Differentiation done correctly: 3. Partial derivatives

Navigation: 1. The derivative | 2. Higher derivatives | 3. Partial derivatives | 4. Inverse and implicit functions | 5. Maxima and minima

While we saw that differentiable maps may be naturally split into component functions when the codomain is a product of Banach spaces, the situation for the domain is more complicated. (This is partly due to the fact that as a topological space, there is no natural injection into a product of Banach spaces.) In this post, we will look at how the existence of partial derivatives relates to differentiability, how the symmetry of higher derivatives (covered in part 2) affects mixed partial derivatives, and finally a short proof of differentiation under the integral sign.

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Differentiation done correctly: 2. Higher derivatives

Navigation: 1. The derivative | 2. Higher derivatives | 3. Partial derivatives | 4. Inverse and implicit functions | 5. Maxima and minima

Last time, we covered the definition of the derivative and its basic properties, which all turn out to be quite similar to their single variable counterparts. Now we are going to explore higher derivatives. In traditional multivariable calculus, true higher derivatives do not exist (except in a specific situation which will be discussed in part 5). Of course, we have so-called “mixed/higher partial derivatives”, which are coordinate-dependent and notationally tricky to work with. As a consequence, the usual statement of Taylor’s theorem in \(\mathbb{R}^n\) ends up being ugly and hard to remember. In reality, Taylor’s theorem for Banach spaces looks almost exactly the same as the single variable Taylor’s theorem!

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Coursera first impressions

I recently signed up for a few Coursera courses, and overall it’s been a fun experience. Here are some of my comments.

Game Theory

So far the instructors have explained normal form games, Nash equilibrium, Pareto optimality, mixed strategies and maxmin strategies. The definitions are pretty clear and plenty of examples follow. The graded problem sets are usually fairly easy and sometimes include interesting examples not found in the videos. However, there are no proofs of any theorems or results. In fact, the theorems themselves are not even written down in any precise way – the instructors seem to be intent on avoiding mathematical notation here for some reason. Verdict: Enrol.

The Modern World: Global History since 1760

Starting near the end of the Commercial Revolution, the instructor explores global history all the way up to the present. There are usually one or two questions at the end of each video to determine whether you have been paying attention. The instructor (Philip Zelikow from the University of Virginia) really shows his enthusiasm for the subject in the videos. Verdict: Enrol!

Image and video processing: From Mars to Hollywood with a stop at the hospital

The instructor begins by giving many applications of image processing, and then gets straight into topics like Huffman coding and the discrete cosine transform (no proofs). I haven’t watched a lot of them yet. Verdict: ???

Introduction to Finance

The instructor is clearly excited about the subject, but keeps going off on tangents. Furthermore, he takes around an hour to explain the compound interest formula \(P(1+r/n)^{nt}\), something that should take 5 minutes at most. Use Khan Academy instead. Verdict: Avoid.

Analytic Combinatorics, Part I

Finally, a course in which the forums aren’t full of complaints about “too much math”. Sedgewick’s voice can be a little boring at times, but the content should be extremely interesting to anyone who knows about generating functions. Verdict: Enrol!

Every continuous open mapping of R into R is monotonic

I’m willing to bet that most students who have used Rudin’s Principles of Mathematical Analysis have encountered this problem:

15. Call a mapping of \(X\) into \(Y\) open if \(f(V)\) is an open set in \(Y\) whenever \(V\) is an open set in \(X\).

Prove that every continuous open mapping of \(\mathbb{R}\) into \(\mathbb{R}\) is monotonic.


Here is one “solution” that is fairly intuitive. It relies on finding a minimum or maximum and considering the image of a small neighborhood around that min/max:

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PAE patch updated for Windows 8

Note: An updated version for Windows 8.1 is available.

This patch allows you to use more than 3/4GB of RAM on an x86 Windows system. Works on Vista, 7, 8, has been tested on Windows Vista SP2, Windows 7 SP0, Windows 7 SP1 and Windows 8 SP0. Instructions and source code included.

Download: PatchPae2.zip (248174 downloads)

Before using this patch, make sure you have fully removed any other “RAM patches” you may have used. This patch does NOT enable test signing mode and does NOT add any watermarks.

Note: I do not offer any support for this. If this did not work for you, either:

  • You cannot follow instructions correctly, or
  • You cannot use more than 4GB of physical memory on 32-bit Windows due to hardware/software conflicts. See the comments on this page for more information.

Free product of free groups and group presentations

Here is Problem 9-4(b) from Introduction to Topological Manifolds by John M. Lee:

Let \(S_1\) and \(S_2\) be disjoint sets, and let \(R_i\) be a subset of the free group \(F(S_i)\) for \(i=1,2\). Prove that \(\langle S_1 \cup S_2 \mid R_1 \cup R_2 \rangle\) is a presentation of the free product group \(\langle S_1 \mid R_1 \rangle * \langle S_2 \mid R_2 \rangle\).

The proof is fairly straightforward if we stick to the universal properties of the free group, free product and quotient group. Here I will write out the full details of the proof, which is more fun to do than to read. There are three basic steps to showing that two things are isomorphic:

  1. Set up the canonical maps and choose names for them. These maps are given to you by the definition of the free group, free product, etc.
  2. Repeatedly use the existence part of the universal properties to derive two maps.
  3. Use the uniqueness part of the universal properties to show that the maps are inverses to each other.

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The “both open and closed” trick for connected spaces

Here is a basic result concerning connected topological spaces.

The only subsets of a connected space \(X\) that are both open and closed are \(\emptyset\) and \(X\) itself.

Clearly, if \(A\) is a nonempty subset of \(X\) that is both open and closed then \(A=X\). We can derive a useful theorem as a consequence of this principle.

Theorem. Let \(X\) be a connected space and let \(\sim\) be an equivalence relation on \(X\). If every \(x \in X\) has a neighborhood \(U\) such that \(p \sim q\) for every \(p,q \in U\), then \(p \sim q \) for every \(p,q \in X\).

Proof. The result clearly holds if \(X\) is empty, so assume otherwise. Let \(p \in X\) and let \(S=\{q \in X : p \sim q\}\). Note that \(S\) is nonempty. If \(q \in S\) then there is a neighborhood \(U\) of \(q\) such that \(q_1 \sim q_2\) for every \(q_1,q_2 \in U\). In particular, for every \(r \in U\) we have \(p \sim q\) and \(q \sim r\) which implies that \(p \sim r\). Then \(U \subseteq S\), which shows that \(S\) is open. If \(q \in X \setminus S\) we can again find a neighborhood \(U\) of \(q\) such that \(q_1 \sim q_2\) for every \(q_1,q_2 \in U\). If \(p \sim r\) for some \(r \in U\) then \(p \sim q\) since \(q \sim r\), which contradicts the fact that \(q \in X \setminus S\). Therefore \(U \subseteq X \setminus S\), which shows that \(S\) is closed. Since \(X\) is connected, \(S=X\). \(\square\)

We can restate this as follows: if \(X\) is connected and \(\sim\) is an equivalence relation on \(X\) such that every equivalence class is open, then there is exactly one equivalence class. This theorem can be used to derive global properties from local ones. Here are some simple applications of this “local-to-global” theorem.

Theorem. A space that is both connected and locally path-connected is path-connected. A space that is both connected and locally piecewise-smooth-path-connected is piecewise-smooth-path-connected.

Proof. Obvious. \(\square\)

A graph is a CW complex of dimension 0 or 1; the 0-cells are the vertices and the 1-cells are the edges.

Theorem. A graph \(G\) is connected in the graph-theoretic sense if and only if it is connected in the topological sense.

Proof. The \(\Rightarrow\) direction is clear. Suppose that \(G\) is topologically connected. Define an equivalence relation \(\sim\) on \(G\) by “\(p \sim q\) if and only if there exists a graph-theoretic path containing the points \(p\) and \(q\)”, and apply the “local-to-global” theorem. \(\square\)

We say that a topological space \(X\) is homogeneous if for any two points \(p,q \in X\) there is a homeomorphism \(\varphi : X \to X\) such that \(\varphi(p)=q\).

Theorem. Every connected topological manifold is homogeneous.

Proof. Let \(X\) be such a manifold. Write \(\mathbb{B}^n\) for the open unit ball in \(\mathbb{R}^n\). It is not too hard to show that for any two points \(p,q \in \mathbb{B}^n\) there is a homeomorphism \(\varphi:\overline{\mathbb{B}^n}\to\overline{\mathbb{B}^n}\) such that \(\varphi(p)=q\) and \(\varphi|_{\partial\mathbb{B}^n}\) is the identity. From this we can use the gluing lemma to show that every point of \(X\) has a neighborhood \(U\) with the property that for any \(p,q \in U\), there is a homeomorphism from \(X\) to itself taking \(p\) to \(q\). (Here we choose \(U\) to be a certain coordinate ball around the point.) Finally, notice that the relation \(\sim\) on \(X\) defined by “\(p \sim q\) if and only if there exists a homeomorphism from \(X\) to itself taking \(p\) to \(q\)” is indeed an equivalence relation, since we can compose two homeomorphisms to get another homeomorphism. Applying the “local-to-global” theorem finishes off the proof. \(\square\)

Theorem. If \(q : C \to X\) is a covering map and \(X\) is connected then the cardinality of \(q^{-1}(\{x\})\) is the same for all \(x \in X\).

Proof. Define an equivalence relation \(\sim\) on \(X\) by declaring that \(x \sim y\) if and only if \(q^{-1}(\{x\})\) and \(q^{-1}(\{y\})\) have the same cardinality. Every point \(x \in X\) has some evenly covered neighborhood \(U\), and obviously \(p \sim q\) for every \(p,q \in U\). Now apply the “local-to-global” theorem. \(\square\)

Of course, there are many more applications of the “open and closed” principle that do not use the equivalence relation theorem stated above. Here are two of them.

Theorem. Every continuous map \(f:X \to Y\) from a nonempty connected space to a discrete space is constant.

Proof. Choose some \(x \in X\). Since \(\{f(x)\}\) is both open and closed in \(Y\), the set \(f^{-1}(\{f(x)\})\) is both open and closed in \(X\). Therefore \(f^{-1}(\{f(x)\})=X\), i.e. \(f\) is constant. \(\square\)

Lemma. Let \(f(X)=\sum a_n X^n\) be a complex power series convergent in a non-discrete set \(E\subseteq\mathbb{C}\). If \(f(z)=0\) for all \(z \in E\) then \(a_n=0\) for all \(n\).

Theorem (The identity theorem). Let \(U\) be a connected open subset of \(\mathbb{C}\) and let \(f:U\to\mathbb{C}\) be an analytic function. If \(f\) is zero on a non-discrete set then \(f=0\) on \(U\).

Proof. Let \(S\) be the set of points \(z \in U\) such that \(f=0\) in a neighborhood of \(z\); then \(S\) is open by definition. Now let \(z \in U\) be a limit point of \(S\). The preceding lemma shows that there exists a neighborhood of \(z\) on which \(f=0\), and therefore \(z \in S\). This shows that \(S\) is closed in \(U\). Finally, our assumption that \(f\) is zero on a non-discrete set implies that \(f=0\) on some open set (by the preceding lemma), so \(S\) is non-empty. Since \(U\) is connected and \(S\) is both open and closed, \(S=U\). \(\square\)

Asus UX31E – setting up the Elantech touchpad driver for Windows 8

I recently upgraded my Asus Zenbook UX31E laptop to Windows 8. Here I’ll show you what I did to update my Elantech drivers to get gestures (swipe from left, top, right, bottom) working. As a bonus, I’ll also show you how to get three finger tap = middle click so you can open new browser tabs easily.

  1. Download the latest Elantech driver for Windows 8 from here: http://support.asus.com/download.aspx?SLanguage=en&m=Touchpad&p=3&s=110. The file name should look like this: Touchpad_Elanteh_Pega_Win7_Win8_….
  2. Install the driver and reboot.
  3. Try out the gestures and test scrolling. If you’re happy with the driver, stop reading here.
  4. Download Process Hacker and run it with administrator privileges.
  5. Press Ctrl+F and search for “elantech”. You should get some results that have Type = Key and Name = HKU\S-1-5-21-<many digits>\Software\Elantech\….
  6. Double-click on one of the results and click Properties. The registry editor should pop up at that registry key.
  7. On the left, make sure you’re at SmartPad: Elantech Registry
  8. To disable annoying behavior where the touchpad freezes up when you’re typing: set the values of DisableWhenType_DelayTime_Gesture, DisableWhenType_DelayTime_Move, DisableWhenType_DelayTime_Tap all to 0.
  9. To disable scrolling inertia: set EGS_InertialScroll_Enable and SC_InertialScroll_Enable both to 0. (I can’t remember which one it is, so you might as well set both.)
  10. To enable three finger tap = middle click: set Tap_Three_Finger to 2 (left click = 0, right click = 1, middle click = 2) and set Tap_Three_Finger_Enable to 1.
  11. Reboot.

Please post a comment if you have any questions.