tag:blogger.com,1999:blog-55733982577220151032024-03-05T01:07:28.259-08:00The Ben Paul Thurston BlogMath, Movies, Science, Computers, and so on...benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.comBlogger351125tag:blogger.com,1999:blog-5573398257722015103.post-3381925735514694372016-05-13T07:35:00.001-07:002016-05-13T07:35:34.115-07:00UnGaussSupposing you had some initial point source locations, perhaps at the integers, and a standard deviation of a gaussian effect, you can undo the gaussian effect like so:<br />
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Set x(i)'s equal to the point source locations, and t^2 equal to the gaussian effect standard deviation...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxdQpgKfor6Ak083aY50bEdkkezfJPGgTYJ6dXYMJmSXNDSuSMWsdWl3jv3LFlMitzOysxbWR8XZON4Z8ux1n6H9QhmmjdYo6-9BZIfPws2Fq2lrOoiwKmXw_TLFueIa-Axfd1KWCdu9U/s1600/gauss1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxdQpgKfor6Ak083aY50bEdkkezfJPGgTYJ6dXYMJmSXNDSuSMWsdWl3jv3LFlMitzOysxbWR8XZON4Z8ux1n6H9QhmmjdYo6-9BZIfPws2Fq2lrOoiwKmXw_TLFueIa-Axfd1KWCdu9U/s1600/gauss1.PNG" /></a></div>
Now setting this sum evaluated at each x(i) equal to the final intensity at xi gives a system of n linear equations in n unknowns that can be solved...<br />
For example this could be the final distribution after a gaussian effect with a standard deviation of 1:<br />
It has values .2, .25, .2 at x(i) = 1,2,3<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiq6gC2x7CQEyKZa_fDWZnTbZJ-9LkSc5B0JtKBIljmj0ZZ4vgV0pK7zQU8XS5q_36c4DFz61BdyYHICPC8MeTQOlvaqHqUXnxIZEFXK8DPiiHWe17tS1Yr0WY_EWm5iZr8PTyfqx4pCk0/s1600/gauss2.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="296" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiq6gC2x7CQEyKZa_fDWZnTbZJ-9LkSc5B0JtKBIljmj0ZZ4vgV0pK7zQU8XS5q_36c4DFz61BdyYHICPC8MeTQOlvaqHqUXnxIZEFXK8DPiiHWe17tS1Yr0WY_EWm5iZr8PTyfqx4pCk0/s320/gauss2.PNG" width="320" /></a></div>
The solve command will look like:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRARY6yVZ8smxy39SRbmmhxJxwlnL45GLtapjRibsXyV-pmu5ckv2IXB5yGCPw6h8IA16wGFl41q89io5Dq4e3THgRkIS24eWub7ks69EBxAuacl0mE1RDxAQO5RzArSeSzU0oxy6Vj-I/s1600/guass3.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRARY6yVZ8smxy39SRbmmhxJxwlnL45GLtapjRibsXyV-pmu5ckv2IXB5yGCPw6h8IA16wGFl41q89io5Dq4e3THgRkIS24eWub7ks69EBxAuacl0mE1RDxAQO5RzArSeSzU0oxy6Vj-I/s1600/guass3.PNG" /></a></div>
Plugging these coefficients into the equation and setting the standard deviation to a lesser value undoes the Gaussian effect, here I've stepped it back to t=.2:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiV_1BdnFt3k8EzYl5fWhQsloxOa6ZudvRswoWFnuvO87_aAuVAPEy5eN036b6nUbQxUg3tA91xM6QIfO0natDs5CYCKuL7BzOyZPQ4XZ4HtYdhgTmMVql9Id_kYUgcIsq3wNhSTXTbE2w/s1600/gauss4.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="344" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiV_1BdnFt3k8EzYl5fWhQsloxOa6ZudvRswoWFnuvO87_aAuVAPEy5eN036b6nUbQxUg3tA91xM6QIfO0natDs5CYCKuL7BzOyZPQ4XZ4HtYdhgTmMVql9Id_kYUgcIsq3wNhSTXTbE2w/s640/gauss4.PNG" width="640" /></a></div>
The lower t is set the sharper the peaks...<br />
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-28281547472254534362016-04-17T11:55:00.001-07:002016-04-25T14:29:43.594-07:00Rectangle warping<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgRSrDZzUGIeFGZROanH41BNNcdlosaZbGFhKUQpzkfVWkQ8-AAkHQQSoS6mbcU19nHwNLQKcdO-o6B0E98gh7w_agoICIVXuYkzozntnMIuqdtsOUl9BpYpDXMX6jKITC4aigq5B4WjjU/s1600/mona.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgRSrDZzUGIeFGZROanH41BNNcdlosaZbGFhKUQpzkfVWkQ8-AAkHQQSoS6mbcU19nHwNLQKcdO-o6B0E98gh7w_agoICIVXuYkzozntnMIuqdtsOUl9BpYpDXMX6jKITC4aigq5B4WjjU/s640/mona.jpg" width="640" /></a></div>
Solving this type of system of linear equations you can do any dimension rectangular grid and warp the picture any way even into 3 dimensions or over time!benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-14289554006276925172016-02-23T06:56:00.004-08:002016-02-23T06:56:50.214-08:00Facebook pageI've collected all the best ideas on this blog and improved the presentation from these past 5 years and posted them over at a Facebook public page:<br />
<a href="https://www.facebook.com/thurstonideas">https://www.facebook.com/thurstonideas</a><br />
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-63494352749885829342016-02-23T02:37:00.000-08:002016-02-23T02:37:24.808-08:00Windmill sail boatJust a thought experiment... I wonder if you have a windmill turning a propeller on a boat if you could sail into the wind? I can kind of make an argument that it would go backwards, stay in one place, or go forward so I guess an experiment maybe with a scale model would have to be done...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3A5xMHfJotCtdzdaNVYj5liKpVcrxLze-BQYYMe1Iq7ug-vJbhQjQB4v9ydnnKyuq7Fxyw5UvGp-FfHE9Q-CH8-77ZB-fesWRMQN4opX30M6F3K_WOO3CZKUjJtJV_4oJlLMzUX9Otjo/s1600/boat.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="297" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3A5xMHfJotCtdzdaNVYj5liKpVcrxLze-BQYYMe1Iq7ug-vJbhQjQB4v9ydnnKyuq7Fxyw5UvGp-FfHE9Q-CH8-77ZB-fesWRMQN4opX30M6F3K_WOO3CZKUjJtJV_4oJlLMzUX9Otjo/s320/boat.png" width="320" /></a></div>
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-43730293089340524872016-02-15T08:40:00.000-08:002016-02-23T01:46:00.122-08:00Perfect pivot quick sortSuppose you have n items to be sorted, for example:<br />
3,7,15,4,8,2,9,13,11,10,6,12,1,0,14,5<br />
The first step is to take the first four items from the list and arrange them in 2 pairs, so that the least and greatest of the four numbers is the first pair and the two in between are the next pair like so:<br />
3,15<br />
4,7<br />
Now the repeating step is to take the next 2 items from the list and arrange the 3 pairs so the six items are in order reading down the left items then up the right like so:<br />
adding 8,2 to<br />
3,15<br />
4,7<br />
becomes<br />
2,15<br />
3,8<br />
4,7 because these six items in order are 2,3,4,7,8,15<br />
Now we can start a pile with 2 and 15 in it, and our four new numbers are just:<br />
3,8<br />
4,7<br />
and repeat until all the numbers are exhausted for example it continues:<br />
9,13 are added to<br />
3,8<br />
4,7<br />
to get<br />
3,13<br />
4,9<br />
7,8<br />
then the 3 and the 13 join the piles so you have 2,15,3,13 and the four new numbers are:<br />
4,9<br />
7,8<br />
to which you add 11,10 to get:<br />
4, 11<br />
7, 10<br />
8, 9<br />
and 4 and 11 get put onto the pile in progress to have 2,15,3,13,4,11<br />
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Now you can see that our bottom two number numbers must eventually be the median numbers of the set because every step found a number less and a number greater than them, and the pile can split in two for items less or greater than those two, this means you will have the perfect pivot to recursively apply the algorithm to the two halves!<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-57913371586729054262016-02-07T03:48:00.001-08:002016-02-15T03:27:01.029-08:00Maximizing CD Vector Magnitude Minimizes Length of Path through pointsFirst defining some terms and a hypothesis for closed paths P on the plane through n points with x,y coordinates x[i], y[i] all in the first quadrant:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1zTO-7n1xrr0joZxHmyVskz-pFx7HWOhegn7QpBUl22Y96sKfn7l_r_vDvWxQMsILmZ1AY6nQB0l7gxdKIFpYK1TyIwhO6vq3QKMkMsjYK6veKjJqmSrF2XmQlzR9huTkOpClbF8G4lc/s1600/cd2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="425" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1zTO-7n1xrr0joZxHmyVskz-pFx7HWOhegn7QpBUl22Y96sKfn7l_r_vDvWxQMsILmZ1AY6nQB0l7gxdKIFpYK1TyIwhO6vq3QKMkMsjYK6veKjJqmSrF2XmQlzR9huTkOpClbF8G4lc/s640/cd2.png" width="640" /></a></div>
c and d so named because they are the cross and dot products of vectors from the origin to successive points on the path, and l is the sum of the lengths of the lines between successive points on the path each squared...<br />
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For example I plotted c,d values for the 720 possible closed paths through 7 random points all with the same starting point and got:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtfT3u3AM20FM9R7HjTQer_FhHaoYHtmY_UykGRW504UJYtak9YsFNLW5n38FaHsakEMCuYHt8qaoY1GTfLtDpUYet2kKjR0jlXrikxOctyevD0RkwnCaceHAGmJF_c2RCeyYW4ylujh8/s1600/dotcross.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="430" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtfT3u3AM20FM9R7HjTQer_FhHaoYHtmY_UykGRW504UJYtak9YsFNLW5n38FaHsakEMCuYHt8qaoY1GTfLtDpUYet2kKjR0jlXrikxOctyevD0RkwnCaceHAGmJF_c2RCeyYW4ylujh8/s640/dotcross.png" width="640" /></a></div>
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P[1] and P[-1] were the two solutions to the travelling salesman problem, or paths minimizing l through the points with the starting point fixed, and they also were the two points that maximized c^2+d^2, P[-1] being just P[1] traversing the points in reverse order...</div>
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**It looked to me even more generally decreasing c^2 +d^2 increased l proportionally but I'll have to investigate further**benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-62869604817946961862016-01-29T16:25:00.002-08:002016-01-29T18:54:13.599-08:00within 6% Vector Coordinate InterpolationFor n points in any number of dimensions with coordinates P(n),<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNAa-VHvLAWHtnQa8A54xCDcrYObzzjE1QYTTGaEdNE3jhj6IeHF82V_0J5UObvAyvH1AUpE19Tts_ysm5QzDTkxcbdOZxBj1CPWVgRwCWDYtyhqCwRiLTkJhkpc9R5sbK5kfT-WGkIHY/s1600/Capture.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="105" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNAa-VHvLAWHtnQa8A54xCDcrYObzzjE1QYTTGaEdNE3jhj6IeHF82V_0J5UObvAyvH1AUpE19Tts_ysm5QzDTkxcbdOZxBj1CPWVgRwCWDYtyhqCwRiLTkJhkpc9R5sbK5kfT-WGkIHY/s200/Capture.PNG" width="200" /></a></div>
The function above interpolates the points within an error range of about 6% of the magnitude of the vector to each point... P(0) in the graph below is A, P(1) is B, up to P(6) being A again...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJO1zTSnPgQ9os3wyHh0h0kmvfDYP1xtixWoVErqq5SF4AxqdCTlv-2nT8-hwyCKLXTh93vFWDe2pLs-6UonWFXjgeVzrdDYVxjVGwoYjtYedMHwo6yH3VBLNbI6b0xL12p-h_ADFihA0/s1600/Capture1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="234" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhJO1zTSnPgQ9os3wyHh0h0kmvfDYP1xtixWoVErqq5SF4AxqdCTlv-2nT8-hwyCKLXTh93vFWDe2pLs-6UonWFXjgeVzrdDYVxjVGwoYjtYedMHwo6yH3VBLNbI6b0xL12p-h_ADFihA0/s320/Capture1.PNG" width="320" /></a></div>
The above is 5 points of 2d vectors being interpolated... The nice thing about this formula is it applies to any number of points in any number of dimensions...<br />
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It goes very nearly through the midpoints between each pair of successive vertices, I think changing the e values in the formula to possibly two different numbers might be able to close in closer to the given points but maybe at the expense of not going as nicely through the midpoints...<br />
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The term inside the sum multiplying with the point vector is a normal distribution, with a value of 1 at a particular integer t value and tapering off on each side so that it's roughly 1/2 at the half integers on either side. The ideal would be a distribution with a peak of 1 at an integer t value and a value of exactly 1/2 at each half integer on either side and going completely to 0 everywhere at the integers on either side and beyond, but I think that's only really possible to approximate with a closed form function... Also if you go all the way to a triangular distribution over the interval so you make perfectly the polygon including the points you lose the nice property that the derivative is defined continuously everywhere...benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-34238271094539324072016-01-24T15:07:00.001-08:002016-01-24T15:07:27.931-08:00Vector Coordinate Parameterized Ellipse in Triangle <div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5FWO6EsSbNt518TJQcIDN5MpqcE-bZ_jcV8yp15ZToOrtTp72mcItkYw8zESDwZ888fImYh5rZiJNUxN83FvWo5zo6DCvjVgEzWhb8q8g9C0fPZxcGIwHBG3brypr59ig8fAwK0JOgXE/s1600/ellipse.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="520" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5FWO6EsSbNt518TJQcIDN5MpqcE-bZ_jcV8yp15ZToOrtTp72mcItkYw8zESDwZ888fImYh5rZiJNUxN83FvWo5zo6DCvjVgEzWhb8q8g9C0fPZxcGIwHBG3brypr59ig8fAwK0JOgXE/s640/ellipse.png" width="640" /></a></div>
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The traced curve varying t goes through the midpoints of the sides of the triangle...benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-6268156115698476412016-01-20T10:50:00.000-08:002016-01-20T10:50:11.301-08:00sphere to modular planeI was thinking some problems that take place on a sphere might be easier thinking of them on the modular plane and some modulo problems might be easier thinking of them on a sphere...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-sVmAiuH0UEBkm9_ggdkfruTvhhYY1CdX_fztqVv1c1GlezSC-e6QI8Y49WPv6hEKDm-i-CMUHGylSmU25urDDEKQzqfI0a7Av_zRZeY4YKmrXKROr8jNVsN2g1YAw_MG7AWg3FKeJGQ/s1600/9grid.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="440" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-sVmAiuH0UEBkm9_ggdkfruTvhhYY1CdX_fztqVv1c1GlezSC-e6QI8Y49WPv6hEKDm-i-CMUHGylSmU25urDDEKQzqfI0a7Av_zRZeY4YKmrXKROr8jNVsN2g1YAw_MG7AWg3FKeJGQ/s640/9grid.png" width="640" /></a></div>
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-61528690087419671112016-01-15T17:44:00.003-08:002016-01-16T00:00:31.348-08:00Point in convex polygon by coefficients times verticesSupposing you have n vertices in the plane forming a convex region chosen to be indexed in some order by i, I found that any point inside that convex polygon can be expressed as the sum of coefficients [0..1] multiplying each vertex with coefficients adding to 1...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj4yNO8lzqjJGA_t-rjq-p5SUfc8X-ublRkumXOAyuf58K9xAXyb6mEDqB51YtRQBAZ_MTizH7tG9jB376S081HUC3H25IFR5DWxURiWG4BjRBvtH-mt3eC_DGhSpCZA4BVLRF0sCGE-C0/s1600/poly.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj4yNO8lzqjJGA_t-rjq-p5SUfc8X-ublRkumXOAyuf58K9xAXyb6mEDqB51YtRQBAZ_MTizH7tG9jB376S081HUC3H25IFR5DWxURiWG4BjRBvtH-mt3eC_DGhSpCZA4BVLRF0sCGE-C0/s1600/poly.png" /></a></div>
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For example if the vertices were V[0], V[1], V[2], V[3], V[4], a point inside the polygon will be </div>
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.2*V[0]+.4*V[1]+.2*V[2]+.1*V[3]+.1*V[4]</div>
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Note that the coefficients add to 1 and are between inclusive 0 and 1!</div>
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I wrote this program to demonstrate. This is plotting points for any coefficients meeting these two criteria in 20 discrete steps between 0 and 1 inclusive and 5 vertices...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3pp2gXi_y_PDblc3xz_Pvb2dIwJABLuuZXbrok0YkqrZjxTjXYjoXgLDgSXEoGTPHjia43_4ch3V1vLZcebZ6aQPae7Uj5l5kGIwKL6HWP4dP6y0ckErMUmDP8WlvldgGfFt_ZOZR-g4/s1600/plot2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="268" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3pp2gXi_y_PDblc3xz_Pvb2dIwJABLuuZXbrok0YkqrZjxTjXYjoXgLDgSXEoGTPHjia43_4ch3V1vLZcebZ6aQPae7Uj5l5kGIwKL6HWP4dP6y0ckErMUmDP8WlvldgGfFt_ZOZR-g4/s400/plot2.png" width="400" /></a></div>
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I've made it so c[0] makes a plotted point redder proportional to how large the coefficient is...</div>
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Below is with 50 discrete values between 0 and 1 of the variables:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVwSX_-4tat1wGBPw10Co77j-J-7tdT1l1dpAELw2lEcceEzHwBZ7DLy-6to8iJerFv4jz4coFlRk4-2nOj-LGgxGv2hV0wEp686-uzvJYASULRvt5Xrzd7O13Mk1MibWxAOFrqBOUWI8/s1600/plot3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="281" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVwSX_-4tat1wGBPw10Co77j-J-7tdT1l1dpAELw2lEcceEzHwBZ7DLy-6to8iJerFv4jz4coFlRk4-2nOj-LGgxGv2hV0wEp686-uzvJYASULRvt5Xrzd7O13Mk1MibWxAOFrqBOUWI8/s400/plot3.png" width="400" /></a></div>
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I think it's clear with fine enough steps every part of the pentagon would be plotted, perhaps uniquely...</div>
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I'm still working on a proof though... I wrote the program just to see if it obviously wasn't true but it looks like it is...</div>
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**Source Code requires Python 3 and Pillow for Python**</h4>
<div>
<div>
from PIL import Image</div>
<div>
<br /></div>
<div>
def descend(n, c, i, s, d, plot, coords):</div>
<div>
if i < n-1:</div>
<div>
m = 1.0</div>
<div>
while(m >= -d/2.0):</div>
<div>
c[i] = m</div>
<div>
if(s+m <=1-d/2.0):</div>
<div>
descend(n, c, i+1, s+m, d, plot, coords)</div>
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m -= d</div>
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else:</div>
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c[n-1] = 1.0-s</div>
<div>
print(c)</div>
<div>
p = [0,0]</div>
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for j in range(0, n):</div>
<div>
p[0] += c[j]*coords[j][0]</div>
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p[1] += c[j]*coords[j][1]</div>
<div>
plot.putpixel((int(p[0]), int(p[1])), (255,255-int(c[0]*255),255-int(c[0]*255)))</div>
<div>
</div>
<div>
def main():</div>
<div>
coords = [[84,232],[104, 424],[342,508],[528,276],[330,212]]</div>
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plot = Image.new("RGB", [800,800])</div>
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c = [0,0,0,0,0]</div>
<div>
descend(5, c, 0, 0, .05, plot, coords)</div>
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plot.save("plot.png")</div>
<div>
main()</div>
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benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-69764409092210268512016-01-15T00:06:00.001-08:002016-01-15T00:06:56.260-08:00recursive cosine square wave? This recursive function might eventually go to a square wave...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHGrc5ZQzuooF3Mt6Rsunm4CjZ99RxFiqJi3_ODuWEXAGCb-DkF1ujGnePfa_nNCRY02HzWqLBVY90Wbg76wznrKbHRAorsdKCwUmvd5qo5ByTFb0Ye7GB8VI3vrbb5A_WdrbInxctaWE/s1600/recursivecos.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="620" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHGrc5ZQzuooF3Mt6Rsunm4CjZ99RxFiqJi3_ODuWEXAGCb-DkF1ujGnePfa_nNCRY02HzWqLBVY90Wbg76wznrKbHRAorsdKCwUmvd5qo5ByTFb0Ye7GB8VI3vrbb5A_WdrbInxctaWE/s640/recursivecos.png" width="640" /></a></div>
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-78157269725242546022016-01-14T22:46:00.000-08:002016-01-14T22:46:12.649-08:00feathery function<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfAU4uAYIjVGL9B7heGEhhY0qE6gMLXifdTx8pFnZcC5lapGq212NmD-rsacec2qlYWJx83IcVZfHw4pCaeYD4J3ah_hx3b-HQuoGjsfkVNxqlKdBstd8Srnr3KfzOQOMxnqP0BCu3Iuc/s1600/thingy.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfAU4uAYIjVGL9B7heGEhhY0qE6gMLXifdTx8pFnZcC5lapGq212NmD-rsacec2qlYWJx83IcVZfHw4pCaeYD4J3ah_hx3b-HQuoGjsfkVNxqlKdBstd8Srnr3KfzOQOMxnqP0BCu3Iuc/s1600/thingy.png" /></a></div>
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-37953502680481098872016-01-14T17:38:00.001-08:002016-01-14T18:21:34.362-08:00Parallel convex hull algorithm #2So supposing you have points on the plane with known coordinates<span style="text-align: center;">...</span>:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvqHrgF_CVhHiUkf0IZWmeetAzDOOUj1CvoOy9ZkXnU-EKEVWyCPKLfiQ8TGsVbyfXgQ0y8vFTYPIAUVKDshEUGvjuaThLsWs9nb3Z6CJqNuDhkhPggKd4y7F09_We-Yzt3mbYY86sBWk/s1600/a.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="338" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvqHrgF_CVhHiUkf0IZWmeetAzDOOUj1CvoOy9ZkXnU-EKEVWyCPKLfiQ8TGsVbyfXgQ0y8vFTYPIAUVKDshEUGvjuaThLsWs9nb3Z6CJqNuDhkhPggKd4y7F09_We-Yzt3mbYY86sBWk/s400/a.png" width="400" /></a></div>
First find the centroid (marked in red) and the four points with extreme values of x and y coordinates...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhBYeciH3ihipja8lAjAXRZmh7McxQBAM6mBqWvi_nUw9z5ZqXr-Sfg60TO9XMcI2hzjVbDxpAnjOWe-LrxX00nXmK8wQgRmODV9Sy7wXLLlg46PrI70ODYkT_n4sA0GUF3vrdiTrXpk8U/s1600/a2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="338" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhBYeciH3ihipja8lAjAXRZmh7McxQBAM6mBqWvi_nUw9z5ZqXr-Sfg60TO9XMcI2hzjVbDxpAnjOWe-LrxX00nXmK8wQgRmODV9Sy7wXLLlg46PrI70ODYkT_n4sA0GUF3vrdiTrXpk8U/s400/a2.png" width="400" /></a></div>
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Any point inside the triangles formed can be removed...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEifbHFNiFNLCFutAEAzShyA6CsZO6t5Qg_4GGLgvR1hPOoAduHQW5rkDhEEi9AqKwY09YRz958rnGgbGRDvMCUWsNWISXhGteXQNMFVaPcFbFWsvPYMiZg12aOEf3Yr5qoMqhRUTaHKIww/s1600/a3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEifbHFNiFNLCFutAEAzShyA6CsZO6t5Qg_4GGLgvR1hPOoAduHQW5rkDhEEi9AqKwY09YRz958rnGgbGRDvMCUWsNWISXhGteXQNMFVaPcFbFWsvPYMiZg12aOEf3Yr5qoMqhRUTaHKIww/s320/a3.png" width="320" /></a></div>
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Now the recursive step is to find from the centroid lines dividing each triangle into two pieces such that the point the line ends on is the farthest from the centroid within that section like so:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsk93yucsEPuViLPCMEMoupjqaq2ac-ufx3Kg85iLvzWUbVsnkOoaUs8m2GatNwMdfhsvsZBnfTZW0WvVmvVC9iqfIQiZg8EXiT8KpP745ghU0UfoQfKQRjMPTguKeKOJOuFSc-iR25GU/s1600/a4.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="338" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsk93yucsEPuViLPCMEMoupjqaq2ac-ufx3Kg85iLvzWUbVsnkOoaUs8m2GatNwMdfhsvsZBnfTZW0WvVmvVC9iqfIQiZg8EXiT8KpP745ghU0UfoQfKQRjMPTguKeKOJOuFSc-iR25GU/s400/a4.png" width="400" /></a></div>
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Now form triangles and remove points again...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_hMjk1AfIg_B1tBOafS7tAwWCcDpLwFdqBkN0V4d1gzEnbPPvXJAsLIE6alOZCGf9kNBWOWossPDFBapLz5qmM0EoNe2L3Vy09_w_A6quptNKdvw30QCRm6LtA8dmKed6vHD7HEET9bE/s1600/a5.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_hMjk1AfIg_B1tBOafS7tAwWCcDpLwFdqBkN0V4d1gzEnbPPvXJAsLIE6alOZCGf9kNBWOWossPDFBapLz5qmM0EoNe2L3Vy09_w_A6quptNKdvw30QCRm6LtA8dmKed6vHD7HEET9bE/s320/a5.png" width="320" /></a></div>
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In this case we were already done with thte last lines added forming the convex hull...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWa6bn-y0LrKe9_uluTWKlG__qYSk95rjscTWL8YSlqqPOUZ5AGkgoPGaqqdQLJ99OcLtstO4SPI-y5ZPez6lpRmfz399iI0CSfOEr8f4KLyeJthMfZc0rqBiL_Cvo9j7cpFTjs9sUgMg/s1600/a6.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWa6bn-y0LrKe9_uluTWKlG__qYSk95rjscTWL8YSlqqPOUZ5AGkgoPGaqqdQLJ99OcLtstO4SPI-y5ZPez6lpRmfz399iI0CSfOEr8f4KLyeJthMfZc0rqBiL_Cvo9j7cpFTjs9sUgMg/s320/a6.png" width="320" /></a></div>
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But the process may need to be iterated until all remaining points are on the polygon...</div>
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It may be useful to use polar coordinates centered at the centroid then the points can be sorted by theta and easily split into sections and the furthest point would be the point with the largest r value in each section...I think it should then be fairly straightforward to extend it to 3 dimensions using spherical coordinates instead...</div>
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benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-43876648900669243542016-01-07T17:54:00.000-08:002016-01-09T09:15:47.862-08:00Notation for Direction and Distance plots on Cartesian plane<div class="separator" style="clear: both; text-align: center;">
I will be considering these two formulas which are good for modelling a situation when you know the angle you want the plot to go in with respect to the x axis from a point and how far in that direction for each of n discrete steps...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHhBez4yME-dDwugaxGjJwa4CBWgfxR6kxayg3tDZub7fm8Y5KYvhxKiLQrrbg4f381DHPxdK1P_EWbAhlrAse0ZFkd7EBKrX4tOJOL1YvS3FKxx45AmGwOWg6iaOsT5_jPzdipAzTp8I/s1600/a1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="266" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHhBez4yME-dDwugaxGjJwa4CBWgfxR6kxayg3tDZub7fm8Y5KYvhxKiLQrrbg4f381DHPxdK1P_EWbAhlrAse0ZFkd7EBKrX4tOJOL1YvS3FKxx45AmGwOWg6iaOsT5_jPzdipAzTp8I/s640/a1.PNG" width="640" /></a></div>
With the meaning that X(n), Y(n) is the nth point in a series...<br />
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And to show with an example first let L(i) = i and theta(i) = i*2*pi/5<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6_H3WtOKbnA3bvW3IR1Xuya1PgJvruCH3JF7PmN9rEhXrcwX8rdvS1a8aewGs-yUWoS7S3s2XCBNGfU0LpK8jfMssw4aHAEDOmpwGYl_lOSM75JlG34ds-zImFFE-tg5pQREeuiWz8eY/s1600/a2.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="324" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6_H3WtOKbnA3bvW3IR1Xuya1PgJvruCH3JF7PmN9rEhXrcwX8rdvS1a8aewGs-yUWoS7S3s2XCBNGfU0LpK8jfMssw4aHAEDOmpwGYl_lOSM75JlG34ds-zImFFE-tg5pQREeuiWz8eY/s640/a2.PNG" width="640" /></a></div>
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Interestingly Maple knew the closed form, but it's not necessary for a limited number of points, it does make it easy to plug in any value of n and find the point without knowing any of the previous points though... </div>
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And we can make a list over n=1, 2, 3,4,5<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgJXXOlSMIFlKbliiGwCKDOw56nu2Wev0wTPn65ySNg-O3ssbP40jMica28nFWeDUF2sxCTUpIBVl7mRQd2h4rja9F-bzba2GVewlAtOzEpkShpBCgf_8lz5ea2OVVa8h9gxrC23u8uG-s/s1600/a3.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="34" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgJXXOlSMIFlKbliiGwCKDOw56nu2Wev0wTPn65ySNg-O3ssbP40jMica28nFWeDUF2sxCTUpIBVl7mRQd2h4rja9F-bzba2GVewlAtOzEpkShpBCgf_8lz5ea2OVVa8h9gxrC23u8uG-s/s640/a3.PNG" width="640" /></a></div>
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And we can let the initial point be [0,0] to get:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2AtaHNcmg8dN9mbcL9NpacOTKAhRQ4ErcYAi22Qpd2u6vzucPKaUmwZ9AIH1VZuCe86LlIwCIFM7ECfmqiVbE3VhL2BXSL1UQYH6DkEgayHPuQrqXxO0Jbz81GNa1WUSDCd_mqxlC1P4/s1600/a4.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="110" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2AtaHNcmg8dN9mbcL9NpacOTKAhRQ4ErcYAi22Qpd2u6vzucPKaUmwZ9AIH1VZuCe86LlIwCIFM7ECfmqiVbE3VhL2BXSL1UQYH6DkEgayHPuQrqXxO0Jbz81GNa1WUSDCd_mqxlC1P4/s640/a4.PNG" width="640" /></a></div>
An then plotting the above, both with natural number n and continuous n:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8wIEOdjsppBHRop9dY3ZaDfqBO_S1i2-VihqZxJmqxn9n2F7QBvrCnquyNLquZ-w99LKK1ulqzEPNIKGPGxH5ujMXmaTqHHOQn-ZGRR5dJjI-ymJSAcA_Z0rmawqsv-hlDC8LUUEPHf4/s1600/spiral2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="388" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh8wIEOdjsppBHRop9dY3ZaDfqBO_S1i2-VihqZxJmqxn9n2F7QBvrCnquyNLquZ-w99LKK1ulqzEPNIKGPGxH5ujMXmaTqHHOQn-ZGRR5dJjI-ymJSAcA_Z0rmawqsv-hlDC8LUUEPHf4/s400/spiral2.png" width="400" /></a></div>
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<br />
In general L(i) can be any discrete function of length of lines for the plot and theta(i) can be any function for the angle the lines make with the x axis...<br />
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**Next is to see what happens when L(i) is a constant with a delta->0 length and theta(i) gets more continuous<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-61474901486613724542015-12-30T18:57:00.002-08:002016-01-04T08:52:23.314-08:00Parameterized Quadratic Conformal transformation<div class="separator" style="clear: both; text-align: center;">
Q(s,t)</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5nOH5br4lyVwf83jXUCP3Y5tx7kDXZGXm4BZyw-Aexx_I-vy3iM5jFHKA4qxLRBD0b27mAJHiSqQ5zCubwaSVfIe5WOjWBKwug9dRVTykyVxkX3g1DT2PnSDLJRL2nCX24AZIkhpvPJQ/s1600/qts.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="176" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5nOH5br4lyVwf83jXUCP3Y5tx7kDXZGXm4BZyw-Aexx_I-vy3iM5jFHKA4qxLRBD0b27mAJHiSqQ5zCubwaSVfIe5WOjWBKwug9dRVTykyVxkX3g1DT2PnSDLJRL2nCX24AZIkhpvPJQ/s640/qts.PNG" width="640" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6NtaK3vhOFJ1GcsMCLi5kxYBW1sQuOzPN5bJFAYm8jvip4zW8DJFaRdtG15qf8EcdU_rtjUA3pu0rQEZnOU8q4I0cGQ8Yl3KC3XrvgapzDeUv7HaJn8psYjOOtcbhj924rnA7DAooSTQ/s1600/qts2+%25282%2529.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6NtaK3vhOFJ1GcsMCLi5kxYBW1sQuOzPN5bJFAYm8jvip4zW8DJFaRdtG15qf8EcdU_rtjUA3pu0rQEZnOU8q4I0cGQ8Yl3KC3XrvgapzDeUv7HaJn8psYjOOtcbhj924rnA7DAooSTQ/s640/qts2+%25282%2529.png" width="428" /></a></div>
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Note that the same formula applies first to the x coordinates, and then to the y coordinates and can work in any number of dimensions 2 or greater, Imagine that Q can be any of X, Y, Z component axis, and s and t vary continuously from -1..1... I'm not sure if there might be a notation using summations or matrices that makes this simpler and nicer to write, but maybe it's not that big of a deal when you can just write Q(s,t) to mean this in other formulas... It can also be fairly easily extended to volumes or higher dimensional objects using more variables than just s and t....<br />
<br />
<div>
It's derived from my earlier post on the E_Bezier:<br />
http://benpaulthurstonblog.blogspot.com/2015/12/ebezier.html...<br />
<br />
One can derive something in a similar way from regular Beziers but then the mapping isn't conformal...<br />
<br />
I made s and t vary from -1..1 so they can shrink to -delta +delta around a point on surface F for a sort of surface approximation, it can be shown that as delta shrinks to 0 in the limit the result is F at that point..Or to make the Q values closer and closer to surface F at a point to get a sort of parameterized surface derivative or tangent surface ( I think! I might explore that in a later post)... Also parameterization is a natural fit for conformal mappings because they are both scale and translation-invariant (also rotation)... A rectangle under a conformal mapping can be achieved by adding a coefficient multiplying one of or both t and s,.</div>
<br />
I think maybe General Relativity might be simpler rewritten with this type of formula in mind, because it seems like the complicated thing is keeping all the axis components sorted out whereas this you can do each variable individually with the same formula, just a guess though I don't know that much about General Relativity...<br />
<br />
<h4>
**Matrix version**</h4>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBVCJ-xUzt57BBxE6pMi-JrilYiC9ewlcBvjG-_xNVHqZJ6TmPQ0-a7xcjqx8_QhxdyBeob7vWdw952TcfWU902E0c4y6Z_U8vQYcKkhcypTNH9OtsJ_oPY6k6t8YJK-yUlc7JsPHSTLY/s1600/m.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="217" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBVCJ-xUzt57BBxE6pMi-JrilYiC9ewlcBvjG-_xNVHqZJ6TmPQ0-a7xcjqx8_QhxdyBeob7vWdw952TcfWU902E0c4y6Z_U8vQYcKkhcypTNH9OtsJ_oPY6k6t8YJK-yUlc7JsPHSTLY/s400/m.PNG" width="400" /></a></div>
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Or using matrices instead of the row and column vector and the tensor product:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1ef7nUparYRiu4pNcqlYRNXm8idzDS0WhsmcwA-i0JbpxuklIXY95dXaOQM4QAZ1ClO3uYkLQz9pE282TDudCENUB5zCBpgXl4VSz0LRkjxBcaUTL-YXt33H7sW7k27uM99lN2pQxtv0/s1600/tensor.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="482" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1ef7nUparYRiu4pNcqlYRNXm8idzDS0WhsmcwA-i0JbpxuklIXY95dXaOQM4QAZ1ClO3uYkLQz9pE282TDudCENUB5zCBpgXl4VSz0LRkjxBcaUTL-YXt33H7sW7k27uM99lN2pQxtv0/s640/tensor.png" width="640" /></a></div>
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<h4>
**Python Program**</h4>
<div>
It's more convenient to use a bit different formula when doing things with computer images, ... It's the function q in the program below and the program takes any image of any size and maps it according to the following, there's also a variable for a value for oversampling; this one below is 2*x oversampled...: This map was actually made with the program, you can see the grid lines only ever cross at right angles, that's the conformal property...</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQZS2SCSUNV2cpLVgPCeTEB1QXawC-Nc30hijwb7RdaVKC4uZuqBPe_xtET7S1EucnqXjiUvotYSsMTff5tIZplFbuMQIxeGnbv0TuQm5mknKSizJN3IVk6QQC28IBOkQF0YYkOzSSICg/s1600/conformal2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQZS2SCSUNV2cpLVgPCeTEB1QXawC-Nc30hijwb7RdaVKC4uZuqBPe_xtET7S1EucnqXjiUvotYSsMTff5tIZplFbuMQIxeGnbv0TuQm5mknKSizJN3IVk6QQC28IBOkQF0YYkOzSSICg/s640/conformal2.png" width="422" /></a></div>
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<div>
**Python Code listing (using Pillow for python)**</div>
<div>
<br /></div>
<div>
from PIL import Image</div>
<div>
<br /></div>
<div>
def q(s,t,a,b,c,d,e,f,g,h):</div>
<div>
p = 1.0*((1/4)*b-(1/2)*s*b+(1/2)*s*s*b+(1/2)*t*t*d+(1/2)*t*d+(1/4)*e+(1/4)*g+</div>
<div>
t*t*s*s*((1/4)*g+(1/4)*b+(1/4)*d+(1/4)*e)-(1/2)*t*t*s*g+(1/2)*s*s*g+</div>
<div>
(1/2)*s*g-s*s*((1/4)*g+(1/4)*b+(1/4)*d+(1/4)*e)-t*t*((1/4)*g+(1/4)*b+</div>
<div>
(1/4)*d+(1/4)*e)-(1/2)*t*t*s*s*e+(1/2)*t*s*s*e+(1/4)*t*s*f+(1/4)*t*s*s*f+</div>
<div>
(1/4)*t*t*s*f+(1/4)*t*t*s*s*f-(1/4)*t*t*s*c-(1/4)*t*s*s*c+</div>
<div>
(1/4)*t*t*s*s*c+(1/4)*t*s*c-(1/2)*t*t*s*s*g-(1/2)*t*e+(1/2)*t*t*e-</div>
<div>
(1/2)*t*t*s*s*b+(1/2)*t*t*s*b+(1/4)*d-(1/2)*t*s*s*d-(1/2)*t*t*s*s*d+</div>
<div>
(1/4)*t*t*s*s*a-(1/4)*t*s*a-(1/4)*t*t*s*a+(1/4)*t*s*s*a-(1/4)*t*s*s*h+</div>
<div>
(1/4)*t*t*s*s*h+(1/4)*t*t*s*h-(1/4)*t*s*h)</div>
<div>
return p</div>
<div>
<br /></div>
<div>
def main():</div>
<div>
grid = Image.open("grid.png")</div>
<div>
output = Image.new("RGB", [700, 1200])</div>
<div>
w, h = grid.size[0], grid.size[1]</div>
<div>
c = [[116, 724],[288,926],[477,1059],[354,622],[579,897], \</div>
<div>
[552,588],[582,717],[634,796]]</div>
<div>
print(w,h)</div>
<div>
oversample = 2</div>
<div>
for i in range(0, oversample*w):</div>
<div>
for j in range(0, oversample*h):</div>
<div>
</div>
<div>
s = (2.0/w)*(i/oversample*1.0)-1</div>
<div>
t = -((2.0/h)*(j/oversample*1.0)-1)</div>
<div>
color = grid.getpixel((i/oversample*1.0,j/oversample*1.0))</div>
<div>
px = q(s,t,c[0][0], c[1][0], c[2][0], c[3][0], c[4][0], c[5][0], c[6][0], c[7][0])</div>
<div>
py = q(s,t,c[0][1], c[1][1], c[2][1], c[3][1], c[4][1], c[5][1], c[6][1], c[7][1])</div>
<div>
px = int(px)</div>
<div>
py = int(py)</div>
<div>
try:</div>
<div>
output.putpixel((px,py), (color, color, color))</div>
<div>
except:</div>
<div>
print((i,j), (px, py))</div>
<div>
output.save("conformal.png")</div>
<div>
main()</div>
</div>
<div>
<br />
<h4>
**Chaining q(s,t)**</h4>
It's also pretty straight forward to chain for example a seamless texture to make a warped texture band...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzj82sEjzq2reHjpNVKgo3b7-3RgCYwxvBOP2XlgpmMfT9NXw5HAEOlnsmac5RCU4xye6G-QA67sGw75Rfsm2s2ORanDglJV1-M61et2Yepvzd9PIW1_95c9NX07YnBZ_YBCYiSvFh_V4/s1600/1.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzj82sEjzq2reHjpNVKgo3b7-3RgCYwxvBOP2XlgpmMfT9NXw5HAEOlnsmac5RCU4xye6G-QA67sGw75Rfsm2s2ORanDglJV1-M61et2Yepvzd9PIW1_95c9NX07YnBZ_YBCYiSvFh_V4/s640/1.JPG" width="426" /></a></div>
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benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com1tag:blogger.com,1999:blog-5573398257722015103.post-9712415871578443122015-12-29T03:51:00.000-08:002015-12-29T04:14:00.635-08:00Interesting non-regular pentagon<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2Ke_-y5MD781_cFiyH8PauAF9T6P7M-t3BsxNtm2gPKwYElwcnC8-QhI9onBqRtzGyjLv6gnIlkrQPwHJVLdfsvBV4zU1EUz2DZS3RaM1z99Cf69KwYRR9GeBBZHJaFo5hDhae2ExNFM/s1600/pentagon.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="528" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2Ke_-y5MD781_cFiyH8PauAF9T6P7M-t3BsxNtm2gPKwYElwcnC8-QhI9onBqRtzGyjLv6gnIlkrQPwHJVLdfsvBV4zU1EUz2DZS3RaM1z99Cf69KwYRR9GeBBZHJaFo5hDhae2ExNFM/s640/pentagon.png" width="640" /></a></div>
Questions for further study: Can you tile the plane with these? Are the angles inside the points of the star the same? Is the center pentagon made by the star regular?benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-40056051667543364532015-12-27T12:22:00.000-08:002015-12-27T15:02:12.882-08:00Tr(n)^2+Tr(n+1)^2=Tr((Tri(n) + Tri(n+1)) for triangular numbersI found this about the triangular numbers, also known as the number of distinct pairwise combinations of n items or the sum of the first n of 1,2,3,4,... this wasn't listed on the Encyclopedia of Integer sequences... , The formula for a triangular number is:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim8urCJVt1eL5W_ja7h2Nq1zEuWbnxke1rInmCa5nmPP-IxHdoJNg6E_K4g5ofXq_wvtvL69SUAglFGPyH7J9m8eS9Fvh-3FjMsd1CBx1ucvCOGIec6hzut9SxVt_K90Eoiy4vtkfNId0/s1600/4.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim8urCJVt1eL5W_ja7h2Nq1zEuWbnxke1rInmCa5nmPP-IxHdoJNg6E_K4g5ofXq_wvtvL69SUAglFGPyH7J9m8eS9Fvh-3FjMsd1CBx1ucvCOGIec6hzut9SxVt_K90Eoiy4vtkfNId0/s1600/4.PNG" /></a></div>
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I found that:<br />
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4S2jFjx49W6Udg9yvu-dTp3pgbr4bHtYgRftSN0V-Y2VDMvrOR_0SVHsObOZsVvTCrrsD1kZ19CbHsP9Dfn44J0YP9IIh1-N1TzmY1DBBmHJ50XlXEn6I1gDBLnnCS9b5Evf7bAv9-XY/s1600/5.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="26" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4S2jFjx49W6Udg9yvu-dTp3pgbr4bHtYgRftSN0V-Y2VDMvrOR_0SVHsObOZsVvTCrrsD1kZ19CbHsP9Dfn44J0YP9IIh1-N1TzmY1DBBmHJ50XlXEn6I1gDBLnnCS9b5Evf7bAv9-XY/s320/5.png" width="320" /></a></div>
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Provable easily by substituting the formula for Tri in each side:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEivsodQiYjHUVT2IFb9KnEtNbajKzI6gOGI0SXL1GeuBhXcgId8JPWotkVrDLHPTH3_bwktQYCoQLZ2J5SHVK02EZ7AMIMLgAmBmEiHeA8hc08v0Yl3efAikTAG0T72mnmvmE708UUWksY/s1600/2.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="302" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEivsodQiYjHUVT2IFb9KnEtNbajKzI6gOGI0SXL1GeuBhXcgId8JPWotkVrDLHPTH3_bwktQYCoQLZ2J5SHVK02EZ7AMIMLgAmBmEiHeA8hc08v0Yl3efAikTAG0T72mnmvmE708UUWksY/s640/2.PNG" width="640" /></a></div>
<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-16574911806761586542015-12-22T23:20:00.001-08:002015-12-23T08:41:36.067-08:00MacLaurin Generating FunctionExponential generating functions follow the form:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEincdSWL2yDxc2c6ZZgTKWub0GiaCJ3NM8aP09n3JT4YyeP9NximK53XuJwnE2C-0GqmfjA8XDjAnNqkg0JZZGhU2Mkb8XsoDn8ICOSFChhIFFdMl9Yo5HXptxa1c_mV8Yd9WrTAS55jKI/s1600/2a.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEincdSWL2yDxc2c6ZZgTKWub0GiaCJ3NM8aP09n3JT4YyeP9NximK53XuJwnE2C-0GqmfjA8XDjAnNqkg0JZZGhU2Mkb8XsoDn8ICOSFChhIFFdMl9Yo5HXptxa1c_mV8Yd9WrTAS55jKI/s1600/2a.PNG" /></a></div>
I was considering a modification of this structure to read:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjlOjtnjvxLcLM__6wGvVyZOu1Wpam84y_fRo6h-gTno6ZjEaOQJAf-fuKtGNOcUmms414T7RoEcg2MmtzZITgT3uMTxO0Y5xfx-wY6sHDneDI1XK2553x8bobzAglTcWrlwfj4kc5wKc/s1600/x3.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjlOjtnjvxLcLM__6wGvVyZOu1Wpam84y_fRo6h-gTno6ZjEaOQJAf-fuKtGNOcUmms414T7RoEcg2MmtzZITgT3uMTxO0Y5xfx-wY6sHDneDI1XK2553x8bobzAglTcWrlwfj4kc5wKc/s1600/x3.PNG" /></a></div>
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Where f_n_(x) at x is both the sequence of coefficients to find the generating function for and the nth derivative at x following the convention of the MacLaurin Series for a function...<br />
If we call this a MacLaurin Generating Function, then we can get some interesting series at x=1...<br />
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqo1ZykNHRzJFglstWntCVcvQgMQnSTE9xfARF5Z8voWsvLohsY4fdgxdTLsG3PWLcDuuuqBnCnoVc3gliGh8pyv0OJVby_aZxQjRznWWWItpDhmvIIjoynnFKHN3od2sJ0FaJ6HDhVdw/s1600/1d.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="492" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqo1ZykNHRzJFglstWntCVcvQgMQnSTE9xfARF5Z8voWsvLohsY4fdgxdTLsG3PWLcDuuuqBnCnoVc3gliGh8pyv0OJVby_aZxQjRznWWWItpDhmvIIjoynnFKHN3od2sJ0FaJ6HDhVdw/s640/1d.PNG" width="640" /></a></div>
It continues after 0,1,2,3 with 8,10,54... and I've just learned that these are the Lehmer-Comtet numbers: <a href="https://oeis.org/A005727">Lehmer-Comtet</a><br />
<br />
Now we can see what other of these types of series might have simple MacLaurin generating functions, below is if you use x^(2*x)<br />
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBJdPmGLguumgJztYUGoQ5DhntMgHiXM0CEas7XV0r0JCPZukXcjgRKvaT88RjbtnUOCheIdJmI9RdmrTsbFOmBCZz9dIf4PaPzxBv6TMhvKjct3b0lOcf_ROd2mCHRJ-XPeWriEhlMJ4/s1600/1e.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBJdPmGLguumgJztYUGoQ5DhntMgHiXM0CEas7XV0r0JCPZukXcjgRKvaT88RjbtnUOCheIdJmI9RdmrTsbFOmBCZz9dIf4PaPzxBv6TMhvKjct3b0lOcf_ROd2mCHRJ-XPeWriEhlMJ4/s1600/1e.PNG" /></a></div>
Maybe not as complicated as it looks when you substitute in x=1 that last 7th derivative is:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizG6VSo79gR2ZaR5pxtqa48IepL2Hmsj2mbo72DVQ8SY4PA6A4jVNHHzxXwnxeZ0iXKQhlKThmS5W7OD8BLlwau35i-S4Als-WUN3ekxHO8fWy13Vl4FlMuuGznT8Tw21kCmYSBkGZ9Dc/s1600/1.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="60" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizG6VSo79gR2ZaR5pxtqa48IepL2Hmsj2mbo72DVQ8SY4PA6A4jVNHHzxXwnxeZ0iXKQhlKThmS5W7OD8BLlwau35i-S4Als-WUN3ekxHO8fWy13Vl4FlMuuGznT8Tw21kCmYSBkGZ9Dc/s640/1.PNG" width="640" /></a></div>
Still pretty complicated but the non power of 2 numbers look to be related to multiples of factorials?<br />
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The series 1, 2, 6, 18, 64, 220, 888 is not in the Encyclopedia of Integer sequences, but looks interesting! In fact <span style="background-color: white; color: #222222; font-family: arial, sans-serif; font-size: 12.8px;">Alois P. Heinz</span> has already pointed out that the exponential generating function for x^2*x is x^2*x itself!<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-36931336093805967482015-12-22T21:38:00.002-08:002015-12-22T22:42:25.025-08:00Factorial sum Maple doesn't knowConsidering the sum:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsaihTI1laKLDl4Mb_8pEKu4lrGkwUBNEJj3B8q9BuqpjX7FklZnNyEwz7zwwmQps2x12odmX8quTSwIb_aLT2n9sMyFAWyLRF4fW5uE3G0HJYTkYOdiO7Gw66Zf-UVX2IteWD8FpSWto/s1600/2c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="324" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsaihTI1laKLDl4Mb_8pEKu4lrGkwUBNEJj3B8q9BuqpjX7FklZnNyEwz7zwwmQps2x12odmX8quTSwIb_aLT2n9sMyFAWyLRF4fW5uE3G0HJYTkYOdiO7Gw66Zf-UVX2IteWD8FpSWto/s640/2c.PNG" width="640" /></a></div>
But it's clear this sum should be equal to one, consider the first 7 terms:<br />
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If we were to increase the last coefficient by 1 it is exactly unity:<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi23MkELPR_H6MXA0UbzVJ4oKaRIPJj1CntxdOLkn9eR5tqWJHwnXNht9MMGk9Yy3ued1GhZBLWJXQKsadSvhshSC4EMCGj2n-T-zEq0sVgSpv5-h0j6rIeKt_BE4uily00fnPC-apUy_8/s1600/1c.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="92" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi23MkELPR_H6MXA0UbzVJ4oKaRIPJj1CntxdOLkn9eR5tqWJHwnXNht9MMGk9Yy3ued1GhZBLWJXQKsadSvhshSC4EMCGj2n-T-zEq0sVgSpv5-h0j6rIeKt_BE4uily00fnPC-apUy_8/s640/1c.PNG" width="640" /></a></div>
This is because 8/8! is 7! which adds one more 7! to 6/7! to make 7/7! which is 6!, etc.. until you have 1/2! plus 1/2! equaling 1...<br />
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It's kind of interesting to think what function would have this as it's MacLaurin series, that at f=1 it is 0, and the first derivative there is 1, and every successive derivative is 1 more than the one before it...<br />
It actually is (x^x)-1, see:<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-80954721284594133462015-12-22T21:18:00.003-08:002015-12-22T21:25:15.502-08:00Alternate numbering systemJust considering this series, here are the first 9 terms:<br />
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Where the next coefficient is chosen so the sum is as large as it can be not exceeding 1... The growth of the coefficients can be solved for by noting the next term should by multiplied by some a so that it is as large as the previous term when it has a coefficient of 1...<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3FZj9_sVbsaukyHINC9ugzTEhtFPssnACc8ISoJ1ynPaATH1cUXuEwDZR2E6zZbJAktUQtHYbNCEuaIDDNCNLB-z0yGzErpmpekcEFtrLjlp-eSz5WFV40AYnipos7LpLlmACuSBdEL0/s1600/1b.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="86" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3FZj9_sVbsaukyHINC9ugzTEhtFPssnACc8ISoJ1ynPaATH1cUXuEwDZR2E6zZbJAktUQtHYbNCEuaIDDNCNLB-z0yGzErpmpekcEFtrLjlp-eSz5WFV40AYnipos7LpLlmACuSBdEL0/s640/1b.PNG" width="640" /></a></div>
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Plotted it is linear:<br />
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So I've noticed that this series is basically as efficient for writing any number between 0 and 1 with regards to number of decimal digits as decimal itself is! By the pigeon hole principle no series can be better for getting k decimal digits than the decimal series a1/10 + a2/100...benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-55815803422975136292015-12-17T13:47:00.003-08:002015-12-17T13:47:51.702-08:00Quadratic rule of eighths<div class="separator" style="clear: both; text-align: center;">
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This follows simply by substituting 1/4 and 3/4 into the E_Bezier formula from last post...benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-61351024435690357782015-12-12T09:57:00.000-08:002015-12-12T09:57:59.898-08:00E_BezierNormally the quadratic Bezier through three points A,B,C is:<br />
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I found an alternative that has some nice properties I'm calling the E_Bezier also through points A,B,C:<br />
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Both the Bezier and the E_Bezier are defined over the interval t=0..1, and equal A at t=0 and C at t=1, but the E_Bezier has the property that at t=1/2, it equals exactly B, which is not the case for the Bezier curve...<br />
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The acceleration of the E_Bezier curve is exactly double the acceleration of the Bezier curve...<br />
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I called it the E_Bezier because it spends "E" qual time between points A and B and points B and C...<br />
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**I might continue with an update for higher order than quadratic E_Bezier**benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-67606626174747527452015-12-09T09:20:00.001-08:002015-12-09T12:35:16.604-08:00Balance scale seriesTrying to get the least amount of elements in a series of natural numbers that can be combined by adding or subtracting unique elements of the series to get as many consecutive natural numbers as possible starting with 1. I came up with:<br />
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1,2,7, 21, 52<br />
For example<br />
13=21-7-1,<br />
14=21-7,<br />
15=21-7+1, etc... Every natural number less than 84 can be reached by adding or subtracting these first 5 numbers...<br />
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The pattern I found is that x-sum(previous) = sum(previous)+1, solve for x, because the previous were able to reach every number up to sum(previous), and you need an x so that subtracting all of those numbers yields every number up to x...<br />
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This isn't in the Encyclopedia of Integer sequences yet, but I don't know if it's the best solution...<br />
The name comes from the fact that on a balancing scale, you could weigh any natural number weight by putting the additions on one side of the balance and the subtractions and the object being weighed on the other...<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-45282781868161311272015-12-07T00:10:00.000-08:002015-12-09T06:29:54.901-08:00Faster method than Newton-Raphson<div>
This will be considering a replacement of the Newton-Raphson method for better performance changing:<br />
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to:<br />
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First recall that the Newton-Raphson method started as a generalization of the Babylonian method for finding square roots... I'll start the same way by showing this method speeds up finding square roots...</div>
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Starting with a reasonably close approximation to the square root of N such as the closest perfect square, one can first use this formula where x is the estimate.<br />
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For example with an N of 61, we can chose x to be 7 because 7^2=49 is close to 61 and then:</div>
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Now recursively use the formula again:<br />
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Which is very close to the square root of 61 compare:</div>
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After 2 iterations it's already accurate to 9 decimal places! Two iterations of the Newton Raphson method with the same initial conditions only correctly estimates 3. In fact it is better than 3 iterations of Newton-Raphson by a couple of decimal places...There are 7 arithmetical steps per iteration of this method and 5 for Newton-Raphson, so two iterations of this method take 14 arithmetic steps and get a better result than when Newton-Raphson takes 15 arithmetical steps, and Newton-Raphson would need 20 arithmetical steps to improve upon two iterations of this method... I think the difference becomes more marked when extreme accuracy is needed or when just the amount of accuracy that this method gives over Newton-Raphson is needed... And of course when very many of these calculations are needed...</div>
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So for another example if f(x) is 2*e^(x)-3 starting with a guess of .5 and doing two iterations we get:<br />
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That is very close to the actual root of:<br />
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Three iterations of Newton's method only correctly gets 5 decimal places right for more arithmetic...<br />
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benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0tag:blogger.com,1999:blog-5573398257722015103.post-56374058095708647952015-12-02T07:48:00.000-08:002015-12-02T09:17:23.227-08:00Possible commutative pairing functionMy conjecture is that given a,b positive integers:<br />
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is a positive integer unique to the pair a,b except for exchanging a and b...<br />
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I tried proving it a few different ways but really I don't know why it seems to work. I looked at many examples where it holds and usually there are 3,6,9, or 12 solutions over the integers but except for a,b and b,a one of the two integers is always negative... Maybe the proof is simple, if 12 is the most solutions there can be and you know two solutions will both be negative integers, and 8 will have 1 negative and 1 positive integer, that only leaves two to be positive, but I wasn't able to get to that... Maybe also there's an argument from the symmetries that have to exist, I don't know...<br />
For 3,4 the solutions are:<br />
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<br />benpaulthurstonhttp://www.blogger.com/profile/09293228754296510996noreply@blogger.com0