Mathematical Analysis « S C H W I G

Convergence of an Interesting Sequence

Wednesday, May 21, 2008

Let

$x_1 = 0, ~ x_2 = 0.\alpha_1, ~ x_3 = 0.\alpha_1 \alpha_2, \hdots, x_n = 0.\alpha_1 \alpha_2 \hdots \alpha_n, \hdots$

be a sequence of finite binary fraction in which each successive fraction is obtained by adjoining a 0 or a 1 to its predecessor. It may sound strange, but the sequence converges. Always.

Let $m>n$ . Then:

$\displaystyle |x_m - x_n| = \left| \frac{\alpha_{n+1}}{2^{n+1}} + \hdots + \frac{\alpha_{m}}{2^{m}} \right| \leq$

$\displaystyle \leq \frac{1}{2^{n+1}} + \hdots + \frac{1}{2^m} =$
$\displaystyle = \frac{\left(\frac{1}{2}\right)^{n+1} - \left(\frac{1}{2}\right)^{m+1}}{1-\frac{1}{2}} =$
$\displaystyle = \frac{1}{2^n}.$

Therefore, $\{x_n\}$ is Cauchy and hence converges.

Would you have guessed?

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Posted by Nikita

Simple Proof of Mixed Derivative Theorem

Tuesday, February 5, 2008

Everybody knows the good old

Theorem (Mixed Derivative Theorem): If the function $f : G \to \mathbb{R}$ is in class $\mathcal{C}^{(2)}$ (id est, twice continuously differentiable) in an open set containing $a \in G \subset \mathbb{R}^n$ , then:

$\dfrac{ \partial^2 f (a)}{\partial x \partial y} = \dfrac{ \partial^2 f (a)}{\partial y \partial x}.$

In other words, the order of differentiation does not matter. The proof of this fact is not completely trivial, though.

I would like to give a much simpler proof of this fact, which I found to be quite interesting and somewhat elegant. It relies on the theorem of Fubini. Before I write out the proof, I state the Fubini’s theorem for a two-dimensional case.

Theorem (Theorem of Fubini): Let $X, Y \subset \mathbb{R}$ be two closed intervals. Let the function $f: X \times Y \to \mathbb{R}$ be continuous and integrable on $X \times Y$ . Then:

$\int\limits_{X \times Y} f(x,y) ~d(x,y) = \int\limits_X \int\limits_Y f(x,y) ~~dy~dx = \int\limits_Y \int\limits_X f(x,y) ~~dx~dy.$

Here $d(x,y)$ denotes really just a product of differentials $dx$ and $dy$, but the order is not specified and not assumed. I will not state the proof (which is fairly tricky and long). Also I would like to note that this is actually one of the corollaries of Fubini’s theorem. The function $f$ may not be continuous, but the theorem of Fubini would still hold, only in a slightly, but a little strange form. And even this generalisation of the state theorem is not exactly the theorem of Fubini, but is more properly known as “a theorem of Fubini type” for a few reasons.

We are now ready to prove the mixed derivative theorem.

Proof of Mixed Derivative Theorem: We need to show that

$\dfrac{ \partial^2 f (a)}{\partial x \partial y} - \dfrac{ \partial^2 f (a)}{\partial y \partial x} = 0.$

We construct the proof by finding a suitable contradiction. Suppose otherwise:

$\dfrac{ \partial^2 f (a)}{\partial x \partial y} - \dfrac{ \partial^2 f (a)}{\partial y \partial x} > 0.$

Since $f \in \mathcal{C}^{(2)}$ , there is a rectangle $X \times Y \ni a$ , such that

$\dfrac{ \partial^2 f }{\partial x \partial y} - \dfrac{ \partial^2 f }{\partial y \partial x} > 0.$

Since this difference is thus a nonnegative function on $X \times Y$ ,

$\int\limits_{X \times Y} \dfrac{ \partial^2 f }{\partial x \partial y} - \dfrac{ \partial^2 f }{\partial y \partial x} ~d(x,y) > 0,$

$\int\limits_{X \times Y} \dfrac{ \partial^2 f }{\partial x \partial y} ~d(x,y) - \int\limits_{X \times Y} \dfrac{ \partial^2 f }{\partial y \partial x} ~d(x,y) > 0.$

Since both of the subintegral functions are continuous, we can apply the theorem of Fubini to obtain:

$\int\limits_Y \int\limits_X \dfrac{ \partial^2 f }{\partial x \partial y} ~dx~dy - \int\limits_X \int\limits_Y \dfrac{ \partial^2 f }{\partial y \partial x} ~dy~dx > 0,$