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I am trying to expand the following binomial with negative terms via the binomial theorem (note that $\Delta V$ is a very small quantity, that is $\Delta V \leq 1$): $$(V-\Delta V)^{\gamma}$$

For that, I rearranged it as: $$(-\Delta V+V)^{\gamma}$$

Now according to the theorem: \begin{eqnarray} (-\Delta V+V)^{\gamma}=\sum_{k=0}^{\infty}\binom \gamma k (-\Delta V)^{\gamma -k}V^k \end{eqnarray}

Now under the assumption that the above equation is correct, I expand $(-\Delta V+V)^{\gamma}$ as: $$\binom \gamma 0 (-\Delta V)^{\gamma}+ \binom \gamma 1 (- \Delta V)^{\gamma -1}\,V + \binom \gamma 2 (- \Delta V)^{\gamma -2}\, V^2 \ldots$$ \begin{eqnarray} 1 \cdot (-\Delta V)^{\gamma}+ \gamma \cdot (- \Delta V)^{\gamma -1}\,V + \frac{\gamma !}{2! (\gamma -2)!} (- \Delta V)^{\gamma -2}\, V^2 \ldots \end{eqnarray} As you can see from the above expression, I cannot simplify it any further and there is also the fact that it does not match the answer given in my text book. In my (physics) textbook it has been given as: $$(V- \Delta V)^{\gamma}=V^{\gamma}\left(1-\frac{\Delta V}{V}\right)^{\gamma}$$ Applying Binomial theorem $$V^{\gamma}\left(1-\frac{\Delta V}{V}\right)^{\gamma}=1- \gamma \frac{\Delta V}{V}+\text{negligible terms}$$ How was this achieved? I tried this above but why don't I get this answer? As a note, I know next to nothing about binomial expansion, especially with negative terms. I read a little about it from here but it was not very helpful: Binomial expansion of $(1-x)^n$

Sillysack Buttowski
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  • You physics textbook does not apply the binomial theorem to $(V - \Delta V)^\gamma$ directly. Instead, it rewrites $(V - \Delta V)^\gamma$ into $V^\gamma (1 - \frac{\Delta V}{V})^\gamma$ and then apply the binomial theorem to $(1 - \frac{\Delta V}{V})^\gamma$. Does this answer your question? – Alex Vong May 20 '17 at 14:28
  • Not really. How did my book get $V^{\gamma}(1-\frac{\Delta V}{V})^{\gamma}$ from $(V-\Delta V)^{\gamma}$? I can't see how someone could take $V^{\gamma}$ common from $(V-\Delta V)^{\gamma}$. And also shouldn't I arrive at the same answer as the book even if I apply the binomial theorem to $(V-\Delta V)^{\gamma}$ directly? – Sillysack Buttowski May 21 '17 at 04:34
  • For your first question, the derivation is $(V - \Delta V)^\gamma = (V(1 - \frac{\Delta V}{V}))^\gamma = V^\gamma (1 - \frac{\Delta V}{V})^\gamma$. The first manipulation factors $V$ out, the second manipulation uses the distributivity of power over multiplication. – Alex Vong May 21 '17 at 07:20
  • Okay, I got that. Now the binomial theorem part? – Sillysack Buttowski May 21 '17 at 15:26

1 Answers1

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For your second question, the problem is that, $\Delta V$ is small when compared to $V$, but $V$ may be huge.

When you expand $$(V - \Delta V)^\gamma = \binom{\gamma}{0} (-\Delta V)^\gamma + \binom{\gamma}{1} (-\Delta V)^{\gamma - 1} V + \binom{\gamma}{2} (-\Delta V)^{\gamma - 2} V^2 + \dots$$

You get all those terms of the form $(-\Delta V)^{\gamma - k} V^k$ which you don't know if they are big or small, so you cannot truncate the series at some point. As a result, this method simply doesn't work.

However, if you expand $$\left(1 - \frac{\Delta V}{V}\right)^\gamma = \binom{\gamma}{0} 1^\gamma + \binom{\gamma}{1} 1^{\gamma - 1} \frac{\Delta V}{V} + \binom{\gamma}{2} 1^{\gamma - 2} \left(\frac{\Delta V}{V}\right)^2 + \dots$$

Since $\Delta V$ is small when compared to $V$, $\frac{\Delta V}{V}$ is close to $0$. In particular, $\frac{\Delta V}{V} < 1$. So we have $\left(\frac{\Delta V}{V}\right)^k = o\left(\frac{\Delta V}{V}\right)$ as $\frac{\Delta V}{V} \to 0$ for any $k > 1$.

Hence $$\left(1 - \frac{\Delta V}{V}\right)^\gamma = \binom{\gamma}{0} 1^\gamma + \binom{\gamma}{1} 1^{\gamma - 1} \left(-\frac{\Delta V}{V}\right) + o\left(\frac{\Delta V}{V}\right)$$ which is what we want.

EDIT-1: Fix a sign error.

Alex Vong
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