自己紹介など

しばらく翻訳をしつつ数学基礎論についての記事を書きたいと思っています

アイデア置き場

すでにどこかに発表されていたりしたら連絡お願いします

ペアノの公理で後者写像を２変数にする

種本

//どういう順番で情報を並べればいいかわかりません
現代数学小事典(ブルーバックス)　寺坂英孝編　第41版　2015年7月3日発行　講談社　ISBN4-06-117925-X
岩波数学辞典　第3版　1986年3月20日発行　岩波書店　ISBN4-00-080016-7
証明作法ー論理の初歩から証明の実践へー　石原哉　2023年3月31日発行　共同出版株式会社　ISBN978-4-320-11489-0

sandbox

テンプレート:User sandbox
参照:https://www.mediawiki.org/wiki/Extension:Math/Syntax/ja https://mathlandscape.com/latex-logic/

$\sum _{k=1}^{n}k={\frac {1}{2}}n\left(n+1\right)$

$\exists u\forall x(x\notin u)$

$\neg \exists u\forall v(v\in u)$

proof:
そのような $u$ が存在するとする

よって

\forall v(v\in u)

したがって $2^{u}\in u$

全単射の写像

f:u\to u

が存在するとする

記事のもと

極限

例題

例題 1 関数 $\lim _{x\to 2}{4x^{3}}$ の極限を求めよ。

この式そのものに関する定理はないため、この問題を単純化する必要があります。上の恒等定理から、 $\lim _{x\to 2}{x}=2$ であることがわかります。べき法則より、 $\lim _{x\to 2}{x^{3}}=2^{3}=8$ です。最後に、スカラー倍の定理より、 $\lim _{x\to 2}{4x^{3}}=4\cdot 8=32$ となります。

例題 2

関数 $\lim _{x\to 2}4x^{3}+5x+7$ の極限を求めよ。

To do this informally, we split up the expression, once again, into its components. As above, $\lim _{x\to 2}4x^{3}=32$ .

この問題を解くために、式をその項ごとに分けます。例題 1より $\lim _{x\to 2}4x^{3}=32$ です。

また、 $\lim _{x\to 2}5x=5\cdot \lim _{x\to 2}x=5\cdot 2=10$

であり、 $\lim _{x\to 2}7=7$ です。これらを足し合わせて、

\lim _{x\to 2}4x^{3}+5x+7=32+10+7=49

となります。

例題 3

//より単純な例は?

次の関数の極限を求めよ: $\lim _{x\to 2}{\frac {4x^{3}+5x+7}{(x-4)(x+10)}}$

From the previous example the limit of the numerator is $\lim _{x\to 2}4x^{3}+5x+7=49.$ The limit of the denominator is

\lim _{x\to 2}(x-4)(x+10)=\lim _{x\to 2}(x-4)\cdot \lim _{x\to 2}(x+10)=(2-4)\cdot (2+10)=-24.

例題 2より、分子の極限は $\lim _{x\to 2}4x^{3}+5x+7=49$ であり、分母の極限は

\lim _{x\to 2}(x-4)(x+10)=\lim _{x\to 2}(x-4)\cdot \lim _{x\to 2}(x+10)=(2-4)\cdot (2+10)=-24

です。

As the limit of the denominator is not equal to zero we can divide which gives

\lim _{x\to 2}{\frac {4x^{3}+5x+7}{(x-4)(x+10)}}=-{\frac {49}{24}}

分母の極限が0でないので、単純に割って答えを出すことができます。よって、

$\lim _{x\to 2}{\frac {4x^{3}+5x+7}{(x-4)(x+10)}}=-{\frac {49}{24}}$

となります。

例題 4

//必要ないのでは?

次の関数の極限を求めよ: $\lim _{x\to 4}$ $(x^{4}-16x+7) \over (4x-5).$

We apply the same process here as we did in the previous set of examples;

\lim _{x\to 4}

(x^{4}-16x+7) \over (4x-5)

=

\lim _{x\to 4}(x^{4}-16x+7) \over \lim _{x\to 4}(4x-5)

=

(\lim _{x\to 4}(x^{4})-\lim _{x\to 4}(16x)+\lim _{x\to 4}(7)) \over \lim _{x\to 4}(4x)-\lim _{x\to 4}(5)

We can evaluate each of these; $\lim _{x\to 4}(x^{4})=256.$ $\lim _{x\to 4}(16x)=64.$ $\lim _{x\to 4}(7)=7.$ $\lim _{x\to 4}(4x)=16.$ $\lim _{x\to 4}(5)=5.$ $={199 \over 11}$

例題 5

Evaluate the limit $\lim _{x\to 0}$ $1-\cos(x) \over x$ .

To evaluate this seemingly complex limit, be aware of your sine and cosine identities. We will also have to use two new facts. First, if f(x) is a trigonometric function (that is, one of sine, cosine, tangent, cotangent, secant or cosecant) and is defined at a, then $\lim _{x\to a}f(x)=f(a)$ . Second, $\lim _{x\to 0}$ $\sin(x) \over (x)$ $=1$ .

To evaluate the limit, recognize that $1-\cos(x)$ can be multiplied by $1+cos(x)$ to obtain $(1-cos^{2}(x))$ which, by our trig identities, is $sin^{2}(x)$ . So, multiply the top and bottom by $1+cos(x)$ (This is allowed because it is identical to multiplying by one). This is a standard trick for evaluating limits of fractions; multiply the numerator and the denominator by a carefully chosen expression which will make the expression simplify somehow. In this case, we should end up with:

$\lim _{x\to 0}$ $1-\cos(x) \over x$ $=$ $1-\cos(x) \over x$ $\cdot$ $1 \over 1$ $=$ $1-\cos(x) \over x$ $\cdot$ $1+\cos(x) \over 1+\cos(x)$ $=$ $[(1-\cos(x))\cdot 1]+[(1-\cos(x))\cdot \cos(x)] \over x\cdot (1+\cos(x))$ $=$ $1-\cos(x)+\cos(x)-\cos ^{2}(x) \over x\cdot (1+\cos(x))$ $=$ $1-\cos ^{2}(x) \over x\cdot (1+\cos(x))$ $=$ $\sin ^{2}(x) \over x\cdot (1+\cos(x))$ $=$ $\sin(x)\cdot \sin(x) \over x\cdot (1+\cos(x))$ $=$ $\sin(x) \over x$ $\cdot$ $\sin(x) \over (1+\cos(x))$

Your next step shall be to break this up into $\lim _{x\to 0}$ $\sin(x) \over (x)$ $\cdot$ $\lim _{x\to 0}$ $\sin(x) \over (1+\cos(x))$ by the limit rule of multiplication. By the fact mentioned above, $\lim _{x\to 0}$ $\sin(x) \over (x)$ $=1$ .

Next, $\lim _{x\to 0}$ $\sin(x) \over (1+\cos(x))$ $=$ $\lim _{x\to 0}\sin x \over \lim _{x\to 0}(1+\cos x)$ $=$ $0 \over 1+\cos 0$ $=0$ .

Thus, by multiplying these two results, we obtain 0.

We will now present an amazingly useful result, even though we cannot prove this yet. We can find the limit of any polynomial or rational function, as above, as long as that rational function is defined at c (so we are not dividing by zero). More precisely, c must be in the domain of the function.

Limits of Polynomials and Rational functions

If f is a polynomial or rational function that is defined at c then

\lim _{x\rightarrow c}f(x)=f(c)

We already learned this for trigonometric functions, so we see that it is easy to find limits of polynomial, rational or trigonometric functions wherever they are defined. In fact, this is true even for combinations of these functions; thus, for example, $\lim _{x\to 1}(\sin(x^{2})+4\cos ^{3}(3x-1))=\sin 1^{2}+4\cos ^{3}(3(1)-1)$ .

The Squeeze Theorem

The squeeze theorem is very important in calculus proofs, where it typically are used to confirm the limit of a function via comparison with with two other functions whose limits are known.

It is called the Squeeze Theorem because it refers to a function f whose values are squeezed between the values of two other function g and h. g and h both have the same limit L at a point a. As f are trapped between the values of the two functions that approach L, the values of f must also approach L.

A more mathematical definition is:

Suppose that $g(x)\leq f(x)\leq h(x)$ hold for all x in some open interval containing a, except possibly at $x=a$ itself. Suppose also that $\lim _{x\rightarrow a}g(x)=\lim _{x\rightarrow a}h(x)=L$ . Then $\lim _{x\rightarrow a}f(x)=L$ also. Similar statements hold for left and right limit