若問一個人從山腳走到山頂，是否他經過了此山的所有高度？可能有人會說題意模糊難於論斷。也可能有人依據常理常義衡量講當然如此。想那山道蜿蜒是『連續』的，人腳步伐落處卻是『離散』的，固然『連續』包含著『離散』！！然而『離散』能近似『連續』耶？？所謂天衣無縫，實數方才完備乎！！

Completeness of the real numbers

Intuitively, completeness implies that there are not any “gaps” (in Dedekind’s terminology) or “missing points” in the real number line. This contrasts with the rational numbers, whose corresponding number line has a “gap” at each irrational value. In the decimal number system, completeness is equivalent to the statement that any infinite string of decimal digits is actually a decimal representation for some real number.

Depending on the construction of the real numbers used, completeness may take the form of an axiom (the completeness axiom), or may be a theorem proven from the construction. There are many equivalent forms of completeness, the most prominent being Dedekind completeness and Cauchy completeness (completeness as a metric space).

因其完備，對『連續函數』而言，介值定理不得不然嗎！！？？

Intermediate value theorem

In mathematical analysis, the intermediate value theorem states that if a continuous function, f, with an interval, [a, b], as its domain, takes values f(a) and f(b) at each end of the interval, then it also takes any value between f(a) and f(b) at some point within the interval.

This has two important corollaries: 1) If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano’s theorem).^[1] 2) The image of a continuous function over an interval is itself an interval.

Intermediate value theorem: Let f be a defined continuous function on [a, b] and let s be a number with f(a) < s < f(b). Then there exists at least one x with f(x) = s

奈何證明比定理更難了解呢？？！！

Proof

The theorem may be proved as a consequence of the completeness property of the real numbers as follows:^[2]

We shall prove the first case, $f(a)<u<f(b)$ . The second case is similar.

Let $S$ be the set of all $x\in [a,b]$ such that $f(x)<u$ . Then $S$ is non-empty since $a$ is an element of $S$ , and $S$ is bounded above by $b$ . Hence, by completeness, the supremum $c=\sup S$ exists. That is, $c$ is the lowest number that is greater than or equal to every member of $S$ . We claim that $f(c)=u$ .

Fix some $\varepsilon >0$ . Since $f$ is continuous, there is a $\delta >0$ such that ${\Big |}f(x)-f(c){\Big |}<\varepsilon$ whenever $|x-c|<\delta$ . This means that

f(x)-\varepsilon <f(c)<f(x)+\varepsilon

for all $x\in (c-\delta ,c+\delta )$ . By the properties of the supremum, there exist $a^{*}\in (c-\delta ,c]$ that is contained in $S$ , so that for that $a^{*}$

f(c)<f(a^{*})+\varepsilon \leq u+\varepsilon

Choose $a^{**}\in [c,c+\delta )$ that will obviously not be contained in $S$ , so we have

f(c)>f(a^{**})-\varepsilon \geq u-\varepsilon

Both inequalities

u-\varepsilon <f(c)<u+\varepsilon

are valid for all $\varepsilon >0$ , from which we deduce $f(c)=u$ as the only possible value, as stated.

The intermediate value theorem is an easy consequence of the basic properties of connected sets: the preservation of connectedness under continuous functions and the characterization of connected subsets of ℝ as intervals (see below for details and alternate proof). The latter characterization is ultimately a consequence of the least-upper-bound property of the real numbers.

The intermediate value theorem can also be proved using the methods of non-standard analysis, which places “intuitive” arguments involving infinitesimals on a rigorous footing. (See the article: non-standard calculus.)

且列直觀論證為比較︰

Intermediate value theorem

As another illustration of the power of Robinson‘s approach, we present a short proof of the intermediate value theorem (Bolzano’s theorem) using infinitesimals.

Let f be a continuous function on [a,b] such that f(a)<0 while f(b)>0. Then there exists a point c in [a,b] such that f(c)=0.

The proof proceeds as follows. Let N be an infinite hyperinteger. Consider a partition of [a,b] into N intervals of equal length, with partition points x_i as i runs from 0 to N. Consider the collection I of indices such that f(x_i)>0. Let i₀ be the least element in I (such an element exists by the transfer principle, as I is a hyperfinite set). Then the real number

$c={\mathrm {st}}(x_{{i_{0}}})$

is the desired zero of f. Such a proof reduces the quantifier complexity of a standard proof of the IVT.

條條大路通羅馬，讀者自己判斷哩。

註︰

數學王子

歐拉 $\gamma$ 推導

軟體語言中的 INT(x) 函數

$0^0$ 未定式 $\lim \limits_{x \to0^+} 0^x = 0$

0/0 未定式 $\lim \limits_{x \to 0} \frac{x}{x^3} = \infty$

0/0 未定式 $\lim \limits_{x \to 0} \frac{\sin(x)}{x} = 1$

《莊子‧雜篇‧天下 》
……
惠施多方，其書五車，其道舛駁，其言也不中。厤物之意，曰：『至大無外，謂之大一﹔至小無內，謂之小一。無厚，不可積也，其大千里。天與地卑，山與澤平。日方中方睨，物方生方死。大同而與小同異，此之謂【小同異】﹔萬物畢同畢異，此之謂【大同異】。南方無窮而有窮。今日適越而昔來。連環可解也。我知天之中央，燕之北、越之南是也。泛愛萬物，天地一體也。』……

一九九零年發射的『哈伯太空望遠鏡』 HST Hubble Space Telescope 是以美國著名的天文學家『愛德溫‧鮑威爾‧哈伯』 Edwin Powell Hubble 為名，是一架在地球軌道上的望遠鏡。由於它位於地球大氣層之上，因此獲得了地上望遠鏡所沒有的好處：影像不受大氣湍流的影響、視相度極好，更無大氣散射造成的背景光干擾，甚至能觀測會被臭氧層吸收的紫外線。『哈伯太空望遠鏡』彌補了地面觀測的不足，幫助天文學家『理解』和『解答』了許多天文學上的『基本問題』，使得人類對『宇宙緣起』有了更深的『認識』。

『約翰‧卡爾‧弗里德里希‧高斯』 Johann Karl Friedrich Gauß 【Gauss】是德國著名數學家、物理學家、天文學家和大地測量學家，生於布倫瑞克，卒於哥廷根。高斯被認為是歷史上最重要的數學家之一，而且有『數學王子』的美譽。一八零八年，在高斯的數學巨著《算術研究》 Disquisitiones Arithmeticae 首度出現了一個形式符號 $[x]$ 它表示等於或小於實數 $x$ 的『最大整數』，也就是說 $x - 1 < [x] \leq x$ 。今天這個『高斯符號』又稱之為『底函數』 floor function $floor(x) = \lfloor x\rfloor$ ，與另一『頂函數』 ── 是指比實數 $x$ 大的『最小整數』 ── ceiling functions $ceiling(x) = \lceil x \rceil$ 成為一對，經常出現於『數學』和『計算機科學』之中。這個『高斯符號』有什麼重要的嗎？通常一個好的『符號』能使人清晰『表達』複雜和困難的『概念』，而且讓人容易『理解』所說的『內容』，因此是十分重要的啊！！

舉例來說，歐拉研究過『調和級數』 harmonic series $\sum \limits_{k=1}^n \frac{1}{k}$ 和『自然對數』 natural logarithm $\ln(a)=\int_1^a \frac{1}{x}\,dx$ 之間的關係，雖然這兩者都是『發散的』 ── 值為無限大 $\infty$ ── 它們的『差值』卻是一個叫做『歐拉-馬歇羅尼常數』的 $\gamma$ 值。它可以定義如下
$\gamma = \lim \limits_{n \rightarrow \infty } \left( \sum \limits_{k=1}^n \frac{1}{k} - \ln(n) \right)$
$=\int_1^\infty\left({1\over\lfloor x\rfloor}-{1\over x}\right)\,dx$ .
，計算後得到
$\gamma = \sum \limits_{k=2}^\infty (-1)^k \frac{ \left \lfloor \log_2 k \right \rfloor}{k}$
$= \tfrac12-\tfrac13 + 2\left(\tfrac14 - \tfrac15 + \tfrac16 - \tfrac17\right) + 3\left(\tfrac18 - \dots - \tfrac1{15}\right) + \dots$
$= 0.57721 56649 \cdots$

。對一個不是『整數』的實數 $x$ ，『高斯函數』也可以表示為

$\lfloor x\rfloor = x - \frac{1}{2} + \frac{1}{\pi} \sum \limits_{k=1}^\infty \frac{\sin(2 \pi k x)}{k}$

。因此說『超實數系』裡也有『超整數』 hyperinteger 這就一點也不奇怪了吧！如果只從『形式定義』上講，一個『超整數』就是一個『超實數』的『整數部份』，也就是說

$[r^{*}] = [ st(r + \delta x)] = [r]$

，可能沒有什麼意思。假使設想『超實數系』既有『無窮小』 $\delta x$ ，那它的『倒數』 reciprocal $\frac{1}{\delta x}$ 就是『無限大』，也可以叫做『巨量』 Huge，一般用 $H$ 表示。如果說『某數』 $K$ 是個『巨量』，就是講 $\frac{1}{K}$ 是『無窮小』數 $\epsilon$ ，這樣 $\frac{\delta x}{\epsilon}$ 、 $\frac{H}{K}$ 、 $H \cdot \epsilon$ 和 $H - K$ 又是些什麼樣的數呢？它們被稱作『未定式』 Indeterminate form，因為假使不知道它們的『來歷』，我們並不能『確定』最終的『運算結果』 ── 是無窮小、有限量或是無限大 ── 。純就『形式』上講，它門是 $\frac{0}{0}$ 、 $\frac{\infty}{\infty}$ 、 $\infty \cdot 0$ 與 $\infty - \infty$ 的計算，然而這可在『代數運算』是不被允許的啊！但是如果 $x > 0$ ，那麼不管說 $x$ 多大多小 $0^{x} = 0$ ，因此即使 $x$ 是『正無窮小』數 $\delta x$ ，也應該得到 $0^{\delta x} = 0$ 的『極限結果』的吧！同樣的 $\frac{\delta x}{{\delta x}^3} = \frac{1}{{\delta x}^2}$ 是『趨近』於『無限大』的啊！也就是說『無窮小』與『無限大』也是有『等級』 Order 的，如果忽略了『這件事』，隨便混談『至大』和『至小』，大概就是『非量』與『非非量』的『迷惑』之所從來的了！！

─── 摘自《【Sonic π】電路學之補充《四》無窮小算術‧中上》

縱想事不贅述，理無虛發︰

Mean value theorems for definite integrals

First mean value theorem for definite integrals

Let f : [a, b] → R be a continuous function. Then there exists c in (a, b) such that

\int _{a}^{b}f(x)\,dx=f(c)(b-a).

Since the mean value of f on [a, b] is defined as

{\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx,

we can interpret the conclusion as f achieves its mean value at some c in (a, b).^[5]

In general, if f : [a, b] → R is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that

\int _{a}^{b}f(x)g(x)\,dx=f(c)\int _{a}^{b}g(x)\,dx.

Proof of the first mean value theorem for definite integrals

Suppose f : [a, b] → R is continuous and g is a nonnegative integrable function on [a, b]. By the extreme value theorem, there exists m and M such that for each x in [a, b], $m\leqslant f(x)\leqslant M$ and $f[a,b]=[m,M]$ . Since g is nonnegative,

m\int _{a}^{b}g(x)\,dx\leqslant \int _{a}^{b}f(x)g(x)\,dx\leqslant M\int _{a}^{b}g(x)\,dx.

Now let

I=\int _{a}^{b}g(x)\,dx.

If $I=0$ , we’re done since

0\leqslant \int _{a}^{b}f(x)g(x)\,dx\leqslant 0

means

\int _{a}^{b}f(x)g(x)\,dx=0,

so for any c in (a, b),

\int _{a}^{b}f(x)g(x)\,dx=f(c)I=0.

If I ≠ 0, then

m\leqslant {\frac {1}{I}}\int _{a}^{b}f(x)g(x)\,dx\leqslant M.

By the intermediate value theorem, f attains every value of the interval [m, M], so for some c in [a, b]

f(c)={\frac {1}{I}}\int _{a}^{b}f(x)g(x)\,dx,

that is,

\int _{a}^{b}f(x)g(x)\,dx=f(c)\int _{a}^{b}g(x)\,dx.

Finally, if g is negative on [a, b], then

M\int _{a}^{b}g(x)\,dx\leqslant \int _{a}^{b}f(x)g(x)\,dx\leqslant m\int _{a}^{b}g(x)\,dx,

and we still get the same result as above.

QED

均值定理又現身。

知其議論依何據︰

積分判別法

通過將調和級數的和與一個瑕積分作比較可證此級數發散。考慮右圖中長方形的排列。每個長方形寬1個單位、高1 / n個單位（換句話說，每個長方形的面積都是1 / n），所以所有長方形的總面積就是調和級數的和：矩形面積和: $= 1 \,+\, \frac{1}{2} \,+\, \frac{1}{3} \,+\, \frac{1}{4} \,+\, \frac{1}{5} \,+\, \cdots.$ 而曲線y = 1 / x以下、從1到正無窮部分的面積由以下瑕積分給出：曲線下面積: $= \int_1^\infty\frac{1}{x}\,dx \;=\; \infty.$ 由於這一部分面積真包含於（換言之，小於）長方形總面積，長方形的總面積也必定趨於無窮。更準確地說，這證明了：