Johnny5

Ok, I'll admit you make some good points, but you seem to have glossed over the possible actual fact, which is that there was at least one species of Dinosaur who "hopped." Let me see if i can find you a site, with the footprint thing i mentioned. And 'biggest' was probably not required, it happens to interest me very much though, this particular difference. This is such an exciting topic for me, I've always loved dinosaurs, allosaurus, stegosaurus, etc etc Let me see if i can find a link. This is a break from me thinking about physics/math all the time. Well, not surprisingly i couldn't find the name of the riverbed, but i saw a documentary on it years ago, i think it was someplace in the center US. Nonetheless, i think my main point has been missed, which is why do modern lizards appear "sqashed" to the ground, when the dinosaurs could 'hop' or certainly run on two legs, probably quite fast. Imagine a hungry giant fanged creature coming at you. You would hop one mile if you could. Regards

It's ok with me, in fact the structure of your proof is good, its just that a lemma would be nice, which goes into the gamma function. in other words prove the following statement Lemma: [math] \Gamma (n+1) = n! [/math]

'ought' 'should' deontic logic... well maybe i should. Who knows.

No, i would be satisfied to see whether or not you ever use integration by parts formula, during any phase of your proof. I have used the gamma function, and proved the thing you used in the theorem above. It's the integration by parts 'thing' you have not supplied. If you don't want to provide the lemma, don't, its ok.

I didn't say it had to be an ordinal. And whose convention? Not mine. Let me google something, see what i find, not that it will change my stance. Well I didn't find anything, but i see absolutely no reason at all to choose some strange minded convention, when i will always be able to use commutativity of multiplication to prove my statement. I am at a loss for what you are saying I guess.

Exactly what are you saying? In the end after you expand a product you have a sequence of terms being multiplied together ABCDEFG And since multiplication is commutative ABCDEFG=GFEDCBA Why in the world would you want the notation to force you to multiply in some specific order, and not allow an equation to be formed such as... [math] \prod_{k=a}^{k=b} f(k) = \prod_{k=b}^{k=a} f(k) [/math] Which statement is proven using nothing more than commutativity of multiplication. Or to rephrase my question another way, prove that the statement above leads to a contradiction.

There are so many things here, I cannot possibly address them all at once so I will do so one at a time. You mention this: [math] \prod_{k=0}^{k=1} \frac{\frac{3}{2}-k}{k} [/math] If you pull k out of the product symbol you have: [math] \frac{1}{n!} \prod_{k=0}^{k=1} \frac{3}{2}-k [/math] Which is fine at k=0. Now, when n=0, n! =1, because 0! is defined to be equal to 1, so consider the following... [math] \prod_{k=0}^{k=n} \frac{\frac{3}{2}-k}{k} [/math] The above is defined when k=n=0, because 0! is defined to be 1. As for product from k=0 to 1, of (3/2-k)/k, i know for a fact that divsion by zero isn't allowed, so that the above isn't defined, and i made a point of pointing out this problem in my post, and showed that it suggests that there is a problem with defining 0!=1. I derived a contradiction somewhere. Actually now i see what you are talking about, where you say i corrected a self introduced mistake. That's not what I did. I am well aware of how to use the symbols i was manipulating. The first 'mistake' was intentional, to reveal a problem with 0!=1, which i know about. The 'fix' as you put it, was merely me writing the actual formula which avoids confusion, by writing the first term of the series explicitely, and the rest of it as a series, without any worry about division by zero issues. I was well aware of what I was doing. Some place you say, "why, tell us?" in reference to me using the fact that multiplication is commutative. Are you asking me to explain to you why multiplication is commutative? I dont understand your question. Next you say that you cannot index the product from 1 to zero, in general the product from k=1 to k=0 is equivalent to the product from k=0 to k=1, since multiplication is commutative. So unless there was some constraint somewhere I missed, they are equivalent. I would need to see the line of my work which is bothering you. I have no idea what the empty product is, so you are right about that. And i dont think that 0! is the product from 0 to 1 or whatever you are saying. In fact, i think there is a problem with 0!=1 and 0^0 = 1, as i recently explained to you. The rest of your post seems to be you misreading something which i subtly glossed over, which is an issue about 0!=1 and 0^0=1, which i no longer wish to belabor. Once upon a time I asked you to prove that 0!=1 and 0^0=1, which proof you never supplied. I am no longer waiting. Regards

But now you have to have proofs regarding Gamma function as a lemma to this as a theorem. Can you prove that one? I guess the reason I am asking, is because there is an integration by parts in the proof i am thinking of, thats all.

I remember that, perhaps we can discuss it again. Let me think about your answer to this first.

Any idea why, or what happened? I did a google search on the name 'geistkiesel' and this person has their own website devoted to overthrowing relativity. He is also a member of several other message boards, and generally presents the same argument there too. There is even an email address, but it's not that important to me. It's just that they told me to read 34.7 in Feynman's lectures, they said it would be very informative, but i never got to find out why. I went to the trouble to get the book, but now i am left hanging as to what was so important about chapter 34, article 7.

Has anyone seen Geistkiesel lately, have they been unbanned yet? I finally got Feynman's lectures on physics, and was reading chapter 34, article 7, it was on w, k, and four vectors, and i wanted to discuss it with them. Will they ever be unbanned?

I realize that the limits as z approaches 1 from the right, or the left are each two. And if f(1) =2, then the function is continuous there. Additionally I am not trying to re-define anything, but i'm giving the whole issue some thought, that's all. If you stop and think for a moment, the logic presents itself to you, and here it is... The following is a field axiom: Let x denote an arbitrary real number. [math] not(x=0) \Rightarrow \exists x^{-1} x \cdot x^{-1} = 1 [/math] Now, consider the following theorem. [math] \text{Theorem: For any real number x: } 0 \cdot x = 0 [/math] Let a,b denote arbitrary real numbers. One of the field axioms provides us with: 0+b = b Now use the following statement: for any real numbers x,y,a: if x=y then ax=ay to obtain a(0+b) = ab The distributive axiom provides us with: a0+ab = ab Now, every real number has an additive inverse, ab has an additive inverse which we will represent by -ab, and the additive inverse of ab is such that ab+(-ab)= 0. Now use use the statement that for any real numbers x,y,b: if x=y then x+b=y+b to obtain (a0+ab) + (-ab) = ab + (-ab) Now use associativity of addition to obtain a0+(ab + (-ab)) = ab + (-ab) From which it follows that: a0+ 0 = 0 From which it follows that a0 = 0 And by commutativity of multiplication we finally have: 0a = 0 Thus, zero times a is equal to zero. But the symbol a represented an arbitrary real number throughout the proof, hence the theorem is true for any real number a. QED Suppose that we remove the restriction that division by zero is not allowed, and adopt the following field axiom: Let x denote an arbitrary real number. [math] \exists x^{-1} x \cdot x^{-1} = 1 [/math] Since zero is a real number, we can use universal instantiation to obtain the following true statement: [math] 0 \cdot 0^{-1} = 1 [/math] But by the previous theorem we know that [math] 0 \cdot 0^{-1} = 0 [/math] as a consequence of the field axioms used in that proof. Now, using the transitive property of equality it follows that: [math] 0=1 [/math] Which is false, since one of the field axioms is [math] not(0=1) [/math] We considered taking the following statement as an axiom: [math] \forall x \in \mathbb{R} \exists x^{-1} \in \mathbb{R}: x \cdot x^{-1} = 1 [/math] Since 0 is an element of the real number system, we can use universal instantiation to obtain the following true statement: [math] \exists 0^{-1} \in \mathbb{R}: 0 \cdot 0^{-1} = 1 [/math] Hence: [math] 0 \cdot 0^{-1} = 1 [/math] Whence we can now erroneously conclude that 0=1. So the real field axiom is this: [math] \forall x \in \mathbb{R} [not(x=0) \Rightarrow \exists x^{-1} \in \mathbb{R}: x \cdot x^{-1} = 1 ][/math] Now consider the original function again: [math] f(z) = \frac{z^2-1}{z-1} [/math] We can write it as: [math] f(z) = (z^2-1)(z-1)^{-1} [/math] We can now factor the z^2-1 term and write: [math] f(z) = (z+1)(z-1)(z-1)^{-1} [/math] Now comes the question, what is f(1)? Putting 1 in the expression above we have: [math] f(1) = (1+1)(1-1)(1-1)^{-1} [/math] From which it follows that: [math] f(1) = (2)(0)(0)^{-1} [/math] And the parenthesis are superfluous, so that we can write: [math] f(1) = 200^{-1} [/math] From which it follows that: [math] f(1) = 00^{-1} [/math] Which we know leads to the conclusion that 0=1, which is false. I would say that Tom is right. The original function cannot be defined for z=1. So you are right zaph, they cannot be the same function, since f(z) = z+1 is defined for all z.

Well, and this is for either of you... In the case I gave, we had an indeterminate form [math] f(z) = \frac{0}{0} [/math] Or to put things differently [math] f(z) = \frac{0}{0} = 00^{-1} [/math] And you object, and now i at least see why. However, 0/0 is indeterminate, which means that we aren't necessarily dealing with division by zero error, upon performing some operations, we can determine that f(z) is equal to some number. So there seems to be some confusion about this. And then there is my proof that if f(z) = (z+1) then f(z) = (z^2-1)/(z-1) Mmm i am starting to think the whole issue boils down to this now is (z-a)(z-a)^-1 = 1 regardless of a? I think it boils down to that one question. And i know your response now. No, because if z=a, then (z-a)^-1 is undefined. Something is still bugging me though.

What do you both say to this... For any real numbers A,B, if not (B=0) then [math] \frac{A}{B} \equiv AB^{-1} [/math] ? Here is one of the field axioms: [math] \forall x \in \mathbb{R} \exists (x^{-1}) \in \mathbb{R} \text{such that } [ \neg(x=0) \Rightarrow xx^{-1}=1] [/math]

I meant the field axioms. The list at wolfram is not only superfluous, it is incomplete. But more importantly, what statement of mine are you referring to, which is false?

Well i see that you say that f(z) is not equivalent to z+1, so that i suppose is where to focus. We have transitive property of equality, which doesnt break down. So we have something like this f(z) = X/Y X= AB f(z) = AB/Y Y=B f(z) =AB/B And by axioms of real numbers AB/B=A, hence using transitive property of equality one more time f(z)=A. But i see you very clearly say no, so where is the error in what i've done? To look at things another way, suppose you start off with: f(z) = z+1 Then multiply both sides by z-1 to obtain: (z-1) f(z) = (z-1)(z+1) As long as not (z-1)=0 we can divide to obtain: [math] f(z) = \frac{(z-1)(z+1)}{(z-1)} [/math] and then one more step gets us to... [math] f(z) = \frac{(z^2-1)}{(z-1)} [/math] So whatever you are saying is either wrong, or involves something subtle.

Do you know why this is? why the hip sockets have changed in the manner they have? Regards

Ok follow up question. How are you going about determining what the domain of the original function is exactly? And also, is or isnt the original function equivalent to z+1??

What do you mean the function isn't defined in 1??

Nope. Just asking whether or not f(1)=2.