Question

已知函数f(x)=x²/(1+x²)

已知函数f(x)=x²/(1+x²)
求f(2)与f(1/2)，你能发现f(n)与f(1/n)有什么关系？证明
求f(1)+f(2)+f(3)+... ...+f(2009)+f(1/2)+f(1/3)+... ...+f(1/2009）=?

Answer 1

f(2)=4/5 f(1/2)=1/5
f(n)=n^2/(1+n^2)
f(1/n)=(1/n)^2/(1+1/n^2)=1/(1+n^2)
f(n)+f(1/n)=1

f(1)+f(2)+f(3)+... ...+f(2009)+f(1/2)+f(1/3)+... ...+f(1/2009）=f(1)+f(2)+f(1/2)+f(3)+f(1/3)+.....+f(2009)+f(1/2009)=f(1)+2008=1/2+2008=2008.5

Answer 2

f(2)=4/(1+4)=4/5
f(1/2)=(1/4)/[1+(1/4)]=1/5
f(2)+f(1/2)=4/5+1/5=1

一般的：
f(x)+f(1/x)
=x^2/(1+x^2)+(1/x^2)/[1+(1/x)^2]第二个分式分子分母同乘以x^2
=x^2/(1+x^2)+1/(1+x^2)
=(x^2+1)/(1+x^2)
=1

f(1)=1/(1+1)=1/2

f(1)+f(2)+...+f(2009)+f(1/2)+f(1/3)+...+f(1/2009)
=f(1)+[f(2)+f(1/2)]+[f(3)+f(1/3)]+...+[f(2009)+f(1/2009)]
=1/2+2008
=2008又1/2

Answer 3

f(2)=4/5 f(1/2)=1/5
f(n)=n^2/(1+n^2)
f(1/n)=(1/n)^2/(1+1/n^2)=1/(1+n^2)
f(n)+f(1/n)=1

f(1)+f(2)+f(3)+... ...+f(2009)+f(1/2)+f(1/3)+... ...+f(1/2009）=f(1)+f(2)+f(1/2)+f(3)+f(1/3)+.....+f(2009)+f(1/2009)=f(1)+2008=1/2+2008=2008.5

Answer 4

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