# 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.

• $g$ is just twice differentiable, how do we know that $g^{(n)}(x)$ is well-defined? Perhaps we can use that $g''(x)=g(x)$, so that $g^{(3)}(x) = g'(x)$, so $g^{(3)}(x)$ exists, and $g^{(4)}(x)=g''(x)$, so that $g^{(2n)}(x)=g(x)$ and $g^{(2n+1)}(x)=g'(x)$ by induction. This gives you all the coefficients of the Taylor expansion, which should give you an idea of the behavior of the function Mar 10, 2014 at 22:28

$$g''(x)-g(x)=0$$ 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 $$\infty$$.

$$g(x)=\frac{1}{2}e^{-x}+\frac{1}{2}e^x$$

$$g^{(n)}(x)= \begin{cases} +\frac{1}{2}e^{-x}+\frac{1}{2}e^x, & \text{if }n\text{ even}\\ -\frac{1}{2}e^{-x}+\frac{1}{2}e^x, & \text{if }n\text{ odd} \end{cases}$$

$$g^{(n)}(\theta x)= \begin{cases} +\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}, & \text{if }n\text{ even}\\ -\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}, & \text{if }n\text{ odd} \end{cases}$$

$$\ +\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}\le \ +\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$$
$$\ -\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}\le |-\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}|$$
$$|-\frac{1}{2}e^{-\theta x}+\frac{1}{2}e^{\theta x}|\le \ +\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$$

$$\forall x\in\mathbb{R}\colon |g^{(n)}(x)|\le M=+\frac{1}{2}e^{-x}+\frac{1}{2}e^{x}$$

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.