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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 13:46
    If i^2= (-1), what is i^i? i to the ith power.
  2. Standard member Palynka
    Upward Spiral
    27 Oct '07 14:35
    i^i = exp(-PI/2+2.n.PI)
  3. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 15:08 / 2 edits
    What is n in that equation?
    And how does PI enter into this? interesting.
    Also, what does exp mean? exponent? if so, what exponent?
  4. Standard member TheMaster37
    Kupikupopo!
    27 Oct '07 15:24 / 1 edit
    Originally posted by sonhouse
    What is n in that equation?
    And how does PI enter into this? interesting.
    Also, what does exp mean? exponent? if so, what exponent?
    He used another way of writing i:

    i = e ^ (PI * i / 2)

    Since 1 = e ^ (2*PI*i) you can add extra factors of e ^ (2*PI*i), hence the n.

    So i^i = e ^(-PI/2) mod e ^ (2*PI)
  5. Subscriber AThousandYoung
    Proud Boys Beware
    27 Oct '07 16:15
    Originally posted by TheMaster37
    He used another way of writing i:

    i = e ^ (PI * i / 2)

    Since 1 = e ^ (2*PI*i) you can add extra factors of e ^ (2*PI*i), hence the n.

    So i^i = e ^(-PI/2) mod e ^ (2*PI)
    i is part of the definition of i? That doesn't make sense.
  6. 27 Oct '07 17:26 / 1 edit
    Originally posted by AThousandYoung
    i is part of the definition of i? That doesn't make sense.
    It's not part of the defiition of i, but it's an extremely useful general result:

    cos(theta) + i sin(theta) = e^(i theta)

    He's using the case where theta = PI/2. Then the left-hand side is equal to i.
  7. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 17:29 / 1 edit
    Originally posted by mtthw
    It's not part of the defiition of i, but it's an extremely useful general result:

    cos(theta) + i sin(theta) = e^(i theta)

    He's using the case where theta = PI/2. Then the left-hand side is equal to i.
    q
    I'm getting PI i'd thinking about this:
    So theta is just any angle and he made that angle PI/2.
    You said 'extremely useful'. can you show some samples of the usefulness of this? Also, are there other identities like this that can be useful?
    Other angles of theta, or equivalences?
    what about definitions in radians? Are there any generalizations to be made there?
  8. Standard member Palynka
    Upward Spiral
    27 Oct '07 17:33 / 1 edit
    You just use the fact that:

    e^(b.i)= cos(b)+i. sin(b) (1)

    This is equal to i (look at the RHS) when b = PI/2 - (or +) 2.n.PI with n being any integer.

    i^i = e^[(b.i)^i]= e^(-b) = e^ (PI/2 - 2.n.PI).i.i = e ^ (-PI/2 + 2.n.PI)
  9. Standard member Palynka
    Upward Spiral
    27 Oct '07 17:35
    Originally posted by mtthw
    It's not part of the defiition of i, but it's an extremely useful general result
    Yep, someone that needs to deal with dynamic optimization should be familiar with it. It's called de Moivre's theorem.
  10. 27 Oct '07 17:45 / 1 edit
    Originally posted by Palynka
    Yep, someone that needs to deal with dynamic optimization should be familiar with it. It's called de Moivre's theorem.
    Or almost any serious applied mathematics. It crops up in wave theory, quantum mechanics, fluid dynamics, instability calculations...
  11. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 17:58
    Originally posted by mtthw
    Or almost any serious applied mathematics. It crops up in wave theory, quantum mechanics, fluid dynamics, instability calculations...
    Nice. Thanks for that. I Goog'd De Moivre and found this neat wiki:
    http://en.wikipedia.org/wiki/De_Moivre's_formula
    This piece goes into it in some detail.
    BTW, my present read is 'The Riemann Hypothesis"
    by Karl Sabbagh. Anyone read this? Great read if you are into maths.
    I think my next maths book will be 'the joy of i'. I think there is a book with something like that title.
  12. 27 Oct '07 18:03 / 1 edit
    Calculating with i ( as sqrt(-1) ) is only an extension of the real number system. What you can do with R you can do with C.

    ...and more:

    With the complex number system (C) you can calculate square root of any number (including complex numbers), not only positive numbers. You can calulate arcsin with any number (including complex numbers), not only numbers between -1 and 1. You can take logaritm of any number (including complex numbers), not only positive numbers. And so on.

    With i = sqrt(-1) you can calculate i^i, sqrt (i), log (i) with 'no problem'. But you cannot calculate i/0 but who can?

    But one pecularity of the complex number is that no complex number is larger or smaller than any other number. Another one is that negative infinity is the same as positive infinity.

    Complex numbers are fun!
  13. Standard member Palynka
    Upward Spiral
    27 Oct '07 18:06 / 1 edit
    Originally posted by sonhouse
    Nice. Thanks for that. I Goog'd De Moivre and found this neat wiki:
    http://en.wikipedia.org/wiki/De_Moivre's_formula
    This piece goes into it in some detail.
    BTW, my present read is 'The Riemann Hypothesis"
    by Karl Sabbagh. Anyone read this? Great read if you are into maths.
    I think my next maths book will be 'the joy of i'. I think there is a book with something like that title.
    I love the fact that -e^(pi.i)=1. (also shown by de Moivre's formula)

    It's one of those things that show you how mathematics can be so beautiful and understand why some cultures saw glimpses of the divine in them. Three of the most important mathematical constants combined into such a simple result.
  14. Subscriber sonhouse
    Fast and Curious
    27 Oct '07 18:09
    But one peculiarity of the complex number is that no complex number is larger or smaller than any other number. Another one is that negative infinity is the same as positive infinity.

    I follow everything but that one. It sounds like you are saying all complex numbers are equal?? And that (+inf) + (-inf)=0? and (+inf) * (-inf)= (-inf)^2?
  15. Standard member wolfgang59
    Infidel
    27 Oct '07 18:51
    Originally posted by FabianFnas
    Calculating with i ( as sqrt(-1) ) is only an extension of the real number system. What you can do with R you can do with C.

    ...and more:

    With the complex number system (C) you can calculate square root of any number (including complex numbers), not only positive numbers. You can calulate arcsin with any number (including complex numbers), not only ...[text shortened]... er one is that negative infinity is the same as positive infinity.

    Complex numbers are fun!
    I didnt know that!

    But isnt it a matter of definition?

    What if I define the 'size' of a complex number C where C = R +iI as

    sqrt(I^2 + R^2) ... does that have a use?