function value

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\hspace{-16}$If $\bf{f(x)=\frac{x^2}{1+x^2}.}$ Then Determine value of the following expression\\\\\\ $\bf{f\left(\frac{1}{2000}\right)+f\left(\frac{2}{2000}\right)+...+f\left(\frac{1999}{2000}\right)+f\left(\frac{2000}{2000}\right)+f\left(\frac{2000}{1999}\right)+...+f\left(\frac{2000}{1}\right)}$

#Mathematics #Calculus

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Shaswata Roy ·2013-02-28 06:48:34

\inline \dpi{200} f(x)+f(\frac{1}{x}) = \frac{x^{2}}{1+x^{2}}+\frac{\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}} = \frac{x^{2}+1}{x^{2}+1} = 1

Required sum = \inline \dpi{200} [f(\frac{1}{2000})+f(\frac{2000}{1})]+[f(\frac{2}{2000})+f(\frac{2000}{2})]......+f(\frac{2000}{2000})

= 1999 + 1/2
=1999.5

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man111 singh Yes .
Upvote·0· Reply ·2013-02-28 07:09:32
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