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Everything posted by MandrakeRoot

  1. You still didnt say what is x. But your algorithm is exponential in the size of your problem instance D. (since you have a lot of terms of the type x^(something with D), like D, D - 1 etc... The bigger D, the more terms you have. The only reason in the last term X^D, you have something like x^D/D! is because you have D! x^n/n!^2 for n =D, so maybe you can estimate the terms before ? Something like the coefficient of the n-1ste term behave at least like .... (something with n) and at most like (something with n), that way you know how your function behave approximately and you can send n to infinity and maybe use some of the assymptotic formula for hypergeometric fucntions. (Remember that they are each a solution to some differential equation). Mandrake
  2. What is x then ? Does it depend on D is some way ? (If x(D) = D! then your above limit argument is flawed). Mandrake
  3. What is D ? If D > 0 and if D is an integer indeed the series will terminate and your hypergeometric function will be a polynomial. What do you mean with x^D/D! converges to zero ? (You should at least keep x fixed in such an argument). Anyway the way you write down the hypergeometric function the D! will be in the upperpart of the fraction, so you have more something like D! x^D then x^D/D! A quick maple calculation shows that hypergeom(-D;2:-x) behaves like : 1 + 1/2 D x + 1/12 D (D - 1) x^2 + 1/144 D (D - 1) (D - 2) x^3 + 1/2880 D (D - 1) (D - 2) (D - 3) x^4 + 1/86400 D (D - 1) (D - 2) (D - 3) (D - 4) x^5 + 1/3628800 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) x^6 + 1/203212800 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) x^7 + 1/14631321600 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) (D - 7)x^8 + 1/1316818944000 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) (D - 7) (D - 8) x ^9 + O(x^10) Your algorithm looks more factorial than anything else i think. Mandrake
  4. Yes indeed, the notation is rather sloppy. It would be better to write y(X), y'(X) and Y(X) for example (and indicating the domain and range of these functions !) Mandrake
  5. I think with integral you mean primitive here ! Mandrake
  6. Can't you use the binomial formula or something like that ? If you can do it at (0,0), i think you are surely pretty close to showing it for every couple (x,y) Mandrake
  7. Yeah that is pretty much a typical discussion with Doron :=) Mandrake
  8. You still dont see the difference between application and theory. It is hopeless. Mandrake
  9. Mathematics is not an empirical science. It is based on logical deduction. When doing applied mathematics it is important to verify your results with observations. It is true that physical laws have restrictions on their domain of use. In pure mathematics restrictions are clearly formulated in the theorems and lemmes. When applying math to the "real world" (whatever you may mean with that) it is up to you to verify that these conditions are satisfied. (Like in your example above that indeed all scientists are aliens). The fact that not all scientists are aliens does not in any case make untrue the conclusion "Einstein is a scientist => Einstein is an alien", under the hypothesis that all scientists are aliens. Mandrake
  10. Some hints : 1) Uniqueness is quite trivial (Suppose there are two functions that satisfy the conditions, look at their difference and use the fact that for every point x in the closure of U there is a sequence entirely in U that converges to x, to show that both functions have to be equal. (Here you need the fact that the difference of two contniuous functions is again a continuous function) 2) => Take any point x in the closure of U and use the above fact about the sequence entirely in U that converges to x. Use uniformy continuity to conclude that f(x_n) is a Cauchy sequence in Y => hence convergent. Define g(x) to be this limit. 3) Check that g is well defined and does not for instance depend on the choice of the sequence x_n above 4) Show that g is the same as f on U 5) Show that g is continuous. Mandrake
  11. Yes i could. Consider the space (Q,d), where Q is the set of fractions and d is absolute distance (like the metric on R in fact) (note that we will pretend that R does not exist and therefore in the above replace epsilon by 1/k) Take the following sequence defined by [math]q_{n+1} = q_n - \frac{q_n^2 - 2}{2q_n}[/math] with q_0 say 2; This is clearly a sequence in Q that should converge to sqrt(2), but well that is not an element of Q. Mandrake
  12. A complete metric space is a metric space, but in this metric space all Cauchy-sequences are convergent. Note that in any metric space all convergent sequences are Cauchy ! (A Cauchy sequence is a sequences that sticks together in the tail : Let {x_n}_n be a sequence in the metric space (X,d), then {x_n}_n is called a Cauchy sequence if for every epsilon > 0 there exists a N (integer) such that d(x_n,x_m) < epsilon for all n,m >= N) Often it is easily shown that a sequences is Cauchy and so using the fact that your space is complete you know that there is some limit for your sequence ! If you would like some hints for the proof i could write them down ? Mandrake Ps: it is not a silly question
  13. No it isnt. When einstein wrote down his theory most of the consequences could not be checked. It was not until much later that some empirical data could be obtained to demonstrate the "applicability" of his theory. Basically you are saying that if we are a primitive culture and cant check let's say Newtons laws' date=' newtons laws are not a proper theory. Yes indeed an infinitisemal instant of time. Many physical formula are derived using this notion, and like you seem to say "if it works it is true", since these formula seem to be supported by empirical data, the abstraction of using "infinitisemal instants" is a not too absurd one. Mandrake
  14. Let me write down the entire theorem Let [math](X,d)[/math] be a metric space and [math](Y,\rho)[/math] a complete metric space. Let [math]f : U \rightarrow Y[/math] be a uniformly continuous function on the subset [math]U \subseteq X[/math] of X. There exists a unique continuous function [math]g : \bar{U} \rightarrow Y[/math] such that [math]g|_{U} =f[/math]. The proof is quite simple really, i am sure you can do it bloodhound. Mandrake
  15. Well the real topological definition of a compact set is a little more complicated. The equivalence closed and bounded and compact only holds in specific topological spaces (like the R^d's). In topological vector spaces compact implies bounded and closed but the inverse is not perse true. A closed set that is not bounded in R would be [0,infty) If i would give you a uniformly continuous function on some open interval, with the above argument it could be immediately extended to a continuous function on the closed interval that is the closure of this open interval, so i dont see how we could give non-trivial exemples ? In fact the whole strenght of the theorem is when we consider functions on some metric space X (not even supposed of finite dimension), then the closure of a set is less trivial and also the extension to go with. Or functions on some general open sets of R (that could be the union of many open intervals) Mandrake
  16. I will give an example. If f :(a,b) -> R is some uniform continuous function, then it is possible to find a continuous function g: [a,b] (the closure in R of (a,b)) -> R such that f(x) = g(x) for all x in (a,b), this g would be the extension of f to the closure of (a,b).... Do you see what i mean ? Mandrake
  17. How about the fact that the Gamma function can help in calculating integrals and simplifying expressions. Not to forget that you can use them to make a short hand notation for the pochhammer symbols. The gamma function often comes about in special function theory and is therefore also usefull in finding analytic solutions of some DE's ! Mandrake
  18. Maybe he wants to know the following. If f is a uniformly continuous from some set A, a subset of a metric space (lets say a subset of R) into some metric space (lets say R again), then f has a continuous extension on the closure of A into R (or the appropriate space). There are quite a lot of different theorems on limits though. I doubt he wants you to check that the space of continours functions from some vectorspace X into some vectorspace Y is a vectorspace. Mandrake
  19. Haha this one is funny. Though i think doron would say 1_1 or 1_0_1 Mandrake
  20. I think you should separate abstraction and application. You can perfectly create the natural numbers from a simple axiom using set theoretical arguments. Now when you would like to apply this system you decide that the unity is an apple regardless of its size, colour, configuration, weight etc...... If you disagree with this decision and find that ambiguous and consider that a red apple is not the same as a green one, then that in no way changes the abstract system. The only comment you are making is on the application of the abstract system to the real world. Mandrake
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