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Consider a population of organisms for which there is a constant supply of food and limited space, and no predators. Many insect populations in the temperate zones fit this description at certain times in their history.（考慮一生物族群，在固定的食物供給及有限空間中，沒有掠食者，在特定區域的許多種類昆蟲的族群適合這樣的描述。）

In order to model the populations in successive generations, let 　 denote the population of the nth generation, and adjust the numbers so that the capacity of the environment equals to 1, which means that .

One formula, called logistic equation, has gained widespread fame: 　where the parameter　 indicates the rate of 　 which the populations grow.

In general, 　If 　　　　　, then 　has a　-cycle for 　　　　　　（Period-doubling bifurcation） It is known, but difficult to prove, that 　for 　　　　and that the sequence 　has a limit 　given by

Feigenbaum number The number is sometimes called the Feigenbaum number for the quadric family, named after the physicist Mitchell Feigenbaum, who in the mid-1970,s had found a very precise value for it. Let ,　　　　　　　for

Feigenbaum found that the sequence converges to a number we will denote　　 , where What is astonishing is that this constant 　 seems to be universal. It is referred to as Feigenbaum constant, because Feigenbaum was the first to discover it and its universality.

Exercise Period-doubling bifurcation Pitchfork bifurcation

Basic ABC 1. 按 Esc 鍵 2. 同時按 Alt 及 f 鍵 3. 選擇 Open 開啟舊檔或選擇 New 建立新檔 4. 編輯檔案 5. 同時按 Alt 及 r 鍵執行程式 6. 同時按 Alt 及 f 鍵 7. 選擇 Save 儲存舊檔或選擇 Save As 儲存成新檔 8. 同時按 Alt 及 f 鍵 9. 選擇 Exit 結束

Basic ABC 整數 -32768 ……..+32767 實數 10E-38 …….10E+38 變數 ABC123 函數 INT( ), SIN( ), COS( ) 運算 + , - , * , / , MOD , ^ 常用指令 PRINT (WRITE) INPUT (READ) IF condition THEN statement IF condition THEN statement END IF IF condition THEN statement ELSE statement GOTO 110 110 statement

Basic ABC 範例: 求 IF x>=0 THEN absx=x ELSE absx=-x FOR I= TO STEP statement NEXT I FOR Y=1973 TO 1988 STEP 1 PRINT Y, “ “ NEXT Y PSET (X,Y)

Basic ABC 範例 logistic map REM logistic map OPEN "d:\chaos\chaos.dat" FOR OUTPUT AS #1 CLS SCREEN 11 x = .123 FOR i = 1 TO 600 b = 1 + i * .005 FOR j = 1 TO 200 x = b * x * (1 - x) IF j > 90 THEN PSET (INT(b * 200) - 200, 480 - INT(x * 400) - 30) PRINT #1, b, x END IF NEXT j NEXT i CLOSE #1

EXCELL EXCELL程式