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$$ N_v = 0.5(t^{2}+2t^{6/7})\ln(1+2t^{-8/7})-t^{6/7} \tag{1} $$ $$ N_v =(0.871+0.125\ln t)^2 \tag{2}$$ Eq(2) is the approximated version of Eq(1). Does anyone know how to derive (2) from (1)? I'd appreciate your intuiton.

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Are you sure these equations are well given?, because if my calculations are correct the 1st one doesn't verify the 2nd. –  johan paul Nov 16 '11 at 17:16
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For what range of $t$ is this approximation supposed to hold? –  JavaMan Nov 16 '11 at 17:18
    
i think range of t is 0.001<t<1 –  angbong Nov 17 '11 at 3:46
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1 Answer

The two formulas are pretty close (within about .03) for $t$ in an interval from about $.0002$ to $0.54$. But there doesn't seem to be any obvious sense in which the second is a best approximation of the first.

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