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 Continuum sum - a new hope kobi_78 Junior Fellow Posts: 11 Threads: 3 Joined: Dec 2009 04/24/2010, 09:01 PM (This post was last modified: 04/24/2010, 09:13 PM by kobi_78.) Hi, I wanted to share with you some things I found in the last few months, that might interest you. These are a few ideas that involve Continuum sum, fixed points, Taylor series and etc. I discovered Ansus's formula independently in the past. The disadvantages of the formula are that we must know the value of the derivative in 0, but we don't. Instead we can try a similar approach: Define $f_0(x) = x$ $f_{n}(x) = a^{f_{n - 1}(x)}$ Then $f_{n}(x) = a^{f_{n - 1}(x)}$ $f'_{n}(x) = a^{f_{n - 1}(x)} \cdot \ln{a} \cdot f'_{n - 1}(x) = f_{n}(x) \cdot \ln{a} \cdot f'_{n - 1}(x)$ $\frac{f'_{n}(x)}{f'_{n - 1}(x)} = f_{n}(x) \cdot \ln{a}$ $\frac{f'_{n}(x)}{f'_{0}(x)} = \ln{a}^n \cdot \prod_{k=1}^{n}{f_{k}(x)}$ $f'_{n}(x) = \left( \ln{a} \right)^n \cdot \prod_{k=1}^{n}{f_{k}(x)}$ Which is similar to Ansus's formula, but has its advantages: If we use the natural definition for continuum product (that can work if the limit exists, i.e. if ${\frac{1}{e}}^{\frac{1}{e}} \le a \le e^{\frac{1}{e}}$): Let $y = f_{\alpha}(x)$ Then: $y' = f'_{\alpha}(x) = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{f_{k}(x)} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \prod_{k=1}^{\infty}{\frac{f_{k}(x)}{f_{k+\alpha}(x)}}$ where $b = \lim_{m \to \infty} {f_m(x)}$ is the fixed point of $a^x = x$ We can rewrite it: $y' = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \prod_{k=1}^{\infty}{\frac{f_{k}(x)}{f_{k+\alpha}(x)}} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \prod_{k=1}^{\infty}{\frac{f_{k}(x)}{f_{k}(f_{\alpha}(x))}} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \prod_{k=1}^{\infty}{\frac{f_{k}(x)}{f_{k}(y)}}$ This is a differential equation that we can solve, but we "don't" know the initial conditions. We can assume that y(b) = b, and solve the equation. I tried to solve the truncated equation for some bases with wolfram alpha and got the following result. It seems, that the solution for that equation is the regular iteration Tetration aka "natural" Tetration, so it seems it is really natural. We get problems at $a = e^{\frac{1}{e}}$, there we get $f_{\alpha}(x) = x$. This is the first idea I wanted to share. The second one is the PDE of $f_{\alpha}(x)$ which you can see here. I will explain how I discovered it later. (It reminds a bit the first idea) The third idea, is an idea that is inspired by Ansus's last post about the tetration's derivative: Let $g_{\alpha}(x) = f'_{\alpha}(x)$. Then $\ln\left({g_{n}(x)}\right) = \ln\left({g_{n - 1}(x)}\right) + \ln{\ln{a}} + \ln\left({f_{n}(x)}\right)$ $\frac{g'_{n}(x)}{g_{n}(x)} = \frac{g'_{n - 1}(x)}{g_{n - 1}(x)} + \frac{f'_{n}(x)}{f_{n}(x)}$ $\frac{f'_{n}(x)}{f_{n}(x)} = f'_{n - 1}(x) \cdot \ln{a}$ $\frac{g'_{n}(x)}{g_{n}(x)} - \frac{g'_{n - 1}(x)}{g_{n - 1}(x)} = f'_{n - 1}(x) \cdot \ln{a} = g_{n - 1}(x) \cdot \ln{a}$ Therefore, $\frac{g'_{n}(x)}{g_{n}(x)} = \ln{a} \cdot \sum_{k = 1}^{n} { g_{k - 1}(x) }$ It may seem useless, but now we try to develop a Taylor series around a fixed point b of $a^x$ and we get: $f_{\alpha}(b) = b$ $f'_{\alpha}(b) = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{f_{k}(b)} = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{b} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha}$ $\frac{g'_{n}(x)}{g_{n}(x)} = \ln{a} \cdot \sum_{k = 1}^{n} { g_{k - 1}(x) }$ $f''_{\alpha}(b) = g'_{\alpha}(b) = g_{\alpha}(b) \cdot \ln{a} \cdot \sum_{k = 1}^{\alpha} { g_{k - 1}(b) } = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \ln{a} \cdot \sum_{k = 1}^{\alpha} { \left( \ln{a} \right)^{k - 1} \cdot b^{k - 1} } = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \ln{a} \cdot \frac{\left( \ln{a} \right)^{\alpha} \cdot b^{\alpha} - 1}{ \ln{a} \cdot b - 1}$ I think we can continue in a similar fashion and find values in higher derivatives order, but I am not sure. I hope you enjoy these ideas and maybe develop them to bigger things... « Next Oldest | Next Newest »

 Messages In This Thread Continuum sum - a new hope - by kobi_78 - 04/24/2010, 09:01 PM RE: Continuum sum - a new hope - by kobi_78 - 04/25/2010, 05:48 AM RE: Continuum sum - a new hope - by bo198214 - 04/25/2010, 10:41 AM RE: Continuum sum - a new hope - by kobi_78 - 05/03/2010, 09:27 PM RE: Continuum sum - a new hope - by bo198214 - 05/09/2010, 11:35 AM RE: Continuum sum - a new hope - by sheldonison - 06/11/2010, 12:34 AM RE: Continuum sum - a new hope - by bo198214 - 06/12/2010, 04:42 AM RE: Continuum sum - a new hope - by mike3 - 06/12/2010, 11:10 AM RE: Continuum sum - a new hope - by bo198214 - 06/13/2010, 06:49 AM RE: Continuum sum - a new hope - by sheldonison - 06/13/2010, 11:23 PM

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