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Posers and Puzzles

Posers and Puzzles

  1. 21 Jan '05 21:50
    0=0
    (x+y)(x-y)=(x+y)(x-y)
    (x+y)(x-y)=x^2-y^2
    (x-y)(1+dy/dx)+(x+y)(1-dy/dx)=2x-2y(dy/dx)
    x+x(dy/dx)-y-y(dy/dx)+x-x(dy/dx)+y-y(dy/dx)=2x-2y(dy/dx)
    2x-2y(dy/dx)=2x-2y(dy/dx)
    0=0

    damn!
  2. Standard member neight
    Cheese log/Beef log
    21 Jan '05 22:29
    what exactly are you trying to show?
  3. Standard member TheMaster37
    Kupikupopo!
    23 Jan '05 14:15
    That equations don't all hold with derivatives?
  4. 25 Jan '05 15:08 / 1 edit
    clearly one can write an "equation" in which all variables will cancle, but:

    can you make an equation in which all variables will cancle, but of which one can find the deriviative?

    i haven't learned integrals (anit-derivatives) yet, but are they any use here?

    it's a bit complex what i'm looking for here, so just show anything you find.
  5. 25 Jan '05 18:26
    That would be like saying
    x=x
    and then trying to find the rate of change of x
  6. Standard member XanthosNZ
    Cancerous Bus Crash
    25 Jan '05 18:28
    Originally posted by iamatiger
    That would be like saying
    x=x
    and then trying to find the rate of change of x
    Is it x?
  7. 26 Jan '05 02:27
    Originally posted by fearlessleader
    clearly one can write an "equation" in which all variables will cancle, but:

    can you make an equation in which all variables will cancle, but of which one can find the deriviative?

    i haven't learned integrals (anit-derivatives) yet, but are they any use here?

    it's a bit complex what i'm looking for here, so just show anything you find.
    If all the variables cancel, you do not have an equation, only a statement. yx=yx is not an equation because there is no parameters for the variables to fall into, an so belong to an endless set, and so one cannot find the derivative for the equation as the derivative is a rate of change for a varible. No parameters = no definition of the varibles change.