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 $|c| < |x|$. 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 $x>0$ 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 $0 < c_1 < x$ and $f(c_1) = -f(x) \cdot 2/x^2 \ne 0$. Then there's a $c_2$ with $0<c_2<c_1$ 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?
 A: 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 $M>0$ 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 $x$.
A: Hint: So I think you've deduced from the given that the Taylor polynomial of degree $n$ at $0$ is $0$ for every $n$. Now deduce that on any interval $[0,a]$ we have a constant $M$ so that $|f^{(n)}(x)|\le M$ for all $x$ and all $n$. Now consider the remainder formula for the error, and let $n\to\infty$.
A: If $f$ is twice differentiable and $f+f''=0$, it follows that $f\in C ^\infty$ and $f^{(k)}(0)=0$. Choose $x\in \mathbb R$. For all $n\in \mathbb N$, expanding around zero, Taylor's theorem gives $$f(x)=\dfrac{f^{(n+1)}(tx)}{(n+1)!}x^{n+1}, \qquad t\in (0,1).$$
Let $M_i=\max _{u^2 \leq x^2} |f^{(i)}(u)|$ for $0\leq i \leq 3$, and $M=\max \{M_i\}$. Then:$$|f(x)|\leq M \frac{x^{n+1}}{(n+1)!}.$$ Letting $n\rightarrow\infty$, you get $|f(x)|\leq 0$.
A: 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?
