tag:blogger.com,1999:blog-54068916531207630342019-01-18T05:21:06.849+01:00The Mathematics Blog by Nilo de RoockOpen University pure maths study and research blognilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.comBlogger1017125tag:blogger.com,1999:blog-5406891653120763034.post-56258584582565634572014-12-21T11:29:00.001+01:002014-12-21T11:43:43.885+01:00Dirichlet ' s Theorem<p dir="ltr">$$p / ( p-1 ) ! + \left(\frac{a}{p}\right)  a^{(p-1)/2}$$</p><p dir="ltr">Proof.<br>Consider the equation $Ax \equiv a \bmod p$ with $A, x \in \{ 1,2, \cdots p-1\}$,</p><p dir="ltr">Case $\left(\frac{a}{p}\right)=-1$<br>In this case $x$ and $A$ are different members of the set $\{ 1,2, \cdots p-1\}$, there are $(p-1)/2$ distinct pairs $(A, x) $ and pairwise multiplication gives the following identity: $( p-1 ) ! = a^{(p-1)/2}$.</p><p dir="ltr">Case $\left(\frac{a}{p}\right)=1$<br>In this case $a$ is a quadratic residue of $p$ so there are two pairs where $x$ and $A$ are equal members of the set $\{ 1,2, \cdots p-1\}$, there are $(p-3)/2$ distinct pairs $(A, x) $ and pairwise multiplication gives the following identity: $\frac {( p-1 ) ! }{k (p-k)}= a^{(p-3)/2}$. <br>Now $k( p-k) = kp - k^2 \equiv -a \bmod p $.  Another pairwise multiplication gives the following identity: $( p-1 ) ! = - a^{(p-1)/2}$.</p><p dir="ltr">Combining both cases and replacing the sign with the Legendre symbol gives $$p / ( p-1 ) ! + \left(\frac{a}{p}\right)  a^{(p-1)/2}.$$</p>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-418549647870102732014-11-12T12:13:00.001+01:002014-11-12T12:13:57.285+01:00Epiphany<p dir="ltr">I had one again. I am still in awe. I used to call them "cognitions", but that's not enough. The right word is epiphany. The protocol for reading maths is to decipher what the author is trying to say. Often mathematical ideas are simple to visualize but notoriously hard to write on paper. When you have deciphered the text the visualisation is implanted and suddenly it all makes sense. And when the idea is particularly beautiful the implant is experienced as an epiphany. <br>( Epiphany : An illuminating realization or discovery, often resulting in a personal feeling of elation, awe, or wonder. )</p>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-64705751194237864542014-10-22T15:57:00.001+02:002014-10-22T16:24:37.903+02:00Exercise<p dir="ltr">Let $ p \in \mathbb {Z}[X] $ and of fifth degree with only the terms for $x^5$, $x^4$ and $x^3$ known, the terms for $x^2$, $x$ and $1$ are not known. $$ p=x^5 - 5x^4 - 35x^3 + \cdots $$ The ( five ) roots of $p $ form an arithmetic sequence. Find the roots of $p $.</p>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-52671072490271308982014-09-15T20:22:00.001+02:002014-10-22T15:49:33.946+02:00Disquisitiones Arithmeticae<p dir="ltr">It's time to read Gauss's original work. More later.<br /><br /><a href="http://lh6.ggpht.com/-WQOOaDbs5z4/VBcuX6XHxNI/AAAAAAAACFM/YEJUfjiB1k0/s1600/Screenshot_2014-09-15-20-19-35.png" imageanchor="1" style="text-align: center; margin-left: 1em; margin-right: 1em;"><img border="0" src="http://lh6.ggpht.com/-WQOOaDbs5z4/VBcuX6XHxNI/AAAAAAAACFM/YEJUfjiB1k0/s640/Screenshot_2014-09-15-20-19-35.png"></a></p>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-71273158551781716652014-08-02T12:35:00.002+02:002014-08-02T12:49:08.006+02:00A Five By Five Magic SquareA few years ago I wanted to find a 5 by five magic square. I gave up on it. See <a href="http://mathematics-diary.blogspot.nl/2011/04/magic-squares-of-type-5-by-5.html">this post</a> for the results I came up with in 2011. I used the wrong tools, I can see that clearly now. <br /><br />Yesterday and today I worked on it. Again without finding a solution. This morning I thought "Should I spend another day on it? What am I doing with my time? I am not going to solve this one ( either )". But problems you can't solve become little traumas in your subconscious. From time to time they remind you that you weren't able to crack them. It hurts. Probably more than you can imagine.<br /><br />There was one thing I could add to my search algorithm, I decided to add that and then stop. I had only one diagonal left that didn't add up to 65. Anyway, the last thing worked and I found one. One hurting trauma less.<br /><br />$$<br />\begin{array}{ccccc} <br />2 & 25 & 5 & 20 & 13 \\ <br />18 & 4 & 17 & 7 & 19 \\ <br />8 & 24 & 22 & 10 & 1 \\ <br />23 & 9 & 6 & 16 & 11 \\ <br />14 & 3 & 15 & 12 & 21 <br />\end{array} <br />$$<br /><br />The integers $1,2 \dots 25$ are laid out in a 5 by 5 matrix such that the rows, columns and the two diagonals total to 65.<br /><br />Now the real fun begins: finding bigger ones, special ones maybe, and analyzing and optimizing my method. Where are the limits? And questions, many questions like how many different 5 by 5s are there? Is there one with the number 13 in the centre?nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-1723556979944134002014-05-18T16:07:00.000+02:002014-05-18T16:07:37.369+02:00Explained: Mathematical Virus type MV/C<h3>ARE YOU INFECTED WITH MV/C ?</h3><h3>What is MV</h3><p>A mathematical virus (MV) is a preconception about the structure, function or method of mathematics which impairs one's ability to do mathematics.</p><br /><br /><h3>What is MV/C?</h3><p>MV/C is a mathematical virus which is easily diagnosed, the infected suffer from the delusion that <b>"Coordinates are essential to calculations"</b>. Physicists and engineers are especially susceptible to this virus, because most of their textbooks are infected, and infected teachers pass it on to their students.</p><br /><p>The first MV was discovered by David Hestenes, a theoretical physicist, best known as chief architect of geometric algebra as a unified language for mathematics and physics.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://physics.asu.edu/sites/default/files/profile/David%20Hestenes/DavidHestenes_pic.jpg?1290368931" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://physics.asu.edu/sites/default/files/profile/David%20Hestenes/DavidHestenes_pic.jpg?1290368931" /></a></div><br /><h3>References</h3><a href="http://download.adamas.ai/dlbase/ebooks/VX_related/Mathematical%20Viruses.pdf">David Hestenes, Mathematical Viruses</a>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-49999905244853912942013-12-15T18:58:00.000+01:002013-12-15T18:58:18.038+01:00( Progress on the ) tiling printer.Uni-color-2D graphics can't be spectaculair, let alone breathtaking. Unless you are in the know. Let me explain. I am now able to - almost - print tilings automatically from just a string of numbers as I wrote about in a previous post.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-pMkUyU-loKY/Uq3n8C4hw4I/AAAAAAAAByw/SnjqVq5bPKw/s1600/progress.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="http://1.bp.blogspot.com/-pMkUyU-loKY/Uq3n8C4hw4I/AAAAAAAAByw/SnjqVq5bPKw/s400/progress.png" width="281" /></a></div>Not spectaculair at all, it's just a collection of 8 tilings containing 72 polygons in total. But if I change ( 4,2 ) by (10,5) and add a colorfunction the program creates immediately the following graphic.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-760gJqU9Zwo/Uq3qmEg707I/AAAAAAAABy8/3pCYun5HHnM/s1600/progress2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="http://2.bp.blogspot.com/-760gJqU9Zwo/Uq3qmEg707I/AAAAAAAABy8/3pCYun5HHnM/s640/progress2.png" width="411" /></a></div>Both are, in fact, representations of the same sequence.{6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4} but the variations from this sequence alone are endless. The program creates a graphic for every sequence input. Not all graphics will be tilings of the plane by definition. The following ( major ) step is to let the program reject sequences that do not lead to tilings of the plane.<br /><br />Tne next part will include an implentation of a 'Point in Polygon' predicate. Many of those exist and have been implemented for many languages. That could be a topic for a more in-depth post next time.nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-86520536373986386492013-12-08T18:50:00.001+01:002013-12-08T18:50:10.295+01:00Challenging the eternal truth of mathematicsI learned that 0,999... = 1. I believe it was in M381 that I learned to prove it. The proof is quite simple actually.<br /><br />Let <br />x = 0,999..., multiply both sides with 10<br />10x = 9,999..., now subtract x from 10x<br />9x = 9, and we have our result<br />x = 1.<br /><br />What I like about mathematics is that it is timeless, i.e. we can still read math books of hundreds of years old and still learn things from them. That's not an advisable strategy hfor any other science than mathematics. Of course, new branches of mathematics appear, new discoveries are made, but they don't invalidate the truths of the past.<br /><br />Today however I found someone who actually is challenging some of the established truths in mathematics, like for example that 0,999... = 1. I don't think that he is a crackpot, although I have no doubt that the mathematical establishment, professors who are 'safe' by all means, will call him like that.<br /><br />The man is arrogant though, he calls the proof above, 'juvenile' for example.<br /><br />Who is he, what are his ideas and how did he disproof that 1 = 0,999? The links below will help you abshereing these questions.<br />- <a href="http://johngabrie1.wix.com/newcalculus">The New Calculus - The first rigorous formulation of calculus in history.</a><br />- <a href="https://www.filesanywhere.com/fs/v.aspx?v=8b6b6a8a596275a7a7a9">Proof that 0.999 not equal 1.pdf</a><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com1tag:blogger.com,1999:blog-5406891653120763034.post-54089216723211846132013-12-08T00:08:00.000+01:002013-12-08T00:20:09.962+01:00Some progress...If you read the previous posts on my tiling printing algorithms you'll understand the reason for this post: I made some progress that unraveled some serious knots in my stomach.<br /><br />{ 4, 2, 4 } -><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-ML39_JRHIJo/UqOgLtUHwHI/AAAAAAAAByg/w4cshHoJbhk/s1600/forpost0A.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" src="http://3.bp.blogspot.com/-ML39_JRHIJo/UqOgLtUHwHI/AAAAAAAAByg/w4cshHoJbhk/s320/forpost0A.png" width="320" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-ZXBfsDwvOfs/UqOgLuHcNMI/AAAAAAAAByU/GZNnvhPfyk4/s1600/forpost0B.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="166" src="http://2.bp.blogspot.com/-ZXBfsDwvOfs/UqOgLuHcNMI/AAAAAAAAByU/GZNnvhPfyk4/s320/forpost0B.png" width="320" /></a></div><br />{ 6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4 } -><br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-uvEHTW1C_3Q/UqOgLumEToI/AAAAAAAAByQ/W7MD--V_Blg/s1600/forpost1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://2.bp.blogspot.com/-uvEHTW1C_3Q/UqOgLumEToI/AAAAAAAAByQ/W7MD--V_Blg/s320/forpost1.png" width="210" /></a><img border="0" height="320" src="http://3.bp.blogspot.com/-1vqn9xVzqxk/UqOgMAF2E9I/AAAAAAAAByc/YtaKAFHvX-M/s320/forpost2.png" width="210" /></div><br />The second white polygon is made from the following points :<br />\begin{array}{cc}<br />-\frac{\sqrt{3}}{2} & \frac{1}{2} \\<br />-\frac{\sqrt{3}}{2} & -\frac{1}{2} \\<br />0 & -1 \\<br />\frac{\sqrt{3}}{2} & -\frac{1}{2} \\<br />\frac{\sqrt{3}}{2} & \frac{1}{2} \\<br />\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\<br />\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\<br />\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\<br />0 & 1+\sqrt{3} \\<br />-\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\<br />\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\<br />\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\<br />\end{array}<br /><br />Ready to enter the next level of the problem. ;-)nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-72993132263735239082013-11-30T11:52:00.004+01:002013-11-30T13:03:34.144+01:00Organizing Research NotesIf only Mindjet's MindManager could handle LaTeX it would be the perfect tool for organizing my notes, unfortunately it can't and their advice is to use stuff like Equation Editor. That's not for me.<br /><br />Perhaps at a subconscious level I find that the notes I produce aren't worth saving, fact is that I still haven't found a way of organizing my study notes that suits me. I regularly search the web for tools and today I landed on this <a href="http://mathoverflow.net/questions/1785/how-do-you-keep-your-research-notes-organized" target="_blank">MathOverflow question.</a> From this page I jumped to <a href="http://drorbn.net/AcademicPensieve/" target="_blank">Dror Bar-Natan's Academic Pensieve</a>, which I invite you to visit because it's unique, impressive and might give you some ideas in organizing your own set of study ( or raw research ) notes.<br /><br />I have tried a Zillion mindmappers but none come close to MindManager. <span style="font-size: x-small;">At work I use FreePlane because it is free ( ( That accurately describes my position with that company ) , the alternative would be to struggle with the tools provided like the damned Word.</span><span style="font-size: small;"> FreePlane does support LaTeX however.</span><span style="font-size: x-small;"><br /></span><br /><span style="font-size: small;"><br /></span><span style="font-size: small;">My little search this morning landed me on <a href="http://www.docear.org/" target="_blank">DocEar</a>. It supposedly is the MindMapper for Scientists, while MindManager is more positioned at creative business people. </span><span style="font-size: small;">If it is any good, I will report on it in a future post. </span><span style="font-size: small;"><br /></span><br /><span style="font-size: small;"><br /></span><span style="font-size: small;">UPDATE:</span><br /><span style="font-size: small;">I am giving DocEar a try. It's based on FreePlane and JabRef tools that have proved themselves. More later.</span>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-81983804125718095582013-11-24T23:04:00.001+01:002013-11-24T23:08:14.245+01:00Open University: TMA Cut-off dateAs an OU student you have to make several TMAs for each course you do. TMA stands for Tutor Marked Assignment. A TMA consists of assignments covering all the booklets you studied in the period prior to the cut-off date of the TMA. The cut-off date is the latest date the work has to be received by your Tutor. This is a period of cut-off dates. Usually you'll find a lot of blog, forum, twitter or facebook posts about TMAs that have been done. Completing a TMA is just part of the study process but for an OU student it's somewhat of an event. Not like an exam, but sending it out gives a feeling of relief, achievement perhaps. Completing a TMA is not something most people do in a few hours, some TMAs take weeks if not months to complete.<br /><br />The average result of the TMA ( after some formula has been applied to it ), is the maximum result you can get from the course because the final result of the course is the minimum of the exam result and the average TMA score. Also, a minimum TMA score is required to be eligible for taking the exam. For example.<br />TMA result: 15/100 not eligible for exam.<br />TMA result: 65/100. Exam result: 100/100. Course result 65/100.<br />TMA result: 100/100. Exam result: 15/100. Course result 15/100 and thus a FAIL.<br />Completing all the TMAs on time, requires regular study, and regular study enhances the chance on a good exam result significantly. I suppose that's the thought behind it all. The second example may seem rather undesirable, but it just is not a realistic scenario. A student with a 100% exam score usually has no problems with the TMAs.<br /><br />Personally, the TMAs are no longer my 'major math challenge'. Slowly but steady I am working on my own mathematical projects. I still need to study, of course, but I am an OU student mainly to justify the time I spend on mathematics to the other stakeholders in my time. - When you say you study mathematics as a hobby, people accept it ( at best ), but explaining that, in fact, you are involved in your own mathematical research is worse than telling that you apply the tools of Scientology in your life. So... I am an OU Student, if you know what I mean. ;-)<br /><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-19819707641583974422013-11-24T16:35:00.000+01:002013-11-24T16:35:43.430+01:00Closed PolyLine - (2)<div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">I developed an algorithm that takes sequences like $\{ 6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4 \}$ and turns them into drawings like this ( in Mathematica, of course ).</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-5RXwqH5Pzdk/UpIYMYgEArI/AAAAAAAABxg/j3ZrpwsBixc/s1600/Tiling1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-5RXwqH5Pzdk/UpIYMYgEArI/AAAAAAAABxg/j3ZrpwsBixc/s400/Tiling1.png" /></a></div><div class="separator" style="clear: both; text-align: left;">In order to determine if the tile above tiles the plane I need to remove all internal lines...</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-uO0x87mZ9as/UpIYMc9ZOjI/AAAAAAAABxo/6D6-u5fbpxg/s1600/Tiling1Edges.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-uO0x87mZ9as/UpIYMc9ZOjI/AAAAAAAABxo/6D6-u5fbpxg/s400/Tiling1Edges.png" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">... which turned out a tad more dificult than expected. I got as far as a prototype algorithm which I am currently testing and improving.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-7r7_AhClaDk/UpIYMROViTI/AAAAAAAABxk/m7OLuWEDxNk/s1600/Tiling1EdgesWithAxes.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-7r7_AhClaDk/UpIYMROViTI/AAAAAAAABxk/m7OLuWEDxNk/s400/Tiling1EdgesWithAxes.png" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">To be continued.</div><div class="separator" style="clear: both; text-align: left;"><br /></div>nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-12544984433976882592013-11-17T21:52:00.001+01:002013-11-17T21:58:30.948+01:00Closed PolyLine<div class="separator" style="clear: both; text-align: left;">The picture ( below) contains two drawings ( created with Mathematica ), the drawing on the left consists of a hexagon, a square, a triangle and again a square. In order to facilitate an algorithm that decides if this drawing tiles the plane I need a closed polyline to determine if a point is within the borders of the drawing. Oddly enough it is more difficult to create the drawing on the right than the original on the left.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-ZE26E7nl65E/UokqypDeDkI/AAAAAAAABxI/_V5K066H7Es/s1600/PolygonPolyLine.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="367" src="http://4.bp.blogspot.com/-ZE26E7nl65E/UokqypDeDkI/AAAAAAAABxI/_V5K066H7Es/s400/PolygonPolyLine.PNG" width="400" /></a></div><br />The drawing on the right is a closed polyline, it consists of a number (11) of line segments and has no begin- and endpoint, it is closed. ( In another project I am working on we call PolyLines, MultiLines but I haven't seen that name in use elsewhere. )nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-29829657977484748082013-11-17T13:16:00.000+01:002013-11-17T13:16:29.819+01:00Geometric cell division Take a square and add a similar square to one of its sides and remove the shared edge. What remains is a rectangle.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-K82u0rmdiQQ/UoixnHcKlDI/AAAAAAAABwU/xGBdv62_8dc/s1600/division4.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://3.bp.blogspot.com/-K82u0rmdiQQ/UoixnHcKlDI/AAAAAAAABwU/xGBdv62_8dc/s320/division4.png" width="164" /></a></div>Take an octagon and add a similar octagon to one of its sides and remove the shared edge. What remains is the following polygon.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-tiDgDJes0d4/Uoix3LqjggI/AAAAAAAABwc/BZq66NwZy_U/s1600/division8.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://2.bp.blogspot.com/-tiDgDJes0d4/Uoix3LqjggI/AAAAAAAABwc/BZq66NwZy_U/s320/division8.png" width="164" /></a></div> Take an n-gon and add a similar n-gon to one of its sides and remove the shared edge. At infinity the n-gons will divide into two circles. Unfortunately my computer is too slow to effectively record this in a video.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-4Xiapqdg7wY/UoiyhseT5tI/AAAAAAAABwo/fDX4QSWK1Fg/s1600/division12.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://1.bp.blogspot.com/-4Xiapqdg7wY/UoiyhseT5tI/AAAAAAAABwo/fDX4QSWK1Fg/s320/division12.png" width="164" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">n=12</td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-Nd3n4lkNgto/Uoiyhu_60bI/AAAAAAAABws/jPKnrxiJWfo/s1600/division16.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://2.bp.blogspot.com/-Nd3n4lkNgto/Uoiyhu_60bI/AAAAAAAABws/jPKnrxiJWfo/s320/division16.png" width="164" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">n=32</td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-Ixd8Lh2i8JA/Uoiyhi2zjGI/AAAAAAAABww/Ihu90Kf1-vc/s1600/division64.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://4.bp.blogspot.com/-Ixd8Lh2i8JA/Uoiyhi2zjGI/AAAAAAAABww/Ihu90Kf1-vc/s320/division64.png" width="164" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">n=64</td></tr></tbody></table><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-48826760481026133332013-11-09T22:47:00.001+01:002013-11-09T22:57:27.977+01:00Example of a polygonal tilingThe tiling in the image below is edge-to-edge, polygonal, but not uniform because it has different vertex types, i.e. (3,3,4,3,4), (3,4,4,6), (3,4,6,4) and (3,6,4,4). It is therefore NOT an Archimedean tiling,<br /><br />With the following piece we can tile the entire plane...<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-lDbW0Dg53Hw/Un6ssQdW1QI/AAAAAAAABss/x09vIejlLDA/s1600/Tiling3464-34334Piece.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://1.bp.blogspot.com/-lDbW0Dg53Hw/Un6ssQdW1QI/AAAAAAAABss/x09vIejlLDA/s320/Tiling3464-34334Piece.png" width="210" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>... as we can see here.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-GHq0BC9I2Ks/Un6swFCxp3I/AAAAAAAABs0/cswGiY_zLmk/s1600/Tiling3464-34334.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="230" src="http://1.bp.blogspot.com/-GHq0BC9I2Ks/Un6swFCxp3I/AAAAAAAABs0/cswGiY_zLmk/s400/Tiling3464-34334.png" width="400" /></a></div><br /><br /><br /><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-68891349469542549512013-11-09T11:03:00.000+01:002013-11-09T13:19:32.726+01:00Regular polygons, exercise.With a drawing program yesterday's pictures are easy to fake, of course. But these drawing programs don't give you the numbers.<br /><br /><span style="font-size: large;"><b>Exercise.</b></span><br /><br />Place a decagon edge-to-edge on a square with sides of length 1 ( see figure ). What is the distance between the two marked points?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-7z-7ym707-Y/Un4HxCy2liI/AAAAAAAABsc/TuHHuZq8nhk/s1600/PolygonExercise.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-7z-7ym707-Y/Un4HxCy2liI/AAAAAAAABsc/TuHHuZq8nhk/s400/PolygonExercise.png" /></a></div><br /><br />( Answer: $ \frac{1}{4} \left(3 \sqrt{10-2 \sqrt{5}}+\sqrt{50-10 \sqrt{5}}+4\right) $ )nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com1tag:blogger.com,1999:blog-5406891653120763034.post-43654592343893904392013-11-08T23:28:00.000+01:002013-11-08T23:30:22.932+01:00Printing polygons edge-to-edge<u>Recipe</u><br />Suppose you have some polygon with cornerpoints { p1, p2, ..., pk } and you want to print a regular polygon with n edges along one of its edges (p(j), p(j+1) then you can simply find the first point of the regular n-gon by rotating p(j+1) with centre p(j) over 360/n degrees. You can continue this process until you have found all points or you can calculate the centre of the regular polygon, its orientation and edge-length which you need to print a regular n-gon. Here are some examples.<br /><br /><u>Examples</u><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-hvu27Zqw_SM/Un1lITdUH6I/AAAAAAAABsE/YhDOaq2r_zg/s1600/PolygonFest.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="505" src="http://1.bp.blogspot.com/-hvu27Zqw_SM/Un1lITdUH6I/AAAAAAAABsE/YhDOaq2r_zg/s640/PolygonFest.PNG" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">3-on-4, 4-on-3 and and 5,7,9,11-on 4 (dark-on-light ).</td></tr></tbody></table><u><br /></u><br /><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-36542696664092836442013-11-08T17:40:00.001+01:002013-11-08T17:40:34.605+01:00The (3,12,12) TilingI made ( what programmers would call ) a 'recipe' for creating a program that prints an Archimedean Tiling. Needless to say that I am talking about Mathematica code. I basically have a program that can distribute any set of Mathematica Graphics objects over a lattice of points. So the recipe basically means calculating the points of the motif, putting them in a set and handing them over to the printer. The last one I did is (3,12,12) Tiling.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-GZCqqn6PK7Q/Un0RZysj18I/AAAAAAAABr0/LToOA31uVxs/s1600/ArchimedeanTiling31212.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="640" src="http://4.bp.blogspot.com/-GZCqqn6PK7Q/Un0RZysj18I/AAAAAAAABr0/LToOA31uVxs/s640/ArchimedeanTiling31212.png" width="464" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Tiling with vertex type (3,12,12).</td></tr></tbody></table>The next step is to abstract and code the recipe itself, i.e. translating lists like (3,12,12) or (3,3,4,3,4) to graphics. In fact, I calculated the points for (3,12,12) with a first version of that program. -nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-6349297644796938232013-11-05T22:06:00.000+01:002013-11-05T23:27:33.869+01:00Johannes Kepler and (3,3,4,3,4)Who doesn't know the name of Johannes Kepler? Kepler (1571 - 1630 ) formulated the laws of planetary motion and his work provided the foundation for Isaac Newton's theory of gravity. My point being that Kepler was a scientific giant in his days and his name will live on forever.<br /><br />My current mathematical project ( personal challenge if you like ) is focused towards tilings of the plane, creating ( Mathematica ) software to print and generate tilings. And ultimately -find- new tilings I haven't seen before. There aren't many textbooks on the subject, the classic work is very recent ( in mathematical terms ), it was published in 1986: Tilings and Patterns, by Branko Grunbaum and Geoffrey C. Shephard<br /><br />Earlier today I was doodling on the tiling (3,3,4,3,4) of which I uploaded a picture.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-t4r9hJDseTc/UnlVmELQWHI/AAAAAAAABrc/1OePdLIG08U/s1600/Untitled+Note+-+5+nov.+2013+21.11+-+Page+5.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="400" src="http://4.bp.blogspot.com/-t4r9hJDseTc/UnlVmELQWHI/AAAAAAAABrc/1OePdLIG08U/s400/Untitled+Note+-+5+nov.+2013+21.11+-+Page+5.png" width="281" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Doodle of (3,3,4,3,4).</td></tr></tbody></table>Then, to my surprise I read in Grunbaum / Shephard that it was Johannes Kepler (!) who started the mathematical research on tilings and patterns. One of the beautiful books Kepler wrote is called the Harmony of the World, originally published in 1619 but recently translated into English by Aiton, Duncan and Field and published the American Philosophical Society.<br /><br />The Harmony of the World consists of 5 books.<br />- Book 1: On the construction of regular figures<br />- Book 2: On the congruence of regular figures<br />- Book 3: On the origins of the harmonic proportions, and on the nature and differences of those things which are concerned with melody<br />- Book 4: Preamble and explanation of the order<br />- Book 5: ( No title ).<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-aSvdWFtPLmE/UnlZCLaXVxI/AAAAAAAABro/labgU2s0lLE/s1600/Kepler104.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="http://2.bp.blogspot.com/-aSvdWFtPLmE/UnlZCLaXVxI/AAAAAAAABro/labgU2s0lLE/s400/Kepler104.PNG" width="393" /></a></div><br />This is part of a picture ( drawing ) from book 2 which has a drawing of (3, 3, 4, 3, 4 ) just like my doodle rype marked as O ( top right ).<br /><br />This proves once more that mathematics, by itself, does not change over time, its timeless. At least this part ( if not all ) of mathematics has to be discovered. The tilings of the plane have always been there, it just takes us to see them so that we can ultimately categorize them.<br /><br />The fact that Kepler worked on this makes him human ( but still a giant of course ) to me, I can imagine the joy and excitement he must have felt drawing the illustrations especially since they take a lot of ( behind the scenes ) calculations.<br /><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-40203345868178353982013-10-27T18:00:00.000+01:002013-10-27T18:00:38.598+01:00Archimedean (4,6,12) tiling - Color patternsOnce you programmed the printing of a tiling pattern it is very easy to add colors to the tiles. Some examples.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-gGqDLENt5ow/Um1GMs9rNVI/AAAAAAAABoc/MhE1QCGXQaY/s1600/ArchimedeanTiling4612-2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="http://1.bp.blogspot.com/-gGqDLENt5ow/Um1GMs9rNVI/AAAAAAAABoc/MhE1QCGXQaY/s640/ArchimedeanTiling4612-2.png" width="640" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-8n48FE15Ol4/Um1GMnXsJyI/AAAAAAAABok/41E_r87_KxA/s1600/ArchimedeanTiling4612-3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="http://4.bp.blogspot.com/-8n48FE15Ol4/Um1GMnXsJyI/AAAAAAAABok/41E_r87_KxA/s640/ArchimedeanTiling4612-3.png" width="640" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-Tizt5LL8h5A/Um1GMi-cGOI/AAAAAAAABog/0PiXkFBdSOM/s1600/ArchimedeanTiling4612-4.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="http://4.bp.blogspot.com/-Tizt5LL8h5A/Um1GMi-cGOI/AAAAAAAABog/0PiXkFBdSOM/s640/ArchimedeanTiling4612-4.png" width="640" /></a></div><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-37529404716982488852013-10-27T17:56:00.000+01:002013-10-27T17:56:03.521+01:00Archimedean (4,6,12) tilingThis is ( part of ) the Archimedean (4,6,12) tiling.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-nAOAzrxa2G8/Um1FOLDtKdI/AAAAAAAABoQ/x6mYfWPPqIY/s1600/ArchimedeanTiling4612-1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="http://1.bp.blogspot.com/-nAOAzrxa2G8/Um1FOLDtKdI/AAAAAAAABoQ/x6mYfWPPqIY/s640/ArchimedeanTiling4612-1.png" width="640" /></a></div><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-21009392579614960292013-10-27T10:54:00.002+01:002013-10-27T10:55:17.758+01:00OU exams more difficult than ever ( ... ) ?I read a rumor on facebook that the Open University exams were harder than ever. No numbers were shown to substantiate the claim however. You may have been aware that the OU rates have been increased dramatically to align them with the rates of the "Brick Unis" ( = how normal universities are called in OU jargon ). Now one of the commenters said that they are doing the same thing with the exams. Suggesting that until now OU exams were much easier than Brick Uni exams. - To be honest I think it's the other way around. Often homework assignments ( for maths at least ) are part of the grade Brick Uni exams while at the OU you get the lowest grade of homework and exam.<br /><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-51704622746727253552013-10-27T09:15:00.000+01:002013-10-27T09:15:30.230+01:00Archimedean (3,4,6,4) tiling - In colorTilings become appealing when they are colored.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-cOddZLnPPqc/UmzLVRe7wyI/AAAAAAAABoA/p0RvTk7vKAA/s1600/3464-1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="406" src="http://2.bp.blogspot.com/-cOddZLnPPqc/UmzLVRe7wyI/AAAAAAAABoA/p0RvTk7vKAA/s640/3464-1.png" width="640" /></a></div><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-58535935806010171372013-10-26T21:59:00.000+02:002013-10-26T21:59:39.751+02:00Archimedean (3,4,6,4) tilingThis is ( part of ) the Archimedean (3,4,6,4) tiling.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-mo9tve8QRsQ/UmwePKSDH5I/AAAAAAAABnw/oVy3STMw1ls/s1600/ArchimedeanTiling3464.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="424" src="http://3.bp.blogspot.com/-mo9tve8QRsQ/UmwePKSDH5I/AAAAAAAABnw/oVy3STMw1ls/s640/ArchimedeanTiling3464.png" width="640" /></a></div><br /><br />The (3,4,6,4) means that at every vertex you'll find four tiles with 3,4,6 and 4 vertices respectively. The Archimedean tilings are vertex-uniform.nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0tag:blogger.com,1999:blog-5406891653120763034.post-46359247171951347582013-10-26T11:39:00.000+02:002013-10-26T11:39:42.498+02:00Video Lectures about Lie Groups<i><b>"... A Lie group is a smooth manifold obeying the group properties and that satisfies the additional condition that the group operations are differentiable. ..." ( Wolfram Site )</b></i><br /><br />The (self-) study of Lie Group theory is hard. I found two aids that helped me going somewhat in the subject. A book called <a href="https://www.goodreads.com/book/show/4419538-naive-lie-theory?from_search=true" target="_blank">Naive Lie Theory by John Stillwell</a> and a series of <a href="http://webmovies.science.uu.nl/WISM414/" target="_blank">weblectures</a> by Erik van den Ban of the University of Utrecht. On <a href="http://www.staff.science.uu.nl/~ban00101/" target="_blank">van den Ban's homepage</a> under Lecture Notes you'll find a Lie Group's prerequisites pdf with explanations of manifolds, tangent maps etc. which appear frequently in texts about Lie Theory.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-KrnOyXM9tY8/UmuNGXTKWAI/AAAAAAAABng/raAPShB9_wY/s1600/E8.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="266" src="http://2.bp.blogspot.com/-KrnOyXM9tY8/UmuNGXTKWAI/AAAAAAAABng/raAPShB9_wY/s320/E8.jpg" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">E8 structure visualized</td></tr></tbody></table><br />nilo de roockhttp://www.blogger.com/profile/15332190734914631351noreply@blogger.com0