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12/26/2012, 09:22 AM
(This post was last modified: 12/26/2012, 09:26 AM by Nasser.)
I have found an approximation for the iterated product:

when x is non-integer value.

this product is important because it is resulted in the first derive of

Now! after some logical sequences I made an approximation as below:

Let

x = u + v , while u is integer and 0 < v < 1
then the approximation for

is

I can varify it is exactly correct for

min v and max v values.
but I can't make sure for

v , if

v for example equal

0.5 because I don't have Measuring tools or software to compare the variances

I need help.

Posts: 1,370

Threads: 335

Joined: Feb 2009

In base c :

Let u be a positive integer.

Let sexp(v)=a , sexp(v')=a' and sexp(u) = b

Let v = 0.99 and v' = -0.01

according to the formula we get : P(x) = P(u+v) = a^b P(u)

Now x = u+v = u + 0.99 = (u+1) - 0.01 = (u+1) + v'

Thus P(x) = P(u+v) = a^b P(u) = P(u+1+v') = a'^(c^b) P(u+1).

This implies => a^b P(u) = a'^(c^b) P(u) (c^b)

Dividing by P(u) on both sides => a^b = a'^(c^b) (c^b) (*)

Although it was stated 0 < v < 1 and v' is negative here.

However if we repeat the above and let v and v' get even closer to 1 and 0 ( with v-v'=1 ) then the above equation (*) becomes in the limit case :

c^b = 1^(c^b) (c^b) which is always true.

So it appears this derivative is not discontinu at integer values.

However this does not necc mean it is diff over the complex or that it is C^1 over the reals. There is no left or right limit proved afterall.

Also note that a second derivative being continu (C^2) would require that the formula for P(x) and its first derivative are consistant with the second ( and first ! see formula derivative of P(x) ! ) derivative of sexp.

Also the integral of P(x) should be C1 sexp(x) +C2 !

Lets investigate further

P(u+0.5+eta)/P(u+0.5)= [sexp(0.5+eta)/sexp(0.5)]^sexp(u)

( we removed times P(u)/P(u) = times 1 )

So for any sexp C^0 around 0.5 : P(u+0.5) is not discontinu.

And hence neither is the derivative.

We conclude that P(x) is valid in the sense that it is nowhere discontinu over the positive reals.

However at the wiki page we note that even uxp(x) ( the linear approximation) is C^1 and there is even a quadratic approximation (C^2).

Thus this method is not yet shown to be 'better'.

Regards

Tommy1729

Posts: 1,370

Threads: 335

Joined: Feb 2009

Btw more about that integral problem :

if x = u + v and u = 0 we get

d sexp(x)/dx = C P(x) = C P(v) = C sexp(v) = C sexp(x)

but the only function that satisfies d f(x) / dx = C f(x) is an exponential one.

Kind of big problem id say. Makes this solution not better than uxp it seems.

regards

tommy1729