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Taylor Expansion of Natural Logarithmby golanor
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#19
Jan1313, 09:20 AM

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"It could be larger" => you cannot use the general version of the lagrangian error to show convergence
It converges, but you cannot use this way to show it. With c arbitrary small, it would diverge for other x as well, by the way. 


#20
Jan1313, 10:16 AM

P: 56

yes, but since c is between x and 100, if x is bigger than 50 or smaller than 200 it won't diverge for any x and for any c.
when i pick x=1, for example, for 1<c<99 the limit diverges. you cannot talk about a series in that case, because the remainder does not necessarily converge to 0. 


#21
Jan1313, 10:39 AM

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Let me rewrite your limit a bit:
$$\lim_{n \to \infty} \frac{1}{c(n+1)} \frac{(x100)^n}{c^n}$$ Consider c=100epsilon with some (small) constant epsilon: The first part goes to 0, the second part, neglecting its sign, can be written as $$\frac{(100x)^n}{(100\epsilon)^n}$$ This is smaller than 1 for epsilon<x, which is easy to satisfy for x<50 (c is close to 100 then). x=50, c=50 gives convergence as well, and you covered the remaining parts. The series can converge to 0. Where is the problem? 


#22
Jan1413, 10:03 AM

P: 56

It's very confusing. If the Lagrange remainder doesn't approach 0 it means you cannot estimate the function correctly, but in this case the calculation of the series does give accurate estimations.
It's very interesting. Does anyone know the answer to this? 


#23
Jan1413, 11:40 AM

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^The Lagrange remainder does approach 0, you have chosen a terrible way to estimate it. Either use a more subtle estimate of the Lagrange remainder or use the Cauchy Remainder or Schlömilch Remainder. Remember your estimate for c is quite crude.



#24
Jan1413, 01:35 PM

P: 56

If i can find even 1 such number for which the remainder does not approach 0, it means that the series is not a correct estimation for the function. 


#25
Jan1513, 04:01 PM

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But I found one. 


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