1 (1+2^-0.5)(1+2^-0.25)(1+2^-0125)2 已知f(x)=a^x/(a^x+根号a),求f(1/10)+f(2/10)+...+f(9/10)的值
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![1 (1+2^-0.5)(1+2^-0.25)(1+2^-0125)2 已知f(x)=a^x/(a^x+根号a),求f(1/10)+f(2/10)+...+f(9/10)的值](/uploads/image/z/2661972-60-2.jpg?t=1+%281%2B2%5E-0.5%29%281%2B2%5E-0.25%29%281%2B2%5E-0125%292+%E5%B7%B2%E7%9F%A5f%28x%29%3Da%5Ex%2F%28a%5Ex%2B%E6%A0%B9%E5%8F%B7a%29%2C%E6%B1%82f%281%2F10%29%2Bf%282%2F10%29%2B...%2Bf%289%2F10%29%E7%9A%84%E5%80%BC)
1 (1+2^-0.5)(1+2^-0.25)(1+2^-0125)2 已知f(x)=a^x/(a^x+根号a),求f(1/10)+f(2/10)+...+f(9/10)的值
1 (1+2^-0.5)(1+2^-0.25)(1+2^-0125)
2 已知f(x)=a^x/(a^x+根号a),求f(1/10)+f(2/10)+...+f(9/10)的值
1 (1+2^-0.5)(1+2^-0.25)(1+2^-0125)2 已知f(x)=a^x/(a^x+根号a),求f(1/10)+f(2/10)+...+f(9/10)的值
=(1+2^-0.5)(1+2^-0.25)(1+2^-0.125)(1-2^-0.125)/(1-2^-0.125)
=(1+2^-0.5)(1+2^-0.25))(1-2^-0.25)/(1-2^-0.125)
=(1+2^-0.5)(1-2^-0.5)/(1-2^-0.125)
=(1-2^-1)/(1-2^-0.125)
=1/(1-2^-0.125)
先求f(x)+f(1-x)=1,再用倒序相加法即可