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Thread: Sum of functions

  1. #1
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    Sum of functions

    Let $\displaystyle f(x)=\frac{x^{2}} {1+x^2}$

    Compute:

    $\displaystyle f(\frac{1}{1}) + f(\frac{1}{2})+...+f(\frac{1}{10}) + f(\frac{2}{1})+f(\frac{2}{2})+...+f(\frac{2}{10})+ f(\frac{3}{1})+f(\frac{3}{2})+...+f(\frac{3}{10})+ ...+f(\frac{10}{1})+f(\frac{10}{2})+...+f(\frac{10 }{10})$

    Thanks in advance!
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  2. #2
    Junior Member RaisinBread's Avatar
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    Notice that $\displaystyle f(\frac{1}{1})=f(\frac{2}{2})=f(\frac{3}{3})=...=f (\frac{10}{10})=f(1)$
    So 10 out of your 100 terms give f(1)
    $\displaystyle f(\frac{1}{2})=f(\frac{2}{4})=f(\frac{3}{6})=f(\fr ac{4}{8})=f(\frac{5}{10})$
    and
    $\displaystyle f(\frac{2}{1})=f(\frac{4}{2})=f(\frac{6}{3})=f(\fr ac{8}{4})=f(\frac{10}{5})$
    5 of these terms give f(0.5) and 5 more give f(2).

    Already, we have regrouped 20 out of 100 of these terms in 3 simple operations: 10f(1)+5f(0.5)+5f(2).

    I suspect the 80 remaining could also be regrouped like this!
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  3. #3
    Junior Member mathfun's Avatar
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    Notice that $\displaystyle $$f\left( x \right) + f\left( {\tfrac{1}{x}} \right) = 1$$$
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