【1991-12-8 分】 将函数 f ( x ) = 2 + ∣ x ∣ ( − 1 ≤ x ≤ 1 ) \displaystyle f(x)=2+|x|(-1 \leq x \leq 1) f ( x ) = 2 + ∣ x ∣ ( − 1 ≤ x ≤ 1 ) ∑ n = 1 ∞ 1 n 2 \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} n = 1 ∑ ∞ n 2 1
【1993-12-3 分】 设函数 f ( x ) = π x + x 2 ( − π < x < π ) \displaystyle f(x)=\pi x+x^{2}(-\pi<x<\pi) f ( x ) = π x + x 2 ( − π < x < π ) a 0 2 + ∑ n = 1 ∞ ( a n cos n x + b n sin n x ) \displaystyle \frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n} \cos n x+b_{n} \sin n x) 2 a 0 + n = 1 ∑ ∞ ( a n cos n x + b n sin n x ) b 3 b_{3} b 3
【1995-12-6 分】 将函数 f ( x ) = x − 1 ( 0 ≤ x ≤ 2 ) \displaystyle f(x)=x-1(0 \leq x \leq 2) f ( x ) = x − 1 ( 0 ≤ x ≤ 2 )
【2003-1-4 分】 设 x 2 = ∑ n = 0 ∞ a n cos n x ( − π ≤ x ≤ π ) \displaystyle x^{2}=\sum_{n=0}^{\infty} a_{n} \cos n x(-\pi \leq x \leq \pi) x 2 = n = 0 ∑ ∞ a n cos n x ( − π ≤ x ≤ π ) a 2 = a_{2}= a 2 =
【2008-1-11 分】 将函数 f ( x ) = 1 − x 2 ( 0 ≤ x ≤ π ) \displaystyle f(x)=1-x^{2}(0 \leq x \leq \pi) f ( x ) = 1 − x 2 ( 0 ≤ x ≤ π ) ∑ n = 1 ∞ ( − 1 ) n − 1 n 2 \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} n = 1 ∑ ∞ n 2 ( − 1 ) n − 1
【2023-1-5 分】 设 f ( x ) f(x) f ( x ) f ( x ) = 1 − x , x ∈ [ 0 , 1 ] \displaystyle f(x)=1-x,\ x \in[0,1] f ( x ) = 1 − x , x ∈ [ 0 , 1 ] f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n π x \displaystyle f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n \pi x f ( x ) = 2 a 0 + n = 1 ∑ ∞ a n cos nπ x ∑ n = 1 ∞ a 2 n = \displaystyle \sum_{n=1}^{\infty} a_{2 n}= n = 1 ∑ ∞ a 2 n =
【2024-1-5 分】 已知 f ( x ) = 1 + x \displaystyle f(x)=1+x f ( x ) = 1 + x f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n x , x ∈ [ 0 , π ] \displaystyle f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x,\ x \in[0, \pi] f ( x ) = 2 a 0 + n = 1 ∑ ∞ a n cos n x , x ∈ [ 0 , π ] lim n → ∞ n 2 sin a 2 n − 1 = \displaystyle \lim _{n \to \infty} n^{2} \sin a_{2 n-1}= n → ∞ lim n 2 sin a 2 n − 1 =
【1988-12-3 分】 设 f ( x ) f(x) f ( x ) ( − 1 , 1 ] (-1,1] ( − 1 , 1 ] f ( x ) = { 2 , − 1 < x ≤ 0 x 3 , 0 < x ≤ 1 \displaystyle f(x)=\begin{cases}2, & -1<x \leq 0 \\ x^{3}, & 0<x \leq 1\end{cases} f ( x ) = { 2 , x 3 , − 1 < x ≤ 0 0 < x ≤ 1 f ( x ) f(x) f ( x ) x = 1 x=1 x = 1
【1989-12-3 分】 设函数 f ( x ) = x 2 \displaystyle f(x)=x^{2} f ( x ) = x 2 0 ≤ x < 1 \displaystyle 0 \leq x<1 0 ≤ x < 1 S ( x ) = ∑ n = 1 ∞ b n sin n π x \displaystyle S(x)=\sum_{n=1}^{\infty} b_{n} \sin n \pi x S ( x ) = n = 1 ∑ ∞ b n sin nπ x − ∞ < x < + ∞ \displaystyle -\infty<x<+\infty − ∞ < x < + ∞ b n = 2 ∫ 0 1 f ( x ) sin n π x d x , ( n = 1 , 2 , ⋯ ) \displaystyle b_{n}=2 \int_{0}^{1} f(x) \sin n\pi x dx,\ (n=1,2, \cdots) b n = 2 ∫ 0 1 f ( x ) sin nπ x d x , ( n = 1 , 2 , ⋯ ) S ( − 1 2 ) \displaystyle S\left(-\frac{1}{2}\right) S ( − 2 1 ) − 1 2 \displaystyle -\frac{1}{2} − 2 1 − 1 4 \displaystyle -\frac{1}{4} − 4 1 1 4 \displaystyle \frac{1}{4} 4 1 1 2 \displaystyle \frac{1}{2} 2 1
【1992-12-3 分】 设 f ( x ) = { − 1 , − π < x ≤ 0 1 + x 2 , 0 < x ≤ π \displaystyle f(x)= \begin{cases}-1, & -\pi<x \leq 0 \\ 1+x^{2}, & 0<x \leq \pi\end{cases} f ( x ) = { − 1 , 1 + x 2 , − π < x ≤ 0 0 < x ≤ π 2 π 2 \pi 2 π x = π x=\pi x = π
【1999-1-3 分】 设 f ( x ) = { x , 0 ≤ x ≤ 1 2 2 − 2 x , 1 2 < x < 1 \displaystyle f(x)= \begin{cases}x, & 0 \leq x \leq \frac{1}{2} \\ 2-2 x, & \frac{1}{2}<x<1\end{cases} f ( x ) = { x , 2 − 2 x , 0 ≤ x ≤ 2 1 2 1 < x < 1 S ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n π x \displaystyle S(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n \pi x S ( x ) = 2 a 0 + n = 1 ∑ ∞ a n cos nπ x − ∞ < x < + ∞ \displaystyle -\infty<x<+\infty − ∞ < x < + ∞ a n = 2 ∫ 0 1 f ( x ) cos n π x d x ( n = 0 , 1 , 2 , ⋯ ) \displaystyle a_{n}=2 \int_{0}^{1} f(x) \cos n \pi x ~d x(n=0,1,2, \cdots) a n = 2 ∫ 0 1 f ( x ) cos nπ x d x ( n = 0 , 1 , 2 , ⋯ ) S ( − 5 2 ) \displaystyle S\left(-\frac{5}{2}\right) S ( − 2 5 ) 1 2 \displaystyle \frac{1}{2} 2 1 − 1 2 \displaystyle -\frac{1}{2} − 2 1 3 4 \displaystyle \frac{3}{4} 4 3 − 3 4 \displaystyle -\frac{3}{4} − 4 3
【2013-1-4 分】 设 f ( x ) = ∣ x − 1 2 ∣ \displaystyle f(x)=\left|x-\frac{1}{2}\right| f ( x ) = x − 2 1 b n = 2 ∫ 0 1 f ( x ) sin n π x d x ( n = 1 , 2 , ⋯ ) \displaystyle b_{n}=2 \int_{0}^{1} f(x) \sin n \pi x ~d x(n=1,2, \cdots) b n = 2 ∫ 0 1 f ( x ) sin nπ x d x ( n = 1 , 2 , ⋯ ) S ( x ) = ∑ n = 1 ∞ b n sin n π x \displaystyle S(x)=\sum_{n=1}^{\infty} b_{n} \sin n \pi x S ( x ) = n = 1 ∑ ∞ b n sin nπ x S ( − 9 4 ) = \displaystyle S\left(-\frac{9}{4}\right)= S ( − 4 9 ) = 3 4 \displaystyle \frac{3}{4} 4 3 1 4 \displaystyle \frac{1}{4} 4 1 − 1 4 \displaystyle -\frac{1}{4} − 4 1 − 3 4 \displaystyle -\frac{3}{4} − 4 3
【2025-1-5 分】 已知函数 f ( x ) = { 0 , 0 ≤ x < 1 2 x 2 , 1 2 ≤ x ≤ 1 \displaystyle f(x)= \begin{cases}0, & 0 \leq x<\frac{1}{2} \\ x^{2}, & \frac{1}{2} \leq x \leq 1\end{cases} f ( x ) = { 0 , x 2 , 0 ≤ x < 2 1 2 1 ≤ x ≤ 1 ∑ n = 1 ∞ b n sin n π x \displaystyle \sum_{n=1}^{\infty} b_{n} \sin n \pi x n = 1 ∑ ∞ b n sin nπ x S ( x ) S(x) S ( x ) ∑ n = 1 ∞ b n sin n π x \displaystyle \sum_{n=1}^{\infty} b_{n} \sin n \pi x n = 1 ∑ ∞ b n sin nπ x S ( − 7 2 ) = \displaystyle S\left(-\frac{7}{2}\right)= S ( − 2 7 ) =