How do I Taylor expand this equation in epsilon?

Question

Suppose $v = 1 - \epsilon$ , $\epsilon \ll 1$. I'm having trouble understanding how to expand $$ t'= \frac{t}{\sqrt{1-v^2}}. $$

I tried writing $$ t' = \frac{t}{\sqrt{1-(1-\epsilon)^2}}, $$ but then the first term goes to infinity.

Answer 1

Just use the expansion around $\nu = 1$ of $\frac{t}{\sqrt{1-\nu^2}}$

Edit

Use Martin's comment as a starting point.

Since you're regarding ad the "new" function in terms of $\epsilon$, you can easily expand your function in power of $\epsilon$ around $0$:

$$\frac{t}{\sqrt{1 - (1-\epsilon)^2}} \approx \frac{t}{\sqrt{2} \sqrt{\epsilon }}+\frac{t \sqrt{\epsilon }}{4 \sqrt{2}}+\frac{3 t \epsilon ^{3/2}}{32 \sqrt{2}}+\frac{5 t \epsilon ^{5/2}}{128 \sqrt{2}}+\frac{35 t \epsilon ^{7/2}}{2048 \sqrt{2}}+O\left(\epsilon ^{9/2}\right)$$

Notice that as $\epsilon \to 0$ your series goes like

$$\frac{t}{\sqrt{1 - (1-\epsilon)^2}} \sim \frac{t}{\sqrt{2} \sqrt{\epsilon }}$$

Calculations are easy, you can do it.

Answer 2

Writing $c$ for $\epsilon$ because lazy,

$\begin{array}\\ t' &= \dfrac{t}{\sqrt{1-(1-c)^2}}\\ &= \dfrac{t}{\sqrt{1-(1-2c+c^2)}}\\ &= \dfrac{t}{\sqrt{2c-c^2}}\\ &= \dfrac{t}{\sqrt{2c}\sqrt{1-c/2}}\\ &= \dfrac{t}{\sqrt{2c}}(1-c/2)^{-1/2}\\ &= \dfrac{t}{\sqrt{2c}}\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}(c/2)^n\\ \end{array} $

To make it look nicer, do

$\begin{array}\\ \binom{-1/2}{n} &=\dfrac{\prod_{k=0}^{n-1} (-\frac12-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1} (-1-2k)}{2^nn!}\\ &=(-1)^n\dfrac{\prod_{k=0}^{n-1} (2k+1)}{2^nn!}\\ &=(-1)^{n}\dfrac{\prod_{k=0}^{n-1} (2k+1)(2k+2)}{2^nn!\prod_{k=0}^{n-1}(2k+2)}\\ &=(-1)^{n}\dfrac{(2n)!}{4^nn!^2}\\ &=(-1)^{n}\dfrac{\binom{2n}{n}}{4^n}\\ \end{array} $

so $t' =\dfrac{t}{\sqrt{2c}}\sum_{n=0}^{\infty}\binom{-1/2}{n}(c/2)^n =\dfrac{t}{\sqrt{2c}}\sum_{n=0}^{\infty}\binom{2n}{n}\dfrac{c^n}{8^n} $