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+--- Thread: Searching for an asymptotic to exp[0.5] (/showthread.php?tid=863)
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/14/2014
(09/13/2014, 11:49 PM)jaydfox Wrote: \(
\exp(x) \approx \sum_{k=-\infty}^{\infty}\frac{x^{k+\beta}}{\Gamma(k+\beta+1)}
\)
Notice that I put "approximately equal". I haven't checked, but I assume it's exactly equal, but only in the sense that it should satisfy the functional equation exp'(x) = exp(x).
Now, set beta = 1/2 and truncate the negative powers of x:
\(
\exp(x) \approx \sum_{k=0}^{\infty}\frac{x^{k+1/2}}{\Gamma(k+3/2)}
\)
Hey I wanted to post that !
Btw the equation is f ' (x) = f(x) + o(1) for x sufficiently large.
I had the exact same proof ...
I knew I was running out of time since it looked so familiar !
Did's answer on MSE is the classical way to prove it.
There are many proofs of this actually.
But Jay posted the shortest I think.
Gauss and Euler must have known this already.
...
However I still notice nobody has given a justification for sheldon's integral.
This is the second time it gives a correct solution.
It would also give the same solution for exp(x)ln(x+1) + exp(x)sqrt(x).
In general using sheldon's integral we can say
If +fake( f ) and +fake ( g ) exist then
+fake( f + g ) = +fake( f ) + +fake( g )
which makes sense.
One could generalize
If +fake( ln f ) and +fake ( ln g ) exist then
+fake( f * g ) = exp ( +fake( ln f ) + +fake( ln g ) )
Although the RHS is not optimal then and not equal to the integral.
I wonder what post 9 means to sums and products of f and g.
Still thinking.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/14/2014
WARNING : CORRECTED IN POST 117 !
---
Using post 9 for exp(x) sqrt(x) :
ln(a_n x^n) < ln(exp(x)sqrt(x))
...
ln(a_n) = min ( exp(x) - (n-1/2) x )
--
d/dx [exp(x) - (n-1/2) x] = exp(x) - (n - 1/2)
=>
exp(x) = n - 1/2
x = ln(n - 1/2)
--
ln(a_n) = exp(ln(n - 1/2)) - (n - 1/2) ln(n - 1/2)
=>
a_n = exp( n - 1/2 - (n - 1/2) ln(n - 1/2) )
a_n = exp( (n - 1/2) (1 - ln(n - 1/2)) )
...
Gamma(n + 1/2) vs exp( - (n - 1/2) (1 - ln(n - 1/2)) )
=>
Loggamma(n + 1/2) vs - (n - 1/2) (1 - ln(n - 1/2))
=>
Loggamma(z) vs - z ( 1 - ln(z) )
Loggamma(z) vs z ( ln(z) - 1 )
=>
Lim loggamma(z) / ( z ( ln(z) - 1 ) ) < Constant ?
From the Stirling series we know that the limit equals 1.
This implies that using post 9 also gives the correct solution up to a multiplicative constant !
So the post 9 method does a good job to estimate the fake exp(x)sqrt(x).
regards
tommy1729
---
WARNING : CORRECTED IN POST 117 !
RE: Searching for an asymptotic to exp[0.5] - sheldonison - 09/15/2014
(09/13/2014, 11:49 PM)jaydfox Wrote: ...
Treating Gamma(k) at negative non-positive integers as infinity, and the reciprocal of such as zero, we can take the limit from negative to positive infinity. And we can replace k with (k+b), where b is zero in the original solution, but can now be treated as any real...
\(
\exp(x) \approx \sum_{k=-\infty}^{\infty}\frac{x^{k+\beta}}{\Gamma(k+\beta+1)}
\)
Notice that I put "approximately equal". I haven't checked, but I assume it's exactly equal, but only in the sense that it should satisfy the functional equation exp'(x) = exp(x).
So we also have a fakeexp(z) function, and I especially like the simplicity of f'(x)=f(x) in the non-converging limit to explain why f(x)~exp(x). Here, going back to k=0.5, for simplicity.... We have to put some bounds on k, since the infinite Laurent series does not converge anywhere.
\(
\exp(x) \sim f(x) = \sum_{k=0}^{\infty}\frac{x^{k+0.5}}{\Gamma(k+0.5+1)} \)
\( \frac{d}{dx}f(x)=\sum_{k=-1}^{\infty} \frac{x^{k+0.5}}{\Gamma(k+1.5)} =f(x) + \frac{x^{-0.5}}{\Gamma(0.5)} \)
As \( \Re(x) \) gets arbitrarily large, the error term becomes more and more insignificant, relative to the exp(z) term, and the number of terms you can include also increases, until the \( \frac{x^{-n+0.5}}{\Gamma(-n+1.5)} \) starts growing in magnitude as n gets bigger negative ....
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/16/2014
Gösta Mittag-Leffler.
That is this post in a nutshell.
When I said it looked familiar it was the Mittag-Leffler function.
And that is also used for Mittag-Leffler summation.
Mittag-Leffler theorem is also relevant.
Mittag-Leffler has occured here on this forum before.
For instance when trying to get a series expansion larger than a radius , with for instance rational functions.
This resembles my next idea ...
( which will be explained in next posts )
Actually my next idea came first , and that is how I suddenly remembered Mittag-Leffler.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/16/2014
Ok my next idea.
Im having ideas to prove the validity of the Cauchy integral for fake function theory.
Some conditions first :
Lets call our real-analytic function f(z) of which we want a +fake.
1) there needs to be an annulus around the origin that contains at most one branch.
2) the Riemann surface needs to be "well-connected".
As example : log(z^3) log(z^5) is not well connected.
Plot it near the origin to see it.
3) f(z) has no essential singularity.
4) f(x) , f ' (x) , f " (x) > 0 for x > 0
Now we use an old idea of me
f(z) = +Taylor_1(z) + +Taylor_2(z/(z+a_1)) + +Taylor_3(z/(z+a_2))
where +Taylor means a Taylor series with positive real coefficients.
a_1,a_2 are selected positive reals.
There series expansions MUST have a +fake described by the Cauchy.
to be continued.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/18/2014
Sheldon , when discussing post 9 , you started using second derivatives as well.
what is the motivation , reasoning and justification of that ?
Now I think
if x>0 => f(x) , f ' (x) > 0
then a_n x^n < f(x) - f(0)
is a good equation.
if we additionally have x > 0 => f " (x) > 0
then
a_n x^n < f(x) - f(0)
n a_n x^(n-1) < f ' (x) - f ' (0)
seems a good system of equations.
Analogue if we also have f "' (x) > 0.
etc etc.
We can find extrema of f ' (x) - f ' (0) by considering f " (x).
( the analogue of the classical consideration of f ' (x) to find the min as done in post 9 )
However that does not seem what you had in mind , or was it ?
I feel a bit silly asking this question.
Its probably trivial.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/18/2014
Forgive me for not using tex again.
On my to do list are considering different solution to
fake(x^x)
fake(x^ln(x))
fake(exp(x) ln(x)^2)
fake(x^sqrt(x))
But for now I was mainly intrested in :
fake(Gamma(x+2))
In particular we can use for instance the the fake log or fake sqrt results here !
fake Gamma(x+2) = integral_0^oo fake( exp(t) t^(x+1) ) exp(-2t) dt
where fake ( exp(t) t^(x+1) ) = Mittag(t,1,x+1) as obtained before.
( fake exp(x)sqrt(x) = Mittag(x,1,1/2) as example )
Notice the almost self-similarity , Mittag depends on the gamma function !
( A tempting idea is to replace the gamma in the Mittag function with fake gamma itself ?! )
Remember that there is analytic continuation for the integral !
Also if we use the Cauchy integral on
integral_0^oo fake( exp(t) t^(x+1) ) exp(-2t) dt
Then we have some fine looking calculus expression imho.
Many tricks from standard calculus could probably be used such as interchanging the integrals , or (interchanging) a sum and an integral , or feynman integration etc. and of course contour integration techniques/theorems.
Combining integral transforms with fake function theory seems intresting both for computation and theory.
Many complicated functions can be given by contour integration of simpler ones , and those contours can be written as simpler integral transforms.
Those transforms can then contain a sqrt or ln or such and therefore we finally arrive at a fake function of the Original complicated one.
- For instance g(z) , the functional inverse of a function f(z) such that g(z) grows fast enough ( faster then poly ) and g(z) cannot be given by elementary functions -
Probably fake( Gamma(x+2) ) has a closed form or its derivatives at 0 have a closed form.
There has already been research done into asymptotics to the gamma function but not like this.
Probably the Gamma function will popularize fake function theory.
Generalizing recursion and/or functional equations into fake function theory might be the next step.
The number of roads this leads too is uncountable !
Next on the list is fake( exp(x) x / ln^2(x) ) exp(-x) without using a fake ln.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/19/2014
Time for tommy's Q9 method.
The term Q refers to q-analogue as will be clear soon.
The 9 refers to post 9.
In post 9 we tried to find a good fake(f(x)) by using
a_n x^n < f(x).
We arrived at a solution that had the property
a_0 >= a_1 >= a_2 >= a_3 >= ... >= 0 (*)
We can use this property to find a better method.
a_0 + a_1 x + a_2 x^2 + ... + a_n x^n < f(x)
Using (*)
a_n + a_n x + a_n x^2 + ... a_n x^n < f(x)
Simplify
a_n (x^(n+1) - 1) / (x-1) < f(x)
a_n (x^(n+1))/(x-1) < f(x) + a_n/(x-1)
For x large we can ignore the a_n/(x-1) term on the RHS , that might give a worse fake for small x , but has almost no effect on the large x.
a_n (x^(n+1))/(x-1) < f(x)
a_n x^n x/(x-1) < f(x)
a_n x^n < f(x) (x-1)/x
ln(a_n) + n ln (x) < ln(f(x)) + ln(x-1) - ln(x)
ln(a_n) < ln(f(x)) + ln(x-1) - (n+1) ln(x)
ln(a_n) < Min( ln(f(x)) + ln(x-1) - (n+1) ln(x) )
This is an improvement.
Celebrate.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 09/29/2014
Im considering fake function theory for parabolic fixpoints.
regards
tommy1729
RE: Searching for an asymptotic to exp[0.5] - tommy1729 - 10/19/2014
An idea that is very very old.
The connection between series multisection and fake function theory.
An example says more than a 1000 pictures.
Consider f(x) = 1 + x + x^2/2! + x^3/3! - x^4/4! + ...
where the sign pattern continues as +,+,+,- such that every multiple of 4 gives a minus sign.
If you ask someone to estimate f(x) for x > 0 , they will likely say
f(x) ~ 1/4 + 1/2 exp(x) + C for some small real C.
Now the logical questions are
how good is this estimate really ... in other words a deeper study.
Clearly this relates to the mittag leffler function and the classic formula for series multisection that uses roots of unity.
But more relevant here is
fake f(x) ~ 1/4 + 1/2 exp(x) + C ??
How close to the truth is that ?
How good does fake function theory estimate here ?
Is fake function theory the ultimate method for this , or is it weak ?
Also notice the alternative estimates
1 + sinh(x)
or
cosh(x)
Who also have positive derivatives.
---
The differential equations
d^n f / d^n x = f(x)
are also often considered because of the natural connection.
---
These questions seems very reasonable and solvable.
Generalized questions and answers are therefore very likely to exist.
regards
tommy1729
" Together we can do more "
tommy1729