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	<title>Francisco Blanco-Silva</title>
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		<title>Sympy should suffice</title>
		<link>http://blancosilva.wordpress.com/2013/06/06/sympy-should-suffice/</link>
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		<pubDate>Fri, 07 Jun 2013 02:00:06 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Geometry]]></category>
		<category><![CDATA[puzzles]]></category>
		<category><![CDATA[sage]]></category>
		<category><![CDATA[Teaching]]></category>
		<category><![CDATA[circumcenter]]></category>
		<category><![CDATA[cuttheknot]]></category>
		<category><![CDATA[geometry]]></category>
		<category><![CDATA[python]]></category>
		<category><![CDATA[rhombus]]></category>
		<category><![CDATA[sympy]]></category>
		<category><![CDATA[triangle]]></category>

		<guid isPermaLink="false">https://blancosilva.wordpress.com/?p=5750</guid>
		<description><![CDATA[I have just received a copy of Instant SymPy Starter, by Ronan Lamy—a no-nonsense guide to the main properties of SymPy, the Python library for symbolic mathematics. This short monograph packs everything you should need, with neat examples included, in about 50 pages. Well-worth its money. To celebrate, I would like to pose a few [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5750&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">I have just received a copy of <a href="http://www.amazon.com/gp/product/B00CYHB3CG/ref=as_li_qf_sp_asin_tl?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=B00CYHB3CG&amp;linkCode=as2&amp;tag=blancosilva-20">Instant SymPy Starter</a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=B00CYHB3CG" width="1" height="1" border="0" alt="" style="border:none !important;margin:0!important;" />, by Ronan Lamy—a no-nonsense guide to the main properties of <tt>SymPy</tt>, the Python library for symbolic mathematics.  This short monograph packs everything you should need, with neat examples included, in about 50 pages.  Well-worth its money.</p>
<p style="text-align:justify;">To celebrate, I would like to pose a few coding challenges on the use of this library, based on a fun geometric puzzle from <a href="http://www.cut-the-knot.org/">cut-the-knot</a>: <a href="http://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/ForcedRhombus.shtml"><b>Rhombus in Circles</b></a></p>
<blockquote style="text-align:justify;font-family:modern;font-size:12pt;"><p>
Segments <img src='http://s0.wp.com/latex.php?latex=%5Coverline%7BAB%7D&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='&#92;overline{AB}' title='&#92;overline{AB}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Coverline%7BCD%7D&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='&#92;overline{CD}' title='&#92;overline{CD}' class='latex' /> are equal. Lines <img src='http://s0.wp.com/latex.php?latex=AB&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='AB' title='AB' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=CD&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='CD' title='CD' class='latex' /> intersect at <img src='http://s0.wp.com/latex.php?latex=M.&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='M.' title='M.' class='latex' /> Form four circumcircles: <img src='http://s0.wp.com/latex.php?latex=%28E%29%3D%28ACM%29%2C+%28F%29%3D%28ADM%29%2C+%28G%29%3D%28BDM%29%2C+%28H%29%3D%28BCM%29.&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='(E)=(ACM), (F)=(ADM), (G)=(BDM), (H)=(BCM).' title='(E)=(ACM), (F)=(ADM), (G)=(BDM), (H)=(BCM).' class='latex' />  Prove that the circumcenters <img src='http://s0.wp.com/latex.php?latex=E%2C+F%2C+G%2C+H&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='E, F, G, H' title='E, F, G, H' class='latex' /> form a rhombus, with <img src='http://s0.wp.com/latex.php?latex=%5Cangle+EFG+%3D+%5Cangle+AMC.&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='&#92;angle EFG = &#92;angle AMC.' title='&#92;angle EFG = &#92;angle AMC.' class='latex' /></p>
<div style="width:100%;margin-left:auto;margin-right:auto;"><a href="http://www.flickr.com/photos/eseprimo/8973244465/" title="rhombusincircles by eseprimo, on Flickr"><img src="http://farm8.staticflickr.com/7350/8973244465_ae778819b0_c.jpg" style="width:100%;border:2px solid black;" alt="rhombusincircles"></a></div>
</blockquote>
<p style="text-align:justify;">Note that if this construction works, it must do so independently of translations, rotations and dilations.  We may then assume that <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='M' title='M' class='latex' /> is the origin, that the segments have length one, <img src='http://s0.wp.com/latex.php?latex=A%3D%282%2C0%29%2C+B%3D%281%2C0%29%2C&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='A=(2,0), B=(1,0),' title='A=(2,0), B=(1,0),' class='latex' /> and that for some parameters <img src='http://s0.wp.com/latex.php?latex=a%3E0%2C+%5Ctheta+%5Cin+%280%2C+%5Cpi%29%2C&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='a&gt;0, &#92;theta &#92;in (0, &#92;pi),' title='a&gt;0, &#92;theta &#92;in (0, &#92;pi),' class='latex' /> it is <img src='http://s0.wp.com/latex.php?latex=C%3D%28a%2B1%29+%28%5Ccos+%5Ctheta%2C+%5Csin%5Ctheta%29%2C+D%3Da+%28%5Ccos%5Ctheta%2C+%5Csin%5Ctheta%29.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='C=(a+1) (&#92;cos &#92;theta, &#92;sin&#92;theta), D=a (&#92;cos&#92;theta, &#92;sin&#92;theta).' title='C=(a+1) (&#92;cos &#92;theta, &#92;sin&#92;theta), D=a (&#92;cos&#92;theta, &#92;sin&#92;theta).' class='latex' />  We let <tt>SymPy</tt> take care of the computation of circumcenters:</p>
<pre class="brush: python; title: ; notranslate">
import sympy
from sympy import *

# Point definitions
M=Point(0,0)
A=Point(2,0)
B=Point(1,0)
a,theta=symbols('a,theta',real=True,positive=True)
C=Point((a+1)*cos(theta),(a+1)*sin(theta))
D=Point(a*cos(theta),a*sin(theta))

#Circumcenters
E=Triangle(A,C,M).circumcenter
F=Triangle(A,D,M).circumcenter
G=Triangle(B,D,M).circumcenter
H=Triangle(B,C,M).circumcenter
</pre>
<p style="text-align:justify;">Finding that the alternate angles are equal in the quadrilateral <img src='http://s0.wp.com/latex.php?latex=EFGH&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='EFGH' title='EFGH' class='latex' /> is pretty straightforward:</p>
<pre class="brush: python; gutter: false; title: ; notranslate">
In [11]: P=Polygon(E,F,G,H)

In [12]: P.angles[E]==P.angles[G]
Out[12]: True

In [13]: P.angles[F]==P.angles[H]
Out[13]: True
</pre>
<p style="text-align:justify;">To prove it a rhombus, the two sides that coincide on each angle must be equal.  This presents us with the first challenge:  Note for example that if we naively ask SymPy whether the triangle <img src='http://s0.wp.com/latex.php?latex=%5Ctriangle+EFG&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;triangle EFG' title='&#92;triangle EFG' class='latex' /> is equilateral, we get a <tt>False</tt> statement:</p>
<pre class="brush: python; gutter: false; title: ; notranslate">
In [14]: Triangle(E,F,G).is_equilateral()
Out[14]: False

In [15]: F.distance(E)
Out[15]: Abs((a/2 - cos(theta))/sin(theta) - (a - 2*cos(theta) + 1)/(2*sin(theta)))

In [16]: F.distance(G)
Out[16]: sqrt(((a/2 - cos(theta))/sin(theta) - (a - cos(theta))/(2*sin(theta)))**2 + 1/4)
</pre>
<p style="text-align:justify;">Part of the reason is that we have not indicated anywhere that the parameter <tt>theta</tt> is to be strictly bounded above by <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' /> (we did indicate that it must be strictly positive).  The other reason is that <tt>SymPy</tt> does not handle identities well, unless the expressions to be evaluated are perfectly simplified.  For example, if we <em>trust</em> the routines of simplification of trigonometric expressions alone, we will not be able to resolve this problem with this technique:</p>
<pre class="brush: python; gutter: false; title: ; notranslate">
In [17]: trigsimp(F.distance(E)-F.distance(G),deep=True)==0
Out[17]: False
</pre>
<p style="text-align:justify;">Finding that <img src='http://s0.wp.com/latex.php?latex=%5Cangle+EFG+%3D+%5Cangle+AMC&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;angle EFG = &#92;angle AMC' title='&#92;angle EFG = &#92;angle AMC' class='latex' /> with <tt>SymPy</tt> is not that easy either.  This is the second challenge. </p>
<p style="text-align:justify;">How would the reader resolve this situation?</p>
<p></p>
<p style="text-align:center;"><a href="http://www.amazon.com/gp/product/B00CYHB3CG/ref=as_li_qf_sp_asin_il?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=B00CYHB3CG&amp;linkCode=as2&amp;tag=blancosilva-20"><img border="0" src="http://ws.assoc-amazon.com/widgets/q?_encoding=UTF8&amp;ASIN=B00CYHB3CG&amp;Format=_SL160_&amp;ID=AsinImage&amp;MarketPlace=US&amp;ServiceVersion=20070822&amp;WS=1&amp;tag=blancosilva-20"></a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=B00CYHB3CG" width="1" height="1" border="0" alt="" style="border:none !important;margin:0!important;" /><br /><a href="http://www.amazon.com/gp/product/B00CYHB3CG/ref=as_li_qf_sp_asin_tl?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=B00CYHB3CG&amp;linkCode=as2&amp;tag=blancosilva-20">Instant SymPy Starter</a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=B00CYHB3CG" width="1" height="1" border="0" alt="" style="border:none !important;margin:0!important;" /></p>
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		<title>A nice application of Fatou&#8217;s Lemma</title>
		<link>http://blancosilva.wordpress.com/2013/06/01/a-nice-application-of-fatous-lemma/</link>
		<comments>http://blancosilva.wordpress.com/2013/06/01/a-nice-application-of-fatous-lemma/#comments</comments>
		<pubDate>Sun, 02 Jun 2013 02:50:28 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Analysis]]></category>
		<category><![CDATA[Teaching]]></category>
		<category><![CDATA[Fatou's Lemma]]></category>
		<category><![CDATA[Measure Theory]]></category>
		<category><![CDATA[Real Analysis]]></category>

		<guid isPermaLink="false">https://blancosilva.wordpress.com/?p=5657</guid>
		<description><![CDATA[Let me show you an exciting technique to prove some convergence statements using exclusively functional inequalities and Fatou&#8217;s Lemma. The following are two classic problems solved this way. Enjoy! Exercise 1 Let be a measurable space and suppose is a sequence of measurable functions in that converge almost everywhere to a function and such that [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5657&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Let me show you an exciting technique to prove some convergence statements using exclusively functional inequalities and Fatou&#8217;s Lemma. The following are two classic problems solved this way. Enjoy!</p>
<blockquote style="text-align:justify;"><p><b>Exercise 1</b> <em> Let <img src='http://s0.wp.com/latex.php?latex=%7B%28X%2C+%5Cmathcal%7BF%7D%2C+%5Cmu%29%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{(X, &#92;mathcal{F}, &#92;mu)}' title='{(X, &#92;mathcal{F}, &#92;mu)}' class='latex' /> be a measurable space and suppose <img src='http://s0.wp.com/latex.php?latex=%7B%5C%7Bf_n%5C%7D_%7Bn%5Cin+%5Cmathbb%7BN%7D%7D%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;{f_n&#92;}_{n&#92;in &#92;mathbb{N}}}' title='{&#92;{f_n&#92;}_{n&#92;in &#92;mathbb{N}}}' class='latex' /> is a sequence of measurable functions in <img src='http://s0.wp.com/latex.php?latex=%7BL_1%28%5Cmu%29%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{L_1(&#92;mu)}' title='{L_1(&#92;mu)}' class='latex' /> that converge almost everywhere to a function <img src='http://s0.wp.com/latex.php?latex=%7Bf+%5Cin+L_1%28%5Cmu%29%7D%2C&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{f &#92;in L_1(&#92;mu)},' title='{f &#92;in L_1(&#92;mu)},' class='latex' /> and such that the sequence of norms <img src='http://s0.wp.com/latex.php?latex=%7B%5Cbig%5C%7B+%5ClVert+f_n+%5CrVert_1+%5Cbig%5C%7D_%7Bn+%5Cin+%5Cmathbb%7BN%7D%7D%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;big&#92;{ &#92;lVert f_n &#92;rVert_1 &#92;big&#92;}_{n &#92;in &#92;mathbb{N}}}' title='{&#92;big&#92;{ &#92;lVert f_n &#92;rVert_1 &#92;big&#92;}_{n &#92;in &#92;mathbb{N}}}' class='latex' /> converges to <img src='http://s0.wp.com/latex.php?latex=%7B%5ClVert+f+%5CrVert_1%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;lVert f &#92;rVert_1}' title='{&#92;lVert f &#92;rVert_1}' class='latex' />. Prove that the sequence of integrals <img src='http://s0.wp.com/latex.php?latex=%7B%5Cint_E+%5Clvert+f_n+%5Crvert%5C%2C+d%5Cmu%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;int_E &#92;lvert f_n &#92;rvert&#92;, d&#92;mu}' title='{&#92;int_E &#92;lvert f_n &#92;rvert&#92;, d&#92;mu}' class='latex' /> converges to the integral <img src='http://s0.wp.com/latex.php?latex=%7B%5Cint_E+%5Clvert+f+%5Crvert%5C%2C+d%5Cmu%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;int_E &#92;lvert f &#92;rvert&#92;, d&#92;mu}' title='{&#92;int_E &#92;lvert f &#92;rvert&#92;, d&#92;mu}' class='latex' /> for every measurable set <img src='http://s0.wp.com/latex.php?latex=%7BE%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{E}' title='{E}' class='latex' />. </em></p></blockquote>
<p> <em>Proof:</em>  Note first that
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Brcl%7D++%5Clvert+f_n+-+f%5Crvert%5C%2C+%5Cboldsymbol%7B1%7D_E+%5Cleq+%5Clvert+f_n+-+f+%5Crvert+%5Cleq+%5Clvert+f_n+%5Crvert+%2B+%5Clvert+f+%5Crvert.+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;lvert f_n - f&#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;leq &#92;lvert f_n - f &#92;rvert &#92;leq &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert. &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;lvert f_n - f&#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;leq &#92;lvert f_n - f &#92;rvert &#92;leq &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert. &#92;end{array} ' class='latex' /></p>
<p> Set then <img src='http://s0.wp.com/latex.php?latex=%7Bg_n+%3D+%5Clvert+f_n+%5Crvert+%2B+%5Clvert+f+%5Crvert+-+%5Clvert+f_n-f+%5Crvert%5C%2C+%5Cboldsymbol%7B1%7D_E%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='{g_n = &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n-f &#92;rvert&#92;, &#92;boldsymbol{1}_E}' title='{g_n = &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n-f &#92;rvert&#92;, &#92;boldsymbol{1}_E}' class='latex' /> (which are non-negative functions) and apply Fatou&#8217;s Lemma to that sequence. We have then
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Bc%7D++%5Cint_X+%5Climinf_n+g_n%5C%2C+d%5Cmu+%5Cleq+%5Climinf_n+%5Cint_X+g_n%5C%2C+d%5Cmu+%5C%5C+%5C%5C+%5Cint_X+%5Climinf_n+%5Cbig%28+%5Clvert+f_n+%5Crvert+%2B+%5Clvert+f+%5Crvert+-+%5Clvert+f_n+-+f+%5Crvert%5C%2C+%5Cboldsymbol%7B1%7D_E+%5Cbig%29%5C%2C+d%5Cmu+%5Cleq+%5Climinf_n+%5Cint_X+%5Cbig%28+%5Clvert+f_n+%5Crvert+%2B+%5Clvert+f+%5Crvert+-+%5Clvert+f_n+-+f+%5Crvert%5C%2C+%5Cboldsymbol%7B1%7D_E+%5Cbig%29%5C%2C+d%5Cmu+%5C%5C+%5C%5C+%5Cint_X+%5Cbig%28+2%5Clvert+f+%5Crvert+-+%5Climsup_n+%5Clvert+f_n-f+%5Crvert%5C%2C+%5Cboldsymbol%7B1%7D_E%5Cbig%29%5C%2C+d%5Cmu+%5Cleq+%5Climinf_n+%5Cint_X+%5Clvert+f_n+%5Crvert%5C%2C+d%5Cmu+%2B+%5ClVert+f+%5CrVert_1+-+%5Climsup_n+%5Cint_E+%5Clvert+f_n+-f+%5Crvert%5C%2C+d%5Cmu+%5C%5C+%5C%5C+2%5ClVert+f%5CrVert_1+%5Cleq+2%5ClVert+f+%5CrVert_1+-+%5Climsup_n+%5Cint_E+%5Clvert+f_n+-+f%5Crvert%5C%2C+d%5Cmu%2C+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{c}  &#92;int_X &#92;liminf_n g_n&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X g_n&#92;, d&#92;mu &#92;&#92; &#92;&#92; &#92;int_X &#92;liminf_n &#92;big( &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n - f &#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;big)&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X &#92;big( &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n - f &#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;big)&#92;, d&#92;mu &#92;&#92; &#92;&#92; &#92;int_X &#92;big( 2&#92;lvert f &#92;rvert - &#92;limsup_n &#92;lvert f_n-f &#92;rvert&#92;, &#92;boldsymbol{1}_E&#92;big)&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X &#92;lvert f_n &#92;rvert&#92;, d&#92;mu + &#92;lVert f &#92;rVert_1 - &#92;limsup_n &#92;int_E &#92;lvert f_n -f &#92;rvert&#92;, d&#92;mu &#92;&#92; &#92;&#92; 2&#92;lVert f&#92;rVert_1 &#92;leq 2&#92;lVert f &#92;rVert_1 - &#92;limsup_n &#92;int_E &#92;lvert f_n - f&#92;rvert&#92;, d&#92;mu, &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{c}  &#92;int_X &#92;liminf_n g_n&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X g_n&#92;, d&#92;mu &#92;&#92; &#92;&#92; &#92;int_X &#92;liminf_n &#92;big( &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n - f &#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;big)&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X &#92;big( &#92;lvert f_n &#92;rvert + &#92;lvert f &#92;rvert - &#92;lvert f_n - f &#92;rvert&#92;, &#92;boldsymbol{1}_E &#92;big)&#92;, d&#92;mu &#92;&#92; &#92;&#92; &#92;int_X &#92;big( 2&#92;lvert f &#92;rvert - &#92;limsup_n &#92;lvert f_n-f &#92;rvert&#92;, &#92;boldsymbol{1}_E&#92;big)&#92;, d&#92;mu &#92;leq &#92;liminf_n &#92;int_X &#92;lvert f_n &#92;rvert&#92;, d&#92;mu + &#92;lVert f &#92;rVert_1 - &#92;limsup_n &#92;int_E &#92;lvert f_n -f &#92;rvert&#92;, d&#92;mu &#92;&#92; &#92;&#92; 2&#92;lVert f&#92;rVert_1 &#92;leq 2&#92;lVert f &#92;rVert_1 - &#92;limsup_n &#92;int_E &#92;lvert f_n - f&#92;rvert&#92;, d&#92;mu, &#92;end{array} ' class='latex' /></p>
<p> which implies
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Brcl%7D++0+%5Cleq+%5Climinf_n+%5Cint_E+%5Clvert+f_n-f+%5Crvert%5C%2C+d%5Cmu+%5Cleq+%5Climsup_n+%5Cint_E+%5Clvert+f_n-f+%5Crvert%5C%2C+d%5Cmu+%5Cleq+0.+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{rcl}  0 &#92;leq &#92;liminf_n &#92;int_E &#92;lvert f_n-f &#92;rvert&#92;, d&#92;mu &#92;leq &#92;limsup_n &#92;int_E &#92;lvert f_n-f &#92;rvert&#92;, d&#92;mu &#92;leq 0. &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{rcl}  0 &#92;leq &#92;liminf_n &#92;int_E &#92;lvert f_n-f &#92;rvert&#92;, d&#92;mu &#92;leq &#92;limsup_n &#92;int_E &#92;lvert f_n-f &#92;rvert&#92;, d&#92;mu &#92;leq 0. &#92;end{array} ' class='latex' /></p>
<p> It must then be <img src='http://s0.wp.com/latex.php?latex=%7B%5Cint_E+%5Clvert+f_n+-+f+%5Crvert%5C%2C+d%5Cmu+%3D+0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='{&#92;int_E &#92;lvert f_n - f &#92;rvert&#92;, d&#92;mu = 0}' title='{&#92;int_E &#92;lvert f_n - f &#92;rvert&#92;, d&#92;mu = 0}' class='latex' />. But this proves the statement, since
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Bc%7D++%5Cbigg%5Clvert+%5Cint_E+%5Clvert+f_n%5Crvert%5C%2C+d%5Cmu+-+%5Cint_E+%5Clvert+f+%5Crvert%5C%2C+d%5Cmu+%5Cbigg%5Crvert+%3D+%5Cbigg%5Clvert+%5Cint_E+%5Cbig%28+%5Clvert+f_n+%5Crvert+-+%5Clvert+f+%5Crvert+%5Cbig%29%5C%2C+d%5Cmu+%5Cbigg%5Crvert+%5C%5C++%5Cleq+%5Cint_E+%5CBig%5Clvert+%5Clvert+f_n+%5Crvert+-+%5Clvert+f+%5Crvert+%5CBig%5Crvert%5C%2C+d%5Cmu+%5Cleq+%5Cint_E+%5Clvert+f_n+-+f+%5Crvert%5C%2C+d%5Cmu+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{c}  &#92;bigg&#92;lvert &#92;int_E &#92;lvert f_n&#92;rvert&#92;, d&#92;mu - &#92;int_E &#92;lvert f &#92;rvert&#92;, d&#92;mu &#92;bigg&#92;rvert = &#92;bigg&#92;lvert &#92;int_E &#92;big( &#92;lvert f_n &#92;rvert - &#92;lvert f &#92;rvert &#92;big)&#92;, d&#92;mu &#92;bigg&#92;rvert &#92;&#92;  &#92;leq &#92;int_E &#92;Big&#92;lvert &#92;lvert f_n &#92;rvert - &#92;lvert f &#92;rvert &#92;Big&#92;rvert&#92;, d&#92;mu &#92;leq &#92;int_E &#92;lvert f_n - f &#92;rvert&#92;, d&#92;mu &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{c}  &#92;bigg&#92;lvert &#92;int_E &#92;lvert f_n&#92;rvert&#92;, d&#92;mu - &#92;int_E &#92;lvert f &#92;rvert&#92;, d&#92;mu &#92;bigg&#92;rvert = &#92;bigg&#92;lvert &#92;int_E &#92;big( &#92;lvert f_n &#92;rvert - &#92;lvert f &#92;rvert &#92;big)&#92;, d&#92;mu &#92;bigg&#92;rvert &#92;&#92;  &#92;leq &#92;int_E &#92;Big&#92;lvert &#92;lvert f_n &#92;rvert - &#92;lvert f &#92;rvert &#92;Big&#92;rvert&#92;, d&#92;mu &#92;leq &#92;int_E &#92;lvert f_n - f &#92;rvert&#92;, d&#92;mu &#92;end{array} ' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5CBox&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;Box' title='&#92;Box' class='latex' /></p>
<blockquote style="text-align:justify;"><p><b>Exercise 2</b> <em> Let <img src='http://s0.wp.com/latex.php?latex=%7B%28X%2C+%5Cmathcal%7BF%7D%2C+%5Cmu%29%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{(X, &#92;mathcal{F}, &#92;mu)}' title='{(X, &#92;mathcal{F}, &#92;mu)}' class='latex' /> be a finite measure space and let <img src='http://s0.wp.com/latex.php?latex=%7B1%3Cp%3C%5Cinfty%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{1&lt;p&lt;&#92;infty}' title='{1&lt;p&lt;&#92;infty}' class='latex' />. Suppose that <img src='http://s0.wp.com/latex.php?latex=%7B%5C%7B+f_n+%5C%7D_%7Bn+%5Cin+%5Cmathbb%7BN%7D%7D%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;{ f_n &#92;}_{n &#92;in &#92;mathbb{N}}}' title='{&#92;{ f_n &#92;}_{n &#92;in &#92;mathbb{N}}}' class='latex' /> is a sequence of measurable functions in <img src='http://s0.wp.com/latex.php?latex=%7BL_p%28%5Cmu%29%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{L_p(&#92;mu)}' title='{L_p(&#92;mu)}' class='latex' /> whose norms are uniformly bounded in <img src='http://s0.wp.com/latex.php?latex=%7Bn%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{n}' title='{n}' class='latex' /> and which converge almost everywhere to a function <img src='http://s0.wp.com/latex.php?latex=%7Bf%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{f}' title='{f}' class='latex' />. Prove that the sequence <img src='http://s0.wp.com/latex.php?latex=%7B%5Cbig%5C%7B+%5Cint_X+f_ng%5C%2C+d%5Cmu+%5Cbig%5C%7D_%7Bn+%5Cin+%5Cmathbb%7BN%7D%7D%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;big&#92;{ &#92;int_X f_ng&#92;, d&#92;mu &#92;big&#92;}_{n &#92;in &#92;mathbb{N}}}' title='{&#92;big&#92;{ &#92;int_X f_ng&#92;, d&#92;mu &#92;big&#92;}_{n &#92;in &#92;mathbb{N}}}' class='latex' /> converges to <img src='http://s0.wp.com/latex.php?latex=%7B%5Cint_x+fg%5C%2C+d%5Cmu%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{&#92;int_x fg&#92;, d&#92;mu}' title='{&#92;int_x fg&#92;, d&#92;mu}' class='latex' /> for all <img src='http://s0.wp.com/latex.php?latex=%7Bg+%5Cin+L_q%28%5Cmu%29%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{g &#92;in L_q(&#92;mu)}' title='{g &#92;in L_q(&#92;mu)}' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=%7Bq%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{q}' title='{q}' class='latex' /> is the conjugate exponent of <img src='http://s0.wp.com/latex.php?latex=%7Bp%7D&amp;bg=F4F5F7&amp;fg=000000&amp;s=0' alt='{p}' title='{p}' class='latex' />. </em></p></blockquote>
<p> <em>Proof:</em>  The proof is very similar to the previous problem. We start by noticing that under the hypotheses of the problem,
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Bc%7D++%5Cbigg%5Clvert+%5Cint_x+f_ng%5C%2C+d%5Cmu+-+%5Cint_X+fg%5C%2C+d%5Cmu+%5Cbigg%5Crvert+%3D+%5Cbigg%5Clvert+%5Cint_X+%28f_n+-f%29g%5C%2C+d%5Cmu+%5Cbigg%5Crvert+%5C%5C+%5Cleq+%5ClVert+%28f_n-f%29g+%5CrVert_1+%5Cleq+%5ClVert+f_n+-+f+%5CrVert_p%5C%2C+%5ClVert+g+%5CrVert_q.+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{c}  &#92;bigg&#92;lvert &#92;int_x f_ng&#92;, d&#92;mu - &#92;int_X fg&#92;, d&#92;mu &#92;bigg&#92;rvert = &#92;bigg&#92;lvert &#92;int_X (f_n -f)g&#92;, d&#92;mu &#92;bigg&#92;rvert &#92;&#92; &#92;leq &#92;lVert (f_n-f)g &#92;rVert_1 &#92;leq &#92;lVert f_n - f &#92;rVert_p&#92;, &#92;lVert g &#92;rVert_q. &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{c}  &#92;bigg&#92;lvert &#92;int_x f_ng&#92;, d&#92;mu - &#92;int_X fg&#92;, d&#92;mu &#92;bigg&#92;rvert = &#92;bigg&#92;lvert &#92;int_X (f_n -f)g&#92;, d&#92;mu &#92;bigg&#92;rvert &#92;&#92; &#92;leq &#92;lVert (f_n-f)g &#92;rVert_1 &#92;leq &#92;lVert f_n - f &#92;rVert_p&#92;, &#92;lVert g &#92;rVert_q. &#92;end{array} ' class='latex' /></p>
<p> If we prove that <img src='http://s0.wp.com/latex.php?latex=%7B%5Clim_n+%5ClVert+f_n-f+%5CrVert_p+%3D+0%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='{&#92;lim_n &#92;lVert f_n-f &#92;rVert_p = 0}' title='{&#92;lim_n &#92;lVert f_n-f &#92;rVert_p = 0}' class='latex' />, we are done.</p>
<p>
We will achieve this by using the convexity of <img src='http://s0.wp.com/latex.php?latex=%7B%5Clvert+%5Ccdot+%5Crvert%5Ep%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='{&#92;lvert &#92;cdot &#92;rvert^p}' title='{&#92;lvert &#92;cdot &#92;rvert^p}' class='latex' />, since in that case it is
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Brcl%7D++%5Cfrac%7B%5Clvert+f_n+-+f+%5Crvert%5Ep%7D%7B2%5Ep%7D+%5Cleq+%5Ctfrac%7B1%7D%7B2%7D+%5Clvert+f_n+%5Crvert%5Ep+%2B+%5Ctfrac%7B1%7D%7B2%7D+%5Clvert+f+%5Crvert%5Ep.+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;frac{&#92;lvert f_n - f &#92;rvert^p}{2^p} &#92;leq &#92;tfrac{1}{2} &#92;lvert f_n &#92;rvert^p + &#92;tfrac{1}{2} &#92;lvert f &#92;rvert^p. &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;frac{&#92;lvert f_n - f &#92;rvert^p}{2^p} &#92;leq &#92;tfrac{1}{2} &#92;lvert f_n &#92;rvert^p + &#92;tfrac{1}{2} &#92;lvert f &#92;rvert^p. &#92;end{array} ' class='latex' /></p>
<p> hence,
<p align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cbegin%7Barray%7D%7Brcl%7D++%5Clvert+f_n+-f+%5Crvert%5Ep+%5Cleq+2%5E%7Bp-1%7D+%5Cbig%28+%5Clvert+f_n+%5Crvert%5Ep+%2B+%5Clvert+f+%5Crvert%5Ep+%5Cbig%29.+%5Cend%7Barray%7D+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;lvert f_n -f &#92;rvert^p &#92;leq 2^{p-1} &#92;big( &#92;lvert f_n &#92;rvert^p + &#92;lvert f &#92;rvert^p &#92;big). &#92;end{array} ' title='&#92;displaystyle  &#92;begin{array}{rcl}  &#92;lvert f_n -f &#92;rvert^p &#92;leq 2^{p-1} &#92;big( &#92;lvert f_n &#92;rvert^p + &#92;lvert f &#92;rvert^p &#92;big). &#92;end{array} ' class='latex' /></p>
<p> Set then <img src='http://s0.wp.com/latex.php?latex=%7Bg_n+%3D+2%5E%7Bp-1%7D+%5Cbig%28+%5Clvert+f_n%5Crvert%5Ep+%2B+%5Clvert+f+%5Crvert%5Ep%5Cbig%29+-+%5Clvert+f_n-f+%5Crvert%5Ep%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='{g_n = 2^{p-1} &#92;big( &#92;lvert f_n&#92;rvert^p + &#92;lvert f &#92;rvert^p&#92;big) - &#92;lvert f_n-f &#92;rvert^p}' title='{g_n = 2^{p-1} &#92;big( &#92;lvert f_n&#92;rvert^p + &#92;lvert f &#92;rvert^p&#92;big) - &#92;lvert f_n-f &#92;rvert^p}' class='latex' /> (which are non-negative functions) and apply Fatou&#8217;s Lemma as before. <img src='http://s0.wp.com/latex.php?latex=%5CBox&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;Box' title='&#92;Box' class='latex' /></p>
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		<title>Have a child, plant a tree, write a book</title>
		<link>http://blancosilva.wordpress.com/2013/04/23/have-a-child-plant-a-tree-write-a-book/</link>
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		<pubDate>Wed, 24 Apr 2013 02:02:58 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Analysis]]></category>
		<category><![CDATA[Approximation Theory]]></category>
		<category><![CDATA[Books]]></category>
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		<description><![CDATA[Or more importantly: rear your children to become nice people, water those trees, and make sure that your books make a good impact. I recently enjoyed the rare pleasure of having a child (my first!) and publishing a book almost at the same time. Since this post belongs in my professional blog, I will exclusively [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5622&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:center;"><a href="http://www.amazon.com/gp/product/1782161627/ref=as_li_qf_sp_asin_tl?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=1782161627&amp;linkCode=as2&amp;tag=blancosilva-20"><img src="http://farm9.staticflickr.com/8535/8693715736_0bb71d0d1d_z_d.jpg" style="margin-left:auto;margin-right:auto;width:35%;"></a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=1782161627" width="1" height="1" border="0" alt="" style="border:none!important;margin:0!important;" /></a></p>
<p style="text-align:justify;">Or more importantly: rear your children to become nice people, water those trees, and make sure that your books make a good impact.</p>
<p style="text-align:justify;">I recently enjoyed the rare pleasure of having a child (my first!) and publishing a book almost at the same time.  Since this post belongs in my professional blog, I will exclusively comment on the latter: <a href="http://www.packtpub.com/learning-scipy-for-numerical-and-scientific-computing/book">Learning SciPy for Numerical and Scientific Computing</a>, published by <a href="http://www.packtpub.com/">Packt</a> in a series of technical books focusing on Open Source software.</p>
<p style="text-align:justify;">Keep in mind that the book is for a very specialized audience: not only do you need a basic knowledge of Python, but also a somewhat advanced command of mathematics/physics, and an interest in engineering or scientific applications.  This is an excerpt of the detailed description of the monograph, as it reads in the publisher&#8217;s page:</p>
<blockquote style="text-align:justify;">
<p>It is essential to incorporate workflow data and code from various sources in order to create fast and effective algorithms to solve complex problems in science and engineering. Data is coming at us faster, dirtier, and at an ever increasing rate. There is no need to employ difficult-to-maintain code, or expensive mathematical engines to solve your numerical computations anymore.  SciPy guarantees fast, accurate, and easy-to-code solutions to your numerical and scientific computing applications.</p>
<p><i>Learning SciPy for Numerical and Scientific Computing</i> unveils secrets to some of the most critical mathematical and scientific computing problems and will play an instrumental role in supporting your research. The book will teach you how to quickly and efficiently use different modules and routines from the SciPy library to cover the vast scope of numerical mathematics with its simplistic practical approach that is easy to follow.</p>
<p>The book starts with a brief description of the SciPy libraries, showing practical demonstrations for acquiring and installing them on your system. This is followed by the second chapter which is a fun and fast-paced primer to array creation, manipulation, and problem-solving based on these techniques.</p>
<p>The rest of the chapters describe the use of all different modules and routines from the SciPy libraries, through the scope of different branches of numerical mathematics. Each big field is represented: numerical analysis, linear algebra, statistics, signal processing, and computational geometry. And for each of these fields all possibilities are illustrated with clear syntax, and plenty of examples. The book then presents combinations of all these techniques to the solution of research problems in real-life scenarios for different sciences or engineering — from image compression, biological classification of species, control theory, design of wings, to structural analysis of oxides.</p>
</blockquote>
<p style="text-align:justify;">The book is also being sold online in <a href="http://www.amazon.com/gp/product/1782161627/ref=as_li_qf_sp_asin_tl?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=1782161627&amp;linkCode=as2&amp;tag=blancosilva-20">Amazon</a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=1782161627" width="1" height="1" border="0" alt="" style="border:none!important;margin:0!important;" />, where it has been received with pretty good reviews.  I have found other random reviews elsewhere, with similar welcoming comments:</p>
<ul>
<li><a href="http://aimotion.blogspot.com.br/2013/04/review-of-book-learning-scipy-for.html">Artificial Intelligence in Motion</a> by Marcel Caraciolo</li>
<li><a href="http://www.johndcook.com/blog/2013/03/29/new-introduction-to-scipy/">The Endeavour</a>, by John D. Cook</li>
</ul>
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		<title>Project Euler with Julia</title>
		<link>http://blancosilva.wordpress.com/2013/01/04/project-euler-with-julia/</link>
		<comments>http://blancosilva.wordpress.com/2013/01/04/project-euler-with-julia/#comments</comments>
		<pubDate>Fri, 04 Jan 2013 07:20:32 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[puzzles]]></category>
		<category><![CDATA[coding]]></category>
		<category><![CDATA[Julia]]></category>
		<category><![CDATA[programming]]></category>
		<category><![CDATA[project Euler]]></category>

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		<description><![CDATA[Disclaimer: Project Euler discourages the posting of solutions to their problems, to avoid spoilers. The three solutions I have presented in my blog are to the three first problems (the easiest and most popular), as a means to advertise and encourage my readers to &#8220;push the envelope&#8221; and go beyond brute-force solutions. Just for fun, [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5432&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;font-size:8pt;width:60%;margin-left:auto;margin-right:auto;border:1px solid red;padding:3px;"><font color="red">Disclaimer: Project Euler discourages the posting of solutions to their problems, to avoid spoilers.  The three solutions I have presented in my blog are to the three first problems (the easiest and most popular), as a means to advertise and encourage my readers to &#8220;push the envelope&#8221; and go beyond brute-force solutions.</font></p>
<p style="text-align:justify;font-family:modern;font-size:12pt;">Just for fun, and as a means to learn <a href="http://julialang.org/">Julia</a>, I will be attempting problems from the <a href="http://projecteuler.net/">Project Euler</a> coding exclusively in that promising computer language.</p>
<p style="width:75%;text-align:center;border:1px black solid;margin-left:auto;margin-right:auto;"><img src="http://farm9.staticflickr.com/8217/8345263814_20aaa0d726_o_d.jpg" width="100%"></p>
<p style="text-align:justify;font-family:modern;font-size:12pt;">The first problem is very simple:</p>
<blockquote style="text-align:justify;font-family:modern;font-size:12pt;"><p>
<b>Multiples of 3 and 5</b><br />
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multip les is 23.</p>
<p>Find the sum of all the multiples of 3 or 5 below 1000.
</p></blockquote>
<p style="text-align:justify;font-family:modern;font-size:12pt;">One easy-to-code way is by dealing with arrays:</p>
<pre class="brush: python; light: true; title: ; notranslate">
julia&gt; sum( filter( x-&gt;((x%3==0) | (x%5==0)), [1:999] ) )
233168
</pre>
<p style="text-align:justify;font-family:modern;font-size:12pt;">A much better way is to obtain it via summation formulas:  Note that the sum of all multiples of three between one and 999 is given by</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=3%2B6%2B%5Cdotsb%2B999+%3D+3+%281%2B2%2B%5Cdotsb%2B333%29+%3D+3+%5Cfrac%7B333%5Ccdot+334%7D%7B2%7D+%3D+166833.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='3+6+&#92;dotsb+999 = 3 (1+2+&#92;dotsb+333) = 3 &#92;frac{333&#92;cdot 334}{2} = 166833.' title='3+6+&#92;dotsb+999 = 3 (1+2+&#92;dotsb+333) = 3 &#92;frac{333&#92;cdot 334}{2} = 166833.' class='latex' /></p>
<p style="text-align:justify;font-family:modern;font-size:12pt;">Similarly, the sum of all multiples of five between one and 995 is given by</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=5%2B10%2B%5Cdotsb%2B995+%3D+5+%281%2B2%2B%5Cdotsb%2B199%29+%3D+5+%5Cfrac%7B199%5Ccdot+200%7D%7B2%7D+%3D+99500.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='5+10+&#92;dotsb+995 = 5 (1+2+&#92;dotsb+199) = 5 &#92;frac{199&#92;cdot 200}{2} = 99500.' title='5+10+&#92;dotsb+995 = 5 (1+2+&#92;dotsb+199) = 5 &#92;frac{199&#92;cdot 200}{2} = 99500.' class='latex' /></p>
<p style="text-align:justify;font-family:modern;font-size:12pt;">Finally, we need to subtract the sum of the multiples of both three and five, since they have been counted twice.  This sum is</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=15%2B30%2B%5Cdotsb%2B990+%3D+15+%281%2B2%2B%5Cdotsb%2B66%29%3D+15+%5Cfrac%7B66%5Ccdot+67%7D%7B2%7D%3D+33165.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='15+30+&#92;dotsb+990 = 15 (1+2+&#92;dotsb+66)= 15 &#92;frac{66&#92;cdot 67}{2}= 33165.' title='15+30+&#92;dotsb+990 = 15 (1+2+&#92;dotsb+66)= 15 &#92;frac{66&#92;cdot 67}{2}= 33165.' class='latex' /></p>
<p style="text-align:justify;font-family:modern;font-size:12pt;">The final sum is then, <img src='http://s0.wp.com/latex.php?latex=166833%2B99500-33165%3D233168.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='166833+99500-33165=233168.' title='166833+99500-33165=233168.' class='latex' /></p>
<ul style="text-align:justify;font-family:modern;font-size:11pt;">
<li><a href="http://blancosilva.wordpress.com/puzzles/project-euler-with-julia/second-problem-even-fibonacci-numbers/">Problem 2: Even Fibonacci Numbers</a></li>
<li><a href="http://blancosilva.wordpress.com/puzzles/project-euler-with-julia/third-problem-largest-prime-factor/">Problem 3: Largest Prime Factor</a></li>
</ul>
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		<title>Seked</title>
		<link>http://blancosilva.wordpress.com/2012/12/16/seked/</link>
		<comments>http://blancosilva.wordpress.com/2012/12/16/seked/#comments</comments>
		<pubDate>Mon, 17 Dec 2012 04:20:14 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[History]]></category>
		<category><![CDATA[puzzles]]></category>
		<category><![CDATA[Egypt]]></category>
		<category><![CDATA[hieroglyphs]]></category>
		<category><![CDATA[history of math]]></category>
		<category><![CDATA[papyrus]]></category>
		<category><![CDATA[trigonometry]]></category>

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		<description><![CDATA[If a pyramid is 250 cubits high and the side of its base 360 cubits long, what is its seked? Take half of 360; it makes 180. Multiply 250 so as to get 180; it makes of a cubit. A cubit is 7 palms. Multiply 7 by The seked is palms [that is, ]. A&#8217;h-mose. [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5396&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:center;margin-left:auto;margin-right:auto;width:100%;border:2px solid black;"><img src="http://farm9.staticflickr.com/8353/8279201319_00f003b2c2_o_d.jpg" width="100%"></p>
<blockquote style="text-align:justify;font-family:modern;font-size:12pt;"><p>
If a pyramid is 250 <i>cubits</i> high and the side of its base 360 <i>cubits</i> long, what is its <i>seked</i>?</p>
<hr />
Take half of 360; it makes 180.  Multiply 250 so as to get 180; it makes <img src='http://s0.wp.com/latex.php?latex=1%2F2+1%2F5+1%2F50&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='1/2 1/5 1/50' title='1/2 1/5 1/50' class='latex' /> of a cubit.  A cubit is 7 palms.  Multiply 7 by <img src='http://s0.wp.com/latex.php?latex=1%2F2+1%2F5+1%2F50%3A&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='1/2 1/5 1/50:' title='1/2 1/5 1/50:' class='latex' /></p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bl%7Clll%7D+1%267%26%26%5C%5C+1%2F2%263%261%2F2%26+%5C%5C+1%2F5+%261%261%2F3%261%2F15%5C%5C+1%2F50%26%261%2F10%261%2F25+%5Cend%7Barray%7D&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='&#92;begin{array}{l|lll} 1&amp;7&amp;&amp;&#92;&#92; 1/2&amp;3&amp;1/2&amp; &#92;&#92; 1/5 &amp;1&amp;1/3&amp;1/15&#92;&#92; 1/50&amp;&amp;1/10&amp;1/25 &#92;end{array}' title='&#92;begin{array}{l|lll} 1&amp;7&amp;&amp;&#92;&#92; 1/2&amp;3&amp;1/2&amp; &#92;&#92; 1/5 &amp;1&amp;1/3&amp;1/15&#92;&#92; 1/50&amp;&amp;1/10&amp;1/25 &#92;end{array}' class='latex' /></div>
<p>The seked is <img src='http://s0.wp.com/latex.php?latex=5+%5Ctfrac%7B1%7D%7B25%7D&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='5 &#92;tfrac{1}{25}' title='5 &#92;tfrac{1}{25}' class='latex' /> palms [that is, <img src='http://s0.wp.com/latex.php?latex=%283%2B1%2F2%29+%2B+%281%2B1%2F3%2B1%2F15%29+%2B+%281%2F10%2B1%2F25%29&amp;bg=F4F5F7&amp;fg=555555&amp;s=0' alt='(3+1/2) + (1+1/3+1/15) + (1/10+1/25)' title='(3+1/2) + (1+1/3+1/15) + (1/10+1/25)' class='latex' />].</p>
<div style="text-align:right;">A&#8217;h-mose. <i>The Rhind Papyrus. 33 AD</i></div>
</blockquote>
<p style="text-align:justify;">The image above presents one of the problems included in the Rhind Papyrus, written by the scribe Ahmes (A&#8217;h-mose) circa 33 AD.  This is a description of all the hieroglyphs, as translated by August Eisenlohr:</p>
<p style="text-align:center;margin-left:auto;margin-right:auto;width:100%;border:2px solid black;"><img src="http://farm9.staticflickr.com/8081/8286659490_1f53a86564_o_d.jpg" width="100%"></p>
<p style="text-align:justify;">The question for the reader, after going carefully over the English translation is: What does <i>seked</i> mean?</p>
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		<title>Nezumi San</title>
		<link>http://blancosilva.wordpress.com/2012/12/14/nezumi-san/</link>
		<comments>http://blancosilva.wordpress.com/2012/12/14/nezumi-san/#comments</comments>
		<pubDate>Fri, 14 Dec 2012 05:52:35 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[History]]></category>
		<category><![CDATA[puzzles]]></category>
		<category><![CDATA[history of math]]></category>
		<category><![CDATA[Japan]]></category>

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		<description><![CDATA[On January first, a pair of mice appeared in a house and bore six male mice and six female mice. At the end of January there are fourteen mice: seven male and seven female. On the first of February, each of the seven pairs bore six male and six female mice, so that at the [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5393&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<div style="border:2px solid black;text-align:center;width:75%;margin-left:auto;margin-right:auto;">
<img src="http://farm9.staticflickr.com/8074/8271874852_45aec78621_o_d.jpg" width="100%">
</div>
<blockquote style="text-align:justify;font-family:modern;font-size:11pt;"><p>
On January first, a pair of mice appeared in a house and bore six male mice and six female mice.  At the end of January there are fourteen mice: seven male and seven female.</p>
<p>On the first of February, each of the seven pairs bore six male and six female mice, so that at the end of February there are ninety-eight mice in forty-nine pairs.  From then on, each pair of mice bore six more pairs every month.</p>
<ol>
<li>Find the number of mice at the end of December.</li>
<li>Assume that the length of each mouse is 4 <i>sun</i>.  If all the mice line up, each biting the tail of the one in front, find the total length of mice</li>
</ol>
<div style="text-align:right;"><i>Jinkō-ki</i>, 1715</div>
</blockquote>
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		<title>Ruthless Thieves Stealing a Roll of Cloth</title>
		<link>http://blancosilva.wordpress.com/2012/12/14/ruthless-thieves-stealing-a-roll-of-cloth/</link>
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		<pubDate>Fri, 14 Dec 2012 05:27:47 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[History]]></category>
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		<description><![CDATA[Some thieves stole a long roll of silk cloth from a warehouse. In a bush far from the warehouse, they counted the length of the cloth. If each thief gets 6 hiki, then 6 hiki is left over, but if each thief takes 7 hiki then the last thief gets no cloth at all. Find [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5391&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
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<blockquote style="text-align:justify;font-family:modern;font-size:11pt;"><p>
Some thieves stole a long roll of silk cloth from a warehouse.  In a bush far from the warehouse, they counted the length of the cloth.  If each thief gets 6 <i>hiki</i>, then 6 <i>hiki</i> is left over, but if each thief takes 7 <i>hiki</i> then the last thief gets no cloth at all.  Find the number of thieves and the length of the cloth.</p>
<div style="text-align:right;"><i>Jinkō-ki</i>, 1643</div>
</blockquote>
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		<title>Which one is the fake?</title>
		<link>http://blancosilva.wordpress.com/2012/12/07/which-one-is-the-fake/</link>
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		<pubDate>Fri, 07 Dec 2012 05:12:31 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Approximation Theory]]></category>
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		<description><![CDATA[&#8220;Crab on its back&#8221; &#8220;Willows at sunset&#8221; &#8220;Still life: Potatoes in a yellow dish&#8221;<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5377&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
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&#8220;Crab on its back&#8221;
</td>
<td style="border:0;width:33%;text-align:center;vertical-align:middle;">
&#8220;Willows at sunset&#8221;
</td>
<td style="border:0;width:33%;text-align:center;vertical-align:middle;">
&#8220;Still life: Potatoes in a yellow dish&#8221;
</td>
<tr>
</tr>
</table>
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		<title>Stones, balances, matrices</title>
		<link>http://blancosilva.wordpress.com/2012/08/07/stones-balances-matrices/</link>
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		<pubDate>Wed, 08 Aug 2012 00:21:31 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Combinatorics]]></category>
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		<description><![CDATA[Let&#8217;s examine an easy puzzle on finding the different stone by using a balance: You have four stones identical in size and appearance, but one of them is heavier than the rest. You have a set of scales (a balance): how many weights do you need to determine which stone is the heaviest? This is [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5083&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
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<p style="text-align:justify;">Let&#8217;s examine an easy puzzle on finding the different stone by using a balance:</p>
<blockquote style="text-align:justify;"><p>You have four stones identical in size and appearance, but one of them is heavier than the rest.  You have a set of scales (a balance): how many weights do you need to determine which stone is the heaviest?</p></blockquote>
<p style="text-align:justify;">This is a trivial problem, but I will use it to illustrate different ideas, definitions, and the connection to linear algebra needed to answer the harder puzzles below.  Let us start by solving it in the most natural way:</p>
<ol style="text-align:justify;">
<li>Enumerate each stone from 1 to 4.</li>
<li>Set stones 1 and 2 on the left plate; set stones 3 and 4 on the right plate.  Since one of the stones is heavier, it will be in the plate that tips the balance.  Let us assume this is the left plate.</li>
<li>Discard stones 3 and 4. Put stone 1 on the left plate; and stone 2 on the right plate.  The plate that tips the balance holds the heaviest stone.</li>
</ol>
<p style="text-align:justify;">This solution finds the stone in two weights.  It is what we call <em>adaptive measures</em>: each measure is determined by the result of the previous.  This is a good point to introduce an algebraic scheme to code the solution.</p>
<ul style="text-align:justify;">
<li><strong>The weights matrix</strong>: This is a matrix with four columns (one for each stone) and two rows (one for each weight).  The entries of this matrix can only be <img src='http://s0.wp.com/latex.php?latex=-1%2C+0&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='-1, 0' title='-1, 0' class='latex' /> or <img src='http://s0.wp.com/latex.php?latex=1%2C&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='1,' title='1,' class='latex' /> depending whether a given stone is placed on the left plate <img src='http://s0.wp.com/latex.php?latex=%281%29&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='(1)' title='(1)' class='latex' />, on the right plate <img src='http://s0.wp.com/latex.php?latex=%28-1%29&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='(-1)' title='(-1)' class='latex' /> or in neither plate <img src='http://s0.wp.com/latex.php?latex=%280%29.&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='(0).' title='(0).' class='latex' />  For example, for the solution given above, the corresponding matrix would be</p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=W+%3D+%5Cbegin%7Bpmatrix%7D+1+%26+1+%26+-1+%26+-1+%5C%5C+1+%26+-1+%26+0+%26+0+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='W = &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 0 &amp; 0 &#92;end{pmatrix}' title='W = &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 0 &amp; 0 &#92;end{pmatrix}' class='latex' /></div>
</li>
<li><strong>The stones matrix</strong>: This is a square matrix with four rows and columns (one for each stone).  Each column represents a different combination of stones, in such a way that the <em>n</em>-th column assumes that the heaviest stone is in the <em>n</em>-th position.  The entries on this matrix indicate the weight of each stone.  For example, if we assume that the heaviest stone weights <em>b</em> units, and each other stone weights <em>a</em> units, then the corresponding <strong>stones matrix</strong> is
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=B+%3D+%5Cbegin%7Bpmatrix%7D+b+%26+a+%26+a+%26+a+%5C%5C+a+%26+b+%26+a+%26+a+%5C%5C+a+%26+a+%26+b+%26+a+%5C%5C+a+%26+a+%26+a+%26+b+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='B = &#92;begin{pmatrix} b &amp; a &amp; a &amp; a &#92;&#92; a &amp; b &amp; a &amp; a &#92;&#92; a &amp; a &amp; b &amp; a &#92;&#92; a &amp; a &amp; a &amp; b &#92;end{pmatrix}' title='B = &#92;begin{pmatrix} b &amp; a &amp; a &amp; a &#92;&#92; a &amp; b &amp; a &amp; a &#92;&#92; a &amp; a &amp; b &amp; a &#92;&#92; a &amp; a &amp; a &amp; b &#92;end{pmatrix}' class='latex' /></div>
</li>
</li>
</ul>
<p style="text-align:justify;">Multiplying these two matrices, and looking at the sign of the entries of the resulting matrix, offers great insight on the result of the measures:</p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsign%7D+%5Cbig%28+W+%5Ccdot+B+%5Cbig%29+%3D+%5Ctext%7Bsign%7D+%5Cbegin%7Bpmatrix%7D+b-a+%26+b-a+%26+a-b+%26+a-b+%5C%5C+b-a+%26+a-b+%26+0+%26+0+%5Cend%7Bpmatrix%7D+%3D+%5Cbegin%7Bpmatrix%7D+%2B+%26+%2B+%26+-+%26+-+%5C%5C+%2B+%26+-+%26+0+%26+0+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;text{sign} &#92;big( W &#92;cdot B &#92;big) = &#92;text{sign} &#92;begin{pmatrix} b-a &amp; b-a &amp; a-b &amp; a-b &#92;&#92; b-a &amp; a-b &amp; 0 &amp; 0 &#92;end{pmatrix} = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; 0 &amp; 0 &#92;end{pmatrix}' title='&#92;text{sign} &#92;big( W &#92;cdot B &#92;big) = &#92;text{sign} &#92;begin{pmatrix} b-a &amp; b-a &amp; a-b &amp; a-b &#92;&#92; b-a &amp; a-b &amp; 0 &amp; 0 &#92;end{pmatrix} = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; 0 &amp; 0 &#92;end{pmatrix}' class='latex' /></div>
<p style="text-align:justify;">Note the columns of this matrix code the behavior of the measures:</p>
<ul style="text-align:justify;">
<li>The column <img src='http://s0.wp.com/latex.php?latex=%5Cbig%28+%5Cbegin%7Bsmallmatrix%7D+%2B+%5C%5C+%2B+%5Cend%7Bsmallmatrix%7D+%5Cbig%29&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;big( &#92;begin{smallmatrix} + &#92;&#92; + &#92;end{smallmatrix} &#92;big)' title='&#92;big( &#92;begin{smallmatrix} + &#92;&#92; + &#92;end{smallmatrix} &#92;big)' class='latex' /> indicates that the balance tipped to the left in both measures (and therefore, the heaviest stone is the first one)</li>
<li>The column <img src='http://s0.wp.com/latex.php?latex=%5Cbig%28+%5Cbegin%7Bsmallmatrix%7D+%2B+%5C%5C+-+%5Cend%7Bsmallmatrix%7D+%5Cbig%29&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;big( &#92;begin{smallmatrix} + &#92;&#92; - &#92;end{smallmatrix} &#92;big)' title='&#92;big( &#92;begin{smallmatrix} + &#92;&#92; - &#92;end{smallmatrix} &#92;big)' class='latex' /> indicates that the heaviest stone is the second one.</li>
<li>Note that the other two measures can&#8217;t find the heaviest stone, since this matrix was designed to find adaptively a stone supposed to be either the first or  the second.</li>
</ul>
<p style="text-align:justify;">Is it possible to design a solution to this puzzle that is not adaptive?  Note the solution with two measures given (in algebraic form) below:</p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsign%7D+%5Cleft%5B+%5Cbegin%7Bpmatrix%7D+1+%26+1+%26+-1+%26+-1+%5C%5C+1+%26+-1+%26+1+%26+-1+%5Cend%7Bpmatrix%7D+%5Ccdot+B+%5Cright%5D+%3D+%5Cbegin%7Bpmatrix%7D+%2B+%26+%2B+%26+-+%26+-+%5C%5C+%2B+%26+-+%26+%2B+%26+-+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;text{sign} &#92;left[ &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 1 &amp; -1 &#92;end{pmatrix} &#92;cdot B &#92;right] = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;end{pmatrix}' title='&#92;text{sign} &#92;left[ &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 1 &amp; -1 &#92;end{pmatrix} &#92;cdot B &#92;right] = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;end{pmatrix}' class='latex' /></div>
<p style="text-align:justify;">Since each column is different, it is trivial to decide after the experiment is done, which stone will be the heaviest.  For instance, if the balance tips first to the right (-) and then to the left (+), the heaviest stone can only be the third one.</p>
<p style="text-align:justify;">Let us make it a big harder: Same situation, but now we don&#8217;t know whether the stone that is different is heavier or lighter.</p>
<p style="text-align:justify;">The solution above is no good: Since we are not sure whether <em>b</em> is greater or smaller than <em>a</em>, we would obtain two sign matrices which are virtually mirror images of each other.</p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Bpmatrix%7D+%2B+%26+%2B+%26+-+%26+-+%5C%5C+%2B+%26+-+%26+%2B+%26+-+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;end{pmatrix}' title='&#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;end{pmatrix}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Bpmatrix%7D+-+%26+-+%26+%2B+%26+%2B+%5C%5C+-+%26+%2B+%26+-+%26+%2B+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;begin{pmatrix} - &amp; - &amp; + &amp; + &#92;&#92; - &amp; + &amp; - &amp; + &#92;end{pmatrix}' title='&#92;begin{pmatrix} - &amp; - &amp; + &amp; + &#92;&#92; - &amp; + &amp; - &amp; + &#92;end{pmatrix}' class='latex' /></div>
<p style="text-align:justify;">In this case, in the event of obtaining that the balance tips twice to the left: which would be the different stone?  The first, which is heaviest, or the fourth, which is lightest?  We cannot decide.</p>
<p style="text-align:justify;">One possible solution to this situation involves taking one more measure.  Look at the algebraic expression of the following example, to realize why:</p>
<div style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsign%7D+%5Cleft%5B+%5Cbegin%7Bpmatrix%7D+1+%26+1+%26+-1+%26+-1+%5C%5C+1+%26+-1+%26+1+%26+-1+%5C%5C+1+%26+-1+%26+-1+%26+1+%5Cend%7Bpmatrix%7D+%5Ccdot+B+%5Cright%5D+%3D+%5Cbegin%7Bpmatrix%7D+%2B+%26+%2B+%26+-+%26+-+%5C%5C+%2B+%26+-+%26+%2B+%26+-+%5C%5C+%2B+%26+-+%26+-+%26+%2B+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;text{sign} &#92;left[ &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; -1 &amp; 1 &#92;end{pmatrix} &#92;cdot B &#92;right] = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;&#92; + &amp; - &amp; - &amp; + &#92;end{pmatrix}' title='&#92;text{sign} &#92;left[ &#92;begin{pmatrix} 1 &amp; 1 &amp; -1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; 1 &amp; -1 &#92;&#92; 1 &amp; -1 &amp; -1 &amp; 1 &#92;end{pmatrix} &#92;cdot B &#92;right] = &#92;begin{pmatrix} + &amp; + &amp; - &amp; - &#92;&#92; + &amp; - &amp; + &amp; - &#92;&#92; + &amp; - &amp; - &amp; + &#92;end{pmatrix}' class='latex' /> or <img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Bpmatrix%7D+-+%26+-+%26+%2B+%26+%2B+%5C%5C+-+%26+%2B+%26+-+%26+%2B+%5C%5C+-+%26+%2B+%26+%2B+%26+-+%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=555555&amp;s=0' alt='&#92;begin{pmatrix} - &amp; - &amp; + &amp; + &#92;&#92; - &amp; + &amp; - &amp; + &#92;&#92; - &amp; + &amp; + &amp; - &#92;end{pmatrix}' title='&#92;begin{pmatrix} - &amp; - &amp; + &amp; + &#92;&#92; - &amp; + &amp; - &amp; + &#92;&#92; - &amp; + &amp; + &amp; - &#92;end{pmatrix}' class='latex' /></div>
<p style="text-align:justify;">In this case there is no room for confusion: if the balance tips three times to the same side, then the different stone is the first one (whether heavier or lighter). The other possibilities are also easily solvable: if the balance tips first to one side, then to the other, and then to the first side, then the different stone is the third one.</p>
<p style="text-align:justify;">The reader will not be very surprised at this point to realize that three (non adaptive) measures are also enough to decide which stone is different (be it heavier or lighter) in a set of twelve similar stones. To design the solution, a good weight matrix with twelve columns and three rows need to be constructed.  The trick here is to allow measures that balance both plates, which gives us more combinations with which to play.  How would the reader design this matrix?</p>
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		<title>Buy my book!</title>
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		<pubDate>Fri, 20 Jul 2012 16:17:36 +0000</pubDate>
		<dc:creator>Francisco Blanco-Silva</dc:creator>
				<category><![CDATA[Approximation Theory]]></category>
		<category><![CDATA[Books]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Imaging]]></category>
		<category><![CDATA[Linear Algebra]]></category>
		<category><![CDATA[Scientific Computing]]></category>
		<category><![CDATA[Statistics]]></category>
		<category><![CDATA[applied mathematics]]></category>
		<category><![CDATA[approximation theory]]></category>
		<category><![CDATA[books]]></category>
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		<category><![CDATA[chemical engineering]]></category>
		<category><![CDATA[coding]]></category>
		<category><![CDATA[computational geometry]]></category>
		<category><![CDATA[denoising]]></category>
		<category><![CDATA[engineering]]></category>
		<category><![CDATA[image processing]]></category>
		<category><![CDATA[mathematical imaging]]></category>
		<category><![CDATA[nonlocal means]]></category>
		<category><![CDATA[numerical analysis]]></category>
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		<category><![CDATA[scanning transmission electron microscopy]]></category>
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		<category><![CDATA[signal processing]]></category>
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		<description><![CDATA[Well, ok, it is not my book technically, but I am one of the authors of one of the chapters. And no, as far as I know, I don&#8217;t get a dime of the sales in concept of copyright or anything else. As the title suggests (Modeling Nanoscale Imaging in Electron Microscopy), this book presents [&#8230;]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=blancosilva.wordpress.com&#038;blog=14494049&#038;post=5070&#038;subd=blancosilva&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;"><a href="http://www.amazon.com/gp/product/146142190X/ref=as_li_tf_il?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=146142190X&amp;linkCode=as2&amp;tag=blancosilva-20"><img border="0" src="http://ws.assoc-amazon.com/widgets/q?_encoding=UTF8&amp;ASIN=146142190X&amp;Format=_SL160_&amp;ID=AsinImage&amp;MarketPlace=US&amp;ServiceVersion=20070822&amp;WS=1&amp;tag=blancosilva-20" style="float:left;margin:4px;"></a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=146142190X" width="1" height="1" border="0" alt="" style="border:none!important;margin:0!important;" />Well, ok, it is not <em>my book</em> technically, but I am one of the authors of one of the chapters.  And <strong>no</strong>, as far as I know, I don&#8217;t get a dime of the sales in concept of copyright or anything else.</p>
<p style="text-align:justify;">As the title suggests (<a href="http://www.amazon.com/gp/product/146142190X/ref=as_li_tf_tl?ie=UTF8&amp;camp=1789&amp;creative=9325&amp;creativeASIN=146142190X&amp;linkCode=as2&amp;tag=blancosilva-20">Modeling Nanoscale Imaging in Electron Microscopy</a><img src="http://www.assoc-amazon.com/e/ir?t=blancosilva-20&amp;l=as2&amp;o=1&amp;a=146142190X" width="1" height="1" border="0" alt="" style="border:none!important;margin:0!important;" />), this book presents some recent advances that have been made using mathematical methods to resolve problems in electron microscopy. With improvements in hardware-based aberration software significantly expanding the nanoscale imaging capabilities of scanning transmission electron microscopes (STEM), these mathematical models can replace some labor intensive procedures used to operate and maintain STEMs. This book, the first in its field since 1998,   covers relevant concepts such as super-resolution techniques (<em>that&#8217;s my contribution!</em>), special de-noising methods, application of mathematical/statistical learning theory, and compressed sensing.</p>
<p style="text-align:justify;">We even got a nice review in <a href="http://www.physicstoday.org/resource/1/phtoad/v65/i5/p54_s2?bypassSSO=1">Physics Today</a> by Les Allen, no less!</p>
<blockquote style="text-align:justify;"><p>
Imaging with electrons, in particular scanning transmission electron microscopy (STEM), is now in widespread use in the physical and biological sciences. And its importance will only grow as nanotechnology and nano-Biology continue to flourish. Many applications of electron microscopy are testing the limits of current imaging capabilities and highlight the need for further technological improvements. For example, high throughput in the combinatorial chemical synthesis of catalysts demands automated imaging. The handling of noisy data also calls for new approaches, particularly because low electron doses are used for sensitive samples such as biological and organic specimens.</p>
<p>Modeling Nanoscale Imaging in Electron Microscopy addresses all those issues and more. Edited by Thomas Vogt and Peter Binev at the University of South Carolina (USC) and Wolfgang Dahmen at RWTH Aachen University in Germany, the book came out of a series of workshops organized by the Interdisciplinary Mathematics Institute and the NanoCenter at USC. Those sessions took the unusual but innovative approach of bringing together electron microscopists, engineers, physicists, mathematicians, and even a philosopher to discuss new strategies for image analysis in electron microscopy.</p>
<p>In six chapters, the editors tackle the ambitious challenge of bridging the gap between high-level applied mathematics and experimental electron microscopy. They have met the challenge admirably. I believe that high-resolution electron microscopy is at a point where it will benefit considerably from an influx of new mathematical approaches, daunting as they may seem; in that regard Modeling Nanoscale Imaging in Electron Microscopy is a major step forward. Some sections present a level of mathematical sophistication seldom encountered in the experimentally focused electron-microscopy literature.<br />
The first chapter, by philosopher of science Michael Dickson, looks at the big picture by raising the question of how we perceive nano-structures and suggesting that a Kantian approach would be fruitful. The book then moves into a review of the application of STEM to nanoscale systems, by Nigel Browning, a leading experimentalist in the field, and other well-known experts. Using case studies, the authors show how beam-sensitive samples can be studied with high spatial resolution, provided one controls the beam dose and establishes the experimental parameters that allow for the optimum dose.</p>
<p>The third chapter, written by image-processing experts Sarah Haigh and Angus Kirkland, addresses the reconstruction, from atomic-resolution images, of the wave at the exit surface of a specimen. The exit surface wave is a fundamental quantity containing not only amplitude (image) information but also phase information that is often intimately related to the atomic-level structure of the specimen. The next two chapters, by Binev and other experts, are based on work carried out using the experimental and computational resources available at USC. Examples in chapter four address the mathematical foundations of compressed sensing as applied to electron microscopy, and in particular high-angle annular dark-field STEM. That emerging approach uses randomness to extract the essential content from low-information signals. Chapter five eloquently discusses the efficacy of analyzing several low-dose images with specially adapted digital-image-processing techniques that allow one to keep the cumulative electron dose low and still achieve acceptable resolution.</p>
<p>The book concludes with a wide-ranging discussion by mathematicians Amit Singer and Yoel Shkolnisky on the reconstruction of a three-dimensional object via projected data taken at random and initially unknown object orientations. The discussion is an extension of the authors’ globally consistent angular reconstitution approach for recovering the structure of a macromolecule using cryo-electron microscopy. That work is also applicable to the new generation of x-ray free-electron lasers, which have similar prospective applications, and illustrates nicely the importance of applied mathematics in the physical sciences.</p>
<p>Modeling Nanoscale Imaging in Electron Microscopy will be an important resource for graduate students and researchers in the area of high-resolution electron microscopy.</p>
<p>(Les J. Allen, Physics Today, Vol. 65 (5), May, 2012)</p></blockquote>
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<td style="text-align:center;border:0;"><a href="https://www.dropbox.com/s/r0km64pjww3t2ad/9781461421900-t1.pdf">Table of contents</a></td>
<td style="text-align:center;border:0;"><a href="https://www.dropbox.com/s/1pht0o9phgrqxs2/9781461421900-p1.pdf">Preface</a></td>
<td style="text-align:center;border:0;"><a href="https://www.dropbox.com/s/jvls6m0u71xg9m1/9781461421900-c1.pdf">Sample chapter</a></td>
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