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Posers and Puzzles

Posers and Puzzles

  1. 10 Feb '05 20:28
    given f(x)=(-1)^x
    evaluate f'(x)
  2. 10 Feb '05 20:33
    unrelated, but posted for similer reasons:

    in this thread:
    http://www.redhotpawn.com/board/showthread.php?threadid=15717
    we determined that 0\0 can take on any one real value for any one situation.

    but can it take on an unreal value?

    f(x)=x/0

    evaluate.
  3. Standard member DoctorScribbles
    BWA Soldier
    10 Feb '05 23:34 / 1 edit
    Originally posted by fearlessleader
    unrelated, but posted for similer reasons:

    in this thread:
    http://www.redhotpawn.com/board/showthread.php?threadid=15717
    we determined that 0\0 can take on any one real value for any one situation.

    but can it take on an unreal value?

    f(x)=x/0

    evaluate.
    What is your definition of the operation denoted by / ?

    If it represents division, what is your definition of division?
    A typical definition of division is this:

    For real numbers w, x, y, and z,

    w/x = y/z if and only if z*w = x*y


    Now, you define your function as f(x) = f(x)/1 = x/0. From this and the definition of division, it follows that you also have 0*f(x) = 1*x. By the rules of multiplication, it follows that you also have 0 = x.

    That is, your function is only defined at one point, namely 0. For any other point, f cannot be defined, for if it were, its own definition would be a contradiction.

    But what use is a function whose domain is a single point?
  4. Donation Acolyte
    Now With Added BA
    11 Feb '05 08:56
    Originally posted by fearlessleader
    given f(x)=(-1)^x
    evaluate f'(x)
    f'(x) = log(-1)*(-1)^x, I suppose.
  5. 13 Feb '05 14:30 / 1 edit
    Originally posted by fearlessleader
    given f(x)=(-1)^x
    evaluate f'(x)
    Is it an analytic complex valued function? If so its derivative is defined everywhere in the complex x-plane. And therefore it should be quite well defined and continuous for all real values of x too.