# Tetration Forum

Full Version: Searching for an asymptotic to exp[0.5]
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(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
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 !
(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 ....
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
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
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
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.

Next on the list is fake( exp(x) x / ln^2(x) ) exp(-x) without using a fake ln.

regards

tommy1729
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
Im considering fake function theory for parabolic fixpoints.

regards

tommy1729
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
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