Is there a closed form for the recurrence $V_{n+1}={V_n+\Delta V\over 1+{V_n\cdot \Delta V/C^2}}$, for constants $\Delta V$ and $C$?

Question

I was wondering if the following recurrence formula has a closed form:

$$V_{n+1}={V_n+\Delta V\over 1+{V_n\cdot \Delta V\over C^2}}$$

where $\Delta V$ and $C$ are positive constants, $V_n$ is the velocity of the $n$-the inertial frame and the primary velocity $V_0$ is given (take it $0$ if needed).

Attempt

This sequence obviously tends to $C$ (the speed-of-light supremum of speeds of observations), so I naturally tried to crack it using $$e_n=V_n-C$$but I failed. Any idea is appreciated.

Note

The above rule determines the Relativistic Velocity-addition Formula where $V_n$ is supposed to be the velocity of the inertial frame $2$ that is moving with respect to us (inertial frame $1$) and $\Delta V$ is an increase in the speed of the moving object (or we can assume it as the relative speed of object in the inertial frame 2). My work basis is the Lorentz Transformation.

Answer 1

Defining $U_n := V_n/C$ and $D := \Delta V/C$, we can write the simpler-looking recurrence $$U_{n+1} = \frac{D + U_n}{1 + D\,U_n} \tag{1}$$ which bears a resemblance to the angle-addition formula for hyperbolic tangent: $$\tanh(a+b) = \frac{\tanh a + \tanh b}{1 + \tanh a\,\tanh b} \tag{2}$$ Thus, if we further define $u_n := \operatorname{arctanh U_n}$ and $d := \operatorname{arctanh D}$, then we have $$\tanh u_{n+1} = U_{n+1} = \frac{D+U_n}{1+D\,U_n} = \frac{\tanh d + \tanh u_n}{1 + \tanh d\tanh u_n} = \tanh(d+u_n) \tag{3}$$ so that $u_{n+1} = d + u_n$. Consequently, taking $V_0 = U_0 = u_0 = 0$, we have $u_n = n d$, which gives

$$V_n= C\,\tanh\left( n \operatorname{arctanh}\frac{\Delta V}{C} \right) \tag{$\star$}$$

As $n$ grows without bound (with $\Delta V/C > 0$), the $\tanh$ factor approaches $1$, so that $V_n$ approaches $C$ (as OP has observed).

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Using the exponential definition of $\tanh$, we can re-write $(\star)$ as $$V_n = C \frac{e^{nd}-e^{-nd}}{e^{nd}+e^{-nd}} = C\frac{\left(e^d\right)^n-\left(e^d\right)^{-n}}{\left(e^d\right)^n+\left(e^d\right)^{-n}} \tag{4}$$ Note that $$e^d = \exp \operatorname{arctanh} D = \exp \left(\frac12\,\log\frac{1+D}{1-D}\right) = \sqrt{\frac{1+D}{1-D}} \tag{5}$$ Substituting $(5)$ into $(4)$ and simplifying ultimately gives

$$V_n = C\frac{(1+D)^n-(1-D)^n}{(1+D)^n+(1-D)^n} = C\frac{(C+\Delta V)^n-(C-\Delta V)^n}{(C+\Delta V)^n + (C-\Delta V)^n} \tag{$\star\star$}$$

which agrees with @Hushus46's answer.

Answer 2

As I have noted in the comment, Wolfram Alpha gives the closed form, with $k = \Delta V$:

$$V_n = \frac{C_1 c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + C_1 (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n}$$

Taking $V_0 = 0$, we can see

$$ 0 = \frac{C_1 c(c+k) - c(c-k)}{(c-k) + C_1 (c+k) }$$

which is undefined if $C_1 = -\frac{c-k}{c+k}$, so let $C_1 \ne -\frac{c-k}{c+k}$, then

$$0 = C_1 c(c+k)-c(c-k)\Rightarrow 0 = C_1(c+k)-c+k \Rightarrow C_1=\frac{c-k}{c+k}$$

Hence $C_1$ is undefined when $k = \Delta V = -C$, so depending on the physical context of special realtivity (which I don't know enough of), this may never happen

Substituting this into the original closed form: \begin{align}V_n &= \frac{(\frac{c-k}{c+k}) c(c+k) \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - c(c-k)\left(-\frac{k}{c(c+k)}\right)^n}{(c-k)\left(-\frac{k}{c(c+k)}\right)^n + (\frac{c-k}{c+k}) (c+k) \left(\frac{1}{k-c}+\frac{1}{c}\right)^n} \\ &=\frac{c(c-k)\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{(c-k)\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\\ &= c \frac{\left[ \left( \frac{1}{k-c} + \frac{1}{c}\right)^n - \left(-\frac{k}{c(c+k)}\right)^n\right]}{\left[\left(-\frac{k}{c(c+k)}\right)^n + \left(\frac{1}{k-c}+\frac{1}{c}\right)^n\right]}\end{align}

Hence in terms of the way you formulated you have

$$ \boxed{V_n=C \frac{\left[ \left( \frac{1}{\Delta V-C} + \frac{1}{C}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{1}{\Delta V-C}+\frac{1}{C}\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$

if one wishes a preferable altenartive form without negatives inside the power terms:

\begin{align}V_n&=C \frac{\left[ \left( \frac{\Delta V}{C(\Delta V - C)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(\frac{\Delta V}{C(\Delta V - C)}\right)^n\right]} \\ &=C \frac{\left[ \left( -\frac{\Delta V}{C(C-\Delta V)}\right)^n - \left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n\right]}{\left[\left(-\frac{\Delta V}{C(C+\Delta V)}\right)^n + \left(-\frac{\Delta V}{C(C- \Delta V)}\right)^n\right]} \\ &=C \frac{\left[ (-1)^n \left(\frac{\Delta V}{C}\right)^n\left( \frac{1}{(C-\Delta V)}\right)^n - (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C+\Delta V)}\right)^n\right]}{\left[(-1)^n \left(\frac{\Delta V}{C}\right)^n\left(\frac{1}{(C+\Delta V)}\right)^n + (-1)^n \left(\frac{\Delta V}{C}\right)^n \left(\frac{1}{(C- \Delta V)}\right)^n\right]} \end{align} which is written \begin{align}V_n&=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \\ &=C \frac{\left[ \left( \frac{1}{C-\Delta V}\right)^n - \left(\frac{1}{C+\Delta V}\right)^n\right]}{\left[\left( \frac{1}{C-\Delta V}\right)^n + \left(\frac{1}{C+\Delta V}\right)^n\right]} \frac{(C+\Delta V)^n (C-\Delta V)^n}{(C+\Delta V)^n (C-\Delta V)^n}\end{align} Finally, $$\boxed{V_n=C \frac{\left[ \left( C+\Delta V\right)^n - \left(C- \Delta V\right)^n\right]}{\left[\left( C+ \Delta V\right)^n + \left(C- \Delta V\right)^n\right]} \quad \Delta V \ne \pm C ,\, C \ne 0}$$

which is in the form:

$$V_n = C \frac{a^n-b^n}{a^n+b^n}$$

which has limit $C$ as $n \to \infty $ (see this if you don't know how to show that), matching the limit in the OP's question.

If anyone knows how to derive the solution Wolfram gets, that would be preferable to this