Integral order

We write the Mathieu equation as

$$\frac{d^2y}{dz^2} + (a-2q\cos(2z))y=0$$

And consider the case when $q=0$ and write $a=m^2$ and have solutions $\pm \cos mz, \pm \sin mz$ . We then suppose that the case when q is nonzero can be taken into account as a series based on this initial solution. Let

$$a= 1 + \sum_{i=1}^{\infty}\alpha_i q^i$$ (5.1)

Then we suppose that

$$y=\cos z + \sum_{i=1}^{\infty}q^{i}c_i(z)$$ (5.2)

We determine the nature of the $c_i$ functions as follows. Plugging our solution into the Mathieu equation, we get

$$y''=-\cos z + \sum_{i=1}^{\infty}q^ic_i''(z)$$

$$ay = \cos z + \sum_{i=1}^{\infty}q^i\left(c_i + \alpha_i \cos z + \sum_{k=1}^{i-1}\alpha_i c_{i-k}\right)$$

Using the identity

$$2\cos (\frac{1}{2}(A+B))\cos (\frac{1}{2}(A-B))$$

$$= \frac{\left(e^{1/2(A+B)}+e^{-1/2(A+B)}\right)\left(e^{1/2(A-B)}+e^{-1/2(A-B))}\right)}{2}$$

$$= \frac{e^A + e^{-A}}{2} + \frac{e^{B} + e^{-B}}{2}$$

$$= \cos A + \cos B$$

$$-(2q\cos 2z)y=-q(\cos(z)+\cos(3z)-2\cos(2z)\sum_{i=1}^{\infty}q^{i+1}c_1$$

Coefficents are matched:

$$q^0 \cos z = cos z = 0$$

$$q^1 c''_1 + c_1 -\cos(3z)+(\alpha_1 -1)\cos z = 0$$

$$q^2 c''_2 + C_2 + \alpha_1 c_1 -2c_1 \cos 2z + \alpha_2 \cos z =0$$

The particular solution corresponding to $(\alpha_1 -1)\cos z$ is $1/2(1-\alpha_1) z \sin z$ which is not bounded, thus we require that $\alpha_1 = 1$ such that

$$c_1''+c_1 = \cos 3z$$

$$w'' + w = A \cos mz \rightarrow w = -\frac{A \cos mz}{(m^2-1)} \ni$$

$$c_1 = -\frac{1}{8}\cos 3z$$

The arguments presented before imply that

$$\alpha_2 = -\frac{1}{8} \ni c''_2 + c_2 = \frac{1}{8}\cos 3z - \frac{1}{8} \cos 5z$$

$$\Rightarrow c_2 = -\frac{1}{64}\cos 3z + \frac{1}{192} \cos 5z$$

Following [4], we write

$$\alpha_3 = -\frac{1}{64}, c_3 = -\frac{1}{152}\left(\frac{\cos 3z}{3} -\frac{4 \cos 5z}{9}+\frac{\cos 7z}{18}\right)$$

and so we find the c functions can be represted by ``cosine-elliptic'' function,

$$ce_1(z,q) = \cos z -\frac{1}{8}q\cos 3z + \frac{1}{64}q^2\left(-\cos 3z + \frac{\cos 5z}{3}\right)-$$

$$\frac{q^3}{512}\left(\frac{\cos 3z}{3}-\frac{ 4 \cos 5z}{9} + \frac{\cos 7z}{18}\right) +O(q^4)$$ (5.3)

$$a=1 + q -\frac{q^2}{8} -\frac{q^3}{64} + O(q^4)$$ (5.4)