Tutorial 2: Question 2
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g3kho Offline
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Tutorial 2: Question 2
[Image: q1.png]

Between the second and third lines of case (i) the denominator goes from (.d1 d2 ... dt dt+1 ...) to simply (.d1 d2 ...). Is there a justification for this or is it a typo?

This would suggest that the bounds are for del = (x - fl(x)) / fl(x) instead of del = (x - fl(x)) / x.

If it is a typo, then i get as a final answer that (|del| = |b^(1-t) / (1 + b^(1-t)|)
2015-04-26 09:12
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ccc Offline
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RE: Tutorial 2: Question 2
(2015-04-26 09:12)g3kho Wrote:  [Image: q1.png]

Between the second and third lines of case (i) the denominator goes from (.d1 d2 ... dt dt+1 ...) to simply (.d1 d2 ...). Is there a justification for this or is it a typo?

This would suggest that the bounds are for del = (x - fl(x)) / fl(x) instead of del = (x - fl(x)) / x.

If it is a typo, then i get as a final answer that (|del| = |b^(1-t) / (1 + b^(1-t)|)

The denominator (.d1 d2 ...) is meant to be (.d1 d2 ... dt dt+1 ...).
But then, it does not matter, as the denominator is replaced
by the smallest possible (normalized) denominator 0.1.
2015-04-27 16:31
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g3kho Offline
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RE: Tutorial 2: Question 2
(2015-04-27 16:31)ccc Wrote:  
(2015-04-26 09:12)g3kho Wrote:  [Image: q1.png]

Between the second and third lines of case (i) the denominator goes from (.d1 d2 ... dt dt+1 ...) to simply (.d1 d2 ...). Is there a justification for this or is it a typo?

This would suggest that the bounds are for del = (x - fl(x)) / fl(x) instead of del = (x - fl(x)) / x.

If it is a typo, then i get as a final answer that (|del| = |b^(1-t) / (1 + b^(1-t)|)

The denominator (.d1 d2 ...) is meant to be (.d1 d2 ... dt dt+1 ...).
But then, it does not matter, as the denominator is replaced
by the smallest possible (normalized) denominator 0.1.

We are also making an assumption that (.dt+1 dt+2 ... ) is maximal (i.e. equals 1xb^-t) .

Is this assumption compatible with the assumption that (.d1 d2 ... dt dt+1 ..) = 1?

That is, if we've already assumed that (.dt+1 dt+2 ... ) = 1xb^-t then shouldn't the minimum value for (.d1 d2 ... dt dt+1 ..) at least be 1+1xb^-t?
2015-04-29 05:44
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ccc Offline
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RE: Tutorial 2: Question 2
(2015-04-29 05:44)g3kho Wrote:  
(2015-04-27 16:31)ccc Wrote:  
(2015-04-26 09:12)g3kho Wrote:  [Image: q1.png]

Between the second and third lines of case (i) the denominator goes from (.d1 d2 ... dt dt+1 ...) to simply (.d1 d2 ...). Is there a justification for this or is it a typo?

This would suggest that the bounds are for del = (x - fl(x)) / fl(x) instead of del = (x - fl(x)) / x.

If it is a typo, then i get as a final answer that (|del| = |b^(1-t) / (1 + b^(1-t)|)

The denominator (.d1 d2 ...) is meant to be (.d1 d2 ... dt dt+1 ...).
But then, it does not matter, as the denominator is replaced
by the smallest possible (normalized) denominator 0.1.

We are also making an assumption that (.dt+1 dt+2 ... ) is maximal (i.e. equals 1xb^-t) .

Is this assumption compatible with the assumption that (.d1 d2 ... dt dt+1 ..) = 1?

That is, if we've already assumed that (.dt+1 dt+2 ... ) = 1xb^-t then shouldn't the minimum value for (.d1 d2 ... dt dt+1 ..) at least be 1+1xb^-t?
We are not making any assumption.
We are only maximizing the numerator and minimizing the denominator.
In this way we get an upper bound.
The maximization of numerator and minimization of denominator
do not have to be compatible to get a bound (just a bound).
2015-04-29 14:18
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