$ 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.
Please, Let me know this approximation
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0i think range of$t$is 0.001
– 2011-11-17
1 Answers
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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.