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Re: How to find the kinks in a list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104363] Re: How to find the kinks in a list
  • From: Bill Rowe <readnews at sbcglobal.net>
  • Date: Thu, 29 Oct 2009 02:56:48 -0500 (EST)

On 10/28/09 at 4:08 AM, sdzbbjhd at hotmail.com (Mather) wrote:

>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

Here is one method. After putting your list of numbers into a
variable, data, I can plot the differences in the y position.
That is

ListPlot[Diiferences@data[[All, 2]]]

Given the uniform sampling on the x-axis, this plot is an
approximate plot of the derivative of your data.

Looking at this plot it is clear there is a big jump at each of
the kinks. So, the position of the kinks in the data list can be
obtained from

In[5]:= kinks =
  Ordering[Abs@Differences@Differences[data[[All, 2]]], -2] + 1

Out[5]= {34,16}

The reason for adding 1, is each application of Differences
decreases the length of the original list by 1.

These positions can be converted to x coordinates by

In[6]:= Rescale[kinks, {1, Length@data}, data[[{1, -1}, 1]]]

Out[6]= {3.31,1.51}

And finally, the plot

ListPlot[data,
  GridLines -> {Rescale[kinks, {1, Length@data}, data[[{1, -1}, 1]]],
    None}]

verifies the solution



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