Is it possible to find a McLaurin Series for the following function?

Question

Since the limit as $y$ approaches $0$ is $\infty$ for the function $g(y$) = $\frac{1}{\sqrt{1+y^2} - \sqrt{1-y^2}}$, can we say the MacLaurin Series in this case does not exist? And if this is case, what methods are there for approximating the function near $0$?

Answer 1

First, we can write

$$\frac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}=\frac{\sqrt{1+y^2}+\sqrt{1-y^2}}{2y^2}\tag 1$$

Then, we can develop the McLaurin Series for the numerator term on the right-hand side of $(1)$ using the Generalized Binomial Theorem, divide by $y^2$, and obtain a series expression in powers of $y$ around zero. The leading term will be $\frac{1}{y^2}$.

And as $x\to 0$

$$\frac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}}-\frac1{y^2}=-\frac{y^2}{8}+O(y^6)$$

provides an "approximation" that is of order $y^6$.

Answer 2

Since, as Mark Viola showed, $f(y) =\dfrac{1}{\sqrt{1+y^2}-\sqrt{1-y^2}} =\dfrac{\sqrt{1+y^2}+\sqrt{1-y^2}}{2y^2} $, you can approximate the function for small $y$ ($y$ away from $0$ is no problem). We can use the generalized binomial theorem in the special case $\sqrt{1+x} =\sum_{n=0}^{\infty} \binom{1/2}{n}x^n =\sum_{n=0}^{\infty} \binom{2n}{n}\dfrac{(-1)^{n+1}x^n}{4^n(2n-1)} =1-\frac12 x-\frac18 x^2+... $.

For small $y$, $f(y) \approx \dfrac{(1+\frac12 y^2-\frac18 y^4)+(1-\frac12 y^2-\frac18 y^4)}{2y^2} = \dfrac{2-\frac14 y^4}{2y^2} = \dfrac{1}{y^2}-\frac18 y^2 $.

You can take as many terms as you want:

$\begin{array}\\ \sqrt{1+y^2}+\sqrt{1-y^2} &=\sum_{n=0}^{\infty} \binom{1/2}{n}((y^{2n}+(-1)^ny^{2n})\\ &=\sum_{n=0}^{\infty} \binom{1/2}{2n}2y^{4n}\\ &=2(1-\frac18 y^4 - \frac{5}{128}y^6+...)\\ \end{array} $

so $f(y) =\frac1{y^2}-\frac18 y^2 - \frac{5}{128}y^4+... $.