函數項級數的性質

文章目錄

• 連續性
• • 證明
• 逐項求積
• 逐項求導
• • 證明

連續性

• ∑ u n ( x ) \sum u_n(x) ∑un​(x)在 [ a , b ] [a,b] [a,b]上(内閉)一緻收斂
• 每一項 u n u_n un​都連續
• ⇒ \Rightarrow ⇒和函數也連續，即 ∑ ( lim ⁡ x → x 0 u n ( x ) ) = lim ⁡ x → x 0 ( ∑ u n ( x ) ) \sum(\lim\limits_{x\to x_0}u_n(x))=\lim\limits_{x\to x_0}(\sum u_n(x)) ∑(x→x0​lim​un​(x))=x→x0​lim​(∑un​(x))

證明

這裡是通過轉化為函數列，然後用一緻收斂函數列的性質證明的，将級數轉化為函數列的一個重要方法就是 → \to → 部分和函數列 { S n ( x ) } \{S_n(x)\} {Sn​(x)}

• 由于每一項 u n u_n un​都連續 ⇒ \Rightarrow ⇒ 每一項 S n ( x ) S_n(x) Sn​(x)連續
• ∑ u n ( x ) \sum u_n(x) ∑un​(x)一緻收斂 ⇒ \Rightarrow ⇒ S n ( x ) ⇉ S ( x ) S_n(x)\rightrightarrows S(x) Sn​(x)⇉S(x)
• 需證明,對 ∀ x 0 ∈ [ a , b ] \forall x_0\in [a,b] ∀x0​∈[a,b]，有 lim ⁡ x → x 0 S ( x ) = S ( x 0 ) \lim\limits_{x\to x_0}S(x)=S(x_0) x→x0​lim​S(x)=S(x0​)
  • ∣ S ( x ) − S ( x 0 ) ∣ ≤ ∣ S ( x ) − S n ( x ) ∣ |S(x)-S(x_0)|\le|S(x)-S_n(x)| ∣S(x)−S(x0​)∣≤∣S(x)−Sn​(x)∣ + ∣ S n ( x ) − S n ( x 0 ) ∣ + ∣ S n ( x 0 ) − S ( x 0 ) ∣ +|S_n(x)-S_n(x_0)|+|S_n(x_0)-S(x_0)| +∣Sn​(x)−Sn​(x0​)∣+∣Sn​(x0​)−S(x0​)∣
  • 由于 S n ( x ) ⇉ S ( x ) S_n(x)\rightrightarrows S(x) Sn​(x)⇉S(x)
  • S n ( x ) S_n(x) Sn​(x)連續
  • { S n ( x ) } \{S_n(x)\} {Sn​(x)}在 x 0 x_0 x0​處收斂

逐項求積

• ∑ u n ( x ) \sum u_n(x) ∑un​(x)在 [ a , b ] [a,b] [a,b]上一緻收斂
• 每一項 u n u_n un​都連續
• ⇒ \Rightarrow ⇒ ∑ ∫ a b u n ( x ) d x = ∫ a b ∑ u n ( x ) d x \sum\int_a^bu_n(x)dx=\int_a^b\sum u_n(x)dx ∑∫ab​un​(x)dx=∫ab​∑un​(x)dx

逐項求導

• 每一項 u n ′ ( x ) u'_n(x) un′​(x)連續
• x 0 ∈ [ a , b ] x_0\in [a,b] x0​∈[a,b]是 ∑ u n ( x ) \sum u_n(x) ∑un​(x)的收斂點
• ∑ u n ′ ( x ) \sum u'_n(x) ∑un′​(x)在 [ a , b ] [a,b] [a,b]上(内閉)一緻收斂
• ⇒ \Rightarrow ⇒ ∑ ( d d x u n ( x ) ) = d d x ( ∑ u n ( x ) ) \sum(\frac d{dx}u_n(x))=\frac d{dx}(\sum u_n(x)) ∑(dxd​un​(x))=dxd​(∑un​(x))

證明

• 設 ∑ u n ′ ( x ) ⇉ S ∗ ( x ) \sum u'_n(x)\rightrightarrows S^*(x) ∑un′​(x)⇉S∗(x)
• 由一緻收斂+每項導函數連續 ⇒ \Rightarrow ⇒ ∑ u n ′ ( x ) \sum u'_n(x) ∑un′​(x)連續
• 由逐項求積定理得，對 ∀ x ∈ [ a , b ] \forall x\in[a,b] ∀x∈[a,b] ∫ a x S ∗ ( t ) d t = ∑ ∫ a x u n ′ ( t ) d t \int_a^x S^*(t)dt=\sum\int_a^x u'_n(t)dt ∫ax​S∗(t)dt=∑∫ax​un′​(t)dt = ∑ u n ( x ) − ∑ u n ( a ) =\sum u_n(x)-\sum u_n(a) =∑un​(x)−∑un​(a) = S ( x ) − S ( a ) =S(x)-S(a) =S(x)−S(a)
• 兩邊求導，得 S ∗ ( x ) = S ′ ( x ) S^*(x)=S'(x) S∗(x)=S′(x)即 ∑ ( d d x u n ( x ) ) = d d x ( ∑ u n ( x ) ) \sum(\frac d{dx}u_n(x))=\frac d{dx}(\sum u_n(x)) ∑(dxd​un​(x))=dxd​(∑un​(x))