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Seems simple but just can't do it

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data
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    A exp(iax)+B exp(ibx) = C exp(icx) for all x that A+B = C and a=b=c where they are all real constants.


    3. The attempt at a solution
    first part easy
    x = 0

    A + B = C done

    second part

    d/dx -> (x=0) a A + b B = c (A + B) how do i make the last step?
     
  2. jcsd
  3. Aug 8, 2007 #2

    matt grime

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    Why have you only put in one value of x? Why didn't you put some more in to get some more constraints? Or differentiate again? In short, why have you stopped there and not tried some more things?
     
  4. Aug 8, 2007 #3

    berkeman

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    I'm not sure that I understand how you are trying to show the general case. Since the left and right are complex numbers, can you try just showing that their magnitudes and phase angles are the same? Or alternatively, show that their real and imaginary components are equal?
     
  5. Aug 8, 2007 #4
    berkeman's method yields...

    A cos(ax) + B cos(bx) = A cos(cx) + B cos(cx)
    A sin(ax) + B sin(bx) = A sin(cx) + B sin(cx)

    -> tan(ax) + tan(bx) = 2 tan(cx)

    i just don't know how to put into words why c = a = b
     
    Last edited: Aug 8, 2007
  6. Aug 8, 2007 #5

    berkeman

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    Maybe try the magnitude and phase angle approach instead...
     
  7. Aug 8, 2007 #6
    i really am lost on this one
     
  8. Aug 8, 2007 #7

    berkeman

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    How do you calculate the magnitude and phase angle of a complex number? You could do a quick search at wikipedia.org if your textbook doesn't cover it. What class is this from, and what is the textbook?
     
  9. Aug 8, 2007 #8
    i think this will do it...

    d/dx: aA exp(iax)+ bB exp(ibx) = cC exp(icx) for all x

    x = 0 gives

    aA + bB = c(A+B) (1

    x = 1 gives

    aA + bB exp(i(b-a)) = c(A+B) exp(i(c-a)) this must return condition 1 if it is for all x thus a=b=c

    good enough?
     
  10. Aug 8, 2007 #9

    Dick

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    It's hard for me to get a sense of what math level you are at, but do you know that exp(iax) and exp(ibx) are linearly independent functions if a is not equal to b? Do you know about inner products on function spaces? If so there is a much more economical way to think about this than putting individual points in. I think this is what berkeman is after, what do you already know?
     
  11. Aug 8, 2007 #10
    I'm not well practiced in linearly independent functions but yes I know what you're talking about and the inner product on function spaces, please continue...
     
  12. Aug 9, 2007 #11

    Dick

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    Good! Suppose you have the equation aA+bB=cC and you know A, B and C are linearly independent vectors. What do you conclude about a,b and c? That's the spirit of reasoning I'm talking about. In your problem, suppose a,b and c are not all equal. Then what?
     
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