$f'' + f =0$: finding $f$ using power series How can one solve the following differential equation 
$$f'' + f =0$$ 
with the usage of power series?
Writing: 
$$f(x) = \displaystyle\sum_{k=0}^{\infty} a_k x^k$$
$$f'(x) = \displaystyle\sum_{k=1}^{\infty} a_k k x^{k-1}$$
$$f'' (x) = \displaystyle\sum_{k=2}^{\infty} k (k-1) a_k x^{k-2}$$
How to continue? 
 A: I would suggest not using summation notation for a while. Let 
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots.$$
Then
$$f''(x)=(2)(1)a_2+(3)(2)a_3 x+(4)(3)a_4x^2+(5)(4)a_5x^3+\cdots.$$
The power series expansion of $f''(x)+f(x)$ is easy to write down using the above equations. Note that all the coefficients must be $0$.
We can't say anything about $a_0$ or $a_1$. But the constant term in the expansion of $f''(x)+f(x)$ must be $0$. Thus
$$a_0+(2)(1)a_2=0,$$
so $a_2=-\dfrac{a_0}{(2)(1)}$.
The coefficient of $x$ in the expansion of $f''(x)+f(x)$ must be $0$. Thus
$$a_1+(3)(2)a_3=0,$$
and therefore $a_3=-\dfrac{a_1}{(3)(2)}$.
The coefficient of $x^2$ must be $0$. So
$$a_2+(4)(3)a_4=0.$$
Thus $a_4=-\dfrac{a_2}{(4)(3)}=\dfrac{a_0}{(4)(3)(2)(1)}$.
The coefficient of $x^3$ must be $0$. So
$$a_3+(5)(4)a_5=0.$$
Thus $a_5=-\dfrac{a_3}{(5)(4)}=\dfrac{a_1}{(5)(4)(3)(2)}$. 
I will stop here. You may want to calculate $a_6$. What is going on will become clear. Then you may want to start using general indices. 
A: First remove the "$(x)$" after $f$ and $f''$ (you might have meant $z$, but nothing will do fine). Then collect like powers of $z$ in $f''-f$, and equate the coefficients to $0$ (since a power series is zero if and only if all its coefficients are zero), giving a recursion relation for the coefficients. Then solve that recursion.
A: Hint: You may choose $a_0=a$ and $a_1=b$, all other $a_i$ are then uniquely determined. This reflects that the solution will be of the form $a\cos(x)+b\sin(x)$.
