E(p)=p×1+(1−p)×(1+E(p+0.05))，when p=0.35

The given equation is:

\[
E(p) = p \times 1 + (1 - p) \times (1 + E(p + 0.05))
\]

We need to find the value of \( E(p) \) when \( p = 0.35 \).

Let's substitute \( p = 0.35 \) into the equation:

\[
E(0.35) = 0.35 \times 1 + (1 - 0.35) \times (1 + E(0.35 + 0.05))
\]

This simplifies to:

\[
E(0.35) = 0.35 + 0.65 \times (1 + E(0.40))
\]

\[
E(0.35) = 0.35 + 0.65 \times (1 + E(0.40))
\]

Now, we need to find \( E(0.40) \) using the same formula:

\[
E(0.40) = 0.40 \times 1 + (1 - 0.40) \times (1 + E(0.45))
\]

\[
E(0.40) = 0.40 + 0.60 \times (1 + E(0.45))
\]

Similarly, \( E(0.45) \) can be found as:

\[
E(0.45) = 0.45 \times 1 + (1 - 0.45) \times (1 + E(0.50))
\]

\[
E(0.45) = 0.45 + 0.55 \times (1 + E(0.50))
\]

This process continues recursively. To solve for \( E(0.35) \), we'd typically need to assume a base case or approximate the value for higher \( p \).

If the recursion is simplified at higher \( p \), like assuming \( E(p) = 0 \) when \( p \) is sufficiently large, we can compute backward.

Would you like me to calculate this recursively for an approximation?

Explore the recursive calculation of the expectation function E(p) for p = 0.35 using the formula E(p) = p × 1 + (1 - p) × (1 + E(p + 0.05)). This detailed solution demonstrates the application of recursion and algebra in expectation calculations, ideal for undergraduate-level students.

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