The question I need to answer is as follows:

The function $g: R\rightarrow R$ is twice differentiable and is such that $g(0)=1, g'(0)=0$ and $ g''(x)-g(x)=0 \space \forall x \in \mathbb{R}.$

(i) Fix $x$ in R. Show that there exists $M>0$ such that $\forall n \in \mathbb{N}$ and all $\theta$ from $0$ to $1$, $$ |g^{(n)}(\theta x)|\leq M $$

(ii) Find the coefficients of the Taylor expansion of $g$ about $0$, and prove that this expansion converges to $g(x)\space \forall x \in \mathbb{R}. $

I've started by showing by induction that $g$ has derivatives of all orders, such that $g(x) = g^{(2n)}(x)$ and $ g'(x) = g^{(2n +1)}(x) \space \forall n \in \mathbb{N} $, and from there I can get the coefficients for (ii), but I really don't know what to do next or how to approach (i).

I'd appreciate any help anyone could offer.

  • 1
    $\begingroup$ You can also solve the differential equation $y'' = y,~ y(0) = 1,~ y'(0) = 0$. The solution is $cosh$ which is a function with all what do you want to prove. $\endgroup$
    – Kroki
    Apr 1, 2018 at 18:38

1 Answer 1


One performs the classical definition of derivative $\lim_{h\rightarrow 0}\dfrac{g^{(2)}(x+h)-g^{(2)}(x)}{h}=\lim_{h\rightarrow 0}\dfrac{g(x+h)-g(x)}{h}=g'(x)$ to show that $g^{(3)}(x)$ exists and equals to $f'(x)$. Do not try to differentiate both sides to the equation $g''(x)-g(x)=0$ to claim that the third derivative exists, this is simply not a good reasoning.

Then one uses induction to claim that all the orders of derivatives of $g$ exist and we have $g^{(2n)}=g$ and $g^{(2n+1)}=g'$.

For a fixed $x$, since $g$ and $g'$ are continuous at $[-|x|,|x|]$, they attain the maximum on the compact interval, so there are $M_{1},M_{2}>0$ such that $|g^{(2n)}(\theta x)|\leq M_{1}$ and $|g^{(2n+1)}(\theta x)|\leq M_{2}$ for all $\theta\in[0,1]$. These $M_{1},M_{2}$ may depend on $x$, but not $\theta$. Letting $M=M_{1}+M_{2}$, then $|g^{(n)}(\theta x)|\leq M$ for all such $\theta$.

Having shown that $g$ has all the orders of derivatives, then for each $n$, there is some $\theta$ such that \begin{align*} g(x)=g(0)+g'(0)x+\dfrac{1}{2!}g''(0)x^{2}+\cdots+\dfrac{1}{n!}g^{(n)}(\theta)x^{n}, \end{align*} so \begin{align*} \left|g(x)-\left(g(0)+g'(0)x+\dfrac{1}{2!}g''(0)x^{2}+\cdots+\dfrac{1}{(n-1)!}g^{(n-1)}(0)x^{n-1}\right)\right|&\leq\dfrac{1}{n!}|g^{(n)}(\theta)||x|^{n}\\ &\leq\dfrac{M|x|^{n}}{n!}\\ &\rightarrow 0 \end{align*} as $n\rightarrow\infty$.

  • $\begingroup$ Thanks - that's really helpful. Do you know how I would go about answering part (i) of the question? $\endgroup$
    – bobbyc
    Apr 1, 2018 at 19:13
  • $\begingroup$ You may take a look of it now. $\endgroup$
    – user284331
    Apr 1, 2018 at 20:01

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