Use Taylor Theorem and Taylor Expansion to Prove Let $g: R\rightarrow R$ be a twice differentiable function satisfying $g(0)=1, g'(0)=0$ and $ g''(x)-g(x)=0$, for all $x$ in R
Fix $x$ in R. Show that there exists $M>0$ such that for all natural number n and all θ from 0 to 1 $$ |g^{(n)}(θx)|\leq M$$
Also, find the coefficients of the Taylor expansion of $g$ about $0$, and prove that this expansion converges to $g(x)$ for all $x$ in R
p.s.
My idea is to start from proving that $g$ has derivatives of all orders, but I am not sure whether it is a correct start and how I can proceed. Any suggestion or attempt is appreciated. 
 A: $g''(x)-g(x)=0$ means $g''(x)=g(x)$. And because $g$ is twice differentiable, $g''(x)$ is twice differentiable and so on. $g$ is infinitely often differentiable therefore.
$$g^{(n)}(0)= 
\begin{cases}
 1, & \text{if }n\text{ even}\\
 0, & \text{if }n\text{ odd}
\end{cases}$$
$$g(x)=\sum_{n=0}^{\infty}\frac{g^{(n)}(0)}{n!}x^n$$
$$g(x)=\sum_{n=0}^{\infty}\frac{1}{(2n)!}x^{2n}$$
According to the ratio test, this infinite series is convergent. Its radius of convergence is $\infty$.
$$g(x)=\frac{1}{2}e^{-x}+\frac{1}{2}e^x$$
$$g^{(n)}(x)= 
\begin{cases}
 +\frac{1}{2}e^{-x}+\frac{1}{2}e^x, & \text{if }n\text{ even}\\
 -\frac{1}{2}e^{-x}+\frac{1}{2}e^x, & \text{if }n\text{ odd}
\end{cases}$$
$$g^{(n)}(\theta x)= 
\begin{cases}
 +\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}, & \text{if }n\text{ even}\\
 -\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}, & \text{if }n\text{ odd}
\end{cases}$$
$\ +\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}\le \ +\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$
$\ -\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}\le |-\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}|$
$|-\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}|\le \ +\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$ 
$$\forall x\in\mathbb{R}\colon |g^{(n)}(x)|\le M=+\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$$ 
A: First Part:
Since $g$ is differentiable on $\mathbb{R}$, it is also continuous on $\mathbb{R}$. Note that every continuous function must attain a maximum and a minimum on a closed interval, so $g$ must attain a minimum and maximum on $[0, x]$. Let $g$ have a minimum at $x_m$ and maximum at $x_M$ for $x_m, x_M \in [0, x]$, and let $M_1 = max(|g(x_m)|, |g(x_M)|)$. Similarly, since $g'$ is continuous on $[0, x]$, let $M_2 = max(|g'(x_m)|, |g'(x_M)|)$, where $x_m, x_M \in [0, x]$ and $g'$ has its minimum at $x_m$ and maximum at $x_M$. 
Now let $M = max(M_1, M_2)$. Then both $|g(\theta x)| \leq M$ and $|g'(\theta x)| \leq M$ for every $\theta \in [0, 1]$. Also, since $g''(x) = g(x)$, we have $|g''(\theta x)| \leq M$ and $|g^{(3)}(\theta x)| \leq M$ (since $g''(x) = g(x)$ implies $g^{(3)}(x) = g'(x)$).
More generally, $g^{(2n + 1)}(x) = g'(x)$ and $g^{(2n)}(x) = g(x)$ for every $n\geq 0$, which means $|g^{(n)}(\theta x)| \leq M$
Second Part:
IV_ does a good job explaining the Taylor series expansion part.
