時間序列︰生成函數‧漸近展開︰指數對數《上》

知道了『大小』,就知道了『次序』,因此數學方法注重『不等』

Inequality (mathematics)

In mathematics, an inequality is a relation that holds between two values when they are different (see also: equality).

  • The notation ab means that a is not equal to b.
It does not say that one is greater than the other, or even that they can be compared in size.

If the values in question are elements of an ordered set, such as the integers or the real numbers, they can be compared in size.

  • The notation a < b means that a is less than b.
  • The notation a > b means that a is greater than b.
In either case, a is not equal to b. These relations are known as strict inequalities. The notation a < b may also be read as “a is strictly less than b“.

In contrast to strict inequalities, there are two types of inequality relations that are not strict:

  • The notation ab means that a is less than or equal to b (or, equivalently, not greater than b, or at most b); “not greater than” can also be represented by the symbol for “greater than” bisected by a vertical line, “not.”
  • The notation ab means that a is greater than or equal to b (or, equivalently, not less than b, or at least b),; “not less than” can also be represented by the symbol for “less than” bisected by a vertical line, “not.”

In engineering sciences, a less formal use of the notation is to state that one quantity is “much greater” than another, normally by several orders of magnitude.

  • The notation ab means that a is much less than b. (In measure theory, however, this notation is used for absolute continuity, an unrelated concept.)
  • The notation ab means that a is much greater than b.

The feasible regions of linear programming are defined by a set of inequalities.

 

,所以有著許多著名的『不等式』

Well-known inequalities

Mathematicians often use inequalities to bound quantities for which exact formulas cannot be computed easily. Some inequalities are used so often that they have names:

 

以為立論『依據』也。

若是

e_n = {\left( 1 + \frac{1}{n} \right)}^n

根本不『收斂』且又或不『唯一』,不知『自然對數

Natural logarithm

The natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational and transcendental number approximately equal to 2.718281828459. The natural logarithm of x is generally written as ln x, loge x, or sometimes, if the base e is implicit, simply log x.[1] Parentheses are sometimes added for clarity, giving ln(x), loge(x) or log(x). This is done in particular when the argument to the logarithm is not a single symbol, to prevent ambiguity.

The natural logarithm of x is the power to which e would have to be raised to equal x. For example, ln(7.5) is 2.0149…, because e2.0149… = 7.5. The natural log of e itself, ln(e), is 1, because e1 = e, while the natural logarithm of 1, ln(1), is 0, since e0 = 1.

The natural logarithm can be defined for any positive real number a as the area under the curve y = 1/x from 1 to a (the area being taken as negative when a < 1). The simplicity of this definition, which is matched in many other formulas involving the natural logarithm, leads to the term “natural”. The definition of the natural logarithm can be extended to give logarithm values for negative numbers and for all non-zero complex numbers, although this leads to a multi-valued function: see Complex logarithm.

The natural logarithm function, if considered as a real-valued function of a real variable, is the inverse function of the exponential function, leading to the identities:

  e^{{\ln(x)}}=x\qquad {\mbox{if }}x>0
  \ln(e^{x})=x.

Like all logarithms, the natural logarithm maps multiplication into addition:

\ln(xy)=\ln(x)+\ln(y).

Thus, the logarithm function is a group isomorphism from positive real numbers under multiplication to the group of real numbers under addition, represented as a function:

  \ln \colon \mathbb{R}^+ \to \mathbb{R}.

Logarithms can be defined to any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter. For instance, the binary logarithm is the natural logarithm divided by ln(2), the natural logarithm of 2. Logarithms are useful for solving equations in which the unknown appears as the exponent of some other quantity. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches of mathematics and the sciences and are used in finance to solve problems involving compound interest.

By Lindemann–Weierstrass theorem, the natural logarithm of any positive algebraic number other than 1 is a transcendental number.

Graph of the natural logarithm function. The function slowly grows to positive infinity as x increases and slowly goes to negative infinity as x approaches 0 (“slowly” as compared to any power law of x); the y-axis is an asymptote.

 

何謂耶?然而『直接』不易說,故而創造『間接』來!

k! > 2^{k-1} , \ k \geq 2 ,因為

k! = k \times k-1 \times \cdots \times 2 \times 1

> 2 \times 2 \times \cdots \times 2 \times 1

= 2^{k-1}

而且如果

0 < x < 1 , \ 0 < y < 1

那麼

x \times y < 1

於是

e_n = {\left( 1 + \frac{1}{n} \right)}^n

= \sum \limits_{k=0}^{n} \left( \begin{array}{ccc} n \\ k \end{array} \right) \frac{1}{n^k}

= \sum \limits_{k=0}^{n} \frac{1}{k!} \left( 1 - \frac{1}{n} \right) \cdots \left( 1 - \frac{k-1}{n} \right)

< 1 + \left( 1 + \frac{1}{2} + \frac{1}{2 \times 2} + \cdots + \frac{1}{2^n -1} \right)

< 3

再從 e_n 之『單調性』,故知僅有『唯一』極限值乎??

Monotone convergence theorem

In the mathematical field of real analysis, the monotone convergence theorem is any of a number of related theorems proving the convergence of monotonic sequences (sequences that are increasing or decreasing) that are also bounded. Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum.

 

同樣的,還可以證明那個 e 是『無理數』也!!

無理數證明

反證法

證明 e 是無理數可以用反證法。假設 e 是有理數,則可以表示成  {\tfrac {a}{b}} ,其中 a,b 為正整數。以 e 的無窮級數展開式可以得出矛盾。

考慮數字

  x = b\,! \left(e-\sum_{i=0}^b {1 \over i\,!}\right)

以下將推導出  x是小於1的正整數;由於不存在這樣的正整數,得出矛盾,所以得證  e是無理數。

  •   x是整數,因為
0 < x = b\,! \left(e - \sum_{i=0}^b {1 \over i\,!}\right) = b\,! \left({a \over b} - \sum_{i=0}^b {1 \over i\,!}\right)
  = a (b-1)! - \sum_{i=0}^b {b\,! \over i\,!}
  =a(b-1)!-\left[1+\sum _{{n=0}}^{{b-1}}b(b-1)\cdots (n+1)\right]
  •   x是小於1的正數,因為
  0 < x = b\,! \sum_{n=b+1}^\infty {1 \over n!}
= \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)} + \cdots
  < \frac{1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3} + \cdots = {1 \over b} \le 1

但是0與1之間(不含0與1)不存在有整數,故原先假設矛盾,得出  e為無理數。