Show that $f$ is the zero function if $f''(x)=f(x)$ and $f(0)=f'(0)=0$ Suppose that $f''(x)=f(x)$ for all real numbers $x$, and that $f(0)=f'(0)=0$. Show that $f$ is the zero function. 
I know that $f''(0)=0$ from the assumptions listed. I want to consider the Taylor series for $f$ centered at $x=0$, which is $\sum$ $[f^{(n)}(0)/n!]x^n$. I am not sure where to go from here. 
 A: For a real function the fact that the Taylor series is $0$ to any order does not mean the function is $0$; the well known example is
$$f(x)=\begin{cases} e^{-1\over x^2}&x\neq 0\\0&x=0\end{cases}$$
One can check it is infinitely derivable at $0$ with $\forall n,\,f^{(n)}(0)=0$.
No to solve your problem you need to use unicity theorems related to ODE (e.g Cauchy Lipschitz) or equivalently knowing that the solution of a linear second order ODE is uniquely determined by two parameters ($y(0)$ and $y'(0)$) and noticing that the zero function verifies the $y''+y=0$, $y(0)=0$ and $y'(0)=0$ and so is the unique solution.
A: We assume that $f$ is twice differentiable. Then, if $f''=f$, then we see inductively that $f$ is $C^\infty$.  We assume that it can be represented in terms of its Taylor series as
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\,x^n \tag 1$$
Furthermore, in its interval of convergence, the series in $(1)$ may be differentiated term-by-term.  Differentiating twice, we obtain
$$\begin{align}
f''(x)&=\sum_{n=2}^\infty\frac{f^{(n)}(0)}{(n-2)!}\,x^{n-2}\\\\
&=\sum_{n=0}^\infty\frac{f^{(n+2)}(0)}{n!}\,x^n \tag 2
\end{align}$$
Since $f=f''$, then by uniqueness of the Taylor series we have from equating $(1)$ and $(2)$
$$f^{(n)}(0)=f^{(n+2)}(0) \tag 3$$ 
for all $n$.
Furthermore, we have from the initial conditions 
$$\begin{align}
\left.\left(\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\,x^n\right)\right|_{x=0}&=f^{(0)}(0)\\\\
&=0 \tag 4
\end{align}$$
$$\begin{align}
\left.\left(\sum_{n=0}^\infty\frac{f^{(n+1)}(0)}{n!}\,x^n\right)\right|_{x=0}&=f^{(1)}(0)\\\\
&=0 \tag 5 
\end{align}$$
Putting together $(3)-(5)$, we find that $f^{(n)}(0)=0$ for all $n$ and we are done!

NOTE:
The function $f(x)= e^{-1/x^2}$ for $x\ne 0$ and $f(0)=0$ is $C^\infty$.  But its Taylor series is $0$ and therefore does not represent $f(x)$ anywhere.  So, the assumption that $f(x)$ can be represented by its Taylor series was a key here.

A: Because $f'' = f$ on $\mathbb R,$ we see that $f\in C^2(\mathbb R).$ Let $M= \sup_{[0,1]}|f|.$ Then $M=|f(x_0)|$ for some $x_0 \in [0,1].$ By Taylor, there is $c\in (0,x_0)$ such that
$$f(x_0) = f(0) + f'(0)x_0 + f''(c)x_0^2/2 = f''(c)x_0^2/2 = f(c)x_0^2/2.$$
Taking absolute values then gives $M\le Mx_0^2/2 \le M/2.$ That implies $M=0.$ Thus $f\equiv 0$ on $[0,1].$ This argument can be continued to the right to give $f\equiv 0$ on $[0,\infty).$ The argument also works to the left, so we have $f\equiv 0$ on $\mathbb R$ as desired.
A: Suppose $f''=f$ and consider $g(x)=(f'(x)+f(x))e^{-x}$. Then
$$
g'(x)=(f''(x)+f'(x))e^{-x}-(f'(x)+f(x))e^{-x}=0
$$
Therefore $g(x)$ is constant. Since
$$
g(0)=0
$$
we have $f'(x)+f(x)=0$, for every $x$. Therefore $f'=-f$. Consider
$$
h(x)=f(x)e^{x}
$$
Then $h'(x)=f'(x)e^x+f(x)e^x=0$ so also $h$ is constant. Since $h(0)=0$, we are done.
A: Since you wanted to extract the result from a series we can claim that if an analytic solution, $A(x)$, exists, it will be of the form
$$
A(x) = \sum_{n=0}^{\infty} a_nx^n
$$
now we have
$$
A''(x) = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n
$$
and of course $A(0) = A'(0) = 0$.
So our differential equation tells us that
$$
\begin{align*}
\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^{\infty} a_nx^n &= 0 \\
\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - a_nx^n &= 0 \\
\sum_{n=0}^{\infty}\left [ (n+2)(n+1)a_{n+2}- a_n \right ]x^n &= 0
\end{align*}
$$
Equating each term with $0$, we have
$$
a_{n+2} = \frac{a_n}{(n+2)(n+1)}
$$
Which with the initial conditions $a_0 = A(0) = 0$ and $a_1 = A'(0) = 0$, clearly we have $a_n = 0$.
