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

1. ## 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

2. Originally Posted by ice_syncer
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 $f\left(\dfrac{r}{m+1}\right)$

3. Also, since there is no "k" on the left, the right cannot be "mk/2".

4. $f(\frac{r}{m+1})$ works.
Write out the sum in two ways,
$\sum f(x)$ and $\sum k-f(1-x)$, add and divide by 2.

5. it is, r/(m+1)

6. Yes, r/(m+1) gives the required result, (r/m)+1 doesn't.

7. so? give the proof

8. $\displaystyle S=\sum^{m}_{r=1}f\left(\frac{r}{m+1}\right)=f\left (\frac{1}{m+1}\right)+f\left(\frac{2}{m+1}\right)+ \dots+f\left(\frac{m}{m+1}\right).$

$\displaystyle x\equiv\frac{r}{m+1}\quad\mbox{ so }\quad 1-x=1-\frac{r}{m+1}=\frac{m+1-r}{m+1
}.$

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

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.

9. Just noticed I missed an $f$ on the last line .