# $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}.$ What is the value of $f(1/2009) + f(2/2009) + … + f(2008/2009)$?

Here's the question.

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

Then what is the value of $f(1/2009) + f(2/2009) + ... + f(2008/2009)$ ?

All I could think of doing was to add and subtract 1 in the numerator of the function, to get the value of the sum as 2008 - {something}.

Hints??

EDIT: please note the correction. $f(x) = \frac {e^{2x-1}} {(1+e^{2x-1})}$ and NOT $\frac {e^{2x-1}} {(1-e^{2x-1})}$

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Hint: Consider $f(x) + f(1-x)$.

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Got the answer! Thanks!!!! – Parth Thakkar Apr 17 '13 at 15:24
Great Hint! +1... – DonAntonio Apr 17 '13 at 15:25
@CalvinLin If you don't mind me asking, what is the difference between brilliant.org and sites such as art of problem solving? – AlanH Jun 3 '13 at 21:55

I'd like to just bring about a curiosity, now that we have Calvin's very clever observation that leads to the exact result. But the sum looks a lot like a Riemann sum to me, so I went forth:

$$\sum_{k=0}^{2008} f\left(\frac{k}{2009}\right) \approx 2009 \int_0^1 dx \: f(x)$$

Note of course that the sum on the LHS goes from $k=0$, and not $k=1$ as specified in the OP. Keep that in mind.

So the integral is easily evaluated:

$$\int_0^1 dx \: f(x) = \frac{1}{2} \log{\left(\frac{1+e}{1+(1/e)}\right)} = \frac{1}{2}$$

Therefore

$$\sum_{k=0}^{2008} f\left(\frac{k}{2009}\right) \approx \frac{2009}{2}$$

But as I included the $k=0$ term in this sum, let's subtract it out to get the correct approximation:

$$\sum_{k=1}^{2008} f\left(\frac{k}{2009}\right) \approx \frac{2009}{2} - \frac{1}{1+e}$$

The exact value of the sum is $1004$, from which the above approximate result has a relative error of

$$\frac{e}{2 (1+e) 2008} \sim 0.02\%$$

Just saying.

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