一、複習求和符號∑
自從約瑟夫·傅立葉於1820年引入求和符號∑(大寫的希臘字母sigma)以來,求和∑以及雙重求和∑∑在數學公式推導,命題證明中被經常使用,掌握它的定義和性質對於提高我們的數學能力是必不可少的。
注意我們在此只討論有限項 的求和。
結合律:
∑ i = 1 n ( a i + b i ) = ∑ i = 1 n a i + ∑ i = 1 n b i
\sum_{i=1}^{n}( a_{i}+b_{i})=\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}
i = 1 ∑ n ( a i + b i ) = i = 1 ∑ n a i + i = 1 ∑ n b i
分配律:
∑ i = 1 n r a i = r ∑ i = 1 n a i ( r 爲 任 意 常 數 )
\sum_{i=1}^{n} r a_{i}=r \sum_{i=1}^{n} a_{i} \quad( r爲任意常數)
i = 1 ∑ n r a i = r i = 1 ∑ n a i ( r 爲 任 意 常 數 )
從函數角度:
∑ i = 1 10 g ( k , l ) f ( i , j ) = g ( k , l ) ∑ i = 1 10 f ( i , j )
\sum_{i=1}^{10} g(k, l) f(i, j)=g(k, l) \sum_{i=1}^{10} f(i, j)
i = 1 ∑ 1 0 g ( k , l ) f ( i , j ) = g ( k , l ) i = 1 ∑ 1 0 f ( i , j )
g(k, l)是與下標i無關的函數
分段:
∑ i = 1 n a i = ∑ i = 1 m a i + ∑ i = m + 1 n a i
\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{m} a_{i}+\sum_{i=m+1}^{n} a_{i}
i = 1 ∑ n a i = i = 1 ∑ m a i + i = m + 1 ∑ n a i
二、二重求和的定義
有一個n行m列的數表:
a 11 , a 12 , a 13 , ⋯  , a 1 m a 21 , a 22 , a 23 , ⋯  , a 2 m a 31 , a 32 , a 33 , ⋯  , a 2 m ⋯ a n 1 , a n 2 , a n 3 , ⋯  , a n m
\begin{array}{l}{a_{11}, a_{12}, a_{13}, \cdots, a_{1 m}} \\ {a_{21}, a_{22}, a_{23}, \cdots, a_{2 m}} \\ {a_{31}, a_{32}, a_{33}, \cdots, a_{2 m}} \\ {\cdots} \\ {a_{n 1}, a_{n 2}, a_{n 3}, \cdots, a_{n m}}\end{array}
a 1 1 , a 1 2 , a 1 3 , ⋯ , a 1 m a 2 1 , a 2 2 , a 2 3 , ⋯ , a 2 m a 3 1 , a 3 2 , a 3 3 , ⋯ , a 2 m ⋯ a n 1 , a n 2 , a n 3 , ⋯ , a n m
數表裏的每個元素都由兩個相互獨立的數i,j決定,即每一項都是i,j的二元函數,一般項爲aij ,i = 1,2…n; j = 1,2…m
這n × m項的和記爲∑ j = 1 m ( ∑ i = 1 n a i j ) \sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij}) ∑ j = 1 m ( ∑ i = 1 n a i j ) 或者∑ i = 1 n ( ∑ j = 1 m a i j ) \sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij}) ∑ i = 1 n ( ∑ j = 1 m a i j )
三、雙重求和∑∑交換求和順序
第i行的元素的和記爲Ri:
R i = ∑ j = 1 m a i j = a i 1 + a i 2 + . . . + a i m
R_{i} = \sum_{j=1}^{m} a_{ij} = a_{i1} + a_{i2} + ... + a_{im}
R i = j = 1 ∑ m a i j = a i 1 + a i 2 + . . . + a i m
一共有n行,所有行元素的和,即數表所有元素的和記爲S:
S = ∑ i = 1 n R i = ∑ i = 1 n ( ∑ j = 1 m a i j )
S = \sum_{i=1}^{n}R_{i} = \sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij})
S = i = 1 ∑ n R i = i = 1 ∑ n ( j = 1 ∑ m a i j )
第j列的元素的和記爲Cj:
C j = ∑ i = 1 n a i j = a 1 j + a 2 j + . . . + a n j
C_{j} = \sum_{i=1}^{n} a_{ij} = a_{1j} + a_{2j} + ... + a_{nj}
C j = i = 1 ∑ n a i j = a 1 j + a 2 j + . . . + a n j
一共有m列,所有列元素的和,即數表所有元素的和記爲S:
S = ∑ j = 1 m C j = ∑ j = 1 m ( ∑ i = 1 n a i j )
S = \sum_{j=1}^{m}C_{j} = \sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij})
S = j = 1 ∑ m C j = j = 1 ∑ m ( i = 1 ∑ n a i j )
所以∑ i = 1 n ( ∑ j = 1 m a i j ) = ∑ j = 1 m ( ∑ i = 1 n a i j ) \sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij}) = \sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij})
i = 1 ∑ n ( j = 1 ∑ m a i j ) = j = 1 ∑ m ( i = 1 ∑ n a i j )
也可以寫成∑ 1 < = i < = n , 1 < = j < = m a i j \sum_{1<=i<=n,1<=j<=m}a_{ij} ∑ 1 < = i < = n , 1 < = j < = m a i j
即二重和的和號(求和次序)可以交換。
但要注意,但求和項數變爲無窮或者(一個或兩個)和號變爲積分號時,往往要求級數收斂或者函數可積,相應的交換和號的結論才能成立。
Example 1
∑ i = 1 4 ∑ j = 1 i f ( i , j ) \sum_{i=1}^{4} \sum_{j=1}^{i} f(i, j) ∑ i = 1 4 ∑ j = 1 i f ( i , j ) 交換求和次序後是什麼樣的呢?
A.∑ j = 1 i ∑ i = 1 4 f ( i , j ) \sum_{j=1}^{i} \sum_{i=1}^{4} f(i, j) ∑ j = 1 i ∑ i = 1 4 f ( i , j )
B.∑ j = 1 4 ∑ i = 1 j f ( i , j ) \sum_{j=1}^{4} \sum_{i=1}^{j} f(i, j) ∑ j = 1 4 ∑ i = 1 j f ( i , j )
C.∑ j = 1 4 ∑ i = j 4 f ( i , j ) \sum_{j=1}^{4} \sum_{i=j}^{4} f(i, j) ∑ j = 1 4 ∑ i = j 4 f ( i , j )
因爲[1<= i <= 4][1<= j<= i] = [1<= j <= i <= 4] = [1<= j<= 4][ j <= i <= 4]
所以 選C,也可以窮舉出所有元素,如果將i作爲行號,j作爲列號,對於∑ i = 1 4 ∑ j = 1 i f ( i , j ) \sum_{i=1}^{4} \sum_{j=1}^{i} f(i, j) ∑ i = 1 4 ∑ j = 1 i f ( i , j ) ,你會發現這些元素的排列類似於下三角 矩陣的形式(按行求和),然後將按行求和切換爲按列求和,也會得到C答案。
Example 2
求∑ k = 1 n ∑ i = 1 k i a i j k ( k + 1 ) \sum_{k=1}^{n} \sum_{i=1}^{k} \frac{ ia_{ij}}{k(k + 1)} ∑ k = 1 n ∑ i = 1 k k ( k + 1 ) i a i j 交換求和次序後的表達式。
同樣的,[1<= k <= n][1<= i<= k] = [1<= i <= k <= n] = [1<= i<= n][ i <= k <= n]
所以,∑ i = 1 n ∑ k = i n i a i j k ( k + 1 ) \sum_{i=1}^{n} \sum_{k=i}^{n} \frac{ ia_{ij}}{k(k + 1)} ∑ i = 1 n ∑ k = i n k ( k + 1 ) i a i j ,如果將i作爲行號,k作爲列號,對於∑ k = 1 n ∑ i = 1 k i a i j k ( k + 1 ) \sum_{k=1}^{n} \sum_{i=1}^{k} \frac{ ia_{ij}}{k(k + 1)} ∑ k = 1 n ∑ i = 1 k k ( k + 1 ) i a i j ,你會發現這些元素的排列類似於上三角 矩陣的形式(按列求和),然後將按列求和切換爲按行求和。
Example 3
∑ k = 1 n ( k ( ∑ i = 1 k i 2 a i ) ( 2 k ( k + 1 ) ) 2 ) \sum_{k=1}^{n}(k( \sum_{i=1}^{k}\frac{ i^2}{a_{i}}) (\frac{2}{k(k + 1)})^2) ∑ k = 1 n ( k ( ∑ i = 1 k a i i 2 ) ( k ( k + 1 ) 2 ) 2 ) = 4 ∑ k = 1 n ∑ i = 1 k i 2 a i k k ( k + 1 ) 2 4\sum_{k=1}^{n} \sum_{i=1}^{k}\frac{ i^2}{a_{i}} \frac{k}{k(k + 1)^2} 4 ∑ k = 1 n ∑ i = 1 k a i i 2 k ( k + 1 ) 2 k = 4 ∑ i = 1 n i 2 a i ∑ k = i n k k ( k + 1 ) 2 4\sum_{i=1}^{n}\frac{ i^2}{a_{i}} \sum_{k=i}^{n} \frac{k}{k(k + 1)^2} 4 ∑ i = 1 n a i i 2 ∑ k = i n k ( k + 1 ) 2 k
注意 ,容易出錯的地方
( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) ≠ ∑ i = 1 5 ∑ i = 1 5 f ( i ) f ( i ) = ∑ i = 1 5 ∑ i = 1 5 f 2 ( i ) \left(\sum_{i=1}^{5} f(i)\right)^{2}=\left(\sum_{i=1}^{5} f(i)\right) *\left(\sum_{i=1}^{5} f(i)\right) ≠ \sum_{i=1}^{5} \sum_{i=1}^{5} f(i) f(i) = \sum_{i=1}^{5} \sum_{i=1}^{5} f^2(i) ( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) ̸ = ∑ i = 1 5 ∑ i = 1 5 f ( i ) f ( i ) = ∑ i = 1 5 ∑ i = 1 5 f 2 ( i )
而是
( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) = ∑ i = 1 5 ∑ j = 1 5 f ( i ) f ( j ) \left(\sum_{i=1}^{5} f(i)\right)^{2}=\left(\sum_{i=1}^{5} f(i)\right) *\left(\sum_{i=1}^{5} f(i)\right) = \sum_{i=1}^{5} \sum_{j=1}^{5} f(i) f(j) ( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) = ∑ i = 1 5 ∑ j = 1 5 f ( i ) f ( j )
–更詳細內容可閱讀《具體數學》第二章 和式