From the 10th Philippine Mathematical Olympiad:
Let $f$ be the function defined by $$f(x) = \frac{2008^{2x}}{2008 + 2008^{2x}}, \qquad x \in \mathbb{R}.$$ Prove that $$f\left(\frac{1}{2007}\right) + f\left(\frac{2}{2007}\right) + \cdots + f\left(\frac{2005}{2007}\right) + f\left(\frac{2006}{2007}\right) = 1003.$$
The given solution was:
We first show that the function satisfies the identity $f(x) + f(1 – x) = 1.$
\begin{align*}f(1 – x) &= \frac{2008^{2(1-x)}}{2008 + 2008^{2(1 – x)}} \\\\ f(1 – x) &= \frac{2008^2 2008^{-2x}}{2008 + 2008^2 2008^{-2x}} \\\\ f(1 – x) &= \frac{2008}{2008^{2x} + 2008} \\\\\\\\ f(x) + f(1 – x) &= \frac{2008^{2x}}{2008 + 2008^{2x}} + \frac{2008}{2008^{2x} + 2008} \\\\ f(x) + f(1 – x) &= 1\end{align*}
Pairing off the terms of the left-hand side of the desired equality into $$\left[f\left(\frac{1}{2007}\right) + f\left(\frac{2006}{2007}\right)\right] + \cdots + \left[f\left(\frac{1003}{2007}\right) + f\left(\frac{1004}{2007}\right)\right],$$ and applying the above identity solve the problem.
Now, this is not easily noticeable, at least for me. What I tried to do was approximate the sum through the integral $$I(a,b,c,d) = \int_c^d \frac{a^x}{a^x + b}\,dx$$ which, when solved, is equal to $$I(a,b,c,d) = \frac{1}{\ln a}\ln\left(\frac{a^d + b}{a^c + b}\right)$$ as I thought that $2006$ subdivisions is enough for the integral to approximate. The result of the integral will be rounded up to compensate for the difference. The given values in the problem are $a = 2008^{2/2007}$, $b = 2008$, $c = 1$, and $d = 2006$. By substitution, it should give us $$I = \frac{1}{\ln(2008^{2/2007})}\ln\left(\frac{2008^{4012/2007} + 2008}{2008^{2/2007} + 2008}\right).$$ Solving this gives us $$I = \frac{2005}{2} = 1002.5$$ and rounding this up gives us $1003$ as desired.1
My questions will be: How valid is this solution, and are there alternative solutions that does not use the given one?
Additional information:
[1] Solution for $I = \frac{1}{\ln(2008^{2/2007})}\ln\left(\frac{2008^{4012/2007} + 2008}{2008^{2/2007} + 2008}\right)$
We know that $\ln (p/q) = \ln p – \ln q$. Hence, $$\ln\left(\frac{2008^{4012/2007} + 2008}{2008^{2/2007} + 2008}\right) = \ln(2008^{4012/2007} + 2008) – \ln(2008^{2/2007} + 2008).$$ Let $u = 2008^{4012/2007} + 2008$ and let $v = 2008^{2/2007} + 2008$. This means that we are solving for $\ln u – \ln v$.
Notice that $u = 2008^{4012/2007} + 2007$ is equivalent to $u = 2008(2008^{2005/2007} + 1)$. Then, $$\ln u = \ln 2008 + \ln(2008^{2005/2007} + 1).$$ Also, $v = 2008^{2/2007} + 2008$ is equivalent to $v = 2008(2008^{2005/2007} + 1)(2008^{-2005/2007})$. Then, \begin{align*}\ln v &= \ln 2008 + \ln(2008^{2005/2007} + 1) – \ln(2008^{-2005/2007}) \\ \ln v &= \ln 2008 + \ln(2008^{2005/2007} + 1) – \frac{2005}{2007}\ln 2008.\end{align*}
Solving for $\ln u – \ln v$, \begin{align*}\ln u – \ln v &= \ln 2008 + \ln(2008^{2005/2007} + 1) – \left(\ln 2008 + \ln(2008^{2005/2007} + 1) – \frac{2005}{2007}\ln 2008\right) \\ \ln u – \ln v &= \ln 2008 + \ln(2008^{2005/2007} + 1) – \ln 2008 – \ln(2008^{2005/2007} + 1) + \frac{2005}{2007}\ln 2008 \\ \ln u – \ln v &= \frac{2005}{2007}\ln 2008\end{align*}
We are solving for $$I = \frac{1}{\ln(2008^{2/2007})}\ln\left(\frac{2008^{4012/2007} + 2008}{2008^{2/2007} + 2008}\right).$$ We know that $$\ln\left(\frac{2008^{4012/2007} + 2008}{2008^{2/2007} + 2008}\right) = \frac{2005}{2007}\ln 2008,$$ hence, \begin{align*}I &= \frac{1}{\ln(2008^{2/2007})}\cdot\frac{2005}{2007}\ln 2008 \\ I &= \frac{2005}{2007}\cdot \frac{1}{\frac{2}{2007}\ln 2008}\cdot \ln 2008 \\ I &= \frac{2005}{2007} \cdot \frac{2007}{2} \\ I &= \frac{2005}{2}. \qquad \blacksquare\end{align*}
Best Answer
My solution is invalid. Implicitly, I am trying to solve for the lower bound of the sum since it is the right Riemann sum of the integral I calculated. Even though the error is less than 1 which makes it okay to round up, it still doesn't prove that the sum is equal to $1003$.
In short, I just proved that $$1002.5 \leq \sum_{i = 1}^{2006}f\left(\frac{i}{2007}\right) \leq 1003$$ but not $$\sum_{i = 1}^{2006}f\left(\frac{i}{2007}\right) = 1003.$$