09/14/2014, 05:07 PM
(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