Re: " tommy's precalculus " (correcting typo , sorry )
- From: tommy1729 <tommy1729@xxxxxxxxx>
- Date: Fri, 25 Jan 2008 16:38:26 EST
david wrote :
On Thu, 24 Jan 2008 16:35:35 EST, tommy1729
<tommy1729@xxxxxxxxx>
wrote:
fixing typo :dx
tommy1729 wrote:therefore exp [ integral 0,2pi LOG[ sin(x) + 5/4 ]
f(n) = product [ sin(n) + 5/4 ] from 1 till n.
f(n) is bounded.
i forgot the log function.
without it , it makes no sence of course
] = 1.
Perhaps I'm misunderstanding your notation.
no i made a typo , sorry.
The indefinite integral is:
integral (sin(x) + 5/4) dx = (5*x)/4 - cos(x)
So the definite integral from 0 to 2*pi equals
(5*pi)/2 and the
exponential of that integral is not 1.
--Mark
together with the original post , it makes sence now
( inserting a log in every integral )
sorry for the confusion.
i hope it is clear now.
Makes more sense than it did, but still
not much.
Speaking of making sense, the notation in your
original
claim
"f(n) = product [ sin(n) + 5/4 ] from 1 till n.
f(n) is bounded."
makes no sense - you meant to say
(*)f(n) = product [ sin(j) + 5/4 ] from 1 till n.
f(n) is bounded.
yes , that's what i meant of course.
You need to explain how you know that f(n) is
bounded.
Then after you do that you need to explain how it
follows that int_0^{2pi} log|sin(t) + 5/4| =
-infinity.
it is the integral analogue ;
since f(n) is bounded , its exponential trend must be 1.
for a set g of elements g_i
product (g) = exp( sum ( log (g) ) )
taking log on both sides : sum (log (g)) = 0 ( on average since as said no exponential growth )
by the law of big numbers it is logical to rewrite the sum as an integral over the period of log(g).
( in this case 2pi )
thus the integral given.
if you doubt the integral , you may check by contour integration.
(dont know why you wrote = - infinity ? i guess a typo too.)
( btw natural log, base e and not base 10 )
regards
tommy1729
regards
tommy1729
.
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- Re: " tommy's precalculus " (correcting typo , sorry )
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- Re: " tommy's precalculus " (correcting typo , sorry )
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