Question #3acdb

1 Answer
Mar 21, 2016

If x = y + h, where h is relatively small in magnitude, then #x*ln(x/y) = x-y#, nearly, neglecting terms of higher order smallness in #h^2, h^3, h^4, ...#

Explanation:

Let x = y + h, where h is relatively small in magnitude.

#x*ln(x/y) = (y + h)* ln((y + h)/y) = (y + h)*ln(1 + h/y) = (y + h)(h/y-(1/2)(h^2/y^2)+(1/3)(h^3/y^3-...)#, using series expansion for #ln(1+(h/y))# in power of #(h/y)#.

This leads to
#x*ln(x/y)# = h + terms of order higher powers of h
= h + #O(h^2)#
= x-y, nearly..