text:
\begin{document}

\title{The Shape Parameter in the Gaussian Function}
\author{Lin-Tian Luh\\Department of Mathematics, Providence University\\Shalu, Taichung, Taiwan\\
Email:ltluh@pu.edu.tw\\Fax:886-4-26324653, Tel:886-4-26328001ext.15126 }
\date{\today}
\maketitle
{\bf Abstract}. This is the fifth of our series of works about the shape parameter. We now explore the parameter $\beta$ contained in the famous gaussian function $e^{-\beta |x|^{2}},\ x\in R^{n}$. In the theory of radial basis functions(RBF), gaussian is frequently used in virtue of its good error bound and numerical tractability. However the optimal choice of $\beta$ has been unknown. RBF people only know that $\beta$ is very influential, but do not have a reliable criterion of its choice. The purpose of this paper is to uncover its mystery.\\
\\
{\bf AMS classification}:41A05,41A25,41A30,41A63,65D10.\\
\\
{\bf Keywords}: radial basis function, gaussian, shape parameter, interpolation.
\section{Introduction}
Let
\begin{equation}
  h(x):=e^{-\beta |x|^{2}},\ x\in R^{n},\ \beta >0.
\end{equation}
For any given data points $(x_{j},y_{j}),\ j=1,\cdots,N$, where $X=\{x_{1},\cdots,x_{N}\}$ is a subset of $R^{n}$ and the $y_{j}'s$ are real or complex numbers, it's well known that there is a so-called $h$ spline interpolant $s(x)$ of these data points, defined by
\begin{equation}
  s(x)=p(x)+\sum_{j=1}^{N}c_{j}h(x-x_{j})
\end{equation}
where $p(x)$ is a polynomial in $P_{m-1}$, $m$ being the order of the conditional positive definiteness of $h$, and $c_{j}'s$ are chosen so that
\begin{equation}
  \sum_{j=1}^{N}c_{j}q(x_{j})=0
\end{equation}
for all polynomials $q$ in $P_{m-1}$ and
\begin{equation}
  p(x_{i})+\sum_{j=1}^{N}c_{j}h(x_{i}-x_{j})=y_{i},\ i=1,\cdots,N.
\end{equation}
Here $P_{m-1}$ denotes the class of polynomials in $R^{n}$ of degree $\leq m-1$, and $P_{m-1}:=\{0\}$ if $m=0$. The linear system of equations (3) and (4) has a unique solution $(c_{1},\cdots,c_{N})$ with $p(x)\equiv 0$ since $m=0$ for (1). Further details can be found in \cite{MN1}.

We use $s(x)$ as the approximating function. For error estimates there are two kinds of error bound, the algebraic type \cite{MN2} and the exponential type \cite{MN3} and \cite{Lu3-1}. The latter is an improved form of the former. In this paper we will show that if the shape parameter $\beta$ is chosen in an appropriate way, the exponential-type error bound will be minimal.
\section{Fundamental Theory}
In the theory of RBF, any conditionally positive definite(c.p.d.) function $h$ of order $m$ induces a function space called {\bf native space}, denoted by {\bf ${\cal C}_{h,m}$}. Its definition and characterization can be found in \cite{Lu1}, \cite{Lu2}, \cite{Lu4}, \cite{MN1}, \cite{MN2} and \cite{We}. We adopt Madych and Nelson's definition which is quite different from that of Wu and Schaback\cite{WS}. Besides the native space, each $f\in {\cal C}_{h,m}$ has a seminorm $\|f\|_{h}$ which plays an important role in our theory. In this paper $h$ is just the function in (1) and $m=0$.

There are two main theorems in this paper. Before introducing the first of our main theorems, we need a basic definition.
\begin{de}
  For any positive integer n, the number $\gamma_{n}$ is defined by $\gamma_{1}=2$ and $\gamma_{n}=2n(1+\gamma_{n-1})$ if $n\geq 2$.
\end{de}
As pointed out in section1, the theoretical cornerstone of our approach is the exponential-type error bound constructed by Madych and Nelson. However some crucial constants in the error bound had been unknown and considered to be a hard question. Fortunately these constants are thoroughly clarified in \cite{Lu3}. In \cite{Lu3} the author presents a complete and lucid exponential-type error bound for gaussian interpolation. In order to develop useful criteria of the optimal choice of $\beta$, we need the following theorem which is cited from \cite{Lu3} directly.
\begin{thm}
Let $h(x)=e^{-\beta \left\vert x\right\vert ^{2}},$ $\beta >0,$ be the
gaussian function in $R^{n}.$ Then, given a positive number $b_{0},$ there are positive
constants $\delta _{0},c,$and $C$ for which the following is true: If $f\in
{\cal C}_{h,m}$ and $s$ is the $h$ spline that interpolates $f$ on a
subset $X$ of $R^{n},$ then
\begin{equation}
\left\vert f(x)-s(x)\right\vert \leq \triangle ^{^{\prime \prime
}}(C_{\delta })^{\frac{c}{\delta }}\cdot \left\Vert f\right\Vert _{h}
\end{equation}
, where $\Delta ''=\pi^{\frac{n-1}{4}}\cdot (n\cdot \alpha_{n})^{\frac{1}{2}}\cdot 2^{\frac{n+1}{4}}\cdot \left( \frac{\sqrt{3}}{e}\right) ^{\frac{n-2}{4}}$ for even n and $\Delta ''=\pi^{\frac{n}{4}}\cdot (n\cdot \alpha_{n})^{\frac{1}{2}}\cdot \left( \frac{\sqrt{3}}{e}\right) ^{\frac{n-1}{4}}$ for odd n, holds for all x in a cube E provided that (a)E has side b and b$\geq b_{0}$,(b)$0<\delta \leq \delta_{0}$, and (c)every subcube of E of side $\delta$ contains a point of X. Here, $\alpha_{n}$ denotes the volume of the unit ball in $R^{n}$.

The number c is equal to $\frac{b_{0}}{8\gamma_{n}}$ where $\gamma_{n}$ was defined in Definition2.1. The number C is equal to $(3^{\frac{3}{4}}\cdot e\cdot \sqrt{2\rho \beta}\sqrt{n}\cdot e^{2n\gamma_{n}})^{4}\cdot b_{0}^{3}\cdot \gamma_{n}$, where $\rho=\frac{\sqrt{3}}{e}$. Moreover, $\delta_{0}$ can be defined by
$$\delta_{0}=\min \left\{ \frac{1}{(3^{\frac{3}{4}}\cdot e\cdot \sqrt{2\rho \beta}\cdot \sqrt{n}\cdot e^{2n\gamma_{n}})^{4}\cdot b_{0}^{3}\cdot \gamma_{n}},\ \delta_{n}\ \right\}$$
, where
$$\delta_{n}=\left\{ \begin{array}{llll}
                      \frac{b_{0}}{2\gamma_{n}},      & \mbox{if $n=1$,}\\
                      \frac{b_{0}}{2\gamma_{n}(n-1)}, & \mbox{if n is odd and $n>1$,}\\
                      \frac{b_{0}}{2\gamma_{n}(n-2)},& \mbox{if n is even and $n>2$,}\\
                      \frac{b_{0}}{2\gamma_{n}},      & \mbox{if $n=2$.}
                    \end{array}  \right.$$
\end{thm}
Note that the error bound (5) tends to zero as $\delta$ tends to zero. This is the key to understanding this seemingly complicated theorem. In (5) the constant $C$ highly depends on $\beta$. It's tempting to think that in (5) only $C$ is influenced by $\beta$. In fact, $\|f\|_{h}$ also changes as $\beta$ changes. The change of $\|f\|_{h}$ cannot be seen in a transparent way. Therefore Theorem2.2 cnnot be used directly to find the optimal $\beta$.

In order to overcome this problem, we introduce two function spaces as follows.
\begin{de}
  For any $\sigma>0$, the class of band-limited functions in $L^{2}(R^{n})$ is
$$B_{\sigma}:=\{f\in L^{2}(R^{n}):\hat{f}(\xi)=0\ if\ |\xi|>\sigma \}$$
, where $\hat{f}$ denotes the Fourier transform of $f$.
\end{de}
\begin{de}
For any $\sigma>0$,
$$G_{\sigma}:=\{f\in L^{2}(R^{n}):\ \int|\hat{f}(\xi)|^{2}e^{\frac{|\xi|^{3}}{\sigma}}d\xi <\infty \}$$
, where $\hat{f}$ denotes the Fourier transform of $f$. For each $f\in G_{\sigma}$,
$$\|f\|_{G_{\sigma}}:=\left\{ \int |\hat{f}(\xi)|^{2}e^{\frac{|\xi|^{3}}{\sigma}}d\xi \right\} ^{1/2}.$$
\end{de}
Let's investigate $B_{\sigma}$ first. For each $f\in B_{\sigma}$, by Corollary3.3 of \cite{MN2},
\begin{eqnarray*}
  \|f\|_{h} & = & \left\{ \int\frac{|\hat{f}(\xi)|^{2}}{(2\pi)^{2n}\hat{h}(\xi)}d\xi\right\}^{1/2}\\
        & = & \frac{1}{(2\pi)^{n}}\left\{\int|\hat{f}(\xi)|^{2}e^{\frac{\|\xi\|^{2}}{4\beta}}d\xi\right\}^{1/2}\\
        & \leq & \frac{1}{(2\pi)^{n}}e^{\frac{\sigma^{2}}{8\beta}}\left\{\int|\hat{f}(\xi)|^{2}d\xi\right\}^{1/2}\\
        & = & \frac{1}{(2\pi)^{n}}e^{\frac{\sigma^{2}}{8\beta}}\|f\|_{L^{2}(R^{n})}.
\end{eqnarray*}
Substituting this result into (5), we obtain the following useful theorem.
\begin{thm}
  For any $\sigma>0$, $f\in B_{\sigma}$ implies $f\in {\cal C}_{h,m}$ and (5) can be transformed into
\begin{equation}
  |f(x)-s(x)|\leq \Delta''(C\delta)^{\frac{c}{\delta}}\cdot (2\pi)^{-n}\cdot e^{\frac{\sigma^{2}}{8\beta}}\cdot \|f\|_{L^{2}(R^{n})}
\end{equation}
, where $h$ is defined as in (1).
\end{thm}
Functions in $G_{\sigma}$ can be treated in a similar way. For any $f\in G_{\sigma}$, we have
\begin{eqnarray*}
  \|f\|_{h} & = & \frac{1}{(2\pi)^{n}}\left\{\int|\hat{f}(\xi)|^{2}e^{\frac{|\xi|^{2}}{4\beta}}d\xi\right\}^{1/2}\\
            & = & \frac{1}{(2\pi)^{n}}\left\{\int|\hat{f}(\xi)|^{2}e^{\frac{|\xi|^{3}}{\sigma}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}d\xi\right\}^{1/2}\\
            & \leq & \frac{1}{(2\pi)^{n}}\left\{ \sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}\|f\|_{G_{\sigma}}.
\end{eqnarray*}
This gives the following theorem.
\begin{thm}
  For any $\sigma>0,\ f\in G_{\sigma}$ implies $f\in {\cal C}_{h,m}$ and (5) can be transformed into
\begin{equation}
  |f(x)-s(x)|\leq \Delta''(C\delta)^{\frac{c}{\delta}}(2\pi)^{-n}\left\{ \sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}\|f\|_{G_{\sigma}}
\end{equation}
, where $h$ is defined as in (1).
\end{thm}
There is an improved exponential-type error bound from which a set of criteria for the choice of the shape parameter can also be developed. In this kind of error bound the data points are not purely scattered and the interpolation happens in an n-simplex.

The definition of n-simplex in $R^{n}$ can be found in \cite{Fl}. 1-simplex is a line segment. 2-simplex is a triangle. 3-simplex is a tetrahedron.

Let $T_{n}$ be an n-simplex in $R^{n}$ with vertices $v_{1},\cdots,v_{n+1}$. Any $x\in T_{n}$ can be written as a convex combination of the vertices:
$$x=\sum_{i=1}^{n+1}c_{i}v_{i}$$
where $\sum_{i=1}^{n+1}c_{i}=1$ and $c_{i}\geq 0$ for all $i$. We call $(c_{1},\cdots,c_{n+1})$ the barycentric coordinate of $x$. Let's define {\bf `evenly spaced' points} of degree $l$ to be those points whose barycentric coordinates are of the form
$$(\frac{k_{1}}{l},\frac{k_{2}}{l},\cdots,\frac{k_{n+1}}{l}),\ k_{i}\ nonnegative\ integers\ with\ \sum_{i=1}^{n+1}k_{i}=l.$$

It's easily seen that the number of such points in $T_{n}$ is exactly $N=dimP_{l}^{n}$. Also, as stated in \cite{Bo}, evenly spaced points form a determining set for $P_{l}^{n}$, the space of polynomials of degree $l$ in n variables.

We can base our criteria of choosing $\beta$ on a crucial theorem which we cite directly from \cite{Lu3-1} with a slight modification.
\begin{thm}
  Let $h(x):=e^{-\beta|x|^{2}}$ be the gaussian function in $R^{n}$. For any positive number $b_{0}$, there are positive constants $\delta_{0},c_{1},c_{2}$, and $c_{3}$ independent of $n$, for which the following is true: If $f\in {\cal C}_{h,m}$ , the native space induced by $h$, and $s$ is the $h$ spline that interpolates $f$ on a subset $X$ of $R^{n}$, then
\begin{equation}
  |f(x)-s(x)|\leq c_{1}\sqrt{\delta}(c_{2}\delta)^{\frac{c_{3}}{\delta}}\cdot \| f\| _{h}
\end{equation}
for all $x$ in a subset $\Omega$ of $R^{n}$, and $0<\delta \leq \delta_{0}$, where $\Omega$ satisfies the property that for any $x$ in $\Omega$ and any number $\frac{b_{0}}{2}\leq r\leq b_{0}$, there is an $n$ simplex $Q$ with diameter $diamQ=r,x\in Q\subseteq \Omega$, such that for any integer $l$ with $\frac{b_{0}}{\delta}\leq l\leq \frac{2b_{0}}{\delta}$, there is on $Q$ an evenly spaced set of centers from $X$ of degree $l-1$.(In fact, the set $X$ can be chosen to consist of these evenly spaced centers in $Q$ only.) Here $\|f\| _{h}$ is the $h$-norm of $f$ in the native space. The numbers $\delta_{0},c_{1},c_{2}$, and $c_{3}$ are given by

$ \delta_{0}:=\min \left\{ b_{0},\ \frac{1}{\rho_{3}^{4}\cdot 3^{3}\cdot 2^{7}\cdot b_{0}^{3}}\right\} \ \ where\ \ \rho_{3}=12^{\frac{1}{4}}\cdot \sqrt{e\beta}\ \ $;
$\left\{
\begin{array}{lll}
 c_{1}:=\left\{ \begin{array}{ll}
                                                                                \Delta''\cdot \frac{1}{\sqrt{16\pi}}\cdot \frac{1}{\sqrt{b_{0}}}\ \ for\ \ odd\ \ n, \\ \Delta''\cdot \frac{1}{\sqrt{16\pi}}\cdot \frac{1}{\sqrt{b_{0}}}\ \ for\ \ even\ \ n,
                \end{array}\right. \  \\ c_{2}:=\rho_{3}^{4}\cdot 3^{3}\cdot 2^{7}\cdot b_{0}^{3}\ \ , \\ c_{3}:=\frac{b_{0}}{4},
 \end{array} \right.$
 where $\Delta''$ is defined by

$\Delta'':=\left\{ \begin{array}{ll}
                     \sqrt{2+\frac{1}{e}}\cdot \pi^{\frac{n-1}{4}}\cdot (n\alpha_{n})^{\frac{1}{2}}\cdot 2^{\frac{n}{4}}\cdot \rho^{\frac{n-1}{4}}\ for\ odd\ n, \\ \pi^{\frac{n-1}{4}}\cdot (n\alpha_{n})^{\frac{1}{2}}\cdot 2^{\frac{n+1}{4}}\cdot \rho^{\frac{n-2}{4}}\ for\ even\ n,\ with\ \rho=\frac{\sqrt{3}}{e}
                   \end{array} \right. $ ,where the number $\alpha_{n}$ denotes the volume of the unit ball in $R^{n}$.
In particular, if the point $x$ in $\Omega$ is fixed, the only requirement for $\Omega$ is the existence of an n simplex $Q$, with $diamQ=r,\ x\in Q\subseteq \Omega$, satisfying the afore-mentioned property of evenly spaced centers, and the centers $x_{1},\cdots, x_{N}$ in $s(x)$, as defined in (2), are just the evenly spaced points in $Q$.
\end{thm}
{\bf Remark}: This seemingly complicated theorem is in fact not difficult to understand. The number $\delta$ is in spirit equivalent to the well known fill-distance. The error bound (8) tends to zero rapidly as $\delta$ tends to zero. In practical application only finitely many interpolation points will be involved. Hence only a finite number of simplices will appear, even if $\Omega$ is unbounded. Note that among $c_{1},c_{2}$ and $c_{3}$, only $c_{2}$ depends on the shape parameter $\beta$. It's tempting to think that the influence of $\beta$ on (8) happens only in $c_{2}$. This is wrong. Be careful that, as in (5), $\|f\|_{h}$ depends on $\beta$ also.

By transforming the norms we obtain the following useful results now.

\begin{cor}
  Suppose $f\in B_{\sigma},\sigma>0$. Then (8) can be transformed into
\begin{equation}
  |f(x)-s(x)|\leq c_{1}\sqrt{\delta}(c_{2}\delta)^{\frac{c_{3}}{\delta}}\cdot (2\pi)^{-n}\cdot e^{\frac{\sigma^{2}}{8\beta}}\cdot \|f\|_{L^{2}(R^{n})}.
\end{equation}
\end{cor}

\begin{cor}
  Suppose $f\in G_{\sigma},\sigma>0$. Then (8) can be transformed into
\begin{equation}
  |f(x)-s(x)|\leq c_{1}\sqrt{\delta}(c_{2}\delta)^{\frac{c_{3}}{\delta}}\frac{1}{(2\pi)^{n}}\left\{\sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}\|f\|_{G_{\sigma}}
\end{equation}.
\end{cor}
\section{Criteria of Choosing $\beta$--the Scattered Type}
In (6) the parts influenced by $\beta$ are $(C\delta)^{\frac{c}{\delta}}$ and $e^{\frac{\sigma^{2}}{8\beta}}$. By the definition of $C$ and $c$, one can easily find that the function
$$MN(\beta):=\beta^{\frac{b_{0}}{4\gamma_{n}\delta}}\cdot e^{\frac{\sigma^{2}}{8\beta}}$$
describes the dependence of the error bound (6) on $\beta$. Let's call this function the MN function and its graph the MN curve. Obviously, finding the optimal $\beta$ is equivalent to finding the value $\beta$ minimizing $MN(\beta)$.

Similarly. in (7) the MN function is
$$MN(\beta):=\beta^{\frac{b_{0}}{4\gamma_{n}\delta}}\left\{sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}.$$

Another important thing which remains to be investigated is $\delta$. Our fundamental theory begins at Theorem2.2 where $0<\delta\leq \delta_{0}$ is a requirement. By the definition of $\delta_{0}$, we find
\begin{eqnarray*}
  \delta\leq \delta_{0} & iff & \delta\leq \delta_{n}\ and\ \delta\leq \frac{1}{(3^{\frac{3}{4}}e\sqrt{2\rho\beta n}e^{2n\gamma_{n}})^{4}b_{0}^{3}\gamma_{n}}\\
                        & iff & \delta\leq \delta_{n}\ and\ \beta^{2}\leq \frac{1}{(3^{\frac{3}{4}}e\sqrt{2\rho n}e^{2n\gamma_{n}})^{4}b_{0}^{3}\gamma_{n}\delta}\\
                        & iff & \delta\leq \delta_{n}\ and\ \beta\leq \frac{1}{3^{\frac{3}{2}}2\rho n e^{4n\gamma_{n}+2}b_{0}^{\frac{3}{2}}\sqrt{\gamma_{n}}\sqrt{\delta}}.
\end{eqnarray*}
This severely restricts the ranges of $\delta$ and $\beta$. However the upper bound of $\beta$ can be made arbitrarily large because $\delta$ can be arbitrarily small, theoretically.

We summarize these results in the following two criteria.\\
\\
{\bf Case1}. \fbox{$f\in B_{\sigma}$} Let $\sigma>0$ and $f\in B_{\sigma}$. Under the conditions of Theorem2.2, for any fixed $b_{0}>0$ and $\delta\leq \delta_{0}$, the optimal value of $\beta$ in the interval $(0,\beta_{0}]$ where $\beta_{0}= \frac{1}{3^{\frac{3}{2}}2\rho n e^{4n\gamma_{n}+2}b_{0}^{\frac{3}{2}}\sqrt{\gamma_{n}}\sqrt{\delta}}$ is the number minimizing
$$MN(\beta):=\beta^{\frac{b_{0}}{4\gamma_{n}\delta}}e^{\frac{\sigma^{2}}{8\beta}}.$$
{\bf Case2}. \fbox{$f\in G_{\sigma}$} Let $\sigma>0$ and $f\in G_{\sigma}$. Under the conditions of Theorem2.2, for any fixed $b_{0}>0$ and $\delta\leq \delta_{0}$, the optimal value of $\beta$ in the interval $(0, \beta_{0}]$ where $\beta_{0}=\frac{1}{3^{\frac{3}{2}}2\rho n e^{4n\gamma_{n}+2}b_{0}^{\frac{3}{2}}\sqrt{\gamma_{n}}\sqrt{\delta}}$ is the number minimizing
$$MN(\beta):=\beta^{\frac{b_{0}}{4\gamma_{n}\delta}}\left\{\sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}$$\\
\\
{\bf Remark}:(a)In both cases $MN(\beta)\rightarrow \infty$ both as $\beta\rightarrow 0^{+}$ and $\beta\rightarrow \infty$. (b)The number $\beta$ minimizing $MN(\beta)$ can be obtained by Mathematica or Matlab.\\
\\
{\bf Numerical Results}:We now present some pictures of the MN function. The lowest point of the MN curve corresponds to the optimal choice of the shape parameter $\beta$. In order to make the graph look better, we sometimes multiply the function by a constant k and call it the modified MN function. Its graph will then be called the modified MN curve.

\begin{figure}[h]
\centering
\includegraphics[scale=1.0]{BI.fifth.eps}
\caption{$f\in B_{\sigma},n=1,\sigma=1,b_{0}=1$}
\end{figure}

\clearpage

\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{BII.fifth.eps}
\caption{$f\in B_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{BIII.fifth.eps}
\caption{$f\in B_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{BIV.fifth.eps}
\caption{$f\in B_{\sigma},n=1,\sigma=1,b_{0}=1$}

\end{figure}
\clearpage

\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{BV.fifth.eps}
\caption{$f\in B_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{GI.fifth.eps}
\caption{$f\in G_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{GII.fifth.eps}
\caption{$f\in G_{\sigma},n=1,\sigma=1,b_{0}=1$}

\end{figure}

\clearpage

\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{GIII.fifth.eps}
\caption{$f\in G_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{GIV.fifth.eps}
\caption{$f\in G_{\sigma},n=1,\sigma=1,b_{0}=1$}

\includegraphics[scale=1.0]{GV.fifth.eps}
\caption{$f\in G_{\sigma},n=1,\sigma=1,b_{0}=1$}
\end{figure}

\clearpage

Although we didn't show the whole curves of the MN or modified MN functions, the crucial parts were presented.\\
\\
{\bf Remark}: In our criteria both $\delta$ and $\beta$ have an upper bound which is quite restrictive, especially for high dimensions. This seems to be an important topic and deserves future research.

\section{Criteria of Choosing $\beta$--the Evenly Spaced Type}
Formula (9) and (10) provide us with a very good theoretical ground to choose $\beta$.

In (9), if we extract the parts influenced by $\beta$, we will get a function of $\beta$, i.e.
$$(\beta^{2})^{\frac{b_{0}}{4\delta}}e^{\frac{\sigma^{2}}{8\beta}}.$$
As before, let's call it the MN function and denote it by $MN(\beta)$. Thus
$$MN(\beta):=\beta^{\frac{b_{0}}{2\delta}}e^{\frac{\sigma^{2}}{8\beta}}.$$
Its graph will be called the MN curve. Finding the optimal $\beta$ is then equivalent to finding the number minimizing $MN(\beta)$.
Similarly, in (10), the MN function is
$$MN(\beta):=\beta^{\frac{b_{0}}{2\delta}}\left\{\sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}.$$
Note that in Theorem2.7 we require that $\delta\leq \delta_{0}$ where $\delta_{0}$ highly depends on $\beta$ and should be treated in a rigorous way.

By the definition of $\delta_{0}$,
\begin{eqnarray*}
  \delta\leq \delta_{0} & iff & \delta\leq b_{0}\ and\ \delta\leq \frac{1}{12e^{2}\beta^{2}27\cdot2^{7}b_{0}^{3}}\\
                        & iff & \delta\leq b_{0}\ and\ \beta^{2}\leq \frac{1}{12e^{2}27\cdot 2^{7}b_{0}^{3}\delta}\\
                        & iff & \delta\leq b_{0}\ and\ \beta\leq \frac{1}{144\sqrt{2}eb_{0}^{\frac{3}{2}}\sqrt{\delta}}.
\end{eqnarray*}
These results can be summarized in the following criteria.\\
\\
{\bf Case1}. \fbox{$f\in B_{\sigma}$} Let $\sigma>0$ and $f\in B_{\sigma}$. Under the conditions of Theorem2.7, for any fixed $\delta,\ 0<\delta\leq b_{0}$, the optimal choice of $\beta$ in the interval $(0,\beta_{0}]$ where $\beta_{0}=\frac{1}{144\sqrt{2}eb_{0}^{\frac{3}{2}}\sqrt{\delta}}$ is the number minimizing
$$MN(\beta):=\beta^{\frac{b_{0}}{2\delta}}e^{\frac{\sigma^{2}}{8\beta}}.$$
{\bf Numerical Results}:\\
\\
Note that in this case the optimal choice of $\beta$ is independent of the dimension. Therefore in our numerical examples we will totally ignore the influence of n.
\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{BI.sixth.eps}
\caption{$f\in B_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.0]{BII.sixth.eps}
\caption{$f\in B_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.0]{BIII.sixth.eps}
\caption{$f\in B_{\sigma}, \sigma =1, b_{0}=1$}

\end{figure}

\clearpage

\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{BIV.sixth.eps}
\caption{$f\in B_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.0]{BV.sixth.eps}
\caption{$f\in B_{\sigma}, \sigma =1, b_{0}=1$}
\end{figure}

Now we begin the second case.\\
\\
{\bf Case2}. \fbox{$f\in G_{\sigma}$} Let $\sigma>0$ and $f\in G_{\sigma}$. Under the conditions of Theorem2.7, for any fixed $\delta,\ 0<\delta\leq b_{0}$, the optimal choice of $\beta$ in the interval $(0,\beta_{0}]$ where $\beta_{0}=\frac{1}{144\sqrt{2}eb_{0}^{\frac{3}{2}}\sqrt{\delta}}$ is the number minimizing
$$MN(\beta):=\beta^{\frac{b_{0}}{2\delta}}\left\{\sup_{\xi\in R^{n}}e^{\frac{|\xi|^{2}}{4\beta}-\frac{|\xi|^{3}}{\sigma}}\right\}^{1/2}.$$
{\bf Numerical Results}:\\
\\
Here, also, the optimal choice of $\beta$ is independent of n.
\begin{figure}[t]
\centering
\includegraphics[scale=1.1]{GI.sixth.eps}
\caption{$f\in G_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.0]{GII.sixth.eps}
\caption{$f\in G_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.1]{GIII.sixth.eps}
\caption{$f\in G_{\sigma}, \sigma =1, b_{0}=1$}

\end{figure}
\clearpage

\begin{figure}[t]
\centering
\includegraphics[scale=1.0]{GIV.sixth.eps}
\caption{$f\in G_{\sigma}, \sigma =1, b_{0}=1$}

\includegraphics[scale=1.0]{GV.sixth.eps}
\caption{$f\in G_{\sigma}, \sigma =1, b_{0}=1$}
\end{figure}

As in section3, although we didn't present the entire MN curves in virtue of the restriction of Mathematica, the crucial parts were presented.\\
\\
{\bf Remark}:(a)In our criteria there is an upper bound $\frac{1}{144\sqrt{2}eb_{0}^{\frac{3}{2}}\sqrt{\delta}}$ for $\beta$. This is a drawback of our theory and deserves future research. (b) Theorem2.7 requires that interpolation happens in an n-simplex with centers(interpolation points) evenly spaced points. It means that the data points are not purely scattered. This is also a drawback. However the shape of the simplex, and hence the distribution of the centers, is very flexible, making this drawback harmless. (c)In section4 the criteria do not depend on the dimension. This is the main advantage over the criteria of section3.


text:
{\bf Problem.} Given the areas of the three squares in the figure, what is the area of the interior triangle? [asy]
/* AMC8 2003 #6 Problem */
draw((0,0)--(12,0)--(12,5)--cycle);
draw((12,0)--(17,0)--(17,5)--(12,5));
draw((0,0)--(0,-12)--(12,-12)--(12,0));
draw((0,0)--(-5,12)--(7,17)--(12,5));
draw((11,0)--(11,1)--(12,1));
label("169", (4, 8));
label("25", (14.5, 2.5));
label("144", (6, -6));
[/asy]
{\bf Level.} Level 2
{\bf Type.} Prealgebra
{\bf Solution.} Each leg of the right triangle is a side of one of the squares.  Therefore, the legs of the right triangle have lengths $\sqrt{25}=5$ and $\sqrt{144}=12$, so the area of the triangle is $\frac12(5)(12) = \boxed{30}$.

set_name: MATH


TITLE: Understanding why $\frac{-n+1}{n+2}<-\frac{1}{2}$
QUESTION [0 upvotes]: I was trying to prove in $\epsilon,N$ language that $a_n=\frac{(-1)^{n}n+1}{n+2}$ is divergent. As I understand, we should prove that for every $L\in\mathbb{R}$ exists $\epsilon>0$ so for every $N\in \mathbb{N}$ there is $n>N$ so $|a_n-L|\geq \epsilon$. The solution says:

Let $L\in \mathbb{R}$. If $L\geq 0$ then we will choose $\epsilon=0.5$. We will check for which $n\in \mathbb{N}$ we get $\frac{-n+1}{n+2}<-\frac{1}{2}$.
$$\frac{-n+1}{n+2}<-\frac{1}{2} \Leftrightarrow -2n+2<-n-2 \Leftrightarrow n>4$$

Why did the solution choose $-\frac{1}{2}$ (why the minus)? I understand from the solution that if $L\geq 0$ then $n$ is odd (but why?). I don't get those two things.
The questions was asked before, but the solution there proves it using other methods.

REPLY [1 votes]: I would first
see what the terms are
for even and odd $n$.
$\begin{array}\\
a_{2n}
&=\frac{(-1)^{2n}2n+1}{2n+2}\\
&=\frac{2n+1}{2n+2}\\
&=\frac{2n+2-1}{2n+2}\\
&=1-\frac{1}{2n+2}\\
\end{array}
$
$\begin{array}\\
a_{2n+1}
&=\frac{(-1)^{2n+1}(2n+2)}{2n+3}\\
&=-\frac{2n+2}{2n+3}\\
&=-\frac{2n+3-1}{2n+3}\\
&=-1+\frac{1}{2n+3}\\
\end{array}
$
Since
$\dfrac1{2n}
\le \dfrac14
$
for $n \ge 2$,
for such $n$,
$a_{2n} \ge \dfrac34$
and
$a_{2n+1} < -\dfrac34$
so
$|a_{2n}-a_{2n+1}|
\ge \dfrac32
$
which makes non-convergence
clear.


TITLE: How to compute an expected value in shorter ways (when taking all possibilities into account isn't plausible.)
QUESTION [6 upvotes]: There is this question on which I have been spending a lot of time, trying to understand how to compute an expected value in a comprehensive way, as sorting out all the possibilities doesn't seem like that right thing to do, nor does it even seem possible.
The question states:

Dan tosses infinitely many standard, independent coins. The coins are tossed one by one. What is the expected number of tosses it will take Dan to arrive at two consecutive heads?

The answer says it is 6, and I didn't understand what to do. This is my attempt:

Firstly, I need to arrive at the first head. That for itself would have a geometric distribution with $\frac12$, which requires at least $2$ expected steps to be made.
Now, I either get another head and I am done, or I get tail, count one step, and then make 2 more expected steps.

Now I am in my 6th move and how do I know I am expected to arrive at head? Is that because last time I arrived at tail? I feel like I am in the right direction, but I don't fully comprehend the properties of expected value.

REPLY [11 votes]: Your English is fine, but just to confirm:  you are looking for E, the expected number of tosses before you have thrown HH. yes?
You have started well, but once you have tossed a few times the expectation may change.  Clearly, if your first toss is T, for example, you realize it will take longer than you thought (since part of your initial expectation relied on the possibility that you might throw HH to start!).
Let's follow your method:  If the first Toss is T (probability = $\frac 12$) then you are back to square 1, so your expectation from that state is E again, making your total expectation (along this route) E + 1.
If your first toss is H ((probability = $\frac 12$) then you throw again.  With probabilty $\frac 12$ you get H again and along this path it took you 2 tosses.  Thus we have a probability $\frac 14$ route to get HH in 2.  Of course you might throw a T for your second toss in which case you are back to the beginning, so we have a probability $\frac 14$ route to E + 2.  Putting this all together we see that: $$ E = \frac 14 (2)+ \frac 14 (E+2) + \frac 12 (E+1)$$ and it is easy to solve this to get E = 6.


TITLE: Absolutly integrable functions are injective to tempered distribution?
QUESTION [1 upvotes]: We had a theorem that$$\mathcal{L}^1(\mathbb{R}^n)\hookrightarrow \mathscr{S}'(\mathbb{R}^n)$$
Where $\mathscr{S}'$ is the space of tempered distributions. In the proof our lecturer constructed a linear map
$i:f\to T_f$ where $T_f$ is regular distribution. It means that
$$T_fg=i(f)g=\int_{\mathbb{R}^n} fg\mathrm{d}\lambda$$
I don't understand, why is this map injective from $\mathcal{L}$, because function $g\simeq f$, meaning that $g$ that is  almost everywhere equal to $f$ maps to the same distribution, doesn't it?

REPLY [1 votes]: Functions that are equal almost everywhere are identified. The Lebesgue spaces are not spaces of functions, but spaces of equivalence classes of functions.
Sometimes one also considers the spaces of functions - which are not Hausdorff in the $\lVert\,\cdot\,\rVert_p$ seminorm - and then distinguishes the spaces by using different fonts for the $L$, e.g. $\mathscr{L}$ for the non-Hausdorff space of functions, and $L$ for its quotient, the normed space of equivalence classes of functions (modulo being equal almost everywhere).
If that is the case in your lecture, and $\mathcal{L}$ denotes the space of functions instead of the space of equivalence classes of functions, then it was a mistake, and the map $\mathcal{L}^1(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ is not injective, and the other font ought to have been used.
If $\mathcal{L}^1(\mathbb{R}^n)$ is the Banach space of equivalence classes of (absolutely) integrable functions, the mapping is injective.

REPLY [0 votes]: $L^1(\mathbb{R^n})$ to be viewed as a normed space is quotiended over the functions that are almost everywhere zero. Knowing this your function $f\mapsto T_f$ become injective.


TITLE: Evaluating $\int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}2}dt$ using the error function
QUESTION [2 upvotes]: Ok guys... I was playing around with math, and I've noticed one thing... Considering this integral
$$\int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}{2}}dt$$
It looks a lot like the error function. I was trying to manipulate it in a way to make it look like an error function, but I didn't find a way.
Anybody has an idea?

REPLY [3 votes]: Hint. Using
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^x e^{-u^2}du$$
you may write
$$
\begin{align}
\int_{-\infty}^{\frac{x^2}{2}} e^{x-\frac{t^2}{2}}dt&=e^{x}\int_{-\infty}^{\frac{x^2}{2}} e^{-\frac{t^2}{2}}dt\\\\
&=\sqrt{2}\:e^{x}\int_{-\infty}^{\frac{x^2}{2\sqrt{2}}} e^{-u^2}du\\\\
&=\sqrt{2}\:e^{x}\left(\int_{-\infty}^{0} e^{-u^2}du+\int_{0}^{\frac{x^2}{2\sqrt{2}}} e^{-u^2}du\right)\\\\
&=e^x \sqrt{\frac{\pi }{2}} \left(1+\text{erf}\left(\frac{x^2}{2 \sqrt{2}}\right)\right).
\end{align}
$$


TITLE: Inverse Laplace transform of $\frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}$ using complex integration
QUESTION [6 upvotes]: I want to find the inverse Laplace transform of
$$F(s)=\frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}$$
using the Bromwich integral
$$f(t)= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{π\cosh(\sqrt s)}{2 s^{3/2} \sinh(\sqrt s)}e^{st}\, ds$$
The residue at the pole  $s=-(n\pi)^2$ is
$$\frac{-1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^2}e^{-(n\pi)^2t}$$
My progress so far has been stunted by the fact that we have a branch point at $s=0$. Any help is appreciated.

REPLY [4 votes]: Inasmuch as the function $F(s)=\frac{\pi\cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}$ is an even function of $\sqrt s$, there is no branch point at $s=0$.  Rather, there is a second order pole at $s=0$.
Hence, we have for $t>0$
$$\begin{align}
\mathscr{L}^{-1}\{F\}(t)&=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st}\,ds\\\\
&=\text{Res}\left(F(s)e^{st}, s=0\right)+\sum_{n=1}^\infty\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right)
\end{align}$$
For the residue at $s=0$ we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=0\right)=\frac\pi2\,\lim_{s\to 0}\frac{d}{ds}\left(\sqrt s\coth(\sqrt s)e^{st}\right)=\frac\pi6+\frac\pi2 t$$
For the residues at $-n^2\pi^2$, $n\ne0$, we have for $t>0$
$$\text{Res}\left(F(s)e^{st}, s=-n^2\pi^2\right)=\lim_{s\to -n^2\pi^2}(s+n^2\pi^2)\frac{\pi \cosh(\sqrt s)}{2s^{3/2}\sinh(\sqrt s)}\,e^{st}=-\frac{e^{-n^2\pi^2t}}{n^2\pi}$$
Hence, we find that
$$\mathscr{L}^{-1}\{F\}(t)=\frac\pi6+\frac{\pi}{2}t-\sum_{n=1}^\infty \frac{e^{-n^2\pi^2t}}{n^2\pi}$$


TITLE: Non-zero derivative and injectivity of a rational function
QUESTION [1 upvotes]: Let $\Omega\subset\mathbb{C}$ be a bounded domain which does not contain any pole of a rational function $R$ and $R'(z)\neq0$ for all $z\in\Omega$. Is it true that $R(z)$ is injective on $\Omega$?

REPLY [1 votes]: No; consider $R(z)=z^n$ for some $n\geq 2$ and $\Omega$ a neighborhood of a sufficiently long arc of the unit circle $\mathbb{S}^1\subset\mathbb{C}$.


TITLE: Mapping of open subsets of product spaces.
QUESTION [0 upvotes]: Let each $(A_i,T_i) = (\{0,2\}, T_{discrete})$ and define $\phi : \prod (A_i, T_i) \rightarrow [0,1]$ with $\phi (<a_1, a_2, ...>) = \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}}$.
If we let $W = \{a_1\} \times \{a_2\} \times ... \{a_N\} \times A_{N+1} \times A_{N+2} \times ...$
What is $\phi (W)$?
It seems to be $\frac {a_1}{2^2} +\frac {a_2}{2^3} + ... + \frac {a_N}{2^{N+1}}  + \frac{\{0,2\}}{2^{N+2}}$ + ... ?  but this doesn't make sense.

REPLY [1 votes]: HINT: You’re right: as you’ve written it, it doesn’t really make sense. $W$ contains infinitely many different points of the product, each of which is a sequence of $0$s and $2$s, and $\varphi[W]$ contains the images under $\varphi$ of all of these sequences. Thus,
$$\varphi[W]=\left\{\frac{a_1}{2^2}+\frac{a_2}{2^3}+\ldots+\frac{a_N}{2^{N+1}}+\sum_{n\ge N+1}\frac{a_n}{2^{n+1}}:a_n\in\{0,2\}\text{ for each }n\ge N+1\right\}\;.$$
Now notice that
$$\frac{a^n}{2^{n+1}}=\begin{cases}
\dfrac0{2^{n+1}}=\dfrac0{2^n},&\text{if }a_n=0\\
\dfrac2{2^{n+1}}=\dfrac1{2^n},&\text{if }a_n=2\;,
\end{cases}$$
so if we set $b_n=\frac{a_n}2$ for $n\ge 1$, we can rewrite $\varphi[W]$:
$$\varphi[W]=\left\{\frac{b_1}{2^1}+\frac{b_2}{2^2}+\ldots+\frac{b_N}{2^N}+\sum_{n\ge N+1}\frac{b_n}{2^n}:b_n\in\{0,1\}\text{ for each }n\ge N+1\right\}\;.$$
At this point it’s very helpful to think of the elements of $\varphi[W]$ interms of their binary representations: for any specific choice of values of the $b_n\in\{0,1\}$, the sum of the infinite series
$$\frac{b_1}{2^1}+\frac{b_2}{2^2}+\ldots+\frac{b_N}{2^N}+\sum_{n\ge N+1}\frac{b_n}{2^n}\tag{1}$$
is simply the real number whose binary representation is
$${0.b_1b_2\ldots b_Nb_{N+1}\ldots\,}_{\text{two}}\;:\tag{2}$$
by definition the binary expansion $(2)$ means the sum of the series $(1)$. Thus, $\varphi[W]$ is the set of all real numbers that have binary expansions starting $0.b_1b_2\ldots b_N$.

What is the smallest such number?
Is there a largest such number? If so, what is it? If not, what is the supremum (least upper bound) of all such numbers?

Once you’ve answered those questions, you should be able to describe $\varphi[W]$ in very simple terms.


TITLE: Use the monotone convergence theorem to calculate $\int_{-\infty}^{+\infty}e^{-|x|}d\bar{\lambda} $
QUESTION [0 upvotes]: For a particular Lebesgue exercise, I am required to work out the integral   $\int_{-\infty}^{+\infty}e^{-|x|}d\bar{\lambda}  $
using the monotone convergence theorem. I am finding it difficult to start, mainly as the monotone convergence theorem generally involves a sequence of functions that converge to a particular function, but in this case, we are given a singular function.

REPLY [0 votes]: I think what you are asked to do is write the integral as the limit of $\int I_{(-n,n)} e^{-|x|}dx$ (which is permissible by Monotone Convergence Theorem) so we get the answer as  $\lim 2\int_0^{n} e^{-x}dx =\lim 2(1-e^{-n})=2$.


TITLE: $L^p$ space, paralellogram law
QUESTION [1 upvotes]: Let $(X,\mathcal{M},\mu)$ be a measure space and $1\leq p\leq\infty$. Suppose $f,g\in L^p(X,\mathcal{M},\mu)$ with $f(x)g(x)=0$ for almost all $x\in X$ and $\Vert f\Vert_p=\Vert g \Vert_p=1$. How to verify by direct calculation that
$$\Vert f+g\Vert_p^2+\Vert f-g\Vert_p^2=2(\Vert f\Vert_p^2+\Vert g\Vert_p^2)$$
if and ony if $p=2$?

REPLY [2 votes]: Let us first consider the norms $\|f+g\|_p$ and $\|f-g\|_p$ for $p\in[1,\infty)$:
$$\|f\pm g\|^p_p=\int_X|f\pm g|^pd\mu=\int_{[fg=0]}|f\pm g|^pd\mu=\int_{[f\neq0]}|f\pm g|^pd\mu+\int_{[g\neq0]}|f\pm g|^pd\mu=$$
$$=\int_{[f\neq0]}|f|^pd\mu+\int_{[g\neq0]}|g|^pd\mu=\int_X|f|^pd\mu+\int_X|g|^pd\mu=\|f\|^p_p+\|g\|^p_p=1+1=2$$
where the third and fourth transition follow from the fact that $fg=0$ (a.e.). Now from the parallelogram law we get that $2^{2/p}+2^{2/p}=4$, which is true if and only if $p=2$.
The case of $p=\infty$ is similar, just use the supremum norm.


TITLE: Understanding branch cut of the following function $(z^3+1)^{1/3}$
QUESTION [0 upvotes]: I'm trying to understand branch cuts in complex analysis. Let's say I have the following function
$$f(z)=(z^3+1)^{1/3}.$$
I see that the branch points in this case are the solutions to the equation $z^3+1=0$ and this is $z=-1\vee z=e^{\pi i/3}\vee z=e^{-\pi i/3}$. I would like to find a suitable branch cut for the function and what I think is the following branch cut $[-1,\infty)$. I take from -1 to infinity since -1 is a branch point but I am not sure about that. I don't understand how to find branch cut very well so what I want is a good explanation on how to find suitable branch cuts and then I will try for some other functions.
NB: I know someone asked a similar question Branch cuts for $(z^2+1)^{1/3}$
but for me I did not get any idea of branch cuts.

REPLY [1 votes]: A small circle around one of the zeros, the angle of $z^3+1$ increases by $2\pi$.  The angle of its cube root increases by $2\pi/3$, so $f(x)$ changes by a factor $e^{2i\pi/3}$.  So $f(x)$ is not well-defined, or at least not continuous, if that path is allowed.
Take a curved cut from $-1$, via $e^{i\pi/3}$, to $e^{-i\pi/3}$.  Then any allowed path that goes around one root goes around all three roots of $z^3+1$.  Each factor of $z^3+1$ increases in angle by $2\pi$, so $z^3+1$ increases in angle by $6\pi$.  Its cube root increases in angle by $2\pi$, so $f(z)$ takes a consistent cube root of $z^3+1$.


TITLE: Ray Sphere Intersect Problom
QUESTION [0 upvotes]: I'm making a 3D game and I have the following function to detect if a gun firing a laser beam (straight line) intersects a spherical object.
boolean intersect(Point3D raydir, Point3D rayorig, Point3D pos, double rad) {
        double a = raydir.X * raydir.X + raydir.Y * raydir.Y + raydir.X * raydir.X;
        double b = raydir.X * 2.0f * (rayorig.X - pos.X) + raydir.Y * 2.0f * (rayorig.Y - pos.Y) + raydir.Z * 2.0f * (rayorig.Z - pos.Z);
        double c = sum(pos, pos) + rayorig.X * rayorig.X + rayorig.Y * rayorig.Y + rayorig.Z * rayorig.Z -
                2.0f*(rayorig.X * pos.X + rayorig.Y * pos.Y + rayorig.Z * pos.Z) - rad * rad;
        double D = b*b + (-4.0f)*a*c;
        return ( D < 0 );
    }

So if my gun is located at (50,50,50) and the sphere is located at (50, 100, 50) with a radius of 5.0 and the weapon fires toward (50,40,50) which is the oposite direction to the sphere the math above return true.
intersect((50, 40, 50) , (50, 50, 50), (50, 100, 50), 5.0); //returns true

REPLY [0 votes]: Let's assume you have
$$\begin{array}
\vec{o} = (x_o, y_o, z_o) & \text{ Ray origin } \\
\vec{d} = (x_d, y_d, z_d) & \text{ Ray direction } \\
\vec{c} = (x_c, y_c, z_c) & \text{ Sphere center } \\
r \gt 0 & \text{ Sphere radius }
\end{array}$$
This means the points $\vec{p}(t)$ on the ray fulfill
$$\vec{p}(t) = \vec{o} + t \vec{d}$$
with $t \ge 0$.

The ray origin is within the sphere if and only if
$$\left( \vec{c} - \vec{o} \right) \cdot \left( \vec{c} - \vec{o} \right) \le r^2$$
In pseudocode:
Boolean IsInsideSphere(Vec3D o, Vec3D c, Real r):
    return ( ( (o.x - c.x)*(o.x - c.x) +
               (o.y - c.y)*(o.y - c.y) +
               (o.z - c.z)*(o.z - c.z) ) <= r*r )

The center of the sphere is in the same hemisphere (halfspace) as the direction of the ray, if and only if
$$\left( \vec{c} - \vec{o} \right) \cdot \vec{d} \ge 0$$
In pseudocode:
Boolean SameHalfspace(Vec3D o, Vec3D d, Vec3D c):
    return ( ( (c.x - o.x) * d.x +
               (c.y - o.y) * d.y +
               (c.z - o.z) * d.z ) >= 0 )

The minimum distance, squared, between the ray (infinitely long, extending to both directions) and the center of the sphere (from Point-Line Distance at Wolfram Mathworld) is
$$r_{min}^2 = \frac{ \left\lvert \vec{d} \times \left( \vec{c} - \vec{o} \right) \right\rvert^2 }{ \left\lvert \vec{d} \right\rvert^2 } = \frac{ \left( \vec{d} \times \left( \vec{c} - \vec{o} \right) \right) \cdot \left( \vec{d} \times \left( \vec{c} - \vec{o} \right) \right) }{ \vec{d} \cdot \vec{d} }$$
In pseudocode:
Real MinimumDistanceSquared(Vec3D o, Vec3D d, Vec3D c)
    Vec3D t = (c.x - o.x, c.y - o.y, c.z - o.z)
    Real dd = d.x*d.x + d.y*d.y + d.z*d.z
    if (dd > 0.0):
        Vec3D p = ( d.y*t.z - d.z*t.y,
                    d.z*t.x - d.x*t.z,
                    d.x*t.y - d.y*t.x )
        return (p.x*p.x + p.y*p.y + p.z*p.z) / dd
    else:
        # No direction, return the distance squared.
        return t.x*t.x + t.y*t.y + t.z*t.z

The complete ray-hits-sphere test is then
Boolean RayHitsSphere(Vec3D o, Vec3D d, Vec3D c, Real r)

    # If you are inside the sphere, you cannot miss it.
    if (IsInsideSphere(o, c, r)):
        return True

    # We cannot hit the sphere, if we are facing away.
    if (not SameHalfSpace(o, d, c)):
        return False

    # Actual intersection test.
    return (MinimumDistanceSquared(o, d, c) <= r*r)

In a real-world game, you could give better score the closer the ray passes to the center of the sphere. Instead of the above Boolean function, you could have it return 0.0 for a miss, and 1.0 for a perfect hit (within some fraction of r of the center), and some value inbetween for imperfect hits.


TITLE: Are $a$ and $n$ relatively prime?
QUESTION [0 upvotes]: Suppose that $a$ and $n$ are integers, $n>1$. Prove that the equation $ax\equiv1(\mod n)$ has a solution if and only if $a$ and $n$ are relatively prime.
How to solving this problem?

REPLY [0 votes]: $\exists x\in\Bbb Z\!:\  ax\equiv c\pmod n\!\iff\! \exists x,y\in\Bbb Z\!:\ ax+ny = c\!\overset{\rm\ Bezout}\iff\!\gcd(a,n)\mid c.\ $ Let $\ c = 1.$


TITLE: Subset of $\mathbb R$ with equal Fourier, Hausdorff and Minkowski dimensions
QUESTION [6 upvotes]: It is a standard fact that for $0\leq s\le1$, there is a compact set $C\subseteq [0,1]$ with Hausdorff and Minkowski dimensions $s$ (by modifying the construction of a Cantor set).
It is also a standard fact that for $0\leq s\le1$, there is a compact set $S\subseteq [0,1]$ with Fourier and Hausdorff dimensions $s$.
My question is: for an arbitrary $0\leq s\le1$, can we find a subset of $\mathbb R$ so that all three dimensions are equal?

REPLY [2 votes]: Yes. We just use a Baire Category argument (a similar technique also works in high dimensions). Consider the complete metric space $X$ of pairs $(E,\mu)$, where $\mu$ is a probability measure supported on $E$ such that
$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi)| |\xi|^{s/2} < \infty, $$
and $E$ is a compact subset of $[0,1]$. We define a distance function
$$ d((E_1,\mu_1),(E_2,\mu_2)) = \max \left( d_H(E_1,E_2), \sup_{\xi \in \mathbf{Z}} |\widehat{\mu_1}(\xi) - \widehat{\mu_2}(\xi)| |\xi|^{s/2} \right) $$
where $d_H$ is the Hausdorff metric between two sets. It is a useful heuristic that a generic set is as `thin as possible' with respect to the Hausdorff metric. It is simple to see that for any $(E,\mu)$ in $X$, the Fourier dimension of $E$ is at least equal to $s$, so we should expect quasi-all elements of $X$ have dimension $s$.
For each $t > s$, $\delta > 0$, and $\varepsilon > 0$, set
$$ A(t,\delta,s) = \{ (E,\mu) \in \mathcal{X} : |E_\delta| < \varepsilon \cdot \delta^s \} $$
where $E_\delta$ is the $\delta$ thickening of $E$. Then $A(t,\delta,s)$ is an open subset of $X$, and
$$ \bigcap_{n = 1}^\infty \bigcap_{m = 1}^\infty \bigcap_{k = 1}^\infty A(s+1/n,1/m,1/k) $$
is the set of all pairs $(E,\mu)$ in $X$ where $E$ has Minkowski dimension $s$. Thus it suffices to argue that $A(t,\delta,\varepsilon)$ is dense in $X$ for all required parameters. It is slightly technical to argue this, but the basic idea is to consider a random construction which, given a pair $(E_0,\mu_0)$, considers the random measure
$$ \mu = \mu_0 \cdot \sum_{k = 1}^K \phi_{\varepsilon_0}(x - x_k) $$
where $x_1,\dots, x_K$ are uniformly distributed on $[0,1]$, $\varepsilon_0 = K^{-1/s}$, and $\phi_{\varepsilon_0}$ is a smooth bump function supported on a ball radius $\varepsilon_0$. One then shows that with high probability that
$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi) - \widehat{\mu_0}(\xi)| = o(1) $$
as $K \to \infty$, and that $d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$.


text:
\begin{document}
\title{Uniform radius and regular stratifications}
\author{K. Bekka \\{IRMAR,
Universit\'e de Rennes1, 35042 Rennes (France)}}

\maketitle
\markboth{Uniform radius }{K.Bekka}

\section{Introduction}

In this paper we investigate how germs of real functions can change
under deformation.
In particular we look at deformations of germs of isolated
singularities from $\mathbb{R}^n$ to $\mathbb{R}^k$ ($n \geq k$)  and the relation
with there natural stratification in some tame categorie (algebraic,
analytic, semi-algebraic,subanalytic, $o$-minimal straucture polynomially bounded).
The word tame in this paper will refer to one of these categories.

We say that two germs are topologically equivalent
$f:(\mathbb{R}^n,0)\to (\mathbb{R}^k,0)$  and $g:(\mathbb{R}^n,0)\to
(\mathbb{R}^k,0)$  are topologically
equivalents if  there exists a germ of  homeomorphism
$h:(\mathbb{R}^n,0)\to \mathbb{R}^n,0)$ such that $g\circ h=f$
the topological type of a germ is its right equivalence class.

A family of germs is the germ at $\{0\}\times \mathbb{R}^p$ of some
 function $F:(\mathbb{R}^n\times \mathbb{R}^p,
\mathbb{R}^p\times\{0\})\to (\mathbb{R}^k,0).$

We shall usually denote a family of germs by
$f_{t}:(\mathbb{R}^n,0)\to (\mathbb{R}^k,0)$, $t\in \mathbb{R}^p$,
where $f_{t}(x)=F(x,t).$

A stratification of a set, e.g. an algebraic variety, is roughly
 speaking, a partition of it into smooth manifolds so that these manifolds
fit together with respect to some regularity condition.

We are interested in regularity condition that insure topological
triviality, which in consequence implies the constancy of the
topological type.e

Stratification theory was introduced
by R.Thom and H. Whitney for algebraic and analytic sets.

(see  {\cite{GM}} and {\cite{PW}} for some examples of
applications of stratification theory).

We consider in the paper the classical Whitney's conditions:

\begin{defn}
   \label{regular}
Let $X,Y$ be disjoint manifolds in $\R^m$, and let $y \in Y \cap \overline{X}$.
A triple $(X,Y,y)$ is called $(a)$ (resp. $(b)$)- regular if

(a) when a sequence $\{x_n\} \subset X$ tends to $y$
and $T_{x_n} X$ tends in the Grassmanian bundle to
a subspace $\tau_y$ of $\R^m$, then
$T_y Y\subset \tau_y$;

(b) when sequences $\{x_n\} \subset X$ and
$\{y_n\} \subset Y$ each tends to $y$, the unit vector
$\frac{(y_n-x_n)}{|y_n-x_n|} $ tends to a vector $v$, and
$T_{x_n}X$ tends to $\tau_y$, then
$v\in \tau_y$.

$X$ is called $(a)$ (resp. $(b)$)- regular over
$Y$ if each triple $(X,Y,y)$ is $(a)$(resp. $(b)$)- regular.

\end{defn}

\begin{defn}
A  stratification of a subset $V$  in $\R^m$
is a disjoint decomposition
$V=\bigsqcup_{i \in I} V_i, \quad V_i\cap V_j =\emptyset \ \ \
\textup{for} \ \ \ i\neq j
$
into smooth submanifolds $\{V_i\}_{i \in I}$,
called strata, is called an $(a) $(resp. $(b)$)-regular stratification
if

1. each point has a neighborhood intersecting only finitely
many strata;

2. the frontier $\overline{V_j}\setminus V_j$ of each stratum $V_j$
is a union of other strata $\bigsqcup_{i \in J(i)} V_i$;

3. any triple $(V_j,V_i,x)$ such that $x \in V_i \subset
\overline{V_j}$ is $(a)$(resp. $(b)$)-regular.

\end{defn}

\begin{thm}
    For any tame set
$V$ in $\R^m$ there is an $a$(resp. $b$)-regular
stratification.

\end{thm}

For the proof of this theorem i.e. the existence of stratifications,
in various tame categories, see
{\cite{Wh}}, {\cite{Th}}, {\cite{Lo}},{\cite{Hi}}, {\cite{BCR},
{\cite {DM}}.

For a family of germs of isolated singularities
$F:(\mathbb{R}^n\times \mathbb{R}^p,
\{0\}\times \mathbb{R}^p)\to (\mathbb{R},0)$, we associate the
canonical stratification
 of $\mathbb{R}^n\times \mathbb{R}^p$ given by the partition \\
 $$\{\mathbb{R}^n\times \mathbb{R}^p-F^{-1}(0), F^{-1}(0)-\{0\}\times \mathbb{R}^p,
 \{0\}\times \mathbb{R}^p\}$$

We shall denote by $\pi$ the projection on the second factor,
$V=F^{-1}(0)$, $Y=\{0\}\times \mathbb{R}^p$ , $X=V-Y$ and
$X_{t}=\{x\in \R^n| F(x,t)=0\}.$

Since   $X_{t}$ has an isolated singularity at $(0,t)$ i.e. the critical set of the restriction of $\pi$ to $V$ is $Y.$

Then $X$ is a smooth manifold , and for each point $(x,t)\in X,$ we
have\\
$\dis T_{(x,t)}X=\{(u,v)\in \C^n\times\C|
\sum_{i=1}^{n}u_{i}\frac{\partial F}{\partial x_{i}}(x,t) +
\sum_{j=1}^{p}v_{j}\frac{\partial F}{\partial
t_{j}}(x,t)=0\}=\left(\mathbb{R}d F\right)^{\perp}.$

we  use the following notation $d F=(\frac{\partial F}{\partial x_{1}},\ldots, \frac{\partial F}{\partial x_{n}},\frac{\partial F}{\partial t})$,
$d_{x} F=(\frac{\partial F}{\partial x_{1}},\ldots, \frac{\partial F}{\partial x_{n}})$

and $\| d_{x} F\|^2=  \sum_{i=1}^{n}\|\frac{\partial F}{\partial x_{i}} \|^2.$

For the canonical stratification associated to  a family of germs of
isolated singularities, to be (a) ( resp. (b)) regular,  can be made
more practical by the following form:

\begin{defn}
  We say that $F$ is Whitney regular at $0 $ if its canonical
  stratification is Whitney regular and this is equivalent to the following conditions are satisfied:

\underline {  condition $(a)$ :}

$$\lim_{(x,t)\to 0\atop (x,t)\in X}\left ( \frac{\frac{\partial
F}{\partial t_{j}}(x,t)}{\| d_{x} F(x,t)\|}\right)=0 \qquad \text{for
each } 1\leq j\leq p.$$

\underline { condition $(b^\prime)$: }

$$\lim_{(x,t)\to 0\atop (x,t)\in X}\left (\frac{\sum_{i=1}^{n}x_{i}\frac{\partial F}{\partial x_{i}}(x,t)}{\|x\|\| d_{x} F(x,t)\|}\right)=0.$$

\end{defn}

\begin{rem}
    \begin{enumerate}
        \item It is known that condition $a + b^\prime$ is equivalent to
    condition $b$. (see \cite{Th}, \cite{Ma})

        \item A Whitney regular family of function germs is
	topologically trivial, then the topological type is constant
	in such family. (see \cite{Th}, \cite{Ma})

    \end{enumerate}

\end{rem}

\section{Uniform radius and vanishing folds}

A family of germs is said to have no coalessing of critical points  (in the sens of
H.King \cite{K}) if there exists a neighbourhood $U$ of
$\{0\}\times\mathbb{R}^p$ in $\mathbb{R}^n\times \mathbb{R}^p$, such
that the restriction of $f_{t}$ to $U\cap
(\mathbb{R}^n-\{0\}\times\{t\} )$ has no critical point (i.e. submersion) for
each $t\in \mathbb{R}^p.$

For example the family $f_{t}(x)=x^3-3tx^2$ has a coalesing, and
$f_{0}$ is topologically equivalent to the identity, but for
$t\not=0$, $f_{t}$ has a maximum or  a minimum at $0$ and then fails
to be topologically equivalent to the identity.

For a germ of isolated singularity $f:(\mathbb{R}^n,0)\to (\mathbb{R},0)$ we denote by
$\mu(f_t)=\dim\frac{\mathcal{O}_n}{(\frac{\partial f}{\partial
x_1},\ldots, \frac{\partial f}{\partial x_n})}$ its the Milnor number.

We say that a family of isolated singularities is $\mu$-constant
family if $\mu(f_{t})=\mu(f_{0})$ for each $t\in \mathbb{R}^p.$

\begin{rem}
      For complex analytic  germs no coalescing of critical
      points is equivalent to  the family is $\mu$-constant.
\end{rem}

\begin{defn}
   Let $f:(\mathbb{R}^n,0)\to (\mathbb{R},0)$ be a tame germ with isolated
singularity. Let $\rho:(\mathbb{R}^n,0)\to \mathbb{R}^+$ be a germ of
a  tame submersion such that $\rho^{-1}(0)=\{0\}$.

The $\rho$-Milnor radius of $f$,$\rho(f)$, is the smallest critical value of
the restriction of $\rho$ to the smooth variety $f^{-1}(0)-\{0\}$
i.e.\\  $\rho(f)= \inf \{\rho(x)| x\in  f^{-1}(0)-\{0\}  \text{ and }
d_{x} f=\lambda d_{x} \rho, \text{ for some } \lambda\in
\mathbb{R}\}$.

If there are no critical values then $\rho(f)=\infty$.

\end{defn}

To extend this notion to tame maps $f:(\mathbb{R}^n,0)\to
(\mathbb{R}^k,0)$  we use the following notation.

\begin{defn}
   Let $f:(\mathbb{R}^n,0)\to (\mathbb{R}^k,0)$ be a tame germ with isolated
singularity. Let $\rho:(\mathbb{R}^n,0)\to \mathbb{R}^+$ be a germ of
a  tame submersion such that $\rho^{-1}(0)=\{0\}$.

The $\rho$-Milnor radius of $f$, $\rho(f)$, is the smallest critical value of
the restriction of $\rho$ to the smooth variety $f^{-1}(0)-\{0\}$
i.e.\\ $\dis \rho(f)= \inf \{\rho(x)| x\in  f^{-1}(0)-\{0\}  \text{
and } \sum_{1\leq i_1<\ldots<i_{m+1}\leq
n} \left|\frac{D(f_1,\ldots,
f_k,\rho)}{D(x_{i_1},\ldots,x_{i_{m+1}})}(x)\right|^2 =0\}.$

If there are no critical values then $\rho(f)=\infty$.

\end{defn}

\begin{defn}[uniform Milnor radius]

    Let $\{f_{t}\}$, $t\in \mathbb{R}^p$ be a  family of germs at
    $\{0\}$ of isolated singularity.

    We say that $\{f_{t}\}$ has uniform $\rho$-Milnor radius if there
    is an $\epsilon>0$ such that $\rho(f_{t})> \epsilon$ for all $t\in
    \mathbb{R}^p.$

\end{defn}

We call such function $\rho$ a control function.
\vskip 1cm

We say that a point $p \in f^{-1}(0) $ is a $\rho$-Kink ( or simply a { Kink})  of $ f^{-1}(0) $
if $p$  is non singular point of  $f$  and if $p$ is a critical point of $\rho$ restricted to the manifold of smooth points of $ f^{-1}(0).$
(see \cite{O})

\vskip 0.5cm
\begin{rem} For $k=1$,

an easy computation shows that a nonsingular $p\in f^{-1}(0)$  is a
kink if and only if $df(p)=\lambda d_x\rho(p)$ for some $\lambda$ in $\mathbb{R}-\{0\} .$
\end{rem}

\vskip 0.5cm

We suppose that for every $t\in\mathbb{R}^p$, $f_t(0)=0$ and $0$ is an isolated critical point of $f_t.$

Let $\gamma:[0,\epsilon] \to\R^n\times[0,1]$ be a real analytic path $\gamma(s)=(x(s),t(s))$  such that:
\begin{itemize}
\item[1)] $\gamma(0)=(0,0)$
\item[2)] $\vert x(s)\vert>0$ and $\vert t(s)\vert>0$ for all $0<s<\epsilon,$ and
\item[3)] $f(x(s),t(s))=0$ for all $0\leq s\leq \epsilon.$
\end{itemize}

\vskip 5pt

\begin{defn}

The path $\gamma$ will be called a { $\rho$-vanishing fold}  of $f$
(centered at $0$ ) if  $x(s)$ is a $\rho$-kink of $f_{t(s)}^{-1}(0)$
 for every  $s\in (0,\epsilon].$
\end{defn}

\begin{prop}
 $\{f_{t}\} $ has a $\rho$-uniform radius if and only if it has no
 $\rho$-vanishing fold in
 $U(\epsilon_0)=\{x\in\K^n: \rho(x)<\epsilon_0\} $  for some $\epsilon_0 >0.$
\end{prop}

\begin{rem}
    In the analytic complex case we have the following theorem
    which relies the jump of Milnor number with the existence of
    vanishing folds.

  We obtain a generalisation of \cite{O}:

\begin{thm} $\K=\C.$

Let $F:(\mathbb{C}^n\times \mathbb{C},
\{0\}\times \mathbb{C})\to (\mathbb{C},0)$ a family of germs of
isolated singularities  and  $X_{t}=\{f_t^{-1}(0)\}$ the
corresponding family of hypersurfaces.
Let $\mu_t\equiv \mu $ be the Milnor number of $f_t$ at the origine
and suppose that $\mu_t=\mu$ is constant for $0<t\leq 1$ and $\mu<\mu_0$ .

Then, the family $\{f_t^{-1}(0)\}$ admits a vanishing fold centered at $0$.
\end{thm}

\end{rem}

\begin{rem}
It is not difficult, to see that any family of isolated singularity
of quasihomogeneous  functions can not have a vanishing fold with
$\rho=\sum_{i=1}^n |x_i|^2.$
i.e. has $\rho$-Milnor uniforme radius.

In fact,  if $\{f_t\}$ is quasihomogeneous  family of type
$(w_1,\ldots,w_n; D)$, the "Euler formula" gives:

$$\sum_{i=1}^n w_i x_i \frac{\partial f_t}{\partial x_i}=D. f_t.$$

And now if $f$ has a kink at $p=(x,t)$  then there exists $\lambda\in
\R^*$ such that

$d_xf(p)=\lambda {d_x\rho(p)}$ i.e. $\frac{\partial f_t}{\partial x_i}=\lambda {x_i}$

Since $p\in f^{-1 }(0)$ the Euler formula gives
$$\lambda\sum_{i=1}^n w_i \vert x_i\vert^2 =0$$ which implies
$p\in\{0\}\times \mathbb{R}^p$
then it's not in the smooth part.

We can also show that it has uniform $\rho$-milnor radius, for
$\rho$ a quasihomogeneous control function with respect to the system
of weights,  for example: $\sum_{i=1}^n  \vert x_i\vert ^{2\frac{D}{w_i}}.$

\end{rem}

\begin{rem}
Having  a uniform $\rho$-Milnor radius for some ``control function''
$\rho$ doesn't  for a family of germs is  in general weaker than whitney regular.

For example,  take the Brian\c{c}on-speder family  $F(x,y,z,t)=
z^5+ty^6z+y^7x+x^{15}$, its a family
quasihomogeneous polynomials of type $(1,2,3; 15)$

The Milnor number is  given   for an isolated  singularity
quasihomogeneous  $f$ of type
$(w_1,\ldots,w_n; D)$ by the formula:

$$\mu(f)=\frac{(D-w_1)(D-w_2)\ldots(D-w_n)}{w_1.w_2.\ldots w_n}$$

which gives in our case $\mu(f_t)=364.$

The family $f_t$ is $\mu$-constant but for a generic hyperplan $H$ in
$\mathbb{R}^3$, its equation can be written
 $z=ax+by$ with $a,b\in\R-\{0\}$, and so the restriction family
 $g_t=f_t\vert _{H} $ is the family of polynomials
 $g_t(x,y)=x^5+txy^6+y^7(ax+by)+(ax+by)^{15}$

 Then, for $t\not=0,$  $g_t$ is semiquasihomogeneous with leading
 term $x^5+txy^6$ is of type $(3,2:15).$
 Using the fact that $\mu(g_t)=\mu(x^5+txy^6)$ we obtains
 $$\mu(H\cap Y_t)=26.$$
 But for $t=0$, $g_0(x,y)=x^5+y^7(ax+by)+(ax+by)^{15}=x^5+by^8 + (axy^7+(ax+by)^{15})$
 is semiquasihomogeneous with leading term $x^5+by^8$ is of type $(8,5:40)$
then:  $$\mu(H\cap Y_0)=28.$$

Since the Milnor number jumps, this family has $\rho$-vanishing fold.

In fact the family of curves we obtain by the intersection with a generic hyperplan must have a vanishing folds.

In the complex, Brian\c{c}on and Speder show that this family is not
Whitney regular ( using the fact that whitney regular family must have
constant milnor numbers after intersection by generic hyperplan ).
\end{rem}

\section{Vanishing folds and Whitney condition}

Let $F$ an analytic function from $\R^n\times \R$ to $\R$, in an neighbourhood of $0$

$$\begin{matrix} F: & \R^n\times \R^p,0 &\to& \R,0 \\ & (x,t)&\mapsto & F(x,t)  \end{matrix}$$
$F(0,t)=0$

We denote by $\pi$ the projection on the second factor,
$V=F^{-1}(0)$, $Y=\{0\}\times \R$ and $X_{t}=\{x\in \R^n/ F(x,t)=0\}.$

We suppose  $X_{t}$ has an isolated singularity at $(0,t)$ i.e. the critical set of the restriction of $\pi$ to $V$ is $Y.$

Then $X=V-Y$ is an analytic complex manifold of dimension $n$, and for each point $(x,t)\in X$ we have

$\dis T_{(x,t)}X=\{(u,v)\in \R^n\times\R^p/
\sum_{i=1}^{n}u_{i}\frac{\partial F}{\partial x_{i}}(x,t)
+\sum_{j=1}^{p}u_{i}v_{j}\frac{\partial F}{\partial t}(x,t)=0\}=\left(\R {d} F\right)^{\perp}.$

 where $d F=(\frac{\partial F}{\partial x_{1}},\ldots, \frac{\partial
 F}{\partial x_{n}},\frac{\partial F}{\partial t})$, $d_{x} F=(\frac{\partial F}{\partial x_{1}},\ldots, \frac{\partial F}{\partial x_{n}}).$

Let $\mathcal{G} $  be the  set of analytic applications germs from
$\R^n\times\R^p,0$ to $\R^n\times\R^p,0$  of the following type :

$\Phi(y,\tau)=(\Psi(y,\tau),\lambda(\tau))=(x,t)$,

where $\Psi$ for small $\tau$ is a germ of automorphisms of $(\R^n,0)$
(i.e. $\det \left( \frac{\partial \Psi}{\partial y}\right)\not=0$ and $\Psi(0,\tau)=0$ ).

We suppose given  $F: \R^n\times\R^p, 0\to \R,0$ is an analytic
deformation of $f=f_{0}$ such that
$F(0,t)=\frac{\partial F}{\partial x_1}=\ldots=\frac{\partial F}{\partial x_n}=0$, $X=F^{-1}(0)$,
$X_t=f_t^{-1}(0)$  and $Y=\{0\}\times\R$.

The following theorem says that Whitney regularity is equivalent to
the stability of the uniform  $\rho$-Milnor property with respect to
families of  linear change of variable in $x$.

\begin{thm}\label{mainthm}
Let $F$ be a $\mu$-constant deformation. The following conditions are equivalent
\begin{itemize}
\item[(i)] $F$ is Whitney regular
\item[(ii)] $\forall \Phi\in \mathcal{G}$, $F\circ \Phi$ has no vanishing fold.
\end{itemize}
\end{thm}

\pf
(i) $\Rightarrow$   (ii)

We have seen that, if a deformation is Whitney regular then it's has no vanishing folds.
We have then only to show that if $F$ is Whitney then so is $F\circ\Phi$ for all $\Phi\in\mathcal{G}.$

By definition $F\circ \Phi(y,\tau)=F(\Psi(y,\tau),\lambda(\tau))$; this suggest to do it in the two following steps:

Firstly, for $\lambda=Id_{\mathbb{R}^p}$, $\Phi_1(y,\tau)=(\Psi(y,\tau),\tau)$, is
then an analytic diffeomorphism of $\mathbb{R}^{n+p}$, since Whitney's
conditions are invariant by diffeomorphism (see \cite{Ma}), if
$F$ is Whitney regular, so is $F\circ\Phi_1$, where
$\Phi_1(y,\tau)=(\Psi(y,\tau),\tau).$

Secondly,  if $F$ is Whitney regular, then so is $F\circ\Phi_2$, where $\Phi_2(y,\tau)=(y,\lambda(\tau))$
and $\lambda: \R^p,0\to\R^p,0.$

In fact, the condition $b^\prime$ is trivially satisfied since it does not make use of the partial derivative
relatively to the parameter.

To check, the $(a)$ condition, we compute
$\dis\frac{\partial F\circ \Phi_2}{\partial
t_{j}}(y,\lambda(\tau))=\sum_{m=1}^p\frac{\partial\lambda_{m}}{\partial t_{j}}(\tau)
\frac{\partial F}{\partial \lambda_{m}}(y,\lambda(\tau))$,
since $F$ satisfy $(a)$ condition, we have
$$\lim_{(y,\tau)\to 0\atop (x,y)\in X-Y}\left ( \frac{\frac{\partial
F\circ\Phi_2}{\partial t_{j}}(y,\tau)}
{\| d_{x} F\circ\Phi_2(y,\tau)\|}\right)=\lim_{(y,\tau)\to 0\atop
(x,y)\in X-Y}\sum_{m=1}^p\frac{\partial\lambda_{m}}{\partial
t_{j}}(\tau)
\left ( \frac{\frac{\partial F\circ\Phi_2}{\partial t_{m}}(y,\tau)}
{\| d_{x} F\circ\Phi_2(y,\tau)\|}\right)=0.$$

Now $(1)$ and $(2)$ implies that for any $\Phi\in \mathcal{G}$,
$F\circ\Phi$ is Whitney regular, then it has no vanishing folds.

(ii) $\Rightarrow$ (i)

Firstly,  since $F$ is a $\mu$-constant deformation in a neighborhood of $0$,
 its satisfy the  $(a)$ regularity condition (  in fact we have more,
 $\mu$-constant implie ``good stratification'' in the sens of Thom, see \cite{LS},
 \cite{BS}, \cite{T}).

 Let us suppose that $b$ fails, which in turn implies  $b'$ fails,
  since $a$ holds.

Let $\Delta(z,\tau)=\frac{\sum_{i=1}^{n}x_{i}\frac{\partial F}{\partial x_{i}}(z,\tau)}{\|x\|\| grad_{x} F(z,\tau)\|}$
where $(z,\tau)\in X-Y.$

Then there exists a real analytic curve
 $\gamma:[0,\epsilon] \to X$, $\gamma(s)=(x(s),t(s))$   and $\delta_0>0$ such that:
\begin{itemize}
\item[1)] $\gamma(0)=(0,0)$
\item[2)] $f(z(s),\tau(s))=0$ for all $0\leq s\leq \epsilon,$ and
\item[3)] $\lim_{s\to 0}  \Delta(z(s),\tau(s))=l\not=0$
\end{itemize}

Let us dnote par $v$ the valuation of $\mathcal{O}_{\R,0}$ associated to $\gamma.$

We will use the following notations:\\
$\displaystyle v(z)= \inf_{1\leq i\leq n} z_i$  for $z\in \R^n,$
$\displaystyle v(\frac{\partial f}{\partial x})= \inf_{1\leq i\leq
n}v(\frac{\partial f}{\partial x_i}).$\\

In this conditions, if we denote $v(z)=p$ and $v(\frac{\partial f}{\partial x})=q$, we can suppose
(change the order of variables if needed) that $v(z_1)=p.$

Since, $\Delta\circ \gamma(s)$ has a non zero limit when $s$ tends to $0$, we may conclude that
$v(<z,\overline{\frac{\partial f}{\partial x}}(z,\tau)>=v(z)+v(\frac{\partial f}{\partial x}(z,\tau))=p+q$

Let us now  denotes  $\gamma(s)=(p_1(s),\ldots,p_n(s), \lambda(s)),$

$\overline{\frac{\partial f}{\partial x}}\circ\gamma(s)=(q_1(s),\ldots,q_n(s) $
and
$v(<z,\overline{\frac{\partial f}{\partial
x}}>\circ\gamma(s))=u(s).$\\
Let us now define\\
$\Phi: \R^n\times \R, 0\to \R^n\times \R, 0 $ by: $\Phi(y_1,\ldots,y_n, \tau)=(\Psi(y, \tau), \lambda(\tau))$

with $ \Psi(y, \tau)=(y_1-\frac{ p_1}{u}h,y_2+\frac{ p_2}{p_1}y_1-\frac{ p_2}{u}h,\ldots,
y_n+\frac{ p_n}{p_1}y_1-\frac{ p_n}{u}h) $

where $h=q_2y_2+q_3y_3+\ldots+q_ny_n.$

We may first check that, $\Phi\in \mathcal{G}:$

1) $\Phi$ is analytic.

We use for this the valuation along $\gamma.$

If $j\not = 1,$ then for $y_j+\frac{ p_j}{p_1}y_1-\frac{ p_j}{u}h$ , we have by hypothesis
$v(\frac{ p_j}{p_1}y_1)\geq v(p_1)+v(y_1)-v(p_1)\geq0$ and
$v(\frac{ p_j}{u}h)= v(p_j)+v(h)-v(u)\geq (p+q)-(p+q)=0.$

If $j=1 $ for $y_1-\frac{ p_1}{u}h$, we have
$v(\frac{ p_1}{u}h)=v(p_1)+v(h)-v(u)=(p+q)-(p+q)=0.$

2) $\Psi(0, \tau)=0$

3) The jacobian of $\Psi$ is invertible in a neighborhood of $0.$

For this we compute the determinant of this Jacobian and show it equals $1.$

Let $\Phi_1(y,\tau)=y_1-\frac{ p_1}{u}h$ and
$\Phi_j(y,\tau)=y_j+\frac{ p_j}{p_1}y_1-\frac{ p_j}{u}h$ for $2\leq j\leq n.$

Then $\frac{\partial \Phi_1}{\partial y_1}=1$ and $\forall j\geq 2,$

$\frac{\partial \Phi_1}{\partial y_j}=
-\frac{ p_1}{u}q_j.$

 $\forall i,j\geq 2, $ $i\not= j,$   $\frac{\partial \Phi_j}{\partial y_i}=
-\frac{ p_j}{u}q_i$

 $\forall i\geq 2, $   $\frac{\partial \Phi_i}{\partial y_i}=1-\frac{ p_i}{u}q_i$
 and   $\frac{\partial \Phi_i}{\partial y_1}=\frac{ p_i}{p_1}.$

\begin{equation}
 \det\left(\frac{\partial \Psi}{\partial y}(y, \tau)\right)=\left\vert\begin{array}{cccccc}
1 & -\frac{ p_1}{u}q_2 & \ldots & -\frac{ p_1}{u}q_i  & \ldots & -\frac{ p_1}{u}q_n\\
\frac{ p_2}{p_1} & 1-\frac{ p_2}{u}q_2 & & \vdots &  &\vdots\\
 \vdots& & \ddots & &  &\vdots \\
 \vdots & & & 1-\frac{ p_i}{u}q_i &  & \vdots\\
 \vdots & & &  &  &  \vdots\\
 \vdots & &  & & \ddots &  \vdots\\
\frac{ p_n}{p_1} & -\frac{ p_n}{u}q_2& \ldots & -\frac{ p_n}{u}q_i &\ldots &1-\frac{ p_n}{u}q_n
 \end{array}\right\vert
 \end{equation}

If we denotes by $C_j$ the jth row, and applied the transformation $C_j=C_j+\frac{ p_iq_i}{u}C_1$ for $j=1\cdots
m$

we see that
\begin{equation}
 \det\left(\frac{\partial \Psi}{\partial y}(y, \tau)\right)=\left\vert\begin{array}{cccccc}
1 & 0 & \ldots & \ldots & \ldots & 0\\
\frac{ p_2}{p_1} & 1 & & \vdots &  &\vdots\\
 \vdots&0 & \ddots &0&  &\vdots \\
 \vdots & & & 1&  & \vdots\\
 \vdots & & &  &  &  \vdots\\
 \vdots & &  & & \ddots &  0\\
\frac{ p_n}{p_1} &0& \ldots & \ldots &0 &1
 \end{array}\right \vert =1
 \end{equation}

We can now conclude that $\Phi\in \mathcal{G}.$

Moreover, by construction we have
$\Phi(p_1(\tau),0\ldots,0,\tau)=\gamma(\tau)$

The computation gives us

$\frac{\partial F\circ\Phi}{\partial y_{i}}(p_1(\tau),0\ldots,0,\tau)=
\sum_{j=1}^{n}\frac{\partial F}{\partial x_{i}}(\gamma(\tau))
\displaystyle \frac{\partial\Phi_j}{\partial y_{i}}(p_1(\tau),0\ldots,0,\tau)$

$\displaystyle =\sum_{j=1}^{n}q_j(\tau)\frac{\partial\Phi_j}{\partial y_{i}}(p_1(\tau),0\ldots,0,\tau)$

$\displaystyle = -\frac{ q_1(\tau)q_i(\tau) p_1(\tau)}{u}+ q_i(\tau)\left(1- \frac{ p_i(\tau)q_i(\tau)}{u}\right)
+\sum_{j\not=1\atop j\not= i}-\frac{ q_j(\tau)q_i(\tau) p_j(\tau)}{u} $

$\displaystyle =-\frac{ q_i(\tau)}{u}\left(p_1(\tau)q_1(\tau) +
p_i(\tau)q_i(\tau)+\sum_{j\not=1\atop j\not= i}p_j(\tau)q_j(\tau)\right)
+q_i(\tau)$

$=-\frac{ q_i(\tau)}{u}.u +q_i(\tau)=0.$

If $i=1,$  $\displaystyle I=q_1(\tau)+\sum_{j=2}^{n}q_j(\tau)\frac{ p_i(\tau)}{ p_1(\tau)} =\frac{u}{ p_1}$

Then, we obtain that
$\frac{\partial F\circ\Phi}{\partial
y_{i}}(p_1(\tau),0\ldots,0,\tau)=\lambda({p_1(\tau)},0,\ldots,0)$
with $\lambda=\frac{u}{\vert p_1\vert^2}$
this means that $ F\circ\Phi$ has a vanishing fold.

\begin{rem}
In this proof we can replace $\mathcal{G}$ by the set

$\mathcal{G}_l=\{\Phi=(\Psi,\lambda):
\R^n\times\R,0 \to \R^n\times\R,0\; \text{such that} \; \Psi(.,\tau)\in Gl(\R^n)\} $
\end{rem}

A consequence of  this theorem is, an example of $\mu$-constant deformation with a vanishing fold gives an example of
non Whitney regular $\mu$-constant deformation.

\subsection{Example}

\underline { The Brian\c{c}on and Speder  example has vanishing
folds  } (see \cite{BS} \cite{Tr})

From the theorem,  Whitney faults is detected by vanishing folds.
So to find a vanishing fold, it suffises to find an arc along which the Whitney regularity fails.

Let  $F(x,y,z,t)= z^5+ty^6z+y^7x+x^{15}$, then $F$ is quasihomogenous $\mu-$constant family of type $(3,2,1;15).$
Then as we saw,  the canonical stratification is $(a)$ regular.

$\left\{\begin{array}{lll} \frac{\partial F}{\partial x}=y^7+15x^{14} \\
 \frac{\partial F}{\partial y}=6ty^5+7y^{6} \\  \frac{\partial F}{\partial z}=5z^4+ty^6 \end{array}\right.$

We shall construct an explicit analytic path
$\gamma(s)=(x(s),y(s),z(s),t(s))$ contained in $V$ along which the $b'$ condition
fails,   that is   $\Delta(x,y,z,t)=\left
(\frac{\sum_{i=1}^{n}x_{i}\frac{\partial F}{\partial
x_{i}}(x,y,z,t)}{\|x\|\| grad_{x} F(x,y,z,t)\|}\right)$ do not tends
to $0$ when $(x,y,z,t)$  tends to $0$ along $\gamma(s).$

Let's take it of the following form
$$\left\{\begin{array}{lll} x(s)=\lambda s^5 \\
y(s)=\alpha s^5\\z(s)=s^8\\t(s)=-\frac{5}{\alpha^6}s^2 \end{array}\right.$$
with  $\alpha\not =0.$\\
We must have $ F(\gamma(s))=(1-\frac{5}{\alpha^6}\alpha^6+\lambda \alpha^7+
 \lambda^{15}s^{35})s^{40}\equiv 0 $ i.e.\\
$G(\lambda,s)=-4+\lambda \alpha^7+ \lambda^{15}s^{35}$

Since $\frac{\partial G}{\partial \lambda}(\lambda,0)=\alpha^7\not=0$,  by the implicit function theorem
$\lambda $ is a smooth function of $s.$

Then we have along $\gamma(s)$  nears $s=0:$

$\left\{\begin{array}{lll} \frac{\partial F}{\partial x}=y^7+15x^{14}=\alpha^7s^{35} + 15\lambda^{15}s^{70}
\sim  \alpha^7s^{35}\\\\

\frac{\partial F}{\partial y}=6ty^5+7y^{6}
=\left(\frac{-30}{\alpha}+7\alpha^6\lambda\right)s^{35}\\\\

 \frac{\partial F}{\partial z}=5z^4+ty^6=5s^{32}- \frac{5}{\alpha^6}\alpha^6s^{32}\equiv 0\end{array}\right.$

The limit of orthogonal secant vectors $\frac{(x,y,z)}{\|(x,y,z)\|}$ is  $(1:\alpha:0)$
and the limit of normal vectors $\left ( \frac{ grad_{x} F(x,y,z,t)}{\| grad_{x} F(x,y,z,t)\|}\right)$ is
$(\alpha^7:\frac{-30}{\alpha}+ 7\alpha^6:0).$

Then $\Delta(\gamma(s))$ tends to $0$ if and only if
$8\alpha^7-30=0$.  We can clearly choose $\alpha\not=0$ such that
$8\alpha^7-30\not=0$, this means that Whitney
condition fails along this curve.

Now the construction of the family analytique automorphisms in the
proof gives \`a ``control function'' $\rho$ such that the family
$\{f_{t}(x,y,z)=z^5+ty^6z+y^7x+x^{15}\}$, has a $\rho$-vanishing fold
i.e. its $\rho$-Milnor radius is not uniform.

The control function is obtained by composing the standard disytnace
function $|x|^2+|y|^2+|z|^2$ by the family of automorphisms obtained
this way.

It may be interesting to establish an analogue of theorem
\ref{mainthm} for the condition $C$
 (see \cite{B}) and the relation with  uniform radii.

\section{tame mappings}
In this section we establish a version of Theorem \ref{mainthm} for
family of germs of tame mappings (i.e. in the $o$-minimal category ).
We will assume the reader familiar with the basic facts about $o$-minimal structure. The standard references are L. Van den Driess \cite{D}, L. Van den Driess and C. Miller \cite{DM}
and M. Coste \cite{C}.

Let us first recall the definition of an $o$-minimal structure extending the field $(\R,+,.)$.

\begin{defn} Let $\cS=\cup_{n\in\N}\cS_{n}$, where for each $n\in \N$, $\cS_{n}$ is a family of subsets of $\R^{n}$.

We say that the collection $\cS$ is an \textit{$o$-minimal structure} on  $(\R,+,.)$ if:
\begin{itemize}
\item[1)] each $\cS_{n}$ is a boolean algebra
\item[2)] if $A\in\cS_{n}$ and $B\in\cS_{m}$, then $A\times B\in \cS_{n+m}$
\item[3)] let $A\in\cS_{n+m}$ and $\pi: \R^{n+m}\ra\R^n$ be the projection on the first $n$ coordinates, then $\pi(A)\in\cS_{n}$.
\item[4)] all algebraic subsets of $\R^{n}$ are in $\cS_{n}$
\item[5)] the elements of $\cS_{1}$ are the finite unions of points and
intervals.
\end{itemize}

A subset $A$ of $\R^{n}$ which belongs to  $\cS_{n}$ is called
a \textit{definable} set in $\mathcal S$.
A map $f:A\ra\R^m$ is \textit{definable} in $\mathcal S$
if its graph is a definable subset of $\R^n \times\R^m$ in $\mathcal S$,
if in addition, it is $C^{k}$ for some $k\in \N$, we call it
a $C^{k}$ definable map in $\mathcal S$.

We call a tame map map definable in some $o$-minimal structure.
\end{defn}

Let $\cS$ be an $o$-minimal structure on $(\R,+,.)$.
We recall  from \cite{DM} the following notation:

\vspace{2mm}

\noindent {\bf Notation:}
Let $p$ be a natural number.
Let $\Phi^p_{\cS}$ denote the set of all odd,
strictly increasing bijections $\phi:\R\ra\R$ $C^p$ definable in $\cS$
and $p$-flat at $0$ (that is $\phi^{(l)}(0)=0$ for $l=0,\ldots, p$).
\vspace{2mm}

\noindent
We quote also from this paper the following lemma (Lemma C.7. page 523):
\begin{lem}\label{Lem0}
Let $f:A\times \R^*\raa \R$ be a definable function in $\cS$,
$A\subset \R^{n}$.
Then, for any $p\in\mathbb{N}$,
there exists $\phi\in \Phi^p_{\cS}$ such that
$\lim_{t\to 0}\phi(t)f(x,t)=0$ for each $x\in A$.
\end{lem}

A $C^k$ version of this lemma is given in the following:

\begin{lem}\label{Lem1}
Let $f:U\times \R^*\raa \R$ be a  $C^k$ definable
function in $\cS$, and let $U$ be an open subset of $ \R^{n}$.
Then, for any $p\in\mathbb{N}$,
there exists $\phi\in \Phi^p_{\cS}$ such that the function
$$
g(x,t)=\begin{cases}
\phi(t)f(x,t)&\, \text{if $t\not=0$},\\
0&\, \text{if $t=0$}.
\end{cases}
$$
is a $C^k$ definable function.
\end{lem}
\begin{proof}
Let $h:U\times \R^*\ra\R$ denote any function from the collection of partial
derivatives:
$$\{D_{(x,t)}^{\alpha}f\ ;\ \,
\alpha=(\alpha_{1},\ldots,\alpha_{n},\alpha_{n+1})\in \N^{n+1}\,\text{and}\, \vert \alpha\vert\le k\}$$

By Lemma \ref{Lem0}, there exists  $\theta\in \Phi^p_{\cS}$ such that  $\lim_{t\to 0}\theta(t)h(x,t)=0$ for each $x\in U$.
Then $\phi:= \theta^{2k+1}$ satisfies the needs.
\end{proof}

\begin{defn} A structure $\cS$ on the field $(\R,+,.)$ is \textit{polynomially
bounded} if for any function $f:\R\raa\R$ definable in $\cS$, there exists
$N\in \N$ such that
$$\vert f(t)\vert\le t^N$$
for all sufficiently large $t$.
\end{defn}
\begin{lem}\label{Lem2}
If $\cS$ is polynomially bounded, then for any $\phi\in \Phi^p_{\cS}$,
there exists $d\in \N$ and real number $\epsilon>0$ such that:
$$\vert \phi(t)\vert \geq t^d$$
for any $t\in (-\epsilon, \epsilon)$.
\end{lem}
\begin{proof}
We take $\displaystyle\theta(s):=\frac{1}{\phi(\frac{1}{s})}$
where $s=\frac{1}{t}$.
Since $\theta$ is definable in a \textit{polynomially bounded} structure,
there exists $d\in \N$ and $M\in \N$ such that $\vert \theta(s)\vert\leq s^d$
for $\vert s\vert > M$.
Therefore $\vert \phi(t)\vert\geq t^d$ for any $t\in (-\epsilon, \epsilon)$ with $\epsilon=\frac{1}{M}$.
\end{proof}
\begin{defn}
 A tame map is a map definable in some polynomially bounded $o$-minimal structure.
\end{defn}

Using the lemmas above, we can show the following theorem which is a
tame version of the main result.

\begin{thm}\label{Th3}

    Given a  tame family of isolated singularities germs \\
$F:(\mathbb{R}^p\times \mathbb{R}^n,
\mathbb{R}^p\times\{0\})\to (\mathbb{R}^k,0).$ We suppose the family
$F$ is a $\mu$-constant deformation of $f_{0}=f$. Then the  following conditions are equivalent
\begin{itemize}
\item[(i)] $F$ is whitney regular.
\item[(ii)] $\forall \Phi\in \mathcal{G}$, $F\circ \Phi$ has no vanishing fold.
\end{itemize}
\end{thm}

config: arxiv


TITLE: $C^0$-approximation by smooth curves with prescribed curvature
QUESTION [3 upvotes]: A theorem by Ghomi (https://link.springer.com/article/10.1007/s10711-007-9151-y) says that any $C^2$ curve $\gamma:[a,b]\to\mathbb R^{n≥3}$ can be $C^1$-approximated by curves with prescribed constant curvature as long as the prescribed curvature is larger than the maximum of the curvature of $\gamma$. I wonder if one could relax the assumption on the prescribed curvature to be constant and weaken the $C^1$-approximability by $C^0$-approximability. More precisely I wonder if the following statement holds true:
Let $\gamma:[a,b]\to\mathbb R^{n≥3}$ be a curve of class $C^2$ with curvature $k$ and let $k':[a,b]\to\mathbb R$ be a smooth function that satisfies $k'>k$ point wise. There is a smooth curve $\delta:[a,b]\to\mathbb R^{n≥3}$ with curvature $k'$ that is $C^0$-close to $\gamma$.
I'd be thankful for any reference.
EDIT: I corrected the question, thanks.

REPLY [3 votes]: This has been established in a recent paper by Micha Wasem:
h-principle for curves with prescribed curvature.
Geom. Dedicata 184 (2016), 135–142.
In particular Wasem showed that one can prescribe curvature as you suggest, and it is not even necessary to give up $C^1$ closeness.


TITLE: Foundations of Differential Calculus
QUESTION [1 upvotes]: In the preface of Foundations of Differential Calculus there's a section that says:

Thus, if the quantity $x$ is given an increment $\omega$, so that
  it becomes $x + \omega$, its square $x^2$ becomes $x^2 + 2x\omega + \omega^2$, and it takes the
  increment $2x\omega + \omega^2$. Hence, the increment of $x$ itself, which is $\omega$, has the
  ratio to the increment of the square, which is $2x\omega+\omega^2$, as $1$ to $2x+\omega$. This
  ratio reduces to $1$ to $2x$, at least when $\omega$ vanishes.

$f(x) = x^2$
$f(x+w) = (x+\omega)^2 = x^2 + 2x\omega + \omega^2$
$(x+\omega)^2 - x^2 = 2x\omega+\omega^2$
And then the part I don't understand where $\omega = 2x\omega+\omega^2$ goes to "as $1$ to $2x+\omega$".

REPLY [0 votes]: The text shaded in khaki has numerator and denominator interchanged in its last part. Instead it should read as follows:
Thus, if the quantity $x$ is given an increment $ω$, so that
it becomes $x + ω$. In this way the square $x^2$ becomes $x^2 + 2xω + ω^2$, so that the square takes the increment $2xω + ω^2$. Hence, the ratio of the increment of the square, which is $2xω+ω^2$, to the increment of $x$ itself, which is $ω$, is $2x+ω$. This ratio reduces  to $2x$ when $ω$ is "negligible", or "goes to $0$".