# calculus

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

• calculus - ,

did this below

• calculus - ,

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).