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Math Help - Complex Fourier Series Question

  1. #1
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    Complex Fourier Series Question

    Hi
    Can someone explain to be how they got the line the arrow is pointing to?

    P.S
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  2. #2
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    Starting from the preceding line:

    \dfrac{1}{2\pi}\left(\dfrac{\left(-1\right)^n\left(e^{\pi} - e^{-\pi}\right)}{1-nj}\right)

    Move \left(e^{\pi} - e^{-\pi}\right) to the numerator of \dfrac{1}{2\pi} (commutative property of multiplication).

    \dfrac{\left(e^{\pi} - e^{-\pi}\right)}{2\pi}\left(\dfrac{\left(-1\right)^n}{1-nj}\right) = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj}

    Move (-1)^n to the far right of the expression (commutative property of multiplication).

    \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj} = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n

    Multiply by \dfrac{1+nj}{1+nj}.

    \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\dfrac{1+nj}{1+nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n

    I assume j = i = \sqrt{-1}, in which case j^2 = i^2 = -1.

    \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2\left(-1\right)}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 + n^2}\left(-1\right)^n
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  3. #3
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    Can you explain why would you multiply by \frac{1+nj}{1+nj}
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    Can you explain why would you multiply by \frac{1+nj}{1+nj}

    That's the way you do to find an expression of the form a+bi for a fraction of the

    form \frac{1}{\alpha+i\beta}\,,\,\,a,b,\alpha,\beta\in\  mathbb{R}.

    The only weird thing is why you people use the letter "j" to denote what the huge majority

    of mathematicians in the world denote by "i" = the imaginary unit, though I found that

    in a very old book...

    Tonio
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone explain to be how they got the line the arrow is pointing to?

    P.S
    All that is happening is that the top and bottom of the right hand side of the previous line are multiplied by (1+nj) which is the complex conjugate of the denominator of the previous line. This renders the denominator real and moves all the complex terms into the numerator. Other than that there is some minor rearrangement of the other terms in the previous expression.

    CB
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  6. #6
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    Quote Originally Posted by tonio View Post
    That's the way you do to find an expression of the form a+bi for a fraction of the

    form \frac{1}{\alpha+i\beta}\,,\,\,a,b,\alpha,\beta\in\  mathbb{R}.

    The only weird thing is why you people use the letter "j" to denote what the huge majority

    of mathematicians in the world denote by "i" = the imaginary unit, though I found that

    in a very old book...

    Tonio
    Who are the "you people" to whom you refer? In engineering the complex unit is universally referred to as "j"

    CB
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Who are the "you people" to whom you refer? In engineering the complex unit is universally referred to as "j"

    CB


    Oh, there you go! Engineers...****sigh**** ....and I bet they have an "excellent"

    explanation why they use that notation, don't they?

    Oh, well...

    Tonio
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by tonio View Post
    Oh, there you go! Engineers...****sigh**** ....and I bet they have an "excellent"

    explanation why they use that notation, don't they?

    Oh, well...

    Tonio
    To avoid confusion with the symbol for current, and pending further research I will bet that its use can be traced to Oliver Heaviside.

    CB
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