Cantor Confusion
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From: David Marcus on 19 Jan 2007 16:19 Andy Smith wrote: > David Marcus writes > > >You have demonstrated what people on sci.math refer to as dyslexia. You > >have switched the order of the operations. As far as the limit is > >concerned, x is a fixed number. Let's try x = 1/2. The question is what > >is the value of > > > > lim n->oo (1/2)^n > > > >? To be more precise, saying > > > > lim n->oo |x|^n = 0, for |x| < 1 > > > >is different from saying > > > > lim n->oo sup_{|x|<1} |x|^n. > > > >"sup" is like max, but is used when the maximum isn't obtained. > > But, if the formula is to be true for all x<1, then don't we need to > show that > > lim n->oo sup_{|x|<1} |x|^n = 0 ? > > Which was what I was trying, unsuccessfully and amateurishly, to > do/disprove? that is exactly the point at issue - there is no question > that > lim n->oo |x|^n = 0 for any |x| that is finitely different from 1. You can delete the word "finitely" from your sentence. It doesn't mean anything here. > The issue is whether there are any wrinkles and whether the limit is > true for > all |x| strictly <1 i.e. all reals in the interval from 0,1 excluding > the exact point at x = 1. The formula that has to be true for all |x| < 1 is lim_{n->oo} |x|^n = 0. In other words, for all x if |x| < 1 then lim_{n->oo} |x|^n = 0. So, the formula only has to be true for each x individually. Not, for all x uniformly (to use the technical jargon). How would you phrase it if you meant each x in the interval individually? You are quite correct that lim_{n->oo} sup_{|x|<1} |x|^n is not zero. In fact, for any n > 0, sup_{|x|<1}|x|^n equals 1. So, lim_{n->oo} ( sup_{|x|<1} |x|^n ) = lim_{n->oo} 1 = 1. -- David Marcus
From: Andy Smith on 19 Jan 2007 18:02 David Marcus writes > >The formula that has to be true for all |x| < 1 is > > lim_{n->oo} |x|^n = 0. > >In other words, for all x > > if |x| < 1 then lim_{n->oo} |x|^n = 0. > >So, the formula only has to be true for each x individually. Not, for >all x uniformly (to use the technical jargon). > >How would you phrase it if you meant each x in the interval >individually? > >You are quite correct that > > lim_{n->oo} sup_{|x|<1} |x|^n > >is not zero. In fact, for any n > 0, sup_{|x|<1}|x|^n equals 1. So, > > lim_{n->oo} ( sup_{|x|<1} |x|^n ) = lim_{n->oo} 1 = 1. > OK, here is a suggested alternative argument to show that lim_{n->oo} |x|^n != 0 for all |x| <0 We can consider the mapping from x to y_n given by y_n = |x|^n. The effect of the exponentiation is to 'stretch' [0,1] non linearly, so that , depending on the size of n, progressively more points uniformly spaced originally in [0,1] are concentrated in the bottom of [0,1] in y_n. But, that doesn't matter - we have plenty of reals. There are as many reals in any interval of y_n in [0,1] as that of x. So it doesn't matter how large n becomes - y_n has the same "density" of reals as x. And that must also apply at the limit n->oo - if it were true that you could 'stretch' the reals by such a power law, so as to map all the reals in 0,1 to either 0 or 1, this would be tantamount to saying that the reals are not continuous on the line - the reals are infinitely expandable? So lim_{n->oo} |x|^n cannot be 0 for all |x|<1 ? -- Andy Smith
From: Andy Smith on 19 Jan 2007 18:13 David Marcus writes >> >> But I would expect some hard time especially over whether it is >> reasonable to talk about m! as m->oo > >Why? If m is 10^90, is it hard to talk about (10^90)! ? > Some time back you showed me that sin(pi/x) has a fundamental discontinuity at x=0. Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can you explain, please? Regards -- Andy Smith
From: David Marcus on 19 Jan 2007 18:49 Andy Smith wrote: > David Marcus writes > > > >The formula that has to be true for all |x| < 1 is > > > > lim_{n->oo} |x|^n = 0. > > > >In other words, for all x > > > > if |x| < 1 then lim_{n->oo} |x|^n = 0. > > > >So, the formula only has to be true for each x individually. Not, for > >all x uniformly (to use the technical jargon). > > > >How would you phrase it if you meant each x in the interval > >individually? > > > >You are quite correct that > > > > lim_{n->oo} sup_{|x|<1} |x|^n > > > >is not zero. In fact, for any n > 0, sup_{|x|<1}|x|^n equals 1. So, > > > > lim_{n->oo} ( sup_{|x|<1} |x|^n ) = lim_{n->oo} 1 = 1. > > > > OK, here is a suggested alternative argument to show that > lim_{n->oo} |x|^n != 0 for all |x| <0 > > We can consider the mapping from x to y_n given by y_n = |x|^n. The > effect of the exponentiation is to 'stretch' [0,1] non linearly, so that > , depending on the size of n, progressively more points uniformly spaced > originally in [0,1] are concentrated in the bottom of [0,1] in y_n. But, > that doesn't matter - we have plenty of reals. There are as many reals > in any interval of y_n in [0,1] as that of x. > > So it doesn't matter how large n becomes - y_n has the same "density" of > reals as x. And that must also apply at the limit n->oo - if it were > true that you could 'stretch' the reals by such a power law, so as to > map all the reals in 0,1 to either 0 or 1, this would be tantamount to > saying that the reals are not continuous on the line - the reals are > infinitely expandable? > > So lim_{n->oo} |x|^n cannot be 0 for all |x|<1 ? Did you read what I wrote? Doesn't seem like it. I'll try again. The sentence lim_{n->oo} |x|^n = 0, for |x| < 1 means lim_{n->oo} |0|^n = 0, lim_{n->oo} |0.1|^n = 0, lim_{n->oo} |0.2|^n = 0, lim_{n->oo} |-0.278|^n = 0, lim_{n->oo} |0.78|^n = 0, lim_{n->oo} |sqrt(2)/2|^n = 0, etc. That's all it means. It does not mean lim_{n->oo} sup_{|x|<1} |x|^n. When we want to say, "lim_{n->oo} sup_{|x|<1} |x|^n", we say, "lim_{n-> oo} sup_{|x|<1} |x|^n". We don't say, "lim_{n->oo} |x|^n = 0, for |x| < 1", because the latter means that lim_{n->oo} |x|^n = 0 for whatever value of x we happen to try as long as |x| < 1. The only way that lim_{n->oo} |x|^n = 0, for |x| < 1 can be false, is if there is a real number y such that |y| < 1 and lim_{n->oo} |y|^n != 0. Is there such a y? -- David Marcus
From: David Marcus on 19 Jan 2007 18:54
Andy Smith wrote: > David Marcus writes > >> > >> But I would expect some hard time especially over whether it is > >> reasonable to talk about m! as m->oo > > > >Why? If m is 10^90, is it hard to talk about (10^90)! ? > > Some time back you showed me that sin(pi/x) has a fundamental > discontinuity at x=0. We said two things. Division by zero is not defined. So, the expression sin(pi/x) is not defined for x = 0. We also said that the function f(x) = sin(pi/x), x != 0 can not be extended to a continuous function that is defined on all of R. > Doesn't the same thing apply to Lim m->oo {cos(m! pi x)} and if not, can > you explain, please? You don't evaluate limits by taking the value you are approaching and sticking it in the expression. And, oo isn't a number, so it wouldn't even make sense in this case. cos(m! pi x) is defined for all real numbers x and all nonnegative integers m. So, it isn't any harder to consider the limit as m -> oo as it would be to consider the limit as x -> oo. -- David Marcus |