User talk:Wcherowi

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Re: Kepler Fraud

Thanks for revising my edit instead of deleting it. :) https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Nontransitive dice

Hi, I would just like to let you know that following our dispute over grammar in the edit section, I have made a Request for Comment on the nontransitive dice talk page. Skewb? (talk) 11:12, 31 December 2016 (UTC)

Permutation Matrix

Hi Please write here a proof that shows that the order of composition is correct as stated in the wiki page. I tried writing such proofs and the order is simply wrong. — Preceding unsigned comment added by 79.178.242.43 (talk) 18:56, 16 January 2017 (UTC)

Please read the page more carefully. Both orders are correct and are given on the page. The difference depends on whether the matrices act on column vectors on the right, or row vectors on the left. The two formulas correspond to these two choices. If you get them mixed up, you end up with nonsense. --Bill Cherowitzo (talk) 04:43, 17 January 2017 (UTC)

Please provide a proof to :$P_{\sigma }P_{\pi }=P_{\sigma \,\circ \,\pi }.$ (Using the definition of the preceding section). Note that this isn't consistent with :$P_{\pi }\mathbf {g} ={\begin{bmatrix}\mathbf {e} _{\pi (1)}\\\mathbf {e} _{\pi (2)}\\\vdots \\\mathbf {e} _{\pi (n)}\end{bmatrix}}{\begin{bmatrix}g_{1}\\g_{2}\\\vdots \\g_{n}\end{bmatrix}}={\begin{bmatrix}g_{\pi (1)}\\g_{\pi (2)}\\\vdots \\g_{\pi (n)}\end{bmatrix}}.$ since by composing twice you get the wrong composition: $P_{\tau }P_{\pi }\mathbf {g} ={\begin{bmatrix}\mathbf {e} _{\tau (1)}\\\mathbf {e} _{\tau (2)}\\\vdots \\\mathbf {e} _{\tau (n)}\end{bmatrix}}{\begin{bmatrix}\mathbf {e} _{\pi (1)}\\\mathbf {e} _{\pi (2)}\\\vdots \\\mathbf {e} _{\pi (n)}\end{bmatrix}}{\begin{bmatrix}g_{1}\\g_{2}\\\vdots \\g_{n}\end{bmatrix}}={\begin{bmatrix}\mathbf {e} _{\tau (1)}\\\mathbf {e} _{\tau (2)}\\\vdots \\\mathbf {e} _{\tau (n)}\end{bmatrix}}{\begin{bmatrix}g_{\pi (1)}\\g_{\pi (2)}\\\vdots \\g_{\pi (n)}\end{bmatrix}}={\begin{bmatrix}g_{\pi (\tau (1))}\\g_{\pi (\tau (2))}\\\vdots \\g_{\pi (\tau (n))}\end{bmatrix}}.$ Please explain this and provide a proof.

The problem is that your last step is incorrect, it should be,

${\begin{bmatrix}g_{\tau (\pi (1))}\\g_{\tau (\pi (2))}\\\vdots \\g_{\tau (\pi (n))}\end{bmatrix}}=P_{\tau \,\circ \,\pi }\mathbf {g} ,$

that is to say, $\pi$ first then $\tau$. It is possible that you are switching from a right action to a left action when you write the composition of the permutations (that is the only way I can make sense out of what you wrote). I know that this can get very annoying. I learned my permutation actions one way, but I have often had to teach them using the other convention. This makes me keenly aware of the problems that arise when you subconsciously switch between the two.--Bill Cherowitzo (talk) 19:48, 17 January 2017 (UTC)

But if we denote ${\begin{bmatrix}h_{1}\\h_{2}\\\vdots \\h_{n}\end{bmatrix}}={\begin{bmatrix}g_{\pi (1)}\\g_{\pi (2)}\\\vdots \\g_{\pi (n)}\end{bmatrix}}$ then we get ${\begin{bmatrix}\mathbf {e} _{\tau (1)}\\\mathbf {e} _{\tau (2)}\\\vdots \\\mathbf {e} _{\tau (n)}\end{bmatrix}}{\begin{bmatrix}g_{\pi (1)}\\g_{\pi (2)}\\\vdots \\g_{\pi (n)}\end{bmatrix}}={\begin{bmatrix}\mathbf {e} _{\tau (1)}\\\mathbf {e} _{\tau (2)}\\\vdots \\\mathbf {e} _{\tau (n)}\end{bmatrix}}{\begin{bmatrix}h_{1}\\h_{2}\\\vdots \\h_{n}\end{bmatrix}}={\begin{bmatrix}h_{\tau (1)}\\h_{\tau (2)}\\\vdots \\h_{\tau (n)}\end{bmatrix}}={\begin{bmatrix}g_{\pi (\tau (1))}\\g_{\pi (\tau (2))}\\\vdots \\g_{\pi (\tau (n))}\end{bmatrix}}.$ How can you explain this?

You can ignore what I said before (and I probably got my lefts and rights mixed up anyway, I usually do). You are correct in that this property does not follow from the given definition in the article, and the reason is that the definition (actually, the statement that is supposed to be equivalent to it) is wrong! I will shortly fix the page, but let me expand on this here. Given a permutation $\pi$ and the standard basis (row) vectors $e_{j}$, the permutation matrix associated to $\pi$ is (correctly given in the article)

${\begin{bmatrix}e_{\pi (1)}\\e_{\pi (2)}\\\vdots \\e_{\pi (n)}\end{bmatrix}}.$

That is, the rows of the identity matrix are permuted according to $\pi$. This means that in the $k^{th}$ column of the permutation matrix a 1 will appear in row $\pi (k)$ (this is where the error is in the article - it gets the row and column mixed up). Now consider multiplying (on the right) by the standard basis (column) vector that I will still call $e_{k}$. It is fairly easy to see that $P_{\pi }e_{k}=e_{\pi (k)}$ since you only need to know which row of the permutation matrix contains a 1 in the $k^{th}$ column. Composition of two permutation matrices on these basis vectors then follows immediately: $P_{\tau }P_{\pi }e_{k}=P_{\tau }(e_{\pi (k)})=e_{\tau (\pi (k))}$. Since this is valid for all basis vectors, it holds for all vectors and we have $P_{\tau }P_{\pi }=P_{\tau \circ \pi }.$ I think that I will add an example to the page in order to get this clearer.--Bill Cherowitzo (talk) 04:25, 18 January 2017 (UTC)

Shouldn't it be

${\begin{bmatrix}e_{\pi ^{-1}(1)}\\e_{\pi ^{-1}(2)}\\\vdots \\e_{\pi ^{-1}(n)}\end{bmatrix}}.$? — Preceding unsigned comment added by 79.178.238.30 (talk) 05:11, 18 January 2017 (UTC)

Yep, you are right. I've been checking various sources to see what is going on here, and the literature is in a mess. There are left - right, row - column ambiguities galore. To get the formula to work (and I have found a source that has this) the permutation matrix needs to be defined by permuting the columns of the identity matrix, so we need to have

$P_{\pi }={\begin{bmatrix}e_{\pi (1)}&e_{\pi (2)}&\cdots &e_{\pi (n)}\end{bmatrix}}$

with the $e_{j}$ being column vectors. Now the statement I made before, namely "in the $k^{th}$ column of the permutation matrix a 1 will appear in row $\pi (k)$" actually holds, and the rest of the argument follows. Your original concern was correct, with the definition given that formula doesn't work. I knew that the formula was correct, having seen it several times, but I didn't pay enough attention to the assumptions that needed to be made to get there. I am supposing that if permuting the rows of the identity is used to define a permutation matrix, then having the matrix act on the right of row vectors (I think that's correct terminology) should give the same composition formula ... but I need to go through the details before I'll commit to that statement. This will have to wait until tomorrow. --Bill Cherowitzo (talk) 07:46, 18 January 2017 (UTC)

Thanks for your patience. — Preceding unsigned comment added by 79.178.238.30 (talk) 22:22, 18 January 2017 (UTC)

I must not have had enough sleep yesterday, I made so many mistakes in the above. The article is ok as it stands, except for the formula for composition. Here is the proof giving the right formula for this definition. With the $e_{k}$ being the standard row basis vectors, the permutation matrix corresponding to permutation $\pi$ is,

$P_{\pi }={\begin{bmatrix}e_{\pi (1)}\\e_{\pi (2)}\\\vdots \\e_{\pi (n)}\end{bmatrix}}.$

This matrix can be obtained by permuting the COLUMNS of the identity matrix according to $\pi$, i.e., column i is permuted to the $\pi (i)$ position. Thus, in row i of $P_{\pi }$ the 1 appears in column $\pi (i)$. Alternatively said, the 1 in column k of $P_{\pi }$ occurs in row $\pi ^{-1}(k)$. Now consider the product with the standard basis column vector, $P_{\pi }e_{k}^{T}$. The result only has a 1 in the position corresponding to the row of $P_{\pi }$ that has a 1 in column k, that is, $P_{\pi }e_{k}^{T}=e_{\pi ^{-1}(k)}^{T}$. From this it follows that for two permutation matrices we have

$P_{\tau }P_{\pi }e_{k}^{T}=e_{\tau ^{-1}(\pi ^{-1}(k))}^{T}=e_{(\pi \circ \tau )^{-1}(k)}^{T}.$

Since this holds for all basis vectors, it is true for all column vectors and we have

$P_{\tau }P_{\pi }=P_{\pi \circ \tau }$.

I'll fix this and also put in a numerical example tonight.--Bill Cherowitzo (talk) 22:58, 18 January 2017 (UTC)

Venn diagram

Hello Wcherowi,

is "de:Mengendiagramm" or any other German interwikilink actually visible in your view of the page? It isn't in mine. Also the referenced Wikidata page cannot list it since it is already used to link to the more general page on Euler and Venn diagrams.

best regards, KaiKemmann (talk) 16:07, 30 January 2017 (UTC)

Hi, the link, under Languages Deutsch, appears in the left-hand panel of the Euler diagram page. The placement may depend on the skin you are using; I'm using MonoBook. I did notice that there is no corresponding link on the Venn diagram page and this makes me think that there might be a problem involving some administrative decision about interlinks (there was a problem involving French interlinks not so long ago which seems to have a similar feel to it). If this is a problem with Wikidata, then I am out of my league ... the solution might be to add this link to the Venn diagram page. I hope this helps. --Bill Cherowitzo (talk) 17:10, 30 January 2017 (UTC)

I just realized that I had reverted just this change ... my apologies, I'll undo that in a moment.--Bill Cherowitzo (talk) 17:34, 30 January 2017 (UTC)

Thanks. The problem is in my view that Wikidata only allows for exactly one interwikilink per article and language. So when more articles on a certain subject matter exist in one language than in the other or the content is structured differently into articles in the various language wikis, this cannot be correctly mapped on Wikidata. best regards, KaiKemmann (talk) 14:48, 1 February 2017 (UTC)

Yes, this is the same issue that I alluded to above, the French have one article while we have two: Fibonacci number and Fibonacci series, if I am not mistaken.--Bill Cherowitzo (talk) 19:59, 1 February 2017 (UTC)

Notation in the proof of bijectivity in Cantor's diagonal argument

You stated that indication of base is unnecessary, because the discourse has already established that we are in base 2. I think that's a fair point, but I respectfully disagree that the notation g(t)=0.t_{(2)} is superfluous. Because t is used as a sequence on the LHS (essentially an element of R^\infty), but as a radix notation quantity on the RHS (where it equals \sum_k a_k2^{-k}), it helps the reader to be reminded of that fact.

(It's really an egregious abuse of notation, but for the sake of simplicity, I support the current usage, provided that the base is explicitly indicated.)

Obviously, some people felt very strongly about how this section should be written, despite its numerous problems in English language and style, so I won't begrudge a few subscripts if you insist on removing them.

Cheers, -Jimmy Alsosaid1987 (talk) 06:57, 14 February 2017 (UTC)

I really do understand what you are trying to do and I am not claiming that it is in any way improper, but I don't think that it is providing the clarity for the reader that you seem to think it does. Part of writing good mathematics is being able to anticipate the level of your audience; missing that mark by too much, in either direction, causes problems. Sometimes, for the sake of getting this right, I have to ask myself, "can I say in words what I would naturally want to include in notation?" and the answer to this question, when it arises here in Wikipedia, is almost invariably "yes, and I should do so." Note that I do not believe in dumbing down the discourse to reach my target audience, I'm just being careful about not packing too much into the symbolism that would otherwise be my natural way of communicating with mathematical professionals. --Bill Cherowitzo (talk) 17:20, 14 February 2017 (UTC)

Pascal's triangle

Hi, I wonder why you reversed the edits on Pascal's triangle? The country calls Iran and people are Iranian. I know before 1936 in western world used word Persia to reference the entire country, the countries authorities requested all foreign nations to call the country Iran and its residents Iranians (what they always used to call themselves). And this is officially accepted. So if you feel this is not convincing please let me know what is the rationales behind your reversion otherwise please undo your reversion.--F4fluids (talk) 19:33, 24 February 2017 (UTC)

Editors do not have the flexibility to make the changes you wish to make since we are bound to use the terminology that is present in the sources that we use, whether or not this is currently politically correct. When alternatives exist, we are to use the most common expressions. The vast majority of English language sources use the terms Persia and Persian in this historical context and to change that would mean that we were not following the sources. If someday the majority of sources use the terms Iran and Iranian to refer to these historical places and people, then we would be obliged to follow that terminology, but this is not the case today.--Bill Cherowitzo (talk) 00:04, 25 February 2017 (UTC)

Hi Bill. I still think this is wrong as the country and its residents wrongly used to be called Persia (which is a province even in current day Iran which is called Fars province) and Persians through a long history. And the two people who are referenced in the article are not from that province. But I understand the policy and if it doesn't give a way to correct this, then I won't protest against it. Thanks for response though.--F4fluids (talk) 06:53, 25 February 2017 (UTC)

Thanks.

Thanks for you edits at Determinant. Isambard Kingdom (talk) 17:43, 13 March 2017 (UTC)

Actually, I was being lazy and your edit forced me to do the right thing. I had noticed that when you first reverted the IP's change that you hadn't gone back far enough. After thinking about it for a while I realized that the IP had gotten confused by the fact that there seemed to be too many indices, so I tried to fix that up in a rather minimalist way. However, you are right, both versions needed to be explicitly given to make this absolutely clear. So, in my book, I should be thanking you. --Bill Cherowitzo (talk) 20:50, 13 March 2017 (UTC)

Hyperbola

Hi Wcherowi,

Anita5192 and I are seeing something different per the following exchange regarding the last bullet near the bottom of the article; can you understand what she just said to me in Imagine taking the triangle..., and flipping it over. ....? I'm an engineer so I must be missing something subtle.

Anita, Thanks for your hawk eyes; I am the 68.98.184.101 and wonder if the following needs fixing; "The distance from either focus to either asymptote is b, the semi-minor axis;..." should IMO per all previous discussion have 'focus' replaced by 'vertex' at minimum, and might be clearer by saying "The distance from either vertex to either of its asymptotes is b,...."? Jedwin 68.98.184.101 (talk) 22:23, 12 March 2017 (UTC) I just changed one line to read correctly, "The distance b (not shown) is the length of the perpendicular segment from either focus to the asymptotes." It is unfortunate that the diagram does not show b or θ. The line which reads, "The distance from either focus to either asymptote is b . . ." is already correct.—Anita5192 (talk) 23:34, 12 March 2017 (UTC) Anita, A*A + B*B == C*C cannot exist if the focus is used because it is not a right triangle; and also, C is already the distance from the center to the focus that is actually A*(E-1) beyond A. — Preceding unsigned comment added by 68.98.184.101 (talk) 01:23, 14 March 2017 (UTC) b is the distance from a focus to an asymptote, i.e., the perpendicular distance, hence the triangle is a right triangle. Imagine taking the triangle with base from C to the vertex and a right angle at the vertex, and flipping it over. If you do the math, you will see that it is correct.—Anita5192 (talk) 01:41, 14 March 2017 (UTC)

THANKS, JEDWIN at 68.98.184.101 68.98.184.101 (talk) 02:19, 14 March 2017 (UTC)

Hi Jedwin, I'll do my best. Anita5192 is correct about the edit, but her last bit of explanation to you was missing a vital piece of information. Here is the diagram that I believe you are thinking of:

Unfortunately, this isn't labelled so the explanation will be a bit long winded. The line segment from a vertex, drawn perpendicular to the major axis and ending at an asymptote has length b. I think that this is what you are thinking of when you want to replace focus with vertex. However, b is not the distance from the vertex to the asymptote because that distance is measured along the line drawn from the vertex that is perpendicular to the asymptote. If that distance was also b you would be in the awkward position of having a right triangle whose hypotenuse was the same length as one of its legs. (If you don't see it, label a vertex V, the end of a vertical line through V on the asymptote Q and the foot of the perpendicular dropped from V to the asymptote P. VPQ is the right triangle with right angle at P. VQ is the hypotenuse with length b and VP is the leg with length b.) Now, when you drop a perpendicular from a focus (say F) to an asymptote, its length will also be b. To see this, consider the right triangle VQO, where O is the origin (I think Anita labelled this point C), the right angle here is at V. Note that Q is at distance c from O (see diagram). Rotate the asymptote that Q is on so that it lies on the major axis and Q coincides with F. Now flip the right triangle VQO over its hypotenuse (OF) and V turns into the foot of the perpendicular drawn from F to the asymptote (in its original position). Since the rotation and flip are rigid motions, the length of that segment is still b. I hope this helps.

It's not my first way of thinking about these things, but perhaps this would be clearer if I did it with coordinates. Let V(a,0) be the vertex, F(c,0) a focus. The line from F that is perpendicular to the asymptote with equation y = b/ax has equation y = −a/bx + ac/b. So, the point of intersection P has coordinates (a²/c, ab/c). Now the distance from P to F can be calculated to be b and the distance from P to O is calculated to be a (with heavy use of the relation a² + b² = c².) --Bill Cherowitzo (talk) 16:32, 14 March 2017 (UTC)

Editor of the Week

		Editor of the Week
		Your ongoing efforts to improve the encyclopedia have not gone unnoticed: You have been selected as Editor of the Week for your work on math articles. Thank you for the great contributions! (courtesy of the Wikipedia Editor Retention Project)

User:Buster7 submitted the following nomination for Editor of the Week:

Editor Wcherowi is a Mathematician. One of the difficulties in writing Math articles is providing hard-to-grasp information with clarity. As he says, "Writing good mathematics is being able to anticipate the level of your audience; missing that mark by too much, in either direction, causes problems". Having a concern for the wide ranging capacity of our reader is vital to the health of WikiPedia. Wcherowi has been most active since 2014, and spends 85% of his edits to writing and improving math article such as Non-Euclidean geometry, Projective Geometry, Conic section, Oval (projective plane), History of Mathematics, and many, many more. He further states:"I do not believe in dumbing down the discourse to reach my target audience, I'm just being careful about not packing too much into the symbolism that would otherwise be my natural way of communicating with mathematical professionals".

You can copy the following text to your user page to display a user box proclaiming your selection as Editor of the Week:

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Thanks again for your efforts! Lepricavark (talk) 19:18, 18 March 2017 (UTC)