Re: New Analytical Functions - Mathematica Verified

*To*: mathgroup at smc.vnet.net*Subject*: [mg66948] Re: New Analytical Functions - Mathematica Verified*From*: dhky at shaw.ca*Date*: Sun, 4 Jun 2006 02:01:38 -0400 (EDT)*References*: <200605280104.VAA23436@smc.vnet.net> <200606011055.GAA20733@smc.vnet.net> <e5osg1$hvp$1@smc.vnet.net> <e5rf0d$h4r$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

How does Mathematica handle the derivative of a function such as x^5? IF it finds the derivative symbolically then evaluates it. [eg. second derivative of x^5 = 20x^3 then at x=12.2 the result is 36316.96] THEN all that happens using the Fr[x] function is that the derivative of Fr[f[x] is the derivative of f[x]. This is as claimed. The fractional part is expressed as a continuous function on an inverval. Hence for the first derivative of Fr[x^5] =(1/2) -(1/pi)arctan[cot [pi x^5]] , Mathematica's symbolic differentiation reduces this, after two chain rule applications and simplification of an identity to the first derivative of x^5- which is then solved. higher order derivatives follow. (as done by Web Mathematica) However Mathematica also finds the first derivative directly without all the intermediate steps. This is true for other functions such as sin[ln[x]]. It appears that this approach simply adds more steps to what can be done directly with less effort and gains nothing in terms of accuracy in the final evaluation of ,say, 20x^3 at x=12.2 and there is no restriction to non-integer values. If this is the case, then what's the point? Try it both ways for several functions. That is, use Fr to find a derivative, including the final numerical evaluation and also find the derivative directly, including the numerical evalution and compare times and accuracy. This may be a fairer comparison than what you have done. Now it appears that the comparison used with the Mathematica definition of fractional part is forcing Mathematica to use numerical differentiation and, not only that but it is only valid for step sizes such that the integer part of f(x) is constant. Hence a typical numeric differentiation routine is given a penalty as shown by the time of 70 seconds which was given. This is slower than what can be done on a 20 year old hand calculator, with hand entered values (to evaluate f"(x) =[f(x-h)+f(x+h)-2f(x)]/h ) to the desired accuracy (certainly not 400 sig figs) in less time (and on a computer, using an old DOS based APL in less time than the Fr function). If Mathematica's numeric differentiation is that slow, there is a problem- I don't believe that it exists. D. H. Kelly Mohamed Al-Dabbagh wrote: > Daniel Lichtblau wrote: > > > They will change the behavior of its derivatives. Actually I think I did > > not want to use the DiracDelta component in that particular definition > > (or else to use further derivatives thereof in the definition for higher > > derivatives of FractionalPart). > > ........................... > > ........................... > > Did you try it? > > > > Dan, > > Your usage of DiracDelta has made some remarkable correction for the > result, but on a very high cost of runtime! I have made some > experiments and wanted to publish them. You should remember that my > paper: > > http://dabbagh2.fortunecity.com/disc > > has proved that the derivative of the fractional part of any function > is the SAME as derivative of that function WITHOUT involving in the > calculation of fractional part! > > When I used your improvement using Dirac Delta, a very substantial > improvement occurred on the numerical results. HOWEVER, this lead to > some very long delays. To the extent that calculating 5th derivative of > the FractionalPart(x^5) to 1000 places of decimal would take about ONE > HOUR!!!! Here are some results I arranged it in a page prepared for > you: > > http://dabbagh2.fortunecity.com/lichtblau/ > > You will see how my formulas are A LOT faster. > > > Mohamed Al-Dabbagh

**References**:**Re: New Analytical Functions - Mathematica Verified***From:*"Mohamed Al-Dabbagh" <mohamed_al_dabbagh@hotmail.com>