4
$\begingroup$

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.

Thanks!

$\endgroup$
  • $\begingroup$ +$1$ for showing your work! $\endgroup$ – Clayton Mar 9 '13 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 '13 at 0:32
  • 1
    $\begingroup$ @Noble. It isn't sarcasm -- it's nice to see that people have actually tried. $\endgroup$ – John Moeller Mar 9 '13 at 0:42
  • $\begingroup$ @JohnMoeller Thanks John! $\endgroup$ – Noble. Mar 9 '13 at 0:59
2
$\begingroup$

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 $g(x) = \cosh x$.

$\endgroup$
  • $\begingroup$ 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 '13 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 '13 at 7:41
  • $\begingroup$ Thanks a lot, it's very much appreciated. $\endgroup$ – Noble. Mar 9 '13 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.