Jump to content

Wikipedia:Reference desk/Archives/Mathematics/2016 February 9

From Wikipedia, the free encyclopedia
Mathematics desk
< February 8 << Jan | February | Mar >> February 10 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


February 9

[edit]

planar graph proof

[edit]

Hi all,
If a planar graph is 3-regular and 2-edge-connected, how can we prove that its chromatic index is at most 3?
Thanks a lot in advance!

The question above was asked by 217.132.96.145 (talk), 9 February 2016‎

Here's the broad outline of a proof:
1. The 2-edge-connected graphs can also be characterized by the absence of bridges. (K-edge-connected_graph#Related_concepts).
2. When χ′(G) = Δ(G) (the chromatic index is equal to the maximum degree), G is said to be of class 1. (Edge_coloring#Vizing.27s_theorem).
3. The four color theorem (proved by Appel & Haken (1976)) on vertex coloring of planar graphs, is equivalent to the statement that every bridgeless 3-regular planar graph is of class one (Tait 1880). (Vizing's_theorem#Planar_graphs). Egnau (talk) 18:54, 12 February 2016 (UTC)[reply]

Fibonacci sequence convergence

[edit]
Trolling
The following discussion has been closed. Please do not modify it.

Quoting this post by Shareef Fahmy (or Sherif Fahmy Eldeeb):

1) [T]ake the infinite Fibonacci sequence 1,1,2,3,5,8,13,21,34, etc ..... the Golden ratio is found by dividing one of the numbers by the one before it , like 34/21 or 21/13. But the two numbers are not the same exactly(34/21 = 1.6190476~ and 21/13=1.615384~) and the deviation between the two steps is 1.6190476~ - 1.615384~=.0036~ .... The ~ symbol means these decimals keep going.

2) So as you go further along the Fibonacci sequence the deviation gets smaller and smaller but it never disappears , and it actually fluctuates like a damped harmonic vibration. But you will find that it converges on Plancks constant which is (6.626 X 10^-34) as you tend to infinity.

Claim #1 above is documented at Fibonacci_number#Relation_to_the_golden_ratio. Is Claim #2 true, false or even falsifiable? AllBestFaith (talk) 21:03, 9 February 2016 (UTC)[reply]

Claim 2 is total nonsense. (Claim 1 is also only partially sensible -- certainly a mathematician would never speak so loosely.) --JBL (talk) 21:16, 9 February 2016 (UTC)[reply]
Planck's constant has units. Its value is 6.626 x 10^-34 J*s, but it's 4.135 x 10^-15 eV*s, or 6.626 x 10^-27 erg*s, etc. So claim 2 is indeed nonsense. Mnudelman (talk) 21:23, 9 February 2016 (UTC)[reply]
If you have observed claim 2 then I suspect that it is just a coincidence caused by rounding errors in your method of division. Dbfirs 21:47, 9 February 2016 (UTC)[reply]
What limit is observable? AllBestFaith (talk) 00:23, 10 February 2016 (UTC)[reply]
What is the sound of one hand clapping? --JBL (talk) 00:26, 10 February 2016 (UTC)[reply]
Please treat the question seriously or not at all. AllBestFaith (talk) 00:55, 10 February 2016 (UTC)[reply]
It is not a serious question: it is a poorly formulated non sequitur, bearing no discernible relation to anything anyone has said. If you want your questions taken seriously you should do a better job asking them. --JBL (talk) 01:39, 10 February 2016 (UTC)[reply]
The ratio of consecutive terms of the Fibonacci series converges to the golden ratio. As a result, the difference between two such ratios tends to zero, not Planck's constant (in any system of units). Sławomir
Biały
01:03, 10 February 2016 (UTC)[reply]
The above is correct. The difference will tends to 0. You can easily prove it. You can also see that at every step the difference is consistently divided by approximately -2.618. (It will reach 10^-34, and then will keep going further down).Dhrm77 (talk) 01:17, 10 February 2016 (UTC)[reply]
The author of the article quoted at the top did not prove his point, but found it by doing a graph. He is most likely victim of rounding errors of whatever tool he was using.Dhrm77 (talk) 01:25, 10 February 2016 (UTC)[reply]
The golden ratio is an irrational number that cannot be evaluated from a ratio of natural numbers that occur in the Fibonacci series, instead one is forced to accept a quantised approximation. The smallest quantisation that is physically possible in a calculating device, whether analog or digital, is not infinitesimal. Planck's constant is the minimal element of energy of an electromagnetic wave, arguably setting a limit on how precisely irrational numbers can be expressed, subtracted one from another or observed. Believing that the difference between irrational numbers (a Cauchy sequence has elements that become arbitrarily close to each other as the sequence progresses) converges leaves open the question "at what rate?". AllBestFaith (talk) 01:55, 10 February 2016 (UTC)[reply]
@AllBestFaith: You're mixing notions from different areas of science, which are very far apart. The minimum possible energy (whether properly defined and determined or not) would have something to do with the precision of a result in analog calculating devices (see Analog computer), which give their output as an electric voltage or a current, possibly transformed into an angle of a pointer by some ammeter. But electronic digital devices work with energies much higher than that, so their precision is limited only by the size of their memory. Of course accuracy of any engineering, scientific or other real-life related calculation is limited by the accuracy of input data (and by the accuracy of models, being used to analyze problems), so too much precision is useless in them. However calculation of numbers like or or can be arbitrarily precise, which is limited only by the room to store results and the time we want to spend on calculations. --CiaPan (talk) 06:46, 10 February 2016 (UTC)[reply]
The real numbers satisfy the completeness property. There is no "smallest quantised approximation". And indeed, using computers we can actually approximate the golden ratio as a decimal value to many more digits of accuracy than Planck's constant (using Fibonacci numbers, or otherwise). For example, in less than a second on a decent computer, we find that
which is already much much smaller than Planck's constant (in any system of units that are actually used). Here denotes the nth Fibonacci number. Of course, nothing is stopping you from using a system of units where Planck's constant is equal to , but we can then get a smaller value of the above difference just by going a bit further out in the sequence. Sławomir
Biały
12:17, 10 February 2016 (UTC)[reply]
The Planck's constant claim is falsifiable and not true. It's most certainly not nonsensical though, and I'd think anyone who understands English and a little math should be able to make out the claim. Or perhaps the above posters mean by "nonsense" that the claim is worthy of ridicule, but that's simply subjective opinion expressed poorly, and we should not engage trading opinions on a reference desk.
Anyway, the second claim reminded me in a loose way of the Gibbs phenomenon, where we do see one type of convergence to zero, while a different notion of distance remains above a positive and finite limit. We also have the notion of damped oscillation there, so perhaps OP will be interested in this somewhat related phenomenon that is true. SemanticMantis (talk) 17:00, 10 February 2016 (UTC)[reply]
This sort of "every statement should be treated seriously, even ones that are in fact just nonsense" mentality is by far the most irritating aspect of the math reference desk. The claim is pure nonsense, totally unrescuable: Plank's constant is not a real number. Treating this kind of thing as if it is serious just encourages pointless crankery. --JBL (talk) 17:19, 10 February 2016 (UTC)[reply]
A claim that a certain sequence approaches a specific limit is not nonsense. Insisting otherwise is not very sensible or professional. The claim in question is in fact quite easy to understand, and even formally defined with some rigor, if I am correct to assume that you know what a sequence is, and what a limit of a sequence is. If you want criticize the claim because Planck's constant is not a real number then that's fine, but don't act like you have no idea that the claim is in reference to the unitless real number that is the magnitude of Planck's constant. The claim is indeed incorrect, but that's not what you said, is it? Calling something nonsense doesn't help anyone. I'm not sure why anyone participates here if not to help people. If you choose to participate here, please at least attempt to be constructive and helpful, and please don't make personal attacks on the OP. SemanticMantis (talk) 19:54, 10 February 2016 (UTC)[reply]
The phrase "unitless real number that is the magnitude of Plank's constant" is itself nonsense. Did you read Mnudelman's comment above? Please try to have some idea what you're talking about. --JBL (talk) 20:06, 10 February 2016 (UTC)[reply]
Wow you keep using that word, but I don't think it means what you think it means! You are a mathematician, no? I really have a hard time believing that you cannot understand what I meant. I think you just mean "I don't like your phrasing", but in case you didn't understand my simple phrase, let me try to explain it more carefully, using words that you can understand. You said Plank's constant is not a real number. In a strict sense you are correct, because the constant has units, and real numbers are simply numbers. But when we say Planck's constant is 6.626× 10^(−34) J⋅s, I think you'd agree that the bit before the units is an expression is widely recognized as representing a real number. Are you with me so far? Now, numbers have magnitude, and so do physical constants that also have units. For example, if I say "This sequence approaches the weight of my cat, which is 10 pounds", you might say "your cat's weight is not a real number" and I would reply, "sure, but it is a physical constant that has magnitude, and that magnitude is conventionally discussed as a real number, and I'm pretty sure if you weren't acting so uncharitably you'd know that I meant 10 as a number, and was not claiming that '10 pounds' is a real number." Do you get it now? So, when I said "unitless real number that is the magnitude of Plank's constant", what I meant was the exactly that, and I'm pretty sure anyone who regularly responds at math desks would know what I'm saying. And even if my phrasing was bad, or if they misunderstood, there are plenty of ways of addressing that without calling it "nonsense". Now, I'm not sure what you're really on about here. Telling OP that something is nonsense without any explanation is simply not helpful, even if it were true. If you wanted to help the OP, you could have explained some of your reasoning. But instead, you came here to be rude, and did so with a poor choice of words, and you then proceeded to insult my too. I suggest you desist, at least until such time as you can be more civil, and perhaps learn to speak with more precision instead of misusing this "nonsense" so much. And yes, I did read the other comment. Like you, he misused the word nonsense, but at least that user explained what they meant. SemanticMantis (talk) 20:23, 10 February 2016 (UTC)[reply]
@CiaPan an operation by, or expression (display output) from, a digital calculating device is quantised in multiples of its Least significant bit (LSB). The size of computing memory is necessarily finite but observations such as Moore's law suggest that will not be an unresistant limit to variable binary wordlengths. I suggest that it is necessary to accomodate the impossibility of a digital device that works with less than a Plank quantum of energy expended in distinguishing between LSB = 0 and LSB = 1. This impossibility is a new hard limit found in quantum mechanics that was not known when the mathematics of convergent series was developed, so we are necessarily using new science with old science. @Slawomir Bialy, what you found on a decent computer is an approximation in which the equals sign is incorrect. It likely calculated with a Floating point number representation which is a trade-off between range and precision, involving two LSBs (its "smallest quantised approximations") in separately expressing the significand and exponent. AllBestFaith (talk) 17:02, 10 February 2016 (UTC)[reply]
That's irrelevant. The mathematics of convergence exists independently of the architecture of any given computer - indeed, of all possible computers. The statement 'this sequence converges' can be proved true with the normal degree of rigour, and that absolutely literally means that given any finite positive separation, no matter how small, there are two members of the sequence whose difference is smaller than that. That is, in fact, how convergence is often defined. AlexTiefling (talk) 17:16, 10 February 2016 (UTC)[reply]
(edit conflict) The convergence of series is not in the domain of science. Please go be a crank somewhere else. --JBL (talk) 17:20, 10 February 2016 (UTC)[reply]
Indeed. ABF - don't confuse mathematics with real life. And even in real life the precision of calculations in computers is not as limited as you seem to think it is - see arbitrary-precision arithmetic. Gandalf61 (talk) 17:25, 10 February 2016 (UTC)[reply]
Thank you for providing the Wikilink which is appropriate to the use of variable binary wordlengths that is already mentioned. Claim#2 "you will find" predicts the result of an operation in real life. AllBestFaith (talk) 19:12, 10 February 2016 (UTC)[reply]
"What you found on a decent computer is an approximation in which the equals sign is incorrect. It likely calculated with a Floating point number representation which is a trade-off between range and precision, involving two LSBs (its "smallest quantised approximations") in separately expressing the significand and exponent" — This really misses the point. The point is that we can very easily calculate that the difference
is far less than Planck's constant. Obviously, we cannot do this with a single 64 bit floating point number, but we can do it with an arbitrary precision library like GMP. The inability to do this in 64 bits has nothing to do with Planck's constant. That's just because of how bits work. A digital computer can return a digital value of this difference that is correct to a far greater accuracy than Planck's constant. It is true that the underlying device must satisfy the laws of quantum mechanics, but this does not by itself impose any a priori upper bound on the accuracy that is achievable, any more than it imposes an upper bound on the accuracy that is achievable my human activities, despite the fact that humans too are bound by the laws of quantum physics. 18:09, 10 February 2016 (UTC)
This statement: '10–417 is far less than Planck's constant' is false.
The number 10–417 is NOT less than Planck's constant. It's also NOT greater than that.
They are just incomparable, because one is a number while the other one is a physical quantity. It's similar to (impossible) comparing to your body weight. --CiaPan (talk) 19:52, 10 February 2016 (UTC)[reply]
See my above post, where I am more careful, and say "in any system of units", etc. I did not bother repeating this, because it seemed like our troll was only concerned about the use of the equals sign. But yes, you're right. To reiterate what I already said above "...much much smaller than Planck's constant (in any system of units that are actually used)... Of course, nothing is stopping you from using a system of units where Planck's constant is equal to , but we can then get a smaller value of the above difference just by going a bit further out in the sequence." Anyway, apparently now that I have left that out to resolve the trolls latest objection, he has latched on to the new objection, that the units are somehow relevant. I was making a very different point though. Since our troll is just interested in stirring up precisely this kind of discord, I have hatted the thread. Sławomir
Biały
20:38, 10 February 2016 (UTC)[reply]
@AllBestFaith: The first sentence (about quantization in term of LSB) is a truism, and adds nothing to your statement except noise. The second one is void, too, in part concerning the memory finiteness (nobody here declares infinite memory, just 'big enough'), and irrelevant in part about Moore's law. What concerns the third one — there is no such thing like 'digital device that works with less than a Plank quantum of energy expended in distinguishing between LSB = 0 and LSB = 1', so everything after that is talking about nothing. As a result any conclusions are meaningless, like counting angels dancing on the point of a needle. This in turn makes me think you're a troll, who finds some weird kind of pleasure in mixing topics just to make others upset. I can't see you are trying to understand what you say, let alone understanding what others say to you. --CiaPan (talk) 19:44, 10 February 2016 (UTC)[reply]
@Slawomir Bialy, ratios of numbers in the Fibonacci sequence are real numbers. The point was made earlier that Plank's constant is a physical constant that has units, so stripping it of the units to call it a real number, be it 6.626 X 10^-34 or whatever, is unjustifiable. Claim#2 makes that error but so does writing the inequality so easily stated above. My point remains that your equals sign is incorrect. Please sign your posts.
@CiaPan, having suggested it is necessary to accomodate the impossibility of a digital device that works with less than a Plank quantum of energy... I can only approve your confirmation that there is no such thing. Significant conclusions about what exists or can exist in the Universe are drawn in quantum mechanics but mutual WP:AGF should keep us from entering religious topics. AllBestFaith (talk) 20:07, 10 February 2016 (UTC)[reply]

Collision, Impact damage.

[edit]

A listeners comment today on the Jeremy Vine Show said that two vehicles both moving at 40 mph involved in a head on collision the impact would be 40mph not 80mph as I (and Jeremy Vine) thought it would be. The listener said it was down to some type of Newton Law. Can this be right? Alanshewan (talk) 21:15, 9 February 2016 (UTC)[reply]

This appears to be a question about science (particularly, physics), not about math. To the extent sense can be made, it surely depends on the meaning of the words "the impact would be 40mph" -- what is meant by this phrase is not clear to me. --JBL (talk) 21:17, 9 February 2016 (UTC)[reply]
It depends on the mass of the vehicles. Where both have similar masses, both will be stopped, and it's like running into a wall at 40 mph. Where one vehicle has a much greater mass than the other (let's say 18-wheeler versus scooter (motorcycle)), the small vehicle will end up going backwards at close to 40 mph, so, for it, it is like an 80 mph crash. The large vehicle will barely decelerate at all, so it would be a minor collision for them. Of course, in real world collisions, two vehicles hitting exactly head on is rare, and you often get an offset hit, sending both spinning. StuRat (talk) 21:23, 9 February 2016 (UTC)[reply]
I disagree. If the two vehicles have the same mass, both will be stopped, and it will be like running into a wall at 80 mph. Robert McClenon (talk) 21:29, 9 February 2016 (UTC)[reply]
Why is that ? Each vehicle will have decelerated from 40 mph to zero, not from 80 mph to zero. StuRat (talk) 21:30, 9 February 2016 (UTC)[reply]
@Robert McClenon: Assuming an ideal alignment of vehicles and equal masses and speeds, when they collide their fore parts will compress while touching along the vertical plane. And that plane does not move – just like an infinitely heavy, ideally hard wall. The total energy used in material destruction and dissipated as sound wave and heat will be twice that of a single car hitting a wall, but that in turn is not equivalent to the single collision at doubled velocity (as the kinetic energy is proportional to the velocity squared, so would be four times bigger, not twice). --CiaPan (talk) 22:17, 9 February 2016 (UTC)[reply]
(edit conflict)(three times) In terms of total impulse (change in momentum), the collision would be equivalent to a single 80mph collision, but in terms of total energy, only half. To first approximation, for each single vehicle, the collision would be the same as a 40mph collision with an immovable object. Dbfirs 21:31, 9 February 2016 (UTC)[reply]
@Robert McClenon:. That is a common, and understandable conclusion, but not true.--S Philbrick(Talk) 21:01, 11 February 2016 (UTC)[reply]
Mythbusters tested this. See video. Mnudelman (talk) 22:06, 9 February 2016 (UTC)[reply]
Thanks for that excellent link. Energy rules! Dbfirs 23:03, 9 February 2016 (UTC)[reply]