- #1

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Or not, as the case may be

Thanks for any replies :)

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- Thread starter cyclogon
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- #1

- 14

- 0

Or not, as the case may be

Thanks for any replies :)

- #2

fresh_42

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What is

? Where is the one placed at?0.00...1

- #3

mfb

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0.00...1 is not a well-defined expression.

0.00... = 0.

0.00... = 0.

- #4

fresh_42

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Here's a more interesting example. Consider the function ##x \longmapsto e^{-(x-n)^2}##. For ##n=10## it looks like

and the greater ##n## is, the more to the right is the bump. The area below the curve is ##\int_{-\infty}^{+\infty} e^{-(x-n)^2}\, dx \,=\, \sqrt{\pi}\,##. If we move it to infinity, then it is still the same bump with the same area ##\lim_{n \to \infty} \int_{-\infty}^{+\infty} e^{-(x-n)^2}\, dx \,=\, \sqrt{\pi}##. However, if we first move the bump to infinity, we get ## \lim_{n \to \infty} e^{-(x-n)^2} = 0## and the area ##\int_{-\infty}^{+\infty} (\lim_{n \to \infty} e^{-(x-n)^2})\,dx ## vanishes with it. This is the same with your ##1##. If we move it to infinity, it vanishes, because wherever we look, we are left from the bump and arbitrary close to zero, the bump is never reached.

and the greater ##n## is, the more to the right is the bump. The area below the curve is ##\int_{-\infty}^{+\infty} e^{-(x-n)^2}\, dx \,=\, \sqrt{\pi}\,##. If we move it to infinity, then it is still the same bump with the same area ##\lim_{n \to \infty} \int_{-\infty}^{+\infty} e^{-(x-n)^2}\, dx \,=\, \sqrt{\pi}##. However, if we first move the bump to infinity, we get ## \lim_{n \to \infty} e^{-(x-n)^2} = 0## and the area ##\int_{-\infty}^{+\infty} (\lim_{n \to \infty} e^{-(x-n)^2})\,dx ## vanishes with it. This is the same with your ##1##. If we move it to infinity, it vanishes, because wherever we look, we are left from the bump and arbitrary close to zero, the bump is never reached.

- #5

Mark44

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- #6

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But keep in mind that is the limit. And that is different than zero.

- #7

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The "1" is at the end of a length of infinitely many zeros

ie. 0.0000000...(infinte zeros)....1

Thanks

- #8

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Yes, I think people probably realized that's what you meant, but there is no such thing which is why folks are saying that it's undefined. If you get to a point where you can put a 1, then you are not yet at infinity so your statement is nonsensical/undefined.hi, thanks for all your replies. Sorry about not explaining clearing enough.

The "1" is at the end of a length of infinitely many zeros

ie. 0.0000000...(infinte zeros)....1

- #9

russ_watters

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This is a self-contradiction: if the string of zeros is infinite, it doesn't have an end. That's why that expression doesn't work isn't used in math.The "1" is at the end of a length of infinitely many zeros

ie. 0.0000000...(infinte zeros)....1

- #10

jbriggs444

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That limit is not different from zero. It is precisely zero. As one can see from the epsilon/delta definition of a limit.

But keep in mind that is the limit. And that is different than zero.

- #11

jbriggs444

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To be picky, one could index the digits in a decimal string over a set of positions with order type omega plus one. The difficulty is not that this is a self-contradiction. The difficulty is that the resulting digit strings do not naturally form an algebraic field.This is a self-contradiction: if the string of zeros is infinite, it doesn't have an end. That's why that expression doesn't work isn't used in math.

- #12

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That limit is not different from zero. It is precisely zero. As one can see from the epsilon/delta definition of a limit.

Perhaps I should have said the limit is zero but the sequence 1/10, 1/100, ... is never precisely zero.

- #13

Mark44

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ThePerhaps I should have said the limit is zero but the sequence 1/10, 1/100, ... is never precisely zero.

- #14

jbriggs444

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I would phrase it that "no term of the sequence is zero".Perhaps I should have said the limit is zero but the sequence 1/10, 1/100, ... is never precisely zero.

Edit: Beaten to it by @Mark44

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