Month: May 2021

We’re trying to find the volume of the container on the left. First, there is a need to create a system of equations representing the volumes. The linked cylinders will be filled with fluid, to neglect their volume. Let the volume of the measured container be V₁, the auxiliary container, V₂. Let their pressures be P₁ and P₂. The atmospheric pressure will be P₀. The volume pushed, and remaining volume in the main and auxiliary containers: α, β, and γ respectively. The shift of the cylinder is Δ(γ). Using the fact that pressure times volume is constant for constant temperature, we end up with the following system of equations:

V_1 \cdot P_1 = \alpha \cdot P_0 \\[8pt] V_2 \cdot P_2 = \beta \cdot P_0 \\[8pt] P_1 + P_2 = 2 P_0 \\[8pt] \operatorname \Delta \gamma = \gamma + \alpha – V_1 = \beta – V_2

We know that Δ(50mL) = 30mL in the case given, but for a full solution, a general formula will be given here. But in the general case, the known variables are γ and Δ(γ). The third equation provides the first step: divide both sides of it by P₀. This is because the value of atmospheric pressure is not relevant, only the ratios to it! We also divide the first and second equations by P₀ to complete the switch, removing the irrelevant variable to enable completion of the problem. You may ask why this variable was included in the solution: for completeness; to demonstrate that it’s unimportant. Next: split the fourth equation as shown below. This easily resolves, separating the volumes from the variables. Note that V₁ = α + Δ(γ) – γ, so removing it at present is no problem–it can easily be found later.

\alpha = V_1\cdot{P_1 \over P_0} = V_1+\operatorname\Delta\gamma-\gamma \\[8pt] \beta = V_2\cdot{P_2 \over P_0} = V_2+\operatorname\Delta\gamma \phantom{-\gamma}\\[16pt] \alpha-\alpha\cdot{P_1 \over P_0}+\operatorname\Delta\gamma = \gamma \\[8pt] \beta -\beta \cdot{P_2 \over P_0}\phantom{+\operatorname\Delta\gamma} = \gamma

Combining the last two equations, it becomes easy to find the one in terms of the other. The equation below describes the simplified relationship without knowing the volume of the auxiliary container. Know that V₂ = β + Δ(γ), and as we know that V₂ = .5L, so β = .5L – Δ(γ). To solve for this specific case–for to only solve generally and not in the specific case would be just as incomplete as would the opposite–we then fill in our known values of Δ(γ) = 30mL, γ = 50mL, and extending to β = 470mL. We get the value of α = (470mL/30mL – 2)(20mL) ≈ 27.33mL. Remembering from before that V₁ = α + Δ(γ) – γ: V₁ = ≈27.33mL – 20mL ≈ 7.333mL. The answer is then ≈7.333mL for the specific case and the equation below for the general case.

\alpha = \left(\frac\beta{\operatorname\Delta\gamma}-2\right)\cdot \left(\gamma-\operatorname\Delta\gamma\right)

We are attempting to find the average number of steps in a certain mathematical game. Let us define a function which represents the average number of steps to completion. We will have this function be called T, for time, and have T(x) be the expected number of remaining steps at the position x. A review of the formal rules of the game and the definition of this function:

n \in \mathbb{N} \\[4pt] \operatorname f : \mathbb R, \mathbb Q \to \mathbb Q \\[8pt] \operatorname f {(p, x)} = \begin{cases} p > x : x-\frac1{2n} \\[4pt] p < x : x+\frac1{2n} \\[4pt] p = x : x \end{cases} \\[48pt] \operatorname T: \mathbb Q \to \mathbb N \\[4pt] \operatorname T(0) = \operatorname T(1) = 0 \\[12pt] \operatorname T(x) = x\operatorname T\left(x-\frac1{2n}\right)\ + \ (1-x)\operatorname T\left(x+\frac1{2n}\right)\ +\ 1

We can rephrase this equation to express T(x+1/2n) – T(x) in terms of T(x) – T(x-1/2n), a much more use-full equation. Consider a T(-1/2n). This extends the space of T beyond that which actually represents f–allowing us to find T(1/2n). T(-1/2n) = -1, so T(0) – T(-1/2n) = 1. Using the equation below: T(1/2n) – T(0) is 2n/(2n-1) + 1/(2n-1) = (2n+1)/(2n-1). This may then be extended further: T(1/n) – T(1/2n) is (2n(2n-1) + 2(2n-0) + 2) / (2n-2)(2n-1). Can we go for a formula in pi and sigma notation: yes. What about a polynomial formula, or at least something less hard to calculate? NO.

\operatorname T(x)-\operatorname T\left(x-\frac1{2n}\right)\ =\ \frac1{1-x} \ +\ \frac x{1-x}\left(\operatorname T\left(x+\frac1{2n}\right) – \operatorname T(x)\right)\\[24pt] \operatorname T(x) – \operatorname T\left(x-\frac1{2n}\right)\ =\ \left((2nx)!+\sum_{k=0}^{2nx}{(2nx)!(2n-k)\over k!}\right)\div \left(\prod_{l=1}^{2nx}2n-l\right)

And finally, we can summate T(k/2n) – T((k-1)/2n) for all k from 1 to n. The answer is, therefore, the following huge formula!

\Large {\Huge\sum_{k=1}^n}\left((2nk)!+\sum_{l=0}^{2nk}{(2nk)!(2n-l)\over l!}\right) \div\left(\prod_{m=1}^{2nk}2n-m\right)