Johnny5

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. 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.

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.

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.

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]

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]

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

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.

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.

I was thinking about this last night. The angle between i and itself is 2pi.

Can you prove this?

I have a question about the following article: Hamiltonian Quaternions In the article you will see the following quote: 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?

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

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

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

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.

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

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?