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matt grime

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Everything posted by matt grime

  1. No it is not. And your response to someone else pointing this out also is incorrect.
  2. A sequence is a sequence is a sequence. They don't have solutions. They have terms, they may even have limits, but they do not in general have solutions. Equationss have solutions, problems have solutions. So what are you asking? Please, make sure you give all information in a post, not just what you assume everyone will know automatically.
  3. The premise is false (t=1).
  4. I'd be intrigued to learn what methods you have *with* logarithms to solve these things.
  5. If you don't know what proportions of chemicals are stable/unstable, then you have no idea what the probability of a given chemical being stable is. Mind you, if you want to play cards with me anytime please feel free, YT.
  6. why would I suppose it? 3^4=81==1 mod 5 (in fact any number prime to 5 raised to the 4th power is 1 mod 5). I still have no idea what it is you're trying to do. Please try to repost with *more words* in complete sentences explaining what it is you want to do. after all 4 divides 4 but 3^4=81 =/=1 mod 5, so what is it you're trying to say? I think words like 'if' and 'then' are missing from your sentences.
  7. Where does anything require you to use any particular base? logs in different bases differ only by constants. This is why at the end you only write exp{kx{ for some constant k. k can be gotten from f(1).
  8. We should point out that you've omitted to mention at least one important property that you're assuming f must have, namely continuity. It is trivial to show that log(f(x)) for rational x is completely determined by f(1) and hence for rational x, f(x)=exp{kx} for some constant k, and thus by continuity f(x) is exponentiation for all x *if* we assume f is continous. If f is not continuous then there are uncountably many distinct f's with this multiplicative property.
  9. what fixed point? you haven't mentioned a fixed point. which of us truly knows the definition of themselves?
  10. When in doubt just google. That link is the first hit.
  11. What you mean is not what you wrote. what is your question here then? Prove what? Do you know any elementary group theory?
  12. Yes, that has been mentioned (by the OP, by me and by shmoe), and doesn't address the fact that it assumes that the limit exists in the first place. You also need to prove that fact. After all, if I assume there is a largest integer N, then N^2 is larger than N but N is the largest, hence N^2=N so N=0 or 1, and since 1 is larger than 0 there are no integers larger than 1. See what problems you have if you assume a false premise?
  13. No. Just try some examples to see why this is wrong. How can b to all powers be congruent to 1 mod a? Just take d=1.
  14. Sorry to be blunt, but get over it. There is nothing to stop you making your own toy examples either (such as 1x1 matrices) and playing about with things by hand.
  15. a needs to be a function of n, obviously, but you should make that explicit or the answer is just aN.
  16. Yes. Think about the ij'th entry if you need to. There is no harm in thinking 'mod h^2', that is put x+hy, where x is the point you're differentiating at, y is a matrix and h is a real number, in to the defining equations and set h^2 (and higher order terms) to be zero. *If* the function is differentiable you will be able to read off the derivative from it. You can then, if you must justify that this is indeed the derivative. For example, differentiate the function det(X) at the identity matrix we put in det(I+hy) and we get det(1)+htr(y)+ terms in h^2 (this is just expanding the determinant). Thus the derivative of det (at the origin) is trace.
  17. 1. Makes no sense. Here's a guess: if f(x) is x+1 is not divisible by 4 and definitely has an integral root, whereas if f(x) is 4 then f is divisible by 4 and has no integer roots (consider it as the constant function). NB, functions don't have solutions, perhaps you could say what f is? 2. Of course.
  18. the method has been given. Let f_n(x) be the result of x^x^x^..^x, with n terms, and let f(x) be the limit as n tends to infinity of f_n(x). This will not a priori exist for all x, but that is immaterial at this stage. Now, assume that there is a solution to f(x)=2, if there is, then it must be root 2, since x^f(x)=x^2, and x^f(x)=f(x)=2. It remains to show that f_n(sqrt(2)) actually converges which is done using shmoe's hints.
  19. What method? You went from x^x^x^x..=2 to x^2=2 without justifying why this is true (it is: you are just raising x to the same power, but you didn't say this explicitly enough, I feel).
  20. I think I misread what you wrote, so ignore me. Apart from the bit where I back shmoe up about using dummy variables properly.
  21. You are using n as a dummy variable in two different ways in one expression, and as shmoe as pointed out that is not allowed.
  22. There is a significant difference between the two things. The first does not actually make sense, now I think about it. Don't let laziness get in the way of doing things properly: these things can make serious differences. note, youe former vector, sum ke_k does not actually make much sense unless you want to talk about infinite dimensional vector spaces, and then you need to worry about what that infinite sum means (does it even exist?): it is certainly not an element of [math]\coprod_{\aleph_0} \mathbb{R}[/math] for instance.
  23. Define F(x) to be a function, such that dF/dx = (sin(x))^{1/2}, we can do this since the function on the right hand side is continous (when positive - let's not go into complex stuff) and hence integrable. Now, F is the answer to your question. If you don't like that answer then I can't help you. It is a perfectly good solid final answer. If we normalize so that F(0)=0, it is uniquely determined, and F(t) can be evaluated for any t to any arbitrary degree of precision. I don't see how that cannot be considered 'the final solid answer'. If you disagree, then it is because you think that you must have an answer written down in some 'nice' form that isn't actually necessary.
  24. Not all people in the world know what AP calc is, nor took the exam.
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