## Packt Publising is celebrating their 2000th title!

Categories: Uncategorized

Another fun problem, as requested by Qinfeng Li in MA598R: Measure Theory

Let $f \in L_2(\mathbb{R})$ so that $g(x)=xf(x) \in L_2(\mathbb{R})$ too.

Show that $f \in L_1(\mathbb{R})$ and $\lVert f \rVert_1^2 \leq 8 \lVert f \rVert_2 \lVert g \rVert_2.$

I got this problem by picking some strictly positive value $\lambda$ and breaking the integral $\int_{\mathbb{R}} \lvert f \rvert\, d\mu$ as follows:

$\begin{array}{rl} \displaystyle{\int_{\mathbb{R}} \lvert f \rvert\, d\mu} &= \displaystyle{\int_{\lvert x \rvert \leq \lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \lvert f \rvert\, d\mu} \\ \\ &= \displaystyle{\int_{-\lambda}^{\lambda} \lvert f \rvert\, d\mu + \int_{\lvert x \rvert \geq \lambda} \tfrac{1}{\lvert x \rvert} \lvert g(x) \rvert\, d\mu} \\ \\ &= \displaystyle{\int_\mathbb{R} \lvert f \rvert \cdot \boldsymbol{1}_{\lvert x \rvert \leq \lambda}\, d\mu + \int_{\mathbb{R}} \lvert g \rvert \cdot \big( \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda}\big) \, d\mu} \\ \\ &\leq \lVert f \rVert_2 \cdot \big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 + \lVert g \rVert_2 \cdot \big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 \end{array}$

Let us examine now the factors $\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2$ and $\big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2$ above:

$\big\lVert \boldsymbol{1}_{\lvert x \rvert \leq \lambda} \big\rVert_2 = \sqrt{2\lambda}$

$\big\lVert \tfrac{1}{\lvert x \rvert} \boldsymbol{1}_{\lvert x \rvert \geq \lambda} \big\rVert_2 = \displaystyle{ \sqrt{ 2\int_\lambda^\infty \dfrac{dx}{x^2}} = \sqrt{ \dfrac{2}{\lambda}} }$

We have thus proven that $f \in L_1(\mathbb{R})$ with $\lVert f \rVert_1 \leq \sqrt{2\lambda} \lVert f \rVert_2 + \sqrt{\tfrac{2}{\lambda}} \lVert g \rVert_2.$ At this point, all you have to do is pick $\lambda = \lVert g \rVert_2 / \lVert f \rVert_2$ (provided the denominator is not zero) and you are done.

Categories: Analysis, puzzles, Teaching

## Areas of Mathematics

For one of my upcoming talks I am trying to include an exhaustive mindmap showing the different areas of Mathematics, and somehow, how they relate to each other. Most of the information I am using has been processed from years of exposure in the field, and a bit of help from Wikipedia.

But I am not entirely happy with what I see: my lack of training in the area of Combinatorics results in a rather dry treatment of that part of the mindmap, for example. I am afraid that the same could be told about other parts of the diagram. Any help from the reader to clarify and polish this information will be very much appreciated.

And as a bonus, I included a $\LaTeX$ script to generate the diagram with the aid of the tikz libraries.

\tikzstyle{level 2 concept}+=[sibling angle=40]
\begin{tikzpicture}[scale=0.49, transform shape]
\path[mindmap,concept color=black,text=white]
node[concept] {Pure Mathematics} [clockwise from=45]
child[concept color=DeepSkyBlue4]{
node[concept] {Analysis} [clockwise from=180]
child {
node[concept] {Multivariate \& Vector Calculus}
[clockwise from=120]
child {node[concept] {ODEs}}}
child { node[concept] {Functional Analysis}}
child { node[concept] {Measure Theory}}
child { node[concept] {Calculus of Variations}}
child { node[concept] {Harmonic Analysis}}
child { node[concept] {Complex Analysis}}
child { node[concept] {Stochastic Analysis}}
child { node[concept] {Geometric Analysis}
[clockwise from=-40]
child {node[concept] {PDEs}}}}
child[concept color=black!50!green, grow=-40]{
node[concept] {Combinatorics} [clockwise from=10]
child {node[concept] {Enumerative}}
child {node[concept] {Extremal}}
child {node[concept] {Graph Theory}}}
child[concept color=black!25!red, grow=-90]{
node[concept] {Geometry} [clockwise from=-30]
child {node[concept] {Convex Geometry}}
child {node[concept] {Differential Geometry}}
child {node[concept] {Manifolds}}
child {node[concept,color=black!50!green!50!red,text=white] {Discrete Geometry}}
child {
node[concept] {Topology} [clockwise from=-150]
child {node [concept,color=black!25!red!50!brown,text=white]
{Algebraic Topology}}}}
child[concept color=brown,grow=140]{
node[concept] {Algebra} [counterclockwise from=70]
child {node[concept] {Elementary}}
child {node[concept] {Number Theory}}
child {node[concept] {Abstract} [clockwise from=180]
child {node[concept,color=red!25!brown,text=white] {Algebraic Geometry}}}
child {node[concept] {Linear}}}
node[extra concept,concept color=black] at (200:5) {Applied Mathematics}
child[grow=145,concept color=black!50!yellow] {
node[concept] {Probability} [clockwise from=180]
child {node[concept] {Stochastic Processes}}}
child[grow=175,concept color=black!50!yellow] {node[concept] {Statistics}}
child[grow=205,concept color=black!50!yellow] {node[concept] {Numerical Analysis}}
child[grow=235,concept color=black!50!yellow] {node[concept] {Symbolic Computation}};
\end{tikzpicture}


## More on Lindenmayer Systems

We briefly explored Lindenmayer systems (or L-systems) in an old post: Toying with Basic Fractals. We quickly reviewed this method for creation of an approximation to fractals, and displayed an example (the Koch snowflake) based on tikz libraries.

I would like to show a few more examples of beautiful curves generated with this technique, together with their generating axiom, rules and parameters. Feel free to click on each of the images below to download a larger version.

Note that any coding language with plotting capabilities should be able to tackle this project. I used once again tikz for $\text{\LaTeX}$, but this time with the tikzlibrary lindenmayersystems.

 name : Dragon Curve axiom : X order : 11 step : 5pt angle : 90 rules : X -> X+YF+ Y -> -FX-Y  name : Gosper Space-filling Curve axiom : XF order : 5 step : 2pt angle : 60 rules : XF -> XF+YF++YF-XF--XFXF-YF+ YF -> -XF+YFYF++YF+XF--XF-YF  name : Quadric Koch Island axiom : F+F+F+F order : 4 step : 1pt angle : 90 rules : F -> F+F-F-FF+F+F-F  name : Sierpinski Arrowhead axiom : F order : 8 step : 3.5pt angle : 60 rules : G -> F+G+F F -> G-F-G  name : ? axiom : F+F+F+F order : 4 step : 2pt angle : 90 rules : F -> FF+F+F+F+F+F-F  name : ? axiom : F+F+F+F order : 4 step : 3pt angle : 90 rules : F -> FF+F+F+F+FF 

Would you like to experiment a little with axioms, rules and parameters, and obtain some new pleasant curves with this method? If the mathematical properties of the fractal that they approximate are interesting enough, I bet you could attach your name to them. Like the astronomer that finds through her telescope a new object in the sky, or the zoologist that discover a new species of spider in the forest.

## Some results related to the Feuerbach Point

Given a triangle $\triangle ABC,$ the circle that goes through the midpoints of each side, $D, E, F,$ is called the Feuerbach circle. It has very surprising properties:

• It also goes through the feet of the heights, points $G, H, I.$
• If $Oc$ denotes the orthocenter of the triangle, then the Feuerbach circle also goes through the midpoints of the segments $OcA, OcB, OcC.$ For this reason, the Feuerbach circle is also called the nine-point circle.
• The center of the Feuerbach circle is the midpoint between the orthocenter and circumcenter of the triangle.
• The area of the circumcircle is precisely four times the area of the Feuerbach circle.

Most of these results are easily shown with sympy without the need to resort to Gröbner bases or Ritt-Wu techniques. As usual, we realize that the properties are independent of rotation, translation or dilation, and so we may assume that the vertices of the triangle are $A=(0,0), B=(1,0)$ and $C=(r,s)$ for some positive parameters $r,s>0.$ To prove the last statement, for instance we may issue the following:

>>> import sympy
>>> from sympy import *
>>> A=Point(0,0)
>>> B=Point(1,0)
>>> r,s=var('r,s')
>>> C=Point(r,s)
>>> D=Segment(A,B).midpoint
>>> E=Segment(B,C).midpoint
>>> F=Segment(A,C).midpoint
>>> simplify(Triangle(A,B,C).circumcircle.area/Triangle(D,E,F).circumcircle.area)
4


But probably the most amazing property of the nine-point circle, is the fact that it is tangent to the incircle of the triangle. With exception of the case of equilateral triangles, both circles intersect only at one point: the so-called Feuerbach point.

## An Automatic Geometric Proof

We are familiar with that result that states that, on any given triangle, the circumcenter, centroid and orthocenter are always collinear. I would like to illustrate how to use Gröbner bases theory to prove that the incenter also belongs in the previous line, provided the triangle is isosceles.

We start, as usual, indicating that this property is independent of shifts, rotations or dilations, and therefore we may assume that the isosceles triangle has one vertex at $A=(0,0)$, another vertex at $B=(1,0)$ and the third vertex at $C=(1/2, s)$ for some value $s \neq 0.$ In that case, we will need to work on the polynomial ring $R=\mathbb{R}[s,x_1,x_2,x_3,y_1,y_2,y_3,z],$ since we need the parameter $s$ free, the variables $x_1$ and $y_1$ are used to input the conditions for the circumcenter of the triangle, the variables $x_2$ and $y_2$ for centroid, and the variables $x_3$ and $y_3$ for the incenter (note that we do not need to use the orthocenter in this case).

We may obtain all six conditions by using sympy, as follows:

>>> import sympy
>>> from sympy import *
>>> A=Point(0,0)
>>> B=Point(1,0)
>>> s=symbols("s",real=True,positive=True)
>>> C=Point(1/2.,s)
>>> T=Triangle(A,B,C)
>>> T.circumcenter
Point(1/2, (4*s**2 - 1)/(8*s))
>>> T.centroid
Point(1/2, s/3)
>>> T.incenter
Point(1/2, s/(sqrt(4*s**2 + 1) + 1))


This translates into the following polynomials

$h_1=2x_1-1, h_2=8sy_1-4s^2+1$ (for circumcenter)
$h_3=2x_2-1, h_4=3y_2-s$ (for centroid)
$h_5=2x_3-1, h_6=(4sy_3+1)^2-4s^2-1$ (for incenter)

The hypothesis polynomial comes simply from asking whether the slope of the line through two of those centers is the same as the slope of the line through another choice of two centers; we could use then, for example, $g=(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1).$ It only remains to compute the Gröbner basis of the ideal $I=(h_1, \dotsc, h_6, 1-zg) \subset \mathbb{R}[s,x_1,x_2,x_3,y_1,y_2,y_3,z].$ Let us use SageMath for this task:

sage: R.<s,x1,x2,x3,y1,y2,y3,z>=PolynomialRing(QQ,8,order='lex')
sage: h=[2*x1-1,8*r*y1-4*r**2+1,2*x2-1,3*y2-r,2*x3-1,(4*r*y3+1)**2-4*r**2-1]
sage: g=(x2-x1)*(y3-y1)-(x3-x1)*(y2-y1)
sage: I=R.ideal(1-z*g,*h)
sage: I.groebner_basis()
[1]


This proves the result.