Posers and Puzzles

Posers and Puzzles

  1. Joined
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    17 Jul '06 00:261 edit
    Why does the natural log of -1 equal Pi(i)? Neither my calculus nor my physics teacher knew the answer to this (I don't know it either, that's why I'm posting the question).
  2. Standard memberXanthosNZ
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    17 Jul '06 00:30
    This relationship is normally stated as e^i*pi = -1.

    You can find a good basic explanation here:
    http://www.math.toronto.edu/mathnet/questionCorner/epii.html
  3. Joined
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    17 Jul '06 01:22
    Thanks a lot. I kind of was expecting you to be there with the answer.
  4. Joined
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    17 Jul '06 10:306 edits
    Originally posted by Ramiri15
    Why does the natural log of -1 equal Pi(i)? Neither my calculus nor my physics teacher knew the answer to this (I don't know it either, that's why I'm posting the question).
    The real exponential function is a map from the set of real numbers to the set of positive real numbers

    x maps to e^x

    and it has an inverse, called the logarithm

    x maps to log(x).

    The exponential function is defined by a power series

    e^x = sum from k=0 to infinity of x^k/k!

    and this also makes sense for complex numbers z instead of x. So we have z going to e^z.

    The question now is does this new "complex" exponential function admit an inverse that extends the real logarithm mentioned above.

    Well, actually it does. Let's call this new complex logarithm log(z). One has to be careful with what z values we put in (the "domain" of the function -- it has to be restricted in some way -- see below). For us we shall simply define

    log(z) = log(|z|) + i*arg(z)

    for non-zero complex numbers z, and insist that, say, the argument arg(z) lies between -pi/2 and 3*pi/2. This is what I meant by restricting the domain -- we're excluding the negative imaginary axis. We have chosen this definition simply because it *is* an inverse for e^z, and extends the original real log(x). The reason for cutting out the negative imaginary axis is because if we didn't the log(z) function would jump suddenly as we went across it (the imaginary part suddenly changes from -pi/2 to 3*pi/2) and this is a bad thing -- we like our functions to be continuous.

    There are of course many other inverses -- add on fixed multiples of 2*pi*i to the above formula. These other inverses are called "branches" of the logarithm.

    If we plug in z=-1 into the definition for log(z) above we get

    log(-1) = log(1) + i*pi

    so that's i*pi, as you wanted. Another eg. would be z=1+i with

    log(1+i) = log(root(2)) + i*pi/4 = log(2)/2 + i*pi/4.
  5. Shanghai
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    17 Jul '06 11:40
    Originally posted by XanthosNZ
    This relationship is normally stated as e^i*pi = -1.

    You can find a good basic explanation here:
    http://www.math.toronto.edu/mathnet/questionCorner/epii.html
    i like the form e^i*pi+1=0 as it includes pi, e, i, 1 and 0. 5 very important constants
  6. London
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    17 Jul '06 15:17
    Originally posted by deriver69
    i like the form e^i*pi+1=0 as it includes pi, e, i, 1 and 0. 5 very important constants
    I knew a mathematics professor who sported a baseball cap that displayed the formula that way around. "The five most important numbers in mathematics" he said.
  7. Joined
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    17 Jul '06 21:57
    i'm pretty sure the definition of e = lim (n->inf.) (1+1/n)^n
    the formula e^x = x^k/k! is the taylor expansion of e^x around 0.
  8. Joined
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    18 Jul '06 09:362 edits
    Originally posted by aginis
    i'm pretty sure the definition of e = lim (n->inf.) (1+1/n)^n
    the formula e^x = x^k/k! is the taylor expansion of e^x around 0.
    Well, you can do it either way, and then prove the other result.

    It's much cleaner (and easier) to define everything via power series though (exp(x), sin(x), cos(x) etc.) because they have a good theory (infinitely differentiable, multiply them etc.). I mean you know that the differential equation exp'=exp holds true immediately from the power series. You can then easily prove exp(x+y)=exp(x)exp(y) etc etc...

    I prefer to think of the (1+x/n)^n limit as a curiosity - it's less useful.
  9. Account suspended
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    18 Jul '06 14:44
    OK now tell me why 1!=1, but 0! also =1. I just never got it.
  10. Shanghai
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    18 Jul '06 15:05
    One answer is because it works, but if you think of factorials as being the number of ways of arranging that many objects, there is only one way of arranging zero objects.
  11. Joined
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    18 Jul '06 15:58
    Originally posted by GinRose
    OK now tell me why 1!=1, but 0! also =1. I just never got it.
    It's just for convenience.

    Like for example you'd like the formula

    (n choose r) = n! / (r! * (n-r)!)

    to make sense when r=0 or r=n.
  12. Standard membergenius
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    19 Jul '06 11:38
    Originally posted by GinRose
    OK now tell me why 1!=1, but 0! also =1. I just never got it.
    the formula for factorial is n!=(n-1)!*n by definition.

    1!=1 (obviously)

    solving formula for 1!,
    1!=(1-1)!*1
    1!=0!*1=1

    =>0!=1

    we can also use the gamma function to show it,

    http://en.wikipedia.org/wiki/Gamma_function

    but that's kinda complicated...
  13. Shanghai
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    19 Jul '06 12:12
    In essence 0! is a special case. Maths works far better if 0!=1. The "because it is" answer occasionally has to be used in maths.
  14. Standard memberPalynka
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    19 Jul '06 12:24
    Originally posted by deriver69
    In essence 0! is a special case. Maths works far better if 0!=1. The "because it is" answer occasionally has to be used in maths.
    It's exactly the same as x^0=1. An empty product equals 1.
  15. Standard memberBowmann
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    19 Jul '06 15:52
    (0!+0!+0!+0!+0!)!
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