# Mathematical theory of everything

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

1. ### Waiter_2001Registered Senior Member

Messages:
459
• 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...

sunshaker likes this.

3. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
sunshaker likes this.

5. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,759

You don't have a clue as to what you're talking about, do you?

sunshaker likes this.

7. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Untrue.

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

$\lim_{h\to 0} \frac{\sin (x + h) - \sin (x)}{h} \\ \quad = \lim_{h\to 0} \frac{\sin (h) \cos(x) + ( \cos(h) - 1 ) \sin (x) }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{\sin (h) }{h} + \sin (x) \lim_{h\to 0} \frac{\cos(h) - 1 }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^k h^{2k + 1} }{(2k+1)!}}{h} + \sin (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^{k+1} h^{2k + 2} }{(2k+2)!}}{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ h }{h} + \sin (x) \lim_{h\to 0} \frac{ - \frac{ h^2 }{2} }{h} \\ \quad = \cos (x) \lim_{h\to 0} 1 + \sin (x) \lim_{h\to 0} \left( - \frac{ h }{2} \right) \\ \quad = \cos (x) \times 1 + \sin (x) \times 0 \\ \quad = \cos(x)$

sunshaker likes this.
8. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
Uh... I still didn't see the square root of minus one in there.

sunshaker likes this.
9. ### rpennerFully WiredValued Senior Member

Messages:
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.

sunshaker likes this.
10. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
$\frac{1}{\frac{1}{0}}=0$

Yes well, I specifically said complex numbers and stated my ignorance about computers...

danshawen and sunshaker like this.
11. ### rpennerFully WiredValued Senior Member

Messages:
4,833
$\frac{1}{0}$ 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
danshawen and sunshaker like this.
12. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
http://www.wolframalpha.com/input/?i=1/(1/0)

danshawen and sunshaker like this.
13. ### rpennerFully WiredValued Senior Member

Messages:
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
danshawen and sunshaker like this.
14. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
No. There's complex infinity, not lim.

When have I ever deliberately antagonized you?

:EDIT:

You're arguing with Mathematica....

Last edited: Dec 31, 2015
danshawen and sunshaker like this.
15. ### rpennerFully WiredValued Senior Member

Messages:
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.

danshawen and sunshaker like this.
16. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429
Be concise. I don't try to bombard others.

danshawen and sunshaker like this.
17. ### rpennerFully WiredValued Senior Member

Messages:
4,833
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.)

danshawen and sunshaker like this.
18. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
4,429

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

danshawen likes this.
19. ### Waiter_2001Registered Senior Member

Messages:
459
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. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,759
Still don't know what you're talking about, do you:

21. ### Waiter_2001Registered Senior Member

Messages:
459
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. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,759
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_2001Registered Senior Member

Messages:
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.