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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.

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  • $\begingroup$ The link is breaking but you need the whole link, i dont know why its not including the rest in my comment above, so heres another link you can try if you can't get the one above to work here $\endgroup$
    – Hushus46
    Commented Mar 23, 2019 at 12:42
  • $\begingroup$ It breaks on asterisk. $\endgroup$ Commented Mar 23, 2019 at 12:44
  • $\begingroup$ Try my latest link, i removed my old comment $\endgroup$
    – Hushus46
    Commented Mar 23, 2019 at 12:45
  • $\begingroup$ Thank you for the link and the final answer. Can you also give me the details for the solution? $\endgroup$ Commented Mar 23, 2019 at 12:48
  • $\begingroup$ Unfortunately for that I can't, because (1) I know nothing about relativity and (2) I don't even know how to arrive to the closed form Wolfram gives. I assume you know the value of $V_1$ and $V_2$ to determine the arbitrary constant given in the closed form. Hopefully atleast knowing the closed form can lead you to the details. I will try a bit more to see how Wolfram is generating it $\endgroup$
    – Hushus46
    Commented Mar 23, 2019 at 12:50

2 Answers 2

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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).


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.

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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

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    $\begingroup$ Note that you can factor-out $(\Delta V/C)^n$ from the numerator and denominator of your "preferable alternative form". Then, multiplying numerator and denominator by $(C-\Delta V)^n (C+\Delta V)^n$ simplifies things further. $\endgroup$
    – Blue
    Commented Mar 23, 2019 at 14:17
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    $\begingroup$ @Blue Good point, I'll edit that in $\endgroup$
    – Hushus46
    Commented Mar 23, 2019 at 14:18
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    $\begingroup$ @Blue Didn't see your edited comment, will add those changes as well. Thanks for the help $\endgroup$
    – Hushus46
    Commented Mar 23, 2019 at 14:24

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