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Tricky equation

  1. Dec 31, 2007 #1
    I was wondering what series approximation I can use to approximate y:

    y=(1-(dx/dy)^2)^1/2

    when dx/dy is not trigonometric, and contains the derivative of a floor function
     
  2. jcsd
  3. Dec 31, 2007 #2
    Hi,
    the floor function is notoriously discontinuous. What do you mean by its derivative?
     
  4. Dec 31, 2007 #3
    y has properties of sin fuction i guess we solve x in terms of y

    x=1/2*(abs(y)(1-y^2)^0.5 +inverse sin(abs(y)))

    you can also get it solved for y
     
    Last edited: Dec 31, 2007
  5. Jan 1, 2008 #4
    okay well an approximation
     
  6. Jan 1, 2008 #5
    what do you meant by that
     
  7. Jan 6, 2008 #6

    epenguin

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    You can transform that into dx/dy = sqrt(1 - y^2) and integrate wrt y to obtain

    x = -.5 {y*srt(y^2 - 1) - ln ABS(y + sqrt(y^2 -1)} + a constant

    give or take a + or - :biggrin:

    which does not mean the solution of sadhu is not right too.

    Did this d.e. emerge from any 'real' problem?
     
  8. Jan 6, 2008 #7

    olgranpappy

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    how about [itex]\sum_n \delta(x-n)[/itex].
     
  9. Jan 6, 2008 #8

    Gib Z

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    I don't exactly know what you mean by that :( But I doubt that's a derivative. Differentiability implies continuity. The floor function, being discontinuous at an infinite number of points, is not differentiable.
     
  10. Jan 7, 2008 #9

    HallsofIvy

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    That uses the delta "function". It's not a function in the true sense, but a distribution or "generalized function". And, of course, the differentiation is in the sense of distributions. Distributions do not have to be continuous in order to be differentiable. (In fact, I am not sure that "continuous" is defined for distributions!)
     
  11. Jan 7, 2008 #10

    olgranpappy

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    [itex]\delta[/itex] is Dirac's delta function and the [itex]n[/itex] are positive integers. You can, of course, check for yourself that
    [itex]\int_0^x dy \sum_n \delta(y-n)=Floor[y][/itex]

    Lah dee dah, lah dee dah, I'm not worried about that...

    But, if it's not differentiable then the derivative doesn't exist. So how did I just write the derivative down if it doesn't exist?... I'd rather not let the fact that something doesn't exist stop me from using it to solve a problem.

    Mathematicians are so cute...
     
  12. Jan 8, 2008 #11

    Gib Z

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    I was never stopping you from solving the problem. Just don't call it a "derivative". Call it something else. You didn't write down the derivative, you wrote down something that satisfies a nice equation in this same manner a derivative does for continuous functions. That doesn't mean it is a derivative.

    Physicists are just sloppy. There is nothing in mathematics that stops the physics being done, but physicists are too lazy to justify their mathematics. Every working physicist I have asked admits that their sloppy mathematics leads them to errors, why not correct the problem?
     
  13. Jan 8, 2008 #12

    olgranpappy

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  14. Jan 8, 2008 #13

    morphism

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    It's simple: you didn't. :rolleyes:
     
  15. Jan 8, 2008 #14

    olgranpappy

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    oh, girls, stop being so silly.
     
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