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数学の独語表現その2

収束の定義を独語,英語および日本語で与えた.
ドイツ語
Definition. $\, \, $Sei ($x_n$) eine Folge reeller Zahlen, Die Folge heißt konvergent gegen $a \in \mathbb{R}$, falls gilt:
Zu jedem $\varepsilon >0$ existiert ein $N \in \mathbb{N}$, so dass
$ |x_n-a| < \varepsilon$ $\, \, $ für alle $ n \geqq N $

Konvergiert ($x_n$) gegen $a$, so nennt man $a$ den Grenzwert oder den Limies der Folge und schreibt
$ \displaystyle{\lim_{n \to \infty}a_n}=a$ $\, \, $ oder kurt $\, \, $ $\lim a_n=a$.
Auch die Schreibweise
$a_n \longrightarrow a$ $\, \, $ für $n \longrightarrow \infty$
ist gebräuchlich.

英語
Definition. $\, \, $Let ($x_n$) be a sequence of real numbers and $a \in \mathbb{R}$. We shall say that the sequence converges to $a$
if given $\varepsilon > 0$ there exists an integer $N$ such that
$| x_n-a| < \varepsilon$ $\, \, $ for all $n \geqq N$

In this case we say that $a$ is the limit of ($x_n$), and we write
$ \displaystyle{\lim_{n \to \infty}a_n}=a$ $\, \, $ or $\, \, $$\, \, $ $\lim a_n=a$.


日本語
定義.($x_n$) を実数からなる数列とする.数列($x_n$)が $a \in \mathbb{R}$ に収束するとは,
任意に与えられた$\varepsilon > 0$ に対して,ある自然数 $N$ が存在して
$| x_n-a| < \varepsilon \, \, $ がすべての $n \geqq N$
で成り立つときをいう.

収束する値を極限値といい,$ \displaystyle{\lim_{n \to \infty}a_n}=a$ $\, \, $または$\, \, $ $\lim a_n=a$で表す.

コメント
今回は数列の極限を例にとり,定義を独語,英語,日本語で書いてみた.数学よりも英語や独語のほうがが難しい.数学的な表現はどの言語でも似ている.

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2014/08/12   解析学     293TB 0   293Com 0  

ラング 続解析入門P84から

Exercise
3.Find the parametric equation of the tangent line to the curve of intersection of the following surfaces at the indicated point.
(a) $\quad \begin{equation}x^2+y^2+z^2=49, \quad \text{and} \quad x^2+y^2=13 \quad \, \, \text{at the point } \, \, \, P= (3,\, 2,\, -6)\end{equation}$
(b) $\quad \begin{equation}xy+z=0, \quad \text{and} \quad x^2+y^2+z^2=9 \quad \, \, \text{at the point} \, \, \, P= (2,\, 1,\, -2)\end{equation}$

Solution (a).
Let \[ f(x,\, y,\, z)=x^2+y^2+z^2, \quad \text{and} \quad g(x,\, y,\, z)=x^2+y^2. \] Then the first surface is defind by the equation $f(x,\, y,\, z)=49.$ A vector $N_1$ perpendicular to the first surface at $P$ is given by \[ N_1= \text{grad}\, \, f(P), \quad \text{where} \quad \text{grad} \, \, f(x,\, y,\, z)=(2x,\, 2y,\, 2z). \] Thus for $P=(3,\, 2,\, -6)$ we find \[ N_1=(6,\, 4,\, -12). \]
Similarly
\[ N_2= \text{grad}\, \, g(P), \quad \text{where} \quad \text{grad} \, \, g(x,\, y,\, z)=(2x,\, 2y,\, 0). \] Thus for $P=(3,\, 2,\, -6)$ we find \[ N_2=(6,\, 4,\, 0) \]
A vector $A=(a,\, b,\, c)$ in the direction of the line of intersection is perpendicular to both $N_1$, and $N_2$. To find $A$, we therefore have to solve the equations
\[ \begin{equation} A \cdot N_1=0 \quad \quad A \cdot N_2 =0 \end{equation}. \] This amounts to
\[ \begin{eqnarray*} \, 6a+4b -12c&=& 0\\ 6a+4b \quad \quad \, \, \, \, \, &=& 0 \end{eqnarray*} \] Let for instance $a=2$. Solving for $b$ and $c$ yields
\[ a=2, \quad b=-3, \quad c=0. \] Thus $A=(2,\, -3,\, 0).$ Finally, the parametric representation of the desired line is \[ P+tA=(3,\, 2,\, -6)+t(2,\, -3,\, 0). \quad \quad \square\]

Solution (b).
Similarly
\[ P+tA=(2,\, 1,\, -2)+t(-5,\, 4,\, -3). \]

2014/06/15   解析学     278TB 0   278Com 0  

偏微分へ

数学というより MathJax で数式を書く練習になりました.また,ドイツ語の数学書の雰囲気も味わってください.

Partielle Ableitungen
Definition(Partielle Ableitung). Sei $U \subset \mathbb{R}^n$ eine offene Teilmenge und $f:U \rightarrow \mathbb{R}$ eine reelle Funktion. $f$ heißt im Punkt $x \in U$ partiell differenzierbar in der $i$-ten Koordinatenrichtung, falls der Limes
$D_if(x):= \displaystyle{\lim_{h \to 0} \frac{f(x+he_i)-f(x)}{h}}$
existiert. Dabei ist $e_i \in \mathbb{R}^n$ der $i$-te Einheitsvektor,
$e_i= (0,0,\dots,0,1,0,\dots,0)$
und für den Limes $h \to 0$ hat man sich auf solche $ h \in \mathbb{R}$ zu beschränken, für die $h \ne 0$ und $x+he_i \in U$.
$D_if(x)$ heißt $i$-te partielle Ableitung von $f$ in $x$.

Definition. Sei $U \subset \mathbb{R}^n$ offen. Eine Funktion $f:U \rightarrow \mathbb{R}$ heißt partiell differenzierbar, falls $D_if(x)$ für alle $x \in U$ und $i=1,\dots,n$ existiert. $f$ heißt stetig partiell differnzierbar, falls zusätzlich alle partiellen Ableitung $D_if:U \rightarrow \mathbb{R}$ stetig sind.
Schreibweise. Statt $D_if$ schreibt man auch $ \displaystyle{\frac{\partial f }{\partial{x_i}}}$. Entsprechend auch
$D_if(x)=\displaystyle{\frac{\partial f }{\partial{x_i}}(x)}=\displaystyle{\frac{\partial{f(x)}}{\partial{x_i}}}$


Definition (Gradient). Sei $U \subset \mathbb{R}^n$ offen und $f : U \rightarrow \mathbb{R}$ eine partiell differnzierbare Funktion. Dann heißt der Vektor
$\text{grad}\, \, f(x) = \left( \displaystyle{\frac{\partial{f}}{\partial{x_1}}(x),\dots, \frac{\partial{f}}{\partial{x_n}}(x) } \right)$
der Gradient von $f$ im Punkt $x \in U$.

Literaturverzeichnis
[1]Otto Forster : Analysis 1. Springer Spektram, 11. Aufl. 2013
[2]Otto Forster : Analysis 2. Springer Spektram, 10. Aufl. 2013

2014/05/21   解析学     273TB 0   273Com 0  

ラング 『 続解析入門 』合成微分律と勾配ベクトル§4問題12から

Sarge Lang " Calculus of several variables"
Exercise 12
Let $f(x,y)=g(r)$, where $r=\sqrt{x^2+y^2}$.Show that
$\displaystyle{\frac{\partial^2f}{\partial x^2}} + \displaystyle{\frac{\partial^2f}{\partial y^2}}$=$\displaystyle{\frac{d^2g}{dr^2}}+\displaystyle{\frac{1}{r} \frac{dg}{dr}}$.

Solution.
First
$\displaystyle{\frac{\partial f}{\partial x }} =g '(r) \displaystyle{\frac{x}{r}}$.Similarly, $\displaystyle{\frac{\partial f}{\partial y }} =g '(r) \displaystyle{\frac{y}{r}}$.

Using the rule for the derivative of a product, and a quotient, we get:
$\displaystyle{\frac{\partial^2f}{\partial x^2}}$=$(g '(r)) ' \displaystyle{\frac{x}{r}+g '(r)( \frac{x}{r}}) '$
=$g ''(r) \displaystyle{\frac{x}{r} \frac{x}{r}}+ g '(r)\displaystyle{\frac{r-x \displaystyle{\frac{x}{r}}}{r^2}}$.

Replace $x$ by $y$ to get $\displaystyle{\frac{\partial^2f}{\partial y^2}}$.
$\displaystyle{\frac{\partial^2f}{\partial y^2}}$=$(g '(r)) ' \displaystyle{\frac{y}{r}+g '(r)( \frac{y}{r}}) '$
=$g ''(r) \displaystyle{\frac{y}{r} \frac{y}{r}}+ g '(r)\displaystyle{\frac{r-y \displaystyle{\frac{y}{r}}}{r^2}}$.

It follows:
$\displaystyle{\frac{\partial^2f}{\partial x^2}}+\displaystyle{\frac{\partial^2f}{\partial y^2}}$
=$g ''(r) \displaystyle{\frac{x}{r} \frac{x}{r}}+ g '(r)\displaystyle{\frac{r-x \displaystyle{\frac{x}{r}}}{r^2}}+g ''(r) \displaystyle{\frac{y}{r} \frac{y}{r}}+ g '(r)\displaystyle{\frac{r-y \displaystyle{\frac{y}{r}}}{r^2}}$
=$g ''(r)(\displaystyle{\frac{y^2}{r^2}} + \displaystyle{\frac{y^2}{r^2}}) + g '(r)(\displaystyle{\frac{r-x \displaystyle{\frac{x}{r}}}{r^2}} +\displaystyle{\frac{r-y \displaystyle{\frac{y}{r}}}{r^2}}) $
=$g ''(r) +g '(r)( \displaystyle{ \frac{r-\displaystyle{\frac{x^2}{r}} +r- \displaystyle{\frac{y^2}{r}}} {r^2}}) $
=$g ''(r) +g '(r) (\displaystyle{ \frac{2r^2-x^2-y^2}{r^3}} )$
=$g ''(r) + g '(r) ( \displaystyle{ \frac{x^2+y^2}{r^3}}) $
=$g '' (r)+ \displaystyle{ \frac{g ' (r)}{r}} $
=$\displaystyle{\frac{d^2g}{dr^2}}+\displaystyle{\frac{1}{r} \frac{dg}{dr}}$. $\quad \quad$ $\square$

Reference
Serge Lang ${\it " CALCULUS \; \, OF \; \, SEVERAL \; \, VARIABLES "}$

2014/04/26   解析学     263TB 0   263Com 0  

ラング 続解析入門から 練習問題 p87

『ラング続解析入門』から練習問題を解きます.("Calculus of several variables" second edition. P88, 89)
Exercises
1. Let $f(x,y,z)=z-e^x \sin y $, and $P=( \log 3, 3 \pi/2, -3)$. Find:
(a) the directional derivative of $f$ at $P$ in the direction of $(1,2,2),$
(b) the maximum and minimum value for the directional derivative of $f$ at $P$.

Solutions.
(a) Let $B=(1,2,2)$. We note that B is not a unit vector. It norm is 3. Let A=$\displaystyle{\frac{1}{3}}B$. Then A is a unit vector having the same direction as B.
Since $f(x,y,z)=z-e^x \sin y$.
$\displaystyle{\frac{\partial f}{\partial x}}=-e^x \sin y$,
$\displaystyle{\frac{\partial f}{\partial y}}=-e^x \cos y$,
$\displaystyle{\frac{\partial f}{\partial z}}=1$
$\text{grad} f = (-e^x \sin y, -e^x \cos y, 1)$
Therefore
$\text{grad} f (P) = (-e^{\log 3} \sin \frac{3 \pi}{2}, -e^{\log 3} \cos \frac{3 \pi}{2}, 1)$
    $= ( 3,0,1)$
The directional derivative is equal to $\text{grad} f (P) \cdot A = \displaystyle{ \frac{1}{3}}( 3 \times 1+0 \times 2+1\times 2)=\displaystyle{\frac{5}{3}}$ $\quad \square$

(b) As $\left\|\text{grad} f(P)\right\|= \sqrt{3^2+0^2+1^2}=\sqrt{10}$
The maximum value is $\sqrt{10}$.
The minimum value is $-\sqrt{10}$.$\quad \square$

8. Suppose the temperature in $( x, y, z )$-space is given by
$f( x, y, z )=x^2y+yz-e^{xy}$.
Compute the rate of change of temperature at the point $( 1, 1, 1 )$ in the direction pointing toward the origin.

Solution.
The direction toward the origin at the point $( 1, 1, 1 )$ is $( -1, -1, -1 )$. Its norm is $\sqrt{3}$.
Let A=$\displaystyle{\frac{1}{\sqrt{3}}}( -1, -1, -1 )$. Then A is a unit vector having to the direction toward
origin.
Since $f( x, y, z )=x^2y+yz-e^{xy}$, it follows that
$\displaystyle{\frac{\partial f}{\partial x}}=2yx-ye^{xy}$,
$\displaystyle{\frac{\partial f}{\partial y}}=x^2+z-xe^{xy}$,
$\displaystyle{\frac{\partial f}{\partial z}}=y$.
Therefore
$\text{grad} f = (2yx-ye^{xy}, x^2+z-xe^{xy}, y)$.
By substituting $x=1$, $\, y=1$, $z=1$
$\text{grad} f(P)= ( 2-e, 2-e, 1)$.
The rate of change of temperature is equal to
$\text{grad} f (P) \cdot A $
$= \displaystyle{ \frac{1}{\sqrt{3}}}( (2-e) \times (-1)+(2-e) \times (-1)+1\times (-1))$
=$\displaystyle{\frac{1}{\sqrt{3}}(2e-5)}$ $\quad \square$.

2014/04/03   解析学     256TB 0   256Com 0  

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