MandrakeRoot

There are many statements about limits of functions. Please state the statement you would like to proof ! Mandrake

Have you ever thought that maybe there is not a unique solution to your system ? But here goes some general advice : Solve equation 1 after c, that will give you something like c= h/(a*b^2) Plug this in everywhere you see c and then simplify => Solve equation 2 after b Equation b after plugging in c has the form G = (a^2 *b^5)/(a*h^2), which gives you something like b = [ (G*h^2)/a ]^(1/5), Plug this in in equation 4 and solve after d or a, and then finally put all in the last equation and solve this after the last unknown (a or d) if you like. That will give you an expression of a in terms of your constants, then you can plug this in everywhere you see a, and obtain the solution. Mandrake

Maybe it is better to try to solve the problem in the following way (or at least approximate the solution). Fix any point in the grid (x1,x2). Now calculate the cost to make a straight-line cable from P to this point and then from this point to F. Depending on your solution you could do the same thing in considering F to be this new point (x1,x2) and/or (x1,x2) the new origin. Ideally you would solve the problem with variable coordinates for F. This could be maybe some easy approximation of the optimal solution. Mandrake

Maybe it is better to try to solve the problem in the following way (or at least approximate the solution). Fix any point in the grid (x1,x2). Now calculate the cost to make a straight-line cable from P to this point and then from this point to F. Depending on your solution you could do the same thing in considering F to be this new point (x1,x2) and/or (x1,x2) the new origin. Ideally you would solve the problem with variable coordinates for F. This could be maybe some easy approximation of the optimal solution. Mandrake

Your solutions of Diff.eq.'s are always easy to verify ! Try taking the appropriate derivatives and see if they actually satisfy your DE ! If that is the case you have indeed found a solution (maybe not all but well....), if they dont satisfy your DE then you havent found a solution. So with this type of exercise you should be able to check your answers properly. Mandrake

Your solutions of Diff.eq.'s are always easy to verify ! Try taking the appropriate derivatives and see if they actually satisfy your DE ! If that is the case you have indeed found a solution (maybe not all but well....), if they dont satisfy your DE then you havent found a solution. So with this type of exercise you should be able to check your answers properly. Mandrake

Ok without loss of generality we can place P in the origin of some grid. Your factory F is then placed at the point (200,400); Now let C^1([0,1];R) be the space of continuously differentiable functions from [0,1] into R. Then your problem can be formulated as : [math]\min_{x,y \in C^1([0,1];\mathbb{R})} \int_0^1 \sqrt{12x'(t)^2 + 6y'(t)^2}dt[/math] Mandrake

Ok without loss of generality we can place P in the origin of some grid. Your factory F is then placed at the point (200,400); Now let C^1([0,1];R) be the space of continuously differentiable functions from [0,1] into R. Then your problem can be formulated as : [math]\min_{x,y \in C^1([0,1];\mathbb{R})} \int_0^1 \sqrt{12x'(t)^2 + 6y'(t)^2}dt[/math] Mandrake

Yeah dave you said it alright. THe writing just misses the mathematical rigor or even the scientific universality .... Mandrake

Yeah dave you said it alright. THe writing just misses the mathematical rigor or even the scientific universality .... Mandrake

Let me help you a bit : The general form of the n-th derivative of the arcsine function has the following form [math]\frac{\partial f^{(n)}}{\partial x^n}(x) = (1 - x^2)^{-\frac{2n-1}{2}}p_n(x)[/math], where p_n is a polynomial that satisfies the following [math]p_{n+1}(x) = (2n-1)xp_n(x) + (1- x^2)p_n'(x)[/math], Here n >= 1 and p_1 is all constant 1 function. Mandrake

I think you are confusing two things here, the definition of a mapping and evaluation of a mapping at some element. S is a mapping from the set of integers into the set of integers and has the following descriptive formula : [math]m \mapsto \frac{1}{2}m(m+1)[/math] Now you can evaluate the mapping S at any integer n or n-1 or whatever and get an expression for their values. Mandrake

Asking questions is always good in order to understand things. Mandrake

f is clearly injective since the image of any two distinct points is distinct (a simple property of the sum that you can without a doubt assume) f is surjective since every element has an original in the case where you consider the function f from R into R ! (try finding which one that would be) When taking the function from Z into Z this is not true since 2 does not have an original ! (In other words 2 is not in the range of your function !) I've tried to explain it rather simple, i hope that worked out ? Mandrake

If i remember correctly a^b for two cardinals a and b is defined to be the cardinality of the set of all functions from a set of cardinality b into a set of cardinality a. With this definition you can show that 2^b is indeed the cardinality of the powerset of b, since {0,1} is a canonical set of cardinality two and so any mapping from a set of cardinality b into {0,1} can be seen as a characteristic function of a subset of (the set of cardinality) b. Mandrake

Zero is a natural number according to most conventions. It all depends on whether or not you find it usefull to include it in this set. Mandrake

[math]2^{\aleph_0}[/math] is an expression coming form cardinal arithmic. You can actually show that a set having this cardinality has the same cardinality as [math]\mathbb{R}[/math]. I dont understand why you talk of base 2, because we are not talking of real arithmic but cardinal arithmic ! Mandrake

IT is exactly this sort of thing that makes it ill writen. I totally agree with you dave. Doron you introduce concepts without defining them and expect people to see them the same way that you do, if we dont think similar about your concepts there is no way we can follow your reasoning. Mandrake

As far as i remember there is no such formula. Try for instance to see the link between the det(A + B), det(A) and det(B) when both A and B are n x n diagonal matrices ? Now add two elements to B on the (1,2) and (2,1) position and see what happens with your formulae... Mandrake

You can always define what you like. The most important question is if what you define makes sense. You can extend the real numbers by adding infinity to the set of real numbers as an element and then define 0*infinity to equal pi, but that would not make much sense. Mandrake

I think the best way to describe topology is by making a contrast with metric spaces. In metric spaces you study the propeties of a space using a notion of distance. In topology you study spaces and their propeties using an imposed structure of "open sets". Any metric space is particularly a topological (vector) space. Topological spaces are far more general and you can easily construct such spaces, wherein there exists sequences converging to every point of the space, something which is impossible in any metric space. Mandrake

I actually seriously tried to read his work but it is not written very clearly ! The guy also posts the same thing over and over haven't you remarked ? Mandrake