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一道数学新型函数题描述:如题 图片:未命名.bmp 望能详细解答 |
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沙发#
发布于:2011-10-25 23:50
f(x) = 1/(2^x + √2)
f(1-x) = 1/[2^(1-x) + √2) 。。。。(分子、分母同时乘以 2^x ) = 2^x/(2 + √2 * 2^x) 。。。。(分母中提取出 √2) = (2^x/√2) * (1/√2 + 2^x) = (2^x/√2) * f(x) f(x) + f(1-x) = (1+ 2^x/√2) * f(x) =[ (√2 + 2^x)/√2 ] * f(x) = [1/√2*f(x)] * f(x) = 1/√2 f(-5)+f(-4)+…+f(0)+…+f(5)+f(6) = [f(-5) + f(6)] + [f(-4) + f(5)] + [f(-3) + f(4)] + [f(-2) + f(3)] + [f(-1) + f(2)] + [f(0) + f(1)] = 6/√2 =3√2 |
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