Wikipedia:Reference desk/Archives/Mathematics/2010 June 6

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June 6

Operator, or transform, as a mapping between classes of functions

Let's use the term "operator", or "transform", for a mapping (whether explicit or implicit) between classes sets of functions. Is there any common term for an operator/transform which is induced by a first-order formula, containing logical symbols only - except for function-symbols for substituting functions which are mapped to one another by that operator/transform? HOOTmag (talk) 11:17, 6 June 2010 (UTC)

You might mean sets instead of classes, working with classes can be awkward and functions are normally defined in terms of sets. In logic you do get classes. A function may have a set of functions as its domain or co-domain. Substituting into a logical formula is also called instantiation. If you are preserving some structure by your mapping then you might be thinking of category theory and functors. Dmcq (talk) 11:56, 6 June 2010 (UTC)

Sorry, but you haven't answered my question. I've been asking whether there is a common term for a very specific mapping, i.e. a mapping induced by a first-order formula which contains logical symbols only - except for function-symbols for substituting functions which are mapped to one another by that formula. Memorandum: this formula induces a mapping whose domain and co-domain contain functions (rather than simple regular atomic points, e.g. numbers and the like.). HOOTmag (talk) 16:13, 6 June 2010 (UTC)

Niggle: given a mapping between two classes of members

What is exactly mapped? Is it one class (the domain) that is mapped into the other class (the co-domain), or is it the members of the domain that are mapped into members in the co-domain (as I've assumed in my previous thread)?? HOOTmag (talk) 11:21, 6 June 2010 (UTC)

Can refer to both and you might have a mapping into or onto or just a mapped to. Precision of language doesn't matter much when you've got the symbols. Hand waving and pointing at the board are other good ways of getting a meaning across :) Dmcq (talk) 12:02, 6 June 2010 (UTC)

I'm looking for the precise expression, and that's why I called that "niggle..." HOOTmag (talk) 16:15, 6 June 2010 (UTC)

Usually we would say "f is a mapping from D to R" where D (the domain) and R (the range) are sets; but we could informally say "f maps x to x²" referring to elements of the domain and range. I.e. $f:\mathbb {R} \to \mathbb {R}$ is the squaring function $f:x\mapsto x^{2}$ (note the differing arrow symbols for "to" and "maps to"). I agree with Dmcq that you might like to study some category theory, which examines this sort of question rather closely. 75.57.243.88 (talk) 19:59, 6 June 2010 (UTC)

Dmcq mentioned category theory with regard to another question of mine in another thread. HOOTmag (talk) 20:18, 6 June 2010 (UTC)

Always possible to find a matrix that transform a matrix into reduced row echelon form?

Dear all,

I want to find an m-times-m matrix, Q, such that QB=E, where B is an m-times-n matrix and E is the reduced row echelon form of B. Is this always possible?

In my case Q must be invertible and m>n. Does this change things?

Thank you in advance, Marc K (talk) 19:53, 6 June 2010 (UTC)

To answer my own question, yes it must be possible, since any row operation is equivalent to multiplying an elementary matrix from the left. Marc K (talk) 21:06, 6 June 2010 (UTC)

Bijection wanted with S_n

Given a natural number n, let denote I_n:={1,2,..,n} the set of the first positive integers.

How to construct a concrete and nice bijection between the two sets (both having n! elements):

R_n:= {f:I_n→I_n such that f(i)≤ i for all i∈I_n}

and the permutation group

S_n:= {f:I_n→I_n : f is bijective}.

?

--84.221.69.158 (talk) 20:52, 6 June 2010 (UTC)

How about this algorithm?

let array A[1..n] = {1, 2, ... n}

for each i from 1 to n do

swap A[i] and A[f(i)]

For each function f it will produce a unique permutation of A. For example, for n=3 these 6 functions give respective permutations of A:

f	A
	i=1	i=2	i=3
(1, 1, 1)	(1, 2, 3)	(2, 1, 3)	(3, 1, 2)
(1, 1, 2)	(1, 2, 3)	(2, 1, 3)	(2, 3, 1)
(1, 1, 3)	(1, 2, 3)	(2, 1, 3)	(2, 1, 3)
(1, 2, 1)	(1, 2, 3)	(1, 2, 3)	(3, 2, 1)
(1, 2, 2)	(1, 2, 3)	(1, 2, 3)	(1, 3, 2)
(1, 2, 3)	(1, 2, 3)	(1, 2, 3)	(1, 2, 3)

CiaPan (talk) 21:22, 6 June 2010 (UTC)

That's really very nice!! --84.221.69.158 (talk) 22:08, 6 June 2010 (UTC)

More formally, each value of f defines a single-swap $(i\ f(i))$ permutation (see Cycle notation), so the function f defines a composition of n single-swap permutations, with the first one being always identity permutation (1 1). Use induction:

• For n=1 the only f function f={ (1, 1) } defines an identity permutation, the only permutation of the one-element set.
• If each function from R_n defines a permutation from S_n then extending f for (n+1)^st argument defines a permutation of (n+1)-items set. Each value of f(n+1) for the same f gives its own permutation, and all (n+1)-items' permutations for different f′s are diferent — so extending a domain by one gives a new set of (n+1)! functions covering a new set of (n+1)! permutations.

So the bijection holds for each natural n. --CiaPan (talk) 07:06, 7 June 2010 (UTC)

PS. If we modify the 'algorithm' to insert A[i] before A[f(i)] and shift respective items upwards, instead of just swapping two items, it would be much easier to trace similarities and correlations between permutations it produces. CiaPan (talk)

f	A
	i=1	i=2	i=3
(1, 1, 1)	(1, 2, 3)	(2, 1, 3)	(3, 2, 1)
(1, 1, 2)	(1, 2, 3)	(2, 1, 3)	(2, 3, 1)
(1, 1, 3)	(1, 2, 3)	(2, 1, 3)	(2, 1, 3)
(1, 2, 1)	(1, 2, 3)	(1, 2, 3)	(3, 1, 2)
(1, 2, 2)	(1, 2, 3)	(1, 2, 3)	(1, 3, 2)
(1, 2, 3)	(1, 2, 3)	(1, 2, 3)	(1, 2, 3)

Very smart... So the inverse procedure (to the first algorithm), to represent a permutation $\sigma \in \ S_{n}$ in your form as product of single-swaps $(i\ f(i))$ with$f\in R_{n}$ should be: start from the last one, and define $f(n):=\sigma (n)$, and this reduces the problem to the one of representing an element of $S_{n-1}$. Is it so? --pma 10:00, 7 June 2010 (UTC)

That's right. Essentially the same applies to the algorithm with insertions, the only difference is one needs to replace the maximum value in the n-elements sequence permutation with its last term when reverting the swapping algorithm or remove the max item form a sequence (i.e. reindex the sequence tail) when reverting the inserting algorithm. These routines describe respective inverse functions (S_n→R_n) of the bijection. --CiaPan (talk) 06:08, 8 June 2010 (UTC)

Is the following article removed?

Is Newton's law rigth?

1-Is F*dt=m*dv rigth? If yes,

2-Is F*r*dt=m*r*dv energy conservation equation rigth? If yes,

3-Is 1/2*m*V^2+m*g*r+1/2*I*w^2=Ct total energy equation rigth? If yes,

4-Is I*w^2=constant as an innate energy? That is I*w0^2=I*wn^2. If yes,

5-Is 1/2*m*V1^2+m*g1*r1= 1/2*m*V2^2+m*g2*r2? (at position #1 and #2). If yes,

6-Is V^2=Vr^2+Vp^2 ? (Vorbital^2=Vradial^2+Vperpandicular^2) (Vectors).If yes,

7-Is Vr1^2+2*g1*r1=Vr2^2+2*g2*r2 rigth? (Vp does not change the equality).If yes,

8-Is Vp1=Vp2=Vp3=....=Vpn=Constant rigth? If yes,

9-Why Wiki says r*Vp=Constant,and deducts Kepler's area law?

10- If (1/2*m*Vr^2+m*g*r+1/2*I*w^2=m*r*dVr) the energy conservation equation rigth ,then

11-Is 1/2*m*(dr/dt)^2+K/dt^2=m*r*d(dr/dt)/dt where (K/dt^2=m*gr*r+1/2*I*w^2)rigth? If yes,

12-Is r'^2+2*K/m/dt^2=2*r*r" the differential equation rigth? If yes,

13-Is r=-a*t^2+a*t*tmax+K where K=-a*tmax^2/4=Constant and this solution is rigth? If yes

14-How to comment r=-a*t*(t-tmax) graph on polar plane coordinates?

15-Does the last equation indicate the orbital motion of the planets? (3 dimension). TASDELEN (talk) 19:40, 6 June 2010 (UTC)

3 is wrong. By mgr you probably meant the gravitational potential energy, but this is only true for constant acceleration due to gravity, and not for inverse square law. -- Meni Rosenfeld (talk) 19:48, 6 June 2010 (UTC)

• g is variable as stated in 5 (g1;g2;...gn) —Preceding unsigned comment added by TASDELEN (talk • contribs) 15:45, 7 June 2010 (UTC) TASDELEN (talk) 15:47, 7 June 2010 (UTC) —Preceding unsigned comment added by TASDELEN (talk • contribs)

Please look at the derivation of gravitational energy. In fact $mgr$ can work, but you need a negative sign. So you have $mv_{1}^{2}/2-mg_{1}r_{1}=mv_{2}^{2}/2-mg_{2}r_{2}$. -- Meni Rosenfeld (talk) 17:27, 7 June 2010 (UTC)

• Okey.Use negative sign.This doesn't change the statement 8. TASDELEN (talk) 18:44, 7 June 2010 (UTC)

7 is wrong, even with a minus sign. Vp most definitely does change the equality. -- Meni Rosenfeld (talk) 19:50, 7 June 2010 (UTC)

• If 3 is correct, with variable g, we write:

1/2*m*V1^2+m*g1*r1+1/2*I*w1^2=1/2*m*V2^2+m*g2*r2+1/2*I*w2^2 where w1^2=w2^2 (innate), then

from statement 6 (V^2=Vr^2+Vp^2) we write Vr1^2+Vp1^2=Vr2^2+Vp2^2, we substitude, we find:

1/2*m*(Vr1^2+Vp1^2)+m*g1*r1=1/2*m*(Vr2^2+Vp2^2)+m*g2*r2

In an attraction field the work done in the perpandicular direction to the direction of the field equal ZERO. (Physics). So,

1/2*m*Vp1^2=1/2*m*Vp2^2 (Vp doesn't modifie the equality, doesn't affect the equation). Statement 7 should be correct. TASDELEN (talk) 08:14, 8 June 2010 (UTC)

The radial direction depends on where the planet is. Even if no force at all (and hence, no work) was applied to the object, at one point its velocity is perpendicular to radial, and at a different point, the very same velocity has a radial and perpendicular component. Thus Vp is different between points, and you can't just ignore it. -- Meni Rosenfeld (talk) 08:25, 8 June 2010 (UTC)

TASDELEN - your fundamental error is in step 8, where you assume that Vp is constant. This is incorrect, except in the special case of a circular orbit. I think your reasoning may be that in a rotating frame of reference that rotates with the particle (i.e. with angular speed Vp/r) then the only forces acting on the particle are gravitational attraction and the centrifugal force, which are both radial, so only Vr changes, not Vp. However, if r is not constant, there will also be a Coriolis force acting perpendicular to the radius, which then means that Vp changes with time as well. As r decreases, Vp increases, and vice versa. The correct conserved quantity is rVp, which we can deduce from first principles or from conservation of angular momentum. Gandalf61 (talk) 09:36, 8 June 2010 (UTC)

TASDELEN, perhaps it may be easier to read your post if you use the Wiki commands _{Type Text Here} and ^{Type Text Here}. PST 12:44, 8 June 2010 (UTC)

• I don't assume Vp=Constant. The energy conservation equation 3 gives Vp=Ct. Will you please write yourself the energy conservation equation for different positions and discover if Vp is variable or constant? On other hand, as long as Vr varies, the orbit is not circular. So statement 8 is not only for circular orbit but for any shape of the orbit (at any position on the orbit). I don't think " As r decreases, Vp increases, and vice versa". In this concern the physical law (In an attraction field the work done in the perpandicular direction to the direction of the field equal ZERO.) should be considered. It may be more easier if you estimate, on Cartesian coordinates, Vp as the x ordinate divided by t. That is dx/dt=Vp. And then dy/dt=Vr. And the shape is a parabola. TASDELEN (talk) 18:14, 8 June 2010 (UTC)

This is wrong, for reasons I've already explained. $|dx/dt|=|v_{p}|$ only at an instant when the planet is directly above or below (that is, in the y-direction from) the sun. As the planet moves, some energy is transferred between radial and perpendicular motion, simply because the identities of "radial" and "perpendicular" change - even without any work being done. -- Meni Rosenfeld (talk) 18:34, 8 June 2010 (UTC)

TASDELEN - I think I see why you think Vp is constant. You think that phrase "the work done in the perpendicular direction to the direction of the field is zero" means that (1/2)m(Vp)² must be constant. You think of kinetic energy as having directional components parallel to and perpendicular to the field. This is incorrect. The phrase means that no work is done on or by the particle due to a displacement that is perpendicular to the field. In general

${\text{Work done on particle}}=\int \mathbf {F} .d\mathbf {x}$

In polar co-ordinates (r, θ) we have

${\text{Work done on particle}}=\int \mathbf {F} .(d\mathbf {r} +rd{\boldsymbol {\theta }})$

If F is radial then it is always perpendicular to dθ, so this reduces to

${\text{Work done on particle}}=\int \mathbf {F} .d\mathbf {r}$

Now, if F is attractive and

$|\mathbf {F} |={\frac {GMm}{r^{2}}}$

then

${\text{Work done on particle}}=-\int {\frac {GMm}{r^{2}}}={\frac {GMm}{r_{2}}}-{\frac {GMm}{r_{1}}}$

We know that energy gained by particle = work done on particle. So if we can ignore any internal energy of the particle (such as its own rotational energy) we have

${\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}={\frac {GMm}{r_{2}}}-{\frac {GMm}{r_{1}}}$

$\Rightarrow {\frac {1}{2}}mv^{2}-{\frac {GMm}{r}}={\text{constant}}$

which is the correct version of your Step 3 equation.

If r is constant we can conclude that v is also constant, and we have a circular orbit at constant speed. If r decreases then v² increases - but you cannot conclude that the change is all in the radial component of v and that the perpendicular component of v (your Vp) is constant. Kinetic energy does not have directional components - it is a scalar. We find, in fact, that the magnitudes of both the radial component and the perpendicular components of v change, because the angular momentum of the particle m(r x v) (which has magnitude mrVp) must also remain constant. From the point of view of the particle, it appears to experience a Coriolis force that is perpendicular to its radial motion, and that changes Vp. Gandalf61 (talk) 09:57, 9 June 2010 (UTC)

Six Easy Pieces by Feynman is a nice popular book that deals with this problem in one of its chapters. Dmcq (talk) 12:12, 9 June 2010 (UTC)

• Thanks for your explanations.I have posted my comments on (User talk:TASDELEN ‎ (→Reference desk)) as follows:

My entry in TASDELEN's discussion page was copied to this place, but without the formulas, so it is incomprehensible and I am deleting it. Bo Jacoby (talk) 22:04, 9 June 2010 (UTC).

My entry in TASDELEN's discussion page was copied here, but I've removed it. TASDELEN is replying to my assumption that he was attempting to overturn Newton & Kepler. Dbfirs 11:48, 11 June 2010 (UTC)

• Sorry for disrupting Wikipedia. I am not overturning Newton,(this is your mis-understanding) but Kepler's laws with the laws of Newton. I say: Newton's laws do not confirme Kepler's laws. That is all.(Is this an overturning of Newton's laws ?). And for this I ask Wiki people to not to copy -paste the old mathematicians math but make the derivations themselves. For example: you mean d(Vp)/dt=0. You write d(r*Vp)/(dt*r)=0.It is not you who write this, you just copy-paste the article (Kepler's_laws#Acceleration_vector). I don't want to consider this as an answer. According me, this is to multiply the inside of the parenthesis with (r) and divide the outside of the parenthesis with (r). So, you say nothing has changed, as you multiply and divide with the same value (r). Then by integrating, you deduct (r*Vp=Constant). Tricky math, hein! Multiply the interior of the parenthesis with (r^2) and then divide the exterior of the parenthesis by (r^2). And deduct (r^2*Vp=Constant). You discover the "Volume's law of Wiki"!!. Good job. TASDELEN (talk) 15:52, 9 June 2010 (UTC)

I have derived Kepler's laws from Newton's from scratch multiple times as, I expect, have the other people trying to explain this to you. It works. I'm not going to waste my time trying to explain exactly where you've gone wrong on each of the many mistakes you have made when you clearly don't actually want to learn. --Tango (talk) 16:55, 9 June 2010 (UTC)

• Last chance: Is d(Vp)/dt=0 according (Kepler's_laws#Acceleration_vector)? Yes or no. TASDELEN (talk) 17:31, 9 June 2010 (UTC)

• • With this post,I point out,where I think there is a mis-derivation of Wiki.d(Vp)/dt=0 doesn't means d(r*Vp)/(dt*r)=0. Mathematical operation inside-outside an integrable parenthesis should not be correct. TASDELEN (talk) 18:15, 9 June 2010 (UTC)

See also Crank (person). Bo Jacoby (talk) 20:33, 9 June 2010 (UTC).

• Nice definitions!!

Canonically you are rigth. Here, canons are Wiki's comments, writings. You can not argue against Wiki's canons. Otherwise, you have to denie, to refuse your position, your prestige, your respects. You may be excommunicated from Wiki, if you approche to approve me. But, you are autorized to think again and again: "are you rigth"?.Canons do not mean the truth, every time, every where. TASDELEN (talk) 11:55, 10 June 2010 (UTC)

Ok. But since we Wikipedians are a bunch of idiot yesmen who cannot think for themselves but only confirm each other for fear of being excommunicated, I have to ask you - why do you care what we have to say? Is your time not better spent convincing more enlightened people? -- Meni Rosenfeld (talk) 13:26, 10 June 2010 (UTC)

Yes that is correct. Wikipedia is bound by WP:No original research. We have to repeat what the books say so bringing new ideas here without getting them published is a waste of time. Dmcq (talk) 13:47, 10 June 2010 (UTC)

If I understand your point of view correctly, you believe that Kepler's laws are wrong while Newton's laws are correct. Well, Newton formulated his laws based on Kepler - not the other way round. He showed that the Kepler orbits are solutions to his equations of motion. Newton's achievement has been admired by mathematicians for 3½ centuries, but you believe that he is wrong and that you are right. In order to convince anybody of that you must communicate with utmost care. It is not impossible! Einstein did it. You must convince that your English is perfect, that your formulas are perfect, and that your logic is perfect. These three point, which ought to be your strong points, are actually your weak points. Certainly this does not prove you wrong, but I doubt that you are taken seriously, and I think that you do qualify as a crank. When you work on your three weak points you will quietly improve you skills as a mathematician and as a communicator, and you may be aware of your errors, as most of the rest of us do every now and then. Good luck! Bo Jacoby (talk) 14:24, 10 June 2010 (UTC).

• Confessions? Or giving up ? In physics and maths, words or statement have a value when they are supported with absolut maths.

I am a mechanic, wrench and hammer on oily hands, not a mathematician, nor a physist. I assumed you are the experts of maths and other sciences. I think you wish to close this discussion due to my "crank logic". Okey. Thank you any how, and sorry if I disturbed your ideas, your believings. TASDELEN (talk) 18:40, 10 June 2010 (UTC)

There is nothing wrong about being a mechanic. Your interest in physics is a good thing, and you are very welcome to ask questions here. But don't lecture us. Bo Jacoby (talk) 22:19, 10 June 2010 (UTC).

As a mechanic, I expect you have an excellent understanding of rotational dynamics about a fixed axis. Planetary rotation has many similarities, but also some important differences. Newton showed that his gravitational law predicts Kepler's laws which were devised to match observations by Tycho Brahe. If you wish to disturb our ideas and believings, you will have to show, as Einstein did, that your theory gives a better match with the observations. Meanwhile, thank you for making us think. Dbfirs 11:48, 11 June 2010 (UTC)

• It seems,I disturbed you enormly. Sorry. Please agree my apologizings. My intention was not to disturb your believings. I believe to the laws of Newton. I considered F*dt*r=m*dv*r , wrote its differential, discovered r=-a*t*(t-tmax)+K. That was the equation of the planetary motion, according me. That was against Kepler's laws, said to be confirmed by Newton (or attributed to Newton by later mathematicians). Two different solutions using Newton's laws; one of the solution should be wrong. I just wanted to share my reasoning with you. You considered it as a lecture. And refused to comment on this equation of motion. According Wiki's canons you are rigth. Forget please all I wrote you. I feel you started to think on this matter. Thanks88.234.182.137 (talk) 20:23, 11 June 2010 (UTC).This is TASDELEN (talk).Automatic sign is wrong.