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)

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)

Relative Extrema
We're studying relative extrema in my calc III class, and I learned the following:

Let $\mathbf v = \begin{bmatrix} a \\ b \end{bmatrix}$.

Let $f: \R^2 \to \R, \begin{bmatrix} x \\ y \end{bmatrix} \mapsto f\left({x,y}\right)$ have continuous second partials in an open region containing $\mathbf v$.

Let $\nabla(\mathbf v) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

Let $d = \begin{vmatrix} f_{xx}(\mathbf v) & f_{xy}(\mathbf v) \\ f_{yx}(\mathbf v) & f_{yy}(\mathbf v) \end{vmatrix} = f_{xx}(\mathbf v)f_{yy}(\mathbf v) - \left({f_{xy}(\mathbf v)}\right)^2 = f_{xx}(\mathbf v)f_{yy}(\mathbf v) - \left({f_{yx}(\mathbf v)}\right)^2$.

If $d > 0, f_{xx}(\mathbf v) > 0$, $f$ has a relative min at $\mathbf v$

If $d > 0, f_{xx}(\mathbf v) < 0$, $f$ has a relative max at $\mathbf v$

Also, if $d > 0$ then $f_{xx}(a,b)$ and $f_{yy}(a,b)$ have the same sign, so you can use either one.

If $d < 0$, $f$ has a saddle point at $\mathbf v$.

If $d = 0$, then you can conclude nothing.

(That took longer than I thought it would to write out). Is there a general test to find out whether a critical point of $f: \R^n \to \R$ yields a max/min/saddle point? I imagine it would get extremely tedious if you have to find determinants by hand, but is this theorem generalizable? --GFauxPas 17:47, 18 April 2012 (EDT)


 * Does that partially answer your question? --abcxyz 00:59, 19 April 2012 (EDT)


 * Thanks abc that's very helpful! I think I'll stay with functions on $\R^2$ until I learn about Jacobians and eigen-stuffs, though; it looks a bit advanced. --GFauxPas 01:10, 19 April 2012 (EDT)

Every Vsp. has Basis
This follows from an application of Zorn's Lemma; I will get to it once I continue processing Conway's book. --Lord_Farin 07:08, 19 April 2012 (EDT)
 * Cool, sounds good. Can I trouble you to put up the formulation of the theorem without proof as a stub article, in case I want to use it? --GFauxPas 08:23, 19 April 2012 (EDT)
 * Cf. Vector Space has Basis. --Lord_Farin 08:50, 19 April 2012 (EDT)
 * Thanks --GFauxPas 09:09, 19 April 2012 (EDT)

Definite Integral Definition
Regarding the "subdivision $P$" in Definition:Definite Integral, what would the subdivision be if it's a function from $\R^n$ to $\R$?

Larson's definitions all involve an alternative definition that is disliked by proofwiki members because convergence is more finicky:


 * $\displaystyle \lim_{\Vert \Delta \Vert \to 0} \sum_a^b f\left({x_i}\right) \ \Delta x_i$

what's the equivalent definition of the supremum of a subdivision in higher dimensions? I.e.,


 * $\displaystyle \int \int \int_Q f\left({x,y,z}\right) \ \mathrm dV = \lim_{\Vert \Delta \Vert \to 0} \sum_a^b f\left({x_i}\right) \ \Delta V_i$

where $\Delta V_i = \Delta x_i \Delta y_i \Delta z_i$, $Q \subset \R^3$

how would you convert that to an definition analogous to what PW has for a definite single integral? --GFauxPas 12:22, 4 May 2012 (EDT)


 * Take a look at Definition:Real Interval at the section that mentions multi-dimensional intervals. But I suspect that a complete analysis of the problem at the same level as done for single-dimension definite intervals may not be the correct way to go. Long time since I did this, but I think beyond an intuitive level (slices, soldiers and croutons) there is no need to go into the same level of detail - having established the result in 1 dimension, expanding it to more dimensions is an inductive process from there, or something. --prime mover 18:07, 4 May 2012 (EDT)

Bases for Matrix Spaces
To prove, for example, that:


 * $\mathcal B_1 = \left({\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} }\right)$

is an ordered basis of $\mathbf M_2\left({\R}\right)$, my professor says it's enough on a test of his to say something like:

"because $T: \mathbf M_2\left({\R}\right) \to \R^4, \begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is an isomorphism..."

and then use that $\left({\mathbf e_i}\right)_1^4$ is an ordered basis, and "isomorph" back.

Q1) Do we have such a theorem up on PW?

Q2) Is that handwaving by PW standards?

--GFauxPas 11:20, 6 May 2012 (EDT)


 * For a start, an isomorphism is with respect to one or more operations. In this context the nature of the operation(s) is unclear. Addition, yes, that's clear and the isomorphism is trivial to prove. But for multiplication this is a different matter altogether.


 * And I still contend that the concept of an "ordered basis" of $\mathbf M_2\left({\R}\right)$ is claptrap. There is no canonical ordering of the basis elements so how can you order the bloody things? --prime mover 11:24, 6 May 2012 (EDT)