Taylor's theorem: $f'' + f = 0, f(0) = f'(0) = 0$. I am having a hard time coming up with a solution to this problem.

Suppose that $f$ is twice differentiable and that $f'' + f = 0$. If $f(0) = f'(0) = 0$, use Taylor's theorem to show that $f = 0$.

The definition of Taylor's theorem we were given uses the Lagrange remainder $f^{(n+1)}(c) \frac{(x-a)^{n+1}}{(n+1)!}$. Technically we don't have to use Taylor's theorem, but my professor said he doesn't know of a way to do it without it.

The obvious use of Taylor's theorem is that for all $x$ we have $f(x) = f''(c) \cdot x^2/2$ for some $|c| &lt; |x|$. And then this means $f(x) = -f(c) \cdot x^2/2$. Then I don't know how to proceed. I can't think of a way except to show that $f(x) \ne 0$ leads to a contradiction. But what shall I contradict? Here are some of the promising ideas I've had that don't seem to lead anywhere: (suppose WLOG that $x&gt;0$ in the following)


I can construct a sequence $(c_n)$ where $f(x) = (-1)^n f(c_n)\cdot x^{2n}/(2n)!$ since it is easy to show that $f = -f'' = f^{(4)} = \cdots$. So I have all these nonzero $f(c_n)$. But where will that take me?
How about another sequence? There is a $c_1$ such that $0 &lt; c_1 &lt; x$ and $f(c_1) = -f(x) \cdot 2/x^2 \ne 0$. Then there's a $c_2$ with $0&lt;c_2&lt;c_1$ and $f(c_2) = -f(c_1) \cdot 2/{c_1}^2 \ne 0$, and so on. This leads to another sequence $(c_n)$ with nonzero image. It converges (since it's strictly decreasing and bounded below by 0), but does anything useful happen there?
Since each of the $c_i$ are distinct, and $f(c_i)$ has opposite sign to $f(c_{i-1})$ for all $i$, with the IVT I can show that $f$ has infinitely many zeros in the interval $(0, x)$. I don't see how this can be manipulated to a contradiction though.


How can I proceed?
From $f''=-f $ we get that $f $ is infinitely differentiable. 

Let us first work on an interval $[-m,m]$. As $f$ is continuous, there exists $M&gt;0$ with $|f(t)|\leq M$ for all $t\in[-m,m]$. 

Now we write the degree-$n$ Taylor expansion:
$$
f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}2+\cdots+f^{(n-1)}(0)\frac{x^{n-1}}{(n-1)!}+f^{(n)}(c)\,\frac{x^n}{n!},
$$
where $c$ is between $0$ and $x$. The condition $f''=-f$, together with $f(0)=f'(0)=0$, guarantees that $f^{(k)}(0)=0$ for all $k$. Then
$$
f(x)=\frac{f^{(n)}(c)}{n!},
$$
and so for $n $ even we have $|f^{(n)}|=|f| $, and then
$$|f(x)|\leq\frac{M}{n!}.$$
As we can get this estimate for all even $n$, we conclude that $f(x)=0$ for all $x\in [-m,m]$. As the choice of $m$ was arbitrary, $f(x)=0$ for all $x$.
Unless I am grossly mistaken, we don't need Taylor's theorem.  This is a second order differential equation with constant coefficients.  The solution is $c_1 \cos(x) + c_2 \sin(x)$. The initial conditions give that $c_1 = c_2 = 0$.  Your Professor doesn't know of a way to solve this without Taylor's theorem?