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+--- Thread: f ' (x) = f(exp(x)) ? (/showthread.php?tid=985)
f ' (x) = f(exp(x)) ? - tommy1729 - 04/15/2015
I was thinking about f ' (x) = f(exp(x))
It reminds me of James recent paper and the Julia equation.
Continuum sums may also be involved and infinite matrices ( carleman equations ) as well.
For instance D^2 f(x) = f '' (x) = f ' (exp(x)) exp(x)
= f(exp(exp(x))) exp(x).
D^3 f(x) = ( f(exp^[2](x)) + f(exp^[3](x)) ) exp(x).
Clearly for D^n f(0) we get a simple expression with using the pascal triangle / binomium identities.
Assuming f(+oo) = Constant We might use tricks such as James Nixon's construction.
Does the generalized binomium analogue hold for the fractional derivatives of f(0) ??
Or does it hold for one solution , assuming there was choice ?
Can we use Ramanujan's master theorem ?
Can we compute tetration from assuming this generalized binomium analogue for the fractional derivative ?
After some confusion and troubles , I also came to consider
f ' (x) = f(exp(x/2)).
And then consider the analogues from above.
( this to have convergeance problems solved )
Many more ideas come to me , but I will see what you guys think.
At least I believe :
D^t f(0) is of the form " Binomial type " * g(t) where g(t) is 1-periodic.
Probably some theorem from fractional calculus for products can solve this part.
As a sidenote I wonder what f ' (x) = f(exp(x)) says about integral f(x) ?
Also I want to note that f ' (x) = f(exp(x)) " probably" cannot hold everywhere for an entire f(x) because exp is chaotic ... probably ...
regards
tommy1729
RE: f ' (x) = f(exp(x)) ? - fivexthethird - 04/17/2015
I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)
\( f \(c\) = \alpha \)
\( f' \(c\) = f\(e^c \) = f\(c\) = \alpha \)
\( f'' \(c\) = e^c f'\(e^c \) = c \alpha \)
\( f^{[3]} \(c\) = c^3 \alpha + c \alpha \)
\( f^{[4]} \(c\) = c^6 \alpha + c^4 \alpha + 3*c^2 \alpha + c \alpha \)
\( f^{[5]} \(c\) = c^{10} \alpha + c^8 \alpha + 9*c^6 \alpha + c^5 \alpha + 6c^4 \alpha + 7*c^3 \alpha + c \alpha \)
and in general the degree of the polynomial appears to be \( \frac{n(n+1)}{2} \) which obviously grows too fast to converge.
RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
(04/17/2015, 02:15 AM)fivexthethird Wrote: I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)
\( f \(c\) = \alpha \)
\( f' \(c\) = f\(e^c \) = f\(c\) = \alpha \)
\( f'' \(c\) = e^c f'\(e^c )\) = c \alpha \)
\( f^{[3]} \(c\) = c^3 \alpha + c \alpha \)
\( f^{[4]} \(c\) = c^6 \alpha + c^4 \alpha + 3*c^2 \alpha + c \alpha \)
\( f^{[5]} \(c\) = c^10 \alpha + c^8 \alpha + 9*c^6 \alpha + c^5 \alpha + + 6c^4 \alpha + 7*c^3 \alpha + c \alpha \)
and in general the degree of the polynomial appears to be \( \frac{n(n+1)}{2} \) which obviously grows too fast to converge.
Sorry Guys but my 3rd derivative in the op was wrong as clearly shown here.
I guess these are named polynomials , not ?
Regards
Tommy1729
RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
I think this equation has a radius 0 everywhere ?
Regards
Tommy1729
RE: f ' (x) = f(exp(x)) ? - tommy1729 - 04/17/2015
Therefore Maybe f ' (exp(x)) = f(x) works better.
Regards
Tommy1729