Johnny5

Can I suggest you first prove that n!^{1/n} diverges, since it is quite straight foward. It may even convince you of somethings.

 

Ok one thing at a time. You say it diverges, and that you argued as much in another thread. I'll go have a look at how your proof ran first, then go from there.

 

Here is your argument, from the other thread.

 

That's a bit messy and hard to follow' date=' but it can be proved directly and easily.

 

Let k be any positive integer, then n! > k^{n-k} for all n usfficiently large.

 

Proof: Assume n>K, then the RHS is k multiplied by itself k-n times, but the LHS is

 

n! = (k)!*(k+1)(k+2)....(k+(n-k)) > k!k^{n-k} >k^{n-k} as required

 

Thus n!^{1/n} > k^{(n-k)/n}

 

but the rhs tends to k as n tends to infinity thus n!^{1/n}> k-1 for n sufficiently large. But k was artbitrary thus n!^{1/n} must tend to infinity.

 

Actually that is rather unnecessary since we know that the sum of k^n/n! is exp{k} it follows that k^n/n! tends to zero, and inparticular k^n/n! <1 for all n sufficiently large, but there's no harm in having two ways of proving something.

 

As for your proof you are abusing the equals sign a lot. Yes

 

[math'] \frac{1}{n}\sum_{r=1}^{n}\log r[/math]

 

does diverge, but I don't see how you can use l'hopital, which deals with real valued differentiable functions to conclude that.

 

Well i had no trouble following your argument which is good. You found a sequence which diverges, and whose terms are less than the terms of the given sequence, therefore the given sequence must also diverge.

 

I followed your argument. It's also not something that easily would spring to mind.

 

What if the randomly chosen positive integer k is equal to one? What happens? Here is your proof with k=1 being the randomly chosen positive integer.

 

 

That's a bit messy and hard to follow' date=' but it can be proved directly and easily.

 

Let 1 be any positive integer, then n! > 1^{n-1} for all n usfficiently large.

 

Proof: Assume n>1, then the RHS is 1 multiplied by itself n-1 times, but the LHS is

 

n! = (1)!*(1+1)(1+2)....(1+(n-1)) > 1!1^{n-1} >1^{n-1} as required

 

Thus n!^{1/n} > 1^{(n-1)/n}

 

but the rhs tends to 1 as n tends to infinity thus n!^{1/n}> 1-1 for n sufficiently large. But 1 was artbitrary thus n!^{1/n} must tend to infinity.

 

[/quote']

 

Now, look at the following line of work that you would have, if the randomly chosen positive integer is 1.

 

n! = (1)!*(1+1)(1+2)....(1+(n-1)) > 1!1^{n-1} >1^{n-1} as required

 

1!1^{n-1} >1^{n-1} as required

 

1! = 1

 

[math] \therefore [/math]

 

1^{n-1} >1^{n-1} as required

 

 

Let X = 1^{n-1} therefore, if the randomly chosen positive integer k is 1, then you have this:

 

X > X as required

 

Which is false. Therefore, your k wasn't arbitrary.

 

I'm not sure yet how that affects your argument, but your claim that k can be any arbitrary positive integer is false.

 

That might not matter, I'll have to think about it a little more.

 

Regards

 

PS: I think all you have to do, is stipulate that k is a positive integer greater than 1, and then your whole argument works.

 

Also you take log(1)+log(2)+..log(n)

 

and diff wrt n and stated that the first terms vanish because there is no n' date=' which is false, since I can rewrite the sum as

 

log(n) +log(n-1)+log(n-2)+...log(n-(n-2)) + log(n-(n-1))

 

in which every term has an n in it.

[/quote']

 

This is worth having a look at, I'll do it later.

 

Regards

(why do you "presume" your answers?)

 

 

Because some of what I do is beyond the scope of what is "generally known".

 

When I said 'presume' i didnt really mean im not sure that this is the answer. What i actually mean is that i can control my definitions so well, that i can force this to be the answer if i so desire. Yet, it might conflict with some accepted mathematics someplace else, though what i've no clue.

 

I've found so many errors in 'accepted' things, i just do everything on my own now.

 

For example, there is an error in Cantor's diagonal proof.

Applying it to the original question you're going to get that n!^{1/n} converges, which is a shame since it is easy to prove that it diverges, as I did in that original thread - a proof you ignored.

 

Well i realize that, because lim n goes to infinity of log(1+1/n) = log(1)=0

 

So that by L'hopitals rule, which you question, which is ok.. we will get

 

log(L) = 0

 

from which it will follow that L=1.

 

So yes i see that too.

 

Why do you still say it diverges. which step am i doing, that you have a problem with?

 

Actually, we get this:

 

[math] e^0 = \frac{0^0}{0!} [/math]

 

from the power series expansion.

Having read your edited and very lng initial post it seems all you want is the difference f(n+1)-f(n) which is simply log((n+1)/n) in the case you care about. Look up difference equations. What use it is is something unclear.

 

I'm not sure what I wanted to do next.

 

Perhaps you are right though, let me go back and look.

 

 

This is wanted:

 

[math]

\frac{df}{dn} \equiv \lim_{\Delta n \to 0} \frac{\Delta f(n)}{\Delta n}

[/math]

 

In the case where f(n) = log (n) = ln (n).

 

In the case of a sequence f(n) delta n = 1, hence we want:

 

[math]

\frac{df}{dn} \equiv \lim_{\Delta n \to 0} \Delta f(n)

[/math]

 

In the case where f(n) = ln (n)

 

For an arbitrary sequence f(n) we have:

 

[math]

\Delta f(n) \equiv f(n+1) - f(n)

[/math]

 

Hence

 

[math]

\Delta f(n) = log (n) = log(n+1) - log(n) = log(\frac{n+1}{n})=log(1+\frac{1}{n})

[/math]

 

So that we have:

 

[math]

\frac{df}{dn} \equiv \lim_{\Delta n \to 0} \Delta f(n)

[/math]

 

Hence:

[math]

\frac{df}{dn}= \lim_{\Delta n \to 0} log(\frac{n+1}{n})

[/math]

 

As there is no delta n term, i presume the answer is:

 

[math]

\frac{df}{dn} = log(\frac{n+1}{n}) = log(1+\frac{1}{n})

[/math]

 

Well the answer isn't 1/n, as I had before. But consider the very original question...

 

Someone wanted to evaluate the limit of:

 

[math] (n!)^{\frac{1}{n}} [/math]

 

as n increases without bound.

 

Assuming there is a limit L, we have:

 

[math] L \equiv \lim_{n \to \infty} (n!)^{\frac{1}{n}} [/math]

Only valid for |x| less than 1 or alpha an integer, before you go too far down whatever line of enquiry this is.

 

No its valid for more than that Matt.

 

And the line of inquiry is supposed to be deriving a series expression for log x. However, i do thank you for saving me time if the series expression is only valid for |X|<1 but or alpha an integer. However you are mistaken, because you can use that series to compute square root of two, as i have done that before, which means it is valid for x=1 and alpha=1/2.

Recall Newton's binomial formula:

 

[math] (1+x)^\alpha = \sum_{i=1}^{i=\infty} x^i \prod_{k=1}^{k=i} \frac{\alpha +1 - k}{k} [/math]

As the statement of l'hoptal's rule does not allow for the kinds of functions you are talking about I am perfectly correct to state it is not valid to use it. You can attempt to define a different rule with different hypotheses that generalizes L'hopital, but it will not be the same theorem.

 

Well ok, but, i don't recall ever seeing a proof of L'Hopitals rule in the first place.

L'hopital's rule applies to the ratio f/g when f and g are smooth functions that have a taylor/laurent series in the neighbouhood in which we are intersted in. So of course l'hopital is invalid for this situation by the very hypothesis of when it applies.

 

That is incorrect reasoning Matt, come on you are a mathematician, deductive thought is what you are best at.

 

Here i'll show you...

 

Let F denote the set of all functions to which L'Hopitals rule applies.

 

As you have said, L'Hopitals rule applies to cases where F= f/g, where f and g are smooth. No argument here.

 

But that does not necessarily mean that it cannot apply to sequences. I mean it may not apply, but without some analysis, based upon the very definition of 'derivative', you should not a priori dismiss the possibility. In fact, if you found a proof that it couldn't be used, that proof itself would be important, and have pedagogical value as well.

 

If I am wrong, then construct a proof, before I construct one proving the opposite.

For anyone wanting to answer this you should know Johnny wished to apply l'hopital's rule to somthing for which l'hoptals rule is not valid' date=' in particular this involved differentiating the series

 

log(1)+log(2)+...+log(n)

 

with respect to n

 

where n is naturally a natural number.[/quote']

 

Can you prove that L'Hopitals rule is invalid here? Rigorously?

 

I actually think it would be neat if you could do what i did in the other thread. I mean really i suppose you could just put it into a computer and see what happens, as n tends to infinity, by steps of one, and that solves the whole issue i would think. Use maple, or mathematica.

In another thread, there was a stage where the following derivative was to be taken:

 

[math]

\frac{d}{dn} log (n) = \frac{1}{n}

[/math]

 

The problem was that n was restricted to the natural number system, not the real number system. But, if there is an error in the differential calculus, or with the limit concept in general, then perhaps there was no error.

 

In this thread, I simply want to investigate whether or not you can take the 'derivative' above.

 

Let f(x) denote an arbitrary function of the variable x.

 

The difference operator is defined as follows:

 

[math] \Delta f(x) \equiv f(x+h) - f(x) [/math]

 

h is called the step size.

 

In the case here, x is restricted to the natural number system, and h=1. Thus, we have:

 

[math] \Delta f(n) \equiv f(n+1) - f(n) [/math]

 

 

Now, the derivative of an arbitrary function f(x) is defined as follows:

 

[math] \frac{df}{dx} \equiv \lim_{h \to 0} \frac{\Delta f(x)}{h} [/math]

 

I am going to do an example, for the case of a simple parabola.

 

Consider the graph of f(x) = x². The graph is a parabola.

 

Here is something else I found on parabolas.

 

In the case of a parabola, with its vertex at the origin (0,0) we have this:

 

[math]

f(x) = x^2 \text{ and } \Delta f(x) \equiv f(x+h) - f(x)

[/math]

 

[math] \therefore [/math]

 

[math] f(x+h) = (x+h)^2 [/math]

 

[math] f(x+h)-f(x) = (x+h)^2 - x^2 [/math]

 

[math] \therefore [/math]

 

[math] \Delta f(x) = (x+h)^2 - x^2 [/math]

 

Therefore:

 

[math] \frac{\Delta f(x)}{h} = \frac{(x+h)^2 - x^2}{h} [/math]

 

Therefore:

 

[math] \lim_{h \to 0} \frac{\Delta f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} [/math]

 

Therefore:

 

[math] \frac{df}{dx} = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} [/math]

 

Therefore:

 

[math] \frac{df}{dx} = \lim_{h \to 0} \frac{(x^2+h^2+2xh - x^2)}{h} [/math]

 

[math] \therefore [/math]

 

[math] \frac{df}{dx} = \lim_{h \to 0} \frac{(h^2+2xh)}{h} [/math]

 

[math] \therefore [/math]

 

[math] \frac{df}{dx} = \lim_{h \to 0} \frac{h(h+2x)}{h} [/math]

 

[math] \therefore [/math]

 

[math] \frac{df}{dx} = \lim_{h \to 0} (h+2x) [/math]

 

[math] \therefore [/math]

 

[math] \frac{df}{dx} = 2x [/math]

 

Which is the formula for the slope of the straight line which is tangent to x^2, for any x.

 

So that is an example of the concept of derivative in action, in the case where f(x) is a smooth curve. Division by zero error was avoided, which was clear when h was canceled out of the denominator, because it could be factored out of the numerator.

 

Now, consider the case where f(x) is actually a sequence f(n) where [math] n \in \mathbb{N} [/math].

 

We still have to use the definition of derivative.

 

First lets examine f(n) = n².

 

We have:

 

[math] \Delta (n^2) = f(n+1)-f(n) = (n+1)^2 - n^2 = n^2+1+2n-n^2 = 1+2n [/math]

 

Now, in the definition of derivative, we have h going to zero, where the domain of h is the real numbers. But here, h is a constant, and is equal to 1.

 

But there is an alternative definition of derivative, which is this:

 

[math] \frac{df}{dx} \equiv \lim_{\Delta x \to 0} \frac{\Delta f(x)}{\Delta x} [/math]

 

Consider again, the definition of the difference operator:

 

[math]

\Delta f(x) \equiv f(x+h) - f(x)

[/math]

 

Consider the case where f(x) = x.

 

In this case we have:

 

[math]

\Delta f(x) = \Delta x = (x+h) - x = h

[/math]

 

So as you can see, the definitions are equivalent, that is, it doesn't matter whether we take the limit as h goes to zero, or the limit as delta x goes to zero.

 

So now, let us look at h, in the case of the sequence n. Here is the sequence, starting with n=1.

 

[math] (1,2,3,4,5,6,7,...) [/math]

 

 

Thus, in the case where f(n) = n, we have:

 

[math]

\Delta n = \Delta f(n) = f(n+1) - f(n) = (n+1)-n = 1

[/math]

 

So, applying the definition of the derivative of f(x), to a sequence f(n), we have:

 

[math]

\frac{df(n)}{dn} \equiv \lim_{\Delta n \to 0} \frac{\Delta f(n)}{\Delta n} = \lim_{\Delta n \to 0} \frac{\Delta n}{\Delta n} = 1

[/math]

 

Now, let us look at the case where f(n) = n^2, so we can compare it to the smooth parabola, f(x)=x^2. We have:

 

[math]

\frac{df}{dn} \equiv \lim_{\Delta n \to 0} \frac{\Delta f(n)}{\Delta n}

[/math]

 

Now, we have already seen that delta n =1, hence we have:

 

[math]

\frac{df}{dn} \equiv \lim_{\Delta n \to 0} \Delta f(n)

[/math]

 

And for the sequence n^2, namely (1,4,9,16,25,36,49,64,81,...) we have:

 

[math] \Delta f(n) = f(n+1)-f(n) = (n+1)^2-n^2 = n^2+1+2n-n^2 = 1+2n [/math]

 

So that we have:

 

[math]

\frac{df}{dn} = \lim_{\Delta n \to 0} (1+2n)

[/math]

 

As there is no delta n term, i assume the answer is:

 

[math]

\frac{df}{dn} = 1+2n

[/math]

 

So, i suppose now i can attempt to evaluate the 'derivative' of log n.

"What kind of multiplication allows you to write:

 

ii=-1

 

?

 

Not the ordinary multiplication of two real numbers that we learn in grade school.

 

I was thinking about this last night.

 

The angle between i and itself is 2pi.

Don't forget [math]\[n! = \int_0^{\infty} x^n e^{-x} \' date=dx\]

[/math]

 

 

Can you prove this?

I have a question about the following article:

Hamiltonian Quaternions

 

In the article you will see the following quote:

 

Rotations in 3-space

 

Let us write

 

[math] U = \mathcal{f} q \in \mathbb{H}\text{ such that }||q|| = 1 \mathcal{g} [/math]

 

With multiplication, U is a group. Let us briefly sketch the relation between U and the group SO3 of rotations (about the origin) in 3-space.

An arbitrary element q of U can be expressed

 

[math] cos(\frac{\theta}{2}) + sin(\frac{\theta}{2}) (ai+bj+ck) [/math], for some real numbers q,a,b,c such that a^2+b^2+c^2=1 . The permutation v--> qv of U thus gives rise to a permutation of the real sphere. It turns out that that permutation is a rotation. Its axis is the line through (0,0,0) and (a,b,c), and the angle through which it rotates the sphere is q. If rotations F and G correspond to quaternions q and r respectively, then clearly the permutation v-->qrv corresponds to the composite rotation F o G .

 

My question is this:

 

If you look up you will see w,x,y,z expressed for the product of two quaternions. There are 12 components.

 

does anyone understand how you go from those 12 components, to a rotation of a sphere?

Thanks.

 

That would be normal dispersion' date=' there are 3 types:

 

Normal: when v[sub']g[/sub] < v_p

None: when v_g = v_p

Anomalous: when v_g > v_p

 

It is this third for which I have no examples, and was searching for some, wikipedia appears to be little help...

 

 

Right here is the formula I was going for...

 

[math] D = -\frac{\lambda }{c} \frac{d^2n}{d\lambda^2} [/math]

 

If D is greater than zero, then the medium exhibits anamalous dispersion.

 

Now, as for examples, look for experiments in crystallography.

 

There are probably some superconducting crystals which exhibit it.

 

Let me see if i can find something.

 

Here is something that might interest you, i dont know.

 

Anamalous dispersion of optical phonons in high temp superconductors

Just for physical examples of anomalous dispersion...

 

I think X-ray diffraction for one.

 

Here is wikipedia's article on dispersion.

 

I haven't even read it, but I know what dispersion is.

 

Look at how the prism separates light into frequency bands.

 

A while ago, i studed rainbows, and read all about this.

 

The idea runs like this...

 

the index of refraction n, varies with w, and the speed of light decreases upon entering a material.

 

[math] v_{phase} = \frac{c}{n} [/math]

 

Since the speed of light in matter v, must be less than the speed of light in vacuum c, n has to be greater than 1.

 

Now, there is a way to understand why a rainbow does what it does, and it is the same answer for why a glass prism does it.

Does anyone know what the celestial ephemeris origin is?

 

Global rotation of the non-rotating origin

Just a quick question came up today in some revision had a quick google and found a couple of things but was wondering if anyone here could give me some...

 

Examples of anomalous dispersion' date=' when said in relation to phase and group velocities of wavlettes and wave packets...[/quote']

 

Exactly what are you asking?

Hey guys' date=' how do you do all those symbols (sum and product, etc.)?

-Uncool-[/quote']

 

Click on any particular latex image, and the code to write it should come up.

 

for example, to write indefinite integral of x squared dx you would write:

 

 

\int x^2 dx

 

But you have to embed it in math tags like so:

 

[ math] \int x^2 dx [ /math]

 

(Just dont insert the extra spaces between the word math, and the [ brace.

 

The printout is:

 

[math] \int x^2 dx [/math]

 

Regards

 

PS: You can always read the Latex tutorial too.

Firstly you must sepcify distributivity on both sides; i think there are examples to show this

 

 

Ok thanks.

 

As for your comments about axiomatization, i liked them. What i just want to do right now, is be able to precisely formulate the axioms of a ring, and be able to understand what theorems emerge from what axioms.

 

Again thank you, your patience is appreciated.

 

Regards

Seeing as i^2=-1=j^2=k^2 then treating them as "imaginary units" would seem better than treating them as vector' date=' though why you can't simply call them quaternions is a mystery. We rarely point out something is a vector anyway, since the fact it is in a vector space implies that already.

 

You assumed i, j and k had length 1 when you called them unit vectors. That is what a unit vector is for heaven's sake: a vector of length 1.[/quote']

 

Well here is my next question.

 

What kind of multiplication allows you to write:

 

ii=-1

 

?

 

Not the ordinary multiplication of two real numbers that we learn in grade school.

 

So what?

 

As I previously showed, in this thread, it cannot be cross product, because then we do not get Q1Q2 = -(dot product Q1,Q2) + (cross product Q1,Q2), like many sources suggest you should.

 

In the case where

 

Q1=(0,x1,y1,z1)

Q2=(0,x2,y2,z2)

 

So where was my error?

 

Treating the unknown multiplcation as vector cross product, i did get part of the answer but then

 

ii=0 rather than -1.

 

So??

As a wild guess to save the work done in the previous post, let me write what was obtained for r1:

 

[math]

r_1 = 3 + 2i \sqrt{\frac{4p}{32}}

[/math]

 

Since i know that complex roots come in complex conjugate pairs, lets assume that the root above is correct, and that a second root is given by:

 

[math]

r_2 = 3 - 2i \sqrt{\frac{4p}{32}}

[/math]

 

Also recall that:

 

[math]

p \equiv 4 + 2\sqrt{3}

[/math]

 

and that

 

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Hence:

 

[math] 6 + r_3 = 3 [/math]

 

From which it follows that:

 

[math] r_3 = -3 [/math]

 

Which is false by inspection.

 

The principle equation was:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

And substituting we have:

 

[math]

(-3)^3-3(-3)^2+5(-3)-\frac{1}{3} =0

[/math]

 

Which is false.

 

The basic idea isn't wrong, but that idea involved not making any assumptions at all, during the process of finding the roots. Even one assumption injects uncertainty into the process. I was hoping to finish tonight, but it's not going to happen.

 

Uncool's method is probably what ultimately gets found anyway.

 

You find the one root r1 first, and then its a routine matter to find the other two.

 

I was hoping to generate three equations in three unknowns, and then be able so say, now you can solve using matrices, but that still hasn't happened yet.

I think assuming the two roots are complex, at this stage, is way too early. I think I would have done better if i would have used r1r2r3=1/3, so let me try that.

 

You are given the following cubic equation, at random, and asked to find all three roots, be they strictly real, or complex.

 

[math]

3n^3-9n^2+15n-1 =0

[/math]

 

The first thing you do, is make the leading coefficient 1.

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

Now, assume that you know r1, and that r1 is strictly real. Thus, you can factor the LHS of equation above, like so:

 

[math] (n-r_1)(n^2+Xn+Y) = 0 [/math]

 

Now, the roots of the quadratic are:

 

[math] r_2 = \frac{-X+\sqrt{x^2-4Y}}{2} [/math]

[math] r_3 = \frac{-X-\sqrt{x^2-4Y}}{2} [/math]

 

If they are complex, they are a complex conjugate pair. Otherwise, all three roots are strictly real.

 

Now, distributing we have:

 

[math] (n-r_1)(n^2+Xn+Y) = n^3+Xn^2+Yn-r_1n^2-Xr_1n-Yr_1 = 0 [/math]

 

Writing terms in descending powers of n, we have:

 

[math] n^3 + (X-r_1)n^2 + (Y-Xr_1)n - Yr_1 = 0 [/math]

 

And we had:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

From which we can rapidly see that:

 

[math] X-r_1 = -3 [/math]

[math] Y-Xr_1 = 5 [/math]

[math] - Yr_1 = -\frac{1}{3} [/math]

 

From the first equation we can see that:

 

[math] X= r_1 -3 [/math]

 

Thus, we can write root 2, and root 3 as follows:

 

 

[math] r_2 = \frac{-(r_1 -3 )+\sqrt{(r_1 -3 )^2-4Y}}{2} [/math]

[math] r_3 = \frac{-(r_1 -3 )-\sqrt{(r_1 -3)^2-4Y}}{2} [/math]

 

From the third equation we can see that:

 

[math] Y = \frac{1}{3r_1} [/math]

 

Thus:

 

[math] 2r_2 = -(r_1 -3 )+\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

[math] 2r_3 = -(r_1 -3 )-\sqrt{(r_1 -3)^2-\frac{4}{3r_1}} [/math]

 

Adding them together we find that:

 

[math] 2r_2 +2r_3 = -2(r_1 -3 ) [/math]

 

From which it follows that:

[math] r_2 +r_3 = -(r_1 -3 ) [/math]

[math] r_2 +r_3 = -r_1 +3 [/math]

[math] r_1 + r_2 + r_3 = 3 [/math]

 

Subtracting one from the other we have:

[math] 2r_2 -2r_3= 2\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

 

From which it follows that:

[math] r_2 -r_3= \sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

 

Squaring both sides we have:

 

[math] (r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1} [/math]

 

Lastly, let us multiply them together. Thus:

 

[math] 2r_2 = -(r_1 -3 )+\sqrt{(r_1 -3 )^2-\frac{4}{3r_1}} [/math]

[math] 2r_3 = -(r_1 -3 )-\sqrt{(r_1 -3)^2-\frac{4}{3r_1}} [/math]

 

 

[math] 4r_2r_3 = -\frac{4}{3r_1} [/math]

 

From which it follows that:

 

[math] r_2r_3 = -\frac{1}{3r_1} [/math]

 

From which it follows that:

 

[math] r_1r_2r_3 = -\frac{1}{3} [/math]

 

Now, earlier we found that:

 

[math]

(r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1}

[/math]

 

Multiplying both sides by r_1 leads to:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1 -3 )^2-\frac{4}{3}

[/math]

 

Notice that the RHS is cubic in r1, and the LHS is quadratic in r2,r3.

 

Recall that:

 

[math]

n^3-3n^2+5n-\frac{1}{3} =0

[/math]

 

And the statement above is true, when n is a root, hence:

 

[math]

r_1^3-3r_1^2+5r_1-\frac{1}{3} =0

[/math]

 

From which we rapidly find that:

 

[math]

r_1^3 = 3r_1^2-5r_1+\frac{1}{3}

[/math]

 

And we had:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1 -3 )^2-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]

r_1(r_2 -r_3)^2= r_1(r_1^2-6r_1+9)-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]

r_1(r_2 -r_3)^2= r_1^3-6r_1^2+9r_1-\frac{4}{3}

[/math]

 

Therefore, we have:

 

[math]

r_1(r_2 -r_3)^2= (3r_1^2-5r_1+\frac{1}{3})-6r_1^2+9r_1-\frac{4}{3}

[/math]

 

From which it follows that:

 

[math]r_1(r_2 -r_3)^2 = -3r_1^2 +4r_1 -1[/math]

 

From which it follows that:

 

[math] r_1(r_2 -r_3)^2 +3r_1^2 -4r_1 +1 = 0[/math]

 

Which we can write as:

 

[math] 3r_1^2 + [(r_2 -r_3)^2 -4] r_1 +1 = 0[/math]

 

Define xi such that

 

[math] \xi \equiv [(r_2 -r_3)^2 -4] [/math]

 

So we have:

 

[math] 3r_1^2 + \xi r_1 +1 = 0[/math]

 

The roots of which are:

 

[math] \text{root 1} = \frac{-\xi+\sqrt{\xi^2-12}}{6} [/math]

[math] \text{root 2} = \frac{-\xi-\sqrt{\xi^2-12}}{6} [/math]

 

If we stipulate that they must be equal, we must have:

 

[math] \sqrt{\xi^2-12} = 0 [/math]

 

From which it follows that:

 

[math] \xi^2=12 [/math]

 

From which it follows that:

 

[math] \xi = \pm \sqrt{12} [/math]

 

Therefore:

 

[math] [(r_2 -r_3)^2 -4] = \pm \sqrt{12} [/math]

 

From which it follows that:

 

[math] (r_2 -r_3)^2 = 4 \pm 2\sqrt{3} [/math]

 

Assume that:

 

[math] (r_2 -r_3)^2 = 4 + 2\sqrt{3} [/math]

 

Thus, it follows that:

 

[math] r_2 -r_3 = \sqrt{4 + 2\sqrt{3}} [/math]

 

 

 

Now, earlier we found that:

 

[math]

(r_2 -r_3)^2= (r_1 -3 )^2-\frac{4}{3r_1}

[/math]

 

And we now know that:

 

[math] 4r_2r_3 = -\frac{4}{3r_1} [/math]

 

Hence, it follows that:

 

[math]

(r_2 -r_3)^2 = (r_1 -3 )^2+4r_2r_3

[/math]

 

Thus, it follows that:

 

[math] 4 + 2\sqrt{3} = (r_1 -3 )^2+4r_2r_3 [/math]

 

Now, recall that

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Therefore:

 

[math]

r_1= 3 -r_2 -r_3

[/math]

 

Hence:

 

[math] 4 + 2\sqrt{3} = ((3 -r_2 -r_3) -3 )^2+4r_2r_3 [/math]

 

To clean things up, define

 

[math] p \equiv 4 + 2\sqrt{3} [/math]

So

[math] p = ((3 -r_2 -r_3) -3 )^2+4r_2r_3 [/math]

 

And expanding the squared term on the RHS, we have:

 

[math] p = (3 -r_2 -r_3)^2+9-6(3 -r_2 -r_3) +4r_2r_3 [/math]

 

Which leads to:

 

[math] p = 9 +(r_2+r_3)^2-6(r_2+r_3) + 9 -18 +6r_2 +6r_3 + 4r_2r_3 [/math]

 

Which simplifies to:

 

[math] p = 9 +(r_2+r_3)^2 + 9 -18 + 4r_2r_3 [/math]

 

Which simplifies to:

 

[math] p = (r_2+r_3)^2 + 4r_2r_3 [/math]

 

Which leads to:

 

[math] p = r_2^2+r_3^2+2r_2r_3 + 4r_2r_3 [/math]

 

Thus we have:

 

[math] r_2^2 +(6r_3)r_2 + (r_3^2- p) = 0 [/math]

 

Using QF we have:

 

[math] r_2 = \frac{-6r_3 \pm \sqrt{36r_3^2 -4(r_3^2- p)}}{2} [/math]

 

Stipulating that r_2 be unique, we must have:

 

[math] \sqrt{36r_3^2 -4(r_3^2- p)} = 0 [/math]

 

From which it follows that:

 

[math] 36r_3^2 -4(r_3^2- p) = 0 [/math]

 

From which it follows that:

 

 

[math] 36r_3^2 -4r_3^2+4p = 0 [/math]

 

From which it follows that:

 

[math] 32r_3^2 +4p = 0 [/math]

 

From which it follows that:

 

[math] r_3^2 = -\frac{4p}{32} [/math]

 

From which it follows that:

 

[math] r_3 = \pm i \sqrt{\frac{4p}{32}} [/math]

 

Now, assume that:

 

[math] r_3 = i \sqrt{\frac{4p}{32}} [/math]

 

Recall that

 

[math]

r_2 = \frac{-6r_3 \pm \sqrt{36r_3^2 -4(r_3^2- p)}}{2}

[/math]

 

And the quantity inside the square root was set to zero, so that:

 

[math]

r_2 = \frac{-6r_3}{2}

[/math]

 

Hence:

 

[math]

r_2 = -3i \sqrt{\frac{4p}{32}}

[/math]

 

Now, recall that:

 

[math]

r_1 + r_2 + r_3 = 3

[/math]

 

Hence:

 

[math]

r_1 + (-3i \sqrt{\frac{4p}{32}} )+(i \sqrt{\frac{4p}{32}}) = 3 [/math]

 

From which it follows that:

 

[math] r_1 + -2i \sqrt{\frac{4p}{32}} = 3 [/math]

 

From which it follows that:

 

[math] r_1 = 3 + 2i \sqrt{\frac{4p}{32}} [/math]

 

So that r1 is necessarily complex, contrary to the stipulation that it be real. Thus, at least one of the assumptions that was made, is false.

 

Let me leave this post here, since it went fairly well. We got the square root of a square root, which is similar to a formula posted earlier in the thread.

if you don't know the definition of quaternion norm how were you able to state that i had length one?Ie how did you decide it was a unit vector if you don't konw what the norm is?

 

I treated i,j,k as a unit vectors. No reasoning went into that.

 

 

If you wre writing complex numbers would you hat the i in a+ib? No' date=' of course not.[/quote']

 

No of course not. Are you suggesting that i treat i,j,k as imaginary numbers, instead of unit vectors? Or i should say, imaginary units, that is:

 

[math] i=j=k=\sqrt{-1} [/math]

 

?

You asked for a question, well I really have one.

 

Must both of the following statements be taken as axioms of any ring theory, or can we get away with just one?

 

a*(b+c) = a*b+a*c

(b+c)*a = b*a+c*a

 

PS: I think the answer is we have to adopt both, unless the ring is commutative. Anyone?