User christian blatter - MathOverflow

Answer by Christian Blatter for Advantages of the sequence definition of limits

2012-08-30T20:36:37Z

In my view there are no advantages in defining function limits via sequences, and this practice should be abolished. Using this definition you would have to test $\aleph_1^{\aleph_0}$ or so sequences to prove a single instance of $\lim_{x\to a} f(x)=\alpha$. Why should one bring all these sequences into the picture?

The idea of "limit of $f(x)$ when $x\to a$" is the answer to the following question: What is the "natural" value of $f$ at the special, maybe "ideal", point $a\ $? Well, it's the value that would make $f$ continuous there.

This brings me to the main point: The primary and sufficiently intuitive notion is the notion of continuity. Unfortunately the simple concept of Lipschitz continuity does not cover all cases we'd like to handle, e.g. $\sqrt{|x|}$ at $x=0$. Therefore we have to dig deeper and come up with $\epsilon$ and $\delta$, and on, and on.

Sequences, on the other hand, are a fundamental tool to construct new objects, like $e$, or $\sqrt{2}$. Of course in passing we would then prove that $\lim_{x\to a}f(x)=\alpha$ iff for all sequences $\ldots$

Answer by Christian Blatter for Characterization of the Poisson law

2012-04-12T11:26:21Z

The following heuristic argument is finitary apart from the sum of the exponential series.

Consider a birth clinic observing on average $\lambda$ male births and $\lambda$ female births per day. Let $p_k$ be the probability that on a given day we have exactly $k$ boys; then this same $p_k$ is also the probability that we have exactly $k$ girls. Consider the days where together $n$ kids are born. On such days the boys and girls are binomially distributed among them, i.e., we have $${\bf P}[B=k\ \wedge\ G=n-k]={1\over 2^n}{n\choose k}\ .$$ As boys and girls are born independently of each other when their time has come the left side of this equation has the value $p_k p_{n-k}/P_n$ where $P_n={\bf P}[B+G=n]$. It follows that $$p_k p_{n-k}={P_n\over 2^n}{n\choose k}$$ and in particular $${p_0 p_n\over p_1 p_{n-1}}={1\over n}\ .$$ Since this holds for any $n\geq1$ we conclude that $$p_n={\mu^n\over n!}p_0\quad (n\geq1)\ ,\qquad \mu:={p_1\over p_0}\ .$$ The condition $\sum_{n\geq 0} p_n=1$ gives $p_0=e^{-\mu}$, so $$p_n={\mu^n\over n!}e^{-\mu}\qquad(n\geq0)\ .$$ Now we observe $$\lambda={\bf E}(B)=\sum_{k=0}^\infty k p_k=e^{-\mu}\sum_{k=1}^\infty {k\over k!}\mu^k =\mu\ ,$$ so that definitively $p_k={\lambda^k\over k!}e^{-\lambda}$.

Answer by Christian Blatter for Volume of Minkowski sum of a ball and an ellipsoid

2011-06-19T19:06:15Z

Because your ellipsoid has all axes but one of equal length we don't need the "full theory":

Let $x_1=:x$ and $(x_2, \ldots, x_n)=:y\in{\mathbb R}^{n-1}$. The Minkowski sum $M$ of the unit ball and the given ellipsoid is rotationally symmetric with respect to the $x$-axis and consists of all points $z=(x'+x'', y'+y'') \in {\mathbb R}^n$ with $${x'^2\over a^2}+|y'|^2\leq 1\ ,\qquad x''^2 +|y''|^2\leq 1\ .$$ A hyperplane $x={\rm const.}$ intersects $M$ in an $(n-1)$-dimensional ball whose radius can be found by maximizing $$|y'+y''|\leq |y'|+|y''|\leq \sqrt{1-x'^2/a^2} + \sqrt{1-x''^2}$$ under the constraint $x'+x''=x$.

Answer by Christian Blatter for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact?

2011-06-04T13:00:40Z

A triangle is possible iff no part is $>{1\over2}$. With probability ${1\over2}$ both cuts are on the same side of the midpoint $M$, in which case no triangle is possible. If the cuts $x$ and $y$, $\ x < y$, are on different sides of $M$ then with probability ${1\over 2}$ the point $x$ is further left in its half than $y$ is in the right half. In this case there is no triangle possible either. It follows that only ${1\over 4}$ of all cuts admit the forming of a triangle.

Answer by Christian Blatter for Is the notion of fractional dimension compatible with considering a dimension a set of n-tuples?

2011-06-03T18:29:22Z

The notion of "dimension" occurs in various contexts and its meaning is not always the same. There are:

(1) Topological dimension. The topological dimension ${\rm dim}(X)$ of a space $X\ne\emptyset$ is an integer $\geq0$ or $\infty$. It is defined inductively or by means of coverings: An $X$ is, e.g., two-dimensional, if any open covering of $X$ can be "refined" to a covering that covers no point more than three times. This concept of "dimension" is not easy to handle and gives rise to strange theorems, e.g., that a space of dimension $n$ can be the union of $n+1$ spaces of dimension $0$.

(2) Dimension of vector spaces $V$. Any vector space $V$ (over an arbitrary field $K$) has a dimension ${\rm dim}(V)$ which is an integer $\geq 0$ or $\infty$. This is an algebraic concept. If ${\rm dim}(V)=n$ then one can choose $n$ vectors $e_1$, $\ldots$, $e_n\in V$ such that any vector $x\in V$ can be written as $x=\sum_{k=1}^n x_k \ e_k$ or $x=(x_1,\ldots, x_n)$ for short. People say that we have $n$ "degrees of freedom" in such a $V$.

When the ground-field $K$ is the field ${\mathbb R}$ of real numbers then the topological dimension of $V$ is equal to its algebraic dimension $n$; in particular, the real line ${\mathbb R}$ is also topologically one-dimensional. Furthermore there is a natural measure of volume for arbitrary subsets $A\subset V$, given that the volume of the unit cube should be $1$. When a set $A$ is stretched by a factor $\lambda>0$ then its volume increases by the factor $\lambda^n$. If in ${\mathbb R}^n$ the euclidean metric $|x|:=\sqrt{\sum_{k=1}^n x_k^2}$ is adopted then also sets $A$ in $d$-dimensional "planes" $\ U\subset {\mathbb R}^n$ get a natural $d$-dimensional volume, and this volume multiplies by $\lambda^d$ under a linear stretching of $A$ by $\lambda$. This implies that also curved $d$-dimensional "surfaces" in ${\mathbb R}^n$ have a natural $d$-dimensional length or area.

(3) "Fractal" dimension. Given euclidean $n$-space with its bodies, surfaces and curves all having an intuitive dimension $d\leq n$ and a $d$-dimensional "volume" scaling in the expected way one may ask whether there is a way of measuring arbitrary sets $A\subset{\mathbb R}^n$ that is able to scale with noninteger exponents $\alpha$, and whether there are sets $A\subset{\mathbb R}^n$ that for such a value $\alpha\notin{\mathbb N}$ would have an $\alpha$-dimensional volume which is neither zero nor infinity.

The so-called Hausdorff measure (invented in 1919) has the necessary flexibility; and indeed there are "crazy" sets $A$, called "fractals", that for precisely one noninteger value $\alpha$ have $\alpha$-dimensional volume $\ne 0, \infty$. This value $\alpha$ is then called the Hausdorff or fractal dimension of $A$. This "dimension" is nothing esoteric but a certain geometrical quantity associated to $A$, as the semi-axes $a$ and $b$ are associated to an ellipse. There is one deep theorem though: The Hausdorff dimension of a set is never larger than its topological dimension. In particular, all "fractals" in ${\mathbb R}^3$ have a Hausdorff dimension $\leq 3$.

There are mathematically defined fractal sets, e.g., the Cantor set or the snowflake curve, and there are such sets found in nature, e.g. "the coast line of Britain", or cumulus clouds in the atmosphere. Given such a "natural" set the essential problem is to find its fractal dimension $\alpha$ by computational means.

Answer by Christian Blatter for How do you calculate the solid angle of a rectangular, axis aligned section of a surface defined by a two dimensional function?

2011-04-27T15:06:19Z

The normal to the paraboloid plays no role in this. A "surface element" ${\rm d}(x,y)$ at the point $(x,y)$ in the $(x,y)$-parameter plane produces via $\vec f$ (or rather $\vec f_*$) a surface element $dS$ at the point $\vec f(x,y)$ on your paraboloid $S$, and then this surface element $dS$ casts a shadow $d\omega$ on the unit sphere $S^2$ through central projection from $O$, i.e., via normalization of $\vec f$. Since $$|\vec f(x,y)|^2=x^2+y^2+{1\over4}(1-x^2-y^2)^2={1\over4}(1+x^2+y^2)^2$$ it follows that the shadow on $S^2$ is produced by the map $$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)\ .$$ This $\vec g$ is nothing else but an (unusual) parametric representation of $S^2$. In order to compute the area of the shadowed part of $S^2$ one has to compute $d\omega=|g_x\times g_y|{\rm d}(x,y)$ and to integrate this over the intended rectangle in the $(x,y)$-plane.

The computation gives, as already remarked by Ben, $$d\omega={4\over(1+x^2+y^2)^2}{\rm d}(x,y)\ .$$ Transforming to polar coordinates one finds for the $[-1,1]^2$-rectangle the exact value $8\sqrt 2\ \arctan(1/\sqrt 2)\doteq 6.96366$.

Answer by Christian Blatter for Efficient algorithm for finding the minima of a piecewise linear function

2011-04-15T18:54:55Z

Looking at a figure one is lead to the following algorithm which traces out the graph of $f$:

One may assume $a_1< a_2 < \ldots < a_n$. If $0 < a_1$ or $a_n < 0$ then $m:=\min_x f(x)=-\infty$. Otherwise for $x$ large negative one has $f(x)=a_1 x + b_1$ and for $x$ large positive one has $f(x)=a_n x+b_n$. Therefore put $x_1:=-\infty$, $j_1:=1$ and for $k\geq1$ define recursively $$ r_i := {b_{j_k}-b_i\over a_i-a_{j_k}} \ (j_k < i \leq n),\quad x_{k+1}:=\min_{j_k < i \leq n}\ r_i\ ,\quad j_{k+1}:=\arg\min_{j_k < i \leq n} r_i\ .$$ When for the first time $a_{j_{k+1}}\geq0$ the minimum is found: One has $m=a_{j_k}x_{k+1}+b_{j_k}$.

As a bonus I propose the following algorithm which seems more sophisticated and is maybe faster:

Put $g_i(x):=a_i x+ b_i$ and denote by $g_i\wedge g_j$ the point of intersection of the two graphs. Assume for simplicity $ a_1 < \ldots a_s<0 < a_{s+1} < \ldots < a_n$. Then apply in succession

Put $l:=1$, $\ r:=n$.
Put $(\xi,\eta):= g_l\wedge g_r\ $.
If $l=s$ and $r=s+1$, goto 7.
Let $\tau:=\max_{l < i < r}\ g_i(\xi)\ ,\quad p:=\arg\max_{l < i < r}\ g_i(\xi).$
If $\tau\leq \eta$ goto 7.
If $p\leq s$ put $l:=p$, else put $r:=p$; then goto 2.
$m=\eta$.

Answer by Christian Blatter for Explicit example of a smooth - but not analytic- closed curve without self-intersections

2011-04-12T12:23:05Z

Consider the curve

$$\gamma:\quad \phi\ \mapsto\ \Bigl(1+\exp{-1\over \pi^2 -\phi^2}\Bigr)\ (\cos\phi,\sin\phi)\qquad(-\pi< \phi< \pi)$$

with filled-in point $(-1,0)$.

Answer by Christian Blatter for How many "different" colorings (excluding exchanges) exist for a given map (graph)?

2011-04-11T11:16:41Z

As a youthful folly I once wrote a paper "On the algebra of the four color problem", Ens. Math 11 (1965), 175-193. It can be found here:

http://retro.seals.ch/digbib/view?rid=ensmat-001:1965:11::337&id=browse&id2=browse5&id3=1

In this paper the permutations of colors are systematically "quotiented out" via a certain homological process.

Answer by Christian Blatter for A Book You Would Like to Write

2011-02-13T12:40:12Z

What this country needs is a successor to Courant/Robbins' "What is mathematics?", first published in 1941. Gowers' wonderful "Princeton companion to mathematics" cannot serve as a modern replacement of this volume, insofar as it addresses a group which is already deeply interested in mathematics and definitely knows what mathematics is all about. Not unlike Gowers' compendium the book I'm dreaming of would be the work of a devoted collective of authors, but in addition it would need a unifying editorship to make it the landmark in the field for decennia to come, as it was the case with Courant/Robbins' book seventy years ago.

Answer by Christian Blatter for This limit converges to the partial derivative?

2011-02-12T12:37:36Z

For $t\ne 0$ one has $${f(x,t)-f(x,0) \over t}- \partial_2 f(0,0)= \int_0^1 (\partial_2 f(x,\tau \thinspace \thinspace t) - \partial_2 f(0,0))\thinspace d\tau ,$$ and here the right side is $<\epsilon$ when $(x,t)$ is in a suitable neighbourhood of $(0,0)$.

For an $f:X\times {\bf R}^n\to {\bf R}^m$ it is enough to consider the $i$-th coordinate function $f_i:X\times {\bf R}^n\to {\bf R}$, again denoted by $f$, and for the latter consider the auxiliary function $\phi(t):=f(x,t \thinspace v)$ on the interval $[0,1]$. One gets $$f(x,v)-f(x,0) =\int_0^1 \phi'(t)\thinspace dt = \int_0^1 \nabla f_2(x,t\thinspace v)\cdot v \thinspace dt,$$ whence $$f(x,v)-f(x,0) = \nabla f_2(x_0,0)\cdot v + o(\|v\|) \qquad ((x,v)\to(x_0,0)).$$

Answer by Christian Blatter for Generalization of Curl to higher dimensions

2011-01-22T19:50:07Z

Let ${\bf K}$ be a vector field in the neighbourhood of ${\bf p}\in{\mathbb R}^n$, and let ${\bf X}$ and ${\bf Y}$ be two tangent vectors at ${\bf p}$. These two vectors span a parallelogram $P$ with one vertex at ${\bf p}$. The "circulation" of ${\bf K}$ around $P$ computes to $$\int_{\partial P}{\bf K}\bullet d{\bf x}= (L.{\bf X})\bullet{\bf Y}- (L.{\bf Y})\bullet{\bf X} + o(|P|^2)$$ with $L:=d{\bf K}({\bf p})$ and $|P|$:= diam$(P)$. It follows that there is a certain skew bilinear function ${\rm Rot}\thinspace{\bf K}({\bf p}):T_{\bf p}\times T_{\bf p}\to{\mathbb R}$ with $$\int_{\partial P}{\bf K}\bullet d{\bf x}={\rm Rot}\thinspace{\bf K}({\bf p}).({\bf X},{\bf Y})+ o(|P|^2) \ \ \ (|P|\to 0).$$ In the case $n=3$ the bilinear form ${\rm Rot}$ can be represented by the vector ${\rm curl}\thinspace{\bf K}$ in the form $$ {\rm Rot}{\bf K}({\bf p}).({\bf X},{\bf Y}) = {\rm curl}\thinspace{\bf K}({\bf p})\bullet({\bf X}\times{\bf Y}).$$

Answer by Christian Blatter for How many edge-disjoint paths go from upper left to lower right in a $4 \times N$ rectangular gridwork of streets?

2011-01-02T15:30:40Z

Let $v_1$, $\ldots$, $v_N$ be the vertical streets and let $h_{1,j}$, $\ldots$, $h_{4,j}$ be the horizontal edges between $v_{j-1}$ and $v_j$. An admissible path $\gamma$ induces a coloring of the horizontal edges as follows: Consider a vertical street $v_j$. The path $\gamma$ uses either one or three of the incoming edges $h_{k,j}$ $(1\le k\le 4)$. If $\gamma$ uses one edge, color it black and the three other edges white. If $\gamma$ uses three edges, two of them are linked to each other by a part of $\gamma$ extending only to the left of $v_j$. Color these two edges red, the third edge black and the unused edge white. In all, there are 16 possible colorings $c:\lbrace 1,2,3,4\rbrace\to\lbrace b, r, w\rbrace$ that can result in this way. There is a $16\times 16$ transition matrix $T$ that encodes the possible matchings between the coloring $c$ of the edges $h_{k,j}$ and the coloring $c'$ of the edges $h_{k,j+1}$ (e.g., circles must be avoided). This matrix $T$ has to be determined "the hard way", i.e., by listing for each $c$ the possible $c'$. The number of admissible paths $\gamma$ is then obtained by applying $T^{N-1}$ to a suitable starting vector; so there is indeed a linear recurrence for the number of these paths.

An example: If $c$ contains one black and two red edges, then using the vertical edges on $v_j$ in an admissible way one may

(a) continue the black and the two red edges into the next column individually, maybe at a different level,

(b) connect the black end of $c$ to either one of the red ends by a vertical segment creating a $\supset$ and continue the other red edge of $c$ into the next column, but as a black edge,

and, if room on $v_j$ permits, one may

Here is a pictorial list (hopefully complete) of the possible transitions $c\to c'$:

http://www.math.ethz.ch/~blatter/grid.pdf

Answer by Christian Blatter for Most intricate and most beautiful structures in mathematics

2010-12-14T11:28:16Z

I have voted for ${\mathbb N}$; but let me nevertheless propose an object living in the analytical realm, namely the Schwartz space ${\cal S}$ of infinitely differentiable functions $f:{\mathbb R}\to{\mathbb C}$ that for $|x|\to\infty$ together with their derivatives go to zero faster than any power $1/|x|^n$. The "intricateness" of this space stems from the many operations you can perform in it and from the fact that these operations are intertwined with each other in miraculous ways. $$ $$ Responding to a comment: You have (a) ordinary multiplication and convolution, (b) "multiplication" with arbitrary polynomials $p(x)$ and operations $p(D)$, (c) multiplication with functions of the form $x\mapsto e^{iax}$ and the translation operator $T_a: f(\cdot)\mapsto f(\cdot-a)$ and (d) scaling of the variable $x$ resp. $\xi$. The Fourier transform $\Phi$ interchanges in each of these three cases the respective operations; and at heart of it all is Gauss' normal distribution $x\mapsto {1\over \sqrt{2\pi}}\int e^{-x^2/2} dx$ which stays fixed under $\Phi$. And, last not least, there is a scalar product which is preserved by $\Phi$.

Answer by Christian Blatter for Examples of using physical intuition to solve math problems

2010-11-22T09:22:24Z

Here is a proof of Pick's area theorem $\mu(P)=i +{b\over2}-1$ "using physical intuition": Assume that at time 0 a unit of heat is concentrated at each lattice point. This heat will be distributed over the whole plane by heat conduction, and at time $\infty$ it is equally distributed on the plane with density 1. In particular, the amount of heat contained in $P$ will be $\mu(P)$. Where does this amount of heat come from? Consider a segment $e$ between two consecutive boundary lattice points. The midpoint $m$ of $e$ is a symmetry center of the lattice, so at each instant the heat flow is centrally symmetric with respect to $m$. This implies that the total heat flux across $e$ is 0. As a consequence, the final amount of heat within $P$ comes from the $i$ interior lattice points and from the $b$ boundary lattice points. To account for the latter, orient $\partial P$ so that the interior is to the left of $\partial P$. The amount of heat going from a boundary lattice point into the interior of $P$ is a half, minus the turning angle of $\partial P$ at that point, measured in units of $2\pi$. Since the sum of all turning angles for a simple polygon is known to be one full turn, we arrive at the stated formula.

Answer by Christian Blatter for Finding an appropriate Riemannian metric $G:\mathbb{R}^3\rightarrow\mathbb{R}^{3\times{3}}$ on $\mathbb{R}^3$

2010-10-26T07:49:17Z

Try $ds^2 = |d{\bf x}|^2/(1+|{\bf x}|^2)$. The resulting metric space is complete since the "horizon" is infinitely far away. But maybe I have misunderstood the notion "not uniformly positive definite".

Answer by Christian Blatter for Newton equations, second order equation and (im)possible motions

2010-10-24T17:31:02Z

This is a very deep (or philosophical) question. It seems that in mechanical systems a "tension" (e.g., a plucking of a string) gives rise to an ${\bf acceleration}$ proportional to the "tension", while, e.g., in heat conduction a "tension" (i.e., a temperature gradient) gives rise to a ${\bf velocity}$ proportional to the "tension".

Answer by Christian Blatter for product of two riemann integrable is riemann integrable

2010-09-29T08:06:48Z

If $f$ and $g$ are Riemann integrable over the interval $[a,b]$ then there is an $M$ such that $|f|$ and $|g|$ are both $\le M$ on $[a,b]$. The Riemann integrability of $f g$ then immediately follows from the inequality $$|f(x)g(x)-f(x')g(x')|\le |f(x)-f(x')||g(x)|+|f(x')||g(x)-g(x')|$$ $$\le M(|f(x)-f(x')| +|g(x)-g(x')|) $$ for all $x, x'\in [a,b]$.

Answer by Christian Blatter for Walking around Santa Cruz, track around the soccer field

2010-08-16T10:38:39Z

Let $s\mapsto(x(s), y(s))$, $0\leq s\leq L(\Gamma)$, be a simply closed curve $\Gamma$ parametrized by arc length. Then the parallel curve $\Gamma_\epsilon$ at distance $\epsilon$ from $\Gamma$ has the parametric representation $s\mapsto (u(s),v(s))$ with $$u(s)=x(s)-\epsilon\dot y(s),\quad v(s)=y(s)+\epsilon \dot x(s).$$ Here $\epsilon$ may have either sign; I omit the discussion of this point. It follows from Frenet's formula that the line element of $\Gamma_\epsilon$ is given by $d\sigma=(1-\epsilon\kappa)ds$, where $\kappa$ denotes the curvature of $\Gamma$ and we assume $\epsilon\kappa(s)<1$ for all $s$. Integrating we obtain $$L(\Gamma_\epsilon)=L(\Gamma)-\epsilon\int_0^{L(\Gamma)}\kappa(s)ds = L(\Gamma)-2\pi \epsilon,$$ the latter equation following from the fact that the total curvature of a Jordan curve in the plane is $2\pi$ (up to sign).

Answer by Christian Blatter for Why fourier transform tell us energy of any frequency of f(t)

2010-08-15T08:24:29Z

If $g$ is the Fourier transform of $f$ then $|g(\tau)|^2$ can be interpreted as $energy \ density$ at the frequency $\tau$, which means that the total energy contained in a small frequency interval $[\tau-\epsilon,\tau+\epsilon]$ around $\tau$ is approximatively given by $2\epsilon\thinspace |g(\tau)|^2$.

Answer by Christian Blatter for Interesting applications (in pure mathematics) of first-year calculus

2010-08-12T10:07:45Z

A nice application of calculus that leads to a surprising and far reaching result, first obtained by the great Gauss himself, is the computation of the Arithmetic-Geometric Mean of two numbers $a > b > 0$. A comparatively short way to this end is presented on the first pages of J. and P. Borweins "Pi and the AGM".

Answer by Christian Blatter for Help me with this proof: Drop a printed map of the land on the land and there must be some common point.

2010-08-12T08:17:22Z

While it does not do any harm to point to Brouwer's or Banach's theorem here one should note that in the problem at hand the existence of the fixed point $p$ has been established by elementary means, so we only have to prove $p\in L$. For this to hold it suffices that the "country" $L$ is closed: Choose any point $x_0\in L$. Then the sequence of iterates $x_n:=T^n(x_0)$ $(n\ge 0)$ is on the one hand contained in $L$ and on the other hand converges to $p$, as the distances $d(x_n,p)$ decrease to 0 exponentially.

Answer by Christian Blatter for Function satisfying $f^{-1} =f'$

2010-08-01T09:48:32Z

Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum_{k \ge 1} c_k t^k$ we obtain from $(*)$ a recursion formula for the $c_k$, and one can show that $0< c_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.

Answer by Christian Blatter for How are these two ways of thinking about the cross product related?

2010-07-30T11:31:57Z

Let $\varepsilon( )$ be the volume form in $\mathbb R^3$. For given vectors ${\bf p}$ and ${\bf q}$ the function $f:{\bf x}\mapsto\varepsilon({\bf p},{\bf q},{\bf x})$ is a linear functional and so is represented by a vector ${\bf r}\in\mathbb R^3$, i.e., one has $f({\bf x})=\langle{\bf r},{\bf x}\rangle$. This vector ${\bf r}$ depends in a skew bilinear way from ${\bf p}$ and ${\bf q}$ and is called the $vector\ product$ of ${\bf p}$ and ${\bf q}$.

Comment by Christian Blatter

2012-10-22T19:31:46Z

The following is only anecdotal: The lowest natural power of $2$ whose decimal representation begins with a '$9$' is $2^{53}$.

Comment by Christian Blatter

2012-08-31T13:00:10Z

@quid: Of course the thing with the cardinality was meant to be a joke. The essential point is that bringing in sequences replaces a simple limit by an infinity of composite ("verschachtelte") limits, over which reigns an additional $\forall$.

Comment by Christian Blatter

2011-11-19T15:48:51Z

Here is a completely different proof of Pick's theorem: mathoverflow.net/questions/46883/…

Comment by Christian Blatter

2011-06-29T18:36:04Z

@Jerabek: Thank you for the clarification.

Comment by Christian Blatter

2011-06-29T17:53:27Z

Flying on a circle of radius $R=1$ you don't see anything.

Comment by Christian Blatter

2011-04-27T22:33:30Z

@href: You were lucky: $$x^2+y^2+{1/over4}(1-x^2-y^2)^2={1/over4}(1+x^2+y^2)^2$.

Comment by Christian Blatter

2011-04-12T13:39:22Z

@Joseph and Louis: The 1+ went lost in the write-up. Sorry for the slip.

Comment by Christian Blatter

2011-03-27T09:02:39Z

Since student days I have the following dream: In the realm of decimal expansions the expansions of rational numbers are distinguished by being periodic, and in the realm of continued fractions the expansions of rationals are finite and the expansions of quadratic irrationals are periodic. Someone should come up with an encoding of the reals that imparts algebraic numbers of arbitrary degree some distinguished feature. All "expansions" not having this feature would then automatically encode a transcendental number.

Comment by Christian Blatter

2011-02-12T20:05:11Z

Your last equation is obviously wrong. See the correct version in my answer.

Comment by Christian Blatter

2011-02-11T19:31:56Z

What do you mean by "$\partial_2 f(x,t)$ exists for all $x\in X$ and is continuous"? Note that $f$ and $\partial_2 f$ are functions of two variables.

Comment by Christian Blatter

2011-01-27T13:30:53Z

Using Ocam's razor I would assume that $n$ crows are independently uniformly distributed on $[0,1]$.

Comment by Christian Blatter

2011-01-27T13:08:12Z

I think you have to add something about the initial conditions. I can easily envisage an instance where there is exactly one collision.

Comment by Christian Blatter

2011-01-23T16:24:20Z

Why this answer got $\geq 4$ votes is beyond me.

Comment by Christian Blatter

2010-11-09T09:26:52Z

I write $A\thinspace )\thinspace(\thinspace B$ for $A\cap B=\emptyset$ and similarly $A\supset\hskip-5pt\subset B$ for $A\cap B\ne\emptyset$.

Comment by Christian Blatter

2010-11-07T19:10:49Z

Having read the answers $1-7$, for me the most interesting problem here is the one posed by Gil Kalai in MO question 4347: Find a function definable in finite terms that for $x\to\infty$ satisfies $f(f(x)\sim e^x$ in some sense.