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I have an expression:

{((-2*m*n + B*R0^2)*Sin[t])/(4*Pi*R0), -1/4*((-2*m*n + B*R0^2)*Cos[t])/(Pi*R0), 0}

And I would like it to recognize this pattern

%/.{-Sin[t],Cos[t],0}-> ET

such that

{((-2*m*n + B*R0^2)*Sin[t])/(4*Pi*R0), -1/4*((-2*m*n + B*R0^2)*Cos[t])/(Pi*R0), 0} /.{-Sin[t],Cos[t],0}-> ET

should result in

-ET (-2*m*n + B*R0^2)/(4*Pi*R0)

or something the like. It however does not work. Who can help?

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2 Answers 2

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I think your result has a typo. Instead of -ET (2*m*n + B*R0^2)/(4*Pi*R0) it should read -ET (-2*m*n + B*R0^2)/(4*Pi*R0), otherwise there is no solution.

A further problem is the minus sign in front of the Cos term. MMA always uses the full form of an expression for matching. Look at:

((-2 m n + B R0^2) Cos[t])/(4 \[Pi] R0) // FullForm
(* Times[Rational[1,4],Power[Pi,-1],Power[R0,-1],Plus[Times[-2,m,n],Times[B,Power[R0,2]]],Cos[t]] *)

and

-(((-2 m n + B R0^2) Cos[t])/(4 \[Pi] R0)) // FullForm
(* Times[Rational[-1,4],Power[Pi,-1],Power[R0,-1],Plus[Times[-2,m,n],Times[B,Power[R0,2]]],Cos[t]] *)

You see, the minus is integrated in the Rational[-1,4]. Therefore, for matching, the minus can not be matched separately. Here is how you can do it:

{((-2*m*n + B*R0^2)*Sin[t])/(4*Pi*R0), -1/4*((-2*m*n + B*R0^2)*Cos[t])/(Pi*R0),  0} /. { x__ Sin[t], y__ Cos[t], 0} :> ET y

enter image description here

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  • $\begingroup$ I have corrected the wrong minus sign, it was typo. That the integration of the sign into the fraction hampers the pattern detection is a bit odd, but I suppose there have to be some limits in pattern detection otherwise any pattern would be everywhere I suppose. Thank you! $\endgroup$ Dec 10, 2022 at 17:33
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Consider vector projection using the dot product. Let v be the input vector and p be the pattern:

v = {((-2*m*n + B*R0^2)*Sin[t])/(4*Pi*R0), -1/4*((-2*m*n + B*R0^2)*Cos[t])/(Pi*R0), 0};
p = {-Sin[t], Cos[t], 0};

Then the parts of the vector which are in line with the given pattern are:

Simplify[v.p/Norm[p], Element[t, Reals]]

(2 m n - B R0^2)/(4 π R0)

You may multiply this by ET, if you wish. For other expression the Simplify may not be necessary to achieve a nice result, and I have assumed t is a real number here for simplicity. If that is not necessarily the case, this method may not be appropriate.

Note also that the non-conforming parts of the vector are given by:

Simplify[v-v.p/Norm[p], Element[t, Reals]]

{0, 0, 0}

In general two vectors will not be perfectly aligned with each other, so this term is normally non-zero. This is the part of your vector which does not co-align with your given pattern.

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  • $\begingroup$ OK, I see, I have also tried with Solve[b p = v, b] which also works. $\endgroup$ Dec 10, 2022 at 13:12

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