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anonymous

  • one year ago

Explaination wanted -> How can e^(-i*smallomega*t) be equal to cos(-smallomega*t) + i*sin(-smallomega*t) ? A big cookie to whomever explains it so I understand the math behind the conversion. I believe it has something to do with Euler's - however I am unable to piece it together myself :)

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  1. anonymous
    • one year ago
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  2. Astrophysics
    • one year ago
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    You can use taylor series

  3. zepdrix
    • one year ago
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    Ivan have you taken any Calculus? You can generate the relationship using integration, it's pretty neat :) Yah Taylor Series is good too.

  4. anonymous
    • one year ago
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    Yes - I just dont recall much of Taylors though. Working on it.. !

  5. Astrophysics
    • one year ago
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    \[e^t = \sum_{n=0}^{\infty} \frac{ t^n }{ n! }, t \in \mathbb{R} \] so you can add an imaginary number \[e^{i t} = \sum_{n=0}^{\infty} \frac{ (i t)^n }{ n! } = \sum_{n=0}^{\infty} \frac{ i^nt^n }{ n! }\] ooh funny thing is if you don't put space in latex between the i and what ever letter, it makes it go away, truly imaginary xD

  6. Astrophysics
    • one year ago
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    Yes you will have to note A = a+bi, I think zep can explain it a better way without Taylor

  7. zepdrix
    • one year ago
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    If you have an complex number: \(\large\rm z=a+b i\) We can write it in terms of trig, yes? So we'll call the magnitude of this complex number \(\large\rm |z|=r\). And we can write it in this form: \(\large\rm z=r(\cos\theta+i\sin\theta)\)

  8. zepdrix
    • one year ago
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    Differentiate this with respect to the argument, theta,\[\large\rm \frac{dz}{d \theta}=r(-\sin \theta+i \cos \theta)\]And then I guess this is where the tricky step comes in. We're going to factor a -1 out of each term:

  9. zepdrix
    • one year ago
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    Woops, I mean factor an \(\large\rm i\) out of each term :) \[\large\rm \frac{dz}{d \theta}=r i(i\sin \theta+\cos \theta)\]Recall that \(\large\rm -1=i\cdot i\)

  10. zepdrix
    • one year ago
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    So what we've shown is that:\[\large\rm \frac{dz}{d \theta}= i r(\color{orangered}{\cos \theta+i \sin \theta})\]Ooo look we've got our z showing up!\[\large\rm \frac{dz}{d \theta}= i r(\color{orangered}{z})\]

  11. zepdrix
    • one year ago
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    Woops woops, little mistake, my bad. the r is part of my z

  12. zepdrix
    • one year ago
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    \[\large\rm \frac{dz}{d \theta}= i (\color{orangered}{z})\]

  13. anonymous
    • one year ago
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    I'm still here - just processing the information and looking everything up and cross referencing in my formula book ... !

  14. zepdrix
    • one year ago
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    Then separation of variables and integrating gives us:\[\large\rm \int\limits \frac{dz}{z}=\int\limits i d\theta\]Leading to,\[\large\rm \ln z=i \theta+c\]

  15. IrishBoy123
    • one year ago
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    brill astro!! historically, Euler found this from comparing Taylor series

  16. zepdrix
    • one year ago
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    Exponentiating gives us:\[\large\rm z=e^{i \theta+c}=e^{i \theta}e^c\] \[\Large\rm z=e^c \color{orangered}{e^{i \theta}}=r(\color{orangered}{\cos \theta+i \sin \theta})\]

  17. Astrophysics
    • one year ago
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    Thanks irishboy xD Nice one zep!

  18. anonymous
    • one year ago
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    Okay. Working on the trig method. In order to understand (and recall old lessons) I am starting from scratch given from the examples above. Am I understanding this correctly thus far?

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  19. anonymous
    • one year ago
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    Thank you so much! I finally got it, zapdrix! Thanks to everyone else as well for your input! :)

  20. anonymous
    • one year ago
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    If anything - the only thing throwing me a bit off is the e^c and r. From the example e^(-iw). Are those just constants we can.... throw away? ooh.. because they are the same we can just "remove" them! Right? No? @zepdrix @Astrophysics

  21. Astrophysics
    • one year ago
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    Yeah it's just a constant, you can leave it alone, or just add it to your calculations after

  22. anonymous
    • one year ago
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    Thanks

  23. zepdrix
    • one year ago
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    Yes, good eye :) It turns out that the \(\large\rm e^c\) is the same \(\large\rm r\) that we started with. I was having trouble illustrating that though.

  24. zepdrix
    • one year ago
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    Ah yes, good with the trig. Sorry I kinda skipped over that at the start >.<

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