I've got the following question

Let $g:\mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function satisfying $g(0)=1, \ g'(0) = 0$ and $g''(x) - g(x) = 0$, for all $x \in \mathbb{R}$

i) Prove that $g$ has derivatives of all orders.

ii) Fix $x \in \mathbb{R}$. Show that there exists $M > 0$ such that for all $n \in \mathbb{N}$ and all $\theta \in (0,1)$

$|g^{(n)}(\theta x)| \leq M$

iii) Find the coefficients of the Taylor expansion of $g$ about $0$, and prove that this expansion converges to $g(x)$ for all $x \in \mathbb{R}$

I've done ( i ) and noted that $g^{(k)}(0) = 1$ if $k$ is even and $g^{(k)}(0) = 0$ if $k$ is odd. I'm struggling with the last two parts. For ( ii ) I've shown that, via Taylor's Theorem, we get

$g(x) = \displaystyle\sum_{k=0}^{n-1} \dfrac{g^{(k)}(0)}{k!} x^k + \dfrac{g^{(n)}(\theta x)}{n!} x^n$

But I don't see how to get an upper bound on $\dfrac{n! g(x)}{x^n} - \dfrac{n!}{x^n} \displaystyle\sum_{k=0}^{n-1} \dfrac{g^{(k)}(0)}{k!} x^k$ for some fixed $x \in \mathbb{R}$

I've also realised that $\displaystyle\lim_{n \to +\infty}\displaystyle\sum_{k=0}^{n-1} \dfrac{g^{(k)}(0)}{k!} x^k = \cosh(x)$

Any help greatly appreciated.


  • $\begingroup$ +$1$ for showing your work! $\endgroup$
    – Clayton
    Mar 9, 2013 at 0:29
  • $\begingroup$ @Clayton Sorry, I wasn't sure if that was sarcasm or not - if you think I should show more working let me know (sorry if it wasn't sarcasm, hard to tell over the internet). $\endgroup$
    – Noble.
    Mar 9, 2013 at 0:32
  • 1
    $\begingroup$ @Noble. It isn't sarcasm -- it's nice to see that people have actually tried. $\endgroup$ Mar 9, 2013 at 0:42
  • $\begingroup$ @JohnMoeller Thanks John! $\endgroup$
    – Noble.
    Mar 9, 2013 at 0:59

1 Answer 1


ii) Let $x \in \mathbb{R}$. Since $g$ is twice differentiable, it is $C^1$. Let $M = \sup_{\lambda \in [0,1]} \max(|g(\lambda x)|,|g'(\lambda x)|)$.

Since $g^{(2)}(x) = g(x)$, we have $g^{(2n)}(x) = g(x)$ for all $n$. Similarly, since $g^{(3)}(x) = g^{(1)}(x)$, we have $g^{(2n+1)}(x) = g^{(1)}(x)$ for all $n$. Consequently it follows that $|g^{(n)}(\theta x)| \le M$ for all $n$ and for all $\theta \in (0,1)$.

iii) The Taylor coefficients about zero follow from the above formula for $g^{(n)}$. In particular, $g^{(2n)}(0) = 1$, $g^{(2n+1)}(x) = 0$.

Following your notation above, to show that the Taylor series converges to $g(x)$, you need to show that for all $\epsilon>0$, we can find $N$ such that for $n \ge N$, $$\left|g(x)-\sum_{k=0}^{n-1} \dfrac{g^{(k)}(0)}{k!} x^k\right| < \epsilon.$$ From your estimate above, we have $$\left|g(x)-\sum_{k=0}^{n-1} \dfrac{g^{(k)}(0)}{k!} x^k\right| = \left|\dfrac{g^{(n)}(\theta x)}{n!} x^n\right| \le M \frac{|x|^n}{n!},$$ and since $\lim_n \frac{|x|^n}{n!} = 0$, it is clear that we can find such a $N$. Hence the Taylor series converges.

And, as you have noted, we have $g(x) = \cosh x$.

  • $\begingroup$ Many thanks! I just have a few questions. Firstly, in your definition for $M$ you said $\sup_{\lambda \in [0,1]}$ is there any real difference between this and $\sup_{\lambda \in (0,1)}$ which is how it's defined in the question (obviously I know the difference is it includes the end points, I suppose I mean, why this instead of it being open)? Also, what ensures that $M \nrightarrow \infty$? Is it the fact that $g$ is continuous over all $\mathbb{R}$ due to it being infinitely differentiable? Thanks again. $\endgroup$
    – Noble.
    Mar 9, 2013 at 0:58
  • 1
    $\begingroup$ @Noble.: Since the functions are continuous, taking $\sup$ over $(0,1)$ or $[0,1]$ is immaterial. Also, continuity ensures that $M < \infty$. $g$ is continuous on $\mathbb{R}$ by assumption (it is assumed to be twice differentiable). $\endgroup$
    – copper.hat
    Mar 9, 2013 at 7:41
  • $\begingroup$ Thanks a lot, it's very much appreciated. $\endgroup$
    – Noble.
    Mar 9, 2013 at 11:41

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