1
$\begingroup$

I am reading a book (Spectral Theory in Riemannian Geometry), which opens with an interpretation of the Laplacian.

To get a better feeling about the Laplacian operator, consider for example a one-dimension $\mathcal{C}^3$ function $u:\mathbb{R} \to \mathbb{R}$. The mean value of $u$ on the compact set $[-h,h]$ is given by $$\bar{u} := \frac{1}{2h}\int_{-h}^{h}u(x)dx.$$ Now, using the Taylor expansion of $u$ around the origin we get: for all $[-h,h]$, $$u(x) = u(0) + u'(0)x + u''(x)\frac{x^2}{2} + u'''(0)\frac{x^3}{6} + o(x^4).$$ Therefore, $$\bar{u} = \frac{1}{2h}\left(\int_{-h}^{h}u(0) + u'(0)x + u''(0)\frac{x^2}{2} + u'''(0)\frac{x^3}{6}\right) + o(h^4)$$ i.e., $\bar{u} = u(0) + \frac{u''(0)}{12}h^2 + o(h^4),$ hence $\bar{u} - u(0) = \frac{u''(0)}{12}h^2 + o(h^4).$ In other words, $$\Delta u(0) = u''(0) = \frac{12}{h^2}(\bar{u} - u(0)) + o(h^2),$$ thus the Laplacian of $u$ measures the difference between the function $u$ at $0$ and the mean value of $u$ on the neighborhood $[-h,h].$

I have a few simple questions mostly pertaining to $o$.
(1) Wikipedia explicitly gives the definition for $f(x)=o(g(x))$ assuming $x\to\infty$, but in our case $x$ is approaching $0$. What is our new definition?
(2) Why is it $o(h^4)$ rather than some other value in the interval?
(3) Why do we add $o(h^2)$?

$\endgroup$
1
$\begingroup$

First, it appears that there’s an error in the quoted material: The expansion of $u(x)$ via Taylor’s formula should have $O(x^4)$ as the error term instead of $o(x^4)$.

(1) $\phi(\mathbf v)\in o(g(\mathbf v))$ if it’s defined in some ball around the origin and for all $\varepsilon>0$ there is a $\delta>0$ such that for all $\mathbf v$, $||\mathbf v||\le\delta\Rightarrow||\phi(\mathbf v)||\le\varepsilon\;||g(\mathbf v)||$. However, what you really need here is the definition of big-Oh: $\psi(\mathbf v)\in O(\mathbf v)$ if there is some constant $k>0$ such that $||\psi(\mathbf v)||\le k\;||\mathbf v||$ for all $\mathbf v$ in a neighborhood of the origin. More generally, for $\psi(\mathbf v)\in O(g(\mathbf v))$ we need $||\psi(\mathbf v)||\le k\;||g(\mathbf v)||$ in the preceding. Informally, $\phi(\mathbf v)\in o(g(\mathbf v))$ means that $\phi(\mathbf v)$ vanishes “faster” than $g(\mathbf v)$, while $\psi(\mathbf v)\in O(g(\mathbf v))$ vanishes “as fast as” $g(\mathbf v)$.

(2) The $O(h^4)$ term comes from integrating the $O(x^4)$ term of the Taylor expansion. The last term in the expression stands for some function $g(x)\in O(x^4)$. Then we have $$ \left|\int_{-h}^hg(x)\;dx\right| \le \left|\int_{-h}^h\left|g(x)\right|\;dx\right| \le \left|\int_{-h}^hk\left|x^4\right|\;dx\right| \le k\left|\int_{-h}^h\left|x^4\right|\;dx\right|,$$ which means that the error term after integrating is $O(h^5)$. Dividing this by $\frac1{2h}$ gives $O(h^4)$.

(3) In the last step, when you solve $\bar{u} - u(0) = \frac{u''(0)}{12}h^2 + O(h^4)$ for $u''(0)$, you get the term $O(h^4)/h^2$, which is $O(h^2)$.

$\endgroup$
  • $\begingroup$ For (2), what about the term $-o((-h)^4=-o(h^4)?$ Does $o(h^4) - o(h^4) = o(h^4)$? $\endgroup$ – user281519 Oct 19 '15 at 0:11
  • $\begingroup$ And I meant $-o((-h)^4)$. $\endgroup$ – user281519 Oct 19 '15 at 0:22
  • $\begingroup$ Before going any further, I believe that there’s a systematic error in the derivation you’ve quoted. The little oh in Taylor’s formula should be a big Oh—the error term in the expansion of $u(x)$ doesn’t vanish faster than $x^4$. That makes more sense, since in general you can’t say anything about the integral of a little oh, but you can move an integral “inside” of a big Oh. See the answer for more detail about (2). $\endgroup$ – amd Oct 19 '15 at 0:44
  • $\begingroup$ Wikipedia says it's little-$o$. Perhaps we are switching to big-$O$ at the point of integration? $\endgroup$ – user281519 Oct 19 '15 at 0:53
  • $\begingroup$ Sounds like there’s a mistake in Wikipedia, then. It can’t be little-oh since the next term of the Taylor expansion is a multiple of $x^4$ plus an error term, which together is clearly $O(x^4)$, not $o(x^4)$. I looked up Taylor’s formula in other sources to make sure, and it is indeed big-Oh everywhere else. $\endgroup$ – amd Oct 19 '15 at 1:15

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.