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

**ice_syncer** · September 29th 2010, 10:30 AM

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

**Plato** · September 29th 2010, 11:08 AM

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 $f\left(\dfrac{r}{m+1}\right)$

**HallsofIvy** · September 30th 2010, 03:46 AM

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

**BobP** · September 30th 2010, 04:41 AM

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

**ice_syncer** · October 4th 2010, 11:18 AM

it is, r/(m+1)

**BobP** · October 5th 2010, 07:36 AM

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

**ice_syncer** · October 10th 2010, 11:44 AM

so? give the proof

**BobP** · October 10th 2010, 12:58 PM

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

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

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

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.

**BobP** · October 10th 2010, 01:06 PM

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

.

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