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Posted by on Sunday, June 16, 2013 at 6:51pm.

estimate ln(e^2+0.1)-ln(e^2)

  • calculus - , Sunday, June 16, 2013 at 7:44pm

    did this below

  • calculus - , Sunday, June 16, 2013 at 7:46pm

    Log(e^2 + 1/10) =

    Log(e^2) + Log[1+e^(-2)/10]

    We can compute this using the series expansion:

    Log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...

    But we can speed up the convergence of the series as follows. Replacing x by
    -x in the series gives:

    Log(1-x) = -x - x^2/2 - x^3/3 - x^4/4 + ...

    Subtract the two series to obtain:

    Log[(1+x)/(1-x)] =

    2[ x + x^3/3 + x^5/5 + x^7/7 +...]

    If we want to compute log(1+u) for small u, we must choose x such that:

    (1+x)/(1-x) = 1+u ------>

    x = u/(2+u). So, we have:

    Log[1+e^(-2)/10] =

    2 e^(-2)/10 / (2 + e^(-2)/10 ) +

    2/3 e^(-6)/10^3 / [(2 + e^(-2)/10 )]^3

    + 2/5 e^(-10)/10^5 / [(2 + e^(-2)/10)]^5

    +...

    The first 3 terms will give the answer to an accuracy of 10^(-16).

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