Finding the Maclaurin Series for $\sqrt{1+x^2}$ I can't find the Maclaurin series for $\sqrt{1+x^2}$.  Every time it try to find it I get the Maclaurin series for $\sqrt {1+x}$.  Can someone explain it to me?
Thanks!
 A: $$\frac{df}{dx}=\frac{d\sqrt{1+x^2}}{dx}=x(1+x^2)^{-1/2}$$
$$\frac{d^2f}{dx^2}=\frac{d^2\sqrt{1+x^2}}{dx^2}=-x^2(1+x^2)^{-3/2}+(1+x^2)^{-1/2}=(1+x^2)^{-3/2}$$
Thus, the first derivative at zero is indeed zero, but the second is non zero, and is equal to one when $x=0$.
Of course, you could have just used the series for $\sqrt{1+x}$, and substituted $x\rightarrow x^2$. In that case the first derivative would be non-zero, but it would now be the coefficient of $x^2$. In both methods the coefficient of $x$ is zero, as it should be since the function is clearly even.
A: The easiest way is to use the binomial theorem. For $n \ge 1$ you have that:
\begin{align}
\binom{1/2}{n}
  &= \frac{(1/2)^{\underline{n}}}{n!} \\
  &= \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)
             \ldots \left(\frac{1}{2} - n + 1\right)}{n!} \\
  &= \frac{1 \cdot (-3) \cdot (-5) \cdot \ldots \cdot (- 2 n - 3)}{2^n n!} \\
  &= \frac{(-1)^{n - 1} (2 n - 2)!}{n! (n - 1)!} \\
  &= (-1)^{n - 1} \frac{1}{2^{2 n - 1} n}\binom{2 n - 2}{n - 1}
\end{align}
Thus:
$$
\sqrt{1 + x^2}
  = 1 + \sum_{n \ge 1}  
          (-1)^{n - 1} \frac{1}{2^{2 n - 1} n}\binom{2 n - 2}{n - 1} x^{2 n}
$$
To prove the (generalized) binomial theorem in turn is rather easy:
$$
\frac{\mathrm{d}^n}{\mathrm{d} u^n} (1 + u)^a
  = a^{\underline{n}} (1 + u)^{a - n} \\ 
$$
Evaluate at $u = 0$, all that is left is $a^{\underline{n}} = a (a - 1) \ldots (a -n + 1)$
