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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?


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

   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

- Doctor Vogler, The Math Forum 

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.
Associated Topics:
High School Logs

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