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March 8

Bouncing ball confinement

When a ball is dropped in vacuum in a uniform gravitational field from a height $h$ onto an immovable surface whose lowest point is at height $0,$ obeying an idealized version of Newton's laws and bouncing back totally elastically, it can theoretically bounce back to its release height, possibly only after many bounces, but will never get higher than $h.$ This is an easy consequence of the preservation of energy, itself a consequence of Newton's laws, combined with the fact that kinetic energy, initially zero, is a nonnegative quantity.

Watching a video illustrating the butterfly effect in a 2D setting, in which 100 balls with imperceptibly different initial positions all very close to $(x_{0},y_{0})$ are simultaneously dropped on the curve given by $y=|x|,$ it seemed to me that the balls basically remained confined to the rectangle with vertices

$(0,0),\left({\frac {x_{0}{+}y_{0}}{2}},{\frac {y_{0}{+}x_{0}}{2}}\right),(x_{0},y_{0}),\left({\frac {x_{0}{-}y_{0}}{2}},{\frac {y_{0}{-}x_{0}}{2}}\right).$

This is a much stronger confinement than $y\leq y_{0}.$ Is there some simple mathematical argument supporting my observation?  --Lambiam 19:58, 8 March 2024 (UTC)

Don't have an answer, but someone else has a video which documents essentially the same phenomenon, this time tracing a path. The video title says "square" but I'm fairly sure that's just an artifact of the ball being dropped close to the apex. GalacticShoe (talk) 07:07, 9 March 2024 (UTC)

Indeed, one of the replies to the video is just the three words, "not a square". When $x_{0}=0$ the trajectory is either undefined or confined to the line $x=0$, but the limit case of the hypothesized rectangle for $x_{0}\to 0$ is a square. (The best way to define the bounce-back when the trajectory hits $(0,0)$ is to have the ball move back in the direction it came from, the limit case of hitting $(\pm \varepsilon ,\varepsilon ).$)  --Lambiam 10:06, 9 March 2024 (UTC)

A few thoughts: The second video shows that the boundary is the envelope of all the possible trajectories. The possible trajectories are parabolas $y=y_{\mathrm {max} }-b(x-x_{0})^{2}$, with the apex at $x_{0}$ (to be shown that this is always so), height given by $y_{\mathrm {max} }=y_{0}-{\tfrac {1}{2}}{\dot {x}}_{0}^{2}$, and $b$ to be expressed in terms of the transverse velocity at the apex ${\dot {x}}_{0}$ as well. This should give the one-parameter family of parabolas, and somebody should be able to determine the envelope. --Wrongfilter (talk) 07:42, 9 March 2024 (UTC)

Thanks. This probably gives enough hints to prove the confinement hypothesis, after which it may not be a trivial exercise to reduce the proof to a much more elementary one based on invariants following from Newton's laws plus elasticity.  --Lambiam 10:06, 9 March 2024 (UTC)

I think the first thing to do is apply the conservation of energy, if the apex of the parabola is at $y_{1}$ then we have:

${\frac {Energy}{m}}=const.=y_{0}g=y_{1}g+{\frac {{\dot {x}}^{2}}{2}}.$

Assume for now that this occurs at $x=x_{0}$; we'll verify this later. Call the value of ${\dot {x}}$ for a specific path $v_{x}$. Parametric equations for the parabola are given by:

$x=x_{0}+v_{x}t,y=y_{1}-{\frac {gt^{2}}{2}}=y_{0}-{\frac {{v_{x}}^{2}}{2g}}-{\frac {gt^{2}}{2}}.$

Eliminate t and write $k={\frac {{v_{x}}^{2}}{g}}$ to get an implicit form for the equation:

$y-y_{0}+{\frac {k}{2}}+{\frac {(x-x_{0})^{2}}{2k}}=0,$

where k is a constant determining the path. The next step is to verify that that if two of these curves intersect on the line $x=y$, then their angles with the line are the same. That will confirm then when the ball bounces off the line the new curve will be in the same family, which is what we assumed at the beginning. Suppose the two curves

$y-y_{0}+{\frac {k}{2}}+{\frac {(x-x_{0})^{2}}{2k}}=0$

$y-y_{0}+{\frac {l}{2}}+{\frac {(x-x_{0})^{2}}{2l}}=0$

intersect at (x, y) with k≠l. Eliminating y:

$kl=(x-x_{0})^{2}.$

The normal vectors to the two curves are given by

$({\frac {x-x_{0}}{k}},1),({\frac {x-x_{0}}{l}},1).$

The reflection of the first vector by the line x=y is

$(1,{\frac {x-x_{0}}{k}})$

and by the previous equation this is a scalar multiple of the second vector, and this proves that the directions are reflections of each other. The corresponding fact for y=-x should be similar. Note that we did not use x=y anywhere, and I think this implies that you could insert short diagonal line segments into the "field of play" and the result would be the same. The balls may bounce around as in a pinball machine but they would still be constrained to be in the family of parabolic paths. Finally, it remains to find the envelope of the family. This is done by setting the partial derivative with respect to k of the implicit equation to 0 and eliminating k. We get:

${\frac {\partial }{\partial k}}(2k(y-y_{0})+k^{2}+(x-x_{0})^{2})=0,$

which gives:

$k=-(y-y_{0}).$

Plug this back into the original equation to produce:

$(x-x_{0})^{2}=(y-y_{0})^{2},$

and this gives the two lines $y-y_{0}=\pm (x-x_{0})$ in question. As far as the original question is concerned I'm not sure if this counts as "simple" or not. There is a lot of messy algebra going on but conceptually it's not that hard. Consider too that this is from someone specializing in math and having little working knowledge of physics. I'm sure someone who has taken a few more physics courses than I have would have more tricks for tackling this sort of thing mare easily, using additional physical invariants etc. --RDBury (talk) 20:09, 9 March 2024 (UTC)

PS. If you don't like the envelope part there is a simpler approach. If

$2k(y-y_{0})+k^{2}+(x-x_{0})^{2}=0$

has a solution in k then the discriminant must be non-negative, in other words

$(y_{0}-y)^{2}\geq (x_{0}-x)^{2}.$

We know that $y_{0}-y\geq 0$ from which we can deduce $y\leq y_{0}-|x_{0}-x|.$ --RDBury (talk) 20:35, 9 March 2024 (UTC)

PPS. (Some additional random thoughts.) If you replace k with z in the above equation, the result is the equation of a surface in space. This turns out to be a cone and slicing the cone by fixing z=k gives you conic sections. These turn out to be the parabolic trajectories. When you project the cone to the x-y plane you get the region $(y_{0}-y)^{2}\geq (x_{0}-x)^{2}.$ as before.

I think this could be generalized as follows. Start with any curve, say x=x(k), y=y(k) parameterized by k. Assume the apex of each parabola is on this curve, then the equation for each parabola can be worked out using the conservation of energy as above. This gives an equation in x, y and k. If this turns out to be quadratic in k then there is a region where each point is crossed by two of the parabolas. You can work out their tangent lines at that point, find the bisectors, and by assuming that a new curve is tangent to one of the bisectors you get a differential equation for the new curves. There are two bisectors so there are two differential equations, and the two sets of solutions would form orthogonal trajectories. In the case where the starting curve is a vertical line the resulting families of curves turn out to be diagonal lines. It's not hard to see that if the starting curve is a horizontal line then you get vertical and horizontal lines. The upshot in that case is that if you drop a ball in to a rectangular well then it will keep bouncing back to original height. I suppose the next easiest case would be when the starting curve is a diagonal line. The more complex the starting curve the more difficult it will be to work out the envelope of the parabolas and solve the differential equations. In the case where g=0 this is related to reflected caustics

Going back to the original problem, instead of saying that the ball bounces off the wall, say that it goes through but now the gravitational force is reflected through the wall. The result would be a gravitational field of the form f=(0, -g) if y>|x|, (-g, 0) if x>|y|, f=(0, g) if y<-|x|, and (g, 0) if x<-|y|. (Sort of like the taxicab metric but with gravity.) This is not a central field so the conservation of angular momentum would not hold, and indeed it's not hard to see that an "orbit" can change direction sometimes. But there is a well defined potential function, namely g⋅max(|x|, |y|).

Finally, are we sure that we're observing an actual butterfly effect here, or is it an artifact of the simulation? Looking at some of the other bouncing ball simulations on the same channel, with most of them the balls become a chaotic mess limited above only by the original height. With g=0 and the container a square, the path will eventually get arbitrarily close to any point in the interior provided that the starting angle is irrational. But that's not chaotic behavior since small changes in the starting position do not result in increasingly larger changes in future positions. Are we sure that something similar wouldn't happen if the simulation was perfectly accurate? --RDBury (talk) 18:33, 11 March 2024 (UTC)

It is hard to determine visually whether the system is chaotic. It may make a difference that we are dealing with a piecewise linear curve. Does it have a positive Lyapunov exponent? (Rather than $|\delta \mathbf {Z} (t)|\approx e^{\lambda t}|\delta \mathbf {Z} _{0}|,$ I think the criterion should be $|\delta \mathbf {Z} (t)|=O(e^{\lambda t}|\delta \mathbf {Z} _{0}|)$ for some $\lambda >0$; "the" Lyapunov exponent should then be the supremum of the range satisfying the condition.) But with a bounded phase space the supremum is necessarily zero, so we need a modified definition, perhaps something related to the notion of entropy.

Not obviously related, but perhaps there is a connection: a version with more balls shows an intriguing phenomenon at the end of 0:41.  --Lambiam 22:52, 11 March 2024 (UTC)

The fact that you can still make out the moose at the end of video makes me suspicious. There is another video comparing walls y=2|x| and y=.5|x|. Those are piecewise linear but the case for those systems being truly chaotic is a lot more believable for me. I suspect that the two lines being at right angles is the critical factor, but I'm not going to write my own simulator just to find out. Note that the rectangular well I mentioned above does not produce chaotic motion. --RDBury (talk) 06:16, 12 March 2024 (UTC)

March 16

Smallest triangular number with prime signature the same as A025487(n)

Smallest triangular number with prime signature the same as (sequence A025487 in the OEIS)(n), or 0 if no such number exists. (there is a similar sequence (sequence A081978 in the OEIS) in OEIS)

For n = 1 through n = 29, this sequence is: (I have not confirmed the n = 15 term is 0, but it seems that it is 0, i.e. it seems that there is no triangular number of the form p^3*q^2 with p, q both primes)

1, 3, 0, 6, 0, 28, 0, 136, 66, 0, 36, 496, 276, 0, 0, 118341, 120, 0, 1631432881, 300, 8128, 210, 0, 528, 0, 29403, 1176, 32896, 630

Is it possible to extend this sequence to n = 100? 1.165.194.85 (talk) 00:51, 16 March 2024 (UTC)

This paper by Cohn implies that there are no prime solutions $(p,q)$ to $p^{3}+1=2q^{2}$ and $2p^{2}+1=q^{3}$. If anyone here can prove that the elliptic curves $y^{2}=x^{3}\pm 4$ have no nontrivial integral points (for $-4$ that would be $(2,\pm 2)$ and $(5,\pm 11)$, while for $4$ that would be $(0,\pm 2)$), then that implies that there are no triangular numbers of the form $p^{3}q^{2}$ at all. GalacticShoe (talk) 06:22, 16 March 2024 (UTC)

I'm not sure what you mean by "at all". If you drop the requirement that q is prime then there are many solutions, for example 242⋅243/2 = 3³33², 12167⋅12168/2 = 23³78². --RDBury (talk) 08:42, 16 March 2024 (UTC)

Sorry, shoulda clarified; no triangular numbers of the form $p^{3}q^{2}$ at all where $p,q$ are prime. Naturally, the two cases mentioned earlier (the latter example of which I hastily posted before remembering that I completely forgot about the prime signature part) are either not covered by Cohn's paper, with the $242$ case being invalid since we spread powers of $3$ among the cube and the square, or they are one of Cohn's special case, as with $12167$. GalacticShoe (talk) 09:51, 16 March 2024 (UTC)

Thanks for the clarification. I don't have access to the Cohn paper, but it seems to me that the p³+1=2q² is trivial to exclude, and similarly for p³-1=2q². That leaves 2p³±1=q². Using brute force I found 2⋅23³+2=156² but none, even when p is only required to be odd and q can be anything, where the difference is one. Note that this generates the 12167 example, but that's not the way I found it. I'm not sure how y²=x³±4 is related, but I'm no expert on elliptic curves so I may have to take your word on that. In any case, given that the p³q² is proving so difficult, it seems unlikely that getting to n=100 is feasible. Finding entries where there is a solution should only require a computer search, but if none are found then proving that there are none to be found can be difficult. It seems ironic that while there is apparently no p³q² solution, there is a p⁵q², though this seems less likely at first glance. RDBury (talk) 18:24, 16 March 2024 (UTC)

The $y^{2}=x^{3}\pm 4$ condition arises because, if we write $x=2p$ and $y=2q$, then an integer solution to $y^{2}=x^{3}\pm 4$ implies $(2q)^{2}=(2p)^{3}\pm 4\Rightarrow 4q^{2}=8p^{3}\pm 4\Rightarrow q^{2}=2p^{3}\pm 1$ is a rational, possibly integer solution to the equalities we are looking at. All integer solutions to $q^{2}=2p^{3}\pm 1$ can be generated this way, so the lack of a nontrivial integer solution to $y^{2}=x^{3}\pm 4$ (or the existence only of solutions where $x,y$ are odd) would rule out any further triangular numbers of the form $p^{2}q^{3}$ with $p,q$ prime. GalacticShoe (talk) 21:33, 16 March 2024 (UTC)

Right, I should have seen that. The 2p³+1=q² case may be easier than I thought; it reduces to 2p³=(q+1)(q-1). You can rule out p=2, q=2, giving q odd, but then by unique factorization q±1 divides p, which is impossible. I thought of applying a similar trick to 2p³-1=q², or 2p³=(q+i)(q-i), and using unique factorization over Gaussian integers. But I quickly got bogged down with this. --RDBury (talk) 02:10, 17 March 2024 (UTC)

A similar disproof is that after proving $p=2$ and $q=2$ impossible by simple casework, $q$ odd implies that $(q+1)(q-1)$ is a multiple of $4$, which $2p^{3}$ cannot be. I imagine that $2p^{3}-1=q^{2}$ is indeed not as simple to pin down. GalacticShoe (talk) 02:45, 17 March 2024 (UTC)

So if $y^{2}=x^{3}\pm 4$ has no integer solution other than $(0,\pm 2)$, $(2,\pm 2)$, $(5,\pm 11)$, than a(15) in the above sequence is 0? How about a(30), a(31), a(32), …? Do you have the status for a(n) for all $n\leq 100$? (sequence A081978 in the OEIS) has the proof for which some prime signatures do not exists for triangular numbers such as $p^{6}q^{2}$ (i.e. a(33) = 0), see the comment of Jinyuan Wang in Aug 22 2020. (You can also check whether a(1) ~ a(29) which I have listed above are correct) (a(n) is the smallest triangular number with prime signature the same as (sequence A025487 in the OEIS)(n), or a(n) is 0 if no such number exists) 61.224.130.152 (talk) 03:21, 18 March 2024 (UTC)

I mentioned this a few paragraphs back. Given that finding the value of a single entry is turning out to be a project in itself, and that it seems unlikely that computations will get any easier going on, I don't see getting to 100 as feasible. A081978 is asking for less information, and even so it took some high powered number theory to get it's values. --RDBury (talk) 15:36, 18 March 2024 (UTC)

March 18

Tautology logical reasoning

Does every tautology have a verbal logical reasoning not requiring any use of truth tables or logical connectives, just like combinatorial identities/formulae (at least many) have combinatorial reasonings?
For example, this tautology here: ${\bigl (}a\to (b\to c){\bigr )}\to {\bigl (}b\to (a\to c){\bigr )}$. יהודה שמחה ולדמן (talk) 18:44, 18 March 2024 (UTC)

Well one can always read out the maths! But it would become very tedious and you might lose attention for anything much more complicated than the above. That though can be transformed by saying: A implies B means A is false or B is true. Then A implies B implies C means the same as A is false or B is false or C is true and that is the same as B is false or A is false or C is true. I mean what are you looking for beyond that? To start using the ancient business of syllogisms with names like Baroco? NadVolum (talk) 19:30, 18 March 2024 (UTC)

I mean literally, how do we deduce with logical reasoning that the initial part $a\to (b\to c)$ implies the final part $b\to (a\to c)$?

Of course I understand that we use symbols to express arguments, such as $a\to b=(\neg a\land \neg b)\oplus (\neg a\land b)\oplus (a\land b)=\neg a\lor b$.

But I am also asking if this is a definition apriori (${\stackrel {\rm {def}}{=}},{\stackrel {\triangle }{=}}$) or a theorem. יהודה שמחה ולדמן (talk) 20:53, 18 March 2024 (UTC)

If I tell you that $(a\,\natural \,(b\,\natural \,c))\,\natural \,(b\,\natural \,(a\,\natural \,c))$ is a tautology without supplying an interpretation of the connective "$\natural$" and ask you to provide verbal reasoning for this tautology, you will not be able to make any inroads. You can't do anything if you don't know the meaning of "$\natural$". So it is not clear to me what kind of "verbal reasoning" might correspond to this tautology without referencing the interpretation of the material conditional connective "$\rightarrow$". To me, this would be like requiring a verbal proof of $a+b=b+a$ without any use of addition. The tautology can be verbalized: "If we know that if A is true, then if B is true then C is true, it follows that if B is true, then if A is true then C is true." If you can wrap your head around it, this should be self-evident, but a proof by natural deduction of the tautology can also be verbalized. ("Assume we know that if A is true, then if B is true then C is true. We want to show that it follows that if B is true, then if A is true then C is true. So assume that B is true. ...") However, unless one has an uncanny ability, one will get lost in keeping track of which assumptions hold at various steps and which have been discharged.  --Lambiam 08:49, 19 March 2024 (UTC)

March 20

Whether it coincides with a simpler function

Is y = sin (arcsin x) (1) the same function as y=x, if we consider all branches of logarithm (of any real number) and all branches of inverse sine function? Or does (1) remain meaningless for any argument outside the range [-1;1] when we restrict it to real value for both the domain and the image, and (1) will coincide with the identity function only when we regard it as a function that map complex numbers to complex number? Does the logarithm of negative numbers lead to the presence of removable singularities for (1)? (In contrast, the function y=x obviously does not contain any singularity). I was able to prove that y = arcsin (sin x) and y = sin (arcsin x) are not always the same, but I still can't settle the aforementioned problems. 2402:800:63AD:81DB:105D:F4F:3B26:74C5 (talk) 14:36, 20 March 2024 (UTC)

A univalued function and a multivalued function $f:X\to Y,$ possibly partial, can be represented by a relation $R_{f}\subseteq X\times Y.$ The total identity function ${\text{id}}:X\to X$ corresponds to the identity relation $I_{X}=\{(x,x)|x\in X\}\subseteq X\times X.$ Function composition corresponds to relation composition: $R_{(g\,\circ f)}=R_{f}\,;R_{g}.$ The multivalued function inverse correspond to relation converse: $R_{(f^{{-}1})}=(R_{f})^{\text{T}}\!.$

Just like the multivalued complex logarithm is the multivalued inverse of the exponential function $\exp :\mathbb {C} \to \mathbb {C}$, the complex $\arcsin$ including all branches is the multivalued inverse of function $\sin :\mathbb {C} \to \mathbb {C} .$ So

$R_{(\sin \circ \arcsin )}=R_{\arcsin }\,;R_{\sin }=(R_{\sin })^{\text{T}};R_{\sin }.$

Generalizing this from the sine function to an arbitrary (univalued) function $f$, we have:

$(x,y)\in (R_{f})^{\text{T}};R_{f}\Leftrightarrow \exists z:((z,x)\in R_{f})\land (z,y)\in R_{f})\Leftrightarrow \exists z:(x=f(z)\land y=f(z)).$

Clearly, this implies $x=y,$ so the composed relation is the identity relation on the range of $f,$ representing the identity function on that range.  --Lambiam 18:21, 20 March 2024 (UTC)

March 22

Sin, cos and ellipses

In this book [1] (linked to the right page), left column towards the bottom. I'm having a problem with the "hence."

I understand that:

A: the point M has the coordinates (x,y), which is also (sin φ, cos φ), no matter how you slide the straight KL, for all φ.

B: the formula for the ellipse. $${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$$

I just don't get how B follows from A. Maybe I'm missing a concept that the authors take for granted. But shouldn't the text have explained something in between? Something like why the ellipse matches the Pythagorean trigonometric identity? Or why the set of all possibles values of M is $\sin ^{2}\theta +\cos ^{2}\theta =1$? Why is that when you have the coordinates (sin, cos) you add them and equal to 1 to produce an ellipse? Grapesofmath (talk) 17:27, 22 March 2024 (UTC)

I think you've misread the text: the point M (x, y) is actually (a sin φ, b cos φ). So x/a = sin φ and y/b = cos φ. Substituting into the identity $\sin ^{2}\phi +\cos ^{2}\phi =1$ (which is true for all φ) gives the formula in (B). AndrewWTaylor (talk) 18:09, 22 March 2024 (UTC)