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# How to simplify an expression? want a specific form.

Posted 10 years ago
 I have an expr. with the form of  Exp[a1*t]*b1+Exp[a1*t]*b2+....+Exp[a1*t]*bn,a1 and bi are all symbols,not numberI want it in the following form: Exp[a1*t]*(b1+b2+...+bn).What should I do? Here Simplify doesn't work
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Posted 10 years ago
 Perhaps Factor can help: In:= Factor[E^(a1 t) b1 + E^(a1 t) b2 + E^(a1 t) b3]Out= (b1 + b2 + b3) E^(a1 t)
Posted 10 years ago
 Thanks~actually the problem is a little bit more complicated, here is the original expression:E^(-k22 t) ((0.5 b2 bb1^2 (-1 + E^(-k22 t)))/k22 - ( 0.5 b2 c1^2 (-1 + E^(k22 t)))/k22 - ( 0.5 a1^2 b2 (-1 + E^((k22 - 2 k33) t)))/(k22 - 2 k33) -...Expand it and we  get:-((0.5 b2 c1^2)/k22) + (0.5 b2 bb1^2 E^(-2 k22 t))/k22 - ( 0.5 b2 bb1^2 E^(-k22 t))/k22 + (0.5 b2 c1^2 E^(-k22 t))/k22 + ( 0.5 a1^2 b2 E^(-k22 t))/(k22 - 2 k33) - ...the items of the sum are all of  the form: b E^[a t] t^nwhere a,b can be very complicated,I want to put those items with the same a and n together, i.e, to simplify the expression to make sure that  E^[a t]*t^n*(b1+b2+..)only appear once in the sum, put the coefficients together in the bracket. Is this possible?
Posted 10 years ago
 Hi,Here is my suggestion. Get all of E^(a t) using Cases. Then use Collect with var = (List of E^(a t)) (see Mathematica help for details). You will need further processing to factor out the factors t^n.Youngjoo Chung
Posted 10 years ago
 Thanks for your suggestion, Collect works on patterns can solve this problem perfectly~