Searching for an asymptotic to exp[0.5]

[tommy1729]

I have considered many more things then I could ever post ...

I probably said it before but Im considering " fake analytic number theory ".

That might take some time to develop.

Here are some other ideas that inspire me and for which Im currently not sure how to continue.

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Jay's function that approximates the binary partition function has become popular here.

I wonder about variants of type \( \Sigma (2^{n*(n-1)*(n+1)/6} n!)^{-1} x^n \)

DO these variants of " cubic type " solve anything ?

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Initially fake function methods start with an OVERESTIMATE.

It is possible to use a method that is both GOOD and starts with an UNDERESTIMATE ?

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Im wondering about fake fourier series and fake integrals.

A logical way , but possibly not the best , is to consider the fake function methods as fake n'th derivatives.

Then the fake integral is computed as the fake first derivative of the true second integral.

Or something like that ...

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Lets try Jay's function J(x).

a_n x^n = J(x)

ln(a_n) + n ln(x) = ln(J(x))

ln(a_n) = min[ ln(J(x)) - n ln(x) ]

d/dx [ ln(J(x)) - n ln(x) ] = J(x/2)/J(x) - n/x

J(x/2)/J(x) = n/x

I really like the shape of this equation.
And ofcourse I wonder how good a_n will be compared to \( (2^{n*(n-1)/2} n!)^{-1} \)

ALthough Im not completely stuck here , Im also not completely sure how to proceed.
Use asymptotics , invent new special function , use contour integrals , numerical methods , ... ?

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Under some trivial conditions I consider some ideas to improve finding a fake function without actually changing the method ...

example

Find fake exp(x).

Note : We already discussed the idea that we already have an entire function with positive derivatives , yet this is intresting.
And just an example , it applies to non-entire functions too.

Instead of
a_n x^n = exp(x)

We solve for

a_n x^n = 2 exp(x) / (1+x)

Then we multiply our result (taylor series with a_n , possibly scaled ) with (1+x)/2.

Notice multiplication by (1+x)/2 does not change the signs of the derivatives and is easy to compute !

The q-variant of this is also possible of course.

I call these methods

sx9(n) and qx9(n).

So for instance \( qx9 := (1+x)/2 * Qfake[ 2 f(x)/(1+x) ] \)

Much more work needs to be done !

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Comments and help is appreciated.

regards

tommy1729

" Formally define useful and useless , but beware : take into account we are plain mortals and your an atheist who claims to be not obsessed by money nor by ego "
tommy1729 @sci.math