in SO(3) you have defined the direction of the axis and the angle of the rotation. To translate that axis through the origin to a parallel axis elsewhere, you only need add two more dimensions, not three. So the space of general 3D rotations is five dimensional. You can use six parameters if you like, but then there is redundancy. Gandalf61 (talk) 07:11, 31 December 2017 (UTC)

Maybe the following observation will be helpful to the OP: a tuple of 3 numbers is not the same thing as a point in three-dimensional space; the correspondence is a consequence of some arbitrary choices (where to draw the axes, etc.). The algebraic language does not depend and cannot distinguish between some of these choices: the result of performing a certain translation (given in coordinates) to a certain point (given in coordinates) has the same coordinates regardless of whether the coordinate system is right- or left-handed, regardless of how your axes are rotated in space, regardless of where the origin is, etc. --JBL (talk) 19:33, 1 January 2018 (UTC)

Why is the interval when a < b and a,b real numbers then [b,a]=[a,a)=(a,a]=(a,a)=empty set?

I don't understand what's probem? — Preceding unsigned comment added by 151.236.179.187 (talk) 09:39, 30 December 2017 (UTC)

Suppose a is 3 and b is 5. How many numbers are 5 or more and 3 or less, i.e. x>=5 and x<=3? How many numbers are 5 or more and up to but not including 5, i.e. x>=5 and x<5? Dmcq (talk) 09:55, 30 December 2017 (UTC)

I was relieved, I was fearing, so every question can be asked here, thank you mathematicians for supporting us you are great! Alireza Badali (talk) 19:59, 30 December 2017 (UTC)

Our relevant articles are:

• Interval (mathematics)#Including or excluding endpoints -- This states the conditions under which an interval represents the empty set, but does not explain the "why". I think it's fine as is without further explanation. That section does link set builder notation.
• Set-builder notation#Sets defined by a predicate -- This did not explicitly state that {x∈E|Φ(x)}=∅ when Φ(x) is false for all x∈E. I first though this should be equally obvious, but formal set theory can be confusing (this very sections mentions Russell's paradox), so I've added, "If the predicate is false for every element of the domain, then the set defined is the empty set."

-- ToE 00:47, 31 December 2017 (UTC)

January 2

Newton approximation numbers for sqrt(3) and special relativity

I got side tracked while answering this question. I want to stick to the math here, but just for context, if you look at a proton from a frame moving at sqrt(3)/2 times the speed of light, it seems to have (nay, it has) twice the mass. If you look at it from a frame moving that fast compared to the last one you looked at it from, it has seven times the mass...

Working it out, I realize that if you number your frames n=0,1,2,3,4... then g(n) is 1,2,7,26,97,362,1351,5042,18817,70226,262087,978122... The rule for adding these is that g(n1+n2) = g(n1)*g(n2) + sqrt(g1^2*g2^2 - g1^2 - g2^2 + 1). To iterate, use n1=1 ; g(1) = 2 so g(n+1) = 2*g(n1) + sqrt(3*(g(n1)^2 - 1).

These turn out to be iterations in a Newton's method for approximating the square root of 3, apparently. [1] I should note that both numerator and denominator appear in the speeds of the proton in all those different frames, which divides the square root of 3 by those approximations to approximate but never reach the speed of light from the relativistic frames.

Despite all these things, I haven't managed to figure out much about the series...

1) I don't know how to calculate g(n) from n without iterating.

2) It is in no way restricted to integers. However, it seems like g(n) is always integral for integer n, which means that g^2-1 is always 3*N^2, where N is an integer. From the iteration I was doing it is not obvious to me why this is the case.

3) This goes beyond the scope, but this value n seems like it should be useful for describing "speed" (in a loose sense) under relativity in a way that can just be added and subtracted between multiple frames of reference.

Wnt (talk) 14:49, 2 January 2018 (UTC)

The OEIS is your friend. This turns out to be A001075 in the OEIS. From the description, you'll see that this is actually a second-order, linear, homogeneous recurrence relation. Finding explicit formulas for these is straightforward, and would probably help shed some light on what's going on. Also, for future reference, please use some kind of markup (whether it's <math>...</math> or just regular wiki markup with templates) when asking questions; it's a bit tough to try to read as is. –Deacon Vorbis (carbon • videos) 15:42, 2 January 2018 (UTC)

You might look at the article Lucas sequence since your sequence is a simple modification (divide by 2) of V_n(4,1), see A003500. (The OEIS isn't always so good at pointing out relationships between sequences.) Lucas sequence are generalizations of Fibonacci & Lucas numbers and many of the interesting properties of those sequences have similar versions for Lucas Sequences. Also, I'm not sure what you meant by saying this was an application of Newton's method, but the link was about the continued fraction convergents of √3, which is related but not the same. Related to this, and point #2 above, is Pell's equation with n=3. Basically the solutions of Pell's equation are given by the continued fraction convergents of √3, which should help explain the 'why' of point #2. Much of this would covered in a course on elementary number theory, but the connection to relativity is new to me. --RDBury (talk) 16:19, 2 January 2018 (UTC)

PS. The addition formula above seems to be related to the hyperbolic addition formula $\cosh(x+y)=\cosh(x)\cosh(y)+\sinh(x)\sinh(y)$ where the square roots come in because $\sinh(u)={\sqrt {\cosh ^{2}(u)-1}}$. So to answer point #3, you can use hyperbolic cosine and it's inverse to convert between ordinary addition and the 'relativistic addition' of the formula. --RDBury (talk) 16:49, 2 January 2018 (UTC)

See the article on rapidity. Rapidity w corresponding to velocity v is defined as $w=\operatorname {artanh} {\frac {v}{c}},$. Rapidities are additive and $\cosh w=\cosh \left(\operatorname {artanh} {\frac {v}{c}}\right)={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}=\gamma$. Also from the article: Rapidities $\mathbf {w} _{1},\mathbf {w} _{2}$ with directions inclined at an angle $\theta$ have a resultant norm $w\equiv |\mathbf {w} |$ (ordinary Euclidean length) given by the hyperbolic law of cosines, $\cosh w=\cosh w_{1}\cosh w_{2}+\sinh w_{1}\sinh w_{2}\cos \theta .$

-Modocc (talk) 17:29, 2 January 2018 (UTC)

By no means do I understand everything here yet, but I should thank @RDBury: for a key step forward. Both the U_n and V_n sequences are relevant here for the (4,1) Lucas series. From a note in the OEIS entry, the V sequence can be written as g(n) = ((2+sqrt(3))^n + (2-sqrt(3))^n)/2 . (I know, I should get better at math tags, but they are so laborious and in the end I type all this into the R console anyway) Wnt (talk) 20:23, 2 January 2018 (UTC)

For your information:

$g(n)={(2+{\sqrt {3}})^{n}+(2-{\sqrt {3}})^{n} \over 2}$

Bo Jacoby (talk) 21:34, 2 January 2018 (UTC).

Continuing, I agree that g = cosh(artanh(x)), where x = v/c. There is some coincidence of form with the above equation, because, using w = artanh(x) = $\ln {\sqrt {\frac {1+x}{1-x}}}$,

$g=coshw={\frac {e^{w}+e^{-w}}{2}}={\frac {e^{\ln {\sqrt {\frac {1+x}{1-x}}}}+e^{\ln {\sqrt {\frac {1-x}{1+x}}}}}{2}}$ = ${\frac {{\sqrt {\frac {1+x}{1-x}}}+{\sqrt {\frac {1-x}{1+x}}}}{2}}$ = ${\frac {1}{\sqrt {1-x^{2}}}}$. Indeed, I must note that for x = ${\frac {\sqrt {3}}{2}}$, the shift between frames used above for iterations, the two components being added are indeed 2 + sqrt(3) and 2 - sqrt(3)! (yes, it took forever to debug the code above) On further consideration, I realize that this definitely is the rapidity. For the value of w where gamma = 2 and v = sqrt(3)/2, we get w = $\ln {\sqrt {\frac {2+{\sqrt {3}}}{2-{\sqrt {3}}}}}$ ... however, by multiplying this by the numerator, we can simplify this to $\ln {(2+{\sqrt {3}})}$. With cosh we take e to that power, so for n*w we get $(2+{\sqrt {3}})^{n}$ for the +w portion. So it checks out, once I remember that hyperbolic angle isn't measured in radians! Wnt (talk) 04:56, 3 January 2018 (UTC)

Just as an FYI, it looks like there are similar series any time the ratio of mass to rest mass is an integer. For ratio 2 get the above 1, 2, 7, 26, 97, but for ratio 3 the series is 1, 3, 17, 99, ... and for ratio 4 the series is 1, 4, 31, 244, ... . All the entries are integers which relates to the fact that cosh nx is a polynomial is cosh x. --RDBury (talk) 18:23, 3 January 2018 (UTC)

References

January 3

50th perfect number

How many digits (I want an exact figure) does the 50th perfect number have?? I know it falls between 46 million and 47 million, but I want an exact figure. Georgia guy (talk) 18:29, 3 January 2018 (UTC)

According to [2], the recently discovered Mersenne prime is 2^77232917-1. This means that the corresponding perfect number is 2^77232916(2^77232917-1). According to my calculations the logarithm is 46498849.03822324341502, so it has 46498850 digits and starts with 1092... . It should be noted this is the 50th known Mersenne prime, not necessarily the 50th is size, as the the search method is not guaranteed to find the primes in order. --RDBury (talk) 18:59, 3 January 2018 (UTC)

I'm not into math much, but a Perfect number is not the same as a prime number is it? RudolfRed (talk) 19:51, 3 January 2018 (UTC)

No but every perfect number is associated with a Mersenne prime and vice versa. See Euclid-Euler theorem. CodeTalker (talk) 20:21, 3 January 2018 (UTC)

Technically, what's been proved is that that's true for every even perfect number. Almost certainly, there are no odd perfect numbers, but this has not been proved. --Trovatore (talk) 20:31, 3 January 2018 (UTC)

A question of Topology by Munkres

• Give an example of a function that has a left inverse but no right inverse.
• Give an example of a function that has a right inverse but no left inverse.
• Can a function have more than one left inverse? more than one right inverse?

Thanks in advance. Alireza Badali (talk) 20:34, 3 January 2018 (UTC)

Have you read Inverse function#Left and right inverses? -- ToE 22:07, 3 January 2018 (UTC)

Thank you, no I hadn't seen it, so Wikipedia has everything, thanks to Wikipedia too! Alireza Badali (talk) 08:30, 4 January 2018 (UTC)