Personal Scratchpad bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/02/2007, 08:42 PM jaydfox Wrote:$ \begin{eqnarray} \lim_{h \to 0^{+}} \frac{\mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z)}{h} & > & 0 \\ \vspace{10}\\ \lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h) & > & \mathcal{E}^{'}(z) \\ \end{eqnarray}$ I am not sure whether your final result is true. However the above reasoning is formally wrong. As $\lim_{h\downarrow 0} f(z+h)=f(z)$ for a continuous function $f$. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 08:54 PM Ah, perhaps I should use another epsilon instead of zero there, with the limit as h goes to epsilon, and taking a limit as epsilon goes to 0 but is still formally NOT zero. It doesn't affect the final result. Seems a bit pedantic as well, but apparently that's part of doing proofs. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 09:18 PM (This post was last modified: 09/02/2007, 09:20 PM by jaydfox.) jaydfox Wrote:And let's extract t. The derivative T' is strictly increasing, so when we extract a positive t, we'll get a positive s. The value will be uninmportant, so long as we know it's positive.Ah, that looks like a mistake. T' is convex, but initially decreasing, so it can't be "strictly increasing". Let me look closer at that and see what effect this will have. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 09:23 PM jaydfox Wrote:jaydfox Wrote:And let's extract t. The derivative T' is strictly increasing, so when we extract a positive t, we'll get a positive s. The value will be uninmportant, so long as we know it's positive.Ah, that looks like a mistake. T' is convex, but initially decreasing, so it can't be "strictly increasing". Let me look closer at that and see what effect this will have. Yes, it does seem that this knocked out the rest of the calculations. Graphically, I can see that the log-convexity requirement is there, so the final result should be the same, once I fix the maths. Back to the drawing board... Oh wait, this is the drawing board... (my scratchpad) ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 09:39 PM jaydfox Wrote:Ah, perhaps I should use another epsilon instead of zero there, with the limit as h goes to epsilon, and taking a limit as epsilon goes to 0 but is still formally NOT zero. It doesn't affect the final result. Seems a bit pedantic as well, but apparently that's part of doing proofs.Hmm, I do see a point to the pedantry. If the double limit, with epsilon formally positive but going to 0, does not converge on the derivative, then the rest of the maths are wrong. However, this happens precisely when the second derivative is not defined. So, um, I guess I should have specified that we want the partial iterates to be infinitely differentiable? If so, then the objection is moot, so far as I can tell. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/02/2007, 09:57 PM jaydfox Wrote:Ah, perhaps I should use another epsilon instead of zero there, with the limit as h goes to epsilon, and taking a limit as epsilon goes to 0 but is still formally NOT zero. It doesn't affect the final result. Seems a bit pedantic as well, but apparently that's part of doing proofs. Pedantic or not, I have to understand what you write and in this case it is not that clear that I would approve it. The proper way of doing the derivation probably may probably be: There is a c>0 such that for each $\epsilon>$ there is a $h>0$ such that $c-\eps < \frac{f(x+h)-f(x)}{h} < c+\eps$ or equivalently $\left|\frac{f(x+h)-f(x)}{h} - c \right| < \eps$. Then you can do the rearrangements without limits. For example $(c-\eps)h+f(x) < f(x+h) < (c+\eps)h+f(x)$. And you can perhaps use the continuity and strict monotony of the log function, namely for each $\eps_2>0$ there is a $\delta>0$ such that $\log(x) < \log(x+\delta) < \log(x) + \eps_2$ where you let $\eps_2=(c+\eps)h$ or so. But in the moment I am to lazy to apply it myself. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 10:14 PM (This post was last modified: 09/02/2007, 10:17 PM by jaydfox.) "There is a ... such that for each ... there is a ...". Epsilon and h are going to zero, much as they would in limits, so why isn't there a quick shorthand as we use with limits? I had assumed I could use the limit notation, since, after all, that's *essentially* what I was doing anyway. I specified I was approaching 0+, and hence indicating that h is, for the purpose of the limit, a positive number that was "approaching" zero, and could get arbitrarily small, and in fact must get arbitrarily small for rest of the maths to hold. I don't quite see why this isn't essentially what a limit is. A limit as h goes to 0 doesn't find the value at 0, it's the value of convergence of values of h that get arbitrarily close to 0. No? Edit: That's not to say that I object to your suggested notation, and in fact I can see some benefits as far as clearing up any possible confusion. I'm just wondering if it's really necessary, or if it's more of a consideration for others? ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 10:24 PM I think I've figured out where my maths went wrong, with respect to my attempt to require the second derivative of the logarithm to be positive. I'll attempt to redo the maths from the start with Henryk's suggested notation, to prevent any confusion and hopefully make it more rigorous. However, this may unfortunately have to wait a day or two, as doing that maths and writing the TeX code takes some time for me, and I'll be busy most of the rest of today, and at the beach for Labor Day tomorrow. Assuming I can fix this up, the next challenge is then to see if I can perform the same "magic" with a derivative of E(x). So far I've tried to show that E(x) is convex. Now I want to see if I can show what is required for E'(x) and/or E''(x) to be convex. (I may skip to the second derivative, depending on what happens with negative iterates, which have alternating signs/curvatures for their derivatives.) ~ Jay Daniel Fox bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/02/2007, 10:27 PM jaydfox Wrote:"There is a ... such that for each ... there is a ...". Epsilon and h are going to zero, much as they would in limits, so why isn't there a quick shorthand as we use with limits? If you can show it by the normal use of limits (for example $\lim_{x\to a} f(x)\lim_{x\to a} g(x)=\lim_{x\to a} f(x)g(x)$), its ok. But you know you can show everything with limits if you are not careful. And such thing like $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} > 0$ is equivalent to $\lim_{h\to 0} f(x+h) > f(x)$ is absolutely forbidden. And the $\eps$,$\delta$ definition is a clear way to specify what you mean. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 09/02/2007, 10:36 PM jaydfox Wrote:I'm just wondering if it's really necessary, or if it's more of a consideration for others? To be clear for yourself is one thing. This is perhaps possible even without writing it down. But if you want it to prove to others a certain rigorousity is needed (especially if you want it to appear in some journal ) and if you even want it to be enjoyable to others then also a somewhat didactic presentation is needed. « Next Oldest | Next Newest »