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#37
May2511, 12:29 PM

P: 98

Isn't it because that means the sum above is the difference of two other sums?



#39
May2511, 12:32 PM

P: 98

one that sums to one, and one that has a zero at the kth entry?



#40
May2511, 12:35 PM

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P: 18,290

Yes, indeed!
So, now we have shown that only (1,1,1,...) has the property that [tex]\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}=1[/tex]. So, can you now describe the ball around (0,0,0,...) with radius 1? 


#41
May2511, 12:40 PM

P: 98

Would it be all entries are 1? But it wouldn't be finite...would it?



#42
May2511, 12:46 PM

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P: 18,290

[tex]\left\{(x_1,x_2,x_3,...)~\mid~\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}<1\r ight\}[/tex] So... 


#43
May2511, 12:52 PM

P: 98

Isn't that just all the (x_1,x_2, ...) minus the entry with (1,1,1,1...)?



#44
May2511, 12:58 PM

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P: 18,290

Indeed, so the ball around (0,0,0,...) with radius 1 is everything except (1,1,1,1,...).
So, now the obvious generalization: what is the ball around [itex](a_1,a_2,a_3,...)[/itex] with radius 1? 


#45
May2511, 01:04 PM

P: 98

would it be everything but (a_1,a_2, a_3, ...)?



#46
May2511, 01:10 PM

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P: 18,290

Uuh, no, I get the impression that you're just guessing here.
Obviously a=(a_{1},a_{2},a_{3},...) will be in the ball, since d(a,a)=0<1. Moreover, we have just calculated that the ball around (0,0,0,...) is everything but (1,1,1,...). Thus (0,0,0,...) is in the ball... 


#47
May2611, 07:16 PM

P: 98

...I think I'll need another hint



#48
May2611, 07:23 PM

Mentor
P: 18,290

For what sequence [itex](x_n)_n[/itex], does
[tex]\sum_{k=1}^{+\infty}{\frac{x_ka_k}{2^k}}=1[/tex] It's basically the same as my last question (when the a_{k} were simply 0). What must hold for x_{k}a_{k} in order for the above series to be 1? 


#49
May2611, 07:33 PM

P: 98

X_ka_k > 0 as k> infinity?



#50
May2611, 07:44 PM

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P: 18,290

You do know that [itex]\frac{1}{2^k}X_ka_k\rightarrow 0[/itex], but I fail to see how it is relevant here... 


#51
May2611, 07:51 PM

P: 98

Oh...does X_ka_k=1 for k=1..infinity?



#52
May2611, 07:54 PM

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P: 18,290

Yes!!! Very good!!
So, if x_{k}a_{k}=1, then what will x_{k} be? (Remember that x_{k} and a_{k} can only take on 0 and 1 here). 


#53
May2611, 07:58 PM

P: 98

x_k=1, a_k=0 or a_k=1,x_k=0 ...is that right?



#54
May2611, 08:05 PM

Mentor
P: 18,290

Remark that this correspons nicely with the previous thingies where a was (0,0,0,...), then x was (1,1,1,...). So, the ball around a with radius 1 is just everything except the "opposite of a")! Now, we're going to make it a little harder. We're going to figure out what the ball around (0,0,0,...) is with radius 1/2. So, for which vectors does [tex]\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}<\frac{1}{2}[/tex] As above, you may remark that it suffices to calculate when equality holds, thus [tex]\sum_{k=1}^{+\infty}{\frac{x_k}{2^k}}=\frac{1}{2}[/tex] Since if x has a distance of 1/2 of 0, then all vectors with less 1's than x, have distance <1/2. And this is exactly what we wanted... 


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