tag:blogger.com,1999:blog-393324845011978943.post5631862207998826146..comments2018-08-07T01:35:02.986-04:00Comments on Behind the Guesses: Quaternions -- Part 1: How many?Eli Lanseyhttps://plus.google.com/106934211296103810300noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-393324845011978943.post-81534939932795741522011-06-29T18:48:30.046-04:002011-06-29T18:48:30.046-04:00Hi, I have read through your posts. Please do NOT ...Hi, I have read through your posts. Please do NOT stop writing if you have any more ideas..<br />These articles are very useful for an undergradute, ( or anyone who wants good examples for "definition is not equal explanation").<br /><br />Adam from HungaryAdamnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-1964801907894635622010-08-31T07:23:11.879-04:002010-08-31T07:23:11.879-04:00Oh, sure! That explains everything, thank you! I s...Oh, sure! That explains everything, thank you! I should have been more careful and interpret the "*" operator the way you defined it above ;) Thanks a lot and sorry for bothering.<br /><br />AndrewAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-68626938781764444492010-08-31T07:17:11.304-04:002010-08-31T07:17:11.304-04:00Hi Andrew,
Thanks for your comment.
I intended the...Hi Andrew,<br />Thanks for your comment.<br />I intended the '*' to represent complex conjugate, not multiplication. But, you're exactly right otherwise. Except, because of the conjugate, the angle of A represents rotation in the opposite direction, thus beta + (-alpha).<br />Hope that clears things up.Elihttps://www.blogger.com/profile/01955234977479398457noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-24775971734776510502010-08-31T06:48:11.681-04:002010-08-31T06:48:11.681-04:00Hi,
very nice article, thank you for sharing your...Hi,<br /><br />very nice article, thank you for sharing your insights :)<br /><br />There is one thing I fail to understand in the section "From Complex Multiplication to Vector Products": why does A*B equal abe^((beta-alpha)i) rather than abe^((alpha+beta)i)? As far as I understood, the product of two complex numbers A and B is given by the rotation of A (resp. B) by the degree which B (resp. A) forms with the real line (plus a stretching of its modulus of course), so the angle of A*B should be the sum, rather than the difference, of the two angles... Am I missing something?<br /><br />Anyway, thanks again for sharing your intuitions, I just ordered the book you recommend!<br /><br />AndrewAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-22647697356320006472009-12-23T01:56:00.872-05:002009-12-23T01:56:00.872-05:00In mathematics, the quaternions are a number syste...In mathematics, the quaternions are a number system that extends the complex numbers. They were first described by Irish mathematician Sir William Rowan Hamilton in 1843 and applied to mechanics in three-dimensional space. A striking feature of quaternions is that the product of two quaternions is noncommutative, meaning that the product of two quaternions depends on which factor is to the left of the multiplication sign and which factor is to the right. Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space or equivalently as the quotient of two vectors. Quaternions can also be represented as the sum of a scalar and a vector. Quaternions find uses in both theoretical and applied mathematics, in particular for calculations involving three-dimensional rotations such as in three-dimensional computer graphics and epipolar geometry, although they have been superseded in many applications by vectors and matrices. I am a college sophomore with a dual major in Physics and Mathematics @ University of California, Santa Barbara. By the way, i came across these excellent <a href="http://www.funnelbrain.com/" rel="nofollow">physics flash cards</a>. Its also a great initiative by the FunnelBrain team. Amazing!!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-14083347800779132192009-12-03T10:12:40.789-05:002009-12-03T10:12:40.789-05:00Nice post!.. Ive been sitting in front of the came...Nice post!.. Ive been sitting in front of the camera and trying to map its 2D Capture back to 3D and yes, funny things happened.<br /><br />We probably believe that assuming i = sqrt(-1) and calling it imaginary will do the job.<br />The tricky part when we try to define root of a negative number. However if we see an argand diagram, its just rotation in the other direction. <br />Due to this I always think that 'i' means 'independent'.<br /> So if we have x + iy and a + ib, x and b shall interact only if b gets rid 'i'. Similarly for a and y.Bharathhttps://www.blogger.com/profile/11625552783909809165noreply@blogger.comtag:blogger.com,1999:blog-393324845011978943.post-45681452033994334182009-12-01T00:17:04.627-05:002009-12-01T00:17:04.627-05:00A nicely written post once again. It will be inte...A nicely written post once again. It will be interesting where you go with this.Peeter Joothttps://www.blogger.com/profile/13747647271625793131noreply@blogger.com