時間序列︰生成函數‧漸近展開︰白努利多項式之根《六》

已知除了 B_1 (x) = x - \frac{1}{2} 之外,我們可將白努力多項式 B_m (x) 歸納成四種

B_{4k}B_{4k+1}B_{4k+2}B_{4k+3 ,兩類

◎ 偶次 2n 類︰ {(-1)}^n \cdot B_{2n} (x)

在閉區間 [0, \frac{1}{2}] 嚴格遞增↗。從 {(-1)}^n \cdot B_{2n} (0) 負→零→正→正極大值 {(-1)}^n \cdot B_{2n} (\frac{1}{2})

在閉區間 [\frac{1}{2}, 1] 嚴格遞減↘。從正極大值 {(-1)}^n \cdot B_{2n} (\frac{1}{2}) →正→零→負→ {(-1)}^n B_{2n}(1)

◎ 奇次 2n+1 類︰ {(-1)}^n \cdot B_{2n+1} (x)

在開區間 (0, \frac{1}{2}) 恆負 {(-1)}^n \cdot B_{2n+1} (x) < 0。從 B_{2n+1} (0) = 0B_{2n+1} (\frac{1}{2}) = 0

在開區間 (\frac{1}{2}, 1) 恆正 {(-1)}^n \cdot B_{2n+1} (x) > 0。從 B_{2n+1} (\frac{1}{2}) = 0B_{2n+1} (1) = 0

我們能夠得出 B_{4k+2} (0) = B_{4k+2} = | B_{4k+2} | 為正,它的負極小值 B_{4k+2} (\frac{1}{2}) = (2^{-(4k+1)} -1) B_{4k+2} 發生在 B_{4k+1} (\frac{1}{2}) = 0 處。B_{4k} (0) = B_{4k} = - | B_{4k} | 為負,它的正極大值 B_{4k} (\frac{1}{2}) = (2^{1 - 4k} - 1) B_{4k} 發生在 B_{4k -1} (\frac{1}{2}) = 0 處。因此在閉區間 [0, 1] 裡,偶次 2n 類之絕對值 | B_{2n} (x) | \le |B_{2n} |

因為我們無法給出在閉區間 [0, 1]B_{2n} (x) = 0 的一般解,所以只能想辦法估計奇次 2n+1 類之絕對值 | B_{2n+1} (x) | 的上界了。由於 B_{2n+1} (0) = B_{2n+1} (\frac{1}{2}) = B_{2n+1} (1) = 0 ,而且在 x = \frac{1}{2} 處,具有奇對稱性 B_{2n+1} (1-x) = - B_{2n+1} (x) ,因此只須考慮 x \in [0, \frac{1}{2}] 即可。藉著

\int_{a}^{x} B_{2n} (t) dt = \frac{B_{2n+1} (x) - B_{2n+1} (a)}{2n+1} ,可得

| \int_{a}^{x} B_{2n} (t) dt | = | \frac{B_{2n+1} (x) - B_{2n+1} (a)}{2n+1} |

\le \int_{a}^{x} | B_{2n} (t) | dt \le | B_{2n} | \ \cdot | \int_{a}^{x} 1 \ dt | = | B_{2n} | \cdot |x - a| 。故而

x \in [0, \frac {1}{4}] , | \int_{0}^{x} B_{2n} (t) dt | = | \frac{B_{2n+1} (x) }{2n+1} | \le | B_{2n} | \cdot |x| = \frac{|B_{2n}|}{4}

x \in [\frac{1}{4}, \frac {1}{2}] , | \int_{x}^{\frac{1}{2}} B_{2n} (t) dt | = | \frac{B_{2n+1} (x) }{2n+1} | \le | B_{2n} | \cdot |\frac{1}{2} -x| = \frac{| B_{2n} |}{4} ,

\therefore | B_{2n+1} (x) | \le \frac{2n+1}{4} | B_{2n} |

 

也可知道通常探究白努利多項式之根 B_n (x) = 0 ,或得依賴『變號法則』,應用『二分逼近法』哩︰

Bisection method

The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. It is a very simple and robust method, but it is also relatively slow. Because of this, it is often used to obtain a rough approximation to a solution which is then used as a starting point for more rapidly converging methods.[1] The method is also called the interval halving method,[2] the binary search method,[3] or the dichotomy method.[4]

A few steps of the bisection method applied over the starting range [a1;b1]. The bigger red dot is the root of the function.

The method

The method is applicable for numerically solving the equation f(x) = 0 for the real variable x, where f is a continuous function defined on an interval [ab] and where f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root since, by the intermediate value theorem, the continuous function f must have at least one root in the interval (a, b).

At each step the method divides the interval in two by computing the midpoint c = (a+b) / 2 of the interval and the value of the function f(c) at that point. Unless c is itself a root (which is very unlikely, but possible) there are now only two possibilities: either f(a) and f(c) have opposite signs and bracket a root, or f(c) and f(b) have opposite signs and bracket a root.[5] The method selects the subinterval that is guaranteed to be a bracket as the new interval to be used in the next step. In this way an interval that contains a zero of f is reduced in width by 50% at each step. The process is continued until the interval is sufficiently small.

Explicitly, if f(a) and f(c) have opposite signs, then the method sets c as the new value for b, and if f(b) and f(c) have opposite signs then the method sets c as the new a. (If f(c)=0 then c may be taken as the solution and the process stops.) In both cases, the new f(a) and f(b) have opposite signs, so the method is applicable to this smaller interval.[6]

 

且讓我們嘗試運用此法更精細定位偶次 2n 類在閉區間 [0, 1] 之根 B_{2n} (x) = 0 吧。此處重複使用

◎ 乘法公式

m^{1-n} \cdot B_n (m \cdot x) = \sum \limits_{k=0}^{m-1} B_n (x + \frac{k}{m} ), \ m \ge 1

◎ 對稱公式

B_n (1-x) = {(-1)}^n B_n (x), \ n \ge 0

 

簡單運算可得︰

B_{2n} (\frac{1}{2}) = (2^{1-2n} -1) B_{2n}

B_{2n} (\frac{1}{2}) + B_{2n} (\frac{1}{2} + \frac{1}{2}) = 2^{1-2n} \cdot B_{2n} (2 \cdot \frac{1}{2})

B_{2n} (1) = B_{2n} (0) = B_{2n}

 

B_{2n} (\frac{1}{4}) = B_{2n} (\frac{3}{4}) = 2^{-2n} B_{2n} (\frac{1}{2}) = 2^{-2n} (2^{1-2n} -1) B_{2n}

B_{2n} (\frac{1}{4}) + B_{2n} (\frac{1}{4} + \frac{1}{2}) = 2^{1-2n} \cdot B_{2n} (2 \cdot \frac{1}{4})

B_{2n} (\frac{1}{4}) = B_{2n}(\frac{3}{4})

 

B_{2n} (\frac{1}{3}) = B_{2n} (\frac{2}{3}) = 2^{-1} (3^{1-2n} - 1) B_{2n}

B_{2n} (\frac{1}{3}) + B_{2n} (\frac{1}{3} + \frac{1}{3}) + B_{2n} (\frac{1}{3} + \frac{2}{3}) = 3^{1-2n} \cdot B_{2n} (3 \cdot \frac{1}{3})

B_{2n} (\frac{1}{3}) = B_{2n} (\frac{2}{3})

 

B_{2n} (\frac{1}{6}) = B_{2n} (\frac{5}{6}) = (2^{1-2n} -1)(3^{1-2n} -1) \frac{B_{2n}}{2}

B_{2n} (\frac{1}{6}) + B_{2n} (\frac{1}{6} + \frac{1}{2}) = 2^{1-2n} \cdot B_{2n} ( 2 \cdot \frac{1}{6})

B_{2n} (\frac{1}{6}) = (2^{1-2n} -1) B_{2n} (\frac{1}{3})

 

因此可以歸結的說︰當 n \ge 1 時, B_{2n} (\frac{1}{2}) 以及 B_{2n} (\frac{1}{3})B_{2n} (\frac{1}{4})B_{2n} 異號。然而 B_{2n} (\frac{1}{6})B_{2n} 同號。故知 B_{2n} = 0 有一根 r 落在 (\frac{1}{6}, \frac{1}4}) 開區間中。因著『偶對稱性』的要求,另一根定然是 1 - r ,且知其落於 (\frac{3}{4}, \frac{5}{6}) 開區間也☆☆