The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
If I understand correctly, "credence" is the same as "subjective probability". Is this term really widely used/preferred? If not, it would make more sense to rename that article (currently "subjective probability" redirects to a section in "Bayesian probability") and link it to d:Q25382706. — Mikhail Ryazanov (talk) 00:46, 26 February 2024 (UTC)[reply]
One relevant question – to which I don't have the answer – is whether an encyclopedic treatment of the notion of "subjective probability" requires the Bayesian interpretation of probability.* The lead of our article on interpretations of probability states that there are two broad categories of probability interpretations, which can be called "physical" and "evidential" probabilities, and then goes on to claim that evidential probability is "also called Bayesian probability". Now I for one am very hesitant to attach the label "evidential probability" to subjective probability, given my subjectively high probability estimate for people happily attaching subjective probabilities to events without any evidential basis. I must confess, though, I have no evidence for this estimate. The actual treatment of the Bayesian interpretation then follows in the section Probability interpretations § Subjectivism, which begins with Main article: Bayesian probability and does not discuss other interpretations. I quote the first two sentences, because they are immediately relevant to the question posed here:
Subjectivists, also known as Bayesians or followers of epistemic probability, give the notion of probability a subjective status by regarding it as a measure of the 'degree of belief' of the individual assessing the uncertainty of a particular situation. Epistemic or subjective probability is sometimes called credence, as opposed to the term chance for a propensity probability.
This is all unsourced. To complicate the matter, the lead of our article on Bayesian probability states that Bayesian probability belongs to the category of evidential probabilities. This is a strange formulation if Bayesian probability is the only member of that category. Evidential probability is a red link, while Epistemic probability redirects to Uncertainty quantification, which is interpretation-agnostic.
There is clearly some need of more terminological precision and clarity here, which should be based on sourced material, while care should be taken not to assume that if one author equates two terms, it represents a generally accepted view. --Lambiam 09:04, 26 February 2024 (UTC)[reply] *One datapoint: the article Субъективная вероятность (Subjective probability) on the Russian Wikipedia does not refer to Bayesian probability in the main text, but only indirectly by referring to I. J. Good's monograph The Estimation of Probabilities: An Essay on Modern Bayesian Methods as a significant contribution to the development of the theory of subjective probability. It states that the first formulation of subjective probability was given by Frank Ramsey in 1926, presumably in his paper Truth and Probability (which does not mention Bayes, but may perhaps nevertheless be Bayesian in its approach).
Can these numbers have these names?
Can these numbers have these names?
(sequence A153501 in the OEIS): Abundant-perfect numbers (compare with (sequence A271816 in the OEIS): Deficient-perfect numbers)
(sequence A005382 in the OEIS): Sophie-Germain primes of the second kind
(sequence A005383 in the OEIS): Safe primes of the second kind
(sequence A112715 in the OEIS): Proth primes of the second kind
(sequence A005105 in the OEIS): Pierpont primes of the second kind
Like "Cunningham chain of the second kind", also like that Woodall primes may be called "Cullen primes of the second kind" 36.233.209.202 (talk) 10:20, 26 February 2024 (UTC)[reply]
Is there a simple proof that all groups with order less than 60 are solvable?
Is there a simple proof that all groups with order less than 60 are solvable? 114.35.106.188 (talk) 10:45, 26 February 2024 (UTC)[reply]
Burnside's theorem means you need at least three prime factors. Then one can use Sylow's theorem - see the section small groups are not simple. And no it isn't simple ;-) NadVolum (talk) 13:21, 26 February 2024 (UTC)[reply]
Burnside is a big help, but its proof uses representation theory so I don't know if using it really qualifies as "simple". As I recall you can use a variety of tricks from elementary group theory, for example that every p-group has a non-trivial center, the Sylow theorems, etc., to reach 60 without Burnside. Either way, if you're just using basic tools then there's a lot to do and it's more like a project than an exercise. --RDBury (talk) 15:12, 26 February 2024 (UTC)[reply]
OEIS: A001034 gives the first few orders for which there is a simple group that is insolvable. NadVolum (talk) 18:25, 26 February 2024 (UTC)[reply]
OK, I saw this sequence, is there a simple proof that all groups with order less than 1092 and divisible by neither 60 nor 168 are solvable? Also I have a question, prove or disprove that all groups with order not divisible by 12 are solvable. 218.187.66.141 (talk) 17:13, 28 February 2024 (UTC)[reply]
In the previous OEIS sequence, the number 29120 is not divisible by 12. However, according to Isaac Saffold's comment, all numbers in the sequence are divisible by 4, so it would appear that all groups with order not divisible by 4 are solvable. GalacticShoe (talk) 17:30, 28 February 2024 (UTC)[reply]
That there are no non-solvable groups of odd order is the Feit–Thompson theorem. That there are no simple groups of order 2n with n odd is provable using elementary methods. Together these prove that all groups with order not divisible by 4 are solvable. There are stronger, somewhat advanced theorems, e.g. a non-Abelian simple group of even order must have order divisible by 12 or 16, but IIRC they require advanced representation theory. I'm not sure what "simple proof" is supposed to mean here. Accessible to a high school student with no exposure to abstract algebra? Accessible to an undergraduate with a semester of elementary group theory? Accessible to a graduate student with several semesters of advanced group theory? Less then a page long? And what are you allowed to assume in this proof? One "simple" proof is to look at the list in the Classification of finite simple groups and verify all the non-cyclic ones have order less that 60; would that count? As it stands I think the question is too vague to be answered. --RDBury (talk) 00:53, 29 February 2024 (UTC)[reply]
Correction, the stronger theorem I mentioned should be that a non-Abelian simple group of even order must have order divisible by 12, 16 or 56. It was actually proved by Burnside in the late 19th century, so the proof is not as advanced as I was thinking. --RDBury (talk) 01:28, 29 February 2024 (UTC)[reply]
February 28
Least base b such that there is a k such that k*b^m+1 (or k*b^m-1) has a covering set of modulus 2*n
For given positive integer n, find the least base b such that there is a k such that k*b^m+1 (or k*b^m-1) has a covering set of modulus n
For even n, b = 2 if and only if (sequence A257647 in the OEIS)(n/2) and (sequence A258154 in the OEIS)(n/2) and (sequence A289110 in the OEIS)(n/2) are nonzero.
For prime n, there is a sequence (sequence A146563 in the OEIS), but unfortunately there is no OEIS sequence for all n (instead of only the prime n)
For n = 1 through n = 16, I found the sequence (the prime n terms are given by (sequence A146563 in the OEIS):
the 8-cover of base 8 can be (3, 5, 17, 241) or (3, 13, 17, 241)
the 9-cover of base 9 is (7, 13, 19, 37, 757)
the 10-cover of base 25 is (11, 13, 41, 71, 521, 9161)
the 12-cover of base 5 is (3, 7, 13, 31, 601)
the 14-cover of base 51 is (13, 29, 43, 71, 421, 463, 11411, 1572887)
the 15-cover of base 9 is (7, 11, 13, 31, 61, 271, 4561)
the 16-cover of base 5 is (3, 13, 17, 313, 11489)
Could you extend this sequence to n = 25 or n = 50? 218.187.66.141 (talk) 17:01, 28 February 2024 (UTC)[reply]
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 = 26, 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)
Is it possible to extend this sequence to n = 100? 218.187.66.141 (talk) 17:12, 28 February 2024 (UTC)[reply]
March 1
Planarise quaternion graph
Cayley Q8 graph showing the six cycles of multiplication by i, j and k – in the SVG file, cycles can be highlighted by hovering over or clicking on them
I drew this diagram of the quaternion multiplication table some time ago, and am wondering if it's possible to make it (more) planar so it's easier to understand. Is that possible? Thanks, cmɢʟee⎆τaʟκ 06:57, 1 March 2024 (UTC)[reply]
It might not be the most elegant solution, but you could have -1 form an entire outer ring, with the three semicircular circles becoming straight lines passing through, sort of like a point at infinity. Given that -1 is "antipodal" to 1 on the three relevant cycles, I think it would make sense intuitively, but again there might be other much better ways to go about it. GalacticShoe (talk) 08:13, 1 March 2024 (UTC)[reply]
Thanks. It's more or less already like that: the dotted semicircles all passing through -1 like a Metro map. cmɢʟee⎆τaʟκ 09:42, 1 March 2024 (UTC)[reply]
What would be clearer—to me, at least—would be to just list out the six cycles without trying to force them into the same graph. (As it stands, if you're trying to show the patterns in the cycles, the graph honestly obfuscates that more than elucidates it.) 2603:8001:4542:28FB:D83C:41C6:8B71:EABA (talk) 17:22, 1 March 2024 (UTC) (Send talk messages here)[reply]
Not planar, but cycles are clearercmɢʟee⎆τaʟκ 10:13, 2 March 2024 (UTC)[reply]
The symmetries are also more perspicuous. The two-step paths represented by pairs of smooth solid arcs passing through the vertices, could be mistaken for one-step arcs passing under the marbles representing the vertices, bypassing these vertices. I expect that letting the "hidden part" of the smooth arcs go more closely to the centres of these marbles will make it clearer that they are meant to proceed through these vertices. Also, in this version the inner and outer arcs hardly interfere; IMO there is little point in expressing a distinction with a solid and a dashed style.
For the reader whose colour perception is challenged, it may help to make the arcs a bit fatter, and also to use three colours whose brightness is more distinct, e.g. #ff8888 , #009900 , #0000cc . --Lambiam 09:58, 3 March 2024 (UTC)[reply]
Good recommendations, @Lambiam: I've updated the diagram using line style as an alternative to colour for the colour-blind. Your red is a bit too intense, so I've retained #cc0000. With the line styles, colour differences are now less important. Cheers, cmɢʟee⎆τaʟκ 13:30, 3 March 2024 (UTC)[reply]
P.S. Would it be better to rotate the diagram 90° clockwise so that the ±1 and ±i orbs are in their respective directions on the complex plane? cmɢʟee⎆τaʟκ 16:28, 3 March 2024 (UTC)[reply]
There is no connection between specifically the basis element of the quaternions and the imaginary unit The roles of and are perfectly interchangeable: we have In fact, any valid algebraic identity remains valid under a cyclic permutation. Take, for example, So then it is also the case that Therefore there is no point in paying special attention to the orientation of You could introduce this basis-element symmetry in the image by rotating the dotted green cycle 60° clockwise and the dashed blue cycle 120° clockwise, with obvious changes for the outer three cycles. Then the vertices and will form the vertices of an equilateral triangle. --Lambiam 17:18, 3 March 2024 (UTC)[reply]
Good point, thanks. I'll keep the current orientation then. I think keeping i, j, k together is clearer than going i, −k, j, −i, k and −j. cmɢʟee⎆τaʟκ 08:08, 5 March 2024 (UTC)[reply]
Because that is the definition of 2. —Tamfang (talk) 00:21, 2 March 2024 (UTC)[reply]
Curiously, this questioner and that of last year geolocate to locations less than 9 miles apart. Perhaps there is something in the air in Minneapolis–Saint Paul that evokes this question. --Lambiam 01:01, 2 March 2024 (UTC)[reply]
"Why does Minneapolis + Saint Paul = Twin Cities?" GalacticShoe (talk) 01:07, 2 March 2024 (UTC)[reply]
It depends on the context. One cloud plus one cloud makes one cloud. One rabbit plus one rabbit makes three or more rabbits. One gunmam plus one gunmand quite possibly makes no gunmen. In the context of schools 1+1 is 2 is the correct answer and marked with a tick - what more do you need? NadVolum (talk) 11:23, 2 March 2024 (UTC)[reply]
Perhaps one might wish to see an explanation of why, in the context of schools, it is "the" correct answer. Was it decided by a conference of school boards? Or legislated after the powerful pressure of a lobby for publishers of books on arithmetic? Or is it because the teachers check the answers on their pocket calculators, most of which show as the result of calculating 1+1=? --Lambiam 14:10, 2 March 2024 (UTC)[reply]
My first suspicion in cases like this is normally the reptilians from Tibet - but I think in this case it's because of a powerful cabal of mathematicians pushing the Peano axioms. S(0) is called 1 and S(1) is called 2 and 1+1 = 1+S(0) = S(1+0) = S(1) = 2 and no one has yet found a secret backdooor in those axioms allowing other values to be assigned to 1+1. That would be equivalent to finding an x so S(x) = 0. NadVolum (talk) 17:17, 2 March 2024 (UTC)[reply]
But how do we know these axioms are correct? Maybe there is a largest number – admittedly veeery large – where you hit a limit with this S. You could call it a fixed point. At least Boolean algebra (see below) keeps it simple. I hear that many mathematicians believe it has some truth value. --Lambiam 21:29, 2 March 2024 (UTC)[reply]
Axioms are not "correct" or "incorrect", they are basic assumptions upon which further logic is built. {The poster formerly known as 87.81.230.195} 176.24.44.161 (talk) 07:09, 6 March 2024 (UTC)[reply]
If reality uses Saturation arithmetic then the Peano axioms don't model it! That sounds like a reasonable way for some supercomputer to simulate u all but I think it's more likely it wouldn't detect overflow and just produce a huge negative number instead 😁 NadVolum (talk) 19:17, 3 March 2024 (UTC)[reply]
Boole seems to have got in a bit of a mess with + and -. He'd have been much beter off using + for exclusive or and forgetting about -. With that 1+1=0, not x is 1+x, and the usual x or y is x + y + xy. NadVolum (talk) 22:10, 2 March 2024 (UTC)[reply]
March 4
For which natural number n, 1^k+2^k+3^k+…+n^k can be prime?
I saw the sequence (sequence A164307 in the OEIS), and now I have a problem: for For which natural number n, 1^k+2^k+3^k+…+n^k can be prime?
n=2, this is exactly the Fermat primes
n=3, numbers cannot be prime since they are divisible by 2
n=4, numbers cannot be prime since they are divisible by 2
n=5, no fixed small divisor of the numbers, but k must be divisible by 20 since they are divisible by 5 if k is not divisible by 4 and they are divisible by 11 if k == 4, 8, 12, 16 mod 20, and indeed the term k=1440 is prime, but are there infinitely many such primes?
n=6, numbers cannot be prime since they are divisible by 7 if k is not divisible by 6, while they are divisible by 13 if k is divisible by 6
n=7, numbers cannot be prime since they are divisible by 2
n=8, numbers cannot be prime since they are divisible by 2
n=9, numbers cannot be prime since they are divisible by 3
n=10, no fixed small divisor of the numbers, but k must be divisible by 20 since they are divisible by 11 if k is not divisible by 10 and they are divisible by 5 if k == 10 mod 20, but is there any such primes?
n=11, numbers cannot be prime since they are divisible by 2
n=12, numbers cannot be prime since they are divisible by 2
n=13, numbers cannot be prime since they are divisible by 13 if k is not divisible by 12, while they are divisible by 3 if k is divisible by 12
n=14, numbers cannot be prime since they are divisible by 7 if k is not divisible by 6, while they are divisible by 5 if k == 6 mod 12, while they are divisible by 13 if k is divisible by 12
n=15, numbers cannot be prime since they are divisible by 2
n=16, numbers cannot be prime since they are divisible by 2 218.187.65.97 (talk) 15:05, 4 March 2024 (UTC)[reply]
These sums have values given by Faulhaber's formula. That article has quite a bit about it but does not touch this question. I'm not sure it'll help much but it's worth a good look. NadVolum (talk) 21:43, 4 March 2024 (UTC)[reply]
Faulhaber should definitely be looked at here. The problem is that it involves those pesky Bernoulli numbers which aren't integers. If a prime is not a factor of the corresponding denominators then you can say something about whether it divides the sum, but otherwise the situation is more complicated. Let S(k, n) denote the sum. There are two variations on the problem here: First, given k, determine n so that S(k, n) is prime, and second, given n, determine k so that S(k, n) is prime. Faulhaber seems more relevant to the first problem, but the problem as stated in the question is more like the second version. For example, in the n=5 case the problem becomes: For which k is 1k+2k+3k+4k+5k prime? Faulhaber doesn't seem to be very useful here since it gives a different formula for each k; you might as well just compute the sum manually. Fermat's little theorem seems more relevant since it implies that, for a fixed n, S(k, n) has period p-1 mod p. So if p|S(k, n) then p|S(j, n) for any j congruent to k mod p-1 (Edit: Assuming p>n). Unfortunately, unless p is very small, this only eliminates a fraction of possible k's (if any), This is the point where I would consult the literature, for which OEIS would be the starting point. But that's where this whole thing began. It would be nice to get some kind of result as in the n=2 case, that k must have a particular form such as 2a, but that seems unlikely for n>2 unless you can eliminate all k as in the n=3 case. --RDBury (talk) 15:26, 5 March 2024 (UTC)[reply]
Note: There is a small error above in the n=6 case: S(12, 6) ≡ 2438235715 is not divisible by 7 or 13. All you can state from divisibility by p=7 and 13 these is that 12|k. But you can eliminate 4|k using p=5. Between p=5, 7 & 13 you can eliminate n=6. --RDBury (talk) 16:18, 5 March 2024 (UTC)[reply]
Also note: You can show that if S(k, 10) is prime then k is a multiple of 60; consider also p=7. I didn't see any other easy improvements though. S(60, 10) is composite having a factor of 137. S(120, 10) exceeded the amount of computation time I'm allowed on WolframAlpha; a primality test on a 120 digit number is not infeasible though. --RDBury (talk) 16:55, 5 March 2024 (UTC)[reply]
Our article on Faulhaber's formula points out that for odd , is a polynomial which has and terms. In general, we can write where is a polynomial with integer coefficients and is a constant positive integer. We can see that if and are integer-valued polynomials, then if is an integer, then it must be composite if and , since some factor must remain of both and after division by . So if and , then cannot be prime. Since if and only if , we see that cannot be prime for odd if . For even an analogous result exists, except replacing with . Of course, this leaves the problem of finding out what actually is. GalacticShoe (talk) 19:19, 5 March 2024 (UTC)[reply]
is given by the sequence OEIS:A064538. I couldn't find an upper bound in the references. It appears to be extraordinarily fast growing. GalacticShoe (talk) 04:25, 6 March 2024 (UTC)[reply]
For odd , when , and for even , when . So really, since is never prime, the only inequality that matters is for odd and for even . GalacticShoe (talk) 05:16, 6 March 2024 (UTC)[reply]
For , all cannot produce primes, and by direct testing is not. So there are no primes.
For , all cannot produce primes, but again by direct testing, is prime (17).
For we can no longer use the limits, since , but for obvious reasons there are no primes.
For we again run into a problem with the limits as , but again it can be seen that is the only prime here.
For we see that is the only prime.
In sum (pun intended), for the only primes are . The set of primes for any fixed value of can be established with a finite but apparently quickly-growing computation. GalacticShoe (talk) 05:30, 6 March 2024 (UTC)[reply]
Coming back to this, for from through (the last term that A064538 has listed), the only primes are , all Fermat primes as the OP pointed out. GalacticShoe (talk) 07:48, 9 March 2024 (UTC)[reply]
March 6
Simple math, or semantics?
I've asked a couple of AIs the following, resulting in either 12 or 13 18: Add together the first five integers, excluding the number three. Perhaps this is more of a semantics problem, but what is the "correct" answer, or is it ambiguous? -- 136.54.106.120 (talk) 00:30, 6 March 2024 (UTC)[reply]
Well, there is no "first integer". But I don't see how they get 13. Bubba73You talkin' to me? 00:56, 6 March 2024 (UTC)[reply]
Oops, my mistake (corrected: 18, strike 13); per AI response: To add together the first five integers, excluding the number three, you would sum 1 + 2 + 4 + 5 + 6, which equals 18. --136.54.106.120 (talk) 01:34, 6 March 2024 (UTC)[reply]
Besides there not being any first integer, it is ambiguous because it isn't clear if the third integer should be excluded from the sum or excluded from the count of the first five integers. Bubba73You talkin' to me? 01:41, 6 March 2024 (UTC)[reply]
So short answer, semantics. The integers are ... -3, -2, -1, 0, 1, 2, 3, ... . So there is no "first" one. To be clear you have to specify "positive integer" meaning 1, 2, 3, ... . Another interpretation would be "natural number" which, depending on who you ask is either the same as "positive integer" or "non-negative integer", meaning 0, 1, 2, 3, ... . In that case the answer would be either 8, 9, 12, or 13, depending on whether you meant to exclude the third in the sequence or the actual number 3, and whether you meant the excluded number was to be counted as one of the five. The upshot is that the question has to be clear and unambiguous to have any hope of getting a meaningful answer. In any case, it's generally not a good idea to ask AI's math questions; they tend to make stuff up and at best only repeat what is said the most often, without questioning if it's actually correct. --RDBury (talk) 02:44, 6 March 2024 (UTC)[reply]
The ambiguity of this ill-stated problem reminds me of stupid postings circulating on social media like "Can you solve this: 8 ÷ 4 × 2 = ? Most people get this wrong!", ignoring that there is no universal convention for interpreting a one-line term involving division followed by multiplication. --Lambiam 09:32, 6 March 2024 (UTC)[reply]
March 8
Bouncing ball confinement
When a ball is dropped in vacuum in a uniform gravitational field from a height onto an immovable surface whose lowest point is at height 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 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 are simultaneously dropped on the curve given by it seemed to me that the balls basically remained confined to the rectangle with vertices
This is a much stronger confinement than Is there some simple mathematical argument supporting my observation? --Lambiam 19:58, 8 March 2024 (UTC)[reply]
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)[reply]
Indeed, one of the replies to the video is just the three words, "not a square". When the trajectory is either undefined or confined to the line , but the limit case of the hypothesized rectangle for is a square. (The best way to define the bounce-back when the trajectory hits is to have the ball move back in the direction it came from, the limit case of hitting ) --Lambiam 10:06, 9 March 2024 (UTC)[reply]
A few thoughts: The second video shows that the boundary is the envelope of all the possible trajectories. The possible trajectories are parabolas , with the apex at (to be shown that this is always so), height given by , and to be expressed in terms of the transverse velocity at the apex 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)[reply]
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)[reply]
I think the first thing to do is apply the conservation of energy, if the apex of the parabola is at then we have:
Assume for now that this occurs at ; we'll verify this later. Call the value of for a specific path . Parametric equations for the parabola are given by:
Eliminate t and write to get an implicit form for the equation:
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 , 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
intersect at (x, y) with k≠l. Eliminating y:
The normal vectors to the two curves are given by
The reflection of the first vector by the line x=y is
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:
which gives:
Plug this back into the original equation to produce:
and this gives the two lines 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)[reply]
PS. If you don't like the envelope part there is a simpler approach. If
has a solution in k then the discriminant must be non-negative, in other words
We know that from which we can deduce --RDBury (talk) 20:35, 9 March 2024 (UTC)[reply]
as the result of calculating 1+1=? --Lambiam 14:10, 2 March 2024 (UTC)
