Division and Subtraction: Logarithm and ...?Date: 05/30/2014 at 04:15:49 From: Youtian Subject: f(x)-f(y)=f(x/y) for all x>0,y>0. Please solve f. Given f(x) - f(y) = f(x/y) for all x > 0, y > 0. Please solve for f. I know that the logarithm (with any positive base) is one of the answers. But I wonder ... 1) How do you deduce the answer step by step? I want to not only know the answer, but also how to get the answer, or the steps of the process that lead to the answer. 2) Will solving for this function involve differential equations? 3) Besides log, will there be any other answer? Thanks. Date: 05/30/2014 at 11:29:29 From: Doctor Vogler Subject: Re: f(x)-f(y)=f(x/y) for all x>0,y>0. Please solve f. Hi Youtian, Thanks for writing to Dr. Math. Define the function (for all real x) g(x) = f(exp(x)) This has g(x) - g(y) = f(exp(x)) - f(exp(y)) = f(exp(x - y)) = g(x - y) Now ask yourself the question: What are all functions that satisfy the following for all real x and y? g(x) - g(y) = g(x - y) Or, setting w = x - y, for all real y and w, we can rearrange this as g(y + w) = g(w) + g(y) Well, you can use induction to prove that this function must satisfy the following for all real numbers x and positive integers n: g(nx) = n*g(x) Then you can combine two of these with the difference formula above to show that it must also satisfy the same equation when n is any integer. Finally, replace x by y/d to show that it must also satisfy that equation when x is any rational number. In particular, whenever x is a rational number, g(x) = x*g(1) If you also add in the requirement that g be a *continuous* function, then since the rational numbers are dense, it follows that g(x) = x*g(1) for all real x. This means that f(x) = g(log x) = (log x)*g(1) = (log x)*f(e) So f(x) must be a constant multiple of the function log(x). But if g is not required to be continuous, then there are other solutions. The other solutions are defined by first choosing a basis for R (the set of real numbers) as a vector space over Q (the set of rational numbers). Such a basis is infinite, and its existence depends on the Axiom of Choice, and describing it explicitly is not really feasible. But you can choose the value of g(x)/x for each basis element x arbitrarily, and then the value of g(x) on all elements is determined by the formula g(qx) = q*g(x) whenever q is rational. Does this make sense? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 05/31/2014 at 00:19:40 From: Youtian Subject: Thank you (f(x)-f(y)=f(x/y) for all x>0,y>0. Please solve f.) Thank you, Dr. Vogler. I appreciate your help very much. Your explanation is great and makes difficult things simple. I think I understood everything up until this paragraph: "But if g is not required to be continuous...." What follows that may require me to ponder for a while to absorb, due to my own limited knowledge. But I'm ready to learn and reference some notions that are new in it, and I'll try to understand the idea in it. That also helps me. Again, thank you, Doctor Vogler and Dr. Math. |
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