[solved] Define x-range for Series expantion

Questions
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1

Hi all,

I am pretty new (i.e. 2 days old) with Mathematica and so far enjoy the power of it.

My question is how-to make a series expantion for a certain x-range of the function.
The problem is that my function has quite some curves and I want to series expand in a local minimum of the function to know the curvature of this local minimum.
Imagine a Sin series and you only want to series expand in a local minimum - this is exactly the same problem.

I assume I can expand another function which is defined only in an interval around the local minimum of the orignal function ? But isn`t there a nice solution to this problem ?

Best!

PS: The assumption argument would be nice to use but this is not the same as defining a range.

2

If possible, can you give an example of the kind of series expansion you are looking for from an example function?

Please take a look at the Series function if you have not already.

3

The function I have is the one in this figure. This function needs to be series expanded at a=-0.15 m. BUT the range of the series expantion is important. I only want to series expand between -0.18 and -0.12, to get the exact curvature of the (almost) ~x^2 well. I had a look in the Series function description already but there is no such thing as boundaries on a series expantion. I assume it is possible though ? Otherwise my strategy would be to define a new function only defined in this interval and then series expand this but this sounds like a detour. You got a better ideas, Sean ?

Potential.png
Potential.png1025×692 15.1 KB

4

I’m not sure I understand what you mean by boundaries in this case. It sounds like you want to use the Series function and only apply it over a certain range of values.

A better question to ask is, how would you go about doing what you want by hand for a simple example?

5

Taylor Expansions are defined at a point, not over a range. the higher the order of the polynomial, the wider the range it will fit…

6

Hi Simon,

Series[] can expand about any point in the domain of the function. The second order term is then the second derivative divided by 2!. This is often a good aproximation to the curvature. This may be what you have in mind:

Output has been deleted, but the code should execute.

(* define a function *)
y[x_] := (x - 2) (x - 3) (x - 4)

(* plot it *)
Plot[y[x], {x, 1, 5}]

(* choose an initial value to find a minimum *)
x0 = x /. FindMinimum[y[x], {x, 3}][[2, 1]]

(* series expand about the minimum *)
expansion = Series[y[x], {x, x0, 2}]

(* select the coefficient on the 2nd order term 
and mulitply be 2 since it was divided by 2! in the expansion *)
term3 = expansion[[3, 3]]; curvature = 2 term3

(* but this is just 2nd derivative at x0, which is a good approximation to the curvature *)
kApp = y''[x0]

(* formally, the curvature can be obtained as below *)

k = y''[x0]/(1 + y'[x0])^(3/2)

(* Since x0 is a min, we know y'[x0]=0, so they don't match only because of numerical error. *)
kApp - k
7

Explore Wolfram Demonstrations on the subject - a lot of free code to download and nice looking applications. For example:

Taylor Polynomials by Harry Calkins

Manipulate[
 Column[{Style[Row[{"series degree = ", deg}], "Label", 12], 
   Plot[Evaluate@{func /. x -> t, 
      MakeSeriesFunction[func, x, deg, pt]}, {t, 0, 2 Pi}, 
    PlotRange -> {{0, 2 Pi}, {-2, 2}}, 
    PlotStyle -> {{GrayLevel[.25, .85]}, {RGBColor[0, 0, .9, .8]}}, 
    Filling -> {1 -> {{2}, Directive[{Opacity[.1], Green}]}}, 
    Epilog -> {Red, PointSize[.02], Point[{pt, func /. x -> pt}], 
      Blue, Point[{pt, 0}], Dashing[{.01}], 
      Line[{{pt, func /. x -> pt}, {pt, 0}}]}, 
    ImageSize -> {500, 350}]}, Center], {{deg, 10, "series degree"}, 
  1, 24, 1}, {{pt, 0, "expansion point"}, 0, 
  2 Pi}, {{func, Sin[x], 
   "function"}, {Sin[x] -> TraditionalForm[Sin[x]], 
   Cos[x*2]/(x + 1) -> TraditionalForm[Cos[x*2]/(x + 1)], 
   1/(1 + x) -> TraditionalForm[1/(1 + x)], 
   E^x/100 -> TraditionalForm[E^x/100], 
   2 E^-(-3 + x)^2 -> TraditionalForm[2 E^-(-3 + x)^2]}}, 
 SaveDefinitions -> True, 
 Initialization :> {MakeSeriesFunction[fn_, var_, deg_, pt_] := 
    Module[{wrkfn},
     wrkfn[tt_] = 
      If[Head[fn] === Symbol && fn =!= var, fn[tt], fn /. var :> tt];
      Normal@Series[wrkfn[t], {t, pt, deg}]
     ]}]
8

Hi all,
Thank you all for the replys, really appreciated.

Unfortunately the correct answer still is not given but maybe because my question was not formulated correctly (?). The series expantion at a point works but I want to series expand in a point BUT only for a given range of my function (see my previous reply in this thread). Thus I want to define this function only in a certain range and then series expand such that I do not get the negative curvature x^4 coefficient.

I found a way around and that is exporting the plot, to an array, only copy the part of the array that I need (which is within the range) to a new array. Then fit this array with a series expantion. However this operation involves a bit of c++ and labview which has an easy series fit. I believe an easier solution is possible with Mathematica ?

One solution I think of now is write my own fitting routine in Mathematica, assume a coefficient, calculate the mismatch between the real function and the fitted coeff in this range, guess again, etc etc… Doesn`t there exist something like this in Mathematica ?

So to summarize:

I have a function but want to series expand in a certain point AND a certain range. In my previous reply this would be in the range: -0.18 and -0.12. In the other range the function should be undefined, i.e. I do not care how the series expantion is done outside of this range.

Is it really not possible to do this in Mathematica ?
EDIT: I`ve uploaded the mathematica file here: https://svangorp.web.cern.ch/svangorp/MaMa_forum.nb The question is thus how to get a series expantion such that expantion is ONLY in the range 0.12 to 0.18

9

Actually the answer is pretty straightforward:

If you want to series expand only over a range you first define a new data-set of the fit and then you FIT this dataset with a series expantion. Like this:

data[R_] = Table[{x, myPot2[R, x]}, {x, -0.02, .02, 0.002}];

fit[R_, x_] = Fit[data[R], {x^0, x^1, x^2, x^3, x^4, x^6}, {x}] ;

In this example -0.02 to 0.02 is the range over which you want to fit and myPot2[R,x] is the function.