Question

For $0 \leq x \leq 1,$ the function $f(x)$ satisfies the relations $f\left(\dfrac x{x+1}\right) = \dfrac{f(x)}2$ and $f(1-x) = 1 - f(x).$ What is the value of the expression
\[f\left(\dfrac23\right) + f\left(\dfrac25\right) + f\left(\dfrac27\right) + \dots + f\left(\dfrac2{2n+1}\right) + \dotsb\,?\]
Express your answer as a common fraction.

Answer 1

To solve for \( f\left(\frac{2}{2n+1}\right) + f\left(\frac{2}{2n-1}\right) + f\left(\frac{2}{2n+1}\right) + \cdots \), we analyze the functional equations provided.

1. **First Equation**:

The equation \( f\left(\frac{x}{x+1}\right) = \frac{f(x)}{2} \) allows us to explore how the function behaves under certain transformations and specific values.

Let's take \( x = 1 \):
\[
f\left(\frac{1}{2}\right) = \frac{f(1)}{2}.
\]

Now, since \( f(1) \) is not explicitly known, let's denote \( f(1) = c \). We have:
\[
f\left(\frac{1}{2}\right) = \frac{c}{2}.
\]

2. **First Equation Evaluation**:

We will evaluate \( f\left(\frac{1}{2}\right) \) again using \( x = 0 \):
\[
f(0) = \frac{f(0)}{2} \implies f(0) = 0.
\]

This indicates \( f(0) = 0 \).

3. **Exploring \( f(1-x) = 1 - f(x) \)**:

Using \( x = \frac{1}{2} \):
\[
f\left(1 - \frac{1}{2}\right) = f\left(\frac{1}{2}\right) \implies f\left(\frac{1}{2}\right) = 1 - f\left(\frac{1}{2}\right).
\]
Adding both sides gives:
\[
2f\left(\frac{1}{2}\right) = 1 \implies f\left(\frac{1}{2}\right) = \frac{1}{2}.
\]
Since \( f(1) = c \) and \( f\left(\frac{1}{2}\right) = \frac{c}{2} \), it follows that \( \frac{c}{2} = \frac{1}{2} \), thus \( c = 1 \) and we have \( f(1) = 1 \).

4. **Finding other values**:

Continuing with \( f\left(\frac{1}{2}\right) = \frac{1}{2} \), substituting back, we have:
\[
f\left(\frac{x}{x+1}\right) = \frac{f(x)}{2} \implies f(1) = 1 \implies f(0) = 0.
\]

Now consider \( x \) values in the form of \( \frac{a}{b} \), where \( a+b \) is constant. For example:
\[
f\left(\frac{2}{3}\right): x = \frac{2}{3} \Rightarrow f\left(\frac{2/3}{(2/3)+1}\right) = f\left(\frac{2/3}{(5/3)}\right) = f\left(\frac{2}{5}\right) = \frac{f\left(\frac{2}{3}\right)}{2}.
\]
And similarly,
\[
f\left(\frac{2}{5}\right) = \frac{2f\left(\frac{2}{3}\right)}{4}.
\]

5. **Values for series**:

Summarizing \( f\left(\frac{2}{2n+1}\right) + f\left(\frac{2}{2m+1}\right) \):

Each step leads to \( f\) being symmetric and averaging out values derived \( f(a) + f(1-a) = 1\).

Finally calculating:
\[
f\left(\frac{2}{3}\right) = \frac{1}{2},
f\left(\frac{2}{5}\right) = \frac{1}{3} \, \text{to derive the series}.
\]

In conclusion, as by symmetry for geometric sums leading to pairs averaging to \( 1 \) towards "halves":
Assuming \( n \to \infty \) the series converges yielding:
\[
\sum f\left(\frac{2}{2n+k}\right) \text{ emphasizing the common pattern, sums lead leading toward } 1.
\]

The series indeed yields:
\(
\text{ Resultant value: } \sum = 1.
\)

Thus the answer for the series calculated is:
\[
\boxed{1}.
\]