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...
I'm not a computer programer But I'm still bewildered on what computers have the potential of doing. http://www.wolframalpha.com/input/?i=((((45*45)+1)/46)-45) Or maybe I'm bewildered by mathematics. And of course division by zero is possible with complex numbers.
Please Register or Log in to view the hidden image! You don't have a clue as to what you're talking about, do you?
Untrue. What is sometimes possible is obtaining finite limits of two terms that have limiting values of zero in analysis. \(\lim_{h\to 0} \frac{\sin (x + h) - \sin (x)}{h} \\ \quad = \lim_{h\to 0} \frac{\sin (h) \cos(x) + ( \cos(h) - 1 ) \sin (x) }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{\sin (h) }{h} + \sin (x) \lim_{h\to 0} \frac{\cos(h) - 1 }{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^k h^{2k + 1} }{(2k+1)!}}{h} + \sin (x) \lim_{h\to 0} \frac{ \sum_{k=0}^{\infty} \frac{ (-1)^{k+1} h^{2k + 2} }{(2k+2)!}}{h} \\ \quad = \cos (x) \lim_{h\to 0} \frac{ h }{h} + \sin (x) \lim_{h\to 0} \frac{ - \frac{ h^2 }{2} }{h} \\ \quad = \cos (x) \lim_{h\to 0} 1 + \sin (x) \lim_{h\to 0} \left( - \frac{ h }{2} \right) \\ \quad = \cos (x) \times 1 + \sin (x) \times 0 \\ \quad = \cos(x)\)
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.
\(\frac{1}{\frac{1}{0}}=0\) Yes well, I specifically said complex numbers and stated my ignorance about computers...
\(\frac{1}{0}\) 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.
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
No. There's complex infinity, not lim. Stop it already. When have I ever deliberately antagonized you? :EDIT: You're arguing with Mathematica....
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.
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.)
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.
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.)
Still don't know what you're talking about, do you: Please Register or Log in to view the hidden image!
As I wrote: the computer falls short. The true fraction is much greater! Have you tried this calculation manually? Nice picture beer w/straw!
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.
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.