Thread Rating:
  • 2 Vote(s) - 4 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Searching for an asymptotic to exp[0.5]
#80
For sufficiently large k the truncated exp_k(x) = 1 + x + x^2/2 + ... x^k/k! Always has zero's with a positive real part.

This might be surprising to some because exp(-oo) = 0 is the only solution of f(z) = 0 and " -oo " means a infinitely large negative real part.

Yet every truncated exp_k(x) (for large k) has zero's with a positive real part.

Now from many theorems about the remainders of Taylor polynomials we get a " radius of good approximation " as I like to call it.

This radius " pushes out " the zero's of exp_k(x).

The divisors of k also have some influence.

But what is also striking is that the zero's tend to lie on a well defined curve.

That curve is like a twisted U shape.

The curve for exp_(k+1) Always seems to be a rescaling of the curve for exp_k.

So a certain shape seems to appear.

It seems the ratio of the range of im(z) to the range of re(z) approaches a small fraction.

Im fascinated by this.

It reminds me of the fake half-iterate f(x).
Mainly because f(f(x)) also has it's zero's on a twisted U shape !!

I conjecture a connection although Im not sure how exactly.

Maybe this has been investigated before and is a classic problem ? Maybe not.

Maybe the idea that the zero's and shapes of f(f(x)) and exp_k are connected comes from the fact that :

exp(x) = lim k -> +oo exp_k(x) = lim m -> +oo f(f(x+m))/exp(m).

Wonder what you guys think.

regards

tommy1729
Reply


Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 08/09/2014, 12:24 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Merged fixpoints of 2 iterates ? Asymptotic ? [2019] tommy1729 1 1,135 09/10/2019, 11:28 AM
Last Post: sheldonison
  Another asymptotic development, similar to 2sinh method JmsNxn 0 2,808 07/05/2011, 06:34 PM
Last Post: JmsNxn



Users browsing this thread: 1 Guest(s)