In the given equation, we need to understand how $\Delta P_1 = \frac{4}{\sqrt{5}}$ is derived from the preceding steps.

Step-by-step analysis:

1. Hamiltonian setup:
   The equation presented is a Hamiltonian function:
   $H(Q_1, Q_2, P_1, P_2) = \frac{2}{5} P_2^2 + 1$
   for the separatrix, which simplifies the system at specific values.

2. Hamiltonian equation: The key equation is:
   $1 = \frac{5}{2} \left(P_1 + \frac{1}{5} P_2 \right)^2 + \cos(Q_1)$ When $Q_1 = \pi$, we know $\cos(\pi) = -1$, so the equation becomes: $1 = \frac{5}{2} \left(P_1 + \frac{1}{5} P_2 \right)^2 - 1$ Simplifying this gives: $2 = \frac{5}{2} \left(P_1 + \frac{1}{5} P_2 \right)^2$ Multiply both sides by $\frac{2}{5}$ to isolate the square: $\left(P_1 + \frac{1}{5} P_2 \right)^2 = \frac{2}{5}$

3. Solving for $P_1$: Taking the square root of both sides yields: $P_1 + \frac{1}{5} P_2 = \pm \frac{2}{\sqrt{5}}$ This shows that the difference $\Delta P_1$, which represents the variation or width in $P_1$, is: $\Delta P_1 = \frac{4}{\sqrt{5}}$ because the range between the positive and negative solutions is $2 \times \frac{2}{\sqrt{5}}$.

Thus, $\Delta P_1 = \frac{4}{\sqrt{5}}$ comes from the fact that the width in $P_1$ is derived from doubling the value of $\frac{2}{\sqrt{5}}$, which arises from solving the quadratic expression.

Do you want more details on any specific part? Here are 5 related questions:

1. How does the separatrix equation affect the system's dynamics?
2. What is the physical meaning of $\Delta P_1$?
3. Can you explain how Hamiltonian systems relate to this problem?
4. Why is $Q_1 = \pi$ a significant point for this solution?
5. How is the saddle point influenced by the determinant condition?

Tip: When solving quadratic equations, remember to account for both the positive and negative square roots to avoid missing critical solutions.