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 Infinite products andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 04/06/2009, 09:20 AM I just remembered a method I tried like 5 years ago, but haven't visited in a while. Now that we have better power series expansions of tetration, for example, at $x=(-1)$ we have $ \text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k$ and using the definition, we can rewrite this as $ \begin{array}{rl} \text{tet}(x) & = \exp\left(\text{tet}(x-1)\right) \\ & = \exp\left(\sum_{k=0}^{\infty}c_kx^k\right) \\ & = \prod_{k=0}^{\infty} \exp(c_k{x^k}) \\ & = \prod_{k=0}^{\infty} a_k^{x^k} \\ \end{array}$ where $a_k = e^{c_k}$ I wonder if this simplifies the coefficients or just makes things more complicated? Andrew Robbins tommy1729 Ultimate Fellow Posts: 1,359 Threads: 331 Joined: Feb 2009 04/06/2009, 09:38 PM andydude Wrote:I just remembered a method I tried like 5 years ago, but haven't visited in a while. Now that we have better power series expansions of tetration, for example, at $x=(-1)$ we have $ \text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k$ and using the definition, we can rewrite this as $ \begin{array}{rl} \text{tet}(x) & = \exp\left(\text{tet}(x-1)\right) \\ & = \exp\left(\sum_{k=0}^{\infty}c_kx^k\right) \\ & = \prod_{k=0}^{\infty} \exp(c_k{x^k}) \\ & = \prod_{k=0}^{\infty} a_k^{x^k} \\ \end{array}$ where $a_k = e^{c_k}$ I wonder if this simplifies the coefficients or just makes things more complicated? Andrew Robbins that could have been my own post i wondered about it too. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/06/2009, 09:55 PM andydude Wrote:$ \begin{array}{rl} \text{tet}(x) = \dots & = \prod_{k=0}^{\infty} a_k^{x^k} \\ \end{array}$ where $a_k = e^{c_k}$ I wonder if this simplifies the coefficients or just makes things more complicated? To calculate the coefficients you either need on both sides a product or on both sides a sum. So I dont see how the product representation can be useful. « Next Oldest | Next Newest »

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