Mathematics desk
< October 15	<< Sep \| October \| Nov >>	October 17 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 16

4

What is the name given to numeric 4's whose top lines do not meet? February 15, 2009 (talk) 02:05, 16 October 2008 (UTC)[reply]

No idea, but the article 4 (number) references it as “with an open top”. GromXXVII (talk) 02:33, 16 October 2008 (UTC)[reply]

As far as I can tell, it's referred to as an "open-top 4", as opposed to the "closed-top 4" with the triangle. It looks like 4 started out as a kind of cross-shape (like a plus sign - see 4 (number)#Evolution of the glyph), and then joined up into a loop with loose ends, somewhat like a rounded version of the closed-top 4, which stuck around and evolved into the closed-top four around the time of printing. The more recent re-introduction of an open-top form seems to have come in mostly from the digital age, and in particular seven-segment displays as seen on calculators, to help distinguish 4 from 9. I found a little reference in a book called "The Hindu-Arabic Numerals", as seen on Google Books. Confusing Manifestation(Say hi!) 02:40, 16 October 2008 (UTC)[reply]

I don't think the open-top form ever went away. It might have disappeared from type, but not from handwritten numerals. That's the way I write them, and I think so do most people. --Trovatore (talk) 02:50, 16 October 2008 (UTC)[reply]

Well, the book I linked claims that "the open top used in writing [is] completely modern", although it doesn't actually say how modern (it is a 1911 book, so I guess my claim that it's from the digital age is a bit off). Confusing Manifestation(Say hi!) 22:56, 16 October 2008 (UTC)[reply]

If you have more information you can also add it to Regional handwriting variation#Arabic numerals. --71.106.183.17 (talk) 20:18, 19 October 2008 (UTC)[reply]

Do you mean a digit written like these:

? --CiaPan (talk) 13:25, 16 October 2008 (UTC)[reply]

Yes. February 15, 2009 (talk) 00:13, 20 October 2008 (UTC)[reply]

Vector problem in MATLAB

I don't get this; where am I going wrong - it's a vector rotation thing:

aa = 10*pi/180; %angle of 10 degrees in radians

V1= [ 0, 0.342020143325670, 0.939692620785908]; % vector I'm interested in using

R_mat=[ cos(aa), 0, -sin(aa) ; 0 ,1, 0; sin(aa), 0, cos(aa) ]; % rotating by 10 degrees around y-axis

V3 = R_mat * V1'; % rotate V1 by 10 degrees

d = dot(V1,V3) / (norm(V1) * norm(V3)); % dot product

a = acos(d); % check angle between them

The answer I get should be 0.1745 (in format short), but I get 0.1640.

To summarise: I have the particular result shown in V1 from an intermediate stage of a larger problem. I rotate it by 10 degrees using the matrix shown. Then, to verify, I use the dot product of the vectors to determine the angle between them.

What have I done wrong? I am not ashamed to be embarrassed at this stage! Fritzpoll (talk) 10:01, 16 October 2008 (UTC)[reply]

Ok, so I now know that the problem is that the above assumes that the vector that has been rotated is perpendicular to the axis of rotation. The question then is, how can I reverse-engineer this (knowing the axis of rotation) so that I can extract the axis of rotation from the two vectors? Fritzpoll (talk) 10:57, 16 October 2008 (UTC)[reply]

The cross product of two vectors has the direction of the axis of rotation. But you want to extract the axis of rotation from the two vectors and the angle of rotation, right? Bo Jacoby (talk) 11:25, 16 October 2008 (UTC).[reply]

Don't you need to have two vectors and their images to determine a unique axis of rotation ? You only seem to have one vector and its image, so I think the axis of rotation is not uniquely determined here. It could be a rotation of acos(V1.V3/|V1||V3|) about an axis perpendicular to the two vectors, or a rotation of 180 degrees about an axis parallel to V1+V3 or any one of an infinite number of other possibilities. Gandalf61 (talk) 12:51, 16 October 2008 (UTC)[reply]

Isn't it sufficient to know a vector, its image, and the rotation angle, as Bo Jacoby wrote? — Emil J. 12:58, 16 October 2008 (UTC)[reply]

Well, yes, if the angle of rotation is a given. But Fritzpoll just said "extract the axis of rotation from the two vectors". I guess Fritzpoll needs to clarify just what is known and what is unknown here. Gandalf61 (talk) 13:11, 16 October 2008 (UTC)[reply]

Solution - solved this recently. What I was trying to do was reverse-engineer the angle of rotation from the two vectors. The solution is, for example for rotation around the y-axis, to map the two 3D vectors to two 2D vectors [x, y, z] ==> [x, z] and then take the dot product, etc. of these vectors. The reason this is necessary is that the angle in the plane of rotation can't be calculated unless the initial vector is perpendicular to the axis of rotation, which isn't the case here. Fritzpoll (talk) 13:36, 16 October 2008 (UTC)[reply]