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Complex exponential function

  1. Mar 16, 2007 #1
    good evening all!

    1. The problem statement, all variables and given/known data

    Determine the exact values of

    2. Relevant equations

    j = sauare root of -1

    3. The attempt at a solution

    stuck :cry: :cry: :cry:
    Last edited by a moderator: Jan 3, 2014
  2. jcsd
  3. Mar 16, 2007 #2


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    What do you know about the complex exponential function? Surely your text has some information you can apply.
  4. Mar 16, 2007 #3
  5. Mar 16, 2007 #4
    i'm still stuck
  6. Mar 16, 2007 #5
    okay this is what i did:


    i know this looks crazy but i still don't know how to use latex properly :redface: ...if you copy it out on a paper you'll understand what i'm trying to write...then i multiplied -1 with the 1/2...i can do that right?
  7. Mar 16, 2007 #6

    matt grime

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    YOu won't get the answer that way (since it doesn't deal with branches properly).

    Complex numbers have representations as r*exp(jt) for r, j in R. Use it.
  8. Mar 16, 2007 #7
    does exp(-2pi) = -1
  9. Mar 16, 2007 #8
    no it doesn't

    wait no one give me any tips i think i can solve it
  10. Mar 16, 2007 #9

    matt grime

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    No. If you must, put it in your calculator, but e^{-2pi} is obviously a positive real number, and something in the region of 1/2^64.
  11. Mar 16, 2007 #10

    matt grime

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    solve what? exp(-2pi) is right. It just isn't anything nicer than that.
  12. Mar 16, 2007 #11
    okay...i was about to say i'm stuck again but now i know that my teacher is crazy...but just to make sure [tex]j^j[/tex] doesn't equal -1...does it?

    ( [tex]j^j[/tex] <--that is almost the only thing i can write in latex form)
    Last edited: Mar 16, 2007
  13. Mar 16, 2007 #12
    yes it does
  14. Mar 16, 2007 #13

    matt grime

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    no it doesn't. j (or i) is exp(j*pi/2) so j^j is..... You can even google the phrase

    i to the i



    and you'll be given the answer by google calculator.

    Interesting googling j^j doesn't yield the answer. Google obviously doesn't employ engineering grads.
  15. Mar 16, 2007 #14
    my teacher is very very clever man and he said that the answer is -1 but he didn't use any of the complex number rules...he did something else! and he got -1

    sorry i used this formula to show that it does equal -1:

    r[cos(O) + jsin(O)] = rexp(jO)

    and in my desperate want to get this question right i didn't see the j next to the O, and so after i substituted the numbers i'm obviously going to get -1...never mind!

    ( O = theta)

    using your advice i went on google and thank god for wikipedia i got the answer...and i understand this time (for a change)

    you know i'm still stuck on that bernoulli number question...for B_3 you're supposed to get zero i actually got a number
  16. Mar 16, 2007 #15

    matt grime

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    Well keep trying 'til you do get zero. x/(e^x-1) differentiate it three times... eugh, no thanks. Do it the smarter way - sub in e^x first, then do the 1/(1-s) expansion. collect terms and you get

    -1/4! + 2/2!3! - (1/2!)^3 = -1/24+1/6-1/8=0

    (and when did 0 cease to be a number?)
  17. Mar 16, 2007 #16

    [tex]j^j=e^{j\cdot jx}=e^{-x}[/tex]

    so what should be the answer? (certainly not -1)
    Last edited: Mar 16, 2007
  18. Mar 16, 2007 #17
    i took your advice ( i kinda knew it anyway) but i'm going to write out in full what i did:

    [tex]e^x[/tex] = 1 + x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

    [tex]e^x[/tex] - 1 = x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...

    x/([tex]e^x[/tex] - 1) = x/(x + [tex]x^2[/tex]/2! + [tex]x^3[/tex]/3! + ...)

    divid top and bottom by x => 1/(1 + x/2! + [tex]x^2[/tex]/3! + ...)

    (1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1)

    let x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +.. = X

    (1 + X)^(-1) = 1- X + [((-1)(-2))/2!] [tex]X^2[/tex] +...

    so (1+ x/2! + [tex]x^2[/tex]/3! + [tex]x^3[/tex]/4! +...)^(-1) = 1 - x/2! - [tex]x^2[/tex]/3! - [tex]x^3[/tex]/3! -...((-1)(-2))/2!(X)^2 +...

    so B_1 =-1/2, B_2= -1/3!, B_3 = -1/4!.

    B_2 and B_3 are obviously wrong but i don't know where i'm going wrong???
    Last edited: Mar 16, 2007
  19. Mar 16, 2007 #18
    yes, okay, cheers, i get the picture :rolleyes:

    :biggrin: (<-- this is my favourite smiley)
  20. Mar 16, 2007 #19

    matt grime

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    Why does B_2=-1/3? What happend to the higher powers of X (a bad choice. Y would be better). You have only take the contribution from X^1, not X^2 or higher.

    and 1/(1+s) is 1-s+s^2-s^3+s^4+...

    Now put s = x/2!+x^2/3!+x^3/4!+...
    Last edited: Mar 16, 2007
  21. Mar 16, 2007 #20
    i still get a negative for B_2 i know what the answer is uppossed to be but i just can't see how...if i don't see why i don't sleep well!

    i'm tired, you know i'm just going to live with the fact that i'm never going to get this question...

    the thing is there is a minus outside the s and s^2 and the rest don't matter...


    thank you very much for your time and help!!!
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