Johnny5

Suppose you are asked to generate all three permutations of the set {a,b,c,d}. You could attempt to generate the list randomly. But how would you be certain you listed all the elements? If you knew a priori, how many elements there are in it, then you could just keep going until you finally had the list. But there is a systematic method to generate all the r-permutations of a set of n objects, and it is called "lexicographic ordering." In the example problem about to be done, you are given the following set of four elements: {a,b,c,d} The distinction between an n-set, and an n-tuple, is that the order that the elements of a set are listed in from doesn't matter, but the order that the elements of an n-tuple are listed in does matter. The way lexicographic order works is this. Even though we are given a set of random things, we assign increasing numerical value to them, from right to left. Thus, in the example here, a has a lower lexicographic value than b, b has a lower lexicographic value than c, and so on. Now, we generate the set of all r-permutations, by listing out the elements in increasing order, as if they were numbers. So for example suppose you were asked to generate the set of all three permutations of the set {a,b,c,d}, then the lowest valued 3-permutation is: (a,b,c) Now, you want to find the next lowest valued 3-permutation. If you think in terms of numerical quantities, you will eventually realize that it is: (a,b,d) The next lowest is: acb And the rest... in lexicographic order are: acd adb adc bac bad bca bcd bda bdc cab cad cba cbd cda cdb dab dac dba dbc dca dcb And dcb is the higest valued 3-permutation so we are done. Now, we can just count, to see how many elements there were, the answer is: 24 The next question is, can we find a formula that would have told us in advance how many there are, so that if desired, we could generate the list at random, until we had the right list? One way to answer this question, is for us to have access to the multiplication principle. I have a book here and they state the principle as follows: Multiplication Principle: If A is a set of p objects and B is a set of q objects, then the number of ordered pairs of the form (a,b) with a an object of A and b and object of B is equal to the number of elements in the cartesian product A X B... which is p times q. So, using the multiplication principle, we can figure out in advance, how many elements will be generated using lexicographic order, before listing them. In the example just done, we were looking for three tuples. (X,Y,Z) where X,Y,Z are all distinct. There are four ways to choose X. For each of these ways, there are three ways to choose Y. And now, for each of these ways, there are two ways to choose Z. So you can draw a tree diagram, which lists them. For example: a /|\ b c d /\ /\ /\ cdbdbc So right here we have six different 3-permutations, beginning with a. They are: (a,b,c),(a,b,d),(a,c,b),(a,c,d),(a,d,b),(a,d,c) Likewise, for 3-permutations beginning with b, we also have six permutations: b /|\ a c d /\ /\ /\ bcadac And each of these is different from all of the first six, because they begin with b, not a. And for 3-permutations beginning with c, there are six 3-permutations, and six more beginning with d. So the total number of unique paths in the tree diagrams, is 6+6+6+6=24, which agrees with our earlier result. Now, we just want to summarize this using a formula. The number of ways to choose the first element is n. Then for each of those ways, the number of ways to choose the next element is (n-1). Then for each of these ways, the number of ways to choose the next element is (n-2). And you keep going until you reach r rows, in your tree diagrams. In the example problem just considered, n=5, and r=3. So that the answer was given by 5(5-1)(5-2) So as you can see, the general formula is: n(n-1)... (n-r+1) And r is constrained to be less than or equal to n. We can now introduce product notation for this. [math] \prod_{k=n-r+1}^{k=n} k [/math] Multiplication is commutative, so the same number could be written as: [math] \prod_{k=n}^{k=n-r+1} k [/math] I generally write the smaller number at the bottom, so in this case I will go with: [math] \prod_{k=n-r+1}^{k=n} k [/math] Now, as I said r is constrained to be less than or equal to n. Suppose that r is equal to n. Then we have: [math] \prod_{k=n-n+1}^{k=n} k [/math] Which is equivalent to: [math] \prod_{k=1}^{k=n} k [/math] Which is the customary defintion of n factorial, written n!. That is: [math] n! = \prod_{k=1}^{k=n} k [/math] Now, consider the case where r=1. In this case we have: [math] \prod_{k=n-1+1}^{k=n} k [/math] Which is equivalent to: [math] \prod_{k=n}^{k=n} k [/math] Which is equivalent to n. That is: [math] n = \prod_{k=n}^{k=n} k [/math] Now consider the strange case where r=0. Thus, we are given a set of n objects, and asked to generate all 0-tuples. In other words, tuples with no elements. Using the formula already obtained, we have: [math] \prod_{k=n-r+1}^{k=n} k [/math] [math] \prod_{k=n-0+1}^{k=n} k [/math] [math] \prod_{k=n+1}^{k=n} k [/math] Here is where we run into problems. Because multiplication is commutative, the previous iterated product is equivalent to [math] \prod_{k=n}^{k=n+1} k [/math] and [math] \prod_{k=n}^{k=n+1} k = n(n+1) [/math] But I think there is a problem with this. Suppose you are given a set of 5 objects, and asked how many 0-permutations there are. Then n=5, and r=0, and using the above blindly you have: [math] \prod_{k=5}^{k=6} k = 5(6) =30 [/math] So somewhere we need to stop this error from occurring.

The proof I am thinking of goes like this. First, obtain the definition that: 0!=1 Based on some combinatorical argument. Next... Analyze the series for e^x. Notice the following: [math] \prod_{k=1}^{k=n} x = x^n [/math] and also [math] \prod_{k=1}^{k=n} \frac{1}{k} = \frac{1}{n!} [/math] Combine them like so: [math] \prod_{k=1}^{k=n} \frac{x}{k} = \frac{x^n}{n!} [/math] Then define e^x as follows: [math] e^x \equiv \sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{x}{k} [/math] Hence it logically follows that: [math] e^x \equiv \sum_{n=0}^{n=\infty} \frac{x^n}{n!} [/math] Now, from memory, there is a thing about the first constant term, but I have to solve a few ordinary differential equations to remember what it was, but it will come back to me. But the point is this: Presuming that the first term is one leads to: [math] e^0 = \frac{0^0}{0!} + 0 +0 +... = \frac{0^0}{0!} [/math] Then using the result based on a combinatorical argument that 0!=1, it follows that: [math] e^0 = \frac{0^0}{0!} = \frac{0^0}{1} = 0^0 [/math] So granted that e^0=1, you would now have a proof that 0^0=1. Now, if e^0 isn't 1, then the first term of the series isn't 1. Regards

Why not? Not bother with what? That black holes cannot emit photons, something about GR, ???

[math] 3x^3 y^{\prime \prime \prime} + 9x y^\prime - y = 0 [/math] Where prime denotes differentiation with respect to x. i.e. y`=dy/dx Thank you Well I will just begin working on my own problem. Assume a power series solution. [math] y(x) = \sum_{n=0}^{n=\infty} C_n x^n [/math] So the first derivative with respect to x is given by: [math] y^\prime = \frac{dy}{dx} = \sum_{n=0}^{n=\infty} nC_n x^{n-1 [/math] The second derivative is given by: [math] y^{\prime \prime} = \frac{d^2y}{dx^2} = \sum_{n=0}^{n=\infty} n(n-1)C_n x^{n-2 [/math] The third derivative is given by: [math] y^{\prime \prime \prime} = \frac{d^3y}{dx^3} = \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3 [/math] Three x cubed times the third derivative is given by: [math] 3x^3 y^{\prime \prime \prime} = 3x^3 \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3 [/math] Nine x times the first derivative is given by: [math] 9x y^\prime = \frac{dy}{dx} = 9x \sum_{n=0}^{n=\infty} nC_n x^{n-1 [/math] And negative one times y is given by: [math] -y = -\sum_{n=0}^{n=\infty} C_n x^n [/math] Adding these three series together yields: [math] 3x^3 y^{\prime \prime \prime} + 9x y^\prime -y = [/math] [math] 3x^3 \sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n-3} + 9x \sum_{n=0}^{n=\infty} nC_n x^{n-1} -\sum_{n=0}^{n=\infty} C_n x^n = 0 [/math] Now, let's perform the necessary indicial operations, so that we can write the LHS as a single series, if possible. The first thing to notice is that we can move x^3 through the series, as well as x, like so: [math] 3\sum_{n=0}^{n=\infty} n(n-1)(n-2)C_n x^{n} + 9 \sum_{n=0}^{n=\infty} nC_n x^{n} -\sum_{n=0}^{n=\infty} C_n x^n = 0 [/math] Actually, in this case no indicial operations need to be performed, because we can factor out x^n now. [math] \sum_{n=0}^{n=\infty} [3n(n-1)(n-2)C_n + 9n C_n - C_n ]x^n = 0 [/math] And we can also factor out C_n like so: [math] \sum_{n=0}^{n=\infty} C_n[3n(n-1)(n-2) + 9n - 1 ]x^n = 0 [/math] The only way for the statement above to be true, given that not(x=0), is if: [math] C_n [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math] Now, C_n is just an arbitrary constant. Let us stipulate that it be nonzero, so that we can divide through by it. Thus: [math] [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math] Now, let us carry out the multiplication above, and write the LHS as a cubic equation, which will have exactly three roots. [math] (n-1)(n-2) = n^2-2n-n+2 = n^2-3n+2 [/math] [math] 3n(n^2-3n+2 ) = 3n^3-9n^2+6n [/math] [math] 3n(n^2-3n+2 )+9n-1 = 3n^3-9n^2+6n +9n-1[/math] Now, [math] 3n(n-1)(n-2) + 9n - 1 = 3n(n^2-3n+2 )+9n-1 [/math] And [math] 3n^3-9n^2+6n +9n-1 = 3n^3-9n^2+15n-1[/math] Therefore, using the transitive property of equality, as well as the previous three lines of work, it follows that: [math] 3n(n-1)(n-2) + 9n - 1 = 3n^3-9n^2+15n-1 [/math] Therefore: If [math] [3n(n-1)(n-2) + 9n - 1 ] = 0 [/math] then [math] 3n^3-9n^2+15n-1 =0 [/math], and conversely. So now, we need to find the roots of the above polynomial. The method I am going to employ is called synthetic division. Synthetic division Lets look at the factors of the constant term. The constant term is -1. [math] 1 \cdot -1 = -1 [/math] Lets look at the factors of the leading coefficient: [math] 1 \cdot 3 = 3 [/math] So we can check for roots from amongst: [math] \pm 1 [/math] [math] \pm \frac{1}{3} [/math] And we are going to do this using synthetic division. Checking -1: [math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math] [math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|-1} [/math] [math] .\ \ \ -3 \ \ 12 \ \ -27[/math] __________________ [math] 3 \ \ -12 \ \ 27 \ \ -28 [/math] Because the remainder, -28, isn't zero, -1 is not a root. Checking 1: [math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math] [math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|1} [/math] [math] .\ \ \ \ 3 \ \ -6 \ \ \ 9[/math] __________________ [math] 3 \ \ -6 \ \ 9 \ \ -8 [/math] Because the remainder, -8, isn't zero, 1 is not a root. Checking 1/3: [math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math] [math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|\frac{1}{3} } [/math] [math] .\ \ \ \ 1 \ \ \ \frac{-8}{3} \ \frac{37}{9} [/math] __________________ [math] 3 \ \ -8 \ \ \frac{37}{3} \ \frac{28}{9} [/math] Because the remainder, 28/9, isn't zero, 1/3 is not a root. Checking -1/3: [math] n^3 \ \ \ n^2 \ \ n^1 \ \ \ n^0 [/math] [math] 3 \ \ -9 \ \ 15 \ \ -1 \underline{|\frac{-1}{3} } [/math] [math] .\ \ \ -1 \ \ \ \frac{10}{3} \ \frac{-55}{9} [/math] __________________ [math] 3 \ \ -10 \ \ \frac{55}{3} \ \frac{-64}{9} [/math] Because the remainder, -64/9, isn't zero, -1/3 is not a root. Therefore, there are no rational roots of the polynomial. But, n is constrained to be a positive integer.

Let it be the case that [math] 1^1 = 1^0 [/math] Take the natural logarith of both sides of the statement above' date=' so the following statement has the same truth value as the previous one: [math'] ln(1^1) = ln(1^0) [/math] Here is a brief tutorial on logarithms in general: Hyperphysics on logarithms Notice a little ways down, they say that log to the base b of N to the y, is equal to y times log to the base b of N, look for: [math] log_b N^y = y log_b N [/math] Using natural logs, that is written as follows: [math] ln N^y = y ln N [/math] Where the base b, has been chosen to be e=2.71828... , and is now implicit instead of explicit. continuing on... [math] ln(1^1) = ln(1^0) [/math] Using the natural log rule it follows that: [math] 1 ln 1 = 0* ln(1) [/math] The natural logarithm of 1, is equal to zero, that is: [math] ln 1 = 0 [/math] Therefore: [math] 1 *0 = 0*0 [/math] And it is provable, using the axioms of the real number system, that zero times anything equals zero, therefore: [math] 0 = 0 [/math] And there is no contradiction. The following steps in your original argument was invalid: And the reason is that Log 1=0, and you cannot divide by zero, because that will contradict the field axioms. Regards If you look down at the link below you will see that log to any base of 1 is equal to zero. Log 1 = 0 regardless of base I usually do most things using the natural log, but you can go from a logarithm of any base, to natural logs using the following relation: [math] log_a x = \frac{ln x}{ln a} [/math] and Log is usually reserved for log to the base ten. Hence: [math] log_{10} x = Log x = \frac{ln x}{ln 10} [/math] Therefore: [math] ln 10 log_{10} x = Log x = ln x [/math] I am presuming you mean log to the base ten, since you wrote Log. Multiply both sides of the statement above by ln 10, to get: [math] ln(10) * (1)Log 1 = ln (10) * (0)Log 1 [/math] Therefore: [math] (1) ln 1 = (0) ln 1 [/math] I only converted from base ten to base e, because I am used to using the natural log, but you don't have to do that. Nonetheless, from memory I know that ln 1 = 0, and you cannot divide by zero, your step here (1)Log 1 = (0)Log 1 1 = 0 is against the field axioms; Log 1 = 0 as well. Regards

The correct way to answer you, is for me to utilize general order logic. I am still working on it, when I figure out something perfect, you will know. Regards I intend to tie everything in to basic combinatorical formulas. If I had to guess now, i would say that it's arbitrary.

if you define x^0 to equal 1, for any number x, then yes. But that would be a definition, not an axiom, nor a theorem. But mathematics has to be internally consistent, obviously, so one should not choose a definition like that without checking to see first, whether or not it can be a theorem, and additionally what if it contradicts something else? x^3=xxx by definition x^2=xx by definition x^1=x (seems to make sense) x^-1 = 1/x definition x^-2 = 1/xx definition x^-3 = 1/xxx definition but the case x^0, setting it equal to 1... for any x... well it is a theorem of the axioms of the real number system that 0*x = 0 for any real number x. So... 0^3=000=0 0^2=00=0 0^1=0 0^0=? (not so clear) 1/0 (will lead to an explicit contradiction) 1/00 (will lead to an explicit contradiction) 1/000 (will lead to an explicit contradiction) I am not sure yet about 0^0, which is why I asked the question. Actually, provided that d/dx(e^x) = e^x, then it must be the case that 0^0=1, but the relation above is what started this in the first place.

Yep, i know.

Matt, doesn't 0! need to be defined to be one, so that certain extremely important combinatorical relations be satisfied? For example the number of 3-permutations of set {a,b,c,d}. P(N,r) = N(N-1)(N-2)... (N-r+1) Or [math] P(N,r) = \sum_{n=1}^{n=N} \frac{N!}{(N-r)!} [/math] So my point is it has to be the case that 0!=1. Do you see why? Regards Example given: Generate all 3-permutations of the set {a,b,c,d}. A three permutation is going to be a three tuple, an ordered triple, with elements chosen from the set {a,b,c,d}. I simply want to list them all. There are four choices for the first component of the tuple, three choices for the second, and two choices for the third, so the total number of 3-permutations is 4*3*2=24. Now I just have to list them out. Wolfram on Lexicographic order Here is an online program that generates r-permutations using lexicographic order: Permutation generator

suppose that you are in a frame in which the particle is at rest, and that the particle isn't being subjected to any forces. Then it will remain at rest in the frame, until such time as something external acts upon it, at which time it will accelerate in that frame. The way physicists use the uncertainty relation, connects to their measurements of the position of it. But if they aren't trying to determine its position, then it will just sit there at rest. It doesn't care whether or not you "know" it is at rest. But if you were to fire something at it with speed c, and then have that thing reflect, and come back to you with speed c, you would be able to calculate how far away it was when it was hit, provided you measure the time it takes for the thing you fired to return to you. Then since you know you fired it at speed c, and you measured t, ct would give the distance it was away from you. But when the thing hits it, you have to have conservation of momentum, and conservation of energy. Some "energy" will be imparted to the particle upon its being 'hit'. So therefore, that which returns to you will have less "energy" then it started with. The point is, it will no longer be at rest. So that by the time you know where it was, you will no longer know where it is. This is the kind of logic inherent in interpreting the uncertainty principle. So to answer your question... yes, you are right... in the customary sense of the interpretation. Regards PS: There is also the question about what is meant by the term particle. If you have a body, then the uncertainty relation would apply to the center of mass of the body, which is a point, not a particle. There are other issues which need to be addressed, but thankfully they are not infinite in number.

If I make an error in what follows, I would like it pointed out. The purpose has something to do with 0!, and 0^0, but I am not going to say what. Preliminary work: There is a superset [math] \mathbb{S} [/math], and any element of that superset is called a number. Axiom A [math] 0 \in \mathbb{S} [/math] Axiom B [math] \forall x \in \mathbb{S}[0+x=x] [/math] Axiom C [math] \forall x \in \mathbb{S}\forall y \in \mathbb{S} [x+y \in \mathbb{S}] [/math] Axiom D [math] \mathbb{N} \subset \mathbb{S} [/math] Undefined binary relation on S: < Definition: [math]\forall x,y \in \mathbb{S} [ x > y \Leftrightarrow y<x ] [/math] Definition (equality of two numbers): [math]\forall x,y \in \mathbb{S} [ x = y \Leftrightarrow not(x<y) \ \& \ not(y<x) ] [/math] Axiomatic development of the set of natural numbers Axiom I [math] 1 \in \mathbb{N} [/math] Axiom II [math] \forall x[x \in \mathbb{N} \Rightarrow x+1 \in \mathbb{N}] [/math] Theorem: [math] 1+1 \in \mathbb{N} [/math] Proof: By axiom 1, 1 is an element of the set of natural numbers. Now, plug 1 into axiom two to get the following true statement: [math] 1 \in \mathbb{N} \Rightarrow 1+1 \in \mathbb{N} [/math] The antecedent of the statement above, is axiom 1, hence true. Therefore, the consequent is also unconditionally true(modus ponens). Therefore: [math] 1+1 \in \mathbb{N} [/math] Which is the theorem. QED It isn't hard to see that using the theorem just proven, together with axiom2, that 1+1+1 is a natural number. And we can keep repeating the same steps ad infinitum. So that much is clear at this point. However, there has not been introduced enough axioms to reach the following conclusion yet: not (1=1+1) And similarly we will want to know that not (1+1=1+1+1), and so on. Axiom III: [math] \forall y \in \mathbb{N} [ y < y+1] [/math] The axiom above is going to be true, regardless of whether we take it as axiom, or theorem. From it, it now follows that 1<1+1, 1+1<1+1+1, etc. And also, it now follows that not(1=1+1), not(1+1=1+1+1), etc. Since 1<1+1, it follows from the definition of equality of two numbers, that it cannot be the case that 1=1+1, because if 1=1+1 then not (1<1+1), etc.

Astronomical orientation of any monument At this site, they say if you enter the azimuth, latitude, longitude, that the software can give you information about what stars (if any) were used to align the monument. Has anyone here used this software, or software like it before? Thanks

Steve, quick question. Must it be the case that 0! = 1? Regards

Well I'm going to learn about them now. Perhaps I could help you this time. Kind regards

Ok I got it, thank you very much Dr Swanson. Quick question, do you know about Euler angles?

In this post, I am going to try and figure out what the heck Euler was trying to do, by relating the harmonic series to log base e. I will try to answer my own question. Ok so... Harmonic Series First start off with the finite series, defined as follows: [math] H(k) = \sum_{n=1}^{n=k} \frac{1}{n} [/math] So, for example, the third harmonic number is: H(3) = 1/1+1/2+1/3=6/6+3/6+2/6=(6+3+2)/6=11/6 Now, in the limit as k goes to infinity, we get the harmonic series: [math] H(\infty) = \sum_{n=1}^{n=\infty} \frac{1}{n} [/math] So why was Euler messing with this? Here is an article on John Napier. He died in 1617, Newton born 1642, so he died 25 years before Newton was born. That article was dumb. In introductory calculus texts, e^x is often introduced as follows: [math] e^x = \lim_{n \to \infty} (1+\frac{x}{n})^n [/math] And prior to this, the student is introduced to limits. So look at the case where x=1. Then you have: [math] e = \lim_{n \to \infty} (1+\frac{1}{n})^n [/math] And you can just make something resembling a harmonic term. What in the world led him to that, that just doesn't pop out of thin air. You need to understand the mathematics of the time. 1707-Leonhard Euler-1783 1501-Giralamo Cardano-1576 The first mention of the square root of -1 in print, appears to have been by Giralamo Cardano, in 1545, however he dismissed them as useless. In 1572, Raphel Bombelli used them in calculations in his work L'algebra. In 1702, Leibniz appears to have mistrusted root -1, and in 1770 Euler argued that root(-2) times root (-3) = root (-6) See this document for a source: Geometry and Complex arithmetic 1698-Colin MacLaurin-1746 MacLaurin's biography The following quote is taken from the article above: 1865-Brook Taylor-1731 Taylor's biography The following quote is taken from the article above: Ok that gives you an idea of the state of mathematics between 1500-1800. And for those who don't know, Cardano published "Cardan's formula" for solving a cubic equation, but it appears that he stole the idea from Niccolo Tartaglia. But what I am trying to figure out, is where Euler got the idea to take the limit of (1+x/n)^n. That's such an odd thing to just do for no reason at all. Here is a good article on e=2.71828... 2.71828...

Well there is no rush to see your proofs, but I would be interested no matter whether they be right or wrong. Obviously, you make a certain number of assumptions, for a starting point, and you say that if your assumptions are true then such and such, so that your reasoning will be correct. So it is the assumptions I am interested in. But that is for some other time. But... you did say something interesting, which is that black holes do not emit light. I was talking to a fellow here, by the name of 'island', several weeks ago, and he got me thinking about that. Obviously, black holes have an immense gravitational pull. Obviously, they emit something, and that something curves the path of something approaching near. Island's point was simply, that the something emitted (if anything) could not be 'light' since light cannot escape a black hole. A conundrum. Regards

Light which is emitted by the sun, doesn't get gravitationally pulled back to the sun. So it makes more sense to say that light has no gravitational mass. The rest frame of a photon is an inertial frame, but that's for another time. Do you happen to have a proof that light has no inertial mass? PS: Another way to see my point, is that photons communicate gravity, and so are not effected by it.

Is it true that e=2.71828... is related to the harmonic series? If so, how exactly? Was Euler looking for a function that is its own derivative?

How did Euler do it? This is a biography: Leonhard Euler If you read down, you will see this: That was 1765. There is also this comment here: And that is a harmonic series.

There is a variable t, for time, used in relativity, and three spatial variables x,y,z. But what do you mean by the term 'dimension' ? Regards

Well time isn't a dimension. Regards