Problem about equality

  • #1
kelvin490
Gold Member
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Suppose we have two equation x1=Aeiωt + Be-iωt and x2=A*e-iωt + B*eiωt . Where A and B are complex number and A* B* are their conjugate correspondingly.

Now if we want to make x1 and x2 exactly equivalent all the time, one way to do it is to have A=B* and B=A* so that x1 and x2 are equivalent. However, if we don't do it by this approach but instead set (A-B*)eiωt=(A*-B)e-iωt, then we have ei2ωt=(A*-B)/(A-B*). I would like to ask if the A and B chosen can satisfy this criteria (even A≠B* and B≠A*), can we still say that x1 ≡ x2 ?

Another thing trouble me is if A=B* and B=A* , then ei2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?
 
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Answers and Replies

  • #2
HallsofIvy
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Suppose we have two equation x1=Aeiωt + Be-iωt and x2=A*e-iωt + B*eiωt . Where A and B are complex number and A* B* are their conjugate correspondingly.

Now if we want to make x1 and x2 exactly equivalent all the time, one way to do it is to have A=B* and B=A* so that x1 and x2 are equivalent. However, if we don't do it by this approach but instead set (A-B*)eiωt=(A*-B)e-iωt, then we have ei2ωt=(A*-B)/(A-B*). I would like to ask if the A and B chosen can satisfy this criteria (even A≠B* and B≠A*), can we still say that x1 ≡ x2 ?
What you are doing is saying that if [itex]Ae^{i\omega t}+ Be^{-i\omega t}= A^*e^{-i\omega t}+ B^*e^{-i\omega t}[/itex] then we must have [itex]e^{2i\omega t}= \frac{A^*- B}{A- B^*}[/itex]. However, the right side of that equation is a constant, independent of t. In order for that to be true, the left side must also be a constant- so what you are doing is equivalent to assuming that [itex]\omega= 0[/itex], a very restrictive condition!

Another thing trouble me is if A=B* and B=A* , then ei2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?
If [itex]A= B^*[/itex] then [itex]B[/itex] must equal [itex]A^*[/itex]- that is not a distinct condition. And in that case you are saying [itex]x_1= Ae^{i\omega t}+ A*e^{-i\omega t}[/itex] and [itex]x_2= A*e^{-i\omega t}+ Ae^{i\omega t}[/itex], that is, that [itex]x_1= x_2[/itex] for all t. That is exactly what you said you wanted. There is no "problem". If you have two identical equation, say "px= qy" and "px= qy" and subtract them you get 0x= 0y, of course. It would make no sense to try to "solve for y" in that case, you would just get 0/0 as happens here.
 
  • #3
mathman
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The two orignal expressions are complex conjugates of each other. Therefore to be equal, the imaginary part must be 0.
 
  • #4
kelvin490
Gold Member
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I think the concept of linear independence can help. Since e-iωt and eiωt are two linearly independent variables, the coefficient of x1 and x2 must be equal so that x1 and x2 are equivalent.

Also from (A-B*)ei2ωt=(A*-B) we can see that the left hand side is a time dependent function (let's say t is time) and the other side is a constant. The only way to make both sides equal all the time is to have A=B* and B=A*.
 

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