Exponential: proof/explanation for Int[dx/x]=ln x
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From: Jim Burns on 20 Jul 2010 16:25 Stefan.Duke(a)gmail.com wrote: > On Jul 20, 4:53 pm, mjc <mjco...(a)acm.org> wrote: >> Another way to look at ln is to start with its functional equation: >> f(xy) = f(x)+f(y). (This also works for exp, atanh, and others.) >> >> Assume the function is differentiable. Then >> f(x+h)-f(x) = f(1+h/x) = f(1+h/x)-f(1) (since f(1)=0) >> so f'(x) = lim_(h->0) (f(x+h)-f(x))/h >> = lim_(h->0) (f(1+h/x)-f(1))/h >> = (lim_(h->0) (f(1+h/x)-f(1))/(h/x) )/x >> = f'(1)/x. >> >> Thus f(x) = integral from 1 to x of f'(1)/t dt. >> >> The natural log is distinguished by f'(1) = 1. > > Thank you all so much! > Things are clearer now. This might not be necessary, given the other fine responses, but I like it, so I'm posting it. If you define a function ln(x) as ln(x) = Int_1^x [du/u], then it is simple to show ln(x) follows the laws of logarithms, just given that definition and the properties of integrals. ln(x*y) = Int_1^(x*y) [du/u] = Int_1^x [du/u] + Int_x^(x*y) [du/u] = Int_1^x [du/u] + Int_1^y [dv/v] { chg var u = x*v } = ln(x) + ln(y) ln(x^y) = Int_1^(x^y) [du/u] = Int_1^x [ (y*v^(y-1)*dv)/(v^y) ] { chg var u = v^y } = y*Int_1^x [dv/v] = y*ln(x) From this definition of ln(x), d(ln(x))/dx = 1/x If you define exp(y) as the inverse of ln(x), then we have d(ln(exp(y)))/dy = d(y)/dy = 1 but, from the chain rule, we also have d(ln(exp(y)))/dy = (1/exp(y))*d(exp(y))/dy Therefore, d(exp(y))/dy = exp(y) This last equation seems to be what sparked your curiosity. I suppose one could say that the exponential function solves linear differential equations "because" of the integral definition of its inverse and the inverse exponential (logarithm) is defined by that integral "because" it acts like a logarithm that way. Jim Burns |