Let $x_0=1$, and let $\delta$ be some constant satisfying $0<\delta<1$. Iteratively, for $n=0,1,2,\dots$, a point $x_{n+1}$ is chosen uniformly form the interval $[0,x_n]$. Let $Z$ be the smallest value of $n$ for which $x_n<\delta$. Find the expected value of $Z$, as a function of $\delta$.
1. Define \( f(x) \) as the expected value of \( Z \) if \( x_0 = x \) rather than \( 1 \). If \( x < \delta \), then \( f(x) = 0 \) because \( Z = 0 \) in this case.
2. For \( x \geq \delta \), we have:
   \[
   f(x) = 1 + \frac{1}{x} \int_{0}^{x} f(\tau) \, d\tau
   \]
   This is because the next point \( x_{n+1} \) is chosen uniformly from the interval \([0, x_n]\), and we add 1 to account for the current step.
3. Differentiate both sides of the equation with respect to \( x \):
   \[
   f'(x) = \frac{d}{dx} \left( 1 + \frac{1}{x} \int_{0}^{x} f(\tau) \, d\tau \right)
   \]
4. Using the Leibniz rule for differentiation under the integral sign, we get:
   \[
   f'(x) = \frac{d}{dx} \left( 1 \right) + \frac{d}{dx} \left( \frac{1}{x} \int_{0}^{x} f(\tau) \, d\tau \right)
   \]
   \[
   f'(x) = 0 + \left( -\frac{1}{x^2} \int_{0}^{x} f(\tau) \, d\tau + \frac{f(x)}{x} \right)
   \]
   \[
   f'(x) = -\frac{1}{x^2} \int_{0}^{x} f(\tau) \, d\tau + \frac{f(x)}{x}
   \]
5. Simplify the expression:
   \[
   f'(x) = -\frac{1}{x} \left( \frac{1}{x} \int_{0}^{x} f(\tau) \, d\tau \right) + \frac{f(x)}{x}
   \]
   \[
   f'(x) = -\frac{1}{x} (f(x) - 1) + \frac{f(x)}{x}
   \]
   \[
   f'(x) = \frac{1}{x}
   \]
6. Integrate both sides with respect to \( x \):
   \[
   \int f'(x) \, dx = \int \frac{1}{x} \, dx
   \]
   \[
   f(x) = \ln x + c
   \]
   where \( c \) is the constant of integration.
7. To find \( c \), use the boundary condition \( f(\delta) = 0 \):
   \[
   f(\delta) = \ln \delta + c = 0
   \]
   \[
   c = -\ln \delta
   \]
8. Therefore, the function \( f(x) \) is:
   \[
   f(x) = \ln x - \ln \delta = \ln \left( \frac{x}{\delta} \right)
   \]
9. Finally, evaluate \( f(1) \) to find the expected value of \( Z \) when \( x_0 = 1 \):
   \[
   f(1) = \ln \left( \frac{1}{\delta} \right) = -\ln \delta + \ln 1 = -\ln \delta
   \]
   \[
   f(1) = 1 - \ln \delta
   \]

The final answer is \(\boxed{1 - \ln \delta}\).