Mathematical theory of everything

Discussion in 'Pseudoscience' started by Waiter_2001, Dec 31, 2015.

  1. Waiter_2001 Registered Senior Member

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    • Please do not post nonsense to the Science subforums.
    Just as computers cannot divide by zero here is another calculation computers cannot fully grasp:

    0+1+2+3+4+5+6+7+8+9=45

    ((((45*45)+1)/46)-45)

    Again it is a division by zero however it uses ALL operators: *+/-

    The answer must be discovered manually (long division) because although it IS a recurring answer the computer runs short...
     
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  3. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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  5. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    You don't have a clue as to what you're talking about, do you?
     
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  7. rpenner Fully Wired Registered Senior Member

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    Untrue.

    What is sometimes possible is obtaining finite limits of two terms that have limiting values of zero in analysis.

    \(\lim_{h\to 0} \frac{\sin (x + h) - \sin (x)}{h} \\ \quad = \lim_{h\to 0} \frac{\sin (h) \cos(x) + ( \cos(h) - 1 ) \sin (x) }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{\sin (h) }{h} + \sin (x) \lim_{h\to 0} \frac{\cos(h) - 1 }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^k h^{2k + 1} }{(2k+1)!}}{h} + \sin (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^{k+1} h^{2k + 2} }{(2k+2)!}}{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ h }{h} + \sin (x) \lim_{h\to 0} \frac{ - \frac{ h^2 }{2} }{h} \\ \quad = \cos (x) \lim_{h\to 0} 1 + \sin (x) \lim_{h\to 0} \left( - \frac{ h }{2} \right) \\ \quad = \cos (x) \times 1 + \sin (x) \times 0 \\ \quad = \cos(x)\)
     
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  8. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    Uh... I still didn't see the square root of minus one in there.
     
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  9. rpenner Fully Wired Registered Senior Member

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    4,833
    Because I was doing real analysis, not complex arithmetic.

    If you want you can define \(\sin x = \frac{e^{ix} - e^{-ix}}{2i} , \; \cos x = \frac{e^{ix} + e^{-ix}}{2}\) and get:
    \(\lim_{h\to 0} \frac{\sin (x +h) - \sin (x)}{h} \\ \quad = \lim_{h\to 0} \frac{e^{ix} \left( e^{ih} - 1 \right) - e^{-ix} \left( e^{-ih} - 1 \right) }{2 i h} \\ \quad = \frac{e^{ix}}{2} \lim_{h\to 0} \frac{e^{ih} - 1}{i h} - \frac{e^{-ix}}{2} \lim_{h\to 0} \frac{e^{-ih} - 1}{i h} \\ \quad = \frac{e^{ix}}{2} \lim_{h\to 0} \frac{\sum_{k=0}^{\infty} \frac{ (ih)^{k+1} }{(k+1)!}}{i h} - \frac{e^{-ix}}{2} \lim_{h\to 0} \frac{\sum_{k=0}^{\infty} \frac{ (-1)^{k+1}(ih)^{k+1} }{(k+1)!}}{i h} \\ \quad = \frac{e^{ix}}{2} \lim_{h\to 0} \sum_{k=0}^{\infty} \frac{(ih)^{k} }{(k+1)!} - \frac{e^{-ix}}{2} \lim_{h\to 0} \sum_{k=0}^{\infty} \frac{ (-1)^{k+1}(ih)^{k} }{(k+1)!} \\ \quad = \frac{e^{ix}}{2} \lim_{h\to 0} \frac{1 }{1!} - \frac{e^{-ix}}{2} \lim_{h\to 0} \frac{ -1 }{1!} \\ \quad = \frac{e^{ix}}{2} \times 1 - \frac{e^{-ix}}{2} \times (-1) \\ \quad = \frac{e^{ix} + e^{-ix}}{2} \\ \quad = \cos (x)\)

    But nowhere was division by zero accomplished.
     
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  10. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    \(\frac{1}{\frac{1}{0}}=0\)

    Yes well, I specifically said complex numbers and stated my ignorance about computers...
     
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  11. rpenner Fully Wired Registered Senior Member

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    \(\frac{1}{0}\) is not a number, let alone a complex number.

    If \(x \equiv \frac{1}{0}\) was a complex number, it would obey the axioms of complex numbers. Thus as with all complex numbers, \( x - x = 0 \) therefore \( x - x = \frac{1}{x} \) therefore \( 0 = x^2 - x^2 = 1\) which proves that \(\frac{1}{0}\) is not a complex number.

    You can't "fix the problem of division by zero" by introducing a reciprocal of zero. That leads not to better mathematics, but to inconsistent mathematics which is junk.
     
    Last edited: Dec 31, 2015
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  12. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    http://www.wolframalpha.com/input/?i=1/(1/0)
     
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  13. rpenner Fully Wired Registered Senior Member

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    4,833
    Means nothing relevant.

    Specifically it means: \(\lim_{h\to 0} \frac{1}{\frac{1}{h}} = \lim_{h\to 0} h = 0\) which is a statement in analysis, not complex arithmetic.

    You can demonstrate that Wolfram is doing analysis and not arithmetic by asking it the logical question:

    http://www.wolframalpha.com/input/?i=(1/0)/(1/0)

    Which means \(\lim_{a,b \to 0} \frac{ \frac{1}{a} }{ \frac{1}{b} } = \lim_{a,b \to 0} \frac{b}{a} \) which is undefined because it matters how you go about approaching zero.

    \(\lim_{a \to 0} \lim_{b \to 0} \frac{b}{a} = \lim_{a \to 0} 0 = 0\) but \(\lim_{b \to 0} \lim_{a \to 0} \frac{b}{a} = \lim_{b \to 0} \frac{b}{0} \) doesn't exist.

    Likewise if there is a relation between how a and b simultaneously approach 0, all you recover is information about how a and b are constrained \( \left. \lim_{a,b \to 0} \frac{b}{a} \right| _{F(a,b) = F(0,0)} = - \frac{F_a(0,0)}{F_b(0,0)} ; F_b(0,0) \neq 0\), where \(F_a\) and \(F_b\) are partial derivatives
     
    Last edited: Dec 31, 2015
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  14. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    No. There's complex infinity, not lim.

    Stop it already.

    When have I ever deliberately antagonized you?

    :EDIT:

    You're arguing with Mathematica....
     
    Last edited: Dec 31, 2015
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  15. rpenner Fully Wired Registered Senior Member

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    4,833
    Complex infinity is not a number -- it's a placeholder for a class of undefined results where the magnitude exceeds that of any number. Here's what Wolfram's Mathworld page on it says:

    http://mathworld.wolfram.com/ComplexInfinity.html

    http://reference.wolfram.com/language/ref/ComplexInfinity.html

    http://reference.wolfram.com/language/ref/DirectedInfinity.html (Gives examples of uses for an infinite number whose complex argument is known.)

    http://reference.wolfram.com/language/tutorial/IndeterminateAndInfiniteResults.html (Gives examples of why ComplexInfinity is a special type of Indeterminate result. Also starts out explaining that when you type a division by zero like 0/0 that Mathematica assumes a statement in analysis is meant. )

    Telling you that you are misguiding others because you are misguided is not deliberately antagonizing you — it's giving you a choice to act like an enlightened human being and ask questions.

    No, I'm arguing that your reliance on Mathematica-based tools outside your sphere of familiarity is no substitute for an education in the fundamentals of analysis and complex numbers.
     
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  16. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    Be concise. I don't try to bombard others.
     
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  17. rpenner Fully Wired Registered Senior Member

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    http://www.wolframalpha.com/input/?i=1/ComplexInfinity ( A statement in analysis -- \(\lim_{x \to +\infty} \frac{1}{e^{i \theta} x} = 0\) no matter what value of the real parameter \(\theta\). )

    http://www.wolframalpha.com/input/?i=ComplexInfinity - ComplexInfinity (But ComplexInfinity, in addition to being determined in magnitude only and not phase, is not a number and one cannot subtract it from itself to get a quantity which is determined in magnitude or phase. This is consistent with analysis, not arithmetic.)

    http://www.wolframalpha.com/input/?i=ComplexInfinity * 0 (Nor does it share properties of reciprocals. The answer is undefined, not 1. This is consistent with analysis, not arithmetic.)
     
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  18. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    Three links? I think I've had enough.

    In nine hours the new year will be ushered in. So, I'd like to celebrate for now.
     
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  19. Waiter_2001 Registered Senior Member

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    Apologies! It should be:

    ((((45*45)+1)/46)-1)

    As to divide by zero you add 1 before the division, 1 after it and then -1 (because 1 has been added to each side.)
     
  20. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Still don't know what you're talking about, do you:

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  21. Waiter_2001 Registered Senior Member

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    As I wrote: the computer falls short. The true fraction is much greater! Have you tried this calculation manually? Nice picture beer w/straw!
     
  22. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Like I said: you don't know what you're talking about.
    The only reason it "falls short" is because the display is set to 3 decimal places - in other words it doesn't "fall short" at all.
    I note that you've completely ignored the failure of your first claim: 0+1+2+3+4+5+6+7+8+9=45.
     
  23. Waiter_2001 Registered Senior Member

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    459
    0+1=1
    1+2=3
    3+3=6
    6+4=10
    10+5=15
    15+6=21
    21+7=28
    28+8=36
    36+9=45

    The calculator I use is not set to a shorter number of decimal places: some calculations produce more decimal places but this calculation does not.
     

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