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DFA-1 on Quadratic trend
Let us suppose now a series of the type
. The integrated time series
is
![\begin{displaymath}
y(i)=A_{\rm Q}\sum_{j=1}^{i}j^{2}=A_{\rm Q}\frac{2i^{3}+3i^{2}+i}{6}
\end{displaymath}](img432.png) |
(38) |
As before, let us call
and
the sizes of the series
and box, respectively. The rms fluctuation function
measuring the rms fluctuation is now defined as
![\begin{displaymath}
F_{\rm Q}(n)=A_{\rm Q}\sqrt{\frac{1}{N_{max}}\sum_{k=1}^{N_...
...\left(\frac{2i^{3}+3i^{2}+i}{6}-(a_{k}+b_{k}i)\right)
^{2} }
\end{displaymath}](img434.png) |
(39) |
where
and
are the parameters of a least-squares fit of
the
-th box of size
. As before,
and
can be determined
analytically, thus giving:
![\begin{displaymath}
a_{k}=\frac{1}{15}n^{3}+n^{3}k^{2}-\frac{7}{15}n^{3}k+\frac...
...ac{2}{3}k^{3}n^{3}-\frac{1}{2}n^{2}k^{
2}+%%
\frac{1}{15}kn
\end{displaymath}](img435.png) |
(40) |
![\begin{displaymath}
b_{k}=\frac{3}{10}n^{2}+n^{2}k^{2}-n^{2}k+kn-\frac{2}{5}n+\frac{1}{10}
\end{displaymath}](img436.png) |
(41) |
Once
and
are known,
can be evaluated, giving:
![\begin{displaymath}
F_{\rm Q}(n)=A_{\rm Q}\allowbreak \frac{1}{1260}\sqrt{-21\l...
...ght) \left(
32n^{2}-6n-81-210N_{max}-140N_{max}^{2}\right) }
\end{displaymath}](img437.png) |
(42) |
As
, the dominant term
inside the square root is given by
, and then one has approximately
![\begin{displaymath}
F_{\rm Q}(n)\approx A_{\rm Q}
\frac{1}{1260}\sqrt{2940n^{4}N_{max}^{2}}=A_{\rm Q}\frac{1}{90}\sqrt{15}N_{max}n^{2}
\end{displaymath}](img440.png) |
(43) |
leading directly to an exponent 2 in the DFA analysis. An interesting
consequence derived from Eq. (43) is that,
depends on the length of signal
, and the DFA line (
vs
) for
quadratic series
of different
does not overlap (as is the case for linear trends).
Next: Bibliography
Up: Appendix
Previous: DFA-1 on linear trend
Zhi Chen
2002-08-28