Was a question now a debate, infinite outcomes in Finite amount of time?

myopic · Jan 10, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> Sorry, but the gibberish is entirely on your part.If you would care to "call me out" then feel free to try.  But you will have to speak precisely and not in the manner that you used thus far.  It makes no sense.Rather than having proven myself wrong, I have shown you the meaning of a specific infinite series and have provided you a rigorous proof that the sum is in fact 1.  OF COURSE there is a limiting operation, that is implicit in the definition of the sum of an infinite series.You have a fundamental problem -- you don't know what you are talking about.  Not a clue.  <br /> Posted by DrRocket</DIV></p><p>Uh, the proof listed actually does not equal one.  Not sure how you can show support with this equation.  I know certain abstract mathematical  theories were put out in the early 40', forget who, and the purpose was to provide a theorem to shore up the unified field theory.  And then I think it came back again with string theory, but I know in your example posted, this is certainly not the case.</p><p> I want to say you are thinking of scalar models and the use of bridging two planes?  I see this quite a bit in logarithms for computational science...building the internet.  It's the same notion as different sizes of infinity, some may be exponentially bound, others scalar.  Mandelbrodt is an example of scalar, much like a model of the www.  </p><p> But not sure how can say denominator in series being (/x-1) where the variable is set as x is less than 1 could yield a result of 1 itself.</p><p>_________</p><p> </p><p> </p>

amaterasu · Jan 10, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>But not sure how can say denominator in series being (/x-1) where the variable is set as x is less than 1 could yield a result of 1 itself._________   <br />Posted by myopic</DIV></p><p>the standard argument for the equality usually goes like this. . .</p><p>   x = 0.999...</p><p>multiply both sides by 10</p><p>   10x = 9.999...</p><p>subtracting the 1st equation from the 2nd gives</p><p>   9x = 9<br /><br />so</p><p>   x = 1</p><p>feeling like cheated?  i know the feeling.<br /><br /><br />but then someone showed me another even simpler version. . .</p><p>   1/3 = 0.333...</p><p>just multiply both sides by 3, and voila. 

</p><p>   1 = 0.999...</p><p>personally i find this one easier to swallow.</p><p><br />not convinced yet?  you might find this essay by Dedekind quite amusing:<br />http://books.google.co.jp/books?id=ztnY1wbBubIC&printsec=frontcover&hl=en</p> <div class="Discussion_UserSignature"> </div>

DrRocket · Jan 10, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>the standard argument for the equality usually goes like this. . .   x = 0.999...multiply both sides by 10   10x = 9.999...subtracting the 1st equation from the 2nd gives   9x = 9so   x = 1feeling like cheated?  i know the feeling.but then someone showed me another even simpler version. . .   1/3 = 0.333...just multiply both sides by 3, and voila. 

   1 = 0.999...personally i find this one easier to swallow.not convinced yet?  you might find this essay by Dedekind quite amusing:http://books.google.co.jp/books?id=ztnY1wbBubIC&printsec=frontcover&hl=en <br />Posted by amaterasu</DIV></p><p>Maybe this will help:</p><p>x + x^2 + x^3 + x^3 +...+ x^n =  [x - x^(n+1)]/(1 - x)</p><p>so if abs x < 1, x + x^2 + x^3 + ....  = x/(1-x)</p><p>Then taking x = 0.1 you have 0.111111111 = 0.1 + (0.1)^2 + (0.1)^3 + ... = 0.1/0.9 = 1/9</p><p>From whence 0.99999.... = 9 x 1/9 =1</p><p>and 0.33333.... = 3 x 1/9 = 1/3.<br /></p> <div class="Discussion_UserSignature"> </div>

SpeedFreek · Jan 10, 2009

I would just like to point out to all concerned that even if DrRocket does not like to point it out himself, he is a retired rocket scientist with a <em>Ph.D in Mathematics</em> and he used to teach the subject. It is always right to be skeptical, but if you do not have the prerequisite skills to analyse a problem, it <em>is</em> best left to the experts. <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

DrRocket · Jan 10, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>  But not sure how can say denominator in series being (/x-1) where the variable is set as x is less than 1 could yield a result of 1 itself._________   <br />Posted by myopic</DIV></p><p>x/(1-x) is where this started.  Let x = 1/2.   (1/2)/( 1 - 1/2) = (1/2)/(1/2) = 1.  I hope you can at least follow that.<br /></p> <div class="Discussion_UserSignature"> </div>

trumptor · Jan 10, 2009

<p>What I was thinking is pretty simple to explain without explaining limits. If you take a ruler and look at the points between 1 and 2 inches, do you agree that there are an infinite amount of points between the two numbers? Now you take an atom and place it on some spot between the 1 and 2, there are an infinite number of places to put the atom in the time it takes to put the atom down.</p> <div class="Discussion_UserSignature"> <p><em><font color="#0000ff">______________</font></em></p><p><em><font color="#0000ff">Caution, I may not know what I'm talking about.</font></em></p> </div>

amaterasu · Jan 12, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Maybe this will help:x + x^2 + x^3 + x^3 +...+ x^n =  [x - x^(n+1)]/(1 - x)so if abs x < 1, x + x^2 + x^3 + ....  = x/(1-x)Then taking x = 0.1 you have 0.111111111 = 0.1 + (0.1)^2 + (0.1)^3 + ... = 0.1/0.9 = 1/9From whence 0.99999.... = 9 x 1/9 =1and 0.33333.... = 3 x 1/9 = 1/3. <br />Posted by DrRocket</DIV></p><p>   0.999... = 9/10 + 9/100 + 9/1000 +... = (9/10)/(1-1/10) = 1  ;-)<br /><br />yes, that makes sense.  thank you for the help.</p> <div class="Discussion_UserSignature"> </div>

myopic · Jan 13, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Maybe this will help:x + x^2 + x^3 + x^3 +...+ x^n =  [x - x^(n+1)]/(1 - x)so if abs x < 1, x + x^2 + x^3 + ....  = x/(1-x)Then taking x = 0.1 you have 0.111111111 = 0.1 + (0.1)^2 + (0.1)^3 + ... = 0.1/0.9 = 1/9From whence 0.99999.... = 9 x 1/9 =1and 0.33333.... = 3 x 1/9 = 1/3. <br /> Posted by DrRocket</DIV></p><p>Sorry, not trying to ride you out on this.  But isn't this an example of what is called a dileneation series?  Much like Pi?  Where we have an infinite number after the decimal point and  use yet another proof in order to quantify that number so it can act like a whole?</p><p>Only saying this because I see x=0.111111 (assuming that number is boundless and goes off into infinity). </p><p>I know this was the approach that they used with the unified field theory and Pi.  I need to go find that work now.  </p><p> </p><p> </p>

myopic · Jan 13, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>What I was thinking is pretty simple to explain without explaining limits. If you take a ruler and look at the points between 1 and 2 inches, do you agree that there are an infinite amount of points between the two numbers? Now you take an atom and place it on some spot between the 1 and 2, there are an infinite number of places to put the atom in the time it takes to put the atom down. <br /> Posted by trumptor</DIV></p><p>Well you're talking about a marriage of an abstract mathematical concept and a real world number.  The concept of infinity in numbers through division and an atom that occupies a very real space that cannot be divided (otherwise it is no longer an atom and now a nuclear explosion).  </p><p> You can divide anything in the world down to its atomic core, and there it stops.  Mathematically though, any number can continue to be divided.  So you  have two different concepts.  In the real world with an atom on a ruler, it is simply occupying space in a 3D coordinate system. </p>

derekmcd · Jan 13, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>You can divide anything in the world down to its atomic core, and there it stops.<br /> Posted by myopic</DIV></p><p>All atomic cores (nuclei) contain protons and most contain both protons and neutrons.   The nucleus of an atom can be divided into it's constituent protons and neutrons.  The protons and neutrons can be further divided into quarks.  Although, you aren't likely to find any free quarks floating around.   Particle accelerators are constantly 'dividing' atomic cores and exploring what becomes of those divisions.</p><p>As far as theories and technologies capable of exploring said theories, it currently appears as if the quark is the final stop. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

DrRocket · Jan 13, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Sorry, not trying to ride you out on this.  But isn't this an example of what is called a dileneation series?  Much like Pi?  Where we have an infinite number after the decimal point and  use yet another proof in order to quantify that number so it can act like a whole?Only saying this because I see x=0.111111 (assuming that number is boundless and goes off into infinity). I know this was the approach that they used with the unified field theory and Pi.  I need to go find that work now.     <br />Posted by myopic</DIV></p><p>This is a rather plain vanilla infinite series.  It has nothing whatever to do with unified field theories.  It has nothing to do with pi and pi has nothing to do with unified field theories.</p><p>Pi is a transcendental real number, and in particular it is irrational.  That implies that its decimal representation non-terminating and non-repeating.<br /></p> <div class="Discussion_UserSignature"> </div>

myopic · Jan 21, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>All atomic cores (nuclei) contain protons and most contain both protons and neutrons.   The nucleus of an atom can be divided into it's constituent protons and neutrons.  The protons and neutrons can be further divided into quarks.  Although, you aren't likely to find any free quarks floating around.   Particle accelerators are constantly 'dividing' atomic cores and exploring what becomes of those divisions.As far as theories and technologies capable of exploring said theories, it currently appears as if the quark is the final stop. <br /> Posted by derekmcd</DIV></p><p> Yes it does?  I guess for his example it could have been any real thing and didn't necessarily have to be an atom.   It was just a point that a mathematical concept and a real world situation can disagree completely. </p>

myopic · Jan 21, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> If fyou understand calculus then you should already have seen this.  X + X^2 +X^3 + ... + X^n = [ X - X^(n+1)]/[1 - X]You ought to be able to prove this for yourself.  There are couple of ways to do it.  One, often seen in high school is to proceed by induction on n.  There is an easier way if you can find it. Take the limit as n increases without bound to conclude that for X<1,X + X^2 + X^3 + ... = X/(1 - X) Substitute X = 1/2 to find that 1/2 + 1/4 + 1/8 + ... = 1Your statement regarding general relativity and lines is roughly correct, in general relativity one deals with geodesics in a curved space-time.  So what ?  Of course lines are an abstract concept.  So are geodesics.  So is a manifold and hence so is space-time.  You statement regarding fractals is nonsense.  Utter gibberish.  Your statement that "fractals equals a whole" is completely meaningless.  A fractal is a topological spece that happens to have a fractional (non-integer) topological dimension.  Just to explain what that means is WAY beyond calculus. <br /> Posted by DrRocket</DIV></p><p>I still can't see how, in this series, if x<1, you could find a proof to show that x will = 1.  Unless now we are going to throw in pure mathematical conepts and use numbers.  But I suppose that is the nice little conundrum we get to play with when dealing with concepts of inifinity in the first place. </p>

origin · Jan 21, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I still can't see how, in this series, if x<1, you could find a proof to show that x will = 1.  Unless now we are going to throw in pure mathematical conepts and use numbers.  But I suppose that is the nice little conundrum we get to play with when dealing with concepts of inifinity in the first place. <br />Posted by myopic</DIV><br /><br />Imaginary numbers have nothing to do with these infinite series.  You would probably term imaginary numbers as 'pure mathematics' and therefore have no relivence to the real world.  However using imaginary numbers is an ideal way to calculate power usage with ac voltage.  Part of the power going to your house is 'imaginary', unfortunately you cannot use imaginary money to pay for it.</p><p> </p> <div class="Discussion_UserSignature"> </div>

trumptor · Jan 21, 2009

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Well you're talking about a marriage of an abstract mathematical concept and a real world number.  The concept of infinity in numbers through division and an atom that occupies a very real space that cannot be divided (otherwise it is no longer an atom and now a nuclear explosion).   You can divide anything in the world down to its atomic core, and there it stops.  Mathematically though, any number can continue to be divided.  So you  have two different concepts.  In the real world with an atom on a ruler, it is simply occupying space in a 3D coordinate system. <br />Posted by myopic</DIV><br /><br />Ok, well the atom is typically presented as a sphere with protons and neutrons surrounded by an electron cloud I believe. Now the atom does not have to move left or right on the ruler in diameter lengths of itself. Why can't it move left by half its diameter, or a third or a millionth? See? It can be placed in an infinite amount of positions unless space itself is composed of some basic unit measure, like pixels on a monitor, which I doubt is the case. <div class="Discussion_UserSignature"> <p><em><font color="#0000ff">______________</font></em></p><p><em><font color="#0000ff">Caution, I may not know what I'm talking about.</font></em></p> </div>

amaterasu · Jan 22, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ok, well the atom is typically presented as a sphere with protons and neutrons surrounded by an electron cloud I believe. Now the atom does not have to move left or right on the ruler in diameter lengths of itself. Why can't it move left by half its diameter, or a third or a millionth? See? It can be placed in an infinite amount of positions unless space itself is composed of some basic unit measure, like pixels on a monitor, which I doubt is the case. <br />Posted by trumptor</DIV></p><p>interesting discussion.  vaguely reminds me of the Riemann hypothesis.<br /></p> <div class="Discussion_UserSignature"> </div>

myopic · Jan 23, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>interesting discussion.  vaguely reminds me of the Riemann hypothesis. <br /> Posted by amaterasu</DIV></p><p>Yes it does.  And actually, this is an absolutely valid argument that is expressed in the science community today.  </p><p>I'm sure some find it arbitrary and trivial.  Our universe is infinite with space, but we occupy a physical place within it.  </p><p><br />"Now you take an atom and place it on some spot between the 1 and 2, there are an infinite number of places to put the atom in the time it takes to put the atom down."  So this is not farfetched at all. </p>

DrRocket · Jan 23, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I still can't see how, in this series, if x<1, you could find a proof to show that x will = 1.  Unless now we are going to throw in pure mathematical conepts and use numbers.  But I suppose that is the nice little conundrum we get to play with when dealing with concepts of inifinity in the first place. <br />Posted by myopic</DIV></p><p>Infinity IS a mathematicala concept. Of COURSE I was using mathematical concepts.  It is not a conundrum at all.  It is simply using the language precisely and applying the relevant tool, which is mathematics.<br /></p> <div class="Discussion_UserSignature"> </div>

DrRocket · Jan 23, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yes it does.  And actually, this is an absolutely valid argument that is expressed in the science community today.  I'm sure some find it arbitrary and trivial.  Our universe is infinite with space, but we occupy a physical place within it.  "Now you take an atom and place it on some spot between the 1 and 2, there are an infinite number of places to put the atom in the time it takes to put the atom down."  So this is not farfetched at all. <br />Posted by myopic</DIV></p><p>I don't know what scientific community it is that you inhabit.  It is apparently not the usual one.</p><p>No one knows if the unverse is spatially "finite" (the correct terminology is "closed") or not.</p><p>Placing an object, atom or anything else, on a ruler does admit of an infinite number of results, but the time taken to make the placement is completely irrelevant.</p><p>Frankly the only part of your statement that makes any sense at all is "we occupy a physical state within it", and while not arbitrary that statement is most certainly trivial.<br /></p> <div class="Discussion_UserSignature"> </div>

vogon13 · Jan 28, 2009

<p> </p><p>Where we might expect an 'infinity' to turn up, but it doesn't, is when we consider a heated object.  Since it is warm it radiates photons, and the hotter it gets, the shorter the wavelength (energy) of the photons  becomes.  Less appreciated, is that as the peak radiation becomes more powerful, more photons are always emitted at the lesser energies too.</p><p> </p><p>Extrapolating backwards, a finite sample, at any temperature above absolute zero, should emit an 'infinite' number of photons of essentially ~ 0 energy.</p><p> </p><p>There is a lower limit of energy for a photon to have, however, and photons of less energy than that are not produced.  The lower limit corresponds to a wavelength comparable to the dimensions of the universe.  The 'impedance' (to borrow a term from the RF folks) of space does not permit transmission of such low energy photons, and their 'creation' is stifled. This lower limit decreases as the universe expands, and therefore, over longish periods of time, the number of such low energy photons emitted by all objects increases enormously (although their collective energy remains constrained).</p><p> </p><p> </p><p> </p><p> </p><p> </p> <div class="Discussion_UserSignature"> <p><font color="#ff0000"><strong>TPTB went to Dallas and all I got was Plucked !!</strong></font></p><p><font color="#339966"><strong>So many people, so few recipes !!</strong></font></p><p><font color="#0000ff"><strong>Let's clean up this stinkhole !!</strong></font> </p> </div>

myopic · Feb 4, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I don't know what scientific community it is that you inhabit.  It is apparently not the usual one.No one knows if the unverse is spatially "finite" (the correct terminology is "closed") or not.Placing an object, atom or anything else, on a ruler does admit of an infinite number of results, but the time taken to make the placement is completely irrelevant.Frankly the only part of your statement that makes any sense at all is "we occupy a physical state within it", and while not arbitrary that statement is most certainly trivial. <br /> Posted by DrRocket</DIV></p><p>Again I have to read this and defend myself against snide, almost passive aggressive comments that are misread, misunderstood, for whatever reason.</p><p> I was referring to space that the known universe inhabits.  Space, in most scientific communities is agreed to infinite.  Our known universe ofcourse inhabits a part of that space, 'closed' or 'open' model, but space itself is accepted to be boundless.  </p><p>So trivial is hardly the word used when physicists begin to explore QM theories and comsomological theories, and some sort of marriage of the two. </p>

SpeedFreek · Feb 4, 2009

<p>How about finite, but boundless? A finite but multiply connected space? These are possibilites too, and if the whole universe is larger than our observable part of it, we may never know one way or the other. Anyone who says the universe is infinite is merely speculating. It might be, or it might not be.</p><p>A cosmic hall of mirrors? </p> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

DrRocket · Feb 25, 2009

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Again I have to read this and defend myself against snide, almost passive aggressive comments that are misread, misunderstood, for whatever reason. I was referring to space that the known universe inhabits.  Space, in most scientific communities is agreed to infinite.  Our known universe ofcourse inhabits a part of that space, 'closed' or 'open' model, but space itself is accepted to be boundless.  So trivial is hardly the word used when physicists begin to explore QM theories and comsomological theories, and some sort of marriage of the two. <br />Posted by myopic</DIV></p><p>Nonsense.  Basically gibberish.  My comments may have been snide, but they most certainly wre not passive aggressive.  They were quite plain.  You have no idea what you are talking about.  None whatever.  Not a clue.</p><p>Space or more correctly space-time is not know to be either open or closed at this time.  It is the case that space-time is a 4-manifold without boundary.  However, "without boundary" is a well-defined mathematical term and has nothing whatever to do with the lay term "boundless".</p><p>A manifold with boundary is not really a manifold at all in the usual sense.  A true manifold is required to be connecyted and to possess a neighborhood about each point that is homeomorphic to some Euclidean space and it follows from topology that the dimension of that Euclidean space is constant for each point.  A manaifold with boundary contains a subset that is in fact a manifold of one lower dimension than the interior of the manifold, that sub-manifold being the boundary, and it can be show that the  boundary has no boundary. Lack of a boundary is quite different from being "boundless" or even "unbounded" which is another and distinct mathematical term. </p> <div class="Discussion_UserSignature"> </div>

Boilermaker · Mar 12, 2009

Re: Was a question now a debate, infinite outcomes in Finite amo

hello, I just watched a video on google videos about the Mandelbrot set, if you google that you can see Arthur C. Clarke on a video explaining and showing this to you. He says that it is a Math Formula and that it is true infinity and you can look at it. It's quite an interesting video and the whole subject is very interesting if you do more searches on it you might come to believe that the Nature of the Universe we live in is determined by Fractal Geometry. I hope this helps, and is relevant.....I am going to get and post a link for you....here:
[color=#FF0000] [url=http://video.g...deo.google.ca/videosearch?q=Ma ... l=en&emb=0[/color][/url]

Was a question now a debate, infinite outcomes in Finite amount of time?

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