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Re: New Analytical Functions - Mathematica Verified

Using your fractional part expression and finding the derivative
symbolically, there is more work done than by simply finding the
derivative of the original function. Hence the use of the fractional
part to do this seems to be a retrograde step.
I asked you to compare the time to find the second derivative of the
fractional part by your function and also by evaluating, directly, the
second derivative-NOT using the fractional part. Can you do this?

This is what you should be comparing your approach to- not a numerical
approach which may be  unnecessarily handicapped.
 I recognise the problems with numerical differentiation (and do have
appropriate texts at different levels, from introductory to graduate,
but, again, use of fractional part for this is not going to provide any
advantage -particularly as f(x) and f(x+h) may differ by more than an
integer- in which case use of the  fractional part- as generally
defined, may lead to numerical nonsense.   Hence the term "handicapped"
as compared to using numerical differentiation of the original
function- even doing it by hand, step by step, is faster than your
result using Mathematica- I know you don't think that this is germane
to the issue, but when you give the result from a numerical analysis
for comparison, it becomes quite germane when a crude numerical
differentiation  gives a satisfactory answer in less time than what you
claim for Mathematica.

Now, as to the chopping and trimming. This is done by well established
bit manipulation at a low level.  The process, at least at the binary
level is really very simple compared to the process of evaluating
trignometric functions or even a single  addition (which also involves
bit shifting). Look at the way that a floating point number is stored
on a computer.  Also look at the process for evaluating such things as
arctan(x). Bit shifting and chopping is much simpler.

While it is nice to have a continuous function over an interval, to
give the fractional part of another function, and has the same
derivative,   it is not actually, as far as I can see, doing anything
useful, that can't be done more easily by other approaches.

Don Kelly

Mohamed Al-Dabbagh wrote:
> dhky at wrote:
> > 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]
> >
> If I was a Mathematica developer, I would prefer finding the derivative
> symbolically first then substitute for the required x. This will give
> perfect solution. The reason is that in numerical analysis, finding
> derivative is much harder and error-prone operation than integration.
> Derivative in numerical analysis is a real nightmare, and errors can
> grow dangerously contrary to integration. You may refer to a numerical
> analysis reference and read more about numerical derivation and its
> disadvantages. The higher the derivative, the more dangerous are the
> numerical errors.
> In our specific problem in hand, Fractional Part is NOT done according
> to a certain formula in Mathematica AND in other algebraic systems. It
> is done by a rather complicated chopping and "trimming" procedure. The
> result I derived will provide the mathematical formula for Fractional
> Part such that there will be no need for processor-side load necessary
> to neutralize sign then isolate the integer from fractional part. Using
> my result that: the derivative of Fractional Part of any function is
> the same as derivative of that function, will further simplify (one)
> operation associated with fractional parts, i.e., providing the
> derivatives. So there will be no need for using fractional part costly
> operations when the derivative is needed!
> Mohamed Al-Dabbagh

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