@james : nice resume of some of your key ideas.
I even see the connection with fractional derivatives.
Just one question: you say that in many cases the difference equation has no analytic solution ?
like in nowhere analytic ?
Like which ones ?
And in the cases which are not , is it always due to or shown with infinite composition or are there other tools and ideas ?
Does that property relate or carry over to differential equations ??
***
@leo mainly :
about those asymptotics at infinity and difference equations ...
First , I have no clue how you got those series of functions you did.
Not that it is beyond me , but it is not clear how YOU exactly got them.
But for asymptotics at infinity and difference equations there are afak only about four or five methods apart from fixpoint methods :
1)
approximate the difference by differentials.
such as replace difference by derivative.
or better approximations with higher derivatives and using taylor.
this could give asymptotics and boundaries that are potentially useful.
2)
use infinite composition like the ideas from the beta method or gaussian method.
3)
use limits to get asymptotics
As an example :
f(0) = 2
f(n+1) = f(n) + ln( f(n) )
to get asympt for f(n) ...
( notice the inverse of Li(n) is a good asymptotic , thereby motivating the derivative idea in 1). idea 2) can also be used but may be harder )
use this limit technique :
> \[\lim_{n\to\infty}\left(\frac{f_n}n-\log n-\log\log n\right)=-1\]
To prove this, consider
\[g_n=n\log n+n\log\log n-n\]
and
\[h_n=f_n-g_n\].
One wants to prove that
\[h_n=o(n)\].
The identity \[f_{n+1}=f_n+\log f_n\] is equivalent to
\[
h_{n+1}=g_n+h_n+\log(g_n+h_n)-g_{n+1}.
\]
Using simple properties of the logarithm, one can show that this implies
\[
h_{n+1}=h_n+\log\left(1+\frac{\log\log n}{\log n}-\frac1{\log n}+\frac{h_n}{n\log n}\right)+O\left(\frac1{\log n}\right).
\]
In particular, if
\[h_n=o(n\log n)\]
, the logarithm in the RHS goes to zero hence
\[h_{n+1}=h_n+o(1)\]
, which implies
\[h_n=o(n)\].
Thus, our task is to prove the easier statement that
\[
f_n=n\log n+o(n\log n).
\]
To do so, first note that
\[f_n\geqslant2\]
for every \[n\geqslant0\]
yields
\[f_{n+1}-f_n\geqslant\log2\]
hence \[f_n\geqslant n\log2\]
for every \[n\geqslant0\]
. Plugging this once again in the recursion
\[f_{n+1}=f_n+\log f_n\]
yields
\[f_{n+1}-f_n\geqslant\log n+\log\log2\]
hence, summing up,
\[f_n\geqslant n\log n+o(n\log n)\].
In the other direction,
\[f_{k+1}-f_k=\log f_k\leqslant\log f_n\]
for every \[k\leqslant n\]
hence
\[f_n\leqslant f_0+n\log f_n\],
which can be seen to imply
\[f_n\leqslant n\log n+2n\log\log n\]
for every \[n\] large enough. This completes the proof.
SECOND EXAMPLE :
https://math.stackexchange.com/questions...v-n-sqrt-7
4) truncated methods. based on series expansions and/or fixpoints. not necc taylor.
5) sometimes series multisection and/or " fake function theory ".
and combinations of those above ofcourse.
other methods would be nice to see.
regards
tommy1729
Tom Marcel Raes
I even see the connection with fractional derivatives.
Just one question: you say that in many cases the difference equation has no analytic solution ?
like in nowhere analytic ?
Like which ones ?
And in the cases which are not , is it always due to or shown with infinite composition or are there other tools and ideas ?
Does that property relate or carry over to differential equations ??
***
@leo mainly :
about those asymptotics at infinity and difference equations ...
First , I have no clue how you got those series of functions you did.
Not that it is beyond me , but it is not clear how YOU exactly got them.
But for asymptotics at infinity and difference equations there are afak only about four or five methods apart from fixpoint methods :
1)
approximate the difference by differentials.
such as replace difference by derivative.
or better approximations with higher derivatives and using taylor.
this could give asymptotics and boundaries that are potentially useful.
2)
use infinite composition like the ideas from the beta method or gaussian method.
3)
use limits to get asymptotics
As an example :
f(0) = 2
f(n+1) = f(n) + ln( f(n) )
to get asympt for f(n) ...
( notice the inverse of Li(n) is a good asymptotic , thereby motivating the derivative idea in 1). idea 2) can also be used but may be harder )
use this limit technique :
> \[\lim_{n\to\infty}\left(\frac{f_n}n-\log n-\log\log n\right)=-1\]
To prove this, consider
\[g_n=n\log n+n\log\log n-n\]
and
\[h_n=f_n-g_n\].
One wants to prove that
\[h_n=o(n)\].
The identity \[f_{n+1}=f_n+\log f_n\] is equivalent to
\[
h_{n+1}=g_n+h_n+\log(g_n+h_n)-g_{n+1}.
\]
Using simple properties of the logarithm, one can show that this implies
\[
h_{n+1}=h_n+\log\left(1+\frac{\log\log n}{\log n}-\frac1{\log n}+\frac{h_n}{n\log n}\right)+O\left(\frac1{\log n}\right).
\]
In particular, if
\[h_n=o(n\log n)\]
, the logarithm in the RHS goes to zero hence
\[h_{n+1}=h_n+o(1)\]
, which implies
\[h_n=o(n)\].
Thus, our task is to prove the easier statement that
\[
f_n=n\log n+o(n\log n).
\]
To do so, first note that
\[f_n\geqslant2\]
for every \[n\geqslant0\]
yields
\[f_{n+1}-f_n\geqslant\log2\]
hence \[f_n\geqslant n\log2\]
for every \[n\geqslant0\]
. Plugging this once again in the recursion
\[f_{n+1}=f_n+\log f_n\]
yields
\[f_{n+1}-f_n\geqslant\log n+\log\log2\]
hence, summing up,
\[f_n\geqslant n\log n+o(n\log n)\].
In the other direction,
\[f_{k+1}-f_k=\log f_k\leqslant\log f_n\]
for every \[k\leqslant n\]
hence
\[f_n\leqslant f_0+n\log f_n\],
which can be seen to imply
\[f_n\leqslant n\log n+2n\log\log n\]
for every \[n\] large enough. This completes the proof.
SECOND EXAMPLE :
https://math.stackexchange.com/questions...v-n-sqrt-7
4) truncated methods. based on series expansions and/or fixpoints. not necc taylor.
5) sometimes series multisection and/or " fake function theory ".
and combinations of those above ofcourse.
other methods would be nice to see.
regards
tommy1729
Tom Marcel Raes