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Euler equations

  1. Apr 18, 2007 #1
    Hi i am reading about signal and systems course . What i want to prove is not a problem that i have to solve is something that the books take for granted and i want to prove it so i ll be able at exams to reprove so i wont have to remember it, (if u dont believe me i can give u the course's pdf that states what i say. Btw i posted here cause i dont know where else i should post problems like that
    1. The problem statement, all variables and given/known data

    I want to prove why
    e^(-j*(theta))=cos(theta)-jsin(theta) (1)
    i know that e^j(theta)=cos(theta)+jsin(theta) Why the minus sign affects the sin and not the cos
    also i want to find out why cos(theta)=1/2[e^(j*theta)+e^(-j*theta)] (2)
    My teachers book say that these things exist. It proves number (2) using Eulers equation.
    I know that the angle theta can be found from theta=1/sin(y/(sqrt(x^2+y^2))
    theta:is the angle of the complex number

    2. Relevant equations

    3. The attempt at a solution
    I have tried to solve the euler equation like that
    I have tried to convert the j(sin(theta)) to something that has inside e^j*theta or something like that but i have failed :(
    I am really very week converting exponential numbers to equations that include cosins and sins. If u have a good book for that that i can read it online plz suggest it to me
  2. jcsd
  3. Apr 18, 2007 #2
    Euler equation:

    First of all:
    e^(-j*(theta)) = e^(j*(-theta)) =... <--- what is the result here using the Euler equation?

    Secondly, you will need to show
    sin(-x) = -sin(x) and
    cos(-x) = cos(x)
    In order to show that, use the definition of sin(x) and cos(x), see here

    For getting equation (2):
    Calculate e^(j*theta)+e^(-j*theta) using
    equation (1) e^(-j*(theta))=cos(theta)-jsin(theta) and
    the Euler equation e^j(theta)=cos(theta)+jsin(theta)
    Last edited: Apr 18, 2007
  4. Apr 18, 2007 #3


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    You can, by the way, derive the Euler equation
    [tex]e^{i\theta}= cos(\theta)+ i sin(\theta)[/tex]
    By looking at the Taylor's series expansions for the functions

    [tex]e^x= 1+ x+ \frac{x^2}{2}+ \frac{x^3}{3!}+ \cdot\cdot\cdot +\frac{x^n}{n!}+\cdot\cdot\cdot[/tex]
    Replace x by [itex]i\theta[/itex] and remember that i2= -1, i3= -i, i4= 1, i5= -i, etc. You get
    [tex]e^{i\theta}= 1+ i\theta- \frac{\theta^2}{2!}- i\frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+ \cdot\cdot\cdot[/tex]
    Separate that into "real" and "imaginary" parts:
    [tex]e^{i\theta}= (1-\frac{\theta^2}{2!}+ \frac{\theta^4}{4!}- \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n}}{(2n)!}+\cdot\cdot\cdot)+ i(\theta- \frac{\theta^3}{3!}+ \frac{\theta^5}{5!}+ \cdot\cdot\cdot+ (-1)^n\frac{\theta^{2n+1}}{(2n+1)!)}+\cdot\cdot\cdot)[/tex]
    and compare those to the Taylor's series for sine and cosine of [itex]\theta[/itex]:
    [tex]sin(\theta)= \theta- \frac{\theta}{3!}+ \frac{\theta^5}{5!}+\cdot\cdot\cdot+ (-1)^{2n}\frac{\theta^{2n+1}}{(2n+1)!}+ \cdot\cdot\cdot[/tex]
    [tex]cos(\theta)= 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}+\cdot\cdot\cdot+ (-1)^n \frac{\theta^{2n}}{(2n)!}+ \cdot\cdot\cdot[/tex]

    Sorry, but I just cannot force myself to write "j" instead of "i"!
  5. Apr 18, 2007 #4


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    It seems unnatural in some way doesn't it?
  6. Apr 18, 2007 #5
    e^(j*(-theta))=cos(-theta)+jsin(-theta) and now i have to prove the next step u mention
    I think that this image found in the site u mentioned
    show that sin(-x) =(...lot of stuff i dont type)(-x)^(2*n+1) (2*n+1 is odd which means that the minus sign gets out of the parenthesis = -........(x)^(2*n+1)
    for the cosin now
    x^2n which means that (-x)^2*n=(x)^2n
    this is the way i have thought to prove that sin(-x)=-sinx and cos(-x)=cos(x)
    Am i correct?

    Ok thx a lot let me try it
    e^(j*theta)+e^(-j*theta)=cos(theta)+jsin(theta)+cos(theta)-jsin(theta) =2*cos(theta)=>
    cos(theta)=[e^(j*theta)+e^(-j*theta)]/2 so i think i have done it with your help guys..plz fix everything that i have done wrongly
  7. Apr 18, 2007 #6
    The way u have just mentioned is for remembering to prove why
  8. Apr 18, 2007 #7
    dervast, your calculations are correct.
    As an exercise, you could try expressing sin(theta) in terms of the e-functions.
  9. Apr 19, 2007 #8
    Ok let me try it
  10. Apr 19, 2007 #9
    That's correct.
  11. Apr 19, 2007 #10

    Gib Z

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    Yes it does! I thought I would be stupid in thinking that as well, and perhaps I only thought so because I was accustomed to that notation, but now that I know people agree with me I feel much better :)

    Which physicist introduced j and disrespected Euler!
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