Power Series of Elliptic Functions I need some help for the following problem: 

Obtain the following power expansions for the elliptic functions: 
  $$sn(u)=u-{(1+k^2) \over3!}u^3+{(1+14k^2+k^4) \over 5!}u^5-...$$
  $$cn(u)=1-{1 \over 2}u^2+{(1+k^2)\over 4!}u^4-...$$
  $$dn(u)=1-{k^2 \over 2!}u^2+{k^2(4+k^2) \over 4!}u^4-...$$

From the pattern given in the problem, I feel that I need to use the power series of $\sin$ and $\cos$ functions, and I know that, if we define $$u=u(\phi)=\int_0 ^{\phi} {d{\theta}\over (1-k^2\sin^2\theta)^{1\over 2}},$$ then $sn(u)=\sin \phi, cn(u)=\cos \phi$. But since $u$ is a function of $\phi$, how could make use of the power series of $\sin$ and $\cos$ to get the desired power series? 
Thanks in advance. 
 A: It is better to use Maclaurin's series and use repeated differentiation to prove each of the series. You need to differentiate $\text{sn}\, u$ five times and other functions 4 times. You should be able to get the desired results by the Maclaurin's series $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2!}f''(0) + \frac{x^{3}}{3!}f'''(0) + \cdots $$
You will need the following formulas $$\frac{d}{du}\text{sn}(u, k) = \text{cn}(u, k)\,\text{dn}(u, k)$$ $$\frac{d}{du}\text{cn}(u, k) = -\text{sn}(u, k)\,\text{dn}(u, k)$$ $$\frac{d}{du}\text{dn}(u, k) = -k^{2}\,\text{sn}(u, k)\,\text{cn}(u, k)$$ $$\text{sn}(0, k) = 0,\, \text{cn}(0, k) = 1, \,\text{dn}(0, k) = 1$$
A: Since $k^2<1$ you can expand the integrand into geometric series $$\frac{1}{1-k^2 \sin^2 \theta}=\sum_{n=0}^\infty  k^{2n} \sin^{2n} \theta $$
Next you could obtain the power series of $\sin^{2n} \theta$: by Euler's theorem $$\sin^{2n} \theta=\left( \frac{e^{i \theta}-e^{-i \theta}}{2i} \right)^{2n} $$
and applying Newton's binomial theorem gives $$\sin^{2n} \theta=\frac{(-1)^n}{2^{2n}} \sum_{k=0}^n \binom{n}{k} e^{i k \theta} e^{-i (n-k) \theta} $$
Now, we can use the familiar exponential series $$e^z=1+z+\frac{z^2}{2!}+\dots $$ for all terms of the series above. This expresses the integrand as a power series (with even powers only) $$\frac{1}{1-k^2 \sin^2 \theta}=\sum_{n=0}^\infty a_{2n}(k) \theta^{2n} .$$ Integrating this, we express $u$ as a power series in $\phi$. Using Lagrange's inversion theorem we can express $\phi$ as a power series in $u$, and composing with the trig functions $\cos,\sin$ should do it.
Remark: This is merely an outline; there are many computations left to do.
