Grumthorn starts an argument or two

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Jora
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Re: Grumthorn starts an argument or two

Post by Jora » 14 Nov 2006 17:38

Grumthorn wrote:10N -N = (90 + 900 + 9000 + ....) - 9 -(90 + 900 + 9000 + ....)

9N = -9
This step would only be valid, if (90 + 900 + 9000 + ....) were absolute convergent, which is not true, as this infinite sum does not have a limit.
Infinite sums are not always commutative. Only for absolute convergent infinite sums the summands can be reordered without altering the result.
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Re: Grumthorn starts an argument or two

Post by Grumthorn » 14 Nov 2006 17:45

Jora wrote:
Grumthorn wrote:10N -N = (90 + 900 + 9000 + ....) - 9 -(90 + 900 + 9000 + ....)

9N = -9
This step would only be valid, if (90 + 900 + 9000 + ....) were absolute convergent, which is not true, as this infinite sum does not have a limit.
Infinite sums are not always commutative. Only for absolute convergent infinite sums the summands can be reordered without altering the result.
IOU 10G

Although a proof that the original series under discussion (0.999...) converges would also be required for full credit :).
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Post by Stormwern » 14 Nov 2006 17:56

0.99999... is not a number, it's a limit.

sin(x-1)/(x-1) for x=1 is not defined
but for x=lim(1-n) n->0 (=0.99999..), the answer is 1

You use the limit to represent the number, so it's half right, only the limit isn't really a number.
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Post by Grumthorn » 14 Nov 2006 18:09

Stormwern wrote:0.99999... is not a number, it's a limit.

sin(x-1)/(x-1) for x=1 is not defined
but for x=lim(1-n) n->0 (=0.99999..), the answer is 1

You use the limit to represent the number, so it's half right, only the limit isn't really a number.
No, a limit is a number.

Lim (x-> oo) (1) = 1

Lim (x-> oo) (1/x) = 0

The = sign here is entirely correct and justified it's not short hand for anything. The whole idea of limits is to do away with such fuzziness.
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Post by Stormwern » 14 Nov 2006 18:21

Yes the = are valid, but that doesn't mean that the limit is a number. There is no number you can put instead of x to make 1/x=0.

In any case, I just gave you a function that has different answers for x=1 and x=0.9999.., isn't that enough?
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Post by Knetter » 14 Nov 2006 18:32

A totally different subject, but I just had this delicious meat roll stuffed with bacon and gruyère. If my bro doesn't come home soon, I might eat his part as well. :P
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Post by Jora » 14 Nov 2006 18:34

The sequence in question is the sequence of the partial sums of sum(n=1,inf) 9/(10^n) = 9/10 + 9/100 + 9/1000 + ...
This sequence is obviously monotonic increasing, as a positive real number is added in each step.
The number we add in each step is also always smaller than the remaining difference to 1, so it will never be larger than 1, which means it is bounded by 1.
A monotonic and bounded sequence converges.
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Post by Stormwern » 14 Nov 2006 18:40

Indeed, converges, but doesn't become.
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Post by Jora » 14 Nov 2006 18:47

Well, in the end it actually depends on what you want to express with 0.9999...
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Post by Grumthorn » 14 Nov 2006 18:50

Jora wrote:Well, in the end it actually depends on what you want to express with 0.9999...
I would argue that no, it doesn't. The fact that there are multiple decimal representations of the same point on the number line is a strange artifact of the decimal system rather than some deep property of numbers themselves.
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Post by Grumthorn » 14 Nov 2006 18:54

Stormwern wrote:Yes the = are valid, but that doesn't mean that the limit is a number. There is no number you can put instead of x to make 1/x=0.

In any case, I just gave you a function that has different answers for x=1 and x=0.9999.., isn't that enough?
No the fact that the equals is valid _does_ entirely imply that the limit is a number. If I write (in a mathematical context) x = y it implies that x is a number, y is a number and that they are the same number.

Lim(x>oo) (1/x) is not arbitrarily close to 0, it does not 'approach 0' it _is_ 0. Without this fact you cannot do calculus, at all.
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Post by Stormwern » 14 Nov 2006 18:55

Maybe you have a point there jora..

oo
> 9/10^x = 1
1

Since a number can't very well be equal to another and not at the same time, I stand by my conclusion that the infinite sum in itself isn't a number.
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Post by Ltuan » 14 Nov 2006 18:56

Grumthorn wrote:Without this fact you cannot do calculus, at all.
That implies we can do it ;-P
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Post by Stuvvie » 14 Nov 2006 19:00

I say maths and fooling around with numbers has been made harder by mankind and serves no real purpose.

Prove me wrong but I don't see a real subject involving those numbers.
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Post by Stormwern » 14 Nov 2006 19:00

Grumthorn wrote:
Stormwern wrote:Yes the = are valid, but that doesn't mean that the limit is a number. There is no number you can put instead of x to make 1/x=0.

In any case, I just gave you a function that has different answers for x=1 and x=0.9999.., isn't that enough?
No the fact that the equals is valid _does_ entirely imply that the limit is a number. If I write (in a mathematical context) x = y it implies that x is a number, y is a number and that they are the same number.

Lim(x>oo) (1/x) is not arbitrarily close to 0, it does not 'approach 0' it _is_ 0. Without this fact you cannot do calculus, at all.
Yes, it IS!! 0, but x is not infinity. Lim(x>oo) is the limit that is applied to the function 1/x. Don't confuse the terms!
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