已知函数fx= log2(1-x)-log2(1+x),求f(1/2014)+f(1/2015)+f(-1/2015)+f(-1/2014)的值
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解由fx= log2(1-x)-log2(1+x)
=log2[(1-x)/(1+x)]
则f(-x)
=log2[(1-(-x))/(1+(-x))]
=log2[(1+x)/(1-x)]
=log2[(1-x)/(1+x)]^(-1)
=-log2[(1-x)/(1+x)]
=-f(x)
故f(x)是奇函数
故f(-x)+f(x)=0
即f(1/2014)+f(1/2015)+f(-1/2015)+f(-1/2014)
=f(1/2014)+f(-1/2014)+f(1/2015)+f(-1/2015)
=[f(1/2014)+f(-1/2014)]+[[f(1/2015)+f(-1/2015)]
=0+0
=0
=log2[(1-x)/(1+x)]
则f(-x)
=log2[(1-(-x))/(1+(-x))]
=log2[(1+x)/(1-x)]
=log2[(1-x)/(1+x)]^(-1)
=-log2[(1-x)/(1+x)]
=-f(x)
故f(x)是奇函数
故f(-x)+f(x)=0
即f(1/2014)+f(1/2015)+f(-1/2015)+f(-1/2014)
=f(1/2014)+f(-1/2014)+f(1/2015)+f(-1/2015)
=[f(1/2014)+f(-1/2014)]+[[f(1/2015)+f(-1/2015)]
=0+0
=0
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