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Calculus is false.

  1. Jun 6, 2015 #1
    • OP warned to not post "noise" again
    Calculus works for linear equations of the 1st order because it is simple geometry, and that cannot be faulted; one does not need calculus for this. Now take a second order equation, the simple function: y = x^2. From rudimentaries, detay y / deltax = 2x + delta x. That is the EXACT equation for the slope. One can make delta x as small as one wants, however the slope changes. Bringing delta x to infinity, and delta y follows, is the manner in which, all calculations are consistent, but not right. Infinity, as is well known, does not exist, for it is a mathematical trick for some aspects of mathematics to exist. Calculus is a manner in which, to normalise calculations throughout the mathematical and engineering field. Comments please.
     
  2. jcsd
  3. Jun 6, 2015 #2

    micromass

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    Are you familiar with epsilon-delta definitions which prove the correctness of the equations?

    Are you aware that there is ton of observational evidence for calculus?
     
  4. Jun 6, 2015 #3
    At least you could try to phrase your ideas coherently if you're trying to refute such a successful mathematical theory...
     
  5. Jun 6, 2015 #4

    OldEngr63

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    Wow, Cruz! That is asking a lot. How can one hope phrase their ideas coherent when doing something fundamentally incoherent? It took a lot of guts for the OP just to put his thoughts up; it is too much to ask that they make sense too!
     
  6. Jun 6, 2015 #5

    phion

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    In before lock.
     
  7. Jun 6, 2015 #6

    phinds

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    :DD
     
  8. Jun 6, 2015 #7
    I meant grammatical coherence :D
     
  9. Jun 6, 2015 #8

    Mark44

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    It's locked now...

    To the OP: A lot of what you wrote was incorrect; in particular, your formulations for ##\frac{\Delta y} {\Delta x}## and your idea that we take ##\Delta x## to ##\infty##. Please do some research before making absurd statements about calculus "being false."
     
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