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Old 2011-06-06, 01:01   #45
CRGreathouse
 
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Quote:
Originally Posted by science_man_88 View Post
1/x=0 when x=\infty

(1/x)*x = 1 if I remember correct.

0*x=1 ? doesn't seem to work out so it looks as though anything based in what I believe is called Peano arithmetic makes 1/\infty = 0 impossible.
\infty\not\in\mathbb{R}. Don't use rules from \mathbb{R} on objects not in it.
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Old 2011-06-06, 01:58   #46
R.D. Silverman
 
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Quote:
Originally Posted by R.D. Silverman View Post
That will teach me to try (too quickly!) to find a simple explanation
for something subtle. Obviously, I had a quaalude moment....

You are correct. They both go to zero. MAJOR Brain damage on my
part. For some reason, I was thinking about 1/[floor(x) - x)]...... Idiot

This is subtle...

Let me look at it again without invoking Euler-MacLauren summation.....
Everyone can award me the dunce-of-the-week award. The integral does
converge. Re-write it as:

limit a-->oo integral from 1 to a of (1/floor(x) - 1/x)dx

Call the kernel g(x). This is then:

limit a-->oo [ integral from 1 to 2 g(x)dx + integral 2 to 3 g(x) dx + ......+
integral a-1 to a g(x) dx]

What happens is that everything TELESCOPES...... It doesn't matter
that one winds up subtracting two infinite quantities. One winds up with
\gamma.

Major stupidity on my part.

Last fiddled with by R.D. Silverman on 2011-06-06 at 01:59 Reason: typo
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