taeto

The question was not about using computers to solve calculus problems anyway. Since computer science is mainly about proving asymptotic behavior of the maximal running time of algorithms, there are lots of results in computer science that cannot or only with some extra difficulty be proved without calculus, e.g. typically using that the limit behavior is given by a suitable Riemann integral. Generally you cannot escape using limits to get general bounds on running time.

Well in that case the thing is just not worth dealing with: what the paper states just does not make sense mathematically. Sorry to have bothered you.

If you really want to write a text on a mathematical topic, you should try to become familiar with concepts, read a fair amount of well written expositions, and then practice a lot on formulating your own thoughts so that they become accessible to someone else when you put them into writing. I actually believe that you did read some of Lagarias's papers and enjoyed it, and you would want to emulate the style of exposition if you could. As wtf also pointed out already, your draft here is of very limited expressive power. Could you try yourself to look at line 6 together with the initial bit of line 7, and think about what a reader will see there. There is an "n", an "f", an "x" and an "m", none of which is explained as being connected to anything else. Yes, "n" is explained as being a positive integer, which is good, but n is not relevant to anything that follows. The name "f" in mathematics is quite different from the name "F". We do not know what x is. And the conjecture does not fix any particular number m to be the number of iterations before reaching the end. In particular the conjecture itself is badly misstated. Since you have read Lagarias's treatments, then why not stay on the safe road and simply state the conjecture in the same way as he does? Apart from your use of "converge", which I take to mostly mean "reach in a number of iterations", the lines 17-18 make no sense, because you cannot have a function G defined only on primes so that you can iterate and compute from a prime p the sequence p,G(p),G(G(p)),G(G(G(p))),... unless G(p) is also always a prime. If p is a prime, then you can compute G(p), that is fine, but if G(p) is not a prime then there is no G(G(p)). In particular, the function G for which G(x)=x+1 does not cut it, since G(p)=p+1 is only a prime in a very limited circumstance, that is, when p is equal to 2. Maybe for now you should just be content with explaining in a simple language, and using no mathematical terminology which you are not sure that you understand completely, how you intend to go about your plan for a proof.

+1 for rediscovering the Goldbach conjecture

Are you sure of that? Pál Erdős said once, and maybe it was more than once, that mathematics in its current state is not yet mature enough to solve a problem like this. But then, the time may come when powerful enough techniques become developed. Maybe soon we will see.

Your code is missing the line of output which says "your input is a counterexample"

Do not worry: he will want to check every detail in his argument first. Diligence is supposed to be the mother of good luck. Possibly a distant aunt, whatever.

Sure, bring it on

The original question does not contain the noun 'infinity', it contains the adjective 'infinite'. The noun 'infinity' is used in mathematics mostly as the name of the symbol \(\infty\), which figures in expressions that involve a limit operation. Additionally 'infinity' is a name given to some element that is added 'extra' to a structure in order to extend it to a larger structure that has some desirable property, e.g. to construct the extended real numbers, or to make a topological space compact, or to make an elliptic curve closed under the operation of the group. In physics I believe you can find both meanings as well. The adjective 'infinite', in this context, seems to refer to the size, or possibly the mass, or the number of particles, or the amount of time of existence. It is hard to answer the question if you are not specific. If the spacial extent of the universe is infinite, then it becomes very hard to imagine that all those other quantities are not infinite as well. If it is finite, then it becomes hard to imagine that the mass is infinite, whereas the number of particles might be in question, and the time extent could be either finite or infinite. At least that is my naive take on it. I am also interested in the implications either way.

Your basic problem might lie in the use of formal logic. It is not about you personally, but I see this quite a lot. It is possible to use formal logic language to, in a kind of silly way, to express ordinary occurrences. Like you would say in ordinary language "every person has a mother", and for fun express the same in a formal logic language "for every person x there exists a person y, such that y is the mother of x". This kind of translation is routine for anyone mathematically trained, and in fact it is necessary to be able to parse such statements in order to function professionally. But you would be amazed at how many I have met who freely take the latter statement phrased in formal logic language to mean the same as "there exists a person y such that for every person x, it is true that y is the mother of x." I leave it to you to translate this back into a statement of ordinary language and notice that it does not mean the exact same. My point is that we are entertaining the idea of one particular ordering \(<^*\) for which something is true for every subset \(S \subseteq \mathbb{R}\). It is a statement of the form "there exists ... so that for all ... something is true".

You have not found a good way to explain what you want to say. First because there actually never are subsets S like that: to say "for any a,b" includes the case a=b, and then you do not have x. But that slip aside, and otherwise, the property that you ask about does hold for the usual order \(<\) in place of the given different order \(<^*.\) But how will you argue if you get asked why it also holds for \(<^*?\) Like, as wtf points out, if you replace \(\mathbb{R}\) by \(\mathbb{N}\) and \(<^*\) by the usual (well-)order, then it does not quite look right.

That is precisely the point: \( (\mathbb{R},<^*) \) by assumption is a particular kind of order relation with the property that each of \( (S, <^*) \) and \( (T,<^*) \), or any other \( (X,<^*) \) for that matter, has a least element wrt \( <^* \) so long as they are nonempty. It seems you are thinking that we have one order for \(S\) and then after removing the least element of \(S\) we have to start making an order for \(T\), which is \(S\) with one element removed. That is not the case. You really do keep the same order.

This would look more meaningful if you also explain why you think there is no least element of T, assuming T is not empty.

Fine, we can try to put the things together that are in various units. Quite pleased to see that we avoid to consider nautical miles, imperial pounds, fluid barrels etc. The original formula reads, when units are included: \[ a^3 = 398600.8 \mbox{ km}^3/\mbox{s}^2 / (2\pi n \mbox{ day}^{-1})^2. \] Using \( 1 \mbox{ km } = 10^3 \mbox{ m }\) and \( 1 \mbox{ day }= 8.64*10^4 \mbox{ s}: \) \[ a^3 = 398600.8 * 10^9 \mbox{ m}^3/\mbox{s}^2 / (2\pi n * ( 8.64*10^4 \mbox{ s})^{-1})^2 \approx 7.53713*10^{22}/n^2 \mbox{ m}^3.\] I assume that since n isn't actually measured in rad/s, also the value that you gave for dn/dt isn't really measured in rad/s^2. Whatever unit it is, if you can figure out da/dn in units of m from this equation, then multiply by dn/dt in units of 1/s (revolutions per second), then the product ought to give the answer in m/s.

The dimensions do not fit together. From the first equation follows that a is a length. Hence da/dt is a speed and should be measured in m/s or km/s as you also added. Note that angles are dimensionless, so although it is nice of you to point out that you measure in radians, the true unit for angular speed n is 1/s, and the unit for angular acceleration is 1/s². With your numbers, in units of km and s, I get \( da/dt \approx -8.75\cdot 10^{-11} n^{-5/3} \) with unit km/s. In your equation for \(a^3\) you have \( (2\pi n)^2 \) in the denominator, are you sure that it should not be just \( n^2, \) since \(n\) is already measured in rad/s?

So you got da/dt is not da/dn x dn/dt? I cannot quickly see through the conversion between seconds and days here, sorry.

you drunk?

Can you present those real numbers?

But then to you, technology would mean how better to swing between the tree branches, while to the rest of us, science means to ascertain what is out there and how to adapt the environment to suit our lives.

Definite +1 for an on-topic Limerick, good work!

Okay. Teach me.

You happen to possess the historical reference for a time when the state of affairs of science was "good", meaning that everything was already understood?

blockquote widget An obvious reason might be: when I show you a definition that is not circular, you immediately complain that it is circular. That tells me quite a bit.

I know what it means, But I am getting pretty convinced that you do not. That is why I ask.

Explain what you mean by "circular", please.