Picard iteration and Taylor series I'm trying to derive the Taylor series for $\cos(t)$ by applying the Picard method to the first-order system corresponding to the second-order initial value problem
$x" = −x; x(0) = 1, x'(0) = 0$
For the above, I was thinking of converting the 2nd order DE to a first order, get a matrix which looks like:
$A = (0 0, 0 -1)^T$, but unsure on the Picard iteration here.

We are given the system: 
$\tag 1 x'' = −x; x(0) = 1, x'(0) = 0.$
To linearize $(1)$, we let $x_1 = x$ and $x_2 = x'$, yielding:
$\tag 2 x'_1 = x' = x_2 ~~~ \text{and} ~~ x'_2 = x'' = -x_1.$
This gives us the matrix: $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}.$
Solving for the eigenvalues and eigenvectors of $A$, yields:
$\lambda_1 = -i, v_1 = (i, 1)$ and $\lambda_2 = i, v_2 = (-i, 1).$
Using the eigenvalues, eigenvectors and initial conditions, we arrive at:
$$\tag 3 x_1(t) = - \cos  t ~~~~ \text{and} ~~~~ x_2(t) = \sin t.$$ 
You should verify that this satisfies the system in $(2)$ and derive it!
Note: Variations of solutions with varying signs are okay as long as they satisfy the linearized system (so, we can have $\cos t$ and $- \sin t$) as solutions too, but the Picard iteration will change accordingly.
We will want to compare the Taylor series for the solution above, so that would be:
$$\tag 4 \sin t = t- \frac{t^3}{6} + \frac{t^5}{120} + O(t^7) \\
\cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \frac{t^6}{720} + O(t^7)$$
Now, we want to form the Picard iterates on the linearized, first order system.
The Picard-Lindelof Iteration is given by:
$$\tag 5 \displaystyle x_0(t) = x_0, ~~x_{n+1}(t) = x_0 + \int^t_{t_0} f(s, x_n(s))ds$$
Of course, we are going to do this using both $x_1(t)$ and $x_2(t)$ as called for by the linearization that we did earlier (two solutions to consider).
The setup uses $x(0) = (x_1(0),x_2(0))^T = (-1, 0)^T$ and then does the Picard iterates using $(5)$, so we have:
$X_{0}=  \begin{bmatrix} -1 \\0 \\ \end{bmatrix}$ and $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$, so
First Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 \\0 \\ \end{bmatrix} ds = \begin{bmatrix} -1 \\ t \\ \end{bmatrix}$
Second Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ s \\ \end{bmatrix} ds = \begin{bmatrix} -1 + \frac{t^2}{2} \\ t \end{bmatrix}$
Third Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 +\frac{s^2}{2} \\ s \end{bmatrix} ds = \begin{bmatrix} -1 +\frac{t^2}{2} \\ t -\frac{t^3}{6} \end{bmatrix}$
Fourth Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix}  -1 + \frac{s^2}{2} \\ s -\frac{s^3}{6} \end{bmatrix} ds = \begin{bmatrix} -1 +\frac{t^2}{2} - \frac{t^4}{24} \\ t - \frac{t^3}{6} \end{bmatrix}$
Notice what this is converging to? Compare these to the series expansions we wrote above in $(4)$. The top term in your $- \cos t$ the bottom is $\sin t$, as expected. 
Although we see that this is converging to the $x_1(t) = -\cos t$ and $x_2(t) = \sin t$ results, you should really prove this using an inductive argument.
For an inductive argument, you would write the vector we are deriving as a function of $n$ and then use induction that it is true and gives the $x_1(t)$ and $x_2(t)$.
