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March 19

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Functional Convergence

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In a recent thread, if I understand correctly, pma says that converges pointwise to as n approaces infinity. How would you prove that? Black Carrot (talk) 07:53, 19 March 2009 (UTC)[reply]

Won't the limit depend on what branch of log you choose for negative arguments? Algebraist 10:23, 19 March 2009 (UTC)[reply]
My apologies: I made a misprint there (now corrected): the change of variables was , with a minus in the exponent (this is consistent with the line below, that had it). So the term is at the denominator, and the argument of log goes to 1 (actually, in that computation it was always positive). Do you see how to do it now?--pma (talk) 12:40, 19 March 2009 (UTC)[reply]
Here it is:
  • Write the second order Taylor expansion for at 0, with remainder in Peano form: so, for all
, as .
  • For any s we only have to consider the integers . Replace in the expansion above, getting
, as , and uniformly for all .
  • Summing over all
, as .
  • Then you may observe that is the Riemann sum for the integral of on [0,1] (or use the formula for ) and conclude that the whole thing is .
Warning: I have re-edited this answer, to make it more simple and clear (hopefully) --pma (talk) 13:40, 19 March 2009 (UTC)[reply]

Differential Equation

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How should one go about solving this equation.

92.9.236.44 (talk) 20:30, 19 March 2009 (UTC)[reply]

The right hand side is . Does that help? —JAOTC 20:48, 19 March 2009 (UTC)[reply]
Ah yes. It seems to yeild a solution of the form does that seem correct? —Preceding unsigned comment added by 92.9.236.44 (talk) 21:01, 19 March 2009 (UTC)[reply]
Check for yourself - differentiate that a couple of times, substitute everything in and see if the two sides match. If they do, you've got it right, if they don't, you haven't! --Tango (talk) 23:05, 19 March 2009 (UTC)[reply]