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Complex exponential function

  • Thread starter sara_87
  • Start date
  • #1
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good evening all!

Homework Statement



Determine the exact values of
[tex]j^j[/tex]

Homework Equations



j = sauare root of -1

The Attempt at a Solution



stuck :cry: :cry: :cry:
 
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Answers and Replies

  • #2
Integral
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What do you know about the complex exponential function? Surely your text has some information you can apply.
 
  • #3
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i=-1^(1/2)
 
  • #4
763
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What do you know about the complex exponential function? Surely your text has some information you can apply.
i'm still stuck
 
  • #5
763
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okay this is what i did:

((-1)^(1/2))^((-1)^(1/2))

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?
 
  • #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.
 
  • #7
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does exp(-2pi) = -1
 
  • #8
763
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does exp(-2pi) = -1
no it doesn't




wait no one give me any tips i think i can solve it
 
  • #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.
 
  • #10
matt grime
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no it doesn't




wait no one give me any tips i think i can solve it
solve what? exp(-2pi) is right. It just isn't anything nicer than that.
 
  • #11
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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)
 
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  • #12
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yes it does
 
  • #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

or

i^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.
 
  • #14
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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
 
  • #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?)
 
  • #16
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ha.

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

so what should be the answer? (certainly not -1)
 
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  • #17
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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???
 
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  • #18
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ha.

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

so what should be the answer? (certainly not -1)
yes, okay, cheers, i get the picture :rolleyes:



:biggrin: (<-- this is my favourite smiley)
 
  • #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!+...
 
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  • #20
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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...

nevermind

thank you very much for your time and help!!!
 
  • #21
matt grime
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well maybe it's a dumb idea and you just have to use derivatives.
 

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