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Re: How to find the kinks in a list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104352] Re: [mg104342] How to find the kinks in a list
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 29 Oct 2009 02:54:41 -0500 (EST)
  • References: <200910280908.EAA14393@smc.vnet.net>

Mather wrote:
> Hi, all
> 
> I want to get the two kinks in the following list, say, x=1.5 and 3.3
> by Mathematica.
> However I have no idea which command to employ.
> Can anyone help? Thanks in advance!
> 
> 0.01	1.544163223920461e-6
> 0.11	2.7525893580512636e-7
> 0.21	6.8152995392824515e-6
> 0.31	4.972729284187191e-6
> 0.41000000000000003	6.886896332247575e-6
> 0.51	5.028506319197025e-7
> 0.61	1.765869426703084e-6
> 0.71	4.084953122120491e-7
> 0.81	8.908881820563147e-6
> 0.91	3.0059189773700746e-7
> 1.01	1.3790439392802757e-6
> 1.11	8.354283708355008e-6
> 1.21	7.190698243278427e-7
> 1.31	3.792514692609191e-7
> 1.41	2.5758219258643325e-6
> 1.51	0.00573372886943746
> 1.61	0.09069025128158258
> 1.71	0.1678921607552329
> 1.81	0.24047592904187015
> 1.9100000000000001	0.3094989860449043
> 2.0100000000000002	0.37555342886640686
> 2.11	0.43898726713721536
> 2.21	0.500041232643779
> 2.31	0.5588795919592797
> 2.41	0.6156217903422918
> 2.5100000000000002	0.6703511870298868
> 2.61	0.7231492006404858
> 2.71	0.7740553083481225
> 2.81	0.8231311485576547
> 2.91	0.8703722465175462
> 3.0100000000000002	0.9158530005108363
> 3.11	0.9595743351972397
> 3.21	1.0015438350017574
> 3.31	1.034334208101605
> 3.41	1.0346205212139916
> 3.5100000000000002	1.0346175133605735
> 3.61	1.0346312322976905
> 3.71	1.0346297067475791
> 3.81	1.0346299273452295
> 3.91	1.0346026798402437

One approach ise to use interpolating functions, and check for jumps in 
the second derivative. In the example below I imported your data (using 
"Table" format), and worked from there as below.

In[138]:= ff = Interpolation[data];
gg[t_] = ff''[t];
d2vals = Table[{t, gg[t]}, Evaluate[{t, Sequence @@ dom, .1}]];
pairs = Partition[d2vals, 2];
bigdiffs = Select[pairs, Abs[#[[2, 2]] - #[[1, 2]]] > 1 &];
Map[Mean[{#[[1, 1]], #[[2, 1]]}] &, bigdiffs]

Out[143]= {1.46, 3.26}

There are probably also some reasonable ways to do this sort of thing 
using Fourier or convolution.

Daniel Lichtblau
Wolfram Research


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