資料收集: 第一集

1-1 等差數列

This post is to introduce those mathematics courses of senior high school.
Now we start introducing them as follows.

1. Arithmetic progression (A.P.) -  an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant and same. 簡單的說,就是將一堆數排成一列,而每個數與其前一個or後一個的數的差都一樣. For example , given an AP as -1,2,5,8,11, ... We may obverse this form and get get several terms from it . Assume let term 2 is a2 , then -1 is 1st term (a1) , ... Or we may represent them as a straight form as below :
a1 = -1
a2 = 2
a3 = 5
...
an = ?

Observing difference of any two terms is 3 . In general , the difference was called 公差 and use an alphabet d to represent it . See above example , we may define it as below:
a1 = -1 1st term
a2 = a1 + d 2nd term
a3 = a1 + 2d 3rd term
.... ...
am = a1 + (m-1)d mth term
...
an = a1 + (n-1)d nth term

d is the difference. Or we also represent it as below
a1 , a1+d , a1+2d , ... , a1+(m-1)d , ... , a1+(n-1)d , ...

if d = 0 , then for the arithmetic progress is a1,a1,a1, ... , a1 , ... and Basically , we may rewrite it as following a,a,a,a,a,a , ... all terms are same .
In general , ai means ith term , i =1,2,3,4, ... or another representation i ∈ N , N is a nature numbers set.
let that
am = a1 + (m-1)d (1)
an = a1 + (n-1) d (2)

(1)- (2) we obtain the following
an - am = (n-1)d - (m-1)d = (n-m)d
So an = am + (n-m)d , n , m ∈ N
Prove that an= a1 + (n-1)d , n ∈ N and d is a constant .
Proof:
Assume that there is a proposition calculus p(n) : an = a1 + (n-1)d , n ∈ N , N is a nature set .
Induction Basis:
n = 1 , p(1) : a1 = a1 + (1-1)d is true .

And we assume n = k , p(n) is also true , that is ak = a1+ (k-1)d ( this is induction conclusion )
Again , we consider n = k +1 case .
p(n) = p (k+1) = a1 + (k+1-1)d = a1+ kd Q.E.D.
Basically ,we also regard an arithmetic progress is a function. When we input 1st term and d , then nth term can be generated. For example , there is an arithmetic progress an = 9n + 4 ,  n ∈ N .
Now we want to obtain any term via above general form , for example , a2 = 9*2 +4 = 22 , a12 = 9*12 + 4 = 112 , and so forth. Observing above result , we can obtain a conclusion :
an is a function a(n) and it can be defined as 9n + 4 . More generally speaking , a(n) = 9n+4 , it is a linear function.
Ex1 .
There exists an arithmetic progression and knows 15th is -25 and 32th is -42 , then find 25th term .
Sol:
Assume that there exists an arithmetic progression , a1 , a2 , a3 , ... , an
We have knew two terms , 15th term and 32th term .
a15 = - 25 , a32 = - 42 . According to above formula , we can calculate the difference d .
a32 = a15 + (32-15)d
- 42 = - 25 + 17d
- 17 = 17d
d = -1 Now we want to find 25th term a25 and it can be got via above formula again.
a25 =  a15 + (25-15)d = -25 + 10*(-1) = -25 -10 = - 35
And another way is to follow standard formula .
Assume that 1st term of this arithmetic progression is a1 and its difference is d , then
a15 =  a1 + (15-1)d =  a1 + 14d = -25 ---- (1)
a32 =  a1+ (32-1)d =  a1 + 31d = -42 ---- (2)

(1) - (2) we obtain 17 = -17d , d = -1 replace it to (1) or (2) then we get a1 = - 11
Basically , we can conclude a result , that is , Any arithmetic progression is a linear and obtain the a result as below:
Assume that  an is a function and it can be expressed as a 一元一次方程式 a(n) and a(n) = kn +b , k and b are constants and n is a nature number. We revisit above formula  an = a1 + (n-1)d .
subtract a1 to both sides of equal sign and we obtain an - a1 = (n-1)d and divide n - 1 to both sides , it becomes (an - a1)/(n-1) = d , d is the slope of function and d is fixed.
Again , we still consider an example as below. The question is that there is an arithmetic progression -23 , -20 , -17 , ... ; Find the following questions
1) values of nth and 10th terms .
2) if an is positive , what is the smallest value of n ?

Sol:
According to our observation , we can assume 1st term a1 = -23 , a2 = -20  , a3 = -17 , ... and so forth. we can calculate the value of difference of sequence (d) is 3 via this expression -20 = -23 + (2-1)d . So nth term an = -23 + (n-1)*3 = -23 +3n -3 = 3n - 26 and a10 = -23 + (10-1) * 3 = -23 + 9*3 = -23 + 27 = 4 ,1st question has been resolved .
Again , 2nd question is to ask what is the minimum value of n that causes an is positive. This is easy to resolve it. Assume that nth term an is positive . According to the formula  we can get an = 3n - 26 .
Because an is positive , then 3n - 26 > 0 . So 3n > 26 implies n > 26/3. Because n is positive integer , the smallest integer that approaches 26/3 is 9 .

Again , we start introducing a new term - Arithmetic mean ( 等差中項 / 算術平均數 ). What is it ?
Assume that there is an arithmetic progression a1,a2,a3, ... , an , ... , we get any three continuous terms ,
they are ai-1 , ai ,and ai+1 , respectively and  ai-1 < ai < ai+1 , i = 1 , 2 , 3 , ... . Then  ai is called Arithmetic mean between ai-1 and ai+1 .
For example , 1,3,5,7,9,11, ... 5 is the Arithmetic mean between 3 and 7 . Now we prove it.
Assume that there exists an arithmetic progression  a1,a2,a3, ... , an , ... prove that ai is the arithmetic mean of ai-1 and ai+1 , for i = 1,2,3, ...
and ai = ( ai-1 + ai+1 ) / 2 .

Pf:
ai-1 = a1 + ( i-2)d
ai+1 = a1 + (i)d

ai-1 + ai+1 = 2a1 + 2 (i-1)d
(ai-1 + ai+1 )/ 2 = a1 + (i-1)d = ai Q.E.D.

We revisit a famous inequality (a+c) / 2 ≥ √ac, a and c are positive .
It is easy to prove . Consider an expression that (√a -√c )²≥ 0 , we expend it as following a - 2√ac + c ≥ 0 , then a + c ≥ 2√ac Q.E.D.
Okay . See some following properties about arithmetic progressions .

A. 存在一個算術數列,將所有的項目加減一個固定的常數k 則其公差仍然相同
pf:

假設有一個算數數列 a1,a2,a3, ... , an , ... 其公差為 d ,
則令 bi = ai + k , i = 1,2,3,4, ... 且公差為 d' , 則 d' = bi+1 - bi = (ai+1+k) - (ai+k) = ai+1 - ai = d
, 同理可證令 bi = ai - k , i = 1,2,3, ... 且公差為d" , 則 d" = bi+1 - bi = (ai+1-k) - (ai-k) = ai+1 - ai = d

B. 存在一個算術數列,將所有的項目乘上一個常數k 且 k ≠ 0 ,則其新的數列仍然是等差數列
並且其公差為原來數列的公差的k倍

pf:

假設有一個算數數列 a1,a2,a3, ... , an , ... 其公差為 d ,
則令 bi = kai , i = 1,2,3,4, ... k ≠ 0 且公差為 d' ,則 d' = bi+1 - bi = kai+1 - kai = k (ai+1 - ai) = kd
所以 b1,b2,b3, ... , bn 仍是等差數列. Q.E.D

C. 存在一個算術數列,將所有的項目除上一個常數k 且 k ≠ 0 ,則其新的數列仍然是等差數列
並且其公差為原來數列的公差的1/k倍

pf:

假設有一個算數數列 a1,a2,a3, ... , an , ... 其公差為 d ,
則令 bi = ai/k , i = 1,2,3,4, ... k ≠ 0 且公差為 d' ,則 d' = bi+1 - bi = ai+1/k - ai/k = (1/k) (ai+1 - ai) = d/k
Q.E.D.

D. 存在一個只有四個項目的等差數列 a,b,c,d , 則 a+d = b+c

pf:
因為此為等差數列,故假設其公差為 d' , 則 b = a+ d' , c= b+d' , d = c+d'
a+d = a+c+d' , b+c = a+d' +c Q.E.D.

E. 存在一個只有五個項目的等差數列 a,b,c,d,e ,則 a+e = b+d = 2c

pf:
因為此為等差數列,故假設其公差為 r , 則 b = a+ r , c = b+ r , d = c + r 及 e = d +r
a+e = a + d+ r
b+d = a+r + d ∴ a+ e = b + d Q.E.D. 現在考量 c , 2c = 2b + 2r = 2(b+ r) = 2(a+r+r) = 2a +4r
∵ b+d = a + d+ r = a+(a+3r)+r = 2a + 4r Q.E.D.

F. 存在一個只有奇數個項目的等差數列, 若項數為 2n+1 項則第(k+1) 項與2k+1項的差為(k-1)/2倍的d , k ≤ n .
pf:
假設有一個算數數列 a1,a2,a3, ... , an , ..., a2n+1 共有 2n+1 項,其公差為 d ,
取任意一項  ak , 1 ≤ k ≤ n , 若 k 奇數, k = 2m+1 , m = 1,2,3, ... ; ak = a2m+1 =  a1 + (k-1)d
=  a1 + (2m)d , am+1 =  a1 + md ,  ak- ak-1 = md = (k-1)d/2 Q.E.D.

G. 存在兩個數 a , b 則在其間插入m個數而a 為首項, b 為第 m+2 項使其成為一個等差數列, 則其公差為何 ?

Sol: 假設其插入的項目為 a1, a2,a3,a4, ..., am , 則

a1 = a + d
a2 = a1 + d = a +2d
a3 = a2 + d = a1 + d + d = a + 3d

a4 = a3 + d = a2 + d + d = a + 4d
...

am = am-1 + d = a + (m-1)d + d = a + md
b = am + d = a + (m+1)d
b - a = (m+1)d
∴ ( b - a ) / m+1 = d

EX. 若 f(x) = 2x - 1 則 f(2) , f(a) , f(b) , f(8) 形成一個等差數列 , 求 f(a), f(b) 及a,b 的等差中項
f(2) = 3
f(8) = 15
令 d 為其公差, 則 d = 15 - 3 / 2+1 = 12/3 = 4 ∴ f(a) = 7 , f(b) = 11
f(a) = 2a - 1 = 7 , a = 4
f(b) = 2b - 1 = 11 , b = 6 故a,b的等差中項為 4+6/2 = 5

Ex. 有兩數 a,b 其積為 15 且等差中項為 4 ,求 a , b
ab= 15 , 4 = (a+b)/2
ab= 15 , a+b = 8
考慮一元二次方程式求根性質, ax^2 + bx +c , 其解有兩個 x1, x2
則 x1+ x2 = -b/a , x1x2 = c/a 其方程式可分解為 (x-x1)(x-x2) 並令其為 0 即可以解出 x1, x2

銜接教材(1) - 等差數列

(1) 8 為等差數列第一項,第二項為 5 , 故公差 d = -3 且觀察出, 8 < 5 ; 即為一個遞減的數列,所以運用等差數列第n項 an = a1 + (n-1)d , 故可以算出,

a3 = a1 + (3-1)(-3) = 8 + (-6) = 2

a4 = a1 + (4-1)(-3) = 8 + (-9) = -1

a5 = a1 + (5-1)(-3) = 8 + (-12) = -4

(2) b 為等差數列第一項 , 第二項為 b - 3 , 所以 b- 3- b = -3 且 b - 3 < b ; 故為

等差數列. 公差d為 -3,所以運用等差數列第n項 an = a1 + (n-1)d , 故可以算出

a3 = a1 + (3-1)(-3) = b + (-6) = b-6

a4 = a1 + (4-1)(-3) = b + (-9) = b-9

a5 = a1 + (5-1)(-3) = b + (-12) = b-12

(3) 7 為第2項, 15為第3項 , 15 - 7 = 8 且 15 > 7 故為一個等差且遞增的數列, 所以運用等差數列第n項 an = a1 + (n-1)d .
a2 = a1 + (2-1)8 = a1 + 8 . 所以 7 = a1 + 8 , a1 = -1
a3 = a1 + (3-1)8 = -1 +24 = 23
a4 = a1 + (4-1)8 = -1 +32 = 31
(4) b 為第2項, b+ 8 為第3項 , d = b+8 - b = 8 所以, b- 8 為第1項,
第n項 an = a1 + (n-1)d
an = b-8 + (n-1)8 = b + (n-1-1)8 = b+ 8(n-2) 為第n項的通式
(5) 0.9 為第三項, 0.3 為第四項,故公差d為 0.3 - 0.9 = - 0.6為一個遞減的數列. 第n項 an = a1 + (n-1)d
第m項 am = a1 + (m-1)d
am - an= (m-n)d , am = an + (m-n)d . ( m > n ) 故第一項 a3 = a1 + (3-1)d
0.9 = a1 + (3-1)(-0.6)
0.9 = a1 -1.2
2.1 = a1 , 其餘項目類推 .

a32 = -42 , a15 = -25 觀察出為遞減數列 ,
a32 = a15 + (32-15)d
- 42 = - 25 + 18d
-18 = 18d
d = -1
a15 = a1 + (15-1)d
-25 = a1 + 14(-1)
-25 = a1 -14
-11 = a1
a25 = a1 + (25-1)(-1) = -11 - 24 = -35

(1) a1 = -23 , d = 3 ; an = a1 + (n-1)d = -23 + 3 (n-1) = -26 + 3n
(2) a10 = -26 + 30 = 4
(3) 假設第k項開始為正, ak > 0 ; ak = -26 + 3k > 0 , 3k > 26 , k > [26/3] = 8 , 即第9項.

12 為 8 與 X 的等差中項, 即 12 = (8+X ) / 2
24 = (8+X) , X = 16

已知第1項為 X ,最後一項為 -8 ; 則令 a1 = X , 又因其後及在-8前又加入 9個數 ,故可令 a11 = - 8 . 所以由求第n項公式得知, - 8 = X + ( 11-1)d , d 是公差.
又知插入第5項為1 則a6 = X + (6-1)d , 1 = X + 5d .
- 8 = X + 10d
1 = X + 5d
- 9 = 5d , d = -9/5
- 8 = X + (-18)
10 = X

f(2) = 4 -1 = 3 , f(a) = 2a -1 , f(b)= 2b -1 , f(8) = 15
4數成等差. 則 3 + d = 2a -1 , 2a-1+d =2b -1 , 2b-1 +d = 15
4 = 2a - d
2a + d - 2b = 0
2b + d = 16
2a+ 2b = 20
2a - 2b = -d
4a = 20 - d
4a = 8 + 2d
8 + 2d = 20 -d
12 = 3d
4 = d , a = 4 , b = 6 所以等差中項為 5 .

假設兩數 a,b 且其等差中項為 4 , (a+b)/2 = 4 , 8 = a+b 同時, ab = 15
則 a, b 不為零.
a = 15/b , 8 = 15/b+b , 8b = 15 + b^2 ;
b^2 -8b +15 = 0
(b-3)(b-5) = 0
b = 3 or b = 5
b = 3 , a = 5
b = 5 , a = 3