Solving the Equation Q 1 - p(p * Q^(z-1))

When faced with the equation Q 1 - p(p * Q^(z-1)), I initially turned to the Taylor series approximations to solve it. However, this approach might not be the most straightforward, especially when there are more than one unknown in the equation. Let's break it down step by step.

Background and Initial Steps

As mentioned in the comments, one of the key conditions is that Q is a probability, so 0 ≤ Q ≤ 1. The equation becomes particularly interesting when we explore its behavior under specific values of p and z.

Exploring the Equation with p 1/2 and z 4

Setting p 1/2 and z 4, the equation transforms into a more complex form:

Q 1/2 * Q^3 / 2

Rewriting this, we obtain:

Q^3 - 2Q - 1 0

Compared to the simple case, we now have multiple potential solutions:

Q 1 Q (sqrt{5} - 1)/2 Q (-sqrt{5} - 1)/2

However, as sqrt{5} is not a rational number, our original formula doesn't work for all solutions. Let's test one of the solutions:

1 - (2(1/2)*4 - 1 - 2) / ((1/2)*4 - 1*4 - 2) 1 - (3 - 2) / (1/2 * 3 * 2) 1 - 1/3 2/3

Derivatives and Multiple Solutions

Let's examine the derivative to understand the number of solutions. The derivative of 1 - p(p * Q^(z-1)) - Q with respect to p is:

z-1 * p * Q^(z-2) - 1

This derivative is monotonously increasing from negative to positive for Q 0 to Q 1 if z-1 * p > 1. This means there will be a unique solution between 0 and 1 when this condition is met. Otherwise, Q 1 is the only solution.

Limit Analysis

As we increase the value of z, the behavior of the equation becomes interesting. For large values of z, the term Q^(z-1) approaches 0 for any Q in the range [0, 1], except when Q 1. Therefore, the equation simplifies to:

displaystylelim_{zrightarrow infty} 1 - p(p * Q^(z-1)) 1

Substituting this into the equation, we get:

displaystylelim_{zrightarrow infty} (1 - (2pz-1 - 2) / (pz-1 * z-2)) 0

Thus, as z approaches infinity, the formula becomes less accurate, approaching 1.

displaystylelim_{zrightarrow infty} 1 - (2pz-1 - 2) / (pz-1 * z-2) 1

Conclusion

This analysis shows that while the formula is useful in specific cases, it is not a robust general solution for more complex problems involving probabilities and higher-order equations. The presence of multiple solutions and the behavior of the equation as z increases indicate that more precise methods are needed for practical applications.