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About taeto

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  1. I need some help with calculus.

    If you did it and got ~ 3195.323, then it looks like you did it correctly. Or are you just referencing the answer known from a solution book?
  2. Strange connection between phi and pi

    If you also want to see an \( e \) you can have \[ e^{\pi i/5} = \frac{\varphi + i\sqrt{3-\varphi } }{2}. \] It is cheating, since both sides are just a \( \sqrt[5]{-1}. \)
  3. Biggest maths fraud in history

    Is the Ramsey quote really correct, did Ramsey write anything that stupid? A "tautology" is usually a statement that is always true. If you build a theory axiomatically, then the axioms are always true, by construction, in that theory. And anyway, supposing you throw out a number of axioms, it of course would make arithmetic statements even more undecidable than they are already. Eventually you might no longer be able to prove that this is so. But if that is your intention, then you could instead also close your eyes, put your fingers in your ears and go LALALALALA and avoid being disturbed by facts that way.
  4. What are two Dimensional objects?

    But it is an actual boundary more than just an interface. A (non-rotating) black hole at a given time has a precise Schwarzschild radius that depends on its mass, and the event horizon surrounds the singularity spherically at that precise distance. I don't think you are supposed to imagine anything being fuzzy. In which case it qualifies as a 2D surface.
  5. Motion Graph

    He means arctan (times a suitable constant). It is different from a logarithm in an important respect. Although both functions will keep increasing, the log is unbounded, whereas arctan will approach an upper limit. So it depends on which of those properties you are aiming for.
  6. What are two Dimensional objects?

    Do you consider the event horizon of a black hole to be a thing that exists in the real world? It might fit the bill pretty close, no?
  7. Computer Science

    That is actually a good question. I have a firm memory that tells me that it at least used to be a very explicit requirement that every proof can be verified within the theory ZFC. The rules were modified in September of 2012, however. I would be very interested to know whether the requirement got changed, and why. It would clearly be absurd to allow a proof in a theory of your own choice, as you might simply add as a new axiom the exact statement that you desire to proof, or at least some more disguised statement which would imply it. But I definitely surprised by your observation, so I salute you for that. Anyway, the official description of the problem is written by Stephen Cook and it does define a Turing Machine in the language of ZFC. That makes it unlikely that a proof outside ZFC is in any way meaningful. But I am open to suggestion. Do you have a theory to propose different from ZFC in which you would suggest to attempt a proof? Scott Aaronson has a discussion of the dependence of the NP=P question on ZFC on page 26-28 of www.scottaaronson.com/papers/pnp.pdf. If you submit anything rubbish, they would obviously turn it down. If you submit an actual solution they would quite obviously be delighted, no matter what degrees you do or do not have, heck, it would not matter whether you finished high school or not. Did you ever submit a paper to a journal, or are you just guessing here? Not sure what you mean by that. You quote an informal description of a DTM that requires the read head to move either backwards or forwards every time it needs to read from a different cell. I also do not know what you mean by saying that a problem "requires an infinite tape". Either example of a TM requires an infinite tape, otherwise it is simply not even possible to accept all legal inputs. If your TM factorizes integers, and its tape only has room for inputs of N bits that are available on your tape, then what will it do with an integer that has to be written with N+1 bits?
  8. I feel dyslectic. There is a conjecture suggesting that every natural number \( \geq 2 \) can be written as a sum of a prime and two Fibonacci numbers, is that right?
  9. How to solve the problem

    It is a good attempt, though only about half right. You should not write "for n = n+1", since obviously n is always different from n+1, except in very peculiar situations, which do not appear in ordinary arithmetic. Instead you should write something like "Assume that the statement is true for n, and consider the statement for n+1" (which means, the statement about n+1 that you get after replacing n by n+1 in the original statement). So you will assume that you have some number n for which you already know \( 2^n \geq n^2 \). The only thing left to show now is \( 2^{n+1} \geq (n+1)^2 \) and you will be done. You should do it by first grabbing the thing \( 2^{n+1} \) that you want to prove something about (namely that it is \( \geq (n+1)^2 \) ). But you should look at the statement that you know that you are allowed to assume, which is \( 2^n \geq n^2 \). You have to ask yourself, what does the fact \( 2^n \geq n^2 \) tell me about how large the number \( 2^{n+1} \) will be? And then, if you can figure this out, how does your new guaranteed smallest possible value for \( 2^{n+1} \), which you just got, compare to the number \( (n+1)^2 \)? Is it a larger number? If it is, then you are done. The kind of statement here seems tricky because it involves an inequality instead of just a formula with an equality sign in it. Maybe you should practice a bit more with arithmetical theorems that can be written as equations. Something like \( 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \) is another standard exercise.
  10. How to solve the problem

    You are getting somewhere: is essentially the basis of the induction proof. You have shown that the inequality is true when n is equal to 2. This is important. However, is not part of the proof by induction. It shows that the inequality is true when \( n \) is equal to 3. The increase in the difference between LHS and RHS is irrelevant. How would you argue that the statement is also true for \( n = 4, \) except by calculating both sides of the inequality for that case as well? You do not know whether the difference between LHS and RHS will suddenly drop enough to make the inequality false for that value of \( n, \) or for some larger value. Did you read the Wikipedia page, or anything else, on how to prove something by induction? Do you know how to prove that \( 2^n \geq n^2 \) holds for every \( n > 3 \) by induction? You should be familiar with this baby example if you want to prove anything as complicated as the problem that you posted. I am puzzled if you are asked to prove anything complicated like that without having seen the simplest examples. Where did you get your problem from? Is there anything to suggest that you should actually try to solve it using Taylor series?
  11. Computer Science

    Sorry, but the rules for the Clay prizes state explicitly that whatever proof you have, it has to be a valid proof in ZFC. Your last remarks seems not quite on point. Depending on what you mean exactly by your condition of requiring an infinite amount of tape. If a problem is in NP, then it follows easily that each YES-instance of it can be solved without using an infinite amount of tape, simply because there has to be some way to find a solution in finite time. For some NO-instances of NP problems you have to use infinite time for searching, and I doubt that it has any bearing on anything whether you also use infinite tape or not. But even if the problem is in P, you cannot in general do with just a finite amount of tape, because there will be instances that require more space just for stating their input. Either way, the TM that solves all instances has to have an infinite amount of tape available. en.wikipedia.org/wiki/Turing_machine#Formal_definition gives the formal definition of a TM as a set in Set Theory. Btw it is possible to represent an ordered pair (a,b) as a set as well: the most famous definitions are by Wiener, Hausdorff and Kuratowski, see en.wikipedia.org/wiki/Ordered_pair Sure. And sorry about the rant. I am newbie and just hoped that in a science forum it would be easier to find those contributions that actually deal with science topics.
  12. How to solve the problem

    It is possible to use Taylor series to get a proof even in the more general case when n is any real number greater than or equal to 2. But it looks like it will be complicated. Do you know the Taylor expansion of the RHS of your inequality, that is, the expansion of 7x as a function of x? It looks more like you are supposed to apply induction on n: en.wikipedia.org/wiki/Mathematical_induction The easiest might be to first show that the result holds when n = 2, and then prove with simple calculus that the difference LHS - RHS is always increasing for larger n (again assuming the variable is real-valued). This would be kind of "induction on real numbers".
  13. Sum of factorials

    It may no longer be useful, but I will try to add some suggestions. Sorry that I cannot make Latex work on this site. It is clear that the sum of 1/n! for n=3,4,... is equal to e - 5/2, since we are missing just the terms for n=0,1,2 in getting the actual expansion of e as the sum of 1/n! over all natural numbers n. We just need to work out the sum of the terms of the form 1/(n! n) over n=3,4,... in your final RHS expression. Call that sum S. Let a function f be defined by f(x) = sum{ xn/(n! n) : n=1,2,... }. Then it is clear that S = f(1) - 5/4, since S is of the form of f(1), except for missing the terms with n=1 and n=2. Also f(0)=0 trivially holds. Each term xn/(n! n) in f is the integral of tn-1/n! dt from t=0 to t=x. By a simple analysis argument it follows that f(x) is equal to value of the integral of (et - 1 - t - t2/2)/t from t=0 to t=x, which is the same as the integral of (et - 1)/t - 1 - t/2 from t=0 to t=x. Your sum S becomes the value of the integral of (et - 1)/t -1 - t/2 from t=0 to t=1 and then subtracting 5/4. And your sum total becomes e - 5/2 - S.
  14. The people who are capable of recognizing when a paper about the Riemann zeta function does not even define the Riemann zeta function properly, and otherwise consists purely of word salad anyway, might however be active in the forum.
  15. The magic i

    In the study of physical oscillating systems, with possible dampening, you naturally obtain second order linear differential equations for their movement, as a consequence of Newton's second law F=ma, where F is force, m is mass and a is acceleration (the second derivative of the position as a function of time). The basic example is a ball suspended on a spring while subject to a viscose environment such as a surrounding air or liquid. The solutions are products of trigonometric functions like sin(at+b) and cos(at+b) at time t, that provide the oscillations, with exponential functions e-wt , that provide the dampening effect. This is not magical, just consequences of the fact that the trigonometric functions are very closely related to the complex exponential function. The exponential function and the trigonometric functions are also the basic functions for which their second derivatives are proportional to themselves. Which is why they appear naturally together in descriptions of how physical systems behave. The precise identity eit = cos(t) + i sin(t) does the rest to establish a connection between e and pi.