User talk:GFauxPas/Archive1

Change to MathWorld citation template
I noticed (based on One-to-One and Strictly Between) that some pages on MathWorld are credited to different authors from Eric Weisstein, and so require that author to be included in the citation.

I have fixed the template (which is now "MathWorld" not "Mathworld", that's just me tidying up) so as to be able to include the author (which, if not given, defaults to the "Weisstein, Eric W." format as per normal).

What you need to do is add "author=author-name" and "authorpage=author-pagename" where "author-name" is the displayname of the author and "author-pagename" is the name of the html file on MathWorld (not including the full path, not including the extension).

An example:

which gives:
 *  



If the page is given as written by "Weisstein, Eric W." then you should not add the "author" and "authorpage" tags.

I have included this info in the usage section of the Template:MathWorld page itself, but I'm bringing it to your attention because I know you've been active in using it.

Chx. --prime mover 02:55, 31 December 2011 (CST)

Vector Arrows
Book: Linear Algebra, 3rd edition, by John B Fraleigh and Raymond A. Beauregard.

Context: $\R^n$ considered as a euclidean space.

O = $\mathbf{0}$.

Visual representation in 3-space: Where the x,y,z axes intersect.

Here's the juicy part, I'll paraphrase some parts.

We are accustomed to visualizing an ordered pair or triple as a point in the plane or in space and denoting it geometrically by a dot...Physicists have found another very useful geometric interpretation in their consideration of forces acting on a body...[stuff about magnitude, direction]...It is natural to represent a force by an arrow...such an arrow is a force vector.

Using a rectangular coordinate system in the plane, note that if we consider a force vector to start from the origin (0,0), then the vector is completely determined by the coordinates of the point at the tip of the arrow. Thus we can consider each $x \in \R^2$ to represent a vector in the plane as well as a point in the plane. When we wish to regard an ordered pair as a vector, we will use [x,y] instead of (x,y).

''Mathematically, there is no distinction between (1,2) and [1,2]. The different notations merely indicate different views of the same element of $\R^2$.'' Each n-tuple can be viewed as both a point and as a vector.

So there you go, it's purely a matter of perspective. He's saying that they're both legitimate ways to view an n-tuple, but ultimately there's no mathematical difference, just different connotations. --GFauxPas 13:52, 26 January 2012 (EST)


 * Brilliant. Way to go. I note the amendments to the Vector page. --prime mover 16:18, 26 January 2012 (EST)


 * For the formal part of it, I quote: ...if we consider a force vector to start from the origin (0,0), then.... Also, in many (particularly mechanical) situations, the starting point of a vector is very significant for its effect on a system (for example, forces on a rigid axis of a wheel do practically nothing; forces on some surface point of the wheel generally make the wheel turn). Therefore, this assumption is quite questionable, especially when thinking about the starting point of a vector like $\mathbf u -\mathbf v$ where $\mathbf u,\mathbf v$ are vectors... I consider this case not closed yet. --Lord_Farin 17:27, 26 January 2012 (EST)


 * At one point during my mid-teens mathematics education, the concept "position vector (of a point)" was encountered, whose meaning was "the vector from the origin to that point", so one can call $\mathbf 0$ the "position vector of the origin" if that helps. --prime mover 17:40, 26 January 2012 (EST)


 * I was under the impression that vectors are not defined by their location, i.e., the vector issuing from $(0,0)$ and ending at $(1,0)$ is the exact same vector as the one starting from $(5,5)$ and ending at $(6,5)$. Certainly if we define a vector as "magnitude and direction" we don't see "location" there. --GFauxPas 17:44, 26 January 2012 (EST)

It is precisely that approach that I am questioning, on mentioned physical grounds. --Lord_Farin 17:49, 26 January 2012 (EST)
 * Well, in Khan Academy, Khan is very sure about that, and this is what my Linear Algebra professor, er, professes. I'll let you know if I find a book that says otherwise. Oh, if you want a physics source, check out http://www.learner.org/resources/series42.html video 5, which also takes this approach, though it may be dated. --GFauxPas 17:56, 26 January 2012 (EST)

LF I looked through my books to see if I can find any clues for you. I'm paraphrasing. They're only dealing with $\R^n$.

Jewett: Physics for Scientists and Engineers
An example of a vector quantity is displacement...The direction of the arrowhead represents the direction of the displacement, and the length of the arrow represents the magnitude of the displacement.

For many purposes, two vectors $\mathbf{a}$ and $\mathbf{b}$ may be defined to be equal if they have the same magnitude and point in the same direction:


 * $\mathbf{a} = \mathbf{b} \iff \left({||\mathbf{a}|| = ||\mathbf{b}|| \land \text{the vector arrows run along parallel lines}}\right)$

This property allows us to move a vector to a position parallel to itself in a diagram without affecting the vector.

Larson
Larson bolds terms that he's defining for the first time. He's using $\R^2$ for the moment, he addresses $\R^n$ later.

A directed line segment is used to represent a vector quantity. The directed line segment $\overrightarrow{PQ}$ has initial point $P$ terminal point $Q$. Directed line segments that have the same length and direction are equivalent.

(He's defining a new term here. Note he doesn't say equals. - GFP)

The set of all directed line segments that are equivalent to a directed line segment $\overrightarrow{PQ}$ is a vector in the plane and is denoted $\mathbf{v} = \overrightarrow{PQ}$...be sure you see that a vector in the plane can be represented by many different line segments--all pointing in the same direction and all the same length.

The component form of $\mathbf{v}$ is given by:


 * $\mathbf{v} = \langle{v_1,v_2}\rangle$, $v_n \in \R$

where it's implied that the initial point is the origin. If both the initial point and the terminal point lie at the origin, then $v$ is called the zero vector and is denoted by $\mathbf{0} = \langle{0,0}\rangle$.

Two vectors $\mathbf{u} = \langle{u_1,u_2}\rangle$ and $\mathbf{v} = \langle{v_1,v_2}\rangle$ are equal iff $u_1 = v_1 \land u_2 = v_2$.

FWIW, note he's defining equal at the end here, he put it in bold. Also, Equality of Ordered Pairs.

I'm not convinced he has the same approach as Khan and Fraleigh. --GFauxPas 07:35, 27 January 2012 (EST)


 * I hallow the distinction between 'directed line segment' and 'vector' (where the latter is an equivalence class of the former). If this can be implemented, I am satisfied. It is good that we have found at least one reference that calls full rigour to arms on this subject. --Lord_Farin 08:12, 27 January 2012 (EST)


 * On closer investigation, Khan treats vectors differently in the Linear Algebra videos than in the physics videos. I think the video I linked to on the definition:vector page is formal enough for your expectations. He distinguishes the arrow representation of a vector from the vector itself. A vector in $\R^n$ is and only is an ordered n-tuple of n elements of $\R^1$. I guess this supports your view of physics as math without rigor. I'll have to watch his linear algebra videos much more carefully, then I'll try to fix the page.

On a side note, is it more common to say n-toople or n-tuh-pl? --GFauxPas 08:56, 27 January 2012 (EST)


 * N-tuh-pl when used as an adjective (i.e. meaning "multiple" but specifically meaning "with $n$ parts"), n-toople for an ordered set of $n$ elements. This is because in this context it comes from the word "tuple". Except in a UK English accent it would be pronounced something like "n-tyoople". (Sharp-eyed observers would then say: "Which UK English accent? There are thousands!" but y'all know wha'mean, innit?) --prime mover 04:38, 29 January 2012 (EST)


 * I always say n-toople, but that is because this is also how it is pronounced in Dutch. Dealing mostly with Dutch students around, I consider this common practice. I actually enjoyed watching the Khan academy video (even though I already knew all the material covered). Especially when he cares to warn about identifying a vector with the arrow interpretation; note that this distinction is consistent with proposed difference between 'directed line segment' and 'vector'. --Lord_Farin 10:14, 27 January 2012 (EST)
 * I'm glad you like it. I find it's worth watching Khan's videos even if you know the material, as it gives me idea to explain it to people who don't know. cf. my intuition sections. Anyway, how about this. I'll do an entry on euclidean n-space, I'll consolidate the stuff about arrows and mention that this is the primary interpretation used in physics even though it's not 100% correct, and you do other vector spaces? --GFauxPas 10:18, 27 January 2012 (EST)


 * I am afraid I don't know what you mean. There hasn't been any occurrence that I have even thought about drawing a vector space of functions with arrows. Arrows only apply to things we can imagine (that is, mathematically, $\R^n, n \le 3$; $n\le4$ for some gifted persons). --Lord_Farin 10:32, 27 January 2012 (EST)


 * FWIW, I approve of the vector/directed line segment distinction, although I'd probably call a directed line segment vector (physics). Honestly, I think from a mathematical standpoint, the Vector page is fine now (up to maybe reordering the material a bit), the question just arises for physics vectors.  Of course, it wouldn't make sense for standard linear algebra vectors to have endpoints since one of the central requirements in a vector field is additive commutativity, and by assigning endpoints it becomes hard to define addition at all (and impossible to preserve commutativity, as far as I can tell...)  Incidentally, I'd say n-tuh-pl, and that's coming from a US English background. Of course, I'm not exactly an expert on pronunciation, but I'd say that either way is fine. --Alec  (talk) 10:48, 27 January 2012 (EST)
 * I think viewing a vector as an arrow is kind of like viewing a function as a graph of the function, am I wrong? The graph and the function are distinct but we view the graph as an interpretation of the function. E.g., if you define the definite integral as the limit of a riemann sum, it can be represented as the signed area bounded by the graph and the x-axis or whatever. Similarly, the real number line is used as a geometric interpretation of $\R$. --GFauxPas 15:01, 27 January 2012 (EST)

That is quite a good analogy, I think. Be sure to look at my contribution over at Definition talk:Vector. --Lord_Farin 17:57, 27 January 2012 (EST)
 * The confusion between vectors and directed line segments seems to arise from the fact that in order to illustrate a vector, the teacher has to draw an arrow on the board somewhere. Therefore the students natural interpretation is to think: "Aha - that's the point at which the vector is applied." In fact, a vector applies to every point in the space simultaneously, and can perhaps better be illustrated by covering the plane with arrows (making it look a bit like a weather map). As this is easier to do in a computer environment than talk-n-chalk, perhaps this is the approach we might want to adopt. --prime mover 04:43, 29 January 2012 (EST)

Intuitionism / Constructivism
The term which I learned as "intuitionism" seems nowadays to be the same as "constructivism". I found this fascinating article just now:
 * Constructivism is Difficult

on a website which we may want to study.

This may give some background into this whole philosophical quagmire. --prime mover 03:22, 12 February 2012 (EST)


 * Following a course on intuitionistic mathematics at the moment; the two might combine quite well. The lecturer said there will be course notes; I will refer to them if they appear in PDF. --Lord_Farin 18:59, 12 February 2012 (EST)

Theorem Holds in All Models
Anyone know the page name to the theorem that if a theorem is a theorem the theorem has to hold in all models theorem theorem theorem? I can't find it theorem --GFauxPas 08:30, 12 February 2012 (EST)


 * That usually goes by the name of 'Soundness Theorem' (i.e., anything you can prove is true (where true means 'true in all models')). --Lord_Farin 18:59, 12 February 2012 (EST)

Attribution of Sum of Reciprocals is Divergent/Proof 1
You left a comment in the "Historical Note" section of Sum of Reciprocals is Divergent (which has now been moved to Sum of Reciprocals is Divergent/Proof 1) to the effect that you have uncovered evidence that it wasn't Bernoulli who discovered this, but it was in fact Oresme (which would have been some 400 years earlier).

Are you able to find out where you found this evidence? It's an interesting snippet of information to add, and it would be good to find a citation for it. --prime mover 05:16, 11 March 2012 (EDT)


 * Larson says:

''One way to show that the harmonic series diverges is attributed to Jakob Bernoulli. He grouped the terms of the harmonic series as follows:''


 * $ 1 + \frac 1 2 + \underbrace{\frac 1 3 + \frac 1 4}_{> \frac 1 2} +  \underbrace{\frac 1 5 + \cdots + \frac 1 8}_{> \frac 1 2} +  \underbrace{\frac 1 9 + \cdots + \frac 1 {16}}_{> \frac 1 2} +  \underbrace{\frac 1 {17} + \cdots + \frac 1 {32}}_{> \frac 1 2} + \cdots$

Larson doesn't finish the proof, that's left as an exercise. But http://mathworld.wolfram.com/HarmonicSeries.html attributes this proof to Oresme. I don't know which is more reliable. --GFauxPas 08:57, 11 March 2012 (EDT)
 * Just pointing out that wolfram mathworld doesn't say which proof Mengoli and Johann Bernoulli and Jakob Bernoulli used, only that they had a proof. Do you have a source that they had the same proof? Is it implied in Mathworld? --GFauxPas 09:28, 11 March 2012 (EDT)


 * I'll take a look in my copy of and see what it says, but I was assuming that (since this is the proof that was being discussed in MathWorld) this is what it is. --prime mover 09:40, 11 March 2012 (EDT)


 * It's also worth pointing out that all the proofs using calculus in some way require results which hadn't been discovered at the time. If there was another simple proof like Proof 1, it would have been documented eagerly by now. --prime mover 10:09, 11 March 2012 (EDT)

Vector-valued functions
Just a mental note that formally, scalar multiplication and addition of vector-valued functions have not been defined; note that it is probably an instantiation of Definition:Induced Structure. However, that page and its associates could do with a rewrite in due time. --Lord_Farin 06:07, 15 March 2012 (EDT)
 * Can we use this? Definition:Vector Sum. And do we have addition of real functions defined? --GFauxPas 09:10, 15 March 2012 (EDT)


 * In fact, Definition:Vector Sum is necessary to make sense of the right-hand $\oplus$ on Definition:Induced Structure. I think we can invoke Mappings to R-Algebraic Structure form Similar R-Algebraic Structure (with $G = \R^n$, $X=\R$), but as mentioned, this particular (intuitively very natural) part of PW needs to be cleaned and made rigorous. It may be best to leave it for now, as the stuff is intuitively overwhelmingly clear. --Lord_Farin 09:23, 15 March 2012 (EDT)


 * Sho' thing. Oh, and the problem is only going to get worse, as I add theorems for
 * $D_x(f(x)\mathbf{r}(x))$ (well, that's covered by scalar multiplication. Maybe.)
 * $D_x(\mathbf{r}(x) \cdot \mathbf{q}(x))$
 * $\mathbf{r,q}:\R \to \R^3, D_x(\mathbf{r}(x) \times \mathbf{q}(x))$ --GFauxPas 09:27, 15 March 2012 (EDT)

Yes, $[\R\to\R^n]$ is a $[\R\to\R]$-module as well ((abelian) group with multiplication by functions $f:\R\to\R$) making matters indeed even worse. The inner product is rapidly approaching the realm of analysis in multiple variables, along with its advanced notions of differentiation (sensing a possible clash of use in the $D$ notation here, btw); advantage of that theory is that it is intrinsic, because in the particular case of the inner product, it is necessary to prove that the result does not depend on the particular basis chosen; generally a painstaking exercise. Again, had I limitless time to spend on PW, I would have a few more books to cover, in particular one addressing all these rigorous foundations for (real) analysis in more variables. --Lord_Farin 09:54, 15 March 2012 (EDT)
 * ...but I can/should still put up the proofs, right? --GFauxPas 09:56, 15 March 2012 (EDT)
 * Sure, all I'm saying is that I hope to eventually reach the point that everything is rigorous, and the (quite short) proofs using analysis in more variables can be added. This could be months at the least, so please, do continue. --Lord_Farin 09:59, 15 March 2012 (EDT)

Good to see this heavyweight vector calculus stuff going in. It's so easy to get bogged down in the foundations when all you want to do is plug in some 3-D vectors and watch it rip! Here's to gradient, divergence and curl ...--prime mover 14:37, 15 March 2012 (EDT)
 * Np, glad to help. I'm just glad all the derivatives of products are of the same form as $D_xf(x)g(x)$, makes it easy to remember. Which reminds me, I still have to do $D_xf(x)\mathbf{r}(x)$...Oh, and you'd help me out by completing Definition:Derivative/Vector-Valued Function, because I'm not good at transbificating. Plus it would be pretty if this definition exactly matched the other definitions of derivatives, which of course means the same author of all pages! This is of course a ridiculous excuse for me not doing it myself, but saying "I'm lazy and I'll do it later" doesn't sound nice. --GFauxPas 14:59, 15 March 2012 (EDT)
 * LF you wrote there might be a problem in the future with more than one use of $D$, what were you referring to? --GFauxPas 17:34, 15 March 2012 (EDT)
 * Currently, we write $D_x$ for $\dfrac{\mathrm d}{\mathrm dx}$. In the language of multidimensional real analysis, $D$ becomes a map 'computing the total derivative'; the derivative is total in the sense that it is independent of the direction you differentiate in (in $\R$ this doesn't arise, obviously). Effectively, it is bilinear map $Df:\R^n\times \R^n\to\R^n$, written $(x,v)\mapsto D_v f(x) \equiv Df(x,v)$ (differentiation of $f$ at the point $x$ in the direction $v$). Final point is, that for $\R$ this comes down to $D_x f(x) = D_1 f(x)$, giving obvious problems as the left hand side now can mean two different things (differing by a factor $x$). Hope that made at least a bit of sense. --Lord_Farin 18:08, 15 March 2012 (EDT)
 * It is important to add that you can think of $Df$ as being the 'matrix of partial derivatives'; in fact, the two can be shown to be the same. It's only that a matrix requires a particular choice, namely of a basis for your vector spaces. --Lord_Farin 18:10, 15 March 2012 (EDT)
 * Out of my league, I'll wait until I learn differentiation of functions of more than one variable --GFauxPas 19:25, 15 March 2012 (EDT)

Is there such a definition for derivative at a point for vector valued functions?


 * $\displaystyle \lim_{x \to c} \frac {\mathbf r \left({x}\right) - \mathbf r \left({c}\right)}{x -c}$

What about for complex functions? --GFauxPas 08:25, 18 March 2012 (EDT)

Oh, and should I create a category like "Vector Calculus" or "Vector-Valued Calculus" or something? --GFauxPas 08:29, 18 March 2012 (EDT)

Differentiability of Functions of >1 variable
Larson's definition of differentiablity for functions of more than one variable is very non-intuitive (I'm going to use $f:x,y \mapsto f(x,y)$ for ease of asking the question, though the question is for any number of variables):


 * f is differentiable at $(x,y) = (x_0,y_0) \iff \exists \Delta z:$


 * $\Delta z = f_x(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y$

such that $\varepsilon_1, \varepsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$.

Is there an equivalent definition that's more intuitive? Why not define "differentiable" as "differentiable iff all partial derivatives exist"? --GFauxPas 12:42, 28 March 2012 (EDT)


 * As to your last question: Because it isn't enough; derivatives in all directions need to exist.
 * A general definition can be given as follows:


 * A mapping $f: \R^n \to \R^p$ (or defined on some subset of $\R^n$) is said to be differentiable at $a \in \R^n$ iff:
 * There exists a linear mapping $Df(a):\R^n\to\R^p$ (that is, simply put, a matrix) such that:
 * $\displaystyle \lim_{\left\Vert{h}\right\Vert\to 0, h \in \R^n} \frac {\left\Vert{f(a+h)-f(a)-Df(a)h}\right\Vert} {\left\Vert{h}\right\Vert} = 0$


 * This comes down to the existence of a linear approximation $Df(a)$ of $f$ near $a$ which is good enough to make the limit zero (for comparison, you can take $n=p=1$, it will reduce to the familiar expression for $f:\R\to\R$). Note that in the fraction, the norm in the numerator is in $\R^p$, while the one in the denominator is in $\R^n$. Note that $Df(a)h$ means 'the mapping $Df(a)$ evaluated at $h \in \R^n$', not your standard multiplication (well, they are the same iff $n=p=1$; alternatively, this is matrix multiplication with a vector)). Note that this is different from existence of all partial derivatives since the $h \in \R^n$ need to be in a sphere around zero, not just on the coordinate axes. If it is not entirely clear, please say so, and I will demonstrate by means of a small example. --Lord_Farin 14:35, 28 March 2012 (EDT)


 * Alternatively, see this, pp.792 --Lord_Farin 14:40, 28 March 2012 (EDT)


 * How incredibly convenient that in today's Linear Algebra class I first learned about linear maps as matrices! An example would be great. --GFauxPas 15:11, 28 March 2012 (EDT)


 * I thought that the existence of derivatives in all directions does not necessarily ensure differentiability. –Abcxyz (talk | contribs) 20:50, 28 March 2012 (EDT)
 * Correct, but they need to exist for differentiability to possibly apply. I will hopefully get to the example later today. --Lord_Farin 04:42, 29 March 2012 (EDT)

Okay, so let $f: \R^{2n}\simeq\R^n \times \R^n \to \R, (x,y)\mapsto \left\langle{x,y}\right\rangle$.

Say we want to know if $f$ is differentiable at $(a,b)\in\R^n\times\R^n$; then let $h = (h_1,h_2)\in\R^{2n}$, and compute:
 * $f(a,b)-f(a-h_1,b-h_2) = \left\langle{a,b}\right\rangle - \left\langle{a-h_1,b-h_2}\right\rangle = \left\langle{h_1,b}\right\rangle + \left\langle{a,h_2}\right\rangle - \left\langle{h_1,h_2}\right\rangle$

Using Cauchy-Schwarz, the last term can be estimated to $\left\Vert{h}\right\Vert^2$ as the norms of $h_1,h_2$ are dominated by that of $h$. What remains is linear in $h$ (a sum of inner products). Thus, putting $Df((a,b)) = (h\mapsto \left\langle{h_1,b}\right\rangle + \left\langle{a,h_2}\right\rangle)$ we compute the limit to go to zero (by the Cauchy-Schwarz argument).

There is a theorem (not too hard) establishing that the linear mapping $Df((a,b))$ is unique; hence conclude that it equals the given expression (compare the case that $n=1$ for further insights). Hopefully, this slightly nontrivial example gives a bit of insight. --Lord_Farin 06:42, 29 March 2012 (EDT)


 * Also, when considering $f:\R\to\R$, the standard derivative $f'$ is obtained by the canonical identification $\operatorname{Lin}(\R,\R)\simeq \R,Df(a)\mapsto Df(a)1 = f'(a)$. Because $Df(a)1$ is also often denoted $D_af(1)$, this is the origin of the possible confusion I expressed earlier. --Lord_Farin 06:45, 29 March 2012 (EDT)


 * This is significantly harder than what we're doing in Calc III but I'm getting something out of it, thanks! I'm not going to say that I get it completely, but I'm okay with that- I haven't even finished Calc III yet. Is this definition equivalent to Larson's for $\R^2 \to \R$? --GFauxPas 09:21, 29 March 2012 (EDT)


 * I would say so. In matrix form, $Df(a)$ will always be the matrix of partial derivatives (the Jacobian) with respect to the chosen basis. That means, for $\R^2\to\R$, that it becomes a row matrix $(f_x(a), f_y(a))$ (which upon multiplication by the column vector $(\Delta x, \Delta y)$ becomes the first part of Larson's expression; the $\varepsilon$s correspond to the term $\left\langle{h_1,h_2}\right\rangle$ in the example). It would be rather awkward had Larson an incompatible definition of something basic like differentiation. --Lord_Farin 09:45, 29 March 2012 (EDT)


 * I have a much better understanding of Larson's def'n now after discussing it with my Linear Algebra professor.


 * Side note: Has anyone seen $f^{\,'}_x(x,y), f^{\,''}_{xy}(x,y)$ for $\dfrac {\partial z}{\partial x}, \dfrac {\partial^2 z}{\partial y \partial x}$? I keep on wanting to put a prime on it --GFauxPas 10:48, 30 March 2012 (EDT)


 * No, that notation isn't used. You have to know what $f$ is derived with respect to, which is why subscripts are used, and it's strictly instead of primes, which is strictly reserved for total derivative, not partial. --prime mover 13:10, 30 March 2012 (EDT)


 * You mean that $f'$ is seriously used for $Df$ (or $df$, if in differential geometry)?! That's new to me. --Lord_Farin 17:09, 30 March 2012 (EDT)


 * Think so. May be wrong. Point is, it is never used for partial drivs. I think I met it in the context of fluid mechanics but I misremember the details. --prime mover 18:09, 30 March 2012 (EDT)