Fuctional eq.

[Primarygun]

f(x+y)= f(x)f(y)

If we take log for both sides,

we have g(x)= g(x) + g(y) by letting g=log f

Then, after a series of calculation and checking,

we have f(x)= a^ (cx)

I have a few questions dealing with the log.

Can I choose any base, for example 10?

Must a=e where it is the natural log.?

Last, I am confused with the use of fixed point, or perhaps I even don't know the complete definiton of me, hence, finding myself incapable of doing this kind of calculations.

EDIT: I am very interested and enthusiastic for functional equation.

However, I can't find any books related to this topic or subtopic.

I've checked algebra, statistics. Does it belong to Discrete maths or analysis or other areas?

[Bignose]

When in doubt, Mathworld http://mathworld.wolfram.com/FunctionalEquation.html

 

Note in the References three books by Kuczma on functional equations.

[matt grime]

When in doubt just google. That link is the first hit.

[matt grime]

f(x+y)= f(x)f(y)

If we take log for both sides' date='

we have g(x)= g(x) + g(y) by letting g=log f[/quote']

 

 

do we? What happens if you subtract g(x) from both sides?

 

 

Then, after a series of calculation and checking,

we have f(x)= a^ (cx)

I have a few questions dealing with the log.

Can I choose any base, for example 10?

Must a=e where it is the natural log.?

Last, I am confused with the use of fixed point,

 

what fixed point? you haven't mentioned a fixed point.

 

or perhaps I even don't know the complete definiton of me...

 

which of us truly knows the definition of themselves?

[matt grime]

We should point out that you've omitted to mention at least one important property that you're assuming f must have, namely continuity.

 

It is trivial to show that log(f(x)) for rational x is completely determined by f(1) and hence for rational x, f(x)=exp{kx} for some constant k, and thus by continuity f(x) is exponentiation for all x *if* we assume f is continous. If f is not continuous then there are uncountably many distinct f's with this multiplicative property.

[Primarygun]

we have g(x)= g(x) + g(y) by letting g=log f

sorry it is g(x+y) for the left-hand side

It is trivial to show that log(f(x)) for rational x is completely determined by f(1) and hence for rational x, f(x)=exp{kx} for some constant k, and thus by continuity f(x) is exponentiation for all x *if* we assume f is continous. If f is not continuous then there are uncountably many distinct f's with this multiplicative property.

Oh I now realize the importance of the continuity.

Must we take e as the base of log?

[matt grime]

Where does anything require you to use any particular base? logs in different bases differ only by constants. This is why at the end you only write exp{kx{ for some constant k. k can be gotten from f(1).

[Primarygun]

OK thanks