Johnny5

Correct... though I could be persuaded to define 0! somehow, if the reason makes combinatorical sense.

What theory? And what do you mean that...."square root of -1 can be obtained by adding up real numbers"??????

It is going to take me some time to sift through everything you said' date=' but this is a good place to start. In another thread, the question as to what (-1)! means came up, but I don't remember which one or how or why. I will try to find it. As for how I define n!, I have adopted the conventional definition. n! = n*(n-1)(n-2)..... 1 And this can be written as follows, using product notation: [math'] n! = \prod_{k=1}^{k=n} k = 1*2*... *n [/math]

Can you explain this a bit more. time evolution operator?? [math] i\hbar \frac{\partial}{\partial t} [/math] ? You say this leads to solutions which wave, and that if the i wasn't there then the solutions would decay. Can you prove that mathematically, in as simple a manner as you know how? Thanks

One at a time. I already thought about how to answer this exact question over the weekend. I was working on really primitive stuff' date=' regarding summation notation. The entire "axiomatic approach to numbers" which I use, begins by developing the natural number system, in a fashion similiar to Peano. Years ago, when I followed Peano's original work, I recall a neat little proof that multiplication is commutative, though I've forgotten how the proof goes now. At any rate, the idea I had this weekend goes like this: Define the natural numbers axiomatically, similiar or identical to the way Peano did it. That being done, let a,b denote arbitrary natural numbers. a is the sum of 'a' one's, and b is the sum of 'b' one's. That is: [math'] a = \sum_{n=1}^{n=a} 1 [/math] [math] b = \sum_{n=1}^{n=b} 1 [/math] Now consider the sum: [math] a + b = \sum_{n=1}^{n=a} 1 + \sum_{n=1}^{n=b} 1 [/math] Now, we will already have proven trichotomy, so it will now follow that either a=b XOR a<b XOR a>b. Let us first consider the case where not(a=b). Without loss of generality, let a>b. Thus: [math] a + b = \sum_{n=1}^{n=b} 1 +\sum_{n=b+1}^{n=a} 1 + \sum_{n=1}^{n=b} 1 [/math] So that: [math] a + b = \sum_{n=b+1}^{n=a} 1 + 2 \sum_{n=1}^{n=b} 1 [/math] And if we define subtraction, we can write: [math] a + b = \sum_{n=1}^{n=a-b} 1 + 2 \sum_{n=1}^{n=b} 1 [/math] Which is nothing but the obvious fact that: [math] a+b = (a-b)+2b [/math] Now consider the case where b=a. In this case we have: [math] a + a = \sum_{n=1}^{n=a} 1 + \sum_{n=1}^{n=a} 1 = 2 \sum_{n=1}^{n=a} 1=2a [/math] And now, consider a+a+a. [math] a + a + a = \sum_{n=1}^{n=a} 1 + \sum_{n=1}^{n=a} 1 +\sum_{n=1}^{n=a} 1 = 3 \sum_{n=1}^{n=a} 1 =3a [/math] Both of which facts come straight out of the distributive law. So, if we have the sum of m a's, we can write the following: [math] a_1+a_2+a_3+...a_m = m \sum_{n=1}^{n=a} 1 =m \cdot a [/math] at least for natural numbers. Now, consider the case where m=a. In this case we have: [math] a_1+a_2+a_3+...a_a = a \sum_{n=1}^{n=a} 1 =a \cdot a [/math] And we introduce the following notation: [math] a \cdot a = a^2 [/math] [math] a \cdot a^2 = a^3 [/math] And so on. So this gets you from pure addition of natural numbers, to exponents; on the field of the natural numbers at least. So now to your question. What makes me think that I can raise a number to the power m, in the case where m isn't a natural number. The answer comes from the Binomial theorem Matt, I do believe. Binomial Theorem: [math] (1+x)^\alpha = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{(\alpha +1 - k)}{k} [/math] Let's see if we get Pascal's triangle, when m is a natural number. Example 1: Let alpha=2, let x=a/b. Hence: [math] (1+a/b)^2 = \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(2 +1 - k)}{k} [/math] Which leads to: [math] \frac{1}{b^2} (b+a)^2 = \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(2 +1 - k)}{k} [/math] Which leads to: [math] (b+a)^2 = b^2 \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(2 +1 - k)}{k} [/math] Now, from memory the answer is: [math] (b+a)^2 = a^2+b^2+2ab [/math] The combination sum/product should also give this answer. [math] \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(2 +1 - k)}{k} = \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} [/math] Starting off at n=0, we have: [math] \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} = 1+ \sum_{n=1}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k}[/math] Then, evaluating at n=1, we have: [math] \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} = 1+2a/b+ \sum_{n=2}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k}[/math] And then evaluating at n=2, we have: [math] \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} = 1+2a/b+ a^2/b^2+ \sum_{n=3}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k}[/math] At this point, the numerator in the iterated product is zero, when k=n=3. And so for any larger value of n, there will be multiplication by zero, hence: [math] \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} = 1+2a/b+ a^2/b^2 [/math] Therefore: [math] b^2 \sum_{n=0}^{n=\infty} \frac{a^n}{b^n} \prod_{k=1}^{k=n} \frac{(3 - k)}{k} = b^2 +2ab+ a^2 =(a+b)^2 [/math]

In order to answer this question, one has to know the orbital slot of the satellite. The satellite is in the equatorial plane of earth, and it is at an altitude such that the satellite is in geostationary orbit. This means it hovers over the same position above the land, 24 hours a day, 7 days a week. That altitude can be deduced, using conservation of energy, but it's been awhile since I actually carried out the calculation. As for the orbital slot of the Intelsat 803, that can be found from Intelsat's website. Actually, I think they changed it to New skies. So change the satellite to Intelsat 901, here is the link Orbital location of IS-901 The orbital location is given as 342 degrees east, which is the same as 18 degrees west. To understand where this is, in relation to the surface of the earth, simply look at a worldmap, and find the point (0,0). That point happens to be six degrees of latitude south of Accra Ghana, which is in Africa. Then move to the west 18 degrees of longitude. Then, imagine a straight line through the center of the earth, through the point 0 N latitude, -18 E longitude, extending all the way into space, and terminating at the position where a satellite would be in geostationary orbit. Then, figure out where the ship would be. The given information was that it was 10 nautical miles east of the Florida Keys. 1 nautical mile is 1852 meters. So figure out the latitude, and longitude of the ship. Then treat the earth as spherical, to a first approximation. Later this can be corrected for, by treating the earth as an ellipsoid. The mean radius of the earth, is something like 6300 kilometers. Then, from there, figure out what the azimuth, and elevation are, for the ships antenna to be aimed at the satellite. If no one answers this in the next few days, I will work on it. My first step would have to be, to calculate the location of geostationary orbit, in say kilometers above the surface of the earth.

You aren't following the logic, which is quite simple, look above where I discuss who is stipulating what to whom, and who isn't agreeing with who. I didn't dictate to the world that 0!=1. However, if I agree with the world that 0! =1, then I can prove to the world that 0=1, which is absurd. Regards

I did not use x^0 for any x' date=' look more carefully please. I used x^0 for some x. Specifically this is the statement that was used in the argument: [math'] \forall x \in \mathbb{R} [ \text{if not(x=0) then } x^0=1 ] [/math] And furthermore, the statement was concluded to be true, at the outset. Very carefully proven I might add. I used the axioms of the real number system. Now, a definition is a statement which is stipulated to be true. But if I don't agree with it, you can 'stipulate' it all you want... though I've not agreed. 0!=1 is a statement which must be agreed to, by both reasoning agents. I've not yet agreed, because I am still working on a combinatorical reason to agree to it. But lets just say that I've agreed, to locate the contradiction. Suppose then, that I have also agreed that: [math] e^x = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac {x^3}{3!}+...+\frac{x^n}{n!}+... [/math] Then in the case where x=0, we have: [math] e^0 = \frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+\frac {0^3}{3!}+...+\frac{0^n}{n!}+... [/math] And the argument began by showing that: 0^1=0 0^2=0 0^3=0 and so on, therefore: [math] e^0 = \frac{0^0}{0!} [/math] And the argument was for any real number x (other than zero), that x^0 must equal 1. Hence e^0=1, and using transitivity we have: [math] 1 = \frac{0^0}{0!} [/math] Now, using the statement that 0!=1, which I've not really agreed to, though we are pretending I have, it now follows that: [math] 1 = \frac{0^0}{1} [/math] Whence it follows that: [math] 1 = 0^0 [/math] Which contradicts the very original theorem, which was that x^0=1 for any x except zero.

No, the argument is not circular, in fact it makes use of someone elses stipulation that 0!=1. I didn't dictate that 0!=1, and wouldn't because it causes contradiction, as the linear reasoning showed. Regards

Matt' date=' because something is bothering me about this whole 0^0, and 0!=1 thing, combined with e^0. I thought about it over the weekend, so that I could explain it here clearly. here is a 'law' of exponents: [math'] a^n a^m = a^{n+m} [/math] So now, consider the case where n=0. [math] a^0 a^m = a^{0+m} = a^m [/math] Where I have made use of the field axiom that 0+m=0 for any real number m. So certainly, the line of work above is true when m is a natural number, since the natural numbers are a subset of the reals. So for m an element of the natural numbers we have: [math] a^m = a_1 a_2 a_3... a_m [/math] In the case where a=0, the RHS is clearly zero, from a theorem (0*x=0 for any real x) which can be proven from the field axioms. So focus on this line here: [math] a^0 a^m = a^{0+m} = a^m [/math] Provided that a isn't zero, a^m isn't zero, and we can divide by it to obtain: [math] a^0 = \frac{a^m}{a^m} [/math] So here is a very important point. In the case where a isn't zero, it must be the case that a^0=1. I am sure you see this. But in the case where a=0, you get 0/0 on the RHS. But now, here comes the very big problem, which I now see clearly. Here is the exponential function: [math] e^x = \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+... [/math] In the case where x=0 we have: [math] e^0 = \frac{0^0}{0!} [/math] Now, e=2.71828... which is clearly not zero, therefore from an earlier theorem, e^0=1, hence by transitive property of equality it follows that: [math] 1 = \frac{0^0}{0!} [/math] And you have repeatedly said that 0! =1, so that we have: [math] 1 = \frac{0^0}{0!} = \frac{0^0}{1} = 0^0[/math] Which contradicts a previous theorem, in which we concluded that 0^0 is indeterminate. So I've found a contradiction. Now what? Regards PS: And by the way, I was working with P(n,r), because I was wondering if there was a good reason for choosing 0!=1, based upon combinatorics. Also, it was good excercise.

I don't mean anything different from what is standard. Lets see... You are given a set of n-elements, chosen at random. [math] n \in \mathcal{f} 0,1,2,3,4,5,... \mathcal{g} [/math] You don't know what you are going to get for n, because it is being chosen by someone else. All you know for sure, is that n is an element of the set of whole numbers, and n is finite. For now, I am ignoring any and all discussion about infinite sets. So first, the other person must choose a number for n. Suppose that the choice made by them is n=3. e.g. {a,b,c} Here are the permuations of length 3: (a,b,c) (a,c,b) (b,a,c) (b,c,a) (c,a,b) (c,b,a) Now, they are all regarded as different, and the list is comprehensive. So all permutations of length 3 have been listed. Here are the permuations of length 2: (a,b) (a,c) (b,a) (b,c) (c,a) (c,b) Now, each of these is regarded as different from any other, and the list is comprehensive. So all permuations of length two have been listed. Here are the permutations of length 1: (a) (b) © Now, each of these is regarded as different from any other, and the list is comprehensive. So all permutations of length one have been listed. And there are no permutations of length zero. So your question, as I understand it, is for me to clearly define what an r-permutation is. Standard notation, for when order matters, is to use parenthesis. Thus, while it is the case that: {a,b,c}={a,c,b}={b,a,c}={b,c,a}={c,a,b}={c,b,a} It is not the case that (a,b,c)=(a,c,b) It is not the case that (a,b,c) = (b,a,c) and so on. Since usage of parenthesis is standard for when order matters, I use them. As for the issue of replacement, too many words can confuse the other reader. Discussion of replacement vs not, doesn't seem essential to the problem. It is clear just by looking at the generated set, and being told that it is comprehensive, that elements cannot be repeated. But you want a concise mathematical definition. I did individual work on this years ago, and I was using vectors. Arbitrary example given [math] (x_1,x_2,x_3,... x_n) [/math] The usage of this notation seemed easiest at the time. So there is your "n-tuple." The range of the elements depends upon the randomly chosen set. I have done some independent work also, on decision theory. The first decision to be made by the other individual is what n is going to be. In the example under study here, they made the decision that n=3. They then chose three different things, to be elements of the set. In the example here, the three objects were c a b So an arbitrary 3-tuple will be represented by: [math] (x_1,x_2,x_3) [/math] The usage of the parenthesis indicates that order matters. Now, if you back away from the problem, to consider a larger class of problems, then in some cases x1=x2 or x1=x3 or x2=x3 or x1=x2=x3, depending on the specific problem. But in the class of problems being considered here, none of those are possibilitites, and this is embedded in the notion of "r-permutation on an n-set." It is clear from the context what was meant, and overly verbose discourse would have only served to confuse the reader, rather than communicate the information. It is deducible, from what I wrote alone, what an r-permutation on an n-set is. No further explanation was necessary. But you want more. Let A denote a set, chosen at random. Let n denote the number of elements of A. Introduce a variable r, which must satisfy the following constraint: [math] 0 \leq r \leq n [/math] So r can also be chosen at random. Now, in the book I have, the notation P(n,r) is used to denote what they call "number of r-permutations on an n-set." I certainly understood the guy. But i do agree that it's best to be as precise as one is capable of. Well ok. While I will use P(n,r) to denote the number of elements in the set I am interested in, P(n,r) is a number, and not the set in question, which must be generated. The number of elements in that set is P(n,r), but the set itself is something else. It is best to be precise, but in the fewest words possible. ------------------------------------------------------------------------ It is several days later, but what follows belongs here. Here is something I worked on years ago, when I first studied combinatorics. Suppose you are asked to find all 3-permutations of a 5-set. Let the given set be: {a,b,c,d,e} Weight the objects, set them in 1-1 correspondence with: {1,2,3,4,5} You are asked to generate all 3-permutations of {a,b,c,d,e}. To express what you are asked for, it's something like this: [math] \sum_{x_1=1}^{x1=5} \sum_{x_2=1}^{x2=5} \sum_{x_3=1}^{x3=5} (x1,x2,x3) [/math] AND [math] \text{not(x1=x2) and not(x1=x3) and not(x2=x3) } [/math] But when you write +, that must be interpreted as XOR. Thus, the summation symbol doesn't mean the mathematical process of addition, it means the logical process of repetitive XOR. At any rate, the notation above gives sufficient instructions on what set is to be generated. The number of elements in that set is P(n,r).

I strongly disagree. I thought r-permutation was entirely clear, based upon the context. Let's see... are you familiar with either of the following: P(n,r) C(n,r) ? Combinations, permutations

Well it's in at least one book of mine. How would you say it? Regards PS: As for the term 'random' I thought it's meaning was clear from the context. Oh, and actually its an r-permuation on an n-set.

Actually, I am currently working on exactly this. Maybe you can help me. Suppose we have a set given to us by someone else, and they chose it at random. So you recieve the set from them. Since this can randomly vary, let n denote the number of elements in the set you recieve. Now, suppose you are further asked to generate all r-permutations of the set. Suppose that you are given a set with n=2. Constraint: r cannot be greater than n, but r can be less than or equal to n. e.g. {a,b} Case I: r=2 (a,b) (b,a) Case II: r=1 (a) (b) Case III: r=0 How many elements in the generated set, with P(n,r) elements in it? Zero. Yes or no?

you might want to check this out: Wolfram on differentiable functions Regards

There is a Navy destroyer, at sea, 10 nautical miles offshore from the florida keys, parallel to the equator. They are trying to locate the intelsat 803 satellite. Does anyone know how to tell them what their azimuth and elevation needs to be in order to locate it. Does anyone know how to construct a formula for az and el? Thank you

They can all disagree with me, that will change nothing. Kind regards Dr Swanson

I can make one for you. What is the question though? You are not the original guy are you? Give me the "setup," then I can use electrodynamics to work out the B field formula. If the set up is complicated, it will take me awhile to get the right answer for you. But i know how. Regards