神马作文网无穷级数求和.为什么设x=1/3和x=根号1/3答案不同
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无穷级数求和
.为什么设x=1/3和x=根号1/3答案不同
.为什么设x=1/3和x=根号1/3答案不同
(1) 1/(1-x) = ∑{0 ≤ n} x^n,求导得1/(1-x)² = ∑{1 ≤ n} n·x^(n-1) = ∑{0 ≤ n} (n+1)·x^n.
因此2/(1-x)²-1/(1-x) = ∑{0 ≤ n} (2n+1)·x^n.
即∑{0 ≤ n} (2n+1)·x^n = 2/(1-x)²-1/(1-x) = (1+x)/(1-x)².
代入x = 1/3得∑{0 ≤ n} (2n+1)/3^n = (1+1/3)/(1-1/3)² = 3.
用∑{0 ≤ n} (2n+1)·x^(2n)求是一样的.
1/(1-x²) = ∑{0 ≤ n} x^(2n),故x/(1-x²) = ∑{0 ≤ n} x^(2n+1).
求导得(1+x²)/(1-x²)² = ∑{0 ≤ n} (2n+1)·x^(2n).
代入x = 1/√3仍得3.
(2) 1/(1+x) = ∑{0 ≤ n} (-1)^n·x^n,故1/(1+x²) = ∑{0 ≤ n} (-1)^n·x^(2n).
积分得arctan(x) = ∑{0 ≤ n} (-1)^n·x^(2n+1)/(2n+1) (取x = 0可知积分常数为0).
于是arctan(x)/x = ∑{0 ≤ n} (-1)^n·x^(2n)/(2n+1).
代入x = 1/√3得∑{0 ≤ n} (-1)^n/((2n+1)·3^n) = √3·arctan(1/√3) = √3·π/6.
因此2/(1-x)²-1/(1-x) = ∑{0 ≤ n} (2n+1)·x^n.
即∑{0 ≤ n} (2n+1)·x^n = 2/(1-x)²-1/(1-x) = (1+x)/(1-x)².
代入x = 1/3得∑{0 ≤ n} (2n+1)/3^n = (1+1/3)/(1-1/3)² = 3.
用∑{0 ≤ n} (2n+1)·x^(2n)求是一样的.
1/(1-x²) = ∑{0 ≤ n} x^(2n),故x/(1-x²) = ∑{0 ≤ n} x^(2n+1).
求导得(1+x²)/(1-x²)² = ∑{0 ≤ n} (2n+1)·x^(2n).
代入x = 1/√3仍得3.
(2) 1/(1+x) = ∑{0 ≤ n} (-1)^n·x^n,故1/(1+x²) = ∑{0 ≤ n} (-1)^n·x^(2n).
积分得arctan(x) = ∑{0 ≤ n} (-1)^n·x^(2n+1)/(2n+1) (取x = 0可知积分常数为0).
于是arctan(x)/x = ∑{0 ≤ n} (-1)^n·x^(2n)/(2n+1).
代入x = 1/√3得∑{0 ≤ n} (-1)^n/((2n+1)·3^n) = √3·arctan(1/√3) = √3·π/6.
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