将函数f（x）=1-x2（0≤x≤π）展开成余弦级数，并求∞n＝1(−1)n−1n的和．

Question

将函数f（x）=1-x²（0≤x≤π）展开成余弦级数，并求

∞

n＝1

((−1)n−1)/(n)的和．

Answer 1

将f（x）作偶周期延拓，则有b_n=0，n=1，2，…
a0＝(2)/(π)

∫

π 0

(1−x2)dx＝2(1−(π2)/(3))
an＝(2)/(π)

∫

π 0

f(x)cosnxdx
=(2)/(π)[

∫

π 0

cosnxdx−

∫

π 0

x2cosnxdx]
=(2)/(π)[0−

∫

π 0

x2cosnxdx]
=(−2)/(π)[(x2sinnx)/(n)

.

π o

−∫

π 0

(2xsinnx)/(n)dx]
=(2)/(π)(2π(−1)n−1)/(n2)
=(4(−1)n−1)/(n2)
所以f(x)＝1−x2＝(a0)/(2)+

∞

n＝1

ancosnx=1−(π2)/(3)+4

∞

n＝1

((−1)n−1)/(n2)cosnx，0≤x≤π
令x=0，有f(0)＝1−(π2)/(3)+4

∞

n＝1

((−1)n−1)/(n2)
又f（0）=1，所以

∞

n＝1

((−1)n−1)/(n2)＝(π2)/(12)