已知除了 之外,我們可將白努力多項式 歸納成四種
、 、 、 ,兩類
◎ 偶次 類︰
在閉區間 嚴格遞增↗。從 負→零→正→正極大值 。
在閉區間 嚴格遞減↘。從正極大值 →正→零→負→ 。
◎ 奇次 類︰
在開區間 恆負 。從 → 。
在開區間 恆正 。從 → 。
我們能夠得出 為正,它的負極小值 發生在 處。 為負,它的正極大值 發生在 處。因此在閉區間 裡,偶次 類之絕對值 。
因為我們無法給出在閉區間 中 的一般解,所以只能想辦法估計奇次 類之絕對值 的上界了。由於 ,而且在 處,具有奇對稱性 ,因此只須考慮 即可。藉著
,可得
。故而
,
。
也可知道通常探究白努利多項式之根 ,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 [a, b] 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 類在閉區間 之根 吧。此處重複使用
◎ 乘法公式
。
◎ 對稱公式
。
簡單運算可得︰
△
△
△
△
因此可以歸結的說︰當 時, 以及 、 與 異號。然而 和 同號。故知 有一根 落在 開區間中。因著『偶對稱性』的要求,另一根定然是 ,且知其落於 開區間也☆☆