tag:blogger.com,1999:blog-88610692622117960572018-03-06T06:30:39.567-08:00Pauca et non maturumGauss had a motto: Pauca Sed Matura (few, but ripe)Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.comBlogger20125tag:blogger.com,1999:blog-8861069262211796057.post-12668912690719964892017-02-04T22:22:00.000-08:002017-02-04T22:23:11.010-08:00Ellipsoidal Moments of Inertia<script type="text/javascript" src="https://www.wolfram.com/cdf-player/plugin/v2.1/cdfplugin.js"></script><script type="text/javascript">var cdf = new cdfplugin(); cdf.setDefaultContent('<a href="http://www.wolfram.com/cdf-player/"><img src="https://drive.google.com/file/d/0B6fZNFG943hiLVZYOTdTTlhEbVE/view?usp=sharing"></a>'); cdf.embed('https://drive.google.com/view?id=0B6fZNFG943hibTlqeHBrQXlsOUU', 439, 320); </script>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-85462002603045693662015-09-01T11:54:00.001-07:002015-09-01T13:23:34.113-07:00Sun TrackerHere is a demonstration that probably won't work, but it was fun to make. It is an image of the path of the corner of my building throughout the year. Each black curve is the path through the day (call it an equi-day curve) and each red curve is the path of the shadow at the same time every day (on your watch) throughout the year (call it an equi-time curve). The green dot should be the current position of the end of a shadow of a vertical pole of height 1 m, and the black line from the origin to the green dot is the shadow. The darker black curve and darker red curve are the current equi-day and equi-time curves (related to the analemma). Note that if you are viewing this when UTC-5 (Eastern Standard Time) is dark, there shouldn't be any figures. Also, you really have to be patient/fight with the CDF plugin to get it to work. But it did work on a random windows machine... Good luck.<br /><br /><div style="text-align: center;"><iframe frameborder="0" height="942" scrolling="no" src="https://dl.dropboxusercontent.com/u/28060922/SunTracker.html" width="804"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-73873944595151446682014-10-07T05:36:00.000-07:002014-11-03T12:20:02.427-08:00Rolling and Slipping Sphere<div style="text-align: center;"><iframe frameborder="0" height="800px" scrolling="no" src="https://dl.dropboxusercontent.com/u/28060922/SlippingSphere.html" width="800px"></iframe></div><br /><br />The above demonstration illustrates the dynamics of a rigid sphere on a hard frictional surface with friction coefficient $\mu$. I started thinking about this problem after a discussion with students about the following problem. Imagine a spinning sphere (where the rotation axis of the sphere is in the plane of the surface) is dropped from a short distance onto the surface with no initial center of mass velocity. How long does it take to roll without slipping? I worked on this problem before and had <a href="http://physnick.blogspot.com/2012/04/i-recently-did-this-for-my-class.html" target="_blank">a nice demonstration of it</a> to aid my thinking, but I thought the generalization would be very difficult.<br /><br />Later in the day, my friends Rich C and Howard A and I talked about what would happen when a ball that was spinning at a $45^\circ$ angle with respect to the perpendicular was dropped on the surface. Eventually with enough thinking, we determined the initial velocity of the contact point and I convinced myself that the ball would, seemingly paradoxically, move in a straight line.<br /><br />Over the next few days, I convinced myself and eventually worked out the following:<br /><br /><ol><li>The force and torque on the sphere is set by the initial velocity of the center of mass and the initial angular velocity and <i>remains constant relative to the space coordinates while the sphere is slipping.</i></li><li>Since the force and torque are in the plane (the black and red arrows in the plane at the origin in the demonstration), the initial angular velocity component perpendicular to the surface never changes. A ball that is perfectly spinning like a top won't move.</li><li>The velocity of the contact point is composed of two velocities: the center of mass velocity and the velocity of the contact point relative to the body. These two vectors change according to the force and torque (perpendicular to each other), but they seemingly collude to eventually produce a zero velocity of the contact point relative to the surface (shrinking blue vector) at the point of rolling without slipping. This seems like magic to see happening, but Rich had a good explanation of why this would happen. More on that later.</li><li>Since the force and torque are constant from the very beginning of the problem, both the center of mass velocity and angular momentum vector change according to KINEMATICS. That's all! The velocity changes linearly (2D kinematics) and the angular velocity changes linearly (2D kinematics). The only complication is integrating the Euler angles forward. Originally I had thought to express all force and torque laws in terms of the Euler angles. This lead to 6 terrible coupled nonlinear second-order differential equations that gave me stability problems due to the coordinate singularities. I eventually realized that (just like the Hamiltonian approach) I could solve for the velocity dynamics first, then extract the Euler angles later.</li><li>The time-till-rolling can be calculated by evaluating the cross product of the center of mass velocity (a linear function of time) and the relative velocity with respect to the body (a linear function of time). There is no real reason to do the cross product, but when they are antiparallel, the velocity of the contact point relative to the surface is zero. The cross product is just a nice way to get a scalar from vectors and to do the derivation without dealing with magnitudes of vectors (which are ugly).</li><li>The torque and force laws must be done in the space frame since the center of mass is accelerating and therefore a non inertial frame. </li></ol>Here is the time it takes the ball to roll without slipping. Here the velocities and angular velocities are dimensionless (basically $r = g = 1$): <br /><ol></ol><div>$$t_{\textrm{roll}} = \frac{-v_x \omega_x - v_y \omega_y}{\mu \left( \beta f_y v_x - \beta f_x v_y + f_x \omega_x + f_y \omega_y \right)}$$</div><div><br /></div><div>After the slipping phase, the sphere rolls without slipping and travels in a straight line.</div><div><br /></div><div>Some comments about the simulation:</div><div><ul><li>$\mu$ is the frictional coefficient.</li><li>$\beta = \frac{m r^2}{I}$ is the ratio of mass times the radius squared of the sphere and the moment of inertia.</li><li>$mag$ and $\theta \phi$ are the magnitude and direction of the initial angular velocity (black vector attached to the sphere's center). Horizontal is $\theta$ and vertical is $\phi$ of the spherical coordinates of the angular velocity vector.</li><li>$v_0$ is the initial velocity of the center of mass. Centered in the square is (0,0).</li><li>The track of the sphere over the surface is plotted in red while it is slipping and blue when it is rolling without slipping.</li><li>The simulation time is displayed in the graphic. You can speed up and slow down the simulation by pressing the double up and double down arrows next to the time slider.</li></ul></div><div>Further comment about the center of mass and the velocity of the point relative to the body colluding to cancel:<br /><br />Rich suggested that from the ball's perspective, the dynamics should be the same as a ball being held by pincers (or pads or parallel compacting planes) on top and on bottom. Since in this instance, the relative velocity of the contact point(s) is the same as a plane (or two compacting parallel planes), the direction of each force will remain the same throughout the dynamics. There is no way for the pincers or planes to do anything other than slow the ball down in the direction that it sees. The advantage of the pads is that now there is no center of mass velocity and it the angular velocity slowly adjusts so that there is no motion at the contact points and thus the ball ends up spinning around the line connecting the contact points.</div><div><br /></div><div><br /></div><div><br /></div><div><br /></div><div><br /></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-19164249781825099802014-09-29T08:31:00.000-07:002014-09-29T12:07:09.461-07:003 Ball Stress CalculatorSome people may have to make sure to allow/unblock ads to see the cdfs. Additionally you may need to download the Wolfram CDF player.<br /><br />This calculates the stress applied in a 3 point ball loading configuration according to the formula:<br /><br />$$\sigma = \frac{-3}{4 \pi} P \frac{\left( X - Y \right)}{b^2}$$<br /><br />where<br />$$X = (1+\nu) \ln \left( \frac{r_2}{r_3}\right)^2 + \frac{1-\nu}{2} \left( \frac{r_2}{r_3}\right)^2,$$<br />$$Y = (1+\nu) \left(1+\ln \left( \frac{r_1}{r_3}\right)^2 \right) + (1-\nu) \left( \frac{r_1}{r_3}\right)^2,$$<br />$P$ is the applied load in Newtons,<br />$r_1$ is the radius of the support circle in mm,<br />$r_2$ is the radius of the loaded area in mm,<br />$r_3$ is radius of the specimen in mm,<br />$b$ is the sample thickness in mm,<br />and $\nu$ is Poisson's ratio.<br /><br />The radius of the support circle is, as far as I can tell, the radius of the circle produced by the points of the 3 balls supporting the sample. The radius of the loaded area is the radius of the piston on top pressing down on the sample. The radius of the specimen is the radius of the circular sample. From what I can tell, $r_2 < r_1 < r_3$ by definition. <br /><br /><div style="text-align: center;"><iframe frameborder="0" height="800px" scrolling="no" src="https://dl.dropboxusercontent.com/u/28060922/stress.html" width="550px"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-4113649982944181782013-10-24T14:03:00.002-07:002013-10-24T14:48:57.810-07:00Tetrahedral Structures ala Bit-PlayerThe wonderful blog entry found <a href="http://bit-player.org/">here</a> by Brian Hayes at Bit-Player was inspirational and led me to an area that I had long thought barren! I recently got my son some of the Geomag sticks and ball bearings and playing with them I got the impression that they were fun, but the models that I could make were limited. Well I was SO wrong. In his blog entry Brian Hayes introduced me to the tetrahelix, but also issued an irresistible challenge concerning a bridge that he built. He said "The result (below) is no longer a helix at all but a weird sort of bridge with an arched spine and two zigzag rails. If you extend the arc further, will the two ends meet to form a closed loop? I don’t have enough Geomags to answer that question." I just couldn't resist. I modelled it on Mathematica and was eventually led to a notation for a chain of attached tetrahedrons. I started by labelling his original figure<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-rtgyqS7xq4o/UmlwmAaFgEI/AAAAAAAAAdY/qP-znC7ql7E/s1600/zigzag-helix-bridge-2982_1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="117" src="http://1.bp.blogspot.com/-rtgyqS7xq4o/UmlwmAaFgEI/AAAAAAAAAdY/qP-znC7ql7E/s400/zigzag-helix-bridge-2982_1.jpg" width="400" /></a></div><br />I started with arbitrarily numbering the vertices of the tetrahedron on the end. From this, I could specify the next tet by writing down the vertex that was reflected through the plane of the other three. For example, in his picture above, 1 was reflected to 1'. I continued this forward to get the sequence:<br /><div style="text-align: center;"><br /></div><div style="text-align: center;">{1,4,3,2,4,1,3,4,2,1,4,3,2,4,1,3,4,2,1,4,3,2,4} </div><div style="text-align: center;"><br /></div><div style="text-align: left;">The great thing is this repeats! The unit is {1,4,3,2,4,1,3,4,2}. The natural questions now just flood in with this notation. For example, I can do random walks using this sequence. Also, I believe the fact that this structure curves is due to the over abundance of the number 4. For example, the unit {1,2,3,4} leads to the tetrahelix (as do all of the permutations of this unit) which makes me believe that there is definitely some deeper math going on here. For example I believe an even permutation leads to a right-handed helix and an odd permutation leads to a left-handed helix. Anyway, I found with 95 steps using the above unit, the ring comes close, but does not close as you can see below.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-wy0lAvYbHmo/UmmDprEXr0I/AAAAAAAAAds/u5m075cDJFo/s1600/CloseBridge.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="326" src="http://4.bp.blogspot.com/-wy0lAvYbHmo/UmmDprEXr0I/AAAAAAAAAds/u5m075cDJFo/s400/CloseBridge.jpg" width="400" /></a></div>This made me wonder if any units can actually close after repeating. I think not, but I haven't checked yet. I imagine that if you are dealing with units, an integral multiple of the unit would have to match up with the full circle. But then, you could meander away and then come back, or you could have a non-repeating number that closed on itself.<br /><br />Anyway here is a short cdf where you can explore units up to 6. </div><div style="text-align: center;"><iframe frameborder="0" height="730" scrolling="no" src="https://dl.dropboxusercontent.com/u/28060922/tetspaceexplorer.html" width="584"></iframe></div><br /><div style="text-align: left;">I have not worried about intersection, and if you relaxed the requirement that consecutive numbers aren't equal (this just flips a tet back to where it started) you can make any tetrahedral structure, you just label the tetrahedra sequentially. When you get to the end of a limb, you reflect the sequence till you get back to the main line. This is not the best way of doing things, but it seems to be complete. Anyway, many more things to explore! I think I have to invest in a few more Geomag sets. </div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-84855186054270368052013-03-26T14:33:00.000-07:002013-03-27T05:54:53.775-07:00Effective Mass Surface of an EllipsoidHere are the ellipsoidal effective mass surfaces.<br /><br />$E = E_0 \pm \frac{\hbar^2 k_{r}^2}{2 m_e} f(\theta, \phi)$ <br /><br /> <div style="text-align: center;"><iframe frameborder="0" height="654" scrolling="no" src="http://dl.dropbox.com/u/28060922/Ellipsoidalfs.html" width="852" altimage = "http://dl.dropbox.com/u/28060922/Ellipsoidalfs.png"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-66118461488778665102013-01-18T10:47:00.002-08:002013-01-18T11:20:43.689-08:00Band Warping<div style="text-align: center;"><iframe frameborder="0" height="1250" scrolling="yes" src="http://dl.dropbox.com/u/28060922/ThreeBandwKittelZTcondS_bc_350.html" width="800"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-87243434026875211892012-09-11T13:14:00.000-07:002013-11-25T14:15:04.660-08:00DFT of Geometric MatricesHere we plot special matrices (where any element can only be 0,i,-1,-i, and 1) and the DFT of those matrices. Click on a cell of the matrix to cycle through element values. To change the color, select from the drop down box. To select a larger size, find select the size then press "Reset". The numerical value of the DFT gets annoying, so hide it by unchecking the checkbox. <br /><div style="text-align: center;"><iframe frameborder="0" height="884" scrolling="yes" src="http://dl.dropbox.com/u/28060922/MatrixDFTAloneHTML.html" width="675"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-7978511820510165532012-09-10T14:47:00.001-07:002012-09-11T09:57:37.136-07:00Two-band Tight-Binding Band Plot z=0Two-band tight-binding model with 3rd NN interactions<br /><br /><div style="text-align: center;"><iframe frameborder="0" height="441" scrolling="no" src="http://dl.dropbox.com/u/28060922/RectangularNonDegenerateBandWarping.html" width="852"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-3075853364612584622012-08-09T14:38:00.000-07:002012-08-09T14:38:06.934-07:00Graphite Q PointHere is a demonstration of the energy dispersion near the Q point of Graphite.<br /><br /><br /><iframe frameborder="0" height="619" scrolling="no" src="http://dl.dropbox.com/u/28060922/TightBindingECurves.html" width="761"></iframe>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-50959573121820251752012-04-25T23:01:00.003-07:002012-04-25T23:03:02.441-07:00Another Attempt. This one is useful.I think the last post was too short and did not explain enough. I am attempting (as are many others) to put Wolfram's CDF documents online. It seems like it is possible to view in major browsers (unless viewing on a Linux machine). If you are using Mac or Windows and you do not see anything for this post and the last, just click on the icon in the middle and download the plugin. This is like the Adobe Flash plugin or any other kind of plugin. You need it to view the content.<br /><br />If this works (and I work out all of the bugs, like re-sizing of the last entry), I hope to put a lot more on here. It is just easier than making movies and trying to explain everything.<br /><br />Here is an attempt at plotting the absorption coefficient for the indirect phonon-assisted transitions in a crystal:<br /><br /><br /><div style="text-align: center;"><iframe frameborder="0" height="534" scrolling="no" src="http://dl.dropbox.com/u/28060922/PhononAssistedTransport.html" width="450"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-1089951661259720392012-04-24T12:14:00.000-07:002013-11-25T14:17:44.021-08:00CDF Example of Rolling DiskI recently did this for my class. This will be a test of Wolfram's CDF plugin. Here is a disk slipping on a surface:<br /><div style="text-align: center;"> <iframe frameborder="0" height="691" scrolling="yes" src="http://dl.dropbox.com/u/28060922/SlidingBall.html" width="853"></iframe></div>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-29572884372797499922011-10-10T20:11:00.000-07:002011-10-10T20:11:00.275-07:00Musical Spiral<a href="http://wheelof.com/whitney/index.php?var=v20">This is fantastic</a>! I found this in a link off of mathpuzzle.com. <br /><br />There is so many different directions you can go with this! I love that if you close your eyes, the music has <i>some </i>structure, but it mostly sounds too complicated to understand. However, if we found a way to graph it, to represent it in another way, we could understand the complexity. That is why representation of data and ideas is so important. The right notation can make all the difference in the world. <br /><br />What other geometric music could be attempted? Something that would be very simple geometrically, but complicated musically. <br /><br />Is there a three dimensional analogue? What is the best scale to use?<br /><br />My favorite is probably <a href="http://wheelof.com/whitney/index.php?var=v20">this one</a> or <a href="http://wheelof.com/whitney/index.php?var=v15">this one</a>.<br /><br />Have fun!Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-18818546223865670272011-10-05T13:00:00.000-07:002011-10-05T13:00:02.279-07:00MorpionRemember that lame paper and pencil game kids use to play where you connected the dots?<br /><br />Well, I ran into a solitaire game that was so much better than this. Why didn't I know this game earlier!<br /><br />I found it on a link at a favorite website by <a href="http://www.mathpuzzle.com/">Ed Pegg, Jr.</a> that I check every so often, that always has good stuff.<br /><br />It is called <a href="http://www.morpionsolitaire.com/">Morpion</a>, check it out!Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-49887052722849707992011-10-01T07:00:00.000-07:002011-10-01T07:00:04.936-07:00Quick Plumbing ProblemThe hot water knob on the downstairs bathroom does not fit and you also had to turn it the wrong way. So I decided to fix it. I should have known it would develop into a whole project.<br /><br />I casually took the handle off, and went to "Home-I-Depo" to replace it. Oh my goodness. It took forever. This was the most complicated system to organize faucets. For example, there might be a part named 2H-1C/H. Are you kidding me?!? What kind of system is that? Anyway, eventually (R almost went crazy) I figured out that the first number is the length, the letter was the style and size of the part that goes into the handle. The next number is the type of threading (?) and C or H or C/H stands for hot or cold or both.<br /><br />The problem was that I didn't know what I had. There were no markings on my faucet. Anyway, after a few missed tries, and I finally got a set that seemed right.<br /><br />When I went to replace it, I started to turn out the shutoff valve right under the bathtub. As soon as I started turning the knob, it started to leak profusely. Oh no. This is not good.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-PYHeP1-tbFw/ToZ-GQjlhNI/AAAAAAAAAWs/jmbgjU-ms0o/s1600/DSC_0018.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://1.bp.blogspot.com/-PYHeP1-tbFw/ToZ-GQjlhNI/AAAAAAAAAWs/jmbgjU-ms0o/s320/DSC_0018.JPG" width="320" /></a></div>I finished the replacement of the faucet, but now I had a bigger problem.<br /><br />Let's see...<br /><br />I had to replace the valve with a new one.<br /><br />I went back to "Home-I-Depo" and got a $20 Benz-O-matic (propane torch) for copper sweating and a new shutoff valve. <br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-GkD8yPkKXIY/ToZ-j-CwbxI/AAAAAAAAAWw/WZYtC0dCX38/s1600/DSC_0020.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://4.bp.blogspot.com/-GkD8yPkKXIY/ToZ-j-CwbxI/AAAAAAAAAWw/WZYtC0dCX38/s320/DSC_0020.JPG" width="320" /></a></div><br />I crawled into the crawlspace (I hate doing this), and turned off the water.<br /><br />It took forever to unsolder the connections, and once it was undone, I couldn't move it anywhere. The way it had been constructed, there was no play in the pipes, and I couldn't separate the connections. I had to cut the pipe and work it off that way.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-lOxMoQ50gng/ToZ_F5iYO2I/AAAAAAAAAW0/IotolwkBMuQ/s1600/DSC_0024.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://1.bp.blogspot.com/-lOxMoQ50gng/ToZ_F5iYO2I/AAAAAAAAAW0/IotolwkBMuQ/s320/DSC_0024.JPG" width="320" /></a></div><br /><br />Eventually, I replaced the valve, but not before making a HUGE mess of solder and copper and almost burning myself a few times. Man, I had forgotten about doing this. I did a lot of soldering when remodeling my kitchen. I don't exactly remember it fondly, but at least I felt the soldering joints were pretty. My soldering looks horrible, but on the other hand, I only had to do it once! After looking at my soldering job, I was sure that I would have to do it again.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-l68JZmzeCFs/ToZ_bB0zjfI/AAAAAAAAAW4/cI6MajgxIl4/s1600/DSC_0022.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://4.bp.blogspot.com/-l68JZmzeCFs/ToZ_bB0zjfI/AAAAAAAAAW4/cI6MajgxIl4/s320/DSC_0022.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-C_uv6TZ8WDA/ToZ_k4fM30I/AAAAAAAAAW8/TW4LkxUwaA0/s1600/DSC_0026.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://2.bp.blogspot.com/-C_uv6TZ8WDA/ToZ_k4fM30I/AAAAAAAAAW8/TW4LkxUwaA0/s320/DSC_0026.JPG" width="212" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-fCUZZXaWDV8/ToaAD8jt4-I/AAAAAAAAAXA/Df-otRAHpm8/s1600/DSC_0025.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://2.bp.blogspot.com/-fCUZZXaWDV8/ToaAD8jt4-I/AAAAAAAAAXA/Df-otRAHpm8/s320/DSC_0025.JPG" width="320" /></a></div><br /><br />Two things to remember for next time:<br /><br />1) Use a male-male adapter that doesn't have a stop. That way, I can slip it on, and still fit very tight fitting.<br /><br />2) Make sure that you get all of the water out if you can, it took forever to heat the pipe when there was still water in the pipe.<br /><br />3) Try not to end up with the red handle for the cold and the blue handle for the hot.Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com1tag:blogger.com,1999:blog-8861069262211796057.post-74629505913666440592011-08-20T22:26:00.000-07:002011-08-20T22:26:46.005-07:00Trunk BoardI was driving home from my brother's house late at night. It was cold and sleeting. I had forgotten my cell phone but my brother's house was only 15 miles away, so I should be alright...<br /><br />I was on 495 heading west when I saw flashing lights ahead. I was trying to see what was happening when I saw a few metal bed frames in my lane and the next. I ran over them. There was nothing I could do. The visibility was pretty low, and I was moving at about 80 mph. Anyway, as soon as I hit them, I knew it did some damage. I just didn't know what. So I turned off the heat and the radio and I listened. Incidentally, the cop had pulled over a truck carrying bed frames...<br /><br />I was listening for any sort of trouble, but I knew it was only a matter of minutes. I decided to go 201 to 193 to go home as opposed to 495 the rest of the way. As soon as I got off the interstate and onto 201, I heard my tire start to thump. So I pulled over where I could and started got out. It was freezing and wet. I went to my trunk (which is ALWAYS full) and started digging out the spare. I couldn't believe my bad luck! Since we had been celebrating something (Christmas?) I was all dressed up with my leather coat on. I was none to happy. I finally got the spare out, and went to get the tire off. I jacked the car up (which was fun), and went to take off the lug nuts. They wouldn't budge. I worked for 45 minutes in the cold and rain eventually putting all my weight onto the crowbar. I kicked them and cursed them. Eventually I got them off, but I have never had difficulty like that before.<br /><br />Anyway, I was so angry that I *tossed* the tire into the trunk cracking the board that makes a flat surface above the spare tire enclosure. Great.<br /><br />I got home alright, very wet and cold, though nobody ever did stop. I guess I wouldn't either. I big man in the dark rain wearing a black leather coat and carrying a tire-iron.<br /><br />After many dollars (actually the tire was still under warranty), I had a new tire, but the trunk was not very functional. Until today!<br /><br />Here is a before picture of the board:<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-Dz59YgK9i8U/TlCUxTE77pI/AAAAAAAAAWY/946G2Bj9lSE/s1600/DSC_0007_023.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://3.bp.blogspot.com/-Dz59YgK9i8U/TlCUxTE77pI/AAAAAAAAAWY/946G2Bj9lSE/s320/DSC_0007_023.JPG" width="320" /></a></div>After a trip to "Home - I - Depo" (as R says), $15, and a few hours, this is what I got:<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-tgIOWv7HUGk/TlCVHzlDQ-I/AAAAAAAAAWc/zTy5zt9GgFY/s1600/DSC_0009_025.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://2.bp.blogspot.com/-tgIOWv7HUGk/TlCVHzlDQ-I/AAAAAAAAAWc/zTy5zt9GgFY/s320/DSC_0009_025.JPG" width="320" /></a></div><br /><br />Two things to note.<br /><br />1) I put in (or really just couldn't find one cheap 4x4 plywood piece) a hinged joint right where spare tire ends. This is nice because it gives me some extra storage space.<br /><br />2) I had to grind off the screws that protruded from the other side of the hinge. I really thought this was going to be a problem, but it turned out to be easy. Also, R really liked the sparks that flew during the grinding.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-2iDgIS-2YSA/TlCVe6x5nFI/AAAAAAAAAWg/IP8LIPFcLGk/s1600/DSC_0010_026.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="212" src="http://4.bp.blogspot.com/-2iDgIS-2YSA/TlCVe6x5nFI/AAAAAAAAAWg/IP8LIPFcLGk/s320/DSC_0010_026.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><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/-EVFuBY8fu5k/TlCVxxfIGyI/AAAAAAAAAWk/EvE06Q3q71Q/s1600/DSC_0012_028.JPG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://1.bp.blogspot.com/-EVFuBY8fu5k/TlCVxxfIGyI/AAAAAAAAAWk/EvE06Q3q71Q/s320/DSC_0012_028.JPG" width="212" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">All better!</td></tr>
</tbody></table>Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-38150227538034695692011-02-16T20:33:00.000-08:002011-02-16T20:33:00.476-08:00MathJaxTo get equations that just work on your site, go to <a href="http://www.mathjax.org/">http://www.mathjax.org/</a>.<br /><br />On those pages, there are several options, since it is a very flexible program, but the simplest is to copy the following text into your blog template or webpage:<br /><br /><pre><span class="nt"><script </span><span class="na">type=</span><span class="s">"text/javascript"</span> <span class="na">src=</span><span class="s">"</span>http://www.mathjax.org/mathjax/MathJax.js<span class="s">"</span><span class="nt">></span><br /> <span class="nx">MathJax</span><span class="p">.</span><span class="nx">Hub</span><span class="p">.</span><span class="nx">Config</span><span class="p">({</span><br /> <span class="nx">extensions</span><span class="o">:</span> <span class="p">[</span><span class="s2">"tex2jax.js"</span><span class="p">],</span><br /> <span class="nx">jax</span><span class="o">:</span> <span class="p">[</span><span class="s2">"input/TeX"</span><span class="p">,</span> <span class="s2">"output/HTML-CSS"</span><span class="p">],</span><br /> <span class="nx">tex2jax</span><span class="o">:</span> <span class="p">{</span><br /> <span class="nx">inlineMath</span><span class="o">:</span> <span class="p">[</span> <span class="p">[</span><span class="s1">'$'</span><span class="p">,</span><span class="s1">'$'</span><span class="p">],</span> <span class="p">[</span><span class="s2">"\\("</span><span class="p">,</span><span class="s2">"\\)"</span><span class="p">]</span> <span class="p">],</span><br /> <span class="nx">displayMath</span><span class="o">:</span> <span class="p">[</span> <span class="p">[</span><span class="s1">'$$'</span><span class="p">,</span><span class="s1">'$$'</span><span class="p">],</span> <span class="p">[</span><span class="s2">"\\["</span><span class="p">,</span><span class="s2">"\\]"</span><span class="p">]</span> <span class="p">],</span><br /> <span class="p">},</span><br /> <span class="s2">"HTML-CSS"</span><span class="o">:</span> <span class="p">{</span> <span class="nx">availableFonts</span><span class="o">:</span> <span class="p">[</span><span class="s2">"TeX"</span><span class="p">]</span> <span class="p">}</span><br /> <span class="p">});</span><br /><span class="nt"></script></span></pre><br />I found this after a much longer search than I thought at the <a href="http://www.mathjax.org/resources/docs/?///////////configuration.html#configuration">configuration webpage</a>. Now what they actually have for the source is <span class="s">"path-to-MathJax/MathJax.js"</span> and encourage you to get your own copy. However, I don't have a place to put the downloaded mathjax stuff, so I just used theirs. I don't think this is what they really want, but I didn't see a way around it.<br /><br />There are lots of LaTeX commands you can use, here is a <a href="http://www.mathjax.org/resources/docs/?/////tex.html">list</a><br /><br />For example, this just works<br /><br />$$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}}}}$$Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-64169949386113880242011-02-09T20:20:00.000-08:002011-02-09T20:20:00.353-08:00Barcodes!For some crazy reason, I wanted to make barcodes. I have no real reason why. I guess I wanted to be able to read barcodes, or make a reader for barcodes. So in order to do that I need to be able to make them.<br /><br />Of course there are TONS of barcode systems. Some are very tempting, but one of the simplest seems to be <a href="http://en.wikipedia.org/wiki/Code_128">Code 128</a>. This is used fairly often, and it seemed pretty robust.<br /><br />Barcodes are basically set up with a few bars in the front to tell you which part of 128 you are using (A, B, or C), then individual characters come along. Each character is 6 stripes long starting with black and alternating. Varying the thickness of the stripes will give you each character. So, for example, "A" is 111323, that is, 'a black stripe that is 1 unit wide, a white stripe that is 1 unit wide, a black stripe that is 1 unit wide, a white stripe that is 3 units wide, a black stripe that is 2 units wide, a white stripe that is 3 units wide. Look at the animation below<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzIXjGmx2I/AAAAAAAAAUA/4TU9P3XY3pk/s1600/letters.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzIXjGmx2I/AAAAAAAAAUA/4TU9P3XY3pk/s320/letters.gif" width="279" /></a></div><br /><br />The end of the bar code is a check number and a stop character that is 7 bars long and unique. Here is the chart that I used from Wikipedia<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzIxpzyHfI/AAAAAAAAAUE/K5jvYMSSTDQ/s1600/tabletext.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzIxpzyHfI/AAAAAAAAAUE/K5jvYMSSTDQ/s1600/tabletext.jpg" /></a></div><br /><br />There are multiple ways to encode a message, so I just used code B throughout. There are Shift keys so that the following character is in the shifted language. Also, you can encode '12' as a '1' followed by a '2' or as a single character '12'. I wonder if I could make something so that given a message it would give me the simplest barcode... hmm...<br /><br />Anyway, I encoded messages using just column B using <i>Mathematica</i>.<br /><br />I first changed a string like "Hello World" to its list of black and white bar notation using<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzLIIbU-sI/AAAAAAAAAUU/0kvc5RqUhK0/s1600/Barcode_5.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzLIIbU-sI/AAAAAAAAAUU/0kvc5RqUhK0/s1600/Barcode_5.gif" /></a><br /><br /><br /><br /><br /><br /><br /><br />So now I can try "Hello World" <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzO-S5ymII/AAAAAAAAAVE/0ZJvUiq7IPA/s1600/Barcode_6.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzO-S5ymII/AAAAAAAAAVE/0ZJvUiq7IPA/s1600/Barcode_6.gif" /></a></div> <br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzKTC8xD4I/AAAAAAAAAUQ/qtNKeslkusw/s1600/Barcode_6.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzKTC8xD4I/AAAAAAAAAUQ/qtNKeslkusw/s1600/Barcode_6.gif" /></a></div><br />encode uses another function to find the check number<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzLSTqMgXI/AAAAAAAAAUY/uj2JCaXqIg0/s1600/Barcode_2.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzLSTqMgXI/AAAAAAAAAUY/uj2JCaXqIg0/s1600/Barcode_2.gif" /></a></div><br />This is not too important, and I won't go into it. Basically, it calculates another character based on all of the characters in your text. This is a standard way of know when you got the right answer when you are reading it. Just do a quick check.<br /><br />Then I plot it (with a bunch of rectangles)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzLt81s2tI/AAAAAAAAAUc/d7Lqv0970eI/s1600/Barcode_8.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzLt81s2tI/AAAAAAAAAUc/d7Lqv0970eI/s1600/Barcode_8.gif" /></a></div>I only plot black rectangles because that is all I need to plot.<br /><br />Here are some examples<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMIL0jAnI/AAAAAAAAAUg/RpcBpdYcIYc/s1600/Barcode_10.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMIL0jAnI/AAAAAAAAAUg/RpcBpdYcIYc/s1600/Barcode_10.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMSHFwvlI/AAAAAAAAAUo/2Vr9gOWad4U/s1600/Barcode_11.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMSHFwvlI/AAAAAAAAAUo/2Vr9gOWad4U/s1600/Barcode_11.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzMVwpAGCI/AAAAAAAAAUs/7rtmmKfThNY/s1600/Barcode_12.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzMVwpAGCI/AAAAAAAAAUs/7rtmmKfThNY/s1600/Barcode_12.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMZY7nlsI/AAAAAAAAAUw/il_NT3Un0dA/s1600/Barcode_13.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMZY7nlsI/AAAAAAAAAUw/il_NT3Un0dA/s1600/Barcode_13.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzMcwZDWgI/AAAAAAAAAU0/EbrL-68qYx8/s1600/Barcode_14.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUzMcwZDWgI/AAAAAAAAAU0/EbrL-68qYx8/s1600/Barcode_14.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMpQ4jPUI/AAAAAAAAAU4/vngmHA_ZtZw/s1600/Barcode_15.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMpQ4jPUI/AAAAAAAAAU4/vngmHA_ZtZw/s1600/Barcode_15.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMusYVP1I/AAAAAAAAAU8/4BS-75WuysQ/s1600/Barcode_16.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUzMusYVP1I/AAAAAAAAAU8/4BS-75WuysQ/s1600/Barcode_16.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzM1iEB7yI/AAAAAAAAAVA/Yd9rblPBCO8/s1600/Barcode_17.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUzM1iEB7yI/AAAAAAAAAVA/Yd9rblPBCO8/s1600/Barcode_17.gif" /></a></div><br />Note that whatever the message is, it always starts with 211214 and always ends with 2331112.<br /><br />Next, on to the barcode reader!Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-68440275400972793782011-02-02T21:58:00.000-08:002011-02-05T07:48:20.767-08:00More envelopes than a post office!<div class="separator" style="clear: both; text-align: center;"></div><br /><span style="font-size: small;">I wish I had more time to explore this. I found a passing comment in S. V. Meleshko's </span><span style="font-size: small; font-style: italic;">Methods for Constructing Exact Solutions of Partial Differential Equations</span><span style="font-size: small;"> about envelopes of curves. I will get to his statement at the end. Basically suppose we have a family of curves like the blue ones below</span><br /><span style="font-size: small;"><br /></span><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUo8MeDifsI/AAAAAAAAATg/nlUZPXdDtHc/s1600/fig1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="202" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUo8MeDifsI/AAAAAAAAATg/nlUZPXdDtHc/s320/fig1.jpg" width="320" /></a></div><span style="font-size: small;"><br /></span><br /><span style="font-size: small;">This is an example of a one parameter family of curves. Specifically, I made this by finding lines where the intercepts always add to 10,</span><br /><span style="font-size: small;"> </span><span style="font-size: small;">$$y(x,a)= \frac{a - 10}{a} x + 10 - a.$$ </span><br /><span style="font-size: small;">This begs the question of finding the envelope of the family of curves shown in black in the figure. What is the equation that describes the envelope? How do we find it?</span><br /><span style="font-size: small;"><br /></span><br /><span style="font-size: small;">I have always been fascinated by these types of curves ever since I did a string art project in middle school. I wondered, as I do now, what types of curves you could produce.</span><br /><span style="font-size: small;"> </span><br /><br /><span style="font-size: small;">This has been addressed before in various places, and I gather that the topic used to be quite common in Calculus courses. However, in case you don’t see it right away, here is a way to obtain the envelope. </span><br /><br /><span style="font-size: small;">Let’s fix $x$. Now for this $x$, say $x_0$, we have a relation between $y$ and $a$ through the equation for a curve in the family. We can think of the envelope as an extremum of the relationship between $y$ and $a$. Illustrating this graphically, if we consider the intersection of the vertical line $x_0$ and our family of curves $y(x_0,a)$, then as we scan through $a$, we will see that there are special points where the intersection point reaches a maximum. This can be seen in the animation below (right click on the image below and select 'open in a new tab').</span><br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUo8nNOjrFI/AAAAAAAAATk/6OTU4k6XZ6g/s1600/animgif_EnvelopesSmaller.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="202" src="http://2.bp.blogspot.com/_ksS-fQHJJrQ/TUo8nNOjrFI/AAAAAAAAATk/6OTU4k6XZ6g/s320/animgif_EnvelopesSmaller.gif" width="320" /></a></div><br /><br /><span style="font-size: small;">As we scan through $a$, we can see the intersection point (in red). When it slows down and changes direction, this is a point of the envelope. </span><br /><br /><span style="font-size: small;"> </span><br /><span style="font-size: small;">This is exactly $g(x_0)$. We can then describe the envelope equation as</span><br /><span style="font-size: small;">$$g(x) = y ( x, a_{*} (x) ) $$</span><br /><span style="font-size: small;">where $a_{*}(x)$ is found by finding the extremum of $y(x,a)$ with respect to $a$. Thus $a_{*}(x)$ is the solution to</span><br /><span style="font-size: small;">$$\frac{\partial y(x,a)}{\partial a}=0.$$</span><br /><span style="font-size: small;">For our case explicitly, the solution is</span><br /><span style="font-size: small;">$$g(x)=10 \pm 2 \sqrt{10} \sqrt{x} + x.$$</span><br /><span style="font-size: small;">A graph of these two curves can be seen in black in the first figure.</span><br /><br /><span style="font-size: small;">Note also from the way we have written the envelope that every point of $g(x)$ necessarily lies on a specific curve in the family of curves. This means that at each point a particular curve from our family is tangent to the envelope. Likewise, we also know that the envelope is tangent to one of our curves in the family. This is a good thing to recognize because from this we may construct a differential equation. We know that</span><br /><span style="font-size: small;"></span><span style="font-size: small;">$$g(x) = \frac{a_{*}-10}{a_{*}} x + 10 - a_{*},$$ </span><br /><span style="font-size: small;">$$g'(x)=\frac{a_{*} - 10}{a_{*}}.$$</span><br /><span style="font-size: small;"></span><br /><span style="font-size: small;">From this, we may eliminate $a_*$ to obtain a differential equation for $g(x)$. In this case we get</span><br /><span style="font-size: small;">$$g'(x)(10-x+x g'(x))=g(x)(g'(x)-1).$$</span><br /><span style="font-size: small;">This is a nonlinear, first-order differential equation. The solution for this differential equation is then the envelope function. If we plug the envelope in, we find that it is, in fact, a solution.</span><br /><br /><span style="font-size: small;">The interesting point is that the two branches of the envelope are not the only solutions! All of the curves in the family are also solutions to this nonlinear differential equation. Since it is not a linear differential equation, there is not a uniqueness or existence theorem. We can see that the one parameter family of curves satisfies this equation since they satisfy the equations that the differential equation was derived from. I was just shocked that a family of lines could satisfy such an unruly differential equation.</span><br /><br /><span style="font-size: small;">Oftentimes we are confronted with a nonlinear differential equation where we are excited when we are able to get one solution to the problem. This is an example of an infinite number of solutions to a complicated looking differential equation. But most (an infinite number) of the solutions are just straight lines!</span><br /><br /><span style="font-size: small;">The converse of the above is actually easier to prove and we take it out of </span><span style="font-size: small;">Meleshko (not the proof -- although it is not hard). He says it in passing, and for two dimensional surfaces. Basically, he says (on page 9) that if you have a family of curves that is a solution of a differential equation, then the envelope is also a solution! Amazing!</span><br /><span style="font-size: small;"><br /></span><br /><span style="font-size: small;">Anyway, I know this is too long for anyone to read, but I will end with some interesting families of curves and their envelopes.</span><br /><span style="font-size: small;"><br /></span><br /><span style="font-size: small;">Here is a family of lines with sin waves on top</span><br /><span style="font-size: small;"> </span><span style="font-size: small;">$$y_{1}(x,a)= \frac{a - 10}{a} x + 10 - a + \sin(10 x).$$</span><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9J965gAI/AAAAAAAAATo/qdRuLFNlO8U/s1600/fig2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="202" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9J965gAI/AAAAAAAAATo/qdRuLFNlO8U/s320/fig2.jpg" width="320" /></a></div><br /><span style="font-size: small;"> </span><br /><span style="font-size: small;"><br /></span><br /><span style="font-size: small;">Here is a family of exponentials<br />$$y_{2}(x,a) = a e^{\frac{x}{a}}$$</span><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUo9QRHBbHI/AAAAAAAAATs/3b2ITUzd_wA/s1600/fig3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="213" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUo9QRHBbHI/AAAAAAAAATs/3b2ITUzd_wA/s320/fig3.jpg" width="320" /></a></div><br /><br />Here is a family of gaussians<br />$$y_{4}(x,a) = a e^{-(x-a)^2}$$<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9WCbFaWI/AAAAAAAAATw/KmHJOniiszE/s1600/fig4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="213" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9WCbFaWI/AAAAAAAAATw/KmHJOniiszE/s320/fig4.jpg" width="320" /></a></div><br /><br />Here is another family of gaussians<br />$$y_{5}(x,a) = \frac{1}{\sqrt{\pi} a} e^{-\left(\frac{x}{a}\right)^2}$$<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUo9bKTWEMI/AAAAAAAAAT0/LOs2Fy0XJRE/s1600/fig5.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="210" src="http://3.bp.blogspot.com/_ksS-fQHJJrQ/TUo9bKTWEMI/AAAAAAAAAT0/LOs2Fy0XJRE/s320/fig5.jpg" width="320" /></a></div><br /><br />Here is one where I was playing with lines again<br />$$y_{5}(x,a) = \frac{a^2 - 1}{a} \left( \frac{x}{a}+ \frac{1}{2} \right) - a$$<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9gccPRtI/AAAAAAAAAT4/jL6L0fVEwEs/s1600/fig6.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TUo9gccPRtI/AAAAAAAAAT4/jL6L0fVEwEs/s320/fig6.jpg" width="315" /></a></div><br />Two and three dimensions should be fairly straightforward to do (although I have not). What I would like to do is to be able to go backwards. I have played enough to make a<br /><br /><u><i>Conjecture</i></u>: any reasonable curve can be an envelope to a family of lines.<br /><br />I want to make this explicit, but that is a topic for another day.Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0tag:blogger.com,1999:blog-8861069262211796057.post-19327571584392341872011-01-16T09:55:00.001-08:002011-01-31T17:34:33.611-08:00Numbers for RI want to give my son R a head start on numbers and the best way I could think of to do that was to show him that there is a pattern to the series of numbers that keeps repeating. He knew 1 through 10, so I decided to make him a rectangular array of digits so that he could see the pattern.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TTM1-IF_BDI/AAAAAAAAATU/1Ru_qxL1Foc/s1600/numbersforroman3.jpeg"><img style="display: block; margin: 0px auto 10px; text-align: center; cursor: pointer; width: 400px; height: 225px;" src="http://1.bp.blogspot.com/_ksS-fQHJJrQ/TTM1-IF_BDI/AAAAAAAAATU/1Ru_qxL1Foc/s400/numbersforroman3.jpeg" alt="" id="BLOGGER_PHOTO_ID_5562849306173441074" border="0" /></a>Here each number has its own color. That way you can see the regular pattern. If you start on the left and move down the column, you get the first 10 numbers as usual. The advantage of seeing it in a rectangular array is that when you go across the top row, you also see the first ten numbers, but with 0's after them. I hoped this would be clear to him.<br /><br />The only problem is that, for whatever reason, the numbers from 11 to 19 do not follow the same convention as the numbers from 20 - 99. I don't know why. I was considering changing them for R's sake. I would call them "Ten - e - one", "Ten - e - two", etc. Then he would see the structure right away with the disadvantage that it is nonstandard. My wife S vetoed this. Oh well.<br /><br />That brings up an interesting question (at least to me). Where do the words "Twenty", "Thirty", etc. come from. My initial uneducated guess is that they are shortened versions of "Two - ten" and "Three - ten". But I don't know. Does anyone know the answer to this?Nicholashttp://www.blogger.com/profile/05760595107091537533noreply@blogger.com0