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A problem of FindRoot with BSplineFunction

Jianhua Yang

Jianhua Yang, ASIPP

Posted 4 years ago

Hi! I have a problem like the following, data={{-1,2.9},{0.5,3.6},{1.0,5.5},{1.5,-4.2},{2.0,3.5}}; f=BSplineFunction[data]; FindRoot[f[t][[1]]==0,{t,0,0.2}] However, FindRoot firstly wokes like f[t][[1]]=t, but not to find the root of the first argument of f[t]. How can I get the root which one I wanted? Thanks!

POSTED BY: Jianhua Yang

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 4 years ago

I have made my own implementation of `BSplineFunction`, to make it more usable: BSplineCurveFunction[pts_?MatrixQ] := Module[{knots, deg}, knots = Rationalize@BSplineFunction[pts][[6, 1]]; deg = BSplineFunction[pts][[3, 1]]; Apply[Function, List@ Sum[ pts[[i + 1]] BSplineBasis[{deg, knots}, i, #], {i, 0, Length[pts] - 1}]]]; data = {{-1, 2.9}, {0.5, 3.6}, {1.0, 5.5}, {1.5, -4.2}, {2.0, 3.5}}; g[t_] = BSplineCurveFunction[data][t][[1]] FindRoot[g[t] == 0, {t, 0, 0.2}]

I have made my own implementation of BSplineFunction, to make it more usable:

BSplineCurveFunction[pts_?MatrixQ] := 
  Module[{knots, deg}, 
   knots = Rationalize@BSplineFunction[pts][[6, 1]]; 
   deg = BSplineFunction[pts][[3, 1]];
   Apply[Function, 
    List@
     Sum[
      pts[[i + 1]] BSplineBasis[{deg, knots}, i, #], {i, 0, 
       Length[pts] - 1}]]];
data = {{-1, 2.9}, {0.5, 3.6}, {1.0, 5.5}, {1.5, -4.2}, {2.0, 3.5}};
g[t_] = BSplineCurveFunction[data][t][[1]]
FindRoot[g[t] == 0, {t, 0, 0.2}]

POSTED BY: Gianluca Gorni

Alexey Popkov

Posted 4 years ago

Interesting implementation, thank you for sharing! Strongly related MMa.SE question with another implementation: Convert BSplineFunction into two Interpolating Functions

POSTED BY: Alexey Popkov

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 4 years ago

I may have misunderstood which root you were trying to find. If you want to find the crossing of the x axis, as Henrik understood, then similarly you need to use First instead of Last in the g function above (or use Henrik's approach). Regards.

POSTED BY: Neil Singer

Jianhua Yang

Jianhua Yang, ASIPP

Posted 4 years ago

It's not important. You two gave me some good ideas to deal with the problem. The key problem is how to find the root of any one component of BSplineFunction such as f[t] with FindRoot. f[t] is a two components function in some sense. Thank you again. Best regards!

POSTED BY: Jianhua Yang

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 4 years ago

The key problem is how to find the root of any one component of BSplineFunction ... Well, you could look for roots of the product of the components: fi = FunctionInterpolation[Times @@ f[t], {t, 0, 1}] (Somehow I could not avoid using `FunctionInterpolation`!) For finding (sort of) all roots in the interval [0, 1] you could try something (ugly) like this: DeleteDuplicates[N@Round[\[FormalT] /. Quiet@FindRoot[fi[\[FormalT]], {\[FormalT], #}], 1^-7] & /@ Range[0, 1, .2]] ( Out: {0.1412204`,0.6872917`,0.8855624`} *)

POSTED BY: Henrik Schachner

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 4 years ago

Here is one more solution - for me it is the first time I can make use of `FunctionInterpolation`: fi = FunctionInterpolation[First[f[t]], {t, 0, 1}]; FindRoot[fi[t], {t, .2}]