Computing the series representation of $ \frac{1}{\sqrt{1+t^2}} $

Question

Essentially the title is the full question. In order to do some other questions I am trying to learn how to naturally figure out series representations of functions.

Computing the Taylor series of this function $ \frac{1}{\sqrt{1+t^2}} = 1- \frac{x^2}{2}+\frac{3x^4}{8}-\frac{5x^6}{16}+\frac{35x^8}{128}+... $

However when trying to see this in series representation I struggle.. I can see that there is a $\sum^{\infty}_{n=0} \frac{(-1)^nx^{2n}}{(2n)!}$ factor.

In fact taking a look now the $(2n)!$ can't be there as it doesnt work with my series... and also how could I include factors that create the numerators?

Perhaps there is a pattern I should be noticing here but I am struggling..

Any help is appreciated.

Answer 1

Hint. By using the standard Newton's generalized binomial theorem one gets $$ (1+t^2)^{-1/2} =\sum_{n\ge0} \binom{-1/2}{n}t^{2n}=\sum_{n\ge0}\frac{(-1)^n}{4^n} \binom{2n}{n}t^{2n} ,\quad |t|<1, $$ with $$ \binom{2n}{n}=\frac{(2n)!}{(n!)^2}. $$

Answer 2

This is a special case of the binomial series: $$ (1+y)^\alpha =\sum_{k=0}^\infty {\alpha \choose k}y^k $$ with convergence for $|y|<1$, where the choose function is interpreted for non integers as $$ {\alpha \choose k}=\frac{(\alpha)(\alpha-1)\dots(\alpha-k+1)}{k!} $$ with clear analogy to the integer form. Substituting $y=t^2$ and $\alpha=-1/2$ in the above gives you the series you are looking for.

In my opinion, this is one of the few series one should just know.

Answer 3

Note that$$\frac1{\sqrt{1+t}}=(1+t)^{-\frac12}=\sum_{n=0}^\infty\binom{-\frac12}nt^n$$and therefore$$\frac1{\sqrt{1+t^2}}=\sum_{n=0}^\infty\binom{-\frac12}nt^{2n}.$$On the other hand,$$\binom{-\frac12}n=\frac{\left(-\frac12\right)\times\left(-\frac32\right)\times\cdots\times\left(-\frac12-(n-1)\right)}{n!}.$$

Answer 4

A fancy approach: in order to compute the Maclaurin series of $\frac{1}{\sqrt{1+x^2}}$ it is enough to compute the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$ and adjust signs. The last problem is equivalent to finding the Maclaurin series of $\arcsin(x)$, and this can be tackled through the Lagrange inversion formula. As shown in the second example in the linked answer, the series reversion of $z-z^2$ leads to

$$ \frac{1-\sqrt{1-4x}}{2x}=\sum_{n\geq 0} C_n x^n = \sum_{n\geq 0}\binom{2n}{n}\frac{x^n}{n+1} $$ hence by applying $\frac{d}{dx}\left(x\cdot\ldots\right)$ to both sides we have $\sum_{n\geq 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$.
By replacing $x$ with $-x^2/4$ we get: $$ \frac{1}{\sqrt{1+x^2}} = \sum_{n\geq 0}\binom{2n}{n}\frac{(-1)^n x^{2n}}{4^n}.$$