# [solved] Define x-range for Series expantion

GROUPS:
 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 isnt 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.
3 years ago
8 Replies
 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 years ago
 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 ?
3 years ago
 Sean Clarke 1 Vote 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?
3 years ago
 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...
3 years ago
 David Keith 1 Vote 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