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May 22

Using identities to integrate

Given $y=f(x)$ , $x=f^{-1}(y)$ use the identities $\int f^{-1}(y)dy=yf^{-1}(y)-\int y\left({\frac {d}{dy}}f^{-1}(y)\right)dy$ and $\int f^{-1}(y)dy=yf^{-1}(y)-\int f(x)dx$ to find two different expressions for $\int \arctan(y)dy$ .
I can use the first identity to get $y\arctan(y)-{\frac {\ln(1+y^{2})}{2}}+C$ (this is just integration by parts in fact and it is what I would do anyway even if I weren't explicitly asked to use this identity). I am a bit unsure about the use of the other identity. I can use it to get $y\arctan(y)+\ln(\cos \arctan y)+C$ but a small bit of algebra can convert it into $y\arctan(y)-{\frac {\ln(1+y^{2})}{2}}+C$ so I'm not sure that it is sufficiently different to call it a "different" expression. Am I missing something? Widener (talk) 07:51, 22 May 2011 (UTC)[reply]

Well, any antiderivative of the function in a given ~~domain~~ interval will be equal to any other up to a constant, and generally this equality can be demonstrated with some algebra. I think they just wanted you to write

y\arctan(y)-{\frac {\ln(1+y^{2})}{2}}+C

and

y\arctan(y)+\ln(\cos \arctan y)+C

. -- Meni Rosenfeld (talk) 08:52, 22 May 2011 (UTC)[reply]

Meni, I think you should phrase as "any antiderivative of the function in a given interval will be equal to any other up to a constant". The "constant" is constant on connected intervals, not on domains that are a union of separated intervals. Michael Hardy (talk) 00:56, 23 May 2011 (UTC)[reply]

Thanks, fixed. -- Meni Rosenfeld (talk) 08:09, 23 May 2011 (UTC)[reply]

"Reflective" property of repeated decimals

I've noticed that many repeated decimals have this property - take 1/7 = 0.142857... as an example:
Take the repeating digits:

Split them in half:

142
857

The each digit of the second half complements the corresponding digit of the first half to 9: 1+8. 4+5, 2+7
The same is true of some longer repeating digits - take 1/49:

0.020408163265306122448979591836734693877551
020408163265306122448
979591836734693877551

Why is this, and in which repeating decimals does it occur? Ratzd'mishukribo (talk) 10:00, 22 May 2011 (UTC)[reply]

Your "split them in half" trick is equivalent to saying (in the first example) that when you multiply by 1000 and add, you get an integer. We get

1/7    =   0.142857142...
1000/7 = 142.857142857...

add them, and you get

1001/7 = 142.999999999... = 143

So this kind of thing must happen in 1/n whenever (10^k+1)/n is an integer (whole number). This means that n divides 10^k+1 for some k~~, which is true (I think) for all n which aren't divisible by 2 or 5.~~ Staecker (talk) 11:28, 22 May 2011 (UTC)[reply]

Sorry, it's also not true when n is divisible by 3. And probably some other cases too. Someone else can probably say exactly which n have this property. Staecker (talk) 11:35, 22 May 2011 (UTC)[reply]

You have to eliminate multiples of 31 too because the order of 10 in

\mathbb {Z} _{31}^{\times }

is 15 which is odd. The list of forbidden prime divisors is http://oeis.org/A186635. The list of numbers satisfying your condition is http://oeis.org/A043292 (you might want to exclude 2 which only divides 10⁰+1). The repeating block of 1/31 is 032258064516129 and has the different properties 03225 + 80645 + 16129 = 99999 and 032 + 258 + 064 + 516 + 129 = 999. 98.248.42.252 (talk) 19:00, 22 May 2011 (UTC)[reply]

See Midy's theorem. Gandalf61 (talk) 19:16, 22 May 2011 (UTC)[reply]

Matrix multiplication

I've been revising my maths, including the procedure for multiplying two matrices together.

My question is, how was this procedure discovered to be the right one, rather than all the other possible procedures that could be done? Thanks 92.15.21.174 (talk) 12:00, 22 May 2011 (UTC)[reply]

The clue is in 'the right one'. Matrix multiplication was devised to perform a task, the transformation of variables that depended linearly on each other. There would only be one correct result so one can easily see how it is produced. I guess it was originally done for variables and then just extended from there. The second cartoon down in [1] by Gary Larson with Einstein at the board and the wife tidying the desk is not how it is done ;-) Dmcq (talk) 12:26, 22 May 2011 (UTC)[reply]

Supplementary question: who first discovered matrix multiplication? 92.15.21.174 (talk) 12:59, 22 May 2011 (UTC)[reply]

Well James Joseph Sylvester was the one who devised the notation in the middle of the ninteenth century. I'd have though other people would have used the method though before then so I wouldn't attribute the whole business to him but he was important in making matrices a field of study. Dmcq (talk) 13:09, 22 May 2011 (UTC)[reply]

I see in Matrix_(mathematics)#History that the Chinese used matrix methods to solve linear equations a couple of thousand years ago, don't know if they got as far as matrix multiplication but it wouldn't be a big step. Dmcq (talk) 13:15, 22 May 2011 (UTC)[reply]

Actually guessing is done to quite an extent and there are some famous cases. In maths the umbral calculus used to be a dirty secret before it was put on a secure foundation and there's lots of cases in quantum mechanics, Planck's law is an especially famous one. Dmcq (talk) 15:13, 22 May 2011 (UTC)[reply]

A simple way of looking at it: suppose

{\begin{aligned}u&=3x+2y\\v&=7x-4y\end{aligned}}

and

{\begin{aligned}p&=13u+8v\\q&=-11u+6v\end{aligned}}

You have two matrices here. Now find p and q as functions of x and y directly. Then you have another matrix. That's the product matrix. Michael Hardy (talk) 00:43, 23 May 2011 (UTC)[reply]

Presumably you meant to write p and q as functions of u and v? --COVIZAPIBETEFOKY (talk) 02:47, 23 May 2011 (UTC)[reply]

Correct. I've fixed it. Michael Hardy (talk) 04:15, 23 May 2011 (UTC)[reply]

question

Proof

Prove that if x and y are points in a simplex σ ,then there is a vertex v of σ such that ||x-y|| ≤ ||x-v||Bcfgjknors123 (talk) 12:34, 22 May 2011 (UTC)[reply]

Could you please show us what you have done to try to solve the problem, or why you wish to solve this problem. The reference desk isn't here to do your homework for you. Sorry. — Fly by Night (talk) 13:30, 22 May 2011 (UTC)[reply]

sorry i had tried in the following way.Let x,y∈σ=⟨a₀,a₁,...a_k⟩ .Then x=∑ג _ia_i and y = ∑p_ia_i ∑ג_i=1 , ∑p_i=1. Then ||x-y||=||∑(ג_i-p_i)a_i|| ≤ ∑(ג_i-p_i)||a_i|| < ∑ג_i||a_i|| -∑p_i||a_i||+∑p_i||a_i||+a₀ <∑ג_i||a_i||-||a₀||=||∑ג_ia_i-a₀||=||x-a₀|| where a₀=v . please help me by making necessary correcctions in these stepsBcfgjknors123 (talk) 17:20, 22 May 2011

Wow, I don't think I've ever seen subscript tags messed up quite that badly before. StuRat (talk) 17:46, 22 May 2011 (UTC)[reply]

Fixed, I think. I haven't actually tried to follow and understand any of it. --COVIZAPIBETEFOKY (talk) 18:06, 22 May 2011 (UTC)[reply]

Thanks. I added a more useful (sub)title. StuRat (talk) 18:28, 22 May 2011 (UTC)[reply]

The set of points v where ||x-y|| > ||x-v|| is convex so if it contained all the vertices it would contain the entire simplex. But v=y shows this is not so.--RDBury (talk) 18:11, 22 May 2011 (UTC)[reply]

I changed ∑ג_i=1 , ∑p_i=1 to ∑ג_i=1 , ∑p_i=1, which probably is what was ment. Bo Jacoby (talk) 21:05, 22 May 2011 (UTC).[reply]

Terminology for the elements of a Polytope

I am currently doing some stuff on 4-dimensional polytopes (Polychora). However, I have some problems figuring out the correct terminology for the elements that make up a Polychoron. The current status is the following:

Vertex: the Peak (n-3)-face of the Polychoron
Edge: the Ridge (n-2)-face of the Polychoron
Cell: the Facet (n-1)-face of the Polychoron

Can someone help with that? Without knowing the correct terminology, it's a real pain to describe which actual element of the Polychoron I am referring to. Therefore I would like to know the correct terminology.

Also I feel something is missing in my above enumeration of the elements, since I think there should be four types of sub-elements in a Polychoron, but somehow I can't see at the moment what exactly is missing.

Thanks. Toshio Yamaguchi (talk) 15:58, 22 May 2011 (UTC)[reply]

The article you linked lists "vertices, edges, faces, cells" as the terminology, here Polychoron#Euler_characteristic. Points are zero-dimensional, so that is what is missing from your list. SemanticMantis (talk) 16:19, 22 May 2011 (UTC)[reply]

That's strange. I've just realized even though I've written little programs for this sort of thing I don't have any consistent naming for the bits an I've sometimes called the n-1 dimensional objects faces for instance when I've had to write text about them and other times referred to cells. Dmcq (talk) 16:38, 22 May 2011 (UTC)[reply]

I think, it actually should be

Vertex: the ??? ((n-4)-face) of the Polychoron
Edge: the Peak ((n-3)-face) of the Polychoron (according to Peak (geometry))
Ridge: the Ridge or Subfacet (according to Polytope) ((n-2)-face) of the Polychoron (according to Face (geometry))
Cell: the Facet ((n-1)-face) of the Polychoron (according to Face (geometry))

The only thing I can't seem to find out is what an (n-4)-face is generally called (in the Polychoron, it is called a vertex, however). Toshio Yamaguchi (talk) 16:47, 22 May 2011 (UTC)[reply]

Remainder term of Taylor series

Hello. I am studying Calculus out of Spivak's book (the big one) and he says that the Lagrange and Cauchy forms of the Taylor remainder ( $R_{k}(x)={\frac {f^{(k+1)}(\xi )}{(k+1)!}}(x-a)^{k+1}$ and $R_{k}(x)={\frac {f^{(k+1)}(\xi )}{k!}}(x-\xi _{C})^{k}(x-a)$ , respectively, for not necessarily the same ξ in [x,a]) are more useful than the general integral form in certain special cases, but he is characteristically mysterious here and goes into no further detail (probably expecting the reader to discover these cases on her own). Unfortunately while I would like to do that my Calc final is on Tuesday and I have done every exercise in the section that looks even remotely related to the remainder (even the ones that were not assigned) and read your wiki article and still not found when these two forms could be preferable to the integral form, so now I'm here to take the easy way out: When is it better to use the Lagrange or Cauchy form of the remainder than the integral form? Thanks. 72.128.95.0 (talk) 17:22, 22 May 2011 (UTC)[reply]

You could have a function whose (k+1)-st derivative fails to be integrable. Then the Lagrange and Cauchy forms of the remainder hold, but the integral form may not hold. It's not clear if this is the kind of situation Spivak was thinking of, or if he was concerned with something else. Sławomir Biały (talk) 18:11, 22 May 2011 (UTC)[reply]

Attempt to calculate the fundamental group of R^3 - a circle

Hello everyone,

I'm trying to calculate the fundamental group of $X=\mathbb {R} ^{3}-{\text{a circle}}$ : I have verified elsewhere that this is the same group as the fundamental group of the wedge of $S^{2}\vee S^{1}$ , but my method for showing that they have the same fundamental group appears to be wrong.

I believe you can argue that the fundamental group of X is the same as the fundamental group of some ball containing the (removed) circle: now I tried to argue that this is this is the same fundamental group as the 2-sphere with north and south poles joined, by taking the infinite cylinder slicing through the sphere and containing the missing circle on the boundary, and shrinking the part of the ball inside this sphere to a line between the poles, and then "flattening" the remaining part of the ball outside of the cylinder onto the 2-sphere: I'm sorry if that's an unclear description: it's a bit like you're coring an apple and then shrinking the core to 2 discs connected by a line and the outside to a hollow ball minus the two discs.

At this point, I argued that you can 'walk' the south pole around to the north pole to obtain the wedge union of a circle and a sphere. I have seen this trick used before for homology groups (https://nrich.maths.org/discus/messages/117730/143718.html?1207215751), but my teacher says that the trick could cause issues: for example, if you take two 2-spheres, each joined to the other at their north poles and their south poles, then this trick would give you a simply-connected wedge of 2 2-spheres, whereas in the original case it is clear we can take a loop via both poles which is not simply connected. Could anyone correct my intuition on the problem? Does the trick only work because in the link it is homology groups which are being calculated and not homotopy groups? Or does it not work at all perhaps, and the link is wrong? Or is it my teacher who is incorrect? The fundamental group for my $X=\mathbb {R} ^{3}-{\text{a circle}}$ example agrees with the space obtained as the wedge of a circle and a 2-sphere, but perhaps this is just fortunate coincidence. I can see why moving one pole to the same point as another would cause issues, as it would be a very hard map to reverse, moving 1 pole and not the other to get back to 2 distinct poles. Since the 'moving the poles' trick seems to come in extremely handy if it does work, I'd be extremely grateful if someone could enlighten me! Many thanks, Otherlobby17 (talk) 22:43, 22 May 2011 (UTC)[reply]

When I first saw this question in my undergrad, my instinct was to 'drag' the north and south poles to meet at a point, giving a map from one space (poles joined by a line) to the other (wedge with a circle). However, I was unable to show that this map was a homotopy equivalence between the spaces (I don't even think it is), so opted for a different approach. It's possible to find a 'larger' space which deformation retracts onto both spaces, confirming that they are homotopy equivalent. To see what that space might be: join the north and south poles by a straight line 'inside' the sphere, and have the wedged circle also sitting inside the sphere. Viewing the spaces that way, I think (hope!) it's quite suggestive what the larger space is.

With your more general question, it really depends on showing that 'moving poles' is a homotopy equivalence. You're actually forming a quotient space when you do so, so it's not as simple a question as it appears. Icthyos (talk) 09:56, 23 May 2011 (UTC)[reply]

Whenever X is a cell complex, and A is a subcomplex of X which is contractible, then the quotient map that collapses A to a point is a homotopy equivalence. See for example page 11 of Hatcher's book. Hatcher applies this to this case, showing that a sphere with an added line between north and south poles is homotopy equivalent to the wedge of a sphere with a circle. He even treats (a slight generalization of) your example of two spheres glued at both poles on the next page. Algebraist 14:38, 23 May 2011 (UTC)[reply]

Ah-hah! Hatcher, of course. Most illuminating. Icthyos (talk) 18:29, 23 May 2011 (UTC)[reply]

You're not possibly insinuating that Hatcher's book is about as useful as a chocolate fire guard, are you? It does look very pretty, but does it actually do the job it set out to do? — Fly by Night (talk) 20:11, 23 May 2011 (UTC)[reply]

I find Hatcher to be useful. It has certainly contained the answer to most questions I've wondered about...but perhaps I just wonder about uninspiring questions! Icthyos (talk) 22:01, 23 May 2011 (UTC)[reply]