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Searching for an asymptotic to exp[0.5]
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.



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

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