Approximation and Taylor Series

Question

Given that

$$ t^{+}=t^{-}+\Delta t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ $$ c=\frac{L}{2}\Biggl(\frac{1}{t^{+}}+\frac{1}{t^{-}}\Biggl)\,\,\,\,\,\,\,\,(2) $$ how eqns (1) and (2) can be reworked to $$ c=\frac{L}{t^{-}}\Biggl[\frac{1+\Delta t/t^{-}}{1+2\Delta t/t^{-}}\Biggl] \,\,\,\,\,\,\,\,(3) $$

and why (3) can be reduced to

$$ c=\frac{L}{t^{-}}[1-\Delta t/t^{-}] \,\,\,\,\,\,\,\,\,\,\,\,\,\,(4) $$

Answer 1

I think there's a misplaced factor of two in your post, it should be

$$ c = \frac{L}{t^-} \left[ \frac{1 + \Delta t / 2t^-}{1 + \Delta t / t^-}\right] $$

Now assume that $\Delta t / t^- \ll 1$, you can use the binomial approximation , keeping up to first order terms you get

\begin{eqnarray} c &=& \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 + \frac{\Delta t}{t^-}\right)^{-1}} \\ &\approx & \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 - \frac{\Delta t}{t^-}\right)} \\ &=& \frac{L}{t^-}\left[1 - \frac{\Delta t}{t^-} + \frac{\Delta t}{2t^-} - \frac{1}{2}\left(\frac{\Delta t}{t^-}\right)^2 \right] \\ &\approx& \frac{L}{t^-}\left[1 - \frac{\Delta t}{2t^-} \right] \end{eqnarray}

Answer 2

Given \begin{align} t^+ &= t^- + \Delta t, \tag1 \\ c &= \frac L2 \left(\frac{1}{t^+} + \frac{1}{t^-}\right), \tag2 \end{align}

we have

\begin{align} c &= \frac{L}{2t^-} \left(\frac{t^-}{t^+} + 1\right) \\ &= \frac{L}{2t^-} \left(\frac{(t^+ - \Delta t) + t^+}{t^+}\right) \\ &= \frac{L}{2t^-} \left(2 - \frac{\Delta t}{t^+}\right) \\ &= \frac{L}{t^-} \left(1 - \frac{\Delta t}{2t^+}\right). \\ \end{align}

If $\Delta t$ is small enough compared to $t^-$, then $\frac{L}{t^-} \left(1 - {\Delta t}/({2t^-})\right)$ might be a reasonable approximation (plus some higher order terms), but $\frac{L}{t^-} \left(1 - {\Delta t}/{t^-}\right)$ just seems like a mistake.

Plugging in actual numbers, if we let $t^- = 9.75,$ $t^+ =10.25,$ and $L=1000$, then $\Delta t = 0.5,$

\begin{align} \frac L2 \left(\frac{1}{t^+} + \frac{1}{t^-}\right) &= \frac{1000}{2}\left(\frac1{10.25}+\frac1{9.75}\right) \approx 100.0625,\\ \frac{L}{t^{-}}\left(\frac{1+\Delta t/t^-}{1+2\Delta t/t^-}\right) &= \frac{1000}{9.75} \left(\frac{1+0.5/9.75}{1+2(0.5)/9.75}\right) \approx 97.7937, \tag A\\ \frac{L}{t^{-}}\left(1-\frac{\Delta t}{t^-}\right) &= \frac{1000}{9.75} \left(1+\frac{0.5}{9.75}\right) \approx 97.3044, \tag B\\ \frac{L}{t^-} \left(1 - \frac{\Delta t}{2t^-}\right) &= \frac{1000}{9.75} \left(1+\frac{0.5}{2(9.75)}\right) \approx 99.9343 \tag C\\ \frac{L}{t^{-}} &= \frac{1000}{9.75} \approx 102.5641. \end{align}

Now let's look at the errors:

\begin{align} 97.7937 - 100.0625 &= -2.2688, \\ 97.3044 - 100.0625 &= -2.7581, \\ 99.9343 - 100.0625 &= -0.1282, \\ 102.5641 - 100.0625 &= 2.5016. \\ \end{align}

So we see that the factor of $2$ in the denominator of $\frac{L}{t^-} \left(1 - {\Delta t}/({2t^-})\right)$ (in Equation $(C)$) reduces the error by an entire order of magnitude, whereas the simple approximation $L/t^-$ has about the same error as either of the approximations in $(A)$ or $(B)$. This is not surprising, because by subtracting $\Delta t/t^-$ when they should be subtracting about half that value, ${\Delta t}/({2t^+}),$ whoever came up with $(A)$ and $(B)$ overcorrected the $L/t^-$ approximation by about the same amount as its error.