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Math Help - Help me with this proof regarding the sum series involving this function f(x)

  1. #1
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    Help me with this proof regarding the sum series involving this function f(x)

    the function f(x) is such that, f(x) + f(1-x) = k where k is a constant, then , prove that SIGMA(r=1,m)f(r/m+1) = mk/2
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  2. #2
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    Quote Originally Posted by ice_syncer View Post
    the function f(x) is such that, f(x) + f(1-x) = k where k is a constant, then , prove that SIGMA(r=1,m)f(r/m+1) = mk/2
    Is the function f\left(\dfrac{r}{m}+1\right) or f\left(\dfrac{r}{m+1}\right)
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  3. #3
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    Also, since there is no "k" on the left, the right cannot be "mk/2".
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  4. #4
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    f(\frac{r}{m+1}) works.
    Write out the sum in two ways,
    \sum f(x) and \sum k-f(1-x), add and divide by 2.
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  5. #5
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    it is, r/(m+1)
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  6. #6
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    Yes, r/(m+1) gives the required result, (r/m)+1 doesn't.
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  7. #7
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    so? give the proof
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  8. #8
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    \displaystyle S=\sum^{m}_{r=1}f\left(\frac{r}{m+1}\right)=f\left  (\frac{1}{m+1}\right)+f\left(\frac{2}{m+1}\right)+  \dots+f\left(\frac{m}{m+1}\right).

    \displaystyle x\equiv\frac{r}{m+1}\quad\mbox{ so }\quad 1-x=1-\frac{r}{m+1}=\frac{m+1-r}{m+1<br />
}.


    \displaystyle f(x)+f(1-x)=k\quad\mbox{ so }\quad f(x)=k-f(1-x),

    so alternatively,

    \displaystyle S=\sum^{m}_{r=1}\left\{k-f\left(\frac{m+1-r}{m+1}\right)\right\}

    \displaystyle=\left\{k-\left(\frac{m}{m+1}\right)\right\}+\left\{k-f\left(\frac{m-1}{m+1}\right)\right\}+\dots+\left\{k-f\left(\frac{1}{m+1}\right)\right\}.

    Now add the two expressions for S together and divide by two.
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  9. #9
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    Just noticed I missed an f on the last line .
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