CraigD Posted February 20, 2007 Report Share Posted February 20, 2007 I think it would be much easier to just say that a series diverges if the ratio between subsequent terms approaches 1 as you get farther and farther out.Though this is easy to say, it’s demonstrably and provably false. For example, for any p-series with [math]p \gt 1[/math], the series converges, and the ration between subsequent terms approaches 1. But that is not what I was talking about.My apologies. Upon rereadingObviously any reduction of 1/10th the previous increase would be a convergent sum... I wonder if this is the threshold.I’ve another guess at your meaning. You appear to be suggesting that the geometric series[math]\sum_{n=0}^{\infty} \frac{a}{10^n}[/math] converges to a finite value for any [math]a[/math]. This is true. It converges to [math]\frac{10a}9[/math]. However, 10 is not a special constant. Any geometric series [math]\sum_{n=0}^{\infty} \frac{a}{b^n}[/math] converges (to [math]\frac{a}{1-\frac1b}[/math]) if and only if [math]b \gt 1[/math]. The important point that I and others in this thread have been driving at is that the base of the numeral system used – base 10, 2 or whatever – has no impact whatever on whether an infinite series converges or diverges. Numeral systems have only to do with how numbers are written, not what the values of those numbers. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 21, 2007 Author Report Share Posted February 21, 2007 Though this is easy to say, it’s demonstrably and provably false. For example, for any p-series with [math]p \gt 1[/math], the series converges, and the ration between subsequent terms approaches 1. In the series that you mentioned, the ratio is not approaching 1 in the same fashion. One of the perks of not being limited to formal mathematical reasoning is that you have more than one way of looking at things. Try to communicate with someone who only plays by the rules and you will get straw man after straw man. I am finished arguing with you. And btw I never argued that a "number" cannot be translated from one numeral system to another. Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted February 21, 2007 Report Share Posted February 21, 2007 In the series that you mentioned, the ratio is not approaching 1 in the same fashion. ...And btw I never argued that a "number" cannot be translated from one numeral system to another.Well, errr..., no one accused you of arguing that a "number" cannot be translated from one base to another. I don't see where you get that. You did make reference to a dependency upon which number system was being used. This is a very common misconception--hell, I assumed it was true myself until my second or third semester in Calculus. It's not something to get upset over. In the series mentioned above, the Harmonic Series and the P Series are identical when the exponent p=1. Therefore the ratio of their subsequent terms is identical. Again, I don't see what has upset you. The comparison of the two series is straight forward. If p=1, the series diverges. If p is ANY value more than 1, the series converges, although it may do so glacially slow. Frankly, I had forgotten all this and was delighted to relearn it. Now I'm ready for the proof that the Harmonic Series is the slowest growing infinite series that diverges to infinity. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 22, 2007 Author Report Share Posted February 22, 2007 Well my experience in this thread felt like some people had an ulterior motive in responding. I do not approach mathematics using convential manipulation of formulas as the only means to reach understanding. Rather I attempt to use logical reasoning as well and this is often problematic because I find that many logical contradictions in mathematics are hidden by convention simply altering any outcome to match what has been discovered in practice. It is also problematic when trying to communicate with others because most mathematicians seem to stick to formal convention which they have memorized quite well. And thus when I attempt to talk about logical explorations what I get is alot of misunderstanding of what I am saying and/or refusal to address my actual claim since it is not formal mathematics... not that some things that I said were not addressed in this thread but certainly not all of them. Most every explanation in the thread I could have given myself and would have had it to do with what I was talking about. Most of them are clearly explained in the book. When I post exploring the logical implications of something the last thing I expect is someone feverishly trying to formally charactarize everything I say (for the most part giving incorrect interpretations of my claim) and then demonstrate why it is wrong. Wow... Although series such as 1/(1^p) + 1/(2^p) follow the rules which Craig outlined, others such as the expansion of e^1 converge without any exponent in the denominator. Although all the terms after the first few in this expansion are smaller than the terms of his p series, it demonstrates that of more logical importance is the numerical threshold between a converging and diverging series. Pyrotex I dont know about such a proof because I opened a web page and saw the divergence of a harmonic series being proven by comparison to the series [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil}[/math] 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8.... which has all smaller terms and yet diverges as you can see by reducing it to simply [math] \sum_{k=1}^\infty \frac{x}{2}[/math] Which I found interesting because that is the direction I was going in by shifting to base 2, where you can more easily see the threshold between diverging and converging series. 1/2 + 1/4 + 1/8 ... is simply .111111111111.. in base 2 which is equivalent to 1.000 the same way .99999 is equivalent to 1.0 in base 10. And therefore a series like 1/2 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 would simply converge to 2 (10.00 base 2) So to diverge you would need a series that had more of each subsequent term which is what the one I found on the webpage has. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 23, 2007 Report Share Posted February 23, 2007 Ahem, <cough> reasoning on binary numbers, I get: [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil} = \sum_{l=1}^\infty 2^{\frac{l-1}{l}}[/math] which is of course divergent, I don't get what your x is meant to be. Rather I attempt to use logical reasoning as well and this is often problematic because I find that many logical contradictions in mathematics are hidden by convention simply altering any outcome to match what has been discovered in practice.e. g.? It is also problematic when trying to communicate with others because most mathematicians seem to stick to formal convention which they have memorized quite well. And thus when I attempt to talk about logical explorations what I get is alot of misunderstanding of what I am saying and/or refusal to address my actual claim since it is not formal mathematics... not that some things that I said were not addressed in this thread but certainly not all of them.I don't know which mathematicians you know, but for the real ones the whole of math is based on logical argument. It's the means by which the things you discuss have been deduced. If people ignore your claims, either they don't make sense or you're not talking to real mathematicians, unless you simply fail to make yourself understood by tuning in to their language. :shrug: Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 23, 2007 Report Share Posted February 23, 2007 Ahem, <cough> reasoning on binary numbers, I get: [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil} = \sum_{l=1}^\infty 2^{\frac{l-1}{l}}[/math] which is of course divergent, I don't get what your x is meant to be. Rather I attempt to use logical reasoning as well and this is often problematic because I find that many logical contradictions in mathematics are hidden by convention simply altering any outcome to match what has been discovered in practice.e. g.? It is also problematic when trying to communicate with others because most mathematicians seem to stick to formal convention which they have memorized quite well. And thus when I attempt to talk about logical explorations what I get is alot of misunderstanding of what I am saying and/or refusal to address my actual claim since it is not formal mathematics... not that some things that I said were not addressed in this thread but certainly not all of them.I don't know which mathematicians you know, but for the real ones the whole of math is based on logical argument. It's the means by which the things you discuss have been deduced. If people ignore your claims, either they don't make sense or you're not talking to real mathematicians, unless you simply fail to make yourself understood by tuning in to their language. :shrug: Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 23, 2007 Report Share Posted February 23, 2007 Pyrotex I dont know about such a proof because I opened a web page and saw the divergence of a harmonic series being proven by comparison to the series [math] \sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil}\,\![/math] 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8.... Ahem, <cough> reasoning on binary numbers, I get: [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil} = \sum_{l=1}^\infty 2^{\frac{l-1}{l}}[/math]Double-ahem <cough> [math]a^{-(\log_a b)} = \frac1b[/math] so [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil} = \frac11 + \frac12 + \frac13 + . . . [/math] It is, the harmonic series. Kriminal, could you provide a link to the webpage you mention? Something here is not making sense. PS: Kriminal, my apologies for any misunderstandings. I’m honestly not trying to set your ideas up as strawmen, just having a hard time figuring out what you’re saying. Phrases like “not approaching 1 in the same fashion”, without additional explanation, leave me mystified as to your meaning. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 23, 2007 Author Report Share Posted February 23, 2007 Just look at the series 1/2 + 1/4 + 1/4 + 1/8 + 1/8 +1/8 + 1/8 + 1/16 .... All the terms are smaller than harmonic series and it obviously diverges since it can be condensed to just 1/2 + 1/2 + 1/2 + 1/2 + 1/2 Also the reciprocals of the prime numbers diverges and is smaller than the harmonic series. Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted February 23, 2007 Report Share Posted February 23, 2007 gosh, what are those bracket thingies around the log base 2 of k? they have bracket flanges at the top but not at the bottom. Does that make any difference? :xparty: Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 23, 2007 Author Report Share Posted February 23, 2007 Ahaha oh right thx, after they said that I looked at it for a minute and thought they were right hahah So instead of 1/3 it would be 1/2 again because whatever the x is in 2^x=3, it rounds up to 2. And so forth for the rest. Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 23, 2007 Report Share Posted February 23, 2007 gosh, what are those bracket thingies around the log base 2 of k? they have bracket flanges at the top but not at the bottom. Does that make any difference?I think Pyro’s guessed it. [math] \lceil x \rceil[/math] could mean “the integer part of x” (the INT(X) function in BASIC). If that’s their meaning, [math]\sum_{k=1}^\infty 2^{-\lceil \log_2 k \rceil}[/math] gives Krim’s series 1/2 + 1/4 + 1/4 + 1/8 + 1/8 +1/8 + 1/8 + 1/16 .... From which a neat, simple proof follows:1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 … (a) has every term less than or equal to every term of1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 … (b), the harmonic series. (a) can be associated (parentheses added) to give1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 …and the parts within the parentheses evaluated to rewrite it1/2 + 1/2 + 1/2 + 1/2 …Which clearly diverges. So the harmonic series, which is term-for-term greater than or equal to it, must diverge. While this proves that the harmonic series diverges to infinity, if doesn’t quite meet Pyro’s challengeNow I'm ready for the proof that the Harmonic Series is the slowest growing infinite series that diverges to infinity.because it only shows that the harmonic series diverges, not that it is the slowest growing divergent series. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 24, 2007 Author Report Share Posted February 24, 2007 If you were confused about what the brackets meant maybe you should have read what you copy and pasted before you opened your mouth. If your not familiar with the symbol the /ceil code in the latex might have given it away. It wasn't intended to be a proof that the harmonic series is the slowest diverging, since it is a diverging series that has smaller terms than the harmonic series it would be a disproof instead... Quote Link to comment Share on other sites More sharing options...
pgrmdave Posted February 24, 2007 Report Share Posted February 24, 2007 If you were confused about what the brackets meant maybe you should have read what you copy and pasted before you opened your mouth. If your not familiar with the symbol the /ceil code in the latex might have given it away. Hey, Krim :(, you don't need to be nasty to help somebody understand something here. Don't do it again. We also explicitly ask in the rules that you not write in all caps, it is considered to be shouting and quite rude. Your post will be edited to remove them. Once more, don't do it again. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 24, 2007 Author Report Share Posted February 24, 2007 Id rather be nasty and straightforward about it than sink to the same level of being passive aggressive while pretending to be polite. That the former is far better than the latter is something that takes great maturity to learn and many people never do. I don't start fights but I will finish them. Quote Link to comment Share on other sites More sharing options...
pgrmdave Posted February 24, 2007 Report Share Posted February 24, 2007 Id rather be nasty and straightforward about it than sink to the same level of being passive aggressive while pretending to be polite. That the former is far better than the latter is something that takes great maturity to learn and many people never do. I don't start fights but I will finish them.Krim, maybe you didn't understand. I didn't ask you for what you think is better. I told you to not be nasty and not be rude. This isn't your home - this is a community with rules external to your personal philosophy. So follow them. Quote Link to comment Share on other sites More sharing options...
Kriminal99 Posted February 24, 2007 Author Report Share Posted February 24, 2007 It is not my home. It is a forum where people are invited under the pretense of being able to discuss various topics (no unbalanced censorship or tribal morality advertised) and then revenue and prestige is gained from these people visiting the site based on what is advertised. I do not have a personal philosophy, that is what you have. I follow truth. I make allowances for the possibillity that my understanding of things deviates from truth by hearing (not just listening) to any arguments that might contradict what I believe rather than sheltering myself from them using and abusing any means or position. A rule is not something that is completely subjective that can be interpreted to mean anything that you do not like people doing such as "Being rude or nasty". A rule is of the form "Do not say the word X, Or do not make an argument of the form Y" Confronting someone for passive aggressive behavior is not being rude or nasty, it is doing the right thing. To enable this type of behavior by failing to identify it would be worse. Your opinion is not signifigant beyond its logical value just because you are a moderator. Were you to ever meet me in a fair and objective debate enviornment you would quickly lose, and if you tried to resort to force there you would quickly regret it. Quote Link to comment Share on other sites More sharing options...
Boerseun Posted February 24, 2007 Report Share Posted February 24, 2007 Were you to ever meet me in a fair and objective debate enviornment you would quickly lose, and if you tried to resort to force there you would quickly regret it.Whilst having had nothing to do with the original post, you have to admit that was rather funny. Krim, did pgrmdave threaten you with violence? I he did, I didn't see it. Or are you referring to that because you don't seem to be holding your own as far as our rules are concerned? Like it was pointed out: This is a private 'club', if you will, where newcomers are welcomed with open arms provided they stick to the rules. And the rules aren't the US Constitution where 'Freedom of Speech' is guaranteed; the rules are the Rules of Hypography. Grin and bear it. If we don't do it, we'll quickly degenerate into a pile of muck like the myriads of unmoderated forums out there. You will NOT lay down the rules to us that we've been successfully applying over the years. And that's about it. All I can say further, is 'Like it or Leave it'. 'Shape up or Ship out' also comes to mind. pgrmdave 1 Quote Link to comment Share on other sites More sharing options...
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