matt grime

you ought to have been told that already. In the first instance you should always read your notes.

(n+1)/(n^2+2n+1) is not (given as) a function therefore it doesn't make sense to talk of its discontinuities. 2^y=x you realize there is nothing to stop you plotting x as a function of y with the axes flipped round....

The answer is also 'no' because 2e^5t is not something that can be 'solved' in the first place, never mind to some number of significant digits.

One way to solve this is to know the order of 3 mod 100, or a multiple of it such as the number of numbers between 1 and 100 and prime to 100 i.e. phi(100) where phi is Euler's totient function.

The first post states that fenders are factor enders, perhaps grammatically fends would be nicer for factor ends, but we'll allow it to be an agent noun. I'm not sure how much more clearly you need to state that the factor's end is its last digit. Everyone else seemed to have understood that (including me who came into the question with exactly the same prior knowledge of them as you). It then gave an example that demonstrates this: it groups the factors according to their end digit.

Exactly what was described in the first post in this thread.

the hint is: work hard at it. I know you want more, but this is your work to do for your credit, not for anyone else to do for you.

You should read that back. It just does not make sense. I don't mean in physics terms, I mean in linguistic terms, the last sentence in particular. Oh dear. Look, the relativistic calculations of m/(sqrt(1-v^2/c^2) do not apply to photons, or massless particles. You cannot cite zero times infinity is zero, that just is not a legitimate piece of mathematics or physics.

experiment, work it out. it really is a question that can be done by working out a few examples (eg, what are the fenders of 24=1.2.3.4, how do they relate to the prime factors) and then thinking about how to make things happen how you want. Ie, you don't just guess, or have 'inspiration': you work out how things behave. If it werew physics, you wouldn't expect to know the outcome of an experiment without doing the experiment, and this is no different.

one observation: sec^2=tan^2+1 has been used to explain how to translate between the two answers, after the integration, but it can be used bbefore hand too. use it in the original expression and you just get tan(x)sec^2(x) to integrate, which is sin(x)/cos^3(x), and that is integrable by inspection as 1/2cos^2(x) = sec^2(x).

Hang over from the old days, laziness, the fact you don't want to explain what e is, easier to comprehend, inaccuracy, the fact that logs to different bases only differ by a constant multiple anyway so it doesn't actually matter arithmetically what one you use (ie the all behave the same arithmetically), the fact that base 10 is what people know and only know at that stage, necessity. Take your pick or invent your own reason.

31 = 3*10 +1, and look two posts up.

The point about modulo arithmetic that is seemingly never emphasised is that you can reduce mod whatever at any time you want if it helps, you don't have to just do it at the very end. So 2^31 is 2*(2^3)^10 = 2*1^10=2 (mod 3). (Yes, knowing other important results like Fermat's little theorem help, but is no replacement for just looking at the problem and thinking for a second.) I also don't buy the reason that because they've not done it for ages that it is hard. Not least because they will have done long division of polynomials for partial fraction reasons very recently. And you'll almost certainly have just seen Euclid's algorithm which is just long division too: like all the really good ideas it is actually very simple. There are many proofs of things that I've seen that I've thought truly simple whilst simultaneously knowing I'd've never thought of it.

I think this highlights almost exactly one of the first things I try to get across to the students I had at university: you are going to go backwards in terms of arithmetic. You start at school with long divisiond and remainders, then you learn about fractions, then you do decimals. Call this the 'peak' at age 13 when you use calcultors for everything. I've even seen books aimed at A-level students which write things like sqrt(2)=1.4 (come on people!). Anyway, you start then to regress when you are told that it is better to have fractions and symbols (i.e. pi not 3.1415....) in your answers, and then you get to university and you're doing remainder arithmetic again. I don't understand why students find modulo arithmetic so hard given that it is the first thing they learned to do with numbers beyond adding and 'times tables'. Oh, and one thing that always seems to escape students attention is when asked to work out 2^31+1 mod 7, say, that they actually work out what 2^31+1 is first as an integer and then reduce it, when you can by inspection see that the answer is 3.

why would you need a mod button? It's just long division which you were doing in primary school.

You seem to have the wrong impression of Lebesgue integration. It doesn't actually tell you how to integrate something like that, it just defines an integral for more functions than Riemann integration. That function is continuous, so it is Riemann integrable. Finding what the integral is is a different matter, but is a very common one. I'm sure you can find it on the internet: it is the Bell curve, or Gauss curve, or the Normal distribution's pdf after all. The simplest way is to define that integral as I(x), and consider I(x)I(y), then do a change of variable to polar coordinates. Nothing to do with measure theory at all.

in what way would 'solid' geometry not fall under the heading geometry which, for some reasonl is classified as applied maths here, which is risible. trig is hardly a large subject in its own right worthy of a full forum. no, they just probably don't rate a course named after them no, definitely not, math olympiad people make good mathematics majors, generally. but talk to people who are in a better position to advise you than strangers. you might end up favouring the purer courses like combinatorics, group theory, discrete mathematics, and graph theory, try looking at books about them.

f(x)=1 if x is irrational, 0 otherwise. Not Riemann integrable, but the lebesgue integral over any inteval [a,b] is b-a.

But 1/9 is, I suspect, a never ending number (in fact almost all rationals have 'never ending' decimal expansions) in your language; it all depends what base you pick, but then that is immaterial to what 0.9recurring represents, and thes symbols are all just representations of things. Now, if I have 9 atoms, then one of them constitutes a ninth of the total, so that never ending number (1/9) must occur 'in reality'. And what mutliples of 'h' are you allowing? In high energy physics electrons are emitted from atoms at 'all' frequencies. But the notion of measurement is getting even further away from what the mathematics of the real numbers are. The continuum is the 'right' thing to use when modelling high energy stuff.

and photons are the only things with energy? Allowed quanta in some Hamiltonian system are the spectra of certain differential operators, or something, aren't they? And high temperature physical systems have continuous spectra or something. Not that I'm at all a physicist, and will happily admit I'm wrong if indeed I am, but this is what physicists I know and trust say. Besides, planck length, and planck time, are the smallest meaningful measurements of distance and time, we cannot 'measure' beyond these scales, which is strictly different from saying that time is quantized. Exactly what the true nature of any model of 'everything' will be is still unclear, from my limited understanding of mathematical physics, and heavily limited by the unobservability of the phenomena physicists are trying to explain. The direct implication of that sentence is that you think 0.9recurring is irrational. You cite the reason for 0.9recurring not being allowed is because irrational quantities are not allowed (in physical systems). Anyway, the statement 0.9 recurring equals one is a statement about complete metric spaces, not anything to do with physics. You picked an unfortunate topic to post on; the number of cranks out there with their own personal pet theories about this particular subject means that it is necessary (if a website such as this is to have any mathematical integrity) to stop such threads as this going off topic. Indeed it would be better if all 0.9recurring is/is not 1 threads were just locked instantly with a standard explanation of analysis. Further, in the question of mine you object to, you avoid explaining what you meant by: what is a 'near ending number'?

Does wherever you got these questions from not explain all of these methods (googling will give you the methods too).

a photon, by definition, in a vacuum travels at 0.99recurring c because 0.99recurring is the same real number as 1. (irrespective of whether you understand why they are the same number 0.9recurring is clearly a rational number since it has a recurrent decimal expansion). what has this to do with anything (eve if it were true)? that is false. certainly there is planck length, and energy in bound systems is quantized, but free energy is not quantized. no discrete quantized model for space-time has yet been proposed that works. what does that even mean and what does it have to do with mathematics? the real numbers are a totally ordered complete field, in such a place 0.99 recurring as a decimal expansion is the same number as 1 and is no more mysterious than why 1/2 and 2/4 are different representations of the same rational number.

Fortunately we have a routine running in the back ground that lets you typeset mathematics on here. [math] \phi^2=\phi+1[/math] html doesn't do maths very well. Click on a piece of maths to see what the code is. There's a also a tutorial somewhere on the forums. The point is that making qualitative judgements about what is 'pretty amazing' is frequently going to lead to a difference of opinion, especially when it is clear from the definition of phi either as an honest to goodness number or simply the larger root of some polynomial why the properties you noticed holds. But it is good that you figured out it would be Fibonacci numbers appearing. See if you can figure out why the continued fraction (google it) of phi is [1;1,1,....]. This in some sense accounts for the appearance of Fibonacci like patterns in nature since this is in the correct interprettion means that phi has the slowest converging rational approximations. You could also try to figure out what the continued fractions are of all of those numbers that have the same part after the decimal point as their reciprocals (remember they satisfy 1/x=x+n for some integer n, and n=-1 gives phi).

I simply don't see why the fact that if you invert a number you get the same digits after the decimal expansion is 'amazing'. There are infintely many numbers with this property: any roots of 1/x=x+n where n is any integer (-1 in this case), and I'm fed up with all the nonsense that people come up with about the golden ratio at the best of times (classical greek pottery); though it should be pointed out that this is not nonsense. Certainly the Fibonacci sequence occurs here: phi is one root of x^2=x+1 which explains all of those observations made about powers, andthe fibonacci sequence satisfies F(n+2)=F(n+1)+F(n). There are some truly amazing things in mathematics (it is not jaded cynicism on my part), but this isn't one of them: someone writes a computer program to compute phi to 5011 places but doesn't look at the equation that defines phi? It takes 2 seconds with a pen and paper to figure out what's going on here. I truly hope that the OP does continue to investigate mathematics (and I admit I have been too harsh, in retrospect), and I will gladly give up free time to help people's investigations. If the OP wants to investigate it some more, consider the Fibonacci sequence. It is a so called difference equation ( like a differential equation) and just like a (homogeneous) differential equation we can do the following: Suppose that we can write F(n) as a function of n, then what form must it take. Well, experience tells us that in this case we should look for constants a,b,r and s such that F(n)= a(s^n)+b(r^n) Now, can you find r and s? They must satisfy a certain equation given by the recurrence F(n+2)=F(n+1)+F(n), or x^2=x+1.... a and b you choose so that F(1)=F(2)=1 So you see, phi actually enables you to write the n'th Fibonacci number the sums of n'th powers.

Oh, please, just look at the damn recurrence relation before posting more 'isn't this amazing' crap. phi satisfies x^2=x+1 (and you could at least try to learn to typeset a little more clearly your maths [powers etc]), and that's all there is to it. Now what is amazing is why this pattern explicably occurs in sunflowers and so on (owing to convergence of continued fractions). That is an amazing theory: that we have an explanation for this phenomenon, not that it happens.