Designing a Tetration Library andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/19/2008, 07:51 PM bo198214 Wrote:I described it here. The principle $f^{\circ t} \circ f=f\circ f^{\circ t}$ with ${f^{\circ t}}_1={f_1}^t$ is common in the regular iteration literature, however the formula is perhaps too dirty that I have seen it explicitely mentioned. Ah, OK. For some reason, when I apply my interpretation of the formula above, I get $t-1$ instead of t, perhaps, I'm missing something. I'll take a look at the thread you mentioned. bo198214 Wrote:Originally I wanted to use "^" via __and__ for the iteration, like I used "**" via __pow__ for the powers. Do you have an idea why this is not recognized? And also the question about how to put the hyperops package into sage that it is automatically recognized is still open for me. Maybe this is because "^" is __xor__, not __and__. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/21/2008, 06:41 PM Also, your usage of __call__ goes against Sage usage. For example, PowerSeries implements coefficients through __getitem__ or f[n] while you implement coefficients through __call__ or f(n), which should probably use __getitem__ instead for this. Also, in PowerSeries, __call__ is associated with what you use .compose for, so __call__ should use this in PowerSeriesI as well. If it is OK, I will try and change these in the git repository... Andrew Robbins bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 05/21/2008, 06:47 PM (This post was last modified: 05/21/2008, 06:52 PM by bo198214.) andydude Wrote:Also, your usage of __call__ goes against Sage usage. For example, PowerSeries implements coefficients through __getitem__ or f[n] while you implement coefficients through __call__ or f(n), which should probably use __getitem__ instead for this. Also, in PowerSeries, __call__ is associated with what you use .compose for, so __call__ should use this in PowerSeriesI as well. If it is OK, I will try and change these in the git repository... No, I got the same idea and just changed it. See attachment here But compose is not expressed via __call__. You dont write f(g), either you write f(g(x)) or you write fog. Accordingly I also changed the compose to o. So that you can now write f.o(g) bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 05/21/2008, 06:50 PM (This post was last modified: 05/21/2008, 06:50 PM by bo198214.) andydude Wrote:If it is OK, I will try and change these in the git repository... So that means you have your git working?! Ok, then I will push my actual to the repository. To push your changes to the repository you need to give me your public ssh key, otherwise you can only read the public repository. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 05/28/2008, 10:43 AM (This post was last modified: 05/28/2008, 10:45 AM by Gottfried.) For the problem of computing the matrix-exponential the following article may be of interest (and may be added to our - to be constructed - virtual library?). I've not read it yet, though. MATRIX EXPONENTIALS AND INVERSION OF CONFLUENT VANDERMONDE MATRICES UWE LUTHER† AND KARLA ROST‡ Abstract. For a given matrix A we compute the matrix exponential e^A under the assumption that the eigenvalues of A are known, but without determining the eigenvectors. The presented approach exploits the connection between matrix exponentials and confluent Vandermonde matrices V . This approach and the resulting methods are very simple and can be regarded as an alternative to the Jordan canonical form methods. The discussed inversion algorithms for V as well as the matrix representation of V^-1 are of independent interest also in many other applications. Key words. matrix exponential, Vandermonde matrix, fast algorithm, inverse. AMS subject classifications. 34A30, 65F05, 15A09, 15A23. Electronic Transactions on Numerical Analysis. Volume 18, pp. 91-100, 2004. Copyright Ó 2004, Kent State University. ISSN 1068-9613. ETNA Kent State University etna@mcs.kent.edu Preprint (2003) online: Luther/Rost Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 05/29/2008, 04:06 PM Gottfried Wrote:For the problem of computing the matrix-exponential Hm, thanks for the link however do we need the matrix-exponential? We just need $A^t$, which perhaps can be also expressed as $\exp(t\log(A))$ but in this case we also need a fast algorithm for matrix log, or are there other direct uses of the matrix exponential? Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 05/29/2008, 04:50 PM bo198214 Wrote:Gottfried Wrote:For the problem of computing the matrix-exponential Hm, thanks for the link however do we need the matrix-exponential? We just need $A^t$, which perhaps can be also expressed as $\exp(t\log(A))$ but in this case we also need a fast algorithm for matrix log, or are there other direct uses of the matrix exponential? Well, Andrew asked for such a routine, if I recall right Gottfried Helms, Kassel bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 06/06/2008, 03:27 PM (This post was last modified: 06/07/2008, 08:52 AM by bo198214.) Gottfried Wrote:bo198214 Wrote:Hm, thanks for the link however do we need the matrix-exponential?Well, Andrew asked for such a routine, if I recall right I now see that the matrix logarithm could be the iterative logarithm, up to multiplicative constant perhaps. If we define $(\text{ilog}(f))_n=\log(CM(f))_{1,n}$ then exactly $\text{ilog}(f^{\circ t})=\log(CM(f)^t)_1=t\text{ilog}(f)$ is satisfied. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 06/06/2008, 06:56 PM bo198214 Wrote:If we define $(\text{ilog}(f))_n=\log(CM(f))_{1,n}$ then exactly $\text{ilog}(f^{\circ t})=\log(CM(f)^t)=t\text{ilog}(f)$ is satisfied. Do you mean $\text{ilog}(f^{\circ t})=(\log(CM(f)^t))_1=t\text{ilog}(f)$? Then yes, that is very nice! Andrew Robbins bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 06/07/2008, 08:51 AM andydude Wrote:Do you mean $\text{ilog}(f^{\circ t})=(\log(CM(f)^t))_1=t\text{ilog}(f)$? Then yes, that is very nice! Yes, I add the subscripted 1 in my post. « Next Oldest | Next Newest »