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]

If I attempt to do geometry in the hyperbolic sense using the laws of euclidean geometry I will not make any good conclusions' date=' will i?[/quote']

 

 

Nope. But before digressing, what is 'wrong' with this:

 

 

[math]

\prod_{k=a}^{k=b} f(k) = \prod_{k=b}^{k=a} f(k)

[/math]

 

 

You don't catch what I'm saying. Since multiplication has to be commutative, the statement above must be true. There is no reason to adopt any other 'convention', since commutativity of multiplication forces the statement above upon us.

 

That's all I'm saying.

Well you're doing mathematics using mathematical conventions, or you ought to be

'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.

you may not reverse the meaning of the upper and lower terms' date=' it is not like integration.

 

[math']\prod_{k=r}^{k=s}x_s[/math]

 

means the product

[math] x_rx_{r+1}\ldots x_s[/math]

 

this is not defined unless r <=s; you are just misusing notation that is all I am saying

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

[math]\[n! = \int_0^{\infty} x^n e^{-x} \' date=dx\]

[/math]

 

check for n = 1

 

[math]\[\int_0^{\infty} x^1 e^{-x} \,dx\ = 1\]

 

\[\ 1 = 1!\][/math]

 

check for n = (k+1)

 

[math]\[\int_0^{\infty} x^{k+1} e^{-x} \,dx\ = \Gamma(k+2)\]

 

\[\Gamma(k+2) = (k+1)!\][/math]

 

yada yada yada, QED

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.

Yes' date=' but that wasn't what you stated (it can easily be extended to get round all the problems I've raised). But then what you do intend to state is often lost in the size of the post.

 

Your original pair of posts, by the way, contains many misunderstandings, mistatements, and mistakes. Quite what that waffle about 0! was intended to do is something that defies description. And please don't start a debate in this thread about 0!, 0^0 and other things whose standard usage you simply do not accept. They are just formal conventions, you are over analysing.

 

And you've still not proven that the power series converges in finding the square root of 2.[/quote']

 

Well, point out the mistakes, and I'll discuss them in another thread.

 

Regards

If you recall I tried to explain removable singularities to you in an attempt to expllain why 0^0 does not have a universal meaning but can be taken to equal 1 when doing Taylor series.

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

 

Seems good' date=' but the conclusion (hopping dinos) is dead wrong, and based on some flawed assumptions (limited lateral and torsional motion of the femur prevents fast running? Has that guy ever seen an ostrich?).

 

Mokele[/quote']

 

What's wrong with a hopping dino? I think there were some footprint fossils found in a riverbed, and from the stride they inferred the thing could 'hop'.

 

Not sure about it. But Coquina is right, the biggest difference between dinosaurs, and lizards of today, is how the hip socket is configured.

 

Look at the pattern:

 

1. snakes

2. turtles

3. crocodiles

4. alligators

5. others

 

They all seem, umm, like pushed to the ground.

"Permanently Banned" doesn´t sounds like it if you ask me.

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?

no. it is not.

 

the limits at z=1 are equal to 2' date=' but that says nothing about the value of f(z) when z=1.

 

[math'] f(z) = \frac{z^2-1}{z-1} [/math]

 

is not the same function as

 

[math] f(z) = (z+1) [/math]

 

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]

Shame on me but I´ve never heard about the "axioms of real numbers". What are they supposed to be? The definition of a field? If "yes", then your statement isn´t true (see http://mathworld.wolfram.com/FieldAxioms.html[/url'])

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?

I determine what the domain of f(z) is by looking at where it is defined. That is, for which z does f(z) return a real number

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.

One of the main differences between today's lizards and dinosaurs is how the legs are joined to the hip. Dinosaurs walked with their legs perpendicular to the ground. Crocodiles and most lizards walk with their legs almost parallel to the ground. It has to do with how the hip socket is formed.

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

Regards

1 is not in the domain of the original function because the original function does not map 1 onto a real number.

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??

Please be aware that Johnny's method is very limited in scope (it cannot work out the square root of any number bigger than 2' date='

[/quote']

 

[math] \sqrt{3} = (1+2)^{\frac{1}{2}} = [2(\frac{1}{2}+1)]^{\frac{1}{2}} = \sqrt{2} (1+ \frac{1}{2})^{\frac{1}{2}} [/math]

 

Given that you know the square root of two, you can now compute the square root of three by using Newton's formula on:

 

[math] (1+ \frac{1}{2})^{\frac{1}{2}} [/math]

 

Just let X=1/2, and alpha = 1/2. Then multiply by root 2.

 

Regards

When you factor the numerator of that function and cancel the common factor' date=' you aren't "simplifying" the function. You are obtaining a new function that agrees with the original function at all but a single point.

 

z=1 is not in the domain of the original function, and so f(1) does not equal 2.[/quote']

 

Ok, if you don't mind i would like to talk about this more.

 

[math] f(z) \equiv \frac{z^2-1}{z-1} [/math]

 

It is provable using the field axioms that:

 

[math] (z+1)(z-1) = z^2 - 1 [/math]

 

So that the LHS is substitutable for the RHS in any expression, and vice versa.

 

Hence:

 

[math] f(z) \equiv \frac{(z+1)(z-1)}{z-1} [/math]

 

Now, permit z to equal 1, so that we have:

 

[math] f(1) = \frac{(1+1)(1-1)}{1-1} = \frac{0}{0} [/math]

 

Thus, the expression is indeterminate, rather than undefined.

 

But we can cancel (z-1) from the numerator/denominator to obtain:

 

[math] f(z) = (z+1) [/math]

 

Now substituting we have:

 

[math] f(1) = (1+1) = 2 [/math]

 

Now, here is what you said...

 

You said that 1 is not in the domain of the original function.

 

Exactly how do you reach that conclusion. I just want to see the logic.

 

I am inclined to say that the expression was ambiguous, and that cancellation changed that, but that f(1) was defined all the while.

 

But i still want to hear your reasoning.

Ah when you simplify the function, it indeed is 2, but then you worked out the fact that the function clearly isnt defined in 1. Actually, as my analysis prof would say, it is absolutely futile to speak of any function without clearly stating on which domain you work.

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

Are you using L' Hopital's rule here, I assume you are?

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